NEET Physics

Dual Nature Of Radiational Matter

Question 1. In a discharge tube ionization of enclosed gas is produced due to collisions between:

1. neutral gas atoms/molecules
2. positive ions and neutral atoms/molecules
3. negative electrons and neutral atoms/molecules
4. photons and neutral atoms/molecules

Answer: 3. negative electrons and neutral atoms/molecules

Negative electrons and neutral atoms collide and cause the ionization of gas enclosed in a discharge tube.

Question 2. J.J. Thomson’s cathode ray tube experiment demonstrated that:

1. cathode rays are streams of negatively charged ions
2. all the mass of an atom is essentially in the nucleus
3. the EM of electrons is much greater than the e/m,
4. the EM ratio of the cathode-ray particle changes when a different gas is placed in the discharge tube

Answer: 3. the EM of electrons is much greater than the e/m,

⇒ \(\left(\frac{e}{m}\right)_{\text {electron }} \gg\left(\frac{e}{m}\right)_{\text {proton }}\)

∴ \({\left[\left(\frac{e}{m}\right)_{\text {enti }}=\frac{1}{1837}\left(\frac{e}{m}\right)_{\text {electron }}\right]}\)

Question 3. Which of the following is not the property of cathode rays?

1. It produces a heating effect
2. It does not deflect in an electric field
3. It casts shadow
4. It produces fluorescence

Answer: 2. It does not deflect in an electric field

Cathode rays do not deflect in an electric field.

Read and Learn More NEET Physics MCQs

Question 4. A source of light is placed at a distance of 50 cm from a photo cells and the stopping potential is found to be F₀. If the distance between the light source and photocells is 25 cm, the new stopping potential will be:

1. \(V_0 / 2\)
2. \(V_0\)
3. \(4 V_0\)
4. \(2 V_0\)

Answer: 2. \(V_0\)

Stopping potential is independent of distance hence new stopping potential will remain unchanged in required.

Stopping potential = \(V_0\)

Question 5. In the photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is:

1. 1.3 V
2. 0.5 V
3. 2.3 V
4. 1.8 V

Answer: 2. 0.5 V

In a photoelectric effect, the maximum kinetic energy of the electron,

Stopping potential = Maximum kinetic energy

e \(\mathrm{~V} =\mathrm{KE}_{\max } \)

∴ \(\mathrm{V} =\frac{\mathrm{KE}_{\max }}{e}=\frac{0.5 e v}{e}=0.5 \mathrm{v}\)

Question 6. Photoelectric emission occurs only when the incident light has more than a certain minimum:

1. wavelength
2. intensity
3. frequency
4. power

Answer: 3. frequency

From Einstein’s photoelectric equation

E =\(h\left(v-v_0\right) \)

⇒ \(\frac{1}{2} m v^2 =h\left(v-v_0\right)\)

v \(\geq v_0\)

Question 7. The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

1. 2.4 V
2. – 1.2 V
3. – 2.4 V
4. 1.2 V

Answer: 4. 1.2 V

We know that the energy of incident light

E=\(e V=6.2 \mathrm{eV}\)

Using Einstein’s photoelectric equation

E =W+e\( V_0 \)

e \(V_0 =E-W=(6.2-5.01)=1.2 \mathrm{eV} \)

∴ \(V_0 =1.2 \mathrm{~V}\)

Question 8. When monochromatic radiation of intensity l falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 21, the number of emitted electrons and their maximum kinetic energy are respectively:

1. A and 2T
2. 2N and T
3. 2N and 2T
4. A and T

Answer: 2. 2N and T

The kinetic energy of photoelectrons depends on the frequency of incident radiation and a number of photoelectrons depends upon the intensity.

Question 9. The figure shows a plot of photocurrent versus anode potential for a photo-sensitive surface for three different radiations. Which one of the following is a correct statement?

1. Curves a and b represent incident radiations of different frequencies and different intensities
2. Curves a and b represent incident radiations of the same frequency but of different intensities
3. Curves b and c represent incident radiations of different frequencies and different intensities
4. Curves b and c represent incident radiations of the same frequency having the same intensity

Answer: 2. Curves a and b represent incident radiations of the same frequency but of different intensities

Curves a and b represent incident radiations of the same frequency but of different intensities.

Question 10. The number of photoelectrons emitted for light of a frequency v (higher than the threshold frequency v₀) is proportional to:

1. v = v₀
2. threshold frequency (v₀)
3. intensity of light
4. frequency of light (v)

Answer: 3. intensity of light

The number of photoelectrons emitted is directly proportional to the intensity of light.

Question 11. A 5-watt source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of:

1. 8
2. 16
3. 2
4. 4

Answer: 4. 4

Intensity is given by \(\frac{p}{4 \pi d^2}\)

Where p = Power of source

d = distance

I \(\propto \frac{1}{d^2} \)

or \(\frac{I_1}{I_2} =\frac{d_2{ }^2}{d_1{ }^2}\)

or \(\frac{I_1}{I_2} =\left(\frac{1}{0.5}\right)^2 \)

⇒ \(\frac{I_1}{I_2}=4\)

or \(I_2 =\frac{I_1}{4}\)

We know (assume) that light spreads out uniformly in all directions i.e. spherical source

Question 12. A photocell employs the photoelectric effect to convert:

1. change in the frequency of light into a change in the electric current.
2. change in the frequency of light into a change in electric voltage.
3. change in the intensity of illumination into a change in photoelectric current.
4. change in the intensity of illumination into a change in the work function of the photocathode.

Answer: 3. change in the intensity of illumination into a change in photoelectric current.

Photocell employs the photoelectric effect to convert the change in the intensity of illumination into a charge in photoelectric current.

Question 13. A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m then:

1. each emitted electron carries one-quarter of the initial energy.
2. number of electrons emitted in half the initial number.
3. each emitted electron carries half the initial energy.
4. number of electrons emitted is a quarter of the initial number.

Answer: 4. number of electrons emitted is a quarter of the initial number.

For a point source, I \(\propto \frac{1}{r^2}\)

Question 14. When the intensity of incident light increases:

1. photocurrent increase
2. photocurrent decreases
3. the kinetic energy of emitted photoelectrons increases
4. the kinetic energy of emitted photoelectrons decreases

Answer: 1. photocurrent increase

Einstein’s photoelectric effect theory states that a single incident photon ejects a single electron. As a result, as the intensity increases, the number of incident photons increases, and the number of emitted electrons increases, the photocurrent increases.

We have, maximum energy of electrons =\(\frac{1}{2} m v^2 max\)

⇒ \(\frac{1}{2} m v^2{ }_{\max }=e V_0 \)

where \(V_0\) is stopping potential.

Thus, since, the stopping potential is unaffected by increasing ray intensity, the maximum kinetic energy of the electrons is independent of the intensity of the incident rays.

Question 15. The cathode of a photoelectric cell is changed such that the work function changes from W₁ to W2 (W₂ > W₁). If the current before and after changes are I₁ and I₂, all other conditions remaining unchanged, then (assuming hv> W₂)

1. \(I_1=I_2\)
2. \(I_1<I_2\)
3. \(I_1>I_2\)
4. \(I_1<I_2<2 I_1\)

Answer: 1. \(I_1=I_2\)

By the work function of a metal, it means that the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. The work function does not affect photoelectric current as long as hv > W₀. The photoelectric current is proportional to the intensity of incident light. Since there is no change in the intensity of light, hence I₁ = I₂

Question 16. The threshold frequency for the photoelectric effect on sodium corresponds to a wavelength of 5000Å. Its work function is:

1. 4 x 10^-19 J
2. 1 J
3. 2 x 10^-19 J
4. 3 x 10^-19 J

Answer: 1. 4 x 10^-19 J

When a photon of light with frequency v strikes a photosensitive metal surface, its energy (hv) is consumed in two ways. A portion of the photon’s energy is utilized to liberate the electron from the metal surface, which is equal to the metal’s work function \(W_0\). metal.

⇒ \(W_0 = hv_0\)

(where, \(v_0\) is threshold frequency)

⇒ \(W_0 =\frac{h c}{\lambda_0}\)

Here,\(\lambda_0 =5000 Å \)

= 5000 \(\times 10^{-10} \mathrm{~m} \)

⇒ \(W_0 =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} \)

= 4 \(\times 10^{-19} \mathrm{~J}\)

Question 17. When two monochromatic light of frequency v and \(\frac{v}{2}\) are incident on a photoelectric metal, their stopping potential becomes \(\frac{V_{\mathrm{s}}}{2}\) and \(V_{\mathrm{s}}\)respectively. The threshold frequency for this metal is:

1. 2v
2. 3v
3. \(\frac{2}{3} v\)
4. \(\frac{3}{2} v\)

Answer: 4. \(\frac{3}{2} v\)

According to the photoelectric effect,

E =\(h v_o+e v_o \)

⇒ \(h v =h v_o+\frac{e v_s}{2}\)  →  Equation 1

⇒ \(\frac{h v}{2} =h v_o+e v_s \)  →  Equation 2

⇒ \(\frac{h v}{2} =\frac{-e v_s}{2}\)

⇒ –\(h v =e v_s \)  → Equation 3

Putting (3) in eq (1), we get

∴ h v=\(h v_o-\frac{h v}{2}\) , So \(v_o=\frac{3}{2} v\)

Question 18. The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photon emission from the substance is (approximately):

1. 3100 nm
2. 3600 nm
3. 31 nm
4. 310 nm

Answer: 4. 310 nm

The work function of the material is

⇒ \(\mathrm{W}=h v=\frac{h c}{\lambda}\)

Here, h= Planck’s constant

h=6.63 \(\times 10^{-34} \mathrm{~J} / \mathrm{s}\)

c = speed of light =\(3 \times 10^8 \mathrm{~m} / \mathrm{s}\)

⇒ \(\lambda\)= wavelength of light

And \(\mathrm{W}=v=4 \times 1.6 \times 10^{-19} \mathrm{~K}\)

From eq. (1) and (2) we can write

4 \(\times 1.6 \times 10^{-19} =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda}\)

⇒ \(\lambda =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.6 \times 10^{-19}}\)

∴ \(\lambda =3.108 \times 10^{-7} \mathrm{~m}=310 \mathrm{~nm}\)

Question 19. When the light 2v₀ (where v₀ is threshold frequency), in an incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5v₀. the maximum velocity of electrons emitted from the same plate is v₂. The ratio of V₁ to v₂ is:

1. 4: 1
2. 1: 4
3. 1: 2
4. 2: 1

Answer: 3. 1: 2

From Einstein’s photoelectric equation, we can write

E=\(W_0+\frac{1}{2} m v^2\)  → Equation 1

Where, \(W_0\)= work function

⇒ \(\frac{1}{2} m v^2\)= Kinetic energy

and E=h v

When the incident frequency of light is v=2\( v_0 \)then eq.

(1) become

h\(\left(2 v_0\right) =h v_0+\frac{1}{2} m v_1^2\)

⇒ \(h v_0 =\frac{1}{2} m v_1^2\) →  Equation 2

When the incident frequency of light is v=2 \(v_0\) then eq.

(1) becomes

⇒ \(h\left(5 v_0\right) =h v_0+\frac{1}{2} m v_2^2 \)

⇒ \(4 h v_0 =\frac{1}{2} m v_2^2\) →  Equation 3

From eq. (2) and eq. (3) we have

⇒ \(\frac{h v_0}{4 h v_0}=\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2{ }^2}\)

⇒ \(\frac{v_1{ }^2}{v_2{ }^2}=\frac{1}{4} \)

∴ \(\frac{v_1}{v_2}=1: 2 \)

Question 20. The photoelectric threshold wavelength of silver is 3250 x 10^-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 x 10^-10 m is (Given, h = 4.14 x 10^-15 eVS and c = 3 x 10⁸ ms^-1):

1. = 6 x 10⁵ ms^-1
2. = 0.6 x 10⁶ ms^-1
3. = 61 x 10³ ms^-1
4. = 0.3 x 10⁶ ms^-1

Answer: 1. = 6 x 10⁵ ms^-1

Given : Threshold wavelength \(\lambda_0=3250 \times 10^{-10} \mathrm{~m}\)

Wavelength of UV light \(\lambda=2536 \times 10^{-10} \mathrm{~m}\)

We know that Einstein’s photoelectric equation is.

E =k+W

h v =\(\frac{1}{2} m v^2+h v_0 \)

W = work function =h \(v_0\)

⇒ \(\frac{1}{2} m v^2 =h v-h v_0=h\left(v-v_0\right)\)

=\(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\)

velocity , v=\(\sqrt{\frac{2 h c}{m_0}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}\)

= \(\sqrt{\frac{2 \times 6.02 \times 10^{-34} \times 3 \times 10^8}{9.1 \times 10^{-31}}\left(\frac{3250-2536}{3250 \times 2536}\right)} \)

= 0.6 \(\times 10^6 \mathrm{~ms}^{-1}\)

= 6 \(\times 10^5 \mathrm{~ms}^{-1}\)

Question 21. Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative C is:

1. – 3 V
2. + 4 V
3. – 1 V
4. – 3 V

Answer: 4. – 3 V

From the photoelectric equation, we have

⇒ \(h v=W+\frac{1}{2} m v_0^2\)

where W= work function

⇒ \(\mathrm{KE}=h v-W\)

putting the given values

⇒ \(2 e \mathrm{~V} =5 e \mathrm{~V}-W \)

W =3 e v

=\(3 \mathrm{~V}\)

⇒ \(V_{\text {cathod }}-V_{\text {anode }} =3 \mathrm{~V}\)

∴ \(V_{\text {anode }}-V_{\text {cathod }} =-3 \mathrm{~V}\)

Question 22. A photoelectric surface is illuminated successively by monochromatic light of wavelength \(\lambda\) and \(\frac{\lambda}{2}\). If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is: (h = Planck’s constant, c = speed of light)

1. \(\frac{h c}{2 \lambda}\)
2. \(\frac{h c}{\lambda}\)
3. \(\frac{2 h c}{\lambda}\)
4. \(\frac{h c}{3 \lambda}\)

Answer: 1. \(\frac{h c}{2 \lambda}\)

From Einstein’s photoelectric equation

h v=W+\(K_{\max }\)

where, W= work function

⇒ \(K_{\max }\)= kinetic energy

⇒ \(K_{\max }\)=h v-W

\(K_{\max }=\frac{h c}{\lambda}-W\) →  Equation 1

In the question, the maximum kinetic of the emitted electron is 3 times that in the first case.

⇒ \(K_{\max }=\frac{h c}{\frac{\lambda}{2}}-W\) →  Equation 2

From eq. (1) and (2),

∴ \(\mathrm{W}=\frac{h c}{2 \lambda}\)

Question 23. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5 eV to 0.8 eV. The work function of the metal is :

1. 0.65 eV
2. 1.0 eV
3. 1.3 eV
4. 1.5 eV

Answer: 2. 1.0 eV

From Einstein’s electrons photoelectric equation

h v =\(W-(\mathrm{KE})_{\max }\)

K.E.\( _{\max } =h v-W\)  →  Equation 1

In the first condition, 0.5=E-W  →  Equation 2

In the second condition, 0.8=1.2 E-W  → Equation 3

From (2) and (3),

-0.3 =-0.2 E

⇒ \(\mathrm{E} =\frac{0.3}{0.2}=1.5 \mathrm{eV}\)

Putting in eq. (2) we get,

0.5 =1.5-W

W =1.5-0.5=1 eV

Question 24. For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass):

1. \(\sqrt{\frac{h v}{(2 m)}}\)
2. \(\sqrt{\frac{h v}{m}}\)
3. \(\sqrt{\frac{2 h v}{m}}\)
4. 2 \(\sqrt{\frac{h v}{m}}\)

Answer: 3. \(\sqrt{\frac{2 h v}{m}}\)

We know that, \(\frac{1}{2} m v_{\max }{ }^2\)=h v

⇒ \(v_{\max }^2=\frac{2 h v}{m}\)

∴ \(v_{\max }=\sqrt{\frac{2 h v}{m}}\)

Question 25. Two radiations of photon energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electron is:

1. 1:4
2. 1: 2
3. 1: 1
4. 1: 5

Answer: 2. 1: 2

⇒ \(v_{\max } =\sqrt{\frac{2}{m}\left(K E_{\max }\right)}\)

= \(\sqrt{\frac{2}{m}\left(E_{p h}-W\right)}\)

∴ \(\frac{v_1}{v_2} =\sqrt{\frac{1-0.5}{25-0.5}}=\sqrt{\frac{0.2}{2}}\)=1: 2

Question 26. The threshold frequency for a photo-sensitive metal is 3.3 x 10¹⁴Hz. If the light of frequency 8.2 x 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is near:

1. 2 V
2. 13 V
3. 5 V
4. 1V

Answer: 1. 2 V

Given,

Threshold frequency \(v_0=3.3 \times 10^{14} \mathrm{~Hz}\) and light of frequency v=8.2 \(\times 10^{14} \mathrm{~Hz}\)

Then we know from Einstein’s photoelectric equation

⇒ \(e V_0=E-V\)

⇒ \(V_0=\frac{E-V}{e}\)

⇒ \(V_0=\frac{h\left(V-V_0\right)}{e}\)

⇒ \(V_0=\frac{6.62 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}\)

⇒ \(V_0=\frac{6.62 \times 10^{-34}}{1.6} \times 4.9 \times 10^{33} \)

⇒ \(V_0=\frac{6.02 \times 4.9 \times 10^{-1}}{1.6}\)

∴ \(V_0\)=2 volt

Question 27. The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiations for which the stopping potential is 5 V lies in the:

1. ultraviolet region
2. visible region
3. infrared region
4. X-ray region

Answer: 1. ultraviolet region

Using Einstein’s photoelectric equation

Where, \(K E_{\max }\)=E-f

f= work function

⇒ \(K E_{\max }\)= Maximum kinetic energy

h v=\(e V_0+f\)

h v=5 \(\mathrm{eV}+6.2 \mathrm{eV}=11.2 \mathrm{eV}\)

⇒ \(\lambda=\left(\frac{12400}{11.2 \mathrm{eV}}\right)=1.104 \times 10^{-7}\)

Hence the radiation lies in the ultraviolet region.

Question 28. When photons of energy hv fall on an aluminum plate (of work function E₀), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:

1. K+hv
2. K + E₀
3. 2K
4. K

Answer: 1. K+hv

According to a given situation,

h v=\(E_0+K \text { and } 2 h v=E_0+K^{\prime} \)

\(K^{\prime}\)=K+h v

Question 29. The work functions for metal A, B, and C is respectively 1.92 eV, 2.0 eV, and 5 eV. According to Einstein’s equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are:

1. none
2. An only
3. A and B only
4. all the three metals

Answer: 3. A and B only

The energy of an electron with an associated wavelength of 4100 Å is:

⇒ \(\frac{h c}{\lambda} =4.845 \times 10^{-19} \mathrm{~J} \)

= 3.024 eV

This incident electron would emit photons from metals whose work potential is less than its energy.

Question 30. The photosensitive metallic surface has a work function hv₀ fall on this surface, the electrons come out with a maximum velocity of 4 x 10⁶ m/s. When the photon energy is increased to 5hv₀. Then, the maximum velocity of photoelectrons will be:

1. 2 x 10⁷ m/s
2. 2 x 10⁶ m/s
3. 8 x 10⁶ m/s
4. 8 x 10⁵ m/s

Answer: 3. 8 x 10⁶ m/s

According to the question

⇒ \(2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6\right)^2\) → Equation 1

⇒ \(5 h v_0=h v_0+\frac{1}{2} m v^2\)  → Equation 2

Divide eq. (2) by eq. (1), we get

⇒ \(v_1^2 =4 v_1^2 \) or \(v_2=2 v_1 \)

⇒  \(v_2 =2 \times 4 \times 10^6 \)

= 8 \(\times 10^6 \mathrm{~m} / \mathrm{s}\)

Question 31. According to Einstein’s photoelectric equation, the graph between the kinetic energy of the photoelectron ejected and the frequency of incident radiation is:

Answer: 4.

∴ \(h v=\phi+k \mathrm{E}_{\max }\)

Question 32. When ultraviolet rays incident on a metal plate then the photoelectric effect does not occur, it occurs by the incidence of:

1. infrared rays
2. X-rays
3. radio wave
4. light wave

Answer: 2. X-rays

Question 33. A light source is at a distance d from a photoelectric cell, then the number of photoelectrons emitted from the cell is n. If the distance of the light source and the cell is reduced to half, then the number of photoelectrons emitted will become:

1. \(\frac{n}{2}\)
2. 2n
3. 4n
4. n

Answer: 3. 4n

We have, the intensity of the light source,

I=\(\frac{1}{d^2}\)

where d is the distance or light source from the cell. So, for two different cases of intensities, we have,

⇒  \(\frac{I_1}{I_2}=\left(\frac{d_2}{d_1}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

⇒  \(I_2=4 I_1\)

A number of photoelectrons emitted is directly proportional to intensity, so the number of photoelectrons emitted will become 4 times, i.e. 4 n

Question 34. Einstein’s work on photoelectric effect gives support to:

1. \(E=m c^2\)
2. E = hv
3. h v=\(\frac{1}{2} m v^2\)
4. E=\(\frac{h}{\lambda}\)

Answer: 2. E = hv

In 1905, Einstein observed that the photoelectric effect could be explained if the energy in light was concentrated in small packets, or photons, instead of spread out over wavefronts. The energy hv is contained in each photon of light with frequency v. Hence, Einstein’s work on the photoelectric effect supports

E = hv

Question 35. The energy of a photon of light is 3 eV. Then the wavelength of the photon must be:

1. 4125 nm
2. 412.5 nm
3. 41250 nm
4. 4nm

Answer: 2. 412.5 nm

If photon energy is expressed in (eV) and wavelength is expressed (in Å), then photon energy is given by

E=\(\frac{h c}{\lambda}=\frac{12375}{\lambda(Å)} \mathrm{eV}\)

⇒ \(\lambda =\frac{12375}{E(\mathrm{eV})}Å [ h c=12375 \mathrm{eV}-Å]\)

=\(\frac{12375}{3(\mathrm{eV})}Å=4125 Å=412.5 \mathrm{~nm}\)

Question 36. If the threshold wavelength for a certain metal is 2000 Å, then the work function of the metal is:

1. 6.2 J
2. 6.2 eV
3. 6.2 MeV
4. 6.2 KeV

Answer: 2. 6.2 eV

The work function is the minimum energy required to remove an electron from a metal surface without releasing any kinetic energy.

⇒ \(W_0=h v_0\)

Given wavelength, \(\lambda=2000 Å=2000 \times 10^{-10} \)

⇒ \(\mathrm{W}_0=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}}\)

=9.9 \(\times 10^{-19} \mathrm{~J}=\frac{9.9 \times 10^{-19}}{1.6 \times 10^{-19}} \)

⇒ \({[1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}]}\)

=6.2\( \mathrm{eV}\)

Question 37. The number of photons per second on average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 x 10^-3 watts will be (h = 6.6 x 10^-34 Js):

1. 10¹⁸
2. 10¹⁷
3. 10¹⁶
4. 10¹⁵

Answer: 3. 10¹⁶

Given, \(\frac{n}{t}\)= ?

⇒ \(\lambda=600 \mathrm{~nm} \)

⇒ \(\lambda=600 \times 10^{-9}\)

⇒ \(\lambda=6 \times 10^{-7} \mathrm{~m} \)

P=3.3 \(\times 10^{-3} watt \)

We know power, P=\(\frac{E}{t}=\frac{n h v}{t}\)

⇒ \(\frac{n}{t} =\frac{\mathrm{P}}{h v}=\frac{\mathrm{P} \lambda}{h c}\)

⇒ \(\frac{n}{t} =\frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^8}\)

∴ \(\frac{n}{t} =10^{-3-7+34-8}=10^{16}\)

Question 38. Light with an energy flux of 25 x 10⁴ Wm^-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm², the average force exerted on the surface is :

1. 1.25 \(\times 10^{-6} \mathrm{~N}\)
2. 2.50 x HT⁶ N
3. 1.20 x 10^-6 N
4. 12.5 x 10^-6 N

Answer: 1. 25 \(\times 10^{-6} \mathrm{~N}\)

We can write,

⇒ \(F_{a v} =\frac{\Delta P}{\Delta t} \)

= \(\frac{2 I}{c . A t} A\)

= \(\frac{2 \times 25 \times 10^4}{3 \times 10^8} \times 15 \times 10^{-4}\)

= \(\frac{10 \times 25 \times 10^4 \times 10^{-4}}{10^8}\)

= 1.25 \(\times 10^{-6} \mathrm{~N}\)

Question 39. A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:

1. 1.5 x 10²⁰
2. 6 x 10¹⁸
3. 62 x 10²⁰
4. 3 x 10¹⁹

Answer: 1. 1.5 x 10²⁰

Using the formula P=\(\frac{N}{t} \times \frac{h c}{\lambda}\)

= 200 \(\times 0.25 \)

⇒\(\frac{N}{t}=50 \times \frac{\lambda}{h c}\)

=\(\frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^8}=1.5 \times 10^{20}\)

Question 40. A source S1 is producing, 1015 photons Å of wavelength 5000 Å. Another source S₁ is producing 1.02 x 10¹⁵ photons per second of wavelength 5100 Å. Then, (power of S₂) (power of S₁) is equal to:

1. 1.00
2. 1.02
3. 1.04
4. 0.98

Answer: 1. 1.00

Number of photons emitted per second

n=\(\frac{P}{\left(\frac{h c}{\lambda}\right)} \)

P=\(\frac{n h c}{\lambda}\)

⇒ \(\frac{P_1}{P_2}=\frac{n_2 \lambda_1}{n_1 \lambda_2}=\frac{1.02 \times 10^{15} \times 5000}{10^{15} \times 5100}\)

=1

Question 41. The momentum of a photon of energy 1 MeV in kg m/s will be:

1. 5 x 10^-22
2. 0.33 x 10⁶
3. 7 x 10^-24
4. 10^-22 

Answer: 1. 5 x 10^-22

The energy of the photon, E=1 MeV

Momentum of photon, P=\(\frac{E}{C}\)

P =\(\frac{E}{C}=\frac{1 \times 10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8} \)

=5\( \times 10^{-22} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

Question 42. The value of Planck’s constant is:

1. 6.63 x 10^-34 Js^-1
2. 6.63 x 10^-34 m²s^-1
3. 6.63 x 10^-34 kg m²s^-1
4. 6.63 x 10^-34 J s^-1

Answer: 3. 6.63 x 10^-34 kg m²s^-1

The value of Planck’s constant

h=6.63 \(\times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}\)

Question 43. The momentum of a photon of wavelength \(\lambda\) is:

1. \(\frac{h}{\lambda}\)
2. zero
3. \(\frac{h \lambda}{c^2}\)
4. \(\frac{h \lambda}{c}\)

Answer: 1. \(\frac{h}{\lambda}\)

We have,

The energy of light \(\mathrm{E}=\mathrm{hc} / \lambda\)  → Equation 1

The energy of light \(\mathrm{E}=\mathrm{mc}^2\) →  Equation 2

From (1) and (2), we get

⇒ \(\frac{h c}{\lambda}=m c^2 \)

⇒ \(\frac{h}{\lambda}\)=m c = Momentum

Thus, the momentum of the photon =\(\frac{h}{\lambda}\)

Question 44. The wavelength of a 1 KeV photon is 1.24 x 10^-9m. What is the frequency of 1 MeV photon?

1. 1.24 x 10¹⁵ Hz
2. 2.4 x 10²⁰ Hz
3. 1.24 x 10¹⁸ Hz
4. 2.4 x 10²³ Hz

Answer: 2. 2.4 x 10²⁰ Hz

We have, E=\(h c / \lambda\)

or E \(\lambda\)=h c= constant

Therefore, \(E_1 \lambda_1=E_2 \lambda_2\)

Putting the given values, we get

1 \(\times 10^3 \mathrm{eV} \times 1.24 \times 10^{-19} =1 \times 10^6 \mathrm{eV} \times \lambda_2 \)

⇒ \(\lambda_2 =1.24 \times 10^{-12} \mathrm{~m}\)

Now, We have,

so, v=\(\mathrm{c} / \lambda \)

v=2.4 \(\times 10^{20} \mathrm{~Hz}\)

Question 45. The graph shows the variation of the de-Broglie wavelength of a particle and its associated momentum!/?) is:

Answer: 4.

According to de-Broglie wavelength

⇒ \(\lambda=\frac{h}{p}\) or \(\lambda \propto \frac{1}{p}\)

The graph will be a rectangular hyperbola. Hence the correct option is (D).

Question 46. An electromagnetic wave of wavelength ‘\(\lambda\)’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wave length \(\lambda_d\), then:

1. \(\lambda=\left(\frac{2 m}{h c}\right) \lambda_d^2\)
2. \(\lambda=\left(\frac{2 m c}{h}\right) \lambda^2\)
3. \(\lambda=\left(\frac{2 m c}{h}\right) \lambda^2{ }_d\)
4. \(\lambda=\left(\frac{2 h}{m c}\right) \lambda^2{ }_d\)

Answer: 3. \(\lambda=\left(\frac{2 m c}{h}\right) \lambda^2{ }_d\)

Given, Wavelength =\(\lambda\)

⇒ \(\mathrm{W}_f\) =0

mass = \(m_e\)

de-Broglie wavelength =\(\lambda_d\)

We know from Einstein’s relation

⇒ \(h v =W+\mathrm{KE}\)

Since W = 0

h v= \(\mathrm{KE}\) → Equation 1

From the de-Broglie hypothesis, we know,

\(\lambda_d=\frac{h}{p}\)

and p=\(\sqrt{2 \mathrm{mkE}}\)

⇒ \(\lambda_d=\frac{h}{\sqrt{2 m k \mathrm{E}}}\)  → Equation 2

From eq. (2) \(\lambda_d{ }^2=\frac{h^2}{2 m \mathrm{KE}}\)From eq. (2)\(\lambda_d{ }^2=\frac{h^2}{2 m \mathrm{KE}}\)

⇒ \(\mathrm{KE}=\frac{h^2}{2 m \lambda d^2}\) →  Equation 3

From eq. (1) \(\frac{h c}{\lambda}=\mathrm{KE}\)

⇒ \(\lambda=\frac{h c}{\mathrm{KE}}\)  → Equation 4

From eq. (3) and (4),

⇒ \(\lambda=\frac{h c}{h^2} 2 m \lambda_d^2 \)

⇒ \(\lambda=\frac{c}{h} 2 m \lambda_d^2 \)

∴ \(\lambda=\left(\frac{2 m c}{h}\right) \lambda_d^2\)

Question 47. The de-Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly :

1. \(102 \times 10^{-3} \mathrm{~nm}\)
2. 102 \(\times 10^{-4} \mathrm{~nm}\)
3. 102\( \times 10^{-5} \mathrm{~nm}\)
4. 102 \(\times 10^{-2} \mathrm{~nm}\)

Answer: 4. 102 \(\times 10^{-2} \mathrm{~nm}\)

The kinetic energy of electron a=ÿ

KE = 144 eV

Means, eV= 144 eV

V= 144 V

de-Broglie wavelength,

⇒ \(\lambda=\frac{1.227}{\sqrt{V}} \)

From eq. (1),

⇒ \(\lambda =\frac{1.227}{144} \)

= \(\frac{1.227}{12} Å=0.102 \)

1.02 \(\mathrm{~nm} =102 \times 10^{-2} \mathrm{~nm}\)

Question 48. A proton and \(\alpha\)particle are accelerated from rest to the same energy. The de-Broglie wavelength \(\lambda_p\) and \(\lambda_\alpha\) are in the ratio:

1. 2: 1
2. 1: 1
3. \(\sqrt{2}\): 1
4. 4: 1

Answer: 1. 2: 1

The de-Broglie wavelength of any moving particle associated with momentum

\(\lambda=\frac{h}{p}\)  →  Equation 1

where h= Planck’s constant

Now K.E. =\(\frac{1}{2} m v^2\)

K.E. =\(\frac{1}{2}(m v)^2=\frac{p^2}{2 m}\)

p =\(\sqrt{2 m K E}\) →  Equation 2

From (1) and (2),

⇒ \(\lambda=\frac{h}{\sqrt{2 m K E}}\)

For Proton, \(\lambda_{\mathrm{P}}=\frac{h}{\sqrt{2 m_P K E_P}}\)  → Equation 3

For \(\alpha – particle, \lambda_\alpha=\frac{h}{\sqrt{2 m_\alpha K E_\alpha}}\)  → Equation 4

From (3) and (4),

⇒ \(\frac{\lambda_{\mathrm{P}}}{\lambda_\alpha}=\frac{\sqrt{2 m_\alpha \mathrm{KE}_\alpha}}{2 \mathrm{M}_{\mathrm{P}}}\)

⇒ \(\mathrm{KE}_{\mathrm{p}} \)

⇒ {Since } \(\mathrm{KE}_\alpha=\mathrm{KE}_{\mathrm{P}} \) and \(m_\alpha=4 \mathrm{~m}_p \)

⇒ \(\frac{\lambda_{\mathrm{P}}}{\lambda_\alpha}=\sqrt{\frac{4 m_{\mathrm{p}}}{m_{\mathrm{P}}}}=\frac{2}{1}\)

∴ \(\lambda_{\mathrm{P}}: \lambda_\alpha\)=2: 1

Question 49. An electron of mass m with a velocity \(v=v_0 \hat{i}\left(v_0>0\right)\) enters an electric field \(E=E_0 \hat{i} \left(E_0=\text { constant }>0\right)\) at t = 0. If \(\lambda_0\) is its de-Broglie wavelength initially, then its de-Broglie wavelength at time 1 is:

1. \(\lambda_0 t\)
2. \(\lambda_0\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)\)
3. \(\frac{\lambda_0}{\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)}\)
4. \(\lambda_0\)[/latex]

Answer: 3. \(\frac{\lambda_0}{\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)}\)

From question

V=\(V_0 \hat{i}\) and \(E=-E_0 \hat{i}\)

Now initial de-Broglie wavelength is \(\lambda_0=\frac{h}{m v_0}\)  → Equation 1

Acceleration of electron

a=\(\frac{e E_0}{m}\)

Velocity after t is V=\(\left(V_0+\frac{e E_0}{m} t\right)\) [This is obtained from \(V=u+a t ]\) → Equation 2

So, \(\lambda=\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e E_0}{m} t\right)}\)

⇒\(\lambda=\frac{h}{m v_0\left(1+\frac{e E_0}{m v_0} t\right)}=\frac{\lambda_0}{\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)}\)

Thus,\(\lambda=\frac{\lambda_0}{\left(1+\frac{e \mathrm{E}_0}{m v_0} \cdot t\right)}\)

Question 50. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is:

1. \(\frac{h}{\sqrt{m k T}}\)
2. \(\frac{h}{\sqrt{3 m k T}}\)
3. \(\frac{2 h}{\sqrt{3 m k T}}\)
4. \(\frac{2 h}{\sqrt{m k T}}\)

Answer: 2. \(\frac{h}{\sqrt{3 m k T}}\)

de-Broglie wavelength

⇒ \(\lambda=\frac{h}{P}=\frac{h}{m v}\)

⇒ \(\lambda=\frac{h}{\sqrt{2 m(K E)}} \)

⇒ \(\lambda=\frac{h}{\sqrt{2 m\left(\frac{3}{2} K T\right)}}\)

∴ \(\lambda=\frac{h}{\sqrt{3 m K T}}\)

Question 51. An electron of mass m and a photon have the same energy E. The ratio of de-Broglie wavelength associated with them is:

1. \(\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)
2. \(C(2 m E)^{1 / 2}\)
3. \(\frac{1}{C}\left(\frac{2 m}{E}\right)^{\frac{1}{2}}\)
4. \(\frac{1}{C}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)

Answer: 4. \(\frac{1}{C}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}\)

For electron \(\lambda_e=\frac{h}{\sqrt{2 m E}}\) for photon, E=PC

⇒ \(\lambda_{p h} =\frac{h c}{E}\)

⇒ \(\frac{\lambda_e}{\lambda_{p h}} =\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c}\)

=\(\left(\frac{E}{2 m}\right)^{\frac{1}{2}} \times \frac{1}{c}\)

Question 52. Electrons of mass m with de-Broglie wavelength X fall on the target in an X-ray tube. The cut-off wavelength \(\left(\lambda_0\right)\) of the emitted X-ray is:

1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
2. \(\lambda_0=\frac{2 h}{m c}\)
3. \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)
4. \(\lambda_0=\lambda\)

Answer: 1. \(\lambda_0=\frac{2 m c \lambda^2}{h}\)

From de-Broglie equation

⇒ \(\lambda =\frac{h}{P}\)

P =\(\frac{h}{\lambda}\)

KE of electrons, E=\(\frac{P^2}{2 m}=\frac{h^2}{2 m \lambda^2}\)

Also in X-ray, \(\lambda_0=\frac{h c}{t}\)

∴ \(\lambda_0=\frac{2 m c \lambda^2}{h}\)

Question 53. Which of the following figures represents the variation of the particle momentum and the associated de-Broglie wavelength?

Answer: 3.

According to de-Broglie wavelength

⇒ \(\lambda =\frac{h}{p}\)

p \(\lambda \)=h

This equation looks like \(y_x\)=c which is shown in the graph (c)

Question 54. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is:

1. \(<2.8 \times 10^{-10} \mathrm{~m}\)
2. \(<2.8 \times 10^{-9} \mathrm{~m}\)
3. \(\geq 2.8 \times 10^{-9} \mathrm{~m}\)
4. \(\leq 2.8 \times 10^{-12} \mathrm{~m}\)

Answer: 3. \(<2.8 \times 10^{-9} \mathrm{~m}\)

Energy of photon, (E) =\(\frac{12400}{5000}\)

=2.48 \(\mathrm{eV}\)

Work function W=2.28 \(\mathrm{eV}\)

Using Einstein’s photoelectric equation

E = \(W+K E_{\max } \)

2.48 =2.28+ \(K.E._{\max } \)

for electron \(K. E ._{\max } =0.20 \mathrm{eV} \)

⇒ \(\lambda =\frac{h}{\sqrt{2 m E}} \)

So \(\lambda =28 Å\)

∴ \(\lambda \geq 2.8 \times 10^{-9} \mathrm{~m}\)

Question 55. If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de- Broglie wavelength of the particle is:

1. 25
2. 75
3. 60
4. 50

Answer: 2. 75

Using de-Broglie hypothesis,

⇒ \(\lambda=\frac{h}{P}\)

Hence, \(\lambda_1=\frac{h}{P}=\frac{h}{\sqrt{2 m k}}\) →  Equation 1

and \(\lambda_2 =\frac{h}{2 m \times 10 k}\)

=\(\frac{h}{4 \sqrt{2 m k}}=\frac{\lambda_1}{4}\) →  Equation 2

Using eq. (1)

⇒ \(\lambda_2=\frac{1}{4} \times x_1=0.25 \% of \lambda_1\)

Thus 75\% changes in the wavelength.

Question 56. The de-Broglie wavelength of neutrons in thermal equilibrium at temperature T is:

1. \(\frac{3.08}{\sqrt{T}} Å\)
2. \(\frac{0.308}{\sqrt{T}} Å\)
3. \(\frac{0.0308}{\sqrt{T}} Å\)
4. \(\frac{30.8}{\sqrt{T}} Å\)

Answer: 4. \(\frac{30.8}{\sqrt{T}} Å\)

Using de-Broglie wavelength

⇒ \(\lambda =\frac{h}{\sqrt{2 m K_B T}}\)

⇒ \(\lambda =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times \mathrm{T}}}\)

= \(\frac{3.08 \times 10^{-34} \times 10^{25}}{\sqrt{T}} m=\frac{30.8}{\sqrt{T}} Å\)

Question 57. An a-particle moves in a circular path of radius 0.83 em in the presence of a magnetic field of 0.25 Wb/m². The de-Broglie wavelength associated with the particle will be:

1. 1 Å
2. 0.1 Å
3. 10 Å
4. 0.01 Å

Answer: 4. 0.01 Å

We know that, R=\(\frac{m v}{q \mathrm{~B}} and \lambda=\frac{h}{m v}\)

⇒ \(\lambda =\frac{h}{m \nu}=\frac{h}{R q B}\)

⇒ \(\lambda =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times T}}\)

=0.01 Å

Question 58. If the momentum of an electron is changed by p, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of the electron will be:

1. 200 p
2. 400 p
3. \(\frac{p}{200}\)
4. 100 p

Answer: 1. 200 p

From de-Broglie hypothesis

P =\(\frac{h}{\lambda} \)

⇒ \(\left|\frac{\Delta P}{P}\right| =\frac{\Delta \lambda}{\lambda}\)

⇒ \(\frac{\Delta P}{P_i} =\frac{\Delta \lambda}{\lambda} \)

⇒ \(P_i =\frac{P}{\frac{0.5}{100}} \)

∴ \(P_i =\frac{1000}{5}\) P=200 P

Question 59. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would:

1. decrease by 2 times
2. decrease by 4 times
3. increase by 4 times
4. increase by 2 times

Answer: 1. decrease by 2 times

We know that,\(\lambda=\frac{12.27}{\sqrt{V}}\)

⇒ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}} \)

⇒ \(\lambda_2=\lambda_1 \sqrt{\frac{V_1}{V_2}} \)

⇒ \(\lambda_2=\lambda_1 \sqrt{\frac{25}{100}} \)

∴ \(\lambda_2=\lambda_1 \sqrt{\frac{1}{4}}=\frac{\lambda_1}{2}\)

Question 60. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity 3 x \(10^6 \mathrm{~ms}^{-1}\). The velocity of the particle is:

1. \(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\)
2. \(9 \times 10^{-2} \mathrm{~ms}^{-1}\)
3. \(3 \times 10^{-31} \mathrm{~ms}^{-1}\)
4. \(2.7 \times 10^{-21} \mathrm{~ms}^{-1}\)

Mass of electron =9.1 \(\times 10^{-31}\) kg

Answer: 1. \(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\)

De-Broglie wavelength, associated with an electron moving with velocity v, \(\lambda=\frac{h}{m v}\)

So,\(\lambda_e=\frac{h}{9.1 \times 10^{-31} \times 3 \times 10^6}\)

The wavelength of a particle of mass 1 \(\mathrm{~kg}\) moving with velocity v.

⇒ \(\lambda_p=\frac{h}{10^{-3} \times v}\)

As given, \(\lambda_e=\lambda_p\)

⇒ \(\frac{h}{10^{-6} \times v} =\frac{h}{9.1 \times 10^{-31} \times 3 \times 10^6}\)

= 2.7 \(\times 10^{-18} \mathrm{~m} / \mathrm{s}\)

Question 61. If particles are moving with the same velocity, then the de Broglie wavelength is maximum for:

1. proton
2. α-particle
3. neutron
4. β-particle

Answer: 4. β-particle

De-Broglie wavelength is maximum for \(\beta\)-particle

Question 62. The wavelength associated with an electron accelerated through a potential difference of 100 V, is of the order of:

1. 1000 Å
2. 100 Å
3. 10.5 Å
4. 1.2 Å

Answer: 4. 1.2 Å

⇒ \(\lambda =\frac{h}{\sqrt{2 m e V}} \)

=\(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}\)

=1.227 \(\times 10^{-10}=1.227 Å\)

Question 63. The de-Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it

1. \(\frac{h}{m v}\)
2. HMV
3. \(\frac{m h}{v}\)
4. \(\frac{m}{h v}\)

Answer: 1. \(\frac{h}{m v}\)

The de-Broglie wavelength associated with the particle of mass m moving with velocity v is

–\(\lambda =\frac{h}{p}\)

⇒ \(\lambda =\frac{h}{m v}\)

Where, m and v are the mass and the velocity of the particle and h is Planck’s constant.

Question 64. The wave nature of electron was experimentally verified by:

1. de-Broglie
2. Hertz
3. Einstein
4. Davisson and Germer

Answer: 4. Davisson and Germer

The wave nature of electrons was experimentally verified by Davisson and Germer.

Question 65. In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by:

1. increasing the filament current
2. decreasing the filament current
3. decreasing the potential difference between the anode and filament
4. increasing the potential difference between the anode and filament

Answer: 4. increasing the potential difference between the anode and filament

In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and filament.

 

Atoms

Question 1. When an α-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:

1. \(\frac{1}{\sqrt{m}}\)
2. \(\frac{1}{m^2}\)
3. m
4. \(\frac{1}{m}\)

Answer: 4. \(\frac{1}{m}\)

At the closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy

Here, K.E=\(\frac{1}{2} m v^2\)

Electrostatic potential energy,

= \(\frac{K Q q}{d} \)

⇒ \(\frac{1}{2} m v^2 =\frac{K Q q}{d}\)

d \(\propto \frac{1}{m}\)

Question 2. In a Rutherford scattering experiment when a projectile charge Z₁ and mass M₁ approaches a target nucleus of charge Z₂ and mass M₂, the distance of the closest approach is r₀. The energy of the projectile is:

1. directly proportional to M₁ x M₂
2. directly proportional to Z₁Z₂
3. inversely proportional to Z₁
4. directly proportional to mass M₁

Answer: 2. directly proportional to Z₁Z₂

The energy of the projectile is directly proportional to Z₁ Z₂

Question 3. An alpha nucleus of energy \(\frac{1}{2} m v^2 \)bombards a heavy nuclear target of charge Ze. Then the distance of the closest approach for the alpha nucleus will be proportional to:

1. \(\frac{1}{\mathrm{Ze}}\)
2. \(v^2\)
3. \(\frac{1}{m}\)
4. \(\frac{1}{v^2}\)

Answer: 3. \(\frac{1}{m}\)

From question,\(\frac{1}{2} m v^2 =\frac{1}{4 \pi \varepsilon_0} \frac{(2 e)(z e)}{r_0}\)

⇒ \(r_0 =\frac{1}{4 \pi \varepsilon_0} \frac{2 z e^2}{\frac{1}{2} m v^2}\)

∴ \(r_0 \propto \frac{1}{m}\)

Question 4. Let T₁ and T₂ be the energy of an electron in the first and second excited states of the hydrogen atom, respectively. According to the Bohr’s model of an atom, the ratio T₁: T₂ is :

1. 1: 4
2. 4: 9
3. 4: 9
4. 9: 4

Answer: 4. 9: 4

For the first excited state, n=2

For the second excited state, n=3

∴ \(\frac{T_1}{T_2}=\frac{r_2^2}{r_1^2}=\frac{3^2}{2^2}=\frac{9}{4}\)

Read and Learn More NEET Physics MCQs

Question 5. The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by :

1. \(\mathrm{E}_n=\frac{13.6}{n^2} e \mathrm{~V}\)
2. \(\mathrm{E}_n=-\frac{13.6}{n^2} \mathrm{eV}\)
3. \(\mathrm{E}_n=-\frac{1.36}{n^2} e \mathrm{~V}\)
4. \(E_n=-13.6 \times n^2 \mathrm{eV}\)

Answer: 2. \(\mathrm{E}_n=-\frac{13.6}{n^2} \mathrm{eV}\)

The energy of an electron in Bohr’s orbit of a hydrogen atom is given by the expression.

⇒ \(E_n=-\frac{2 \pi^2 m e^4 z^2}{n^2 h^2\left(4 \pi \varepsilon_0\right)^2}=-13.6 \frac{z^2}{n^2} \mathrm{eV}\)

Since, Z=1 for the hydrogen atom.

So, the above equation further simplifies to,

∴ \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)

Question 6. For which one of the following, the Bohr model is not valid?

1. Singly ionised helium atom (He⁺)
2. Deuteron atom
3. Singly ionised neon atom (Ne⁺)
4. Hydrogen atom

Answer: 3. Hydrogen atom

Bohr’s model is not valid for singly ionised neon K \(\left(\mathrm{Ne}^{+}\right)\).

Question 7. The total energy of an electron in an orbit is – 3.4 eV. Its kinetic and potential energies are, respectively :

1. – 3.4 eV, – 6.8 eV
2. – 3.4 eV, – 6.8 eV
3. – 3.4 eV, – 3.4 eV
4. – 3.4 eV, – 3.4 eV

Answer: 2. – 3.4 eV, – 6.8 eV

In Bohr’s model of H -atom,

⇒ \(\mathrm{K} . \mathrm{E} =|\mathrm{TE}|=\frac{|\mathrm{U}|}{2}\)

⇒ \(\mathrm{K} . \mathrm{E} =3.4 \mathrm{eV} \)

∴ \(\mathrm{U} =-6.8 \mathrm{eV}\)

Question 8. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 A and its ground state energy equals – 13.6 eV. If the electron in the hydrogen atom is replaced by muon (μ^-1) [Charge same as electron and mass is 207 m_e, the first Bohr radius and ground state energy will be :

1. 0.53 x 10^-13 m, – 36 eV
2. 25.6 x 10^-13 m, – 2.8 eV
3. 2.56 x 10^-13 m, – 2.8 KeV
4. 2.56 x 10^-13 m, – 13.6 eV

Answer: 3. 2.56 x 10^-13m, – 2.8 KeV

⇒ \(r_n \propto \frac{1}{m}\)

⇒ \(r_r=\frac{0.51}{207}=2.50 \times 10^{-13} \mathrm{~m}\)

Since,\(\mathrm{E} \propto m_e\)

u=\(-13.6 \times 207=-2.8 \mathrm{KeV}\)

Question 9. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is:

1. 2: – 1
2. 1: – 1
3. 1: 1
4. 1: – 2

Answer: 2. 1: – 1

We have that,\(\mathrm{KE}\)=- (total energy)

So kinetic energy: total energy

=1:-1

Question 10. An electron in a hydrogen atom makes a transition n₁ → n₂ where n₁ is the principal quantum number of the two states. Assuming Bohr’s model to be valid, the period of the electron in the initial state is eight times that in the final state. The possible values of n₁ and n₂ are:

1. n₁ = 6 and n₂ = 2
2. n₁ = 8 and n₂ = 1
3. n₁ = 8 and n₂ = 2
4. n₁ = 4 and n₂ = 2

Answer: 4. n₁ = 4 and n₂ = 2

The period of an electron in Bohr orbit is,

T= \(\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \)

T \(\propto n^3 \)

⇒ \(\frac{T_1}{T_2}=\frac{n_1^3}{n_2^3} \)

As, \(T_1=8 T_2\) then

⇒ \((\frac{8 T_2}{T_2})=\frac{n_1^3}{n_2^3} \)

⇒ \((\frac{n_1}{n_2})^3=\left(2^2\right)^3\)

⇒ \(\frac{n_1}{n_2}=2^2 \)

∴ \(n_1=2 n_2 \)

Question 11. The electron in a hydrogen atom first jumps from a third excited state to the second excited state and then from the second excited to the first excited state. The ratio of the wavelengths λ₁: λ₂ emitted in the two cases is:

1. \(\frac{7}{5}\)
2. \(\frac{27}{20}\)
3. \(\frac{27}{5}\)
4. \(\frac{20}{7}\)

Answer: 4. \(\frac{20}{7}\)

We know that,

Energy, E =\(\frac{h c}{\lambda}=13.6\left[\frac{1}{h_2^2}-\frac{1}{h_1^2}\right]\)

=13.6\(\left[\frac{1}{h_1^2}-\frac{1}{h_2^2}\right]\)  → Equation 1

According to the question,

⇒ \(E_1=\frac{h c}{\lambda_1}=13.6\left[\frac{1}{3^2}-\frac{1}{4^2}\right]\)

And \(E_2 =\frac{h c}{\lambda_2} \)

=13.6\(\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\) →  Equation 2

From eq. (1) and (2),

⇒ \(\frac{\lambda_1}{\lambda_2} =\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{10}}=\frac{20}{7}\)

=20: 7

Question 12. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom of mass m acquired as a result of photon emission will be: (R Rydberg constant and h Planck’s constant)

1. \(\frac{24 h R}{25 m}\)
2. \(\frac{25 h R}{24 m}\)
3. \(\frac{25 m}{24 h R}\)
4. \(\frac{24 m}{25 h R}\)

Answer: 1. \(\frac{24 h R}{25 m}\)

According to the question,

Electron passes from fifth energy level to ground level means;-

⇒ \(E_5-E_1 =\frac{h c}{\lambda} \)

⇒ \(\frac{R h c}{25}-R h c =\frac{h c}{\lambda} \)

⇒ \(\frac{24}{25} R =\frac{1}{\lambda}\)

But, p=\(\frac{h}{\lambda}\)

and v=\(\frac{h}{m \lambda}=\frac{24}{25} \frac{\mathrm{R} h}{m}\)

Question 13. Monochromatic radiation emitted when an electron on a hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is:

1. 4 x 10¹⁵ Hz
2. 5 x 10¹⁵ Hz
3. 1.6 x 10¹⁵ Hz
4. 2.5 x 10¹⁵ Hz

Answer: 3. 1.6 x 10¹⁵ Hz

According to the question,

Energy released =(-3.4)-(-13.6)=10.2 \(\mathrm{eV}\)

Using Einstein’s photoelectric equation

W = E-e V=h v

v = \(\frac{E-e V}{h}=\frac{(10.2-3.57) e}{6.67 \times 10^{-34}} \)

v = \(\frac{6.03 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} \)

= 1.6 \(\times 10^{15} \mathrm{~Hz}\)

Question 14. The energy of a hydrogen atom in the ground state is – 13.6 eV. The energy of a He⁺ ion in the first excited state will be:

1. – 13.6 eV
2. – 27.2 eV
3. – 54.4 eV
4. – 6.8 eV

Answer: 1. – 13.6 eV

We know that, E=\(\frac{-13.6}{n^2} z^2\) for first excited state n=2 and for \(\mathrm{He}^{+}=2\)

E=\(\frac{-13.6}{2^2} \times 2^2=-13.6 \mathrm{eV}\)

Question 15. The electron in the hydrogen atom jumps from its excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in

1. 5.1V
2. 21.1V
3. 17.2
4. 7 V

Answer: 4. 7 V

For n=1, \(E_1=\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}\)

For n=3, \(E_7=\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}\)

So, required energy, E=\(E_3-E_1 =-(1.51)-(-13.6)=12.09 \mathrm{eV}\)

=W+e V

e V =E-W

e V =(12.09-5.1) e

V =7 volt

Question 16. The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength:

1. n = 3 to n = 2 states
2. n = 3 to n = 1 states
3. n = 2 to n = 1 states
4. n = 4 to n = 3 states

Answer: 4. n = 4 to n = 3 states

We know that the number of spectral lines

N =\(\frac{n(n-1)}{2}\)

⇒ \(\frac{n(n-1)}{2}\) =6

⇒ \(n^2\)-n-12 =0

(n-4)(n+3) =0

n =4

Question 17. The ground state energy of the hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:

1. 3.4 eV
2. 6.8 eV
3. 10.2 eV
4. zero

Answer: 3. 10.2 eV

Ground state energy of hydrogen atom

⇒ \(E_1=-13.6 \mathrm{eV}\)

The energy of an electron in a first excited state (n=2)

⇒ \(E_2=-\frac{13.6}{(2)^2} \mathrm{eV}\)

Excitation energy, \(\Delta E =E_2-E_1\)

= \(-\frac{13.6}{4}-(-13.6)\)

= -3.4+13.6=10.2 \(\mathrm{eV}\)

Question 18. The total energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:

1. 6.8 eV
2. 13.6 eV
3. 1.7 eV
4. 3.4 eV.

Answer: 4. 3.4 eV.

Energy of \(n^{\text {th }}\) orbit of hydrogen atom is given by

⇒ \(E_n=-\frac{13.6}{(2)^2} \mathrm{eV}\)

for ground state n=1

⇒ \(E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}\)

for excited state n=2

∴ \(E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}\)

Question 19. The total energy of an electron in the first excited state of a hydrogen atom is about – 3.4 eV. Its kinetic energy in this state is:

1. 3.4 eV
2. 6.8 eV
3. – 3.4 eV
4. – 6.8 eV

Answer: 1. 3.4 eV

K.E.=\(\left|\frac{1}{2} P . E\right|\)

But P.E is negative,

Total energy =\(\left|\frac{1}{2} P \cdot E\right|-P . E \)

= \(-\frac{P . E}{2}=-3.4 \mathrm{eV}\)

∴ \(\mathrm{K} . \mathrm{E} =+3.4 \mathrm{eV}\)

Question 20. The Bohr model of atoms:

1. assumes that the angular momentum of electrons is quantized “
2. uses Einstein’s photo-electric equation
3. predicts continuous emission spectra for atoms
4. predicts the same emission spectra for all types of atoms

Answer: 1. assumes that the angular momentum of electrons is quantized ”

Both modes of atoms assume that the angular momentum of electrons is quantized.

∴ \(q= \pm h e\)

Question 21. When a hydrogen atom is in its first excited level, its radius is:

1. four times, it’s ground state radius
2. twice, it’s ground state radius
3. same as its ground state radius
4. half of its ground state radius

Answer: 1. four times, it’s ground state radius

The radius of \(n^{\text {th }}\) orbit of hydrogen

⇒ \(r_{\mathrm{n}}=\frac{\varepsilon_0 n^2 h^2}{\pi m e^2 Z}\)

⇒ \(r_{\mathrm{n}}=\frac{n^2 a_0}{Z} \text { or } r_{\mathrm{n}} \propto \frac{n^2}{Z}\)

For ground state, n=1

Atomic number, Z=1

For the first excited state, n=2

⇒ \(\frac{r_2}{r_1}=\left(\frac{2}{1}\right)^2=4\)

⇒ \(r_2=4 r_1\)

Hence, the radius of the first excited state is four times the radius in the ground state.

Question 22. Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de-Broglie wavelength \(\lambda\) of that electron as:

1. \((0.529) n \lambda\)
2. \(\sqrt{n \lambda}\)
3. \((13.6) \lambda\)
4. n \(\lambda\)

Answer: 4. n \(\lambda\)

In terms of the wavelength of the wave corresponding with an electron, the circumference of an orbit in an atom is given by,

∴ \(2 \pi r_n=n \lambda\)[\(r_n\)=radius of any n orbit]

Question 23. To explain his theory, Bohr used:

1. conservation of linear momentum
2. conservation of angular momentum
3. conservation of quantum frequency
4. conservation of energy

Answer: 2. conservation of angular momentum

Bohr used quantization of angular momentum. For stationary orbits, Angular momentum,

⇒ \(\mathrm{I} \omega(\text { omega })=\frac{n h}{2 \pi}\) .

Where,n=1,2,3,.. etc.

Question 24. If an electron in a hydrogen atom jumps from the 3^rd orbit to the 2^nd orbit, it emits a photon of wavelength \(\lambda\). When it jumps from the 4^th orbit to the 3^rd orbit, the corresponding wavelength of the photon will be:

1. \(\frac{16}{25} \lambda\)
2. \(\frac{9}{16} \lambda\)
3. \(\frac{20}{7} \lambda\)
4. \(\frac{20}{13} \lambda\)

Answer: 3. \(\frac{20}{7} \lambda\)

Using Rydberg formula we have,

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

= \(R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}\)

and \(\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{7 R}{144} \)

⇒ \(\frac{\lambda^{\prime}}{\lambda}=\frac{5 R}{36}-\frac{144}{7 R}=\frac{20}{7}\)

∴ \(\lambda^{\prime}=\frac{20}{7} \lambda \)

Question 25. The transition from the state n = 3 to n = 1 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

1. 2 → 1
2. 3 → 2
3. 4 → 2
4. 4 → 3

Answer: 4. 4 → 3

Infrared radiation is found in the Paschen, Bracket and p fund series and it is obtained when ‘e’ transitions from a high energy level to a minimum 3^rd energy level.

Question 26. The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of a hydrogen-like ion is:

1. 4
2. 1
3. 2
4. 3

Answer: 3. 2

We know that, the Lyman series for \(\mathrm{H}ion\)

⇒ \(\frac{h c}{\lambda}=\mathrm{R} h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

and for H-like atoms.

⇒ \(\frac{h c}{\lambda}=\mathrm{Z}^2 \mathrm{R} h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

compare we get,

⇒ \(\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=Z^2\left(\frac{1}{2}-\frac{1}{16}\right)\)

Solving we get Z=2

Question 27. An electron in the hydrogen atom jumps from the exited state n to the ground state. The wavelength so emitted illuminates a photo-sensitive material having a work function of 2.75 eV. If the stopping potential of the photo-electron is 10 V, the value of n is:

1. 3
2. 4
3. 5
4. 2

Answer: 1. 3

Using Einstein’s photoelectric equation

E =\(W+K E_{\max }=W+e V_0\)

=2.75+10=12.75 \(\mathrm{eV}\)

Difference of =4 and n = 1

the energy level is 19.75 eV

Here Higher energy level is n = 4 and the lower energy level is n = 3

Question 28. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?

1. 1.9 eV
2. 11.1eV
3. 13. eV
4. 0.65 eV

Answer: 2. 11.1eV

But 11.1 eV is not present in the diagram.

So, 11.1 eV is not possible.

Question 29. The ionization potential of a hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:

1. one
2. two
3. three
4. four

Answer: 3. three

According to question, \(12.1 =13.6\left(\frac{1}{l^2}-\frac{1}{n^2}\right)\)

⇒ \(n^2 =\frac{13.6}{1.5}\)=9

n =3

Number of spectral lines emitted

= \(\frac{n(n-1)}{2}=\frac{3 \times 2}{2}\)

= 3

Question 30. Energy levels A, B and C of a certain action correspond to increasing values of energy i.e. \(E_A<E_B<E_C\). If \(\lambda_1, \lambda_2 \) and \(\lambda_3\) are the wavelength of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

1. \(\lambda_3=\lambda_1+\lambda_2\)
2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)
3. \(\lambda_1+\lambda_2+\lambda_3=0\)
4. \(\lambda_3{ }^2=\lambda_1{ }^2+\lambda_2{ }^2\)

Answer: 2. \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

⇒ \(E_A<E_B<E_C\)

E =\(\frac{h c}{\lambda}\)

⇒ \(E_{C A} =E_{C B}+E_{B A} \)

⇒ \(\frac{h c}{\lambda_3} =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \)

⇒ \(\frac{1}{\lambda_3} =\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \)

∴ \(\lambda_3 =\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 31. Energy E of a hydrogen atom with principal quantum number n is given by E = \(\frac{-13.6}{n^2} \mathrm{eV}\). The energy of a photon ejected when the electron jumps for n = 3 state n = 2 state of hydrogen is approximately

1. 1.5 eV
2. 0.85 eV
3. 3.4 eV
4. 1.9 eV

Answer: 4. 1.9 eV

Energy of photon=\(E_2-E_2\)

= \(\frac{13.6}{9}-\left(\frac{-13.6}{9}\right)\)

= \(\frac{5}{30} \times 13.6 \)

= 1.9 \(\mathrm{eV}\)

Question 32. When electron jumps from n = 4 to n = 2 orbit, we get:

1. the second line of the Lyman series
2. the second line of the Balmer series
3. the second line of the Paschen series
4. an absorption line of the Balmer series

Answer: 2. second line of Balmeer series

Jump to the second orbit leads to the Balmer series. When an electron jumps from the 4^th orbit to the 2^nd, orbit shall give rise to the second line of the Balmer series.

Question 33. The spectrum obtained from a sodium vapour lamp is an example of:

1. band spectrum
2. continuous spectrum
3. emission spectrum
4. absorption spectrum

Answer: 3. emission spectrum

A spectrum is noticed, when light coming straightforwardly from a source is analysed with a spectroscope. The spectrum acquired from a sodium vapour lamp is the emission spectrum.

Hence, The spectrum obtained from a sodium vapour lamp is an example of the transmission spectrum.

 

Nuclei

Question 1. When an a-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:

1. \(\frac{1}{\sqrt{m}}\)
2. \(\frac{1}{m^2}\)
3. m
4. \(\frac{1}{m}\)

Answer: 4. \(\frac{1}{m}\)

At the closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy

Here K.E=\(\frac{1}{2} m v^2\)

Electrostatic potential energy,

= \(\frac{K Q q}{d}\)

⇒ \(\frac{1}{2} m v^2 =\frac{K Q q}{d}\)

d \(\propto \frac{1}{m}\)

Question 2. The nuclei of which one of the following pairs of nuclei are isotones?

1. \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ga}^{71}\)
2. \({ }_{38} \mathrm{Sr}^{84},{ }_{38} \mathrm{Ga}^{86}\)
3. \({ }_{42} \mathrm{Se}^{92},{ }_{40} \mathrm{Ga}^{92}\)
4. \({ }_{20} \mathrm{Se}^{40},{ }_{16} \mathrm{Ga}^{12}\)

Answer: 1. \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ga}^{71}\)

The Isotones mean several neutrons remain the same.

Question 3. A nucleus represented by the symbol \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}\) has:

1. Z neutrons and A-Z protons
2. Z protons and A – Z neutrons
3. Z protons and A neutrons
4. Z protons and Z – A neutrons

Answer: 2. Z protons and A – Z neutrons

Z= Number of protons

The total number of protons and the number

.’. Number of neutrons, N =A- Z

Read and Learn More NEET Physics MCQs

Question 4. In the nucleus of \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}\), the number of protons, neutrons, and electrons are:

1. 11,12,0
2. 23, 12, 11
3. 12, 11, 0
4. 23, 11, 12

Answer: 1. 11,12,0

Z= 11 i.e., number of protons = 11, A = 23

Number of neutrons = A – Z= 12

Number of electron= 0 (No electron in the nucleus)

Therefore 11,12,0.

Question 5. A nucleus of mass number 189 splits into two nuclei having mass numbers 125 and 64. The ratio of the radius of two daughter nuclei respectively is:

1. 1: 1
2. 4 : 5
3. 5:4
4. 25: 16

Answer: 3. 5:4

Nuclear radius, R=\(R_0 A^{\frac{1}{3}}\)

R \(\alpha(A)(A)^{\frac{1}{3}} \)

⇒ \(\frac{R_1}{R_2} =\left(\frac{125}{64}\right)^{1 / 3}\),

∴ \(\frac{R_1}{R_2} =\frac{5}{4}\)

Question 6. If radius of the \({ }_{13}^{27} \mathrm{Al}\) nucleus is taken to be \(R_{A l}\) then the radius of \({ }_{13}^{125} \mathrm{Te}\) nucleus is nearly:

1. \(\left(\frac{53}{13}\right)^{\frac{1}{3}} R_{A l}\)
2. \(\frac{5}{3} R_{A l}\)
3. \(\frac{3}{5} R_{A l}\)
4. \(\left(\frac{13}{53}\right)^{\frac{1}{3}} R_{A l}\)

Answer: 2. \(\frac{5}{3} R_{A l}\)

The radius of the nucleus is,

R =\(R_0 A^{1 / 3}\)

R \(\propto A^{1 / 3} \)

⇒ \(\frac{R_{\mathrm{A} l}}{R_{\mathrm{Te}}} =\left(\frac{A_{\mathrm{A} l}}{A_{\mathrm{Te}}}\right)^{\frac{1}{3}}=\left(\frac{27}{125}\right)^{\frac{1}{3}}=\frac{3}{5} \)

∴ \(R_{\mathrm{Te}} =\frac{5}{3} R_{\mathrm{Al}}\)

Question 7. If the nuclear radius of \({ }^{27} \mathrm{Al}\) is 3.6 Fermi, the approximate nuclear radius of \({ }^{64} \mathrm{Cu}\) in Fermi is:

1. 2.4
2. 1.2
3. 4.8
4. 3.6

Answer: 3. 4.8

We know that, For \(\mathrm{A}_1\) Nuclear Radius r \(\propto A^{1 / 3}\)

where, \(\mathrm{A}\)= mass number

⇒ \(r=r_0 A^{1 / 3} \)

= \(r_0(27)^{1 / 3}=3 r_0 \text { since nuclear radius is } 27 \text { (given) }\)

⇒ \(r_0=\frac{3.6}{3}=1.2 \mathrm{fm}\)

For \(\mathrm{Cu} r=r_0 \mathrm{~A}^{1 / 3}=1.2 \mathrm{fm}(6.4)^{1 / 3}=4.8 \mathrm{fm}\)

Question 8. Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be:

1. 1:3
2. 3: 1
3. \((3)^{1 / 3}: 1\)
4. 1: 1

Answer: 4. 1: 1

The density of nuclear matter is independent of mass number so the required ratio is 8: 1.

Question 9. If the nucleus \({ }_{13}^{27} \mathrm{Al}\)has a nuclear radius of about 3.6 fm, then \({ }_{32}^{125} \mathrm{Te}\) would have its radius approximately as:

1. 9.6 fm
2. 12.0 fm
3. 4.8 fm
4. 6.0 fm.

Answer: 4. 6.0 fm.

R=\(R_0 A^{1 / 3}\)

⇒ \(\frac{R_{\overparen{T e}}}{R_{A l}}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5^3}{3^3}\right)^{1 / 3} \)

=\(\frac{5}{3} \)

∴ \(R_{T e}=\frac{5}{3} \times 3.6=6 \mathrm{fm} \)

Question 10. The radius of germanium (Ge) nuclide is measured to be twice the radius of 9 Be . The number of nucleons in Ge is:

1. 72
2. 73
3. 74
4. 75

Answer: 1. 72

Nuclear radii, R =\(R_0 A^{1 / 3}\)

⇒ \(R_0 =1.2 \mathrm{fm}\)

R \(\propto(A)^{1 / 3}\)

Where, \(R_0=1.2 \mathrm{fm}\)

or R \(\propto(A)^{1 / 3}\)

⇒ \(\frac{R_{H_e}}{R_{G_e}}=\frac{(a)^{\frac{1}{3}}}{(A)^{\frac{1}{3}}}\)

or \(\frac{R_{H_e}}{2 R_{B^*}}=\frac{(q)^{\frac{1}{3}}}{(\lambda)^{\frac{1}{3}}}\)

Given \(R_{G e} =2 R_{B C} \)

⇒ \((A)^{1 / 3} =2 \times(9)^{1 / 3} \)

A =23 \(\times 9=8 \times\) 9=72

or Number of nuclei is Ge is 72

Question 11. The volume occupied by an atom is greater than the volume of the nucleus by a factor of about:

1. 10¹
2. 10⁵
3. 10¹⁰
4. 10¹⁵

Answer: 4. 10¹⁵

∴ \(\frac{\text { Volume of atom }}{\text { Volume of nucleus }}=\frac{\frac{4}{3} \pi\left(10^{-10}\right)^3}{\frac{4}{3} \pi\left(10^{-15}\right)^3}\)

Question 12. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:

1. 0.9 MeV
2. 9.4 MeV
3. 804 MeV
4. 216 MeV

Answer: 2. 9.4 MeV

Given, \(A_1\) =240

⇒ \(A_2 =A_3\)=120

⇒ \(\mathrm{BE}_1 =7.6 \mathrm{MeV}\)

⇒ \(\mathrm{BE}_2 =\mathrm{BE}_3=8.5 \mathrm{MeV} \)

⇒ \(\Delta \mathrm{BE}^2 =\mathrm{BE}_2+\mathrm{BE}_3-\mathrm{BE}_1 \)

= 8.5+8.5-7.6

= 17-7.6

= 9.4 \(\mathrm{MeV}\)

Question 13. The energy equivalent of 0.5 g of a substance is:

1. 4.5 x 10¹³ J
2. 1.5 x 10¹³J
3. 0.5 x 10¹³ J
4. 45 x 10¹⁶ J

Answer: 1. 4.5 x 10¹³ J

Given, m=0.5 \(\mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}\)

Einstein’s mass-energy equivalence relation,

E=m \(c^2\)

Where, c= velocity of light =3 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

⇒ \(\mathrm{E} =0.5 \times 10^{-3} \times\left(3 \times 10^8\right)^2 \)

=4.5 \(\times 10^{13} \mathrm{~J}\) .

Question 14. The binding energy per nucleon of \({ }_3^7 \mathrm{Li}\) and \({ }_2^4 \mathrm{He}\) nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ’ \({ }_2^4 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\), the value of energy Q released is:

1. 19.6 MeV
2. -2.4 MeV
3. 8.4 MeV
4. 17.3 MeV

Answer: 4. 17.3 MeV

The value of energy,

Q = 2(4 x 7.06) — 7 x (5.60)

2 = (8 x 7.06-7 x 5.60)

2 = 56.48-39.2 2= 17.28 MeV 2 = 17.3 MeV

Question 15. How does the Binding Energy per nucleon vary with the increase in the number of nucleons?

1. Decrease continuously with mass number.
2. First, it decreases and then increases with an increase in mass number.
3. First, it increases and then decreases with an increase in mass number.
4. Increases continuously with mass number.

Answer: 3. First increase and then decrease with an increase in mass number.

Question 16. The mass of \({ }_3^7 \mathrm{Li}\) nucleus is 0.042M less than the sum of the masses of all its nucleons. The binding energy per nucleon of \({ }_3^7 \mathrm{Li}\) nucleus is nearly:

1. 46 MeV
2. 5.6 MeV
3. 3.9 MeV
4. 23 MeV

Answer: 2. 5.6 MeV

According to the question if,

m=1 u, c=3 \(\times 10^8 \mathrm{~ms}^{-1} \)

E=931 \(\mathrm{MeV}, 1 u=931 \mathrm{MeV}\)

Binding energy =0.042 \(\times 931 \)

= 39.10 \(\mathrm{MeV}\)

Binding energy per nucleon

= \(\frac{39.10}{7}=5.58=5.6 \mathrm{MeV}\)

Question 17. If M(A, Z) M_P and M_n denote the masses of the nucleus \({ }_Z^A X\) proton and neutron respectively in units of n \(\left(1 u=931.5 \mathrm{MeV} / c^2\right)\) and BE represents its binding in MeV, then:

1. \(M(A, Z)=Z M_P+(A-Z) M_n-B E / c^2\)
2. \(M(A, Z)=Z M_P+(A-Z) M_n+B E\)
3. \(M(A, Z)=Z M_P+(A-Z) M_n-B E\)
4. \(M(A, Z)=Z M_p+(A-Z) M_n+B E / c^2\)

Answer: 1. \(M(A, Z)=Z M_P+(A-Z) M_n-B E / c^2\)

Mass defect, \(A_m=2 M_P+(A-Z) H_n-M(A, Z)\)

Binding Energy, \(\mathrm{BE}=\Delta m c^2 \)

B E=\(\left[Z M_P+(A-Z) M_n-M(A, Z)\right] C^2 \)

M(A, Z)=\(Z M_P+(A-Z) M_n-\frac{B E}{C^2} \)

Question 18. A nucleus \({ }_Z^A X\) has mass represented by M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then:

1. B. E. =\(\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)
2. B. E. =\(\left[Z M_p+A M_n-M(A, Z)\right] c^2\)
3. B. E. =\(M(A, Z)-Z M_p-(A-Z) M_n\)
4. B. E. =\(\left[M(A, Z)-Z M_p-(A-Z) M_n\right] c^2\)

Answer: 1. B. E. =\(\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)

Question 19. The binding energy of deuteron is 2.2 MeV and that of \({ }_2^4 \mathrm{He}\) is 28 MeV If two deuterons are fused to form one \({ }_2^4 \mathrm{He}\) then the energy released is:

1. 30.2 MeV
2. 25.8 MeV
3. 23.6 MeV
4. 19.2 MeV

Answer: 3. 23.6 MeV

Energy released, \(\mathrm{Q} =(\mathrm{B} . \mathrm{E}) \text { Product }- \text { (B.E. })_{\text {reactant }} \)

⇒ \(\mathrm{Q}=(28-4.4) \mathrm{MeV}\)

=23.6 \(\mathrm{MeV}\)

Question 20. In the reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 n \text {, }\), if the binding energies of \({ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H}\) and \({ }_2^4 \mathrm{He}\) are respectively a, b and c (in Me V), then the energy (in MeV) released in this reaction is:

1. a + b + c
2. a + b-c
3. c-a-b
4. c + a-b

Answer: 3. c-a-b

Energy in given reaction = BE of products – B.E. of reactants

= c – (a + b)

= c-a-b

Question 21. Mp denotes the mass of a proton and M_n of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light):

1. M(N, Z) = NM_n + ZMp – BC²
2. M(N, Z) = NM_n + ZMp – BC²
3. M(N, Z) = NM_n + ZMp – B/C²
4. M(N, Z) = NM_n + ZMp + B/C²

Answer: 3. M(N, Z) = NM_n + ZMp – B/C²

⇒ Binding, B =\([N M_n+2 M_P-M(N_1^2] C^2\).

∴ \(M\left(N_1^2\right) =N M_n+Z M_P-\frac{B}{C^2}\)

Question 22. The energy equivalent of one atomic mass unit is:

1. 1.6 \(\times 10^{-19} \mathrm{~J}\)
2. 6.02 \(\times 10^{23} \mathrm{~J}\)
3. 931 \(\mathrm{Mev}\)
4. 9.31 \(\mathrm{MeV}\)

Answer: 3. 931 \(\mathrm{Mev}\)

⇒ According to Einstein,

Mass-energy equation, \(E=m c^2\).

Taking, mass, \(m =1 \mathrm{amu} \)

= \(1.66 \times 10^{-27} \mathrm{~kg}\),

c =\(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)

and c=3.0 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

Therefore, E =\(\left(1.66 \times 10^{-27}\right) \times\left(3 \times 10^8\right)^2 \mathrm{~J} \)

= \(1.49 \times 10^{-10} \mathrm{~J}\)

= \(\frac{1.49 \times 10^{-10}}{1.6 \times 10^{-13}} \mathrm{MeV} \)

⇒ \((1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}) \)

=931.25 \(\mathrm{MeV}\)

Hence, 1 \(\mathrm{amu} \approx 931 \mathrm{MeV}\)

Question 23. In the given nuclear reaction, the element X is : \({ }_{11}^{22} \mathrm{Na} \rightarrow \mathrm{X}+\mathrm{e}^{+}+v\)

1. \(\frac{1}{2}\)
2. \(\frac{1}{2 \sqrt{2}}\)
3. \(\frac{2}{3}\)
4. \(\frac{2}{3 \sqrt{2}}\)

Answer: 3. \(\frac{2}{3}\)

Using the law of charge conservation,

⇒ \({ }_{11}^{22} \mathrm{Na} \rightarrow \mathrm{X}+\mathrm{e}^{+}+\mathrm{v} \)

∴ \({ }_{11}^{22} \mathrm{Na} \longrightarrow{ }_{10}^{22} \mathrm{Ne}+\mathrm{e}^{+}+\mathrm{v}\)

Question 24. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:

1. \(\frac{1}{2}\)
2. \(\frac{1}{2 \sqrt{2}}\)
3. \(\frac{2}{3}\)
4. \(\frac{2}{3 \sqrt{2}}\)

Answer: 2. \(\frac{1}{2 \sqrt{2}}\)

Given, \(\mathrm{T}_{\% 0}^{\prime}=100 \text { hours. }\)

For \(\mathrm{N}_0,\)

⇒ \(\mathrm{T}=150 \text { hours }\)

⇒ \(\mathrm{N}=\text { ? }\)

We know, \(\mathrm{N}(t)=\mathrm{N}_0\left(\frac{1}{2}\right)^{\frac{t}{t_{112}}}\)

⇒ \(\mathrm{N}(t)=\mathrm{N}_0\left(\frac{1}{2}\right)^{\frac{150}{100}} \)

⇒ \(\mathrm{~N}(t)=\mathrm{N}_0\left(\frac{1}{2}\right)^{\frac{3}{2}}\)

⇒ \(\frac{\mathrm{N}(t)}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\frac{3}{2}}\)

=\(\frac{1}{2^{\frac{3}{2}}}=\frac{1}{2 \cdot \sqrt{2}} \)

Question 25. A radioactive nucleus \({ }_Z^A X\) undergoes spontaneous decay in the sequence. \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \rightarrow \mathrm{Z}-{ }_1 \mathrm{~B} \rightarrow \mathrm{Z}-{ }_3 \mathrm{C} \rightarrow \mathrm{Z}-{ }_2 \mathrm{D}\). where Z is the atomic number of element X. The possible decay particles in the sequence are:

1. \(\alpha, \beta^{-}, \beta^{+}\)
2. \(\alpha, \beta^{+}, \beta^{-}\)
3. \(\beta^{+}, \alpha, \beta^{-}\)
4. \(\beta^{-}, \alpha, \beta^{+}\)

Answer: 3. \(\beta^{+}, \alpha, \beta^{-}\)

Given, Possible decayed particles

⇒ \(\text { (a) } Z \rightarrow Z-1 \text { means loss of } \beta^{+}\)

Z-1\( \rightarrow Z-3 \text { means loss of } \alpha\)

Z-3 \(\rightarrow Z-2 \text { means loss of } \beta^{-} \)

Question 26. What happens to the mass number and atomic number of an element when it emits γ-radiation?

1. The mass number decreases by four and the atomic number decreases by two.
2. Mass number and atomic number remain unchanged.
3. The mass number remains unchanged, while the atomic number decreases by one.
4. Mass number increases by four and atomic number increases by two.

Answer: 1. Mass number decreases by four and atomic number decreases by two.

When an atom emits \(\gamma\)-radiation from its nucleus then the mass number and atomic number of the nucleus remain unchanged so that no new element is formed.

Question 27. The half-life of a radioactive sample undergoing a-decay is 1.4 x 10¹⁷ s. If the number of nuclei in the sample is 2.0 x 10²¹. The activity of the sample is nearly:

1. 10⁴Bq
2. 10⁵ Bq
3. 10⁶Bq
4. 10³ Bq

Answer: 1. 10⁴Bq

From the question \(T_{1 / 2}=1.4 \times 10^{17} \mathrm{~s} \)

and no. of sample N=2.0 \(\times 10^{21}\)

Activity of the sample is =\(\lambda \mathrm{N}\)

⇒ \(\lambda \mathrm{N} =\frac{0.693}{\mathrm{~T}_{\mathrm{yz}}} \times 2 \times 10^{21}\)

=\(\frac{0.693}{1.4 \times 10^{17}} \times 2 \times 10^{21}\)

=0.99\( \times 10^4 \approx 10^4 \mathrm{~Bq}\)

Question 28. \(\alpha\)-particle consists of:

1. 2 electrons, 2 protons and 2 neutrons
2. 2 electrons and 4 protons only
3. 2 protons only
4. 2 protons and 2 neutrons only.

Answer: 1. 2 electrons, 2 protons, and 2 neutrons

∴ \(\alpha\)Particle consists of 2 protons and 2 neutrons only.

Question 29. The rate of radioactive disintegration at an instant for a radioactive sample of half-life 2.2 x 10⁹ s is 10¹⁰ s^-1. The number of radioactive atoms in the sample at that instant is:

1. 3.17 x 10²⁰
2. 3.17 x 10¹⁷
3. 3.17 x 10¹⁸
4. 3.17 x 10¹⁹

Answer: 1. 3.17 x 10²⁰

Here \(\mathrm{T}_{1 / 2} =\frac{0.693}{\lambda} \)

2.2 \(\times 10^9 =\frac{0.693}{\lambda} \)

⇒ \(\lambda =\frac{0.693}{2.2 \times 10^9}=3.15 \times 10^{-10} \)

Now ,R =\(\lambda N \)

N =\(\frac{R}{\lambda}=\frac{10^{10}}{3.15 \lambda \times 10^{-10}}\)

= 3.17 \(\times 10^{20}\)

Question 30. For a radioactive material, the half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is:

1. 30
2. 10
3. 20
4. 15

Answer: 3. 20

According to the question,

Initially, the number of nuclei \(N_0\)=600.

After disintegration, the number of nuclei \(N^{\prime}\)=450.

The remaining number of nuclei =600-450=150

Half-life \(t_{1 / 2}=10 \mathrm{~mm}\)

We can write, \(\frac{N}{N_0} =\left(\frac{1}{2}\right)^{\frac{1}{2}} \)

⇒ \(\left(\frac{150}{600}\right) =\left(\frac{1}{2}\right)^{\frac{1}{2}}\)

⇒ \(\left(\frac{1}{2}\right)^{\frac{1}{2}} =\left(\frac{1}{2}\right)^{\frac{1}{2}} \)

t =2=2 \(\times 10=20 minute \)

Question 31. Radioactive material A has a decay constant of 8 λ, and material B has a decay constant of λ. Initially, they have same number of nuclei. After what time the ratio of number of nuclei of material B to that A will be \(\frac{1}{e}\)?

1. \(\frac{1}{\lambda}\)
2. \(\frac{1}{7 \lambda}\)
3. \(\frac{1}{8 \lambda}\)
4. \(\frac{1}{9 \lambda}\)

Answer: 2. \(\frac{1}{7 \lambda}\)

Suppose initial nuclear nuclei in material A and B is \(N_{\mathrm{o}}\)

The number of nuclei in material A after time t is

⇒ \(N_A=N_0 e^{-8 \lambda t}\)  → Equation 1

And nuclear nuclei in material B after time t is

⇒ \(N_B=N_0 e^{-\lambda t}\)  → Equation 2

The ratio of nuclear nuclei of material B to A is \(\frac{1}{e}\)

⇒ \(\frac{N_A}{N_B}=\frac{1}{e}\)  → Equation 2

From eq. (2) and (3)

⇒ \(\frac{e^{-8 \lambda t}}{e^{-\lambda t}} =\frac{1}{e}\)

⇒ \(e^{-1} =e^{-7 \lambda t} \)

-7\( \lambda t \)=-1

t =\(\frac{1}{7 \lambda}\)

Question 32. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is:

1. 15
2. 30
3. 45
4. 60

Answer: 4. 60

Using the formula, Half-life of a radioactive substance

⇒ \(\mathrm{T}_{1 / 2} \propto \log \left(\frac{N_{\mathrm{o}}}{N}\right)\)

Here , \(N_1 =0.6 \mathrm{~N}_0 \text { because } 40 \% \text { decay } \)

⇒ \(N_2 =0.15 \mathrm{~N}_0 \text { because } 85 \% \text { decay } \)

⇒ \(\frac{N_2}{N_1} =\frac{0.15}{0.6 \mathrm{~N}_{\mathrm{o}}}=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)

Time =2 \(\times t_{1 / 2}=2 \times 30 \)

= 60 minutes

Question 33. A nucleus of uranium decays at rest into nuclei of thorium and helium, then :

1. the helium nucleus has more kinetic energy than the thorium nucleus
2. the helium nucleus has less momentum than the thorium nucleus
3. the helium nucleus has more momentum than the thorium nucleus
4. The helium nucleus has less kinetic energy than the thorium nucleus.

Answer: 1. the helium nucleus has more kinetic energy than the thorium nucleus

⇒ \(\dot{\mathrm{U}}=\stackrel{\alpha}{\stackrel{\mathrm{Th}}{\longleftrightarrow}}\)

Using the law of conservation of linear momentum

⇒ \(\vec{P}_U=\vec{P}_\alpha+\vec{P}_{T H}\)

O=\(\vec{P}_\alpha+\vec{P}_{T H}\)

⇒ \(\left|\vec{P}_\alpha\right|=\left|\vec{P}_{T H}\right|=P \)

⇒ \(\mathrm{KE} \text { of } \alpha=\frac{P^2}{2 M_\alpha} \)

and \(\tilde{\mathrm{KE}} \text { of } \mathrm{Th}=\frac{P^2}{2 M_{T H}}\)

⇒ \(M_\alpha<M_{T H} \)

K E of \(\alpha>\) K E of T h

Question 34. A radioisotope X with a half-life of 1.4 x 10⁹ yr decays of Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1: 7. The age of the rock is :

1. 1.96 x 10⁹ yr 
2. 3.93 x 10⁹yr
3. 4.20 x 10⁹yr
4. 8.40 x 10⁹vr

Answer: 3. 4.20 x 10⁹yr

It is given X: Y=1: 7

Means, \(\frac{M_X}{M_Y}=\frac{1}{7}\)

7\( M_X=M_Y\)

And initial ratio, M =\(M_X+M_Y \)

= \(\frac{M_Y}{7}+M_Y\)

= \(\frac{8}{7} M_Y\)

⇒ \(M_Y =\frac{7}{8} M\)

This means \(\frac{1}{8}\) parts remains

Time is taken to become \( \frac{1}{8}\) unstable parts.

= \(3 T_{1 / 2}=3 \times 1.4 \times 10^9 \)

T =4.2 \(\times 10^9 \mathrm{yr}\)

Question 35. The half-life of a radioactive isotope X is 20 yr. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio 1: 7 in a sample of a given rock. The age of the rock is estimated to be:

1. 40 yr
2. 60 yr
3. 60 yr
4. 100 yr

Answer: 2. 60 yr

⇒ \(\frac{N}{N_Y}=\frac{1}{7} \)

⇒ \(\frac{N_X}{N_X+N_Y}=\frac{\mathrm{N}}{\mathrm{N}_0}=\frac{1}{8} \)

By using, \(N=N_0 e^{-\lambda t}\)

we have, \(\frac{N_0}{8}=N_0 e^{-\lambda t} \)

t=3 \(\times 20 \text { years }=60 \text { years. } \)

Question 36. \(\alpha\)-particle, \(\beta\)-particle and \(\gamma\)-rays are all having same energy. Their penetrating power in a given medium in increasing order will be:

1. \(\gamma, \alpha, \beta\)
2. \(\alpha, \beta, \gamma\)
3. \(\beta, \alpha, \gamma\)
4. \(\beta, \gamma, \alpha\)

Answer: 1. \(\gamma, \alpha, \beta\)

∴ \(\gamma\)-rays have the highest penetrative power α-rays have the least penetrative power.

Question 37. A mixture consists of two radioactive materials A₁ and A₂ with half-lives of 20 s and 10 s respectively. Initially, the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after:

1. 60 s
2. 80 s
3. 20 s
4. 40 s

Answer: 4. 40 s

According to the question for 40 \(\mathrm{~g} \) amount

40 \(\mathrm{~g} \rightarrow[\text { half life }]{20 \mathrm{~S}} 20 \mathrm{~g} \rightarrow{20 \mathrm{~S}} 10 \mathrm{~g}\)

For 160 \(\mathrm{~g}\) amount

160 \(\mathrm{~g} \rightarrow[\text { half life }]{10 \mathrm{~S}} 80 \mathrm{~g} \rightarrow{10 \mathrm{~S}} 40 \mathrm{~g} \rightarrow{10 \mathrm{~S}} 20 \rightarrow{\text { 10S }} 10 \mathrm{~g}\)

Hence, after 40\( \mathrm{~s}, \mathrm{~A}_1 and \mathrm{A}_2 \)remain same.

Question 38. The half-life of a radioactive nucleus is 50 days. The 9-time interval {t₂ — t₁) between the time t₂ when \(\frac{2}{3}\) of it has decayed and the time tj when \(\frac{1}{3}\) of it had decayed is:

1. 30 days
2. 50 days
3. 60 days
4. 15 days

Answer: 2. 50 days

Radioactive substance decays \(2 / 3^{\text {rd }}\) in time \(t_1\) and \(1 / 3^{\text {rd }}\) in time \(t_1=\frac{1}{3} N_o=N_{\mathrm{o}} e^{-\lambda t}\)  →  Equation 1

It becomes 1 / 2 when t disintegrates \(2 / 3^{\text {rd }} to 1 / 3^{\text {rd }}\). So time \(t_2-t_1\) is the half-life of the substance.  → Equation 2

\(\frac{2}{3} N_{\circ}=N_{\circ} e^{-\lambda t}\),

Divide (1) by (2),

we get,\(\frac{1}{2} =\frac{e^{-\lambda t_2}}{e^{-\lambda t_1}}\)

⇒ \(\lambda\left(t_2-t_1\right) =\ln 2 \)

⇒ \(t_2-t_1 =\ln 2 / \lambda \)

∴ \(\mathrm{T}_{1 / 2} =50 \text { days }\)

Question 39. A radioactive nucleus of mass M emits a photon of frequency v and the nucleus recoils, the recoil energy will be:

1. \(\frac{h^2 v^2}{2 M c^2}\)
2. zero
3. h v
4. \(M c^2-h v\)

Answer: 1. \(\frac{h^2 v^2}{2 M c^2}\)

We know that momentum of photon \(\mathrm{P}=\frac{h v}{c} \)and Energy,

E =\(\frac{p^2}{2 m}\)

= \(\frac{\left(\frac{h v}{c}\right)^2}{2 m}=\frac{h^2 v^2}{2 m C^2}\)

Question 40. The half-life of a radioactive X to 50 yr. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1: 15 in a sample of a given rock. The age of the rock was estimated to be:

1. 200 yr
2. 250 yr
3. 100 yr
4. 150 yr

Answer: 1. 200 yr

⇒ \(\mathrm{X} \longrightarrow \mathrm{Y}\)

⇒ \(\mathrm{X}: \mathrm{Y}\)=1: 15

And \(\frac{N}{N_0} =\left(\frac{1}{2}\right)^{t / t / 2} \)

⇒ \(N_0 \)=16

⇒ \(\frac{1}{16} =\left(\frac{1}{2}\right)^{t / 50}\)

⇒ \(\left(\frac{1}{2}\right)^4 =\left(\frac{1}{2}\right)^{t / 50} \)

⇒ \(\frac{t}{50}\) =4

t =200 years

Question 41. A nucleus \({ }_n^m \mathrm{X}\) emits one a-particle and two \(\beta^{-}\) particles. The resulting nucleus is:

1. \({ }_n^{m-6} \mathrm{Z}\)
2. \({ }_n^{m-4} \mathrm{X}\)
3. \({ }_{n-2}^{m-4} \mathrm{Y}\)
4. \({ }_{n-4}^{m-6} \mathrm{Z}\)

Answer: 2. \({ }_n^{m-4} \mathrm{X}\)

∴ \(\mathrm{X} \rightarrow{\alpha}{ }_{n-2} \mathrm{X}^{m-4} \rightarrow{2 \beta}{ }_n \mathrm{X}^{m-4}\)

Question 42. Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, several P species are 4N0 and that of Q is N0. The half-life of P (for conversion to R) is 1 min whereas that of ft is 2 min. Initially, there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be:

1. \(3 N_0\)
2. \(\frac{9 N_0}{2}\)
3. \(\frac{5 N_0}{2}\)
4. \(2 N_0\)

Answer: 2. \(\frac{9 N_0}{2}\)

According to the question,

Initially P\( \longrightarrow 4 N_0\)

and G \(\longrightarrow N_0\)

Half-life \(T_P \longrightarrow 1 \mathrm{~min}\)

and \(T_Q \longrightarrow 2 \mathrm{~min}\)

After time t nuclear of nuclei of \(\mathrm{P}\) and \(\mathrm{Q}\) are equal then,

⇒ \(\frac{4 N_0}{2^{t / 2}}=\frac{4 N_0}{2^{t / 2}}\)

4 =2^{t / 2} [/latex]

2^2 =2^{t / 2} [/latex]

⇒ \(\frac{t}{2}\) =2

t =4 \(\mathrm{~min}\)

Now Disactive nuclear Nuclei of

⇒ \(\mathrm{R}=\left(4 N_0-\frac{4 N_0}{2^4}\right)+\left(N_0-\frac{N_0}{2^2}\right)\)

⇒ \(\mathrm{R}=4 N_0-\frac{4 N_0}{4}+N_0-\frac{N_0}{4}\)

∴ \(\mathrm{R}=5 N_0-\frac{N_0}{2}=\frac{9}{2} N_0\)

Question 43. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N₀/t counts per minute t =5 min. The time (in minutes) at which the activity reduces to half its value is:

1. \(\log _e \frac{2}{5}\)
2. \(\frac{5}{\log _e 2}\)
3. \(5 \log _{10} 2\)
4. \(5 \log _e, 2\)

Answer: 4. \(5 \log _e, 2\)

Given, \(t_{\text {avg }}=5, t_{1 / 2}=\text { ? } \)

Now ,\(t_{1 / 2}=\log \left(\frac{1}{\lambda}\right)\)

= 5 \(\log \left(\frac{1}{\lambda}\right)=5 \log _e / 2\)

Question 44. The decay constant of a radioisotope is λ. If A₁ and A₂ are its activities at time t₁ and t₂ respectively the number of nuclei that have decayed during the time (t₁ -t₂):

1. \(A_1 t_1-A_2 t_2\)
2. A_1-A_2
3. \(\frac{\left(A_1-A_2\right)}{\lambda}\)
4. \(\lambda\left(A_1-A_2\right)\)

Answer: 3. \(\frac{\left(A_1-A_2\right)}{\lambda}\)

From question,

⇒ \(\mathrm{A}_1 =\lambda \mathrm{N}_1\)

⇒ \(\mathrm{~A}_2 =\lambda \mathrm{N}_2 \)

∴ \(\mathrm{N}_1-\mathrm{N}_2 =\frac{\mathrm{A}_1-\mathrm{A}_2}{\lambda}\)

Question 45. In the nuclear decay given below:\({ }_Z^A A \rightarrow{ }_{Z+1}^A Y \rightarrow{ }_{z-1}^{A-4} B^* \rightarrow{ }_{z-1}^{A-4} B,\) the particles emitted in the sequence are:

1. \(\beta, \alpha, \gamma\)
2. \(\gamma, \beta, \alpha\)
3. \(\beta, \gamma, \alpha\)
4. \(\alpha, \beta, \gamma\)

Answer: 1. \(\beta, \alpha, \gamma\)

Radioactive transition will be given by,

⇒ \({ }_2^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{2+1}^{\mathrm{A}} \mathrm{Y}+\beta_{-1}^0\)

⇒ \({ }_{2+1}^{\mathrm{A}} \mathrm{X} \longrightarrow{ }_{2-1}^{\mathrm{A}-4} \mathrm{Y}+\alpha_2^4\)

∴ \({ }_{2+1}^{\mathrm{A}-4} \mathrm{X} \longrightarrow{ }_{2-1}^{\mathrm{A}-4} \mathrm{Y}+v_0^0\)

Question 46. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:

1. isobar of parent
2. isomer of parent
3. isotone of parent
4. isotope of parent.

Answer: 4. isotope of parent.

According to the question,

⇒ \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \rightarrow{-\alpha}{ }_{2-2}^{\mathrm{A}-4} \mathrm{X} \rightarrow{-2 \beta}{ }_{\mathrm{z}}^{\mathrm{A}-4} \mathrm{X}\)

The atoms of an element having the same atomic number but different mass numbers are called isotopes

∴ \({ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X}\) and \({ }_{\mathrm{Z}}^{\mathrm{A}-4} \mathrm{X}\) are isotopes

Question 47. A radioactive nucleus X converts into a stable nucleus Y. Half-life of X is 50 yr. Calculate the age of the radioactive sample when the ratio of X and Y is 1: 15.

1. 200 Years
2. 150 Years
3. 100 Years
4. 250 Years

Answer: 1. 200 Years

From question, \(\frac{X}{Y}=\frac{1}{15}\)

The ratio of the nucleus of atoms remains undecayed

⇒ \(\frac{X}{X_0}=\frac{1}{16}\)

Active fraction \(\frac{X}{X_0}\)

or \(\frac{N}{N_0}=\frac{1}{16}\)

and \(\frac{N}{N_0}=\left(\frac{1}{20}\right)^{\frac{t}{T}}\)

⇒ \(\left(\frac{1}{20}\right)^{\frac{1}{\mathrm{~T}_1}} =\frac{1}{16}=\left(\frac{1}{2}\right)^4 \)

or \(\frac{t}{\frac{T_1}{2}}\) =4

or New Half-life =50 years

t=4 \(\times \)50=200 year

Question 48. Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei X₁ to that of X₂ will be \(\frac{1}{e}\) after a time:

1. \(\lambda\)
2. \(\frac{1}{2} \lambda\)
3. \(\frac{1}{4 \lambda}\)
4. \(\frac{e}{\lambda}\)

Answer: 3. \(\frac{1}{4 \lambda}\)

If N =Number of radioactive nuclei present at some instant.

N =\(N_0 e^{-\lambda t}\)

⇒ bow \(\frac{N_1}{N_2}=\frac{e^{-\lambda_1 t}}{e^{-\lambda_2 t}}=\frac{e^{-5 \lambda t}}{e^{-\lambda t}}=e^{-4 \lambda t}\)

⇒ \(\frac{N_1}{N_2} =\frac{1}{e}\)

⇒ \(\frac{1}{e} =\frac{1}{e^{4 \lambda t}} \)

4 \(\lambda\) t =1

t =\(\frac{1}{4 \lambda}\)

Question 49. Two radioactive substances A and B have decay constants 5λ and λ respectively. At t = 0, they have the same number of nuclei. The ratio of several nuclei of A to those of B will be (1/e)2 after a time interval.

1. \(4 \lambda\)
2. 2 \(\lambda\)
3. \(\frac{1}{2} \lambda\)
4. \(\frac{1}{4} \lambda\)

Answer: 3. \(\frac{1}{2} \lambda\)

According to the question,

⇒ \(\frac{N_A}{N_B}=\frac{1}{e^{4 \lambda t}} \)

⇒ \(\frac{1}{e^{4 \lambda t}}=\frac{1}{e^2} \)

t=\(\frac{1}{2 \lambda} \)

Question 50. In a radioactive decay process, the negatively charged emitted β-particles are:

1. the electrons produced as a result of the decay of neutrons inside the nucleus
2. the electrons produced as a result of collisions between atoms.
3. the electrons orbiting around the nucleus
4. the electrons present inside the nucleus.

Answer: 1. the electrons produced as a result of the decay of neutrons inside the nucleus

In \(\beta^{-}\)decay a neutron is transformed into a proton and an electron is emitted with the nucleus along with an antineutrino.

∴ \(p+\bar{e}+\bar{v}\) where \(\bar{v}\)= antineutrino

Question 51. In a radioactive material the activity at time ft is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is A, then:

1. \(R_1=R_2\)
2. \(R_1=R_2 e^\lambda\left(t_1-t_2\right)\)
3. \(R_1=R_2 e^\lambda\left(t_1-t_2\right)\)
4. \(R_1=R_2=\left(t_2-t_1\right)\)

Answer: 2. \(R_1=R_2 e^\lambda\left(t_1-t_2\right)\)

⇒ \(R_1 =R_0 e^{-\lambda 4} \text { and } R_2=R_0 e^{-\lambda t_2}\)

⇒ \(\frac{R_1}{R_3} =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}\)

= \(e^{-\lambda\left(t_1-t_2\right)} \)

∴ \(R_1 =R_2 e^{-\lambda\left(t_1-t_2\right)}\)

Question 52. The half-life of radium is about 1600 years. Of 100 g of radium existing now 25 g will remain unchanged after L:

1. 4800 years
2. 6400 years
3. 2400 years
4. 3200 years

Answer: 4. 3200 years

Applying the formula

N =\(N_0\left(\frac{1}{2}\right)^n \)

⇒ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^n \)

⇒ \(\frac{25}{100} =\left(\frac{1}{2}\right)^n \)

n =2

Total time is 2 \(\times\) 1600=3200 year

Question 53. A sample of radioactive element has a mass of 10 g at an instant t= 0. The approximate mass of this element in the sample after two mean lives is:

1. 1.35 g
2. 2.50 g
3. 3.70 g
4. 6.30 g

Answer: 1. 1.35 g

⇒ \(\mathrm{N}=\mathrm{N}_0 e^{-\lambda t} n =n_0 e^{-\lambda t}=n_0 e^{-\lambda(2 / \lambda)} \)

= \(\frac{10}{e^2}=1.35 \mathrm{gm}\)

Question 54. A sample of radioactive elements contains 4 x 10¹⁶ active nuclei. The half-life of the element is 10 decays, then the number of decayed nuclei after 30 days is:

1. 0.5 x 10¹⁶
2. 2 x 10¹⁶
3. 3.5 x 10¹⁶
4. 1 x 10¹⁶

Answer: 3. 3.5 x 10¹⁶

Half-life number is given by n=\(\frac{1}{T}\)

= \(\frac{30}{10}\)=3

Nuclear of active nuclei left after \(\mathrm{n}\) half-life is

N =\(N_0\left(\frac{1}{2}\right)^n\)

Decay nuclei =\(N_0-N\)

= 4 \(\times 10^{16}-0.5 \times 10^{16} \)

= 3.5 \(\times 10^{16}\)

Question 55. A deuteron is bombarded on \({ }_8 \mathrm{~O}^{16}\) nuclei then α-particle is emitted then the product nuclei are:

1. \({ }_7 \mathrm{~N}^{13}\)
2. \({ }_5 \mathrm{~B}^{10}\)
3. \({ }_4 \mathrm{Be}^9\)
4. \({ }_7 \mathrm{~N}^{14}\)

Answer: 4. \({ }_7 \mathrm{~N}^{14}\)

⇒ \({ }_8 \mathrm{O}^{16}+{ }_1 \mathrm{H}^1 \longrightarrow{ }_2 \mathrm{X}^4+{ }_2 \mathrm{He}^4\)

Use conservation of charge and mass, The conservation of mass number gives,

16+2=A+4

A=14

The conservation of atomic number gives.

8+1 =Z+2

Z =7

Therefore, \({ }_2 \mathrm{X}^4 is { }_7 \mathrm{~N}^{14}\).

Question 56. In compound X\((n, \alpha) \rightarrow{ }_3 \mathrm{Li}^7\), the element A is:

1. \({ }_2 \mathrm{He}^4\)
2. \({ }_5 \mathrm{~B}^{10}\)
3. \({ }_5 \mathrm{~B}^9/\)
4. \({ }_4 \mathrm{Be}^{11}\)

Answer: 2. \({ }_5 \mathrm{~B}^{10}\)

We can write the given reaction, as \({ }_2 X^{\mathrm{A}}+{ }_0 n^1 \rightarrow{ }_3 \mathrm{Li}^7+{ }_2 \mathrm{He}^4\)

According to the conservation of mass number, we have,

A+1=7+4

A=10

According to the conservation of charge number, we have

A+0 =Z+3

Z =5

Hence, the nucleus of the element will be \({ }_5 \mathrm{~B}^{10} \).

Question 57. The half-life of a radioactive substance is 12.5 h and its mass is 256 g. After what time, the amount of remaining substance is 1g?

1. 75 h
2. 100 h
3. 125 h
4. 150 h

Answer: 2. 100 h

We have, No. of half-lives required =\(\frac{\text { Original mass }}{2^x}\)

⇒ \(\frac{256}{2^x}\) =1 gm

⇒ \(2^x\) =256=28

x =8

Since, Half-life =12.5 hours.

Total time of reduce 256 \(\mathrm{gm} \)to 1 gm,

=12.5 \(\times\) 8=100 hours.

Question 58. Complete the equation for the following fission process\({ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{38} \mathrm{Sr}^{90}+\ldots \ldots\)

1. \({ }_{54} \mathrm{Xe}^{143}+3{ }_0 n^1\)
2. \({ }_{54} \mathrm{Xe}^{145}\)
3. \({ }_{57} \mathrm{Xe}^{142}\)
4. \({ }_{54} \mathrm{Xe}^{142}+{ }_0 n^1\)

Answer: 1. \({ }_{54} \mathrm{Xe}^{143}+3{ }_0 n^1\)

⇒ \({ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{38} \mathrm{Sr}^{90}+{ }_{54} \mathrm{Xe}^{143}+3{ }_0 n^1\)

Total atomic mass,

LHS =92+0=92

RHS =38+54+0=92

Total mass number: LHS =235+1=236

RHS =90+143+3 \(\times\) 1=236

So, option (1) is correct.

Question 59. The activity of a radioactive sample is measured as 9750 counts/min at t = 0 and as 975 counts/min at t = 5 min. The decay constant is approximately:

1. 0.922/m in
2. 0.691/min
3. 0.461/min
4. 0.230/min

Answer: 3. 0.461/min

According to radioactivity law,

⇒ \(\frac{N}{N_0}=e^{-\lambda t} \)

⇒ \(\frac{N_0}{N}=e^{-\lambda t}\)

N=textinal concentration

⇒ \(N_0\)= initial concentration

⇒ \(\lambda\)= decay constant

Take logarithm on both sides of Eq. (1), we get

⇒ \(\log _e\left(\frac{N_0}{N}\right) =\log _e\left(e^{\lambda t}\right) \)

= \(\lambda t \log _e e=\lambda\) t

We have, \(\log _e x=2.3026 \log _{10} x\)

By making substitutions, we get

⇒ \(\lambda=\frac{2.3026 \log _{10}\left(\frac{9750}{975}\right)}{5}\)

⇒ \([N_0 \)=9750. counts \(  \mathrm{min} \)and N=975 counts \(  \mathrm{min}]\)

= \(\frac{2.3026}{5} \log _{10}\)

10 =\(\frac{2.3026}{5} \mathrm{~min}^{-1}\)

= 0.461 \(\mathrm{~min}^{-1}\)

Question 60. The most penetrating radiation out of the following is:

1. γ-rays
2. α-particles
3. β-rays
4. X-rays

Answer: 1. γ-rays

Penetrating is directly proportional to the energy of radiations.

The most penetrating radiation out of the following is y-rays.

Question 61. The nucleus \({ }_{48} \mathrm{Cd}^{115}\), after two successive P-decay will give:

1. \({ }_{46} \mathrm{~Pa}^{115}\)
2. \({ }_{49}{In}^{114}\)
3. \({ }_{50} \mathrm{Sn}^{113}\)
4. \({ }_{50} \mathrm{Sn}^{115}\)

Answer: 4. \({ }_{50} \mathrm{Sn}^{115}\)

The emission of a beta beta-negative particle decays a neutron into a proton (increases the atomic number by l), electron, and electron anti-neutrino, but the mass number remains the same. So two beta negative particles will increase the atomic number by 2 and \({ }^{48}\mathrm{Cd}_{115}\) will convert to \({ }^{50} \mathrm{Sn}_{115}\).

Question 62. When a uranium isotope \({ }_{92}^{235} \mathrm{U}\) is bombarded with a neutron. It generates \({ }_{92}^{89} \mathrm{Kr}\), three neutrons and:

1. \({ }_{40}^{91} \mathrm{Zr}\)
2. \({ }_{36}^{10} \mathrm{Kr}\)
3. \({ }_{36}^{103} \mathrm{Kr}\)
4. \({ }_{56}^{144} \mathrm{Ba}\)

Answer: 4. \({ }_{56}^{144} \mathrm{Ba}\)

The reaction of Uranium is:

⇒ \({ }_{92}^{235} \mathrm{U}+{ }_0 n \rightarrow{ }_{36}^{89} \mathrm{Kr}+3 b_0 n+\mathrm{X}_{\mathrm{X}}^{\mathrm{A}}\)

According to the law of conservation

⇒ \([latex](\text { Atomic number })_{\text {Reaction }}=(\text { Atomic number })_{\text {Product }}\)

92+0=36+Z

Z=92-36=56

⇒ \(\text { (Atomic mass }_{\text {Reactant }}=(\text { Atomic mass })_{\text {Product }}\)

235+1=89+\(\mathrm{A}+3 \times 1\)

Similarly, The element is \({ }_{56}^{144} \mathrm{Ba}\)

Question 63. A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in the fusion reaction is 0.02866 u. The energy liberated per u is (given In = 931 MeV):

1. 2.67 MeV
2. 26.87 MeV
3. 6.675 MeV
4. 13.35 MeV

Answer: 3. 6.675 MeV

Energy released per u=\(\left(\frac{0.02866}{4}\right)(931 \mathrm{MeV})\)

= 6.675 \(\mathrm{MeV}\)

Question 64. The power obtained in a reactor using U²³⁵ disintegration is 100 kW. The mass decay of U²³⁵ per hour is:

1. 20 μg
2. 40 μg
3. 1 μg
4. 10 μg

Answer: 2. 40 μg

From the question,

Here, power \(\mathrm{P}=1000 \mathrm{~W}\)

Energy per hour =1000 \(\times 3600 \mathrm{~J}\)

Energy per fission =200 \(\mathrm{MeV}\)

= 200 \(\times 1.6 \times 10^{-13} \mathrm{~J}\)

No. of fission per hour n=\(\frac{1000 \times 3600}{200 \times 1.6 \times 10^{-13}}\)

Mass per hour =\\(frac{n}{\mathrm{~N}} \times 235 \)

= \(\frac{1000 \times 3600 \times 235}{200 \times 1.6 \times 10^{-13} \times 6.02 \times 10^{23}} \)

=43.9 \(\times 10^{-6} \mathrm{~g}=40 \mu \mathrm{g}\)

Question 65. Fusion reaction takes place at high temperatures because:

1. atoms get ionized at high temperature
2. kinetic energy is high enough to overcome the coulomb repulsion between nuclei
3. molecules break up at high temperature
4. nuclei break up at high temperatures.

Answer: 2. kinetic energy is high enough to overcome the coulomb repulsion between nuclei

Thermal K.E. \(\geq \)Electrostatic P.E.

Question 66. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:

1. 23.6 MeV
2. 2.2 MeV
3. 28.0 MeV
4. 30.2 MeV

Answer: 1. 23.6 MeV

Mass of \({ }_1 \mathrm{H}^2\)=2.014478 amu

Mass of\( { }_2 \mathrm{He} e^4=4.00388 amu\)

Mass of deuterium =\(2 \times 2.01478\)=4.02956 amu

Energy equivalent \({ }_2 \mathrm{H}^2=4.02956 \times 1.2\)=4.4 MeV

Energy equivalent to \({ }_2 \mathrm{He}^4=4.00388 \times\) 7=28 MeV

Energy released =28-4.4=23.6 MeV

Question 67. In any fission process the \(\text { ratio: } \frac{\text { mass of fission products }}{\text { mass of parent nucleus }}\) is:

1. equal to 1
2. greater than 1
3. less than 1
4. depends on the mass of the parent nucleus.

Answer: 3. less than 1

The mass of fission products is always less than the mass of the parent nucleus.

Question 68. Fission of nuclei is possible because of the binding energy per nucleon in them:

1. increase with mass number at low mass numbers
2. decrease with mass number at low mass numbers
3. increase with mass number at high mass numbers
4. decreases with mass numbers at high mass numbers.

Answer: 1. increase with a mass number at low mass numbers

For nuclei having A > 56 binding energy nuclear gradually decreases.

Question 69. In a nuclear fusion process, the masses of the fusing nuclei are m₁ and m₂ and the mass of the resultant nucleus is m₂, then:

1. \(m_3=m_1+m_2\)
2. \(m_3=\left|m_1+m_2\right|\)
3. \(m_3<\left(m_1+m_2\right)\)
4. \(m_3>\left(m_1+m_2\right)\)

Answer: 3. \(m_3<\left(m_1+m_2\right)\)

Product is more stable so that its mass is less than the sum masses of reactants i.e. \(m_3<(m_1+m_2)\).

Question 70. Solar energy is mainly caused by to:

1. burning of hydrogen in the oxygen
2. fission of uranium present in the Sun
3. fusion of protons during the synthesis of heavier elements
4. gravitational contraction.

Answer: 3. fusion of protons during synthesis of heavier elements

Solar energy → fusion of protons into helium.

Question 71. Which of the following are suitable for the process?

1. Light nuclei
2. Heavy nuclei
3. Element must be lying middle of the periodic table
4. Middle elements which are lying on the binding energy curve

Answer: 1. Light nuclei

In nuclear fusion, lighter nuclei merge to form a heavier nucleus.

Question 72. Nuclear fission can be explained by:

1. proton-proton cycle
2. liquid drop model of the nucleus
3. independent of the nuclear particle model
4. nuclear shell model.

Answer: 2. liquid drop model of the nucleus

On the basis of a liquid drop model of the nucleus, Neil Bohr and J.A.

Question 73. Heavy water is used as a moderator in a nuclear reactor. The function of the moderator is:

1. to control energy released in the reactor
2. to absorb neutrons and stop chain reaction
3. to cool the reactor
4. to slow down the neutrons to thermal energies.

Answer: 4. to slow down the neutrons to thermal energies.

A moderator’s function is to slow down the fast-moving secondary neutrons created during fission because only slow-moving neutrons can initiate the fission reaction. The moderator should be made of a light substance that does not absorb neutrons. Heavy water, graphite, deuterium, paraffin, and other materials can act as moderators.

Semiconductor Electronics: Materials, Devices, And Simple Circuits

Question 1. Sodium has body-centered packing. The distance between the two nearest atoms is 3.7Å. The lattice parameter is:

1. 6.8Å
2. 4.3 Å
3. 3.0Å
4. 8.6Å.

Answer: 2. 4.3 Å

We have, a neighbor distance of a body-centered cubic cell

d=\(\frac{\sqrt{3}}{2}\) a,

where a is the lattice parameter.

3.7=\(\frac{\sqrt{3 a}}{2}\)

a=\(\frac{2 \times 3.7}{\sqrt{3}}=4.3 Å\)

Question 2. Choose the only false statement from the following:

1. In conductors, the valence and the conduction bands overlap.
2. Substances with an energy gap of the order of 10 eV are insulators.
3. The resistivity of a semiconductor increases with an increase in temperature.
4. The conductivity of a semiconductor increases with an increase in temperature.

Answer: 3. The resistivity of a semiconductor increases with an increase in temperature.

Question 3. Carbon, silicon, and germanium atoms have four valence electrons each. Their valence and conduction bands are separated energy band gaps represented \(\left(E_g\right)_c,\left(E_g\right)_{s i}\), and \(\left(E_g\right)_{G e}\) respectively. Which one of the following relationships is true in their case?

1. \(\left(\mathrm{E}_g\right)_C>\left(\mathrm{E}_g\right)_{S i}\)
2. \(\left(\mathrm{E}_g\right)_C<\left(\mathrm{E}_g\right)_{S i}\)
3. \(\left(\mathrm{E}_g\right)_C=\left(\mathrm{E}_g\right)_{S i}\)
4. \(\left(\mathrm{E}_g\right)_C>\left(\mathrm{E}_g\right)_{G e}\)

Answer: 1. \(\left(\mathrm{E}_g\right)_C>\left(\mathrm{E}_g\right)_{S i}\)

The band gap of carbon is 5.5 V while that of silicon is 1.1

∴ \(\mathrm{eV}\left(E_g\right)_C>\left(E_B\right)_S\)

Read and Learn More NEET Physics MCQs

Question 4. In semiconductors at room temperature:

1. The valence band is partially empty and the condition band is partially filled
2. The valance band is completely filled and the condition band is partially filled
3. The valence band is completely filled
4. The conduction band is completely empty.

Answer: 3. The valence band is completely filled

In semiconductors at room temperature, the valence band is partially empty and the conductor band is partially filled.

Question 5. In the bcc structure of lattice constant a, the minimum distance between atoms is:

1. \(\frac{\sqrt{3}}{2} \mathrm{a}\)
2. \(\sqrt{2 \mathrm{a}}\)
3. \(\frac{\mathrm{a}}{\sqrt{2}}\)
4. \(\frac{a}{2}\)

Answer: 1. \(\frac{\sqrt{3}}{2} \mathrm{a}\)

In a bcc structure, the position vectors of the nearest neighbors are \(\left( \pm \frac{\mathrm{a}}{2} \hat{\mathrm{i}}, \pm \frac{\mathrm{a}}{2} \hat{\mathrm{j}}, \pm \frac{\mathrm{a}}{2} \hat{\mathrm{k}}\right)\).

In bcc, there will be atoms at the body center and at corners. So the distance between the two nearest atoms is

∴ \(\sqrt{\left(\frac{\mathrm{a}}{2}\right)^2+\left(\frac{\mathrm{a}}{2}\right)^2+\left(\frac{\mathrm{a}}{2}\right)^2}=\sqrt{\left(\frac{3 \mathrm{a}^2}{4}\right)}=\frac{\sqrt{3}}{2} \mathrm{a}\)

Question 6. Si and Cu are cooled to a temperature of 300K, then resistivity:

1. For Si increases and for Cu decreases.
2. For Cu increases and Si decreases.
3. Decreases for both Si and Cu.
4. Increases for both Si and Cu.

Answer: 1. For Si increases and for Cu decreases.

Due to the positive temperature coefficient, metal resistivity is directly proportional to temperature, but due to the negative temperature coefficient, the semiconductor resistivity is inversely proportional to temperature. It means that as temperature drops, metal resistivity drops but semiconductor resistivity rises. Si is a semiconductor, whereas Cu is a metal, in this case. As a result, Si resistivity increases while Cu resistivity falls.

Question 7. C and Si both have the same lattice structure, having 4 bonding electrons in each. However, C is an insulator whereas Si is an intrinsic semiconductor. This is because:

1. In the case of C, the valence band is not completely filled at absolute zero temperature.
2. In the case of C, the conduction band is partly filled even at absolute zero temperature.
3. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.
4. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.

Answer: 3. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si, they lie in the third.

Tetravalent atom → C , Si , Ge , Sn

Their outermost shell has four electrons

In Si, electrons have a loosely bond but in C, they tightly bond. Therefore C is used.

Question 8. The electron concentration in an n-type semiconductor is the same as the hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them:

1. Current in n-type = current in p-type
2. Current in p-type > current in n-type
3. Current in n-type > current in p-type
4. No current will flow in p-type, current will only flow in n-type.

Answer: 1. Current in n-type = current in p-type

Since electron and hole concentrations are the same, applying the same field will give the same current.

Question 9. An intrinsic semiconductor is converted into an n-type extrinsic semiconductor by doping it with

1. Phosphorous
2. Aluminium
3. Silver
4. Germanium.

Answer: 1. Phosphorous

For an n-type semiconductor pentavalent impurity is needed and this is phosphorous.

Question 10. For a p-type semiconductor, which of the following statements is true?

1. Holes are the majority carriers and trivalent atoms are the dopants.
2. Holes are the majority carriers and pentavalent atoms are the dopants.
3. Electrons are the majority carriers and pentavalent atoms are the dopants.
4. Electrons are the majority carriers and trivalent atoms are the dopants.

Answer: 1. Holes are the majority carriers and trivalent atoms are the dopants.

p-type Semi-Conductor: To prepare a p-type semiconductor, a trivalent impurity with Si and G. Such an impurity atom wants to accept an electron from the crystal lattice. Thus effectively each dopant atom provides a hole. Thus holes are the majority carriers penta valent atoms are the dopants.

Question 11. In a n-type semiconductor, which of the following statement is true?

1. Electrons are majority carriers and trivalent atoms are dopants
2. Electrons are minority carriers and pentavalent atoms are dopants
3. Holes are minority carriers and pentavalent atoms are dopants
4. Holes are majority carriers and trivalent atoms are dopants

Answer: 3. Holes are minority carriers and pentavalent atoms are dopants

n-type semiconductors can be obtained by doping impurities such as pentavalent atoms (phosphores) etc.

Question 12. If a small amount of antimony is added to the germanium crystal

1. The antimony becomes an acceptor atom.
2. There will be more free electrons than holes in the semiconductor.
3. Its resistance is increased
4. It becomes a p-type semiconductor.

Answer: 2. There will be more free electrons than holes in the semiconductor.

By adding pentavalent impurity only n-type semiconductors are constructed.

Question 13. Pure Si at 500 K has an equal number of electron (n_e) and hole (n_h) concentrations of 1.5 x 10¹⁶ m^-3, Doping by indium increases nh to 4.5 x 10²² m^-3. The doped semiconductor is of:

1. n-type with electron concentration \(n_e=5 \times 10^{22} \mathrm{~m}^{-3}\)
2. n-type with electron concentration \(n_e=2.5 \times 10^{10} \mathrm{~m}^{-3}\)
3. n-type with electron concentration \(n_e=2.5 \times 10^{23} \mathrm{~m}^{-3}\)
4. n-type with electron concentration \(n_e=5 \times 10^9 \mathrm{~m}^{-3}\)

Answer: 4. n-type with electron concentration \(n_e=5 \times 10^9 \mathrm{~m}^{-3}\)

See the question,\(n_i{ }^2 =n_e n_n\)

⇒ \(n_e =\frac{\left(n_i\right)^2}{n_n}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.5 \times 10^{22}}\)

⇒ \(\left\{\right. Here n_i=1.5 \times 10^{16} and \left.n_n=4.5 \times 10^{22}\right\}\)

⇒ \(n_e=5 \times 10^9 \mathrm{~m}^{-3}\)

∴ \(n_n \gg n_e\) The semi-conductor is p-type.

Question 14. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is:

1. An insulator
2. A metal
3. An u-type semiconductor
4. A p-type semiconductor.

Answer: 4. A p-type semiconductor.

That material is a p-type semiconductor.

Question 15. When arsenic is added as an impurity to silicon, the resulting material is:

1. u-type semiconductor
2. p-type semiconductor
3. u-type conductor
4. Insulator.

Answer: 1. u-type semiconductor

An n-type semiconductor is formed when a small amount of pentavalent impurity is added to a pure semiconductor. Arsenic (33) is the Pentavalent impurity. The presence of a pentavalent impurity in the semiconductor crystal results in a large number of free electrons.

Question 16. The increase in the width of the depletion region in a p-n junction diode is due to:

1. Reverse bias only
2. Both forward bias and reverse bias
3. Increase in forward current
4. Forward bias only.

Answer: 1. Reverse bias only

A p-n Junction diode is formed by fusing one p-type and one n-type. This happens because when substrate we apply reverse bias voltage, the electrons drift away from the junction and hence conduction is not possible. So, the width of the depletion region in a p-n junction diode is increased by reverse bias.

Question 17. In an unbiased p-n junction holes from the p-region to the n-region because of:

1. The attraction of the free electrons of the n-region.
2. The higher hole concentration in the p-region than that in the n-region.
3. The higher concentration of electrons in the n-region than that in the p-region.
4. The potential difference across the p-n junction.

Answer: 2. The higher hole concentration in the p-region than that in the n-region.

Question 18. In forward biasing of the p-n junction:

1. The positive terminal of the battery is connected to the n-side and the depletion region becomes thin
2. The positive terminal of the battery is connected to the n-side and the depletion region becomes thick
3. The positive terminal of the battery is connected to the p-side and the depletion region becomes thin
4. The positive terminal of the battery is connected to the p-side and the depletions region becomes thick.

Answer: 3. The positive terminal of the battery is connected to the p-side and the depletion region becomes thin

In forward biasing, the width of the deption layer is decreased.

Question 19. Application of a forward bias to a p-n junction:

1. Widens the depletion zone
2. Increases the potential difference across the depletion zone
3. Increases the number of donors on the N-side
4. Decreases the electric field in the depletion zone.

Answer: 4. Decreases the electric field in the depletion zone.

Forward biased to p-n Junction decreases the electric field in the depletion zone

Question 20. In ap-n junction:

1. High potential is at the n-side and low potential at the p-side
2. High potential is at the p-side and low potential at the n-side
3. P and both are at the same potential
4. Undetermined.

Answer: 1. High potential is at the n-side and low potential at the p-side

In a p-n junction high potential is at w-side and law potential at p-side.

Question 21. The depletion layer in the p-n junction region is caused by:

1. Drift of holes
2. Diffusion of charge carriers
3. Migration of impurity ions
4. Drift of electrons.

Answer: 2. Diffusion of charge carriers

When a p-n junction is formed, some of the free electrons in the n region diffuse across the junction and combine with holes to form negative ions. In doing so, they leave behind positive ions at the donor impurity sites. Similarly, holes from the p side diffuse to the n side and thus form a layer called diffusion layer at the junction.

So, the depletion region is caused by the diffusion of charge carriers.

Question 22. Out of the following which one is a forward-biased diode?

Answer: 4.

Question 23. In a p-n junction diode, change in temperature due to heating:

1. Does not affect the resistance of the p-n junction
2. Affects only forward resistance
3. Affects only reverse resistance
4. Affects the overall V-I characteristics of the p-n junction.

Answer: 4. Affects the overall V-I characteristics of the p-n junction.

Due to heating, the number of electron-hole pairs will increase. So, the overall resistance of the diode will change. Due to this forward biasing and reverse biasing both are changed.

Question 24. Which one of the following represents the forward bias diode?

Answer: 1.

In forward bias, the p-type semiconductor is at higher potential w.r.t. n-type semiconductor, and in reverse bias p – type semiconductor is at lower potential w.r.t. p-type semiconductor.

Question 25. Consider the junction diode as ideal. The value of current flowing through AB is:

1. \(10^{-2} \mathrm{~A}\)
2. \(10^{-1} \mathrm{~A}\)
3. \(10^{-3} \mathrm{~A}\)
4. \(0 \mathrm{~A}\)

Answer: 1. \(10^{-2} \mathrm{~A}\)

Current, I =\(\frac{V_A-V_B}{R} \)

since V=I R

= \(\frac{4-(-6)}{1 \mathrm{~K} \Omega}=\frac{10}{1 \times 10^3}=10^{-2} \mathrm{~A}\)

Question 26. The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R₁ will be:

1. 2.5 A
2. 10.0 A
3. 1.43 A
4. 3.13 A

Answer: 1. 2.5 A

Current will not flow through \(D_1\) as it is reverse biased. Current will flow through cell, \(R_1, R_2\) and \(R_3\) is

I=\(\frac{10}{2+2}=2.4 \mathrm{~A}\)

Question 27. If in a p-n junction, a square input single of 10 V is applied as shown in the given figure: Then the output across RL will be:

Answer: 4.

Because of + 5 V, it supports the forward bias of the diode and blocks the 5 V.

Question 28. In the given figure a diode D is connected to an external resistance R = 100 Ω and an E.M.F. of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:

1. 30 mA
2. 40 mA
3. 20 mA
4. 35 mA

Answer: 1. 30 mA

According to the question,

External resistance, R=100 \(\Omega\)

emf=3.5 V

Potential difference on

R, \(V_R =3.5 V-0.5 \mathrm{~V}=3 \mathrm{~V}\)

Current I =\(\frac{V_R}{R}=\frac{3}{100}=0.03 \mathrm{~A}\)

= 30 mA

Question 29. The given graph represents V-I characteristics for semiconductor devices. Which of the following statements is correct?

1. It is V-I characteristics to the solar cell where point A represents open circuit voltage and point B short circuit current
2. It is for a solar cell and points A and B represent open circuit voltage and current respectively
3. It is for a photodiode and points A and B represent open circuit voltage and current respectively
4. It is for an LED and points A and B represent open circuit voltage and short circuit current respectively

Answer: 1. It is V-I characteristics of solar cells where point A represents open circuit voltage and point B short circuit current

Question 30. The barrier potential of the p-n junction depends on:

1. Type of semiconductor material
2. Amount of doping
3. Temperature

Which of the following is correct?

1. (1) and (2) only
2. (2) only
3. (2) and (3) only
4. (1) (2) and (3)

Answer: 4. (1) (2) and (3)

The barrier potential of a p-n junction depends on:

(1) Type of Semiconductor method: The potential barrier for a silicon p-n junction is 0.7 V at room temperature while for a generation p-n junction is 0.3 V.

(2) Amount of doping: Due to the difference in concentration, electrons diffuse from the n-side to the p-side and holes diffuse from the ft-side to the p-side and holes diffuse from the p-side to the n-side.

(3) Temperature: According to the temperature, the number of majority and minority carriers depends minority will also change

Question 31. Two ideal diodes are connected to a battery as shown in the circuit. The  current supplied by the battery is:

1. 0.75 A
2. zero
3. 0.2s A
4. 0.5 A

Answer: 4. 0.5 A

Here \(D_1\) is at forward bias and \(D_2\) is at reverse bias, current flows only in \(D_1\). Then the diagram shows the Current flowing in the battery:

∴ \(\mathrm{I}=\frac{V}{R}=\frac{5 V}{10 \Omega}=0.5 \mathrm{~A}\)

Question 32. In the following figure, the diodes, which are forward-biased, are:

1. 3 only
2. 3 and 1
3. 2 and 4
4. 1, 2, and4

Answer: 2. 3 and 1

(1) and (3) both are in forward bias for forward bias p-type should be higher potential and in n-type at lower potential.

Question 33. In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be:

1. Zero
2. 30 Hz
3. 60 Hz
4. 120 Hz

Answer: 3. 60 Hz

For half-wave rectification, the frequency remains the same as 60 Hz.

Question 34. forward biased diode is:

Answer: 1.

A diode is bound to be forward biased if if-type semi conducts of the p-n junction is at high potential w.r.t. n-type semiconductor of the p-n junction. It is so for the circuit (a).

Question 35. On the diodes shown in the following diagrams, which one is reverse biased:

Answer: 3.

A diode is at reverse bias if the p-side of the semiconductor of the p-n junction is at low potential w.r.t. n-type semiconductor of the p-n junction.

Question 36. Reverse bias applied to a junction diode:

1. Lowers of the potential barrier
2. Raises the potential barrier
3. Increases the majority carrier’s current
4. Increases the minority carrier’s current

Answer: 2. Raises the potential barrier

Reverse bias increases the potential barrier.

Question 37. For the given circuit of the p-n junction diode which is correct?

1. In forward bias the voltage across R is V.
2. In reverse bias the voltage across R is V.
3. In forward bias, the voltage across R is 2 V.
4. In reverse bias, the voltage across R is 2 V.

Answer: 1. In forward bias the voltage across R is V.

Question 38. If the internal resistance of the cell is negligible, then the current flowing through the circuit is:

1. 3/50 A
2. 5/50 A
3. 4/50 A
4. 2/50 A

Answer: 2. 5/50 A

Diode \(D_1\) is forward biased in the circuit, while diode \(D_2\) is reverse biased. As a result, no current flows via the \(D_2\)-containing arm, whereas all current flows through the \(D_1\)-containing arm.

Thus, current flowing through the circuit is I=\(\frac{V}{R_{e q}}=\frac{5}{20+30}=\frac{5}{20} \mathrm{~A}\)

Question 39. A semiconducting device is connected in a series in a circuit with a battery and a resistance. A current is allowed to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be:

1. P-n junction
2. An intrinsic semiconductor
3. A p-type semiconductor
4. An n-type semiconductor

Answer: 1. P-n junction

We have, the current in a p-n junction on the order of milli-ampere in forward biasing and on the order of micro-ampere in reverse biasing (negligible). As a result, the device is a p-n junction.

Question 40. The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with the diode, for obtaining maximum current l?

1. 200
2. 6.67
3. 5
4. 1.5

Answer: 3. 5

We have, current in the circuit,

i =\(\frac{P}{V_d}=\frac{100 \times 10^{-3}}{0.5}\)

=200\( \times 10^{-3} \mathrm{~A}\)

The voltage across resistance R,

⇒ \(V^{\prime}=1.5-0.5=1.0 \mathrm{~V}\)

Thus, resistance R =\(\frac{V^{\prime}}{I} \)

= \(\frac{1}{200 \times 10^{-3}}=5 \Omega\)

Question 41. In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be:

1. Zero
2. 30 Hz
3. 60 Hz
4. 120 Hz

Answer: 3. 60 Hz

For half-wave rectification, the frequency remains the same as 60 Hz.

Question 42. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without a filter is 10 V. The D.C. component of the output voltage is:

1. \(\frac{10}{\sqrt{2}} V\)
2. \(\frac{10}{\pi} \mathrm{V}\)
3. \(10 \mathrm{~V}\)
4. \(\frac{20}{\pi} V\)

Answer: 2. \(\frac{10}{\pi} \mathrm{V}\)

∴ \(V_a=\frac{V_m}{\pi}=\frac{10}{\pi} \mathrm{V}\)

Question 43. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be:

1. 25 Hz
2. 50 Hz
3. 70.7 Hz
4. 100 Hz

Answer: 4. 100 Hz

In a full wave rectifier, the fundamental frequency in ripple is twice of input frequency.

Question 44. Consider the following statements (1) and (2) and identify the correct answer:

1. A zener diode is connected in reverse bias when used as a voltage regulator.

2. The potential barrier of the p-n junction lies between 0.1 V to 0.3 V.

1. (1) and (2) both are correct
2. (1) and (2) both are incorrect
3. (1) is correct and (2) is incorrect
4. (1) is incorrect but (2) is correct

Answer: 3. (1) is correct and (2) is incorrect

When the zener diode is forward biased, it behaves like an ordinary diode and when reverse biased, it will be a voltage stabilizer. Also, at room temperature, the voltage across the depletion layer for silicon is about 0.6- 0.7 V, and for germanium is about 0.3- 0.35 V.

Question 45. An LED is constructed from a p-n junction diode using Ga As P. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to:

1. \(10.4 \times 10^{-26} \mathrm{~m}\)
2. 654 nm
3. \(654 Ål\)
4. \(654 \times 10^{-11} \mathrm{~m}\)

Answer: 2 . 654 nm

The energy of light of wavelength \(\lambda\) is given by,

E =h v=\(\frac{h c}{\lambda}\)

⇒ \(\lambda =\frac{h c}{E}\)

Here h= Planck’s constant =6.63 \(\times 10^{-34} \mathrm{~J}.s.\)

c= speed of light =3 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

⇒ \(\mathrm{E}= energy gap =19 \mathrm{eV}\)

= \(1.9 \times 10^6 \times 10^{-19} \mathrm{~J}\)

Substituting the given values in Eq. (1), we get

⇒  \(\lambda =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}}\)

=6.54 \(\times 10^{-7} \mathrm{~m} \approx 654 \mathrm{~nm}\)

Thus, the wavelength of light emitted from LED will be 654 nm

Question 46. A Zener diode, having a breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in the figure. The current through the diode is:

1. 10 mA
2. 15 mA
3. 20 mA
4. 5mA

Answer: 4. 5mA

The actual diagram becomes

for 1 \(\mathrm{~K} \Omega, i_1=\frac{15}{1}=15 \mathrm{~mA}\)

for 250 \(\Omega,i_{250}=\frac{20-15}{250}=\frac{5}{250}\)

= \(\frac{20}{1000}=20 \mathrm{~mA}\)

∴ \(i_{\text {zener }}=20-15=5 \mathrm{~mA}\)

Question 47. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength:

1. 6000 Å
2. 4000 nm
3. 6000 nm
4. 4000 Å

Answer: 4. 4000 Å

Using de-Broglie equation

E =h v=\(\frac{h c}{\lambda} \)

⇒  \(\lambda =\frac{h c}{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}}\)

=5000Å

As 4000 \(Å <5000 Å\)

The signal of wavelength is 4000 Å can be detected by the photodiode.

Question 48. A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed in the material is nearly:

1. \(10 \times 10^{14} \mathrm{~Hz}\)
2. \(5 \times 10^{14} \mathrm{~Hz}\)
3. \(1 \times 10^{14} \mathrm{~Hz}\)
4. \(20 \times 10^{14} \mathrm{~Hz}\)

Answer: 2. \(5 \times 10^{14} \mathrm{~Hz}\)

The frequency corresponding to 2 eV is given by, E =h v

2 \(\times 1.6 \times 10^{-19} =66 \times 10^{-36} \times v \)

v =\(\frac{3.2 \times 10^{19}}{6.0+10^{-34}}=5 \times 10^{14} \mathrm{~Hz}\)

Question 49. Zener diode is used for:

1. Amplification
2. Rectification
3. Stabilization
4. Producing oscillations in an oscillator

Answer: 3. Stabilisation

The Zener diode is used for stabilization while the p-n junction divide is used for notification.

Question 50. In a p-n junction photocell, the value of the photo electromotive force by monochromatic light is proportional to:

1. The barrier voltage at the p-n junction
2. The intensity of the light falling on the cell
3. The frequency of the light falling on the cell
4. The voltage applied at the p-n junction

Answer: 2. The intensity of the light falling on the cell

In a p-n junction photocell, the value of the photo electromotive force produced by monochromatic light is proportional to the intensity of the light falling on the cell.

Question 51. The truth table for the given logic circuit is:

The Truth Table For The Given Logic Circuit Is: 

Answer: 3.

The output of the given circuit,

C=\(\overline{A B}+\overline{\bar{A} B} \)

C=\(\bar{A}+\bar{B}+\overline{\bar{A}}+\bar{B}\)

⇒  \(\mathrm{C}=\bar{A}+\bar{B}+A+\bar{B} \)

∴ \(\mathrm{C}=\overline{\mathrm{B}}\)

Question 52. For the given circuit, the input digital signals are applied at terminals A, B, and C. What would be the output at the terminal?

Answer: 3.

So, the output at y is high (1) i.e. V0 = 5 V.

Hence, option (3) is correct

Question 53. An n-p-n transistor is connected in a common emitter configuration (see figure) in which the collector voltage drop across load resistance (800 Ω) connected to the collector circuit is 0.8 V. The collector current is:

1. 2 mA
2. 0.1 mA
3. 1 mA
4. 0.2 mA

Answer: 3. 1 mA

From the given circuit diagram,

⇒  \(R_c =800 \Omega\)

⇒  \(V_{e c} =8 \mathrm{~V}\)

Now, the voltage drop across \(R_C,\)

⇒  \(I_C R_C =0.8 \)

∴ \(I_C =\frac{0.8}{R_C}=\frac{0.8}{800} 10^{-3} \mathrm{~A}=1 \mathrm{~mA} \).

Question 54. For transistor action, which of the following statements is correct?

1. Base, emitter, and collector regions should have the same size.
2. Both, the emitter junction as well as the collector junction are forward-biased.
3. The base region must be very thin and lightly doped.
4. Base, emitter, and collector regions should have the same doping concentrations.

Answer: 3. The base region must be very thin and lightly doped.

For transistor action, the base region must be very thin and lightly doped.

Question 55. Which of the following gates is called a universal gate?

1. OR gate
2. AND gate
3. NAND gate
4. NOR gate

Answer: 3. NAND gate

NAND and NOR gates are universal gates.

Question 56. For the logic circuit shown, the truth table is

Answer: 4.

The logic gate becomes,

Here, Y=\(\overline{\bar{A}+\bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}\)

This Is In Option 4

Question 57. The correct Boolean operation represented by the circuit diagram drawn is

1. OR
2. NAND
3. NOR
4. AND

Answer: 4. AND

From the given logic circuit LED will glow when the voltage across the LED is high.

This is the output of the NAND gate.

Question 58. The circuit diagram shown here corresponds to the logic gate:

1. NOR
2. OR
3. AND
4. NAND

Answer: 1. NOR

From the circuit diagram given below, it can be seen that the current will flow to the ground if any of the switches are closed.

Thus, the truth table for the circuit diagram can be formed as the circuit diagram is given a solution.

The output (Y) is equivalent to that of the NOR gate.

Question 59. In the circuit shown in the figure, the input voltage \(V_{i}=0\) is 20 V, \(V_{B E}=0\), and \(V_{C E}=0\). The values of \(I_B, I_C\)and \(\beta\) given by:

1. \(I_{\mathrm{B}}=20 \mu \mathrm{A}, I_{\mathrm{c}}=5 \mathrm{~mA}, \beta=250\)
2. \(I_{\mathrm{B}}=25 \mu \mathrm{A}, I_{\mathrm{c}}=5 \mathrm{~mA}, \beta=200\)
3. \(I_{\mathrm{B}}=40 \mu \mathrm{A}, I_{\mathrm{c}}=5 \mathrm{~mA}, \beta=250\)
4. \(I_{\mathrm{B}}=40 \mu \mathrm{A}, I_{\mathrm{c}}=5 \mathrm{~mA}, \beta=125\)

Answer: 4. \(I_{\mathrm{B}}=40 \mu \mathrm{A}, I_{\mathrm{c}}=5 \mathrm{~mA}, \beta=125\)

From question, \(V_{B E} =0 \mathrm{~V} \)

⇒  \(V_{C B} =0 \mathrm{~V}\)

⇒  \(V_i =20 \mathrm{~V}\)

and Applying Kirchhoff’s law,

⇒  \(V_I=I_B R_B+V_{B E}\)

Putting the values,

20 =\(I_B \times\left(500 \times 10^3\right)+0\)

⇒  \(I_B=40 \times 10^{-6}=40 \mu \mathrm{A}\) →  Equation 1

Similarly, \(V_{C C}=l_C R_C+V_{E C}\)

Again, putting the values

20=\(I_C \times\left(4 \times 10^3\right)+0\)

⇒  \(I_C=5 \times 10^{-3}=5 \mathrm{~mA}\)

Again we know that current \(\beta=\frac{I_C}{I_B}\)  → Equation 2

⇒  \(\beta =\frac{5 \times 10^{-3}}{40 \times 10^{-6}}\)

=\(0.125 \times 10^3\)=125

∴ \(I_B=40 \mu \mathrm{A}, I_C=5 \mathrm{~mA}, \beta\)=125

Question 60. In a common emitter transistor amplifier, the audio signal voltage across the collector is 3 V. The resistance of the collector is 3Ω). If the current gain is 100 and the base resistance is 2Ω), the voltage and power gain of the amplifier are;

1. 200 and 1000
2. 15 and 200
3. 150 and 15000
4. 20 and 2000

Answer: 3. 150 and 15000

Current gain,\((\beta)\)=100

Voltage gain, \(\left(A_V\right)=\beta \frac{R_c}{R_b}=100 \times \frac{3}{2}\)=150

Now Powers gain, P =\(A_V \cdot b\)

=150 \(\times 100\)=15000

Question 61. In the combination of the following gates, the output Y can be written in terms of inputs A and B as:

1. \(\overline{\mathrm{A} \cdot \mathrm{B}}+\mathrm{A} \cdot \mathrm{B}\)
2. \(\mathrm{A} \cdot \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}\)
3. \(\overline{\mathrm{A} . \mathrm{B}}\)
4. \(\overline{\mathrm{A}+\mathrm{B}}\)

Answer: 2. \(\mathrm{A} \cdot \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}\)

The combined figures for inputs and outputs are given below:

Question 62. The given electrical network is equivalent to:

1. AND gate
2. OR gate
3. NOR gate
4. NOT gate

Answer: 3. NOR gate

Truth Table Is:

Y→ NOR – gate

Question 63. A n-p-n transistor is connected in a common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it a 0.6 V. If the current amplification factor is 0.96 and the input resistance of the circuits is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be:

1. 3.69,3.84
2. 4.4
3. 4,3.69
4. 4,3.84

Answer: 4. 4,3.84

Here, \(R_0 =800 \Omega, R_i=192 \Omega\)

⇒ \(\beta\) =0.96

Voltage gain =0.96 \(\times \frac{800}{192}\)=4

Power gain =0.96 \(\times 4\)=3.84

Question 64. For the CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kD. is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kfi then the input signal voltage is:

1. 100 mv
2. 20 mV
3. 30 mV
4. 15 mV

Answer: 2. 20 mV

We Know The Formula,

Voltage amplification,

⇒ \(A_V=\beta \frac{\mathrm{R}_{\text {out }}}{R_n}=\frac{\left(\mathrm{V}_{\text {out }}\right)_{A C}}{\left(V_{\text {in }}\right)_{A C}}\)

Where,\( R_{\text {out }}\) =2 K W= collector resistance

B =100=amplification factor

⇒ \(R_n =1 \mathrm{KW}\)= Base resistance

V =4 \(\mathrm{~V}\)= output voltage

putting these values in equation (1)

⇒ \(A_V=\beta \frac{R_{\text {out }}}{R_n}=100 \times \frac{2 \mathrm{~K} \Omega}{1 \mathrm{~K} \Omega}\)

⇒ \(A_V=200\)

Now,\( A_V=\frac{\left(\mathrm{V}_{\text {out }}\right)_{\mathrm{AC}}}{\left(\mathrm{V}_{\text {in }}\right)_{\mathrm{AC}}}\)

∴ \(\left(V_{i n}\right)_{A C}=\frac{4}{200}=20 \mathrm{mV} \)

Question 65. What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1?

1. 0, 1
2. 0, 0
3. 1,0
4. 1, 1

Answer: 3. 1,0

for A=B = C = 0, Y=1

for A = B = C=1,Y=0

Y is(1,0)

Question 66. To get output 1 for the following circuit, the correct choice for the input is:

1. A=1,B=0,C=0
2. A=1,B=1,C=0
3. A=1,B=0,C=1
4. A=0,B=1,C=0

Answer: 3. A=1,B=0,C=1

From the figure, the boolean algebra expression of the logic gate is

Y = (A+B).C

Consider different cases :

Case (1): If = 0,5 = 0, C= 0

7 = (6 + 0).0 = 0

Case (2): If A = 1 B = 1 C=0

Y= 1.0 => Y=0

Case (3): If A( = 1 5 = 0 C = 1

Y = (1 + o),1 = 1.1 = Y=1

Case (4): If ÿ = 05 = 1 C = 0

7 = (1 + 1). 0 = 0

From the above four cases, it is confirmed that output 1 is obtained in option (3)

A =1,5 = 0, C= 1

Question 67. The input signal given to a CE amplifier having a voltage v gain of 150 is \(V_1=2 \cos \left(15 t+\frac{\pi}{3}\right)\). The corresponding output signal will be:

1. \(300 \cos \left(15 t+\frac{\pi}{3}\right)\)
2. \(75 \cos \left(15 t+\frac{2 \pi}{3}\right)\)
3. \(2 \cos \left(15 t+\frac{5 \pi}{3}\right)\)
4. \(300 \cos \left(15 t+\frac{4 \pi}{3}\right)\)

Answer: 4. \(300 \cos \left(15 t+\frac{4 \pi}{3}\right)\)

We know that \(V_0=\beta V_i\) and phase difference of \(\pi\)

⇒ \(V_0 =150 \times 2 \cos \left(15 t+\frac{\pi}{3}+\pi\right)\)

=300 \(\cos \left(15 t+\frac{4 \pi}{3}\right)\)

Question 68. Which logic gate is represented by the following combination of logic gates?

1. OR
2. NAND
3. AND
4. NOR

Answer: 3. AND

Its output Y shows that it is an AND-Gate.

Question 69. In a common emitter (CE) amplifier having a voltage gain of G, the transistor used has a transconductance of 0.03 mho and a current gain of 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will:

1. 2/3 G
2. 1.5G
3. 1/3 G
4. 5/4 G

Answer: 1. 2/3 G

Voltage gain,\(\mathrm{A}_{\mathrm{V}}=\beta \times \frac{R_{\text {out }}}{R_{\text {in }}}\)

Transconductance,

⇒ \(g_m =\frac{\beta}{R_{i n}} \text { or } R_{\text {in }}=\frac{\beta}{g_m}\)

⇒ \(\mathrm{A}_{\mathrm{V}} =g_m \mathrm{R}_{\text {out }}\)

For 1st Case, \(\mathrm{G} =0.03 \mathrm{R}_{\text {out }}\)

For 2nd Case, \(G^{\prime} =0.02 \mathrm{R}_{\text {out }} \)

Now, \(\frac{G^{\prime}}{G^0} =\frac{2}{3} or G^{\prime}=\frac{2}{3} G\)

Question 70. One way in which the operation of an n-p-n transistor differs from that of ap-n-p.

1. The emitter junction injects minority carriers into the base region of the p-n-p.
2. The emitter junction injects minority carriers into the base region of the p-n-p.
3. The emitter injects holes into the base of n-p-n.
4. The emitter junction is reversed-biased in n-p-n.

Answer: 2. The emitter junction injects minority carriers into the base region of the p-n-p.

The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n.

Question 71. The output (X) of the logic circuit shown in the figure will be:

1. \(\mathrm{X}=\overline{\overline{\mathrm{A}}} \cdot \overline{\mathrm{B}}\)
2. \(\mathrm{X}=\overline{\mathrm{A} . \mathrm{B}}\)
3. \(X=A \cdot B\)
4. \(\mathrm{X}=\overline{\mathrm{A}+\mathrm{B}}\)

Answer: 3. \(X=A \cdot B\)

∴ \(\mathrm{X}=\overline{\mathrm{AB}}=\mathrm{A} \cdot \mathrm{B}\) i.e. And gate.

Question 72. The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a:

1. AND gate
2. NOR gate
3. OR gate
4. NOT gate

Answer: 1. AND gate

Output is C=\(\overline{\bar{A} \cdot B}=A \cdot B \rightarrow\) This is AND-gate

Question 73. In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1 kΩ and the current amplification of the transistor is 100, the input signal voltage is:

1. 0.1 V
2. 1.0 V
3. l mV
4. 10 mV

Answer: 4. 10 mV

It is given that,

⇒ \(\beta =100=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}} \)

⇒ \(R_{\mathrm{C}} =2 \mathrm{k} \Omega\)

⇒ \(V_{\mathrm{L}} =2 \mathrm{~V} \)

⇒ \(R_B =1 \mathrm{k} \Omega \)

⇒ \(I_C =\frac{V_C}{R_C}=\frac{2}{2}=1 \mathrm{~mA} \rightarrow\)collector current

⇒ \(I_B =\frac{V_B}{R_B}=\frac{V_B}{I}=\mathrm{V}_{\mathrm{B}} \mathrm{mA} \rightarrow\) Base current

⇒ \(\beta =100=\frac{1}{\mathrm{~V}_{\mathrm{B}}}\)

⇒ \(V_B =\frac{1}{100}=0.01 \mathrm{~V}=10^{-2} \mathrm{~V}\)

∴ \(V_B =10 \mathrm{mV}\)

Question 74. Transfer characteristic [output voltage (V₀) vs input •voltage (V₁)] for a base-biased transistor in CE configuration is as shown in the figure. For using a transistor as a switch, it is used:

1. In region 3
2. Both in the region (1) and (3)
3. In region 2
4. In region 1

Answer: 2. Both in regions (1) and (3)

For using a transistor as a switch it is used in

(1) cut off state

(2) Saturation state only.

They are both in region (1) and (2)

Question 75. The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 μA which results in a change in collector current by 2 mA. This transistor is used as a common-emitter amplifier with a load resistance of 4kμ. The voltage gain of the amplifier is:

1. 2000
2. 3000
3. 4000
4. 1000

Answer: 1. 2000

Current gain, \(\beta=\frac{\Delta \mathrm{I}_e}{\mathrm{I}_\beta}=\frac{2 \times 10^{-3}}{40 \times 10^{-6}}\)

⇒ \(\beta\)=50

∴ Voltage gain, \(A_{\mathrm{V}}=\beta\left(\frac{\mathrm{R}_{\text {out }}}{\mathrm{R}_{\text {in }}}\right)\)=2000

Question 76. The figure shows a logic circuit with two inputs A and B and the output C. The voltage waveforms across A, B, and C are as given. The logic circuit gate is:

1. OR gate
2. No gate
3. AND gate
4. NAND gate

Answer: 1. OR gate

From the given logic gate the truth table is :

This table is the table of the ‘OR’ gate.

Question 77. To get output Y = 1 in the given circuit which of the following inputs will be correct:

Answer:  2. 

Question 78. A transistor is operated in a common emitter configuration at VQ = 2 V such that a change in the base current from 100 μA to 300 μA produces changes in the collector current from 10 mA to 20 mA. The current gain is:

1. 75
2. 100
3. 25
4. 50

Answer: 4. 50

⇒ \(\beta =\frac{\Delta I_C}{\Delta I_B} \)

= \(\frac{(20-10) \times 10^{-3}}{(300-100) \times 10^{-6}}\)=50

Question 79. The symbolic representation of four logic gates is shown as:

Pick out which ones are for AND, NAND, and NOT gates respectively.

1. (3), (2), and (1)
2. (3), (2), and (4)
3. (2), (4), and (3)
4. (2), (3), and (4)

Answer: 3. (2), (4) and (3)

In the given figure

(1) → OR-gate (2)  → AND-gate

(3) →  NOT-gate (4) → NAND-gate

Hence option (C) is correct

Question 80. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω, and an output impedance of 200 Ω. The power gain in the amplifier is:

1. 500
2. 1000
3. 1250
4. 50

Answer: 3. 1250

We know that, voltage gain =\(\beta \times\) impedance gain

50 =\(\beta \times \frac{200}{100} \)

= 25

Power gain =\(\beta^2 \times\) impedance gain

= \((25)^2 \times \frac{200}{100}\)=1250

Question 81. For transistor action:

1. Base, emitter, and collector regions should have similar sizes and doping concentrations.
2. The base region must be very thin and lightly doped.
3. The emitter base junction is forward and the base collector junction is reverse-biased.
4. Both the emitter-base junction as well as the base-collector junction are forward-biased

Which one of the following pairs of statements is correct?

1. (4) and (1)
2. (1) and (2)
3. (2) and (3)
4. (3) and (4)

Answer: 3. (2) and (3)

For transistor action

(1) The base region must be very thin and lightly doped.

(2) The emitter-base is forward-biased and the base-collector junction is reverse-biased.

Question 82. To get output Y = 1 from the circuit shown below, the input must be:

Answer:  4.

Gate 1 OR-gate, \(y^{\prime}=\mathrm{A}+\mathrm{B}\)

Gate 2 AND-gate, \(y=y^{\prime}\). c

∴ \(\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0 \Rightarrow \)Y=1

Question 83. The following figure shows a logic gate circuit with two inputs A and B and output Y. The voltage waveforms of A, B, and Y are as given:

The logic gate is:

1. NOR gate
2. OR gate
3. AND gate
4. NAND gate

Answer: 4. NAND gate

The truth table for logic gate i.e.

This is NAND-gate

Question 84. The device that can act as a complete electronic circuit is:

1. Junction diode
2. Integrated circuit
3. Junction transistor
4. Zener diode

Answer: 2. Integrated circuit

The device that can act as a complete electronic circuit is an integrated circuit.

Question 85. A transistor is operated in a common-emitter configuration at V_c=2 volt such that a change in the base current from 100 μA to 200 μA produces changes in the collector current from 5 mA to 10 mA. The current gain is:

1. 75
2. 100
3. 150
4. 50

Answer: 4. 50

According to the question

For the transistor, \(I_E=I_B+I_C \)

⇒ \(I_E\)= Emitter current

⇒ \(I_B\)= case current =100 \(\mu \mathrm{A} \)

⇒ \(I_C\)= Collector current =5 \(\mathrm{~mA}\)

and current gain, \(\beta=\frac{\Delta I_C}{\Delta I_B}\)

∴ \(\beta=\frac{5 \times 10^{-3}}{100 \times 10^{-6}}\)=50

Question 86. The symbolic representation of four logic gates:

The logic symbols for OR, NOT, and NAND gates are respectively;

1. 3, 4, 2
2. 4, 1, 3
3. 4, 2, 1
4. 1, 3, 4

Answer: 3. 4, 2, 1

(1) NAND Gate

(5) NOT Gate

(4) OR Gate

The correct options would be (3).

Question 87. The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be:

1. 90
2. 10
3. 1.25
4. 100

Answer: 4. 100

Voltage gain with feedback is

⇒ \(A_{v f}=\frac{A_V}{1+\beta A_V}\) →  Equation 1

Where, \(A_V\)= voltage gain without feedback and \(\beta\)= negative feedback

From the question,

⇒ \(A_{v f}=10, \beta=9 \%=\frac{9}{100}\)

Putting in eq. (i)

10=\(\frac{A_V}{1+\frac{9}{100} A_V}\)

Putting in eq. (1)

10 =\(\frac{A_V}{1+\frac{9}{100} A_V}\)

or \(10+\frac{9}{10} \mathrm{~A}_{\mathrm{V}} =A_V\)

0.1 A_V =10

∴ \(A_V\) =100

Question 88. For a transistor circuit, the base current is 10μA, and the collector current is 5.2 μA. Can this transistor circuit be used as an amplifier? Your answer must be supported with a proper explanation.

1. 0.5 v
2. 0.1 v
3. 0.3 v
4. 0 v

Answer: 3. 0.3 v

From the question,

⇒ \(V_{B E} =V_{C C}-I_B R_B \)

= 5.5-10 \(\times 10^{-6} \times 500 \times 10^3 \)

=0.5 \(\mathrm{~V}\)

⇒ \(V_{C E} =V_C-I_C R_C \)

=5.5-5.2 \(\times 10^{-3} \times 1 \times 10^3\)

=0.3 V

Both the E-B function and the C-E function are forward bias, they can not be used as an amplifier.

Question 89. The circuit is equivalent to:

1. AND gate
2. NAND gate
3. NOR gate
4. OR gate

Answer: 3. NOR gate

The output is of the NOR gate hence the combination will act as a NOR gate.

Question 90. A common emitter amplifier has a voltage gain of 50, an input impedance of 100Ω, and an output impedance of 200 Ω. The power gain of the amplifier is:

1. 1000
2. 1250
3. 100
4. 500

Answer: 3. 100

Voltage gain =\((\beta)\left(\frac{\mathrm{R}_0}{\mathrm{R}_i}\right)\)=50

⇒ \(\beta=\frac{50 \times 100}{200}\)=25

Power gain =\((125)^2 \times \frac{200}{100} \)

= \((125)^2 \times \frac{200}{100} \)

= 625 \(\frac{200}{100}\)

= 1250

Question 91. In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table.

Answer: 3.

∴ \(y^{\prime}=\overline{\mathrm{A}+\mathrm{B}}, y=p+e+v=\mathrm{A}+\mathrm{B}\)

Question 92. A transistor is operated in a common-emitter configuration at V_c= 2 volt such that a change in the base current from 100 μA to 200 μA produces changes in the collector current from 5 mA to 10 mA. The current gain is:

1. 0.8 v
2. 0.1 v
3. 0.3 v
4. 0 v

Answer: 4. 0 v

Current gain =\(\beta=\frac{\Delta I_C}{\Delta I_B}\)

=\(\frac{(10-5) \mathrm{mA}}{(150-100) \mu \mathrm{A}}=\frac{5 \times 10^{-3}}{50 \times 10^{-6}} \)

=100

Question 93. The following figure shows a logic gate circuit with two inputs A and B and the output C. The voltage waveforms of A, B, and C are as shown below:

The logic circuit gate is:

1. OR Gate
2. AND Gate
3. NAND Gate
4. NOR Gate

Answer: 2. AND Gate

The circuit is AND- gate

Question 94. The output of the OR gate is 1:

1. If both inputs are zero
2. If either of both inputs is 1
3. Only if both inputs are 1
4. If either input is zero

Answer: 2. If either of both inputs are 1

means if either or both inputs are 1.

Question 95. An-p-n transistor conducts when:

1. Both collector and emitter are positive with respect to the base
2. The collector is positive and the emitter is negative with respect to the base
3. The collector is positive and the emitter is at the same potential as the base
4. Both collector and emitter are negative with respect to the base

Answer: 2. Collector is positive and emitter is negative with respect to the base

In the active region Emitter base p-n function is in FB and the base collector p-n function is in RB.

Question 96. For transistor \(\frac{I_C}{I_{E_c}}=0.96\), then the current gain for common emitter configuration :

1. 12
2. 6
3. 48
4. 24

Answer: 4. 24

For the common emitter configuration, the current transfer ratio is,

⇒ \(\alpha =\frac{I_C}{I_B}\)=0.96

Current gain, =\(\frac{\alpha}{1-\alpha}\)

= \(\frac{0.96}{1-0.96}\)=24

Question 97. For which logic gate, the given truth table is shown?

1. NAND
2. XOR
3. NOR
4. OR

Answer: 1. NAND

It is a NAND-gate

Question 98. The following diagram performs the logic function of

1. OR gate
2. AND gate
3. XOR gate
4. NAND gate

Answer: 2. AND gate

Let the output of the first NAND gate be chosen as X as shown in the diagram below.

Output of 1st NAND gate, X=\(\overline{A \cdot B}\)

Using De-Morgan’s theorem

⇒ \(\overline{A \cdot B}=\bar{A}+\bar{B}\)

So, X=\(\bar{A}+\bar{B}\)

Now, output of \(2^{\text {nd }}\) NAND gate,

Again, Y =\(\bar{X}=\overline{\bar{A}+\bar{B}} \)

⇒ \(\overline{\bar{A}+\bar{B}} =\bar{A} \cdot \bar{B}=A \cdot B\)

Y =A \(\cdot B\)

Hence, This is the AND gate’s logic function.

Question 99. If cr and B are current gains in common-base and common-emitter configurations of a transistor, then B is equal to:

1. \(\frac{1}{\alpha}\)
2. \(\frac{\alpha}{1+\alpha}\)
3. \(\frac{\alpha}{1-\alpha}\)
4. \(\alpha-\frac{1}{\alpha}\)

Answer: 3. \(\frac{\alpha}{1-\alpha}\)

The current gain in common-base configuration,

⇒ \(\alpha=\left(\frac{\Delta i_c}{\Delta i_e}\right)_{V_{c b}}\)

The current gain in common-emitter configuration,

Also, \(\beta=\left(\frac{\Delta i_c}{\Delta i_b}\right)_{v_\sigma}\)

Also, \(i_b =i_e-i_c\)

⇒ \(\Delta i_b =\Delta i_e-\Delta i_c\)

⇒ \(\beta=\frac{\Delta i_c}{\Delta i_b}=\frac{\Delta i_c}{\Delta i_e} \times \frac{\Delta i_e}{\Delta i_b}\)

⇒ \(\beta=\alpha \times \frac{\Delta i_e}{\Delta i_e-\Delta i_c}\)

⇒ \(\beta=\alpha \times \frac{1}{1-\frac{\Delta i_c}{\Delta i_e}}\)

∴ \(\beta=\frac{\alpha}{1-\alpha}\)

Question 100. The truth table given below represents

1. AND gate
2. NOR gate
3. OR gate
4. NAND gate

Answer: 4. NAND gate

A NAND gate gives output as 1 when either of the input signals is low, i.e., O, and gives output as 0 only when better the input signals are high, i.e., Hence the above truth table is for the NAND gate.

Thus, the NAND gate will give an output of 1

Question 101. The transfer ratio p of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is 1kΩ. The peak value of the collector. AC current for an AC input voltage of 0.01 V peak is:

1. 100 μA
2. 0.01 mA
3. 0.25 mA
4. 500 μA

Answer: 4. 500 pA

Input current or We have,

B current, \(i_{\mathrm{b}}=\frac{\Delta V_b}{R_b}=\frac{0.01}{10 \rho 0}=10^{-5} \mathrm{~A}\)

⇒  (\(\Delta V_b\) = input voltage; \(\Delta R_{\mathrm{b}}\) = base resistance)

Also Current gain, \(\beta=\frac{i_c}{i_b}\)

⇒  \(i_{\mathrm{c}}=\beta i_{\mathrm{c}}=50 \times 10^{-5} \mathrm{~A}\)

= \(500 \times 10^{-6} \mathrm{~A}=500 \mu \mathrm{A}\)

Question 102. The part of the transistor which is heavily doped to produce large number of majority carriers is:

1. Emitter
2. Base
3. Collector
4. Any of the above depending upon the nature of the transit

Answer: 1. Emitter

The transistor has three regions, namely emitter, base, and collector. The base is much thinner than the emitter, while the collector is wider than both. The emitter is heavily doped so that it can inject a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and very thin, it passes most of the emitter-injected charge carriers to the collector. The collector is moderately doped.

Question 103. To use a transistor as an amplifier,

1. The emitter-base junction is forward biased and the base-collector junction is reversed biased
2. No bias voltage is required
3. Both junctions are forward-biased
4. Both junctions are revealed as biased

Answer: 1. The emitter-base junction is forward biased and the base-collector junction is reversed biased

A transistor is used as an amplifier in a common emitter (CE) configuration. In this configuration, the base-emitter junction is forward-biased and the collector-base junction is reverse-biased

Question 104. The following truth table corresponds to the logical gate

1. NAND
2. OR
3. AND
4. XOR

Answer: 2. OR

The given truth table shows that, if any or all of the inputs are high, then the output is high. The only method to get a low output is to set all of the inputs to zero. If any or all of the inputs are high, the OR gate’s output 1 is high. As a result, the tiris truth table is an OR gate truth table.

Wave Optics

Question 1. In a Young’s double slit experiment, a student observed 8 fringes in a certain segment of the screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is

1. 6
2. 8
3. 9
4. 12

Answer: 4. 12

The wavelength of light is inversely proportional to the number of fringes observed

So, \(n_1 \lambda_1 =n_2 \lambda_2\)

∴ \(n_2=\frac{n_1 \lambda_1}{\lambda_2}\)

= \(\frac{600 \times 8}{400}=12\)

Question 2. Two coherent sources of light interfere and produce a fringe pattern on a screen. For central maximum, the phase difference between the two waves will be

1. Zero
2. \(\pi\)
3. \(\frac{3 \pi}{2}\)
4. \(\frac{\pi}{2}\)

Answer: 1. Zero

From Young’s double slit experiment phase difference = \(\frac{2 \pi}{\lambda}\) x path difference. But for central maximum, path difference is zero for both coherent sources in interference.

Which means the path difference = 0

Phase difference = \(\frac{2 \pi}{\lambda}\) x 0 = 0

Question 3. In Young’s double-slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes

1. Half
2. Four times
3. One fourth
4. Double

Answer: 2. Four times

From Young’s double slit experiment fringe width

⇒ \(\beta =\frac{\lambda D}{d}\)

⇒  \(\frac{\beta_2}{\beta_1} =\frac{\frac{\lambda_2 D_2}{d_2}}{\frac{\lambda_1 D_1}{d_1}}\)

Given, \(\lambda_1=\lambda_2, D_2=2 D_1\), and \(d_2=\frac{d_1}{2}\) putting these values in above equation,

⇒ \(\frac{\beta_2}{\beta 1}=\overline{\left(\frac{2 D_1}{2}\right)} \times \frac{d_1}{\mathrm{D}_1}=4\)

⇒ \(\beta_2=4 \beta_1\)

Fringe width becomes four times.

Read and Learn More NEET Physics MCQs

Question 4. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μ water = 4/3)

1. 0.15°
2. 0.051°
3. 0.1°
4. 0.266°

Answer: 1. 0.15°

Angular width, \(\theta=0.2^{\circ}\)

Now, \(\theta=\frac{\beta}{D}=0.2^{\circ}\)

In water, \(\theta^{\prime}=\frac{\beta}{\mu D}\)

⇒ \(\theta^{\prime}=\frac{\theta}{\mu}=\frac{0.2}{4 / 3}\)

= \(0.15^{\circ}\)

Question 5. In Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has a path difference

1. \(5 \frac{\lambda}{2}\)
2. \(10 \frac{\lambda}{2}\)
3. \(9 \frac{\lambda}{2}\)
4. \(11 \frac{\lambda}{2}\)

Answer: 3. \(9 \frac{\lambda}{2}\)

In young’s double slit experiment \(\Delta x=\text { path difference }=(2 n-1) \frac{\lambda}{2}\)

So, Here n=5 means for \(5^{\text {th }}\) minima

∴ \(\Delta x=(2 \times 5-1) \frac{\lambda}{2}=\frac{9}{2} \lambda\)

Question 6. In Young’s double slit experiment, the separation d between the slits is 2 mm, the wavelength X of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with the same λ and D) the separation between the slits needs to be changed to

1. 2.1 mm
2. 1.9 mm
3. 1.8 mm
4. 1.7 mm

Answer: 2. 1.9 mm

In Young’s Double slit experiment Angular width of fringe is \(\theta=\frac{\lambda}{d}\)

⇒  \(\theta \propto \frac{1}{d}\)

or \(\frac{\theta_1}{\theta_2}=\frac{d_2}{d_1}\)…..(1)

In question, \(\theta_1=0.21^{\circ}\)

⇒  \(d_1=2 \mathrm{~mm}\)

Then eq. (1) becomes, \(\frac{0.20^{\circ}}{0.21^{\circ}}=\frac{d_2}{2 \mathrm{~mm}}\)

⇒  \(d_2=2 \times \frac{0.20}{0.21}=\frac{0.40}{0.21}\)

= \(1.90 \mathrm{~mm}\)

Question 7. Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that the 5^th dark fringe in the air is equal to the 8^th bright fringe that lies in the medium. The refractive index of the medium is nearly :

1. 1.25
2. 1.59
3. 1.69
4. 1.78

Answer: 4. 1.78

From the question, it is clear that 5^th dark fringe in air = 8th bright fringe in medium

⇒ \((2 \times 5-1) \frac{\lambda_{\text {air }} D}{2 d}=8 \frac{\lambda_m D}{d}\)

⇒ \(9 \frac{\lambda_{\text {air }} D}{2 d}=8 \frac{\lambda_m D}{d}\)

⇒ \(\mu =\frac{\lambda_{\text {air }}}{\lambda m}\)

⇒ \(\mu =\frac{8 \times 2}{9}=\frac{16}{9}\)

Refractive index of medium \(\mu=\frac{16}{9} \cong 1.78\)

Question 8. The intensity at the maximum in Young’s double slit experiment is I0. The distance between slits is d = 5λ, where X is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?

1. \(\frac{\mathrm{I}_0}{4}\)
2. \(\frac{3}{4} \mathrm{l}_0\)
3. \(\frac{1_0}{2}\)
4. \(\mathrm{I}_0\)

Answer: 3. \(\frac{1_0}{2}\)

= \(S_2 P-S_1 P\)

= \(\sqrt{D^2+d^2}-D\)

= \(D\left(1+\frac{1}{2} \frac{d^2}{D^2}\right)-D\)

= \(D\left[1+\frac{d^2}{2 D^2}-1\right]=\frac{d^2}{2 D}\)

Now \(\Delta x=\frac{d^2}{2 \times 10 d}=\frac{d}{20}\)

= \(\frac{5 \lambda}{20}=\frac{\lambda}{4}\)

⇒ \(\Delta \phi=\frac{2 \lambda}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}\)

⇒ \(\Delta \phi=\frac{2 \lambda}{\lambda}, \frac{\lambda}{4}=\frac{\pi}{2}\)

(because phase difference = \(\frac{2 \pi}{\lambda}\). path difference)

So, the intensity at the desired point is \(I=I_0 \cos ^2 \frac{\phi}{2}=I_0 \cos ^2 \frac{\pi}{4}=\frac{I_0}{2}\)

Question 9. The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio \(\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}\) will be:

1. \(\frac{\sqrt{n}}{n+1}\)
2. \(\frac{2 \sqrt{n}}{n+1}\)
3. \(\frac{\sqrt{n}}{(n+1)^2}\)
4. \(\frac{2 \sqrt{n}}{(n+1)^2}\)

Answer: 2. \(\frac{2 \sqrt{n}}{n+1}\)

Let [\(\frac{I_1}{I_2}=\frac{n}{1}\)

\(\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}\)

= \(\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2-\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2+\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}\)

= \(\frac{4 \sqrt{I_1 I_2}}{2\left(I_1+I_2\right)}\)

Dividing numerator and denominator by 2.

Ratio will be \(\frac{\sqrt[2]{\frac{I_1}{I_2}}}{\left(\frac{I_1}{I_2}+1\right)}=\frac{2 \sqrt{n}}{n+1}\)

Question 10. In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of a single slit pattern?

1. 0.2 mm
2. 0.1 mm
3. 0.5 mm
4. 0.02 mm

Answer: 1. 0.2 mm

Given : d = \(1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}\)

D = \(1 \mathrm{~m}\)

⇒ \(\lambda =5 \times 10^{-7} \mathrm{~m}\)

As width of central maxima = width of 10 maxima

⇒ \(\frac{2 \lambda D}{a}=10\left(\frac{\lambda D}{d}\right)\)

a = \(\frac{d}{5}=\frac{10^{-3}}{5}=0.2 \times 10^{-3} \mathrm{~m}\)

a = \(0.2 \mathrm{~mm}\)

Question 11. Two slits in Young’s experiment have widths in the ratio 1: 25. the ratio of intensity at the maxima and minima in the interference pattern \(\frac{I_{\max }}{I_{\min }}\) is

1. 9/4
2. 121/49
3. 49/121
4. 4/9

Answer: 1. 9/4

In Young’s double slit experiment, two slits of width are in the ratio 1: 25,

Ratio of intensity, \(\frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{1}{25}\)

⇒ \(\frac{I_2}{I_2}=\frac{1}{25}\)…..(1)

From the question, \(\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_2}+\sqrt{I_1}\right)^2}{\left(\sqrt{I_2}-\sqrt{I_1}\right)^2}\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\left[\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}\right]^2\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\left[\frac{5+1}{5-1}\right]^2\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{6^2}{4^2}\)

⇒ \(\frac{I_{\mathrm{mix}}}{I_{\min }}=\frac{36}{16}\)

∴ \(\frac{I_{\mathrm{mix}}}{I_{\min }}=\frac{9}{4}\)

Question 12. In Young’s double-slit experiment, the intensity of light at a point on the screen (where the path difference is A,) is K, (λ, being the wavelength of light used). The intensity a point where the path difference is λ/4, will be:

1. K
2. K/4
3. K/2
4. Zero

Answer: 3. K/2

We know that I = \(I_0 \cos ^2\left[\frac{k\left(r_1-r_2\right)}{2}\right]\)

and \(r_1-r_2 =I \Rightarrow I=K\)

K = \(I_0 \cos ^2\left[\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\right]\)

K = \(I_0 \cos ^2 \pi\)

K = \(I_0\)….(1)

( \(\cos ^2 \pi=(-1)^2=1\))

Now again, \(I=I_0 \cos ^2\left[\frac{k\left(r_1-r_2\right)}{2}\right]\)

and \(r_1-r_2=\frac{\pi}{4}\)

and \(K=\frac{2 \pi}{4}\)

I = \(I_2 \cos ^2\left[\frac{2 \pi}{\pi} \times \frac{\pi}{8}\right]\)

= \(I_0 \frac{\cos ^2(\pi)}{4}=\frac{I_0}{2}\)

(because \(I_0=K\) from eq. (1))

\(I=\frac{K}{2}\)

Question 13. In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ₁ = 12000A and λ₂ = 10000 A. At what minimum distance from the common central bright fringe on the screen, 2 m from the slit will a bright fringe from one interference pattern coincide with a bright from the other?

1. 8 mm
2. 6 mm
3. 4 mm
4. 3 mm

Answer: 2. 6 mm

According to the question, \(n_1 \lambda_1=n_2 \lambda_2\)

So, \(\frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}=\frac{10000}{12000}=\frac{5}{6}\)

So maximum \(n_1\) and \(n_2\) are 5 and 6 respectively

⇒ \(X_{m n}=\frac{n_1 \lambda_1 D}{d}=\frac{5\left(12000 \times 10^{-2}\right)(2)}{2 \times 10^{-3}}\)

= \(6 \times 10^{-3} \mathrm{~m}=6 \mathrm{~mm} .\)

Question 14. In Young’s double-slit experiment, the distance between the slits and the screen is doubled. The separation between the slits was reduced to half. As a result the fringe width:

1. Is halved
2. Becomes four times
3. Remains unchanged
4. Is doubled

Answer: 2. Becomes four times

In Young’s double slit experiment fringe width \(\beta=\frac{\lambda D}{d}\)…(1)

where, D = distance between slit and screen

d = distance between slit

According to the question, D is double → 2D and d is half → d/2

Their New fringe width will be \(\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta\)

[A = 4p i.e., becomes four times

Question 15. A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since:

1. A large aperture contributes to the quality and visibility of the images.
2. A large area of the objective ensures better light-gathering power.
3. A large aperture provides a better resolution
4. All of the above

Answer: 4. All of the above

A large focal length and large aperture of an objective of an astronomical telescope ensures better light-gathering power, and resolution and contributes to the quality and visibility of the images of the far-off objects.

Question 16. The Brewsters angle ib for an interface should be:

1. \(30^{\circ}< i_{\text {b }}<45^{\circ}\)
2. \(45^{\circ}<\mathrm{i}_{\mathrm{b}}<90^{\circ}\)
3. \(\mathrm{i}_{\mathrm{b}}=90^{\circ}\)
4. \(0^{\circ}<\mathrm{i}_{\mathrm{b}}<30^{\circ}\)

Answer: 2. \(45^{\circ}<\mathrm{i}_{\mathrm{b}}<90^{\circ}\)

According to Brewster’s law, when a beam of unpolarised light is reflected from a transparent medium of refractive index (μ₂), the reflected light is completely polarised at a certain angle of incidence called the angle of polarization (i_b).

⇒ \(\tan i_b=\frac{\mu_2}{\mu_1}\)

For air, \(\mu_1=1\)

⇒ \(\tan i_b=\mu_2>1\)

∴ \(\tan i_b >1 \Rightarrow 90^{\circ}>i_b>45^{\circ}\)

Question 17. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is

1. \(1.83 \times 10^{-7} \mathrm{rad}\)
2. \(7.32 \times 10^{-7} \mathrm{rad}\)
3. \(6.00 \times 10^{-7} \mathrm{rad}\)
4. \(3.66 \times 10^{-7} \mathrm{rad}\)

Answer: 4. \(3.66 \times 10^{-7} \mathrm{rad}\)

Given: Wavelength, \(\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\)

and diameter of the objective, d = 2m

Limit of resolution of the telescope,

⇒ \(\theta=\frac{1.22 \lambda}{d}\)

⇒ \(\theta=\frac{1.22 \times 600 \times 10^{-9}}{2}\)

∴ \(\theta=3.66 \times 10^{-7} \mathrm{rad}\)

Question 18. The angular width of the central maxima in the Fraunhofer diffraction for A = 6000 Å is θ₀. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is :

1. 1800 A
2. 4200 A
3. 6000 A
4. 420 A

Answer: 2. 4200 A

As, \(\theta_0=\frac{2 \lambda}{a}\)

⇒ \(\theta_0=\frac{2 \times 6000}{a}\)

⇒  \(\frac{\theta^{\prime}}{\theta_0}=\frac{\lambda^{\prime}}{6000}\)

⇒ \(\lambda^{\prime}=0.7 \times 6000\) as \(\theta^{\prime}=0.7 \theta_0)\)

= 4200 Å

Question 19. The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is:

1. 9: 4
2. 3: 2
3. 16:81
4. 8: 27

Answer: 2. 3: 2

Resolving power of a microscope \(=\frac{2 \mu \sin \theta}{\lambda}\)

i.e. \(R \propto \frac{1}{\lambda} \text { or } \frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1}\)

Given that the two wavelengths, \(\lambda_1=4000\) Å and \(\lambda_2=6000 Å\)

∴ \(\frac{R_1}{R_2}=\frac{6000 Å}{4000Å }=\frac{3}{2}\)

Question 20. In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle of 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of:

1. \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2. \(\sin ^{-1}\left(\frac{1}{2}\right)\)
3. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4. \(\sin ^{-1}\left(\frac{1}{4}\right)\)

Answer: 3. \(\sin ^{-1}\left(\frac{3}{4}\right)\)

According to the question first minimum is observed at an angle of 30° at a single slit of width a means

n=1, \(\theta=30^{\circ}\)

From Bragg’s law \(a \sin \theta=n \lambda\)

a \(\sin 30^{\circ} =\lambda\)

a = \(2 \lambda\)  → (1)

And the first secondary maxima \(a \sin \theta_1=\frac{3 \lambda}{2}\)

⇒ \(\sin \theta_1=\frac{3 \lambda}{2 a}\)  → (2)

From eq. (1) and (2), \(\sin \theta_1=\frac{3 \lambda}{2(2 \lambda)}\)

⇒ \(\sin \theta_1=\frac{3}{4}\)

∴\(\theta_1=\sin ^{-1} \frac{3}{4}\)

Question 21. A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 x 10^-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is:

1. 0.10 cm
2. 0.25 cm
3. 0.20 cm
4. 0.15 cm

Answer: 4. 0.15 cm

Here f = D = 60 cm = 0.6 m for first minima,

y = \(\frac{\lambda D}{a}=\frac{5 \times 10^{-7} \times 0.6}{2 \times 10^{-2} \times 10^{-2}}\)

= \(15 \times 10^{-4} \mathrm{~m}=0.15 \mathrm{~cm}\)

Question 22. For a parallel beam of monochromatic light of wavelength ‘λ’ diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be:

1. \(\frac{2 \mathrm{D} \lambda}{a}\)
2. \(\frac{\mathrm{D} \lambda}{a}\)
3. \(\frac{\mathrm{D} a}{\lambda}\)
4. \(\frac{2 \mathrm{D} a}{\lambda}\)

Answer: 1. \(\frac{2 \mathrm{D} \lambda}{a}\)

From diagram \(\sin \theta=\frac{\lambda}{a}\)

⇒ \(\sin \theta=\theta=\frac{Y}{D}\) (for small angle)

⇒ \(\frac{Y}{D}=\frac{\lambda}{a}\)

⇒ \(\mathrm{Y}=\frac{\lambda D}{a}\)

Width of central maxima = \(2 \mathrm{Y}\)

= \(\frac{2 \lambda D}{a}\)

Question 23. At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygens wavelength from the edge of the slit and the wavelength from the midpoint of the slit is:

1. \(\frac{\pi}{4}\) radian
2. \(\frac{\pi}{2}\) radian
3. \(\pi\) radian
4. \(\frac{\pi}{8}\) radian

Answer: 3. \(\pi\) radian

Phase difference \(\Delta \phi=\frac{\Delta x}{\lambda} \times 2 \pi=\frac{(a / 2) \sin \theta}{\lambda} \times 2 \pi\)

∴ \(\Delta \phi=\frac{\lambda}{2 \lambda} \times 2 \pi=\pi \text { radian }\)

Question 24. A beam of light of X = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is :

1. 1.2 cm
2. 1.2 mm
3. 2.4 cm
4. 2.4 mm

Answer: 4. 2.4 mm

For a dark fringe to form \(\frac{d y}{\mathrm{D}}=\lambda\)

y = \(\frac{D \lambda}{d}=\frac{2 \times 600 \times 10^{-9}}{10^{-3}}=1.2 \mathrm{~mm}\)

distance between the first dark fringes on either side of the central bright fringe is 2 x y = 2.4 mm.

Question 25. A parallel beam of fast-moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, then of the electrons is increased, then which of the following statements is correct?

1. The diffraction pattern is not observed on the screen in the case of electrons
2. The angular width of the central maximum of the diffraction pattern will increase
3. The angular width of the central maximum will decrease
4. The angular width of the central maximum will be unaffected

Answer: 2. The angular width of the central maximum of the diffraction pattern will increase

As the speed of electrons is increased the wavelength of electrons will decrease. Therefore, the angular width (ocX) of the central maximum of the diffraction pattern will decrease

Question 26. A parallel beam of light of wavelength X is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is:

1. 2π
2. 3π
3. 4π
4. 5λ

Answer: 3. 4π

We know that the path difference for the second minimum = 2λ

Phase difference = \(\frac{2 \pi}{\lambda} \times 2 \lambda\)

= \(4 \pi\)

Question 27. The periodic waves of intensities \(I_1\) and \(I_2\) pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1. \(I_1+I_2\)
2. \(\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
3. \(\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
4. \(2\left(\mathrm{I}_1+\mathrm{I}_2\right)\)

Answer: 4. \(2\left(\mathrm{I}_1+\mathrm{I}_2\right)\)

The resultant intensity of two periodic waves at a point is given by

⇒ \(\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cdot \cos \phi\)

Resultant intensity is maximum if \(\cos \phi=-1\)

i.e. \(I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}\)

Resultant intensity is minimum if \(\cos \phi=+1\)

i.e, \(I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2}\)

Therefore, the sum of the maximum and minimum intensities is \(\mathrm{I}_{\max }+\mathrm{I}_{\min }\)

= \(I_1+I_2+2 \sqrt{I_1 I_2}+I_1+I_2-2 \sqrt{I_1 I_2}\)

= \(2\left(I_1+I_2\right)\)

Question 28. The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 Å is of the order of:

1. \(10^6 \mathrm{rad}\)
2. \(10^{-2} \mathrm{rad}\)
3. \(10^{-4} \mathrm{rad}\)
4. \(10^{-6} \mathrm{rad}\)

Answer: 4. \(10^{-6} \mathrm{rad}\)

⇒ \(\delta \phi=1.22 \frac{\lambda}{D}\)

= \(1.22 \frac{5000 \times 10^{-10}}{10 \times 10^{-2}}\)

= \(6.1 \times 10^{-6}\)

Order = \(10^{-6}\)

Question 29. A paper, with owo marks having separation d, is held normal to the line of sight of an observer at a distance of 50m. The diameter of the eye-lens of the observer is 2 mm. Which of the following is the least value of d. so that the marks can be seen as separate? The mean wavelength of visible light may be taken as 5000 Å.

1. 1.25 m
2. 12.5 cm
3. 1.25 cm
4. 2.5 mm

Answer: 2. 12.5 cm

Angular limit ofresolution of eye, \(\theta=\frac{\lambda}{d} \text {, }\)

where d is the diameter of the eye lens.

Also, if Y is the minimum separation between two objects at distance D from the eye then \(\theta=\frac{Y}{D}\)

⇒ \(\frac{Y}{D}=\frac{\lambda}{d} \Rightarrow Y=\frac{\lambda D}{d}\)…..(1)

Here, wavelength X = 5000Å = 5 x 10-7 m

D = 50m

Diameter of eye lens = 2 mm = 2 x 10^-3 m

From eq. (1), minimum separation is Y = \(\frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}}=12.5 \times 10^{-3} \mathrm{~m}=12.5 \mathrm{~cm}\)

Question 30. Colours appear on a thin soap film and on soap bubbles due to the phenomenon of:

1. Refraction
2. Dispersion
3. Interference
4. Diffraction

Answer: 3. Interference

We know that the colours for which the condition of constructive interference is satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.

Question 31. Colours of thin soap bubbles are due to:

1. Refraction
2. Dispersion
3. Interference
4. Diffraction

Answer: 3. Interference

When white light strikes a soap bubble, it is reflected partially from the upper surface and partially from the lower surface. Interference is caused by the superposition of these two reflected beams. Colours that satisfy the condition of maxima are visible in reflected light, hence the colour of the soap. bubbles are caused by interference.

Question 32. In a Fresnel biprism experiment, the two positions of the lens give separation between the slits as 16 cm and 9 cm respectively. What is the actual distance of separation?

1. 12.5 cm
2. 12 cm
3. 13 cm
4. 14 cm

Answer: 2. 12 cm

Separation between slits are \(\left(r_1\right)=16 \mathrm{~cm} \text { and }\left(r_2\right)=9 \mathrm{~cm} \text {. }\)

Actual distance of separation = \(\sqrt{r_1 r_2}\)

= \(\sqrt{16 \times 9}=12 \mathrm{~cm}\)

Question 33. Interference was observed in the interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see:

1. No interference
2. Interference with brighter bands
3. Interference with dark bands
4. Interference fringe with larger width

Answer: 4. Interference fringe with a larger width

In vacuum, X increases very slightly compared to that in air. As p oc X, therefore, the width of the interference fringe increases slightly.

Question 34. In Young’s double slit experiment carried out with light of wavelength (λ) = 5000Å, The distance between the slits is 0.2 mm and the screen is 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to:

1. 1.67 cm
2. 1.5 cm
3. 0.5 cm
4. 5.0 cm

Answer: 2. 1.5 cm

x = \((n) \lambda \frac{D}{d}\)

x = \(3 \times 5000 \times 10^{-10} \times \frac{2}{0.2 \times 10^{-3}}\)

x = \(1.5 \times 10^{-2} \mathrm{~m}\)

x = \(1.5 \mathrm{~cm}\)

Question 35. If the yellow light emitted by the sodium lamp in Young’s double-slit experiment is replaced by a monochromatic blue light of the same intensity.

1. Fringe width will decrease
2. Fringe width will increase
3. Fringe width will remain unchanged
4. Fringes will become less intense

Answer: 1. Fringe width will decrease

As \(\beta=\frac{\lambda D}{d}\) and \(\lambda_b<\lambda_y\),

Fringe width \(\beta\) will decrease

Question 36. In Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringe is observed one metre away. If it produces a second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be:

1. \(60 \times 10^{-4} \mathrm{~cm}\)
2. \(10 \times 10^{-4} \mathrm{~cm}\)
3. \(10 \times 10^{-5} \mathrm{~cm}\)
4. \(6 \times 10^{-5} \mathrm{~cm}\)

Answer: 4. \(6 \times 10^{-5} \mathrm{~cm}\)

For dark fringe,

x = \((2 n-1) \frac{\lambda D}{2 d}\)

⇒ \(\lambda=\frac{2 x d}{(2 n-1) D}\)

⇒ \(\lambda =\frac{2 \times 10^{-3} \times 0.9 \times 10^{-3}}{(2 \times 2-1) \times 1}\)

⇒ \(\lambda =0.6 \times 10^{-6} \mathrm{~m}\)

∴ \(\lambda=6 \times 10^{-5} \mathrm{~cm}\)

Question 37. The ratio of intensities of the two waves is given by 4: 1, and Then the ratio of the amplitudes of the two waves is

1. 2: 1
2. 1:2
3. 4: 1
4. 1:4

Answer: 1. 2: 1

⇒ \(\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{4}{1}\) (therefore \(\frac{a_1}{a_2}=\frac{2}{1}\))

If \(w_1\) and \(w_2\) are widths of two slits from which intensities of light \(I_1\) and \(I_2\) emanate, then \(\frac{I_1}{I_2}=\frac{a^2}{b^2}\)

= \(\frac{w_1}{w_2}\) ; a and b are amplitudes.

Question 38. Young’s double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then:

1. x (blue) = x (green)
2. x (blue) > x (green)
3. x (blue) < x (green)
4. \(\frac{x(\text { blue })}{x(\text { green })}=\frac{5460}{4360}\)

Answer: 3. x (blue) < x (green)

Distance of \(n^{\text {th }}\) maxima, \(x=(n) \lambda \frac{D}{d} \propto \lambda\)

As, \(\lambda_b<\lambda_g\)

As, \(x_{\text {blue }}<x_{\text {green }}\)

Question 39. In Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, without disturbing the geometrical arrangement, the new fringe width will be

1. 0.30 mm
2. 0.40 mm
3. 0.53 mm
4. 450 microns

Answer: 1. 0.30 mm

∴ \(\beta^{\prime}=\frac{\beta}{\mu}=\frac{0.4}{\frac{4}{3}}=0.3 \mathrm{~mm}\)

Question 40. Interference is possible in:

1. Light waves only
2. Sound waves only
3. Both light and sound waves
4. Neither light nor sound waves

Answer: 3. Both light and sound waves

Interference is a wave phenomenon shown by both light waves and sound waves.

Question 41. Which one of the following phenomena is not explained by Huygens construction of wavefront

1. Refraction
2. Reflection
3. Diffraction
4. Origin of spectra

Answer: 4. Origin of spectra

Huygens’s construction of wavefront does not apply to the origin of spectra which is explained by quantum theory.

Question 42. Which of the following phenomena is not common to sound and light waves?

1. Interference
2. Diffraction
3. Coherence
4. Polarisation

Answer: 4. Polarisation

Sound waves can not be polarised as they are longitudinal. Light waves can be polarised as they are transverse.

Question 43. Unpolarised light is incident from air on a plane surface of a material of refractive index ‘μ’. At a particular angle of incidence, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1. \(i=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
2. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
3. Reflected light is polarised with its electric vector parallel to the plane of incidence
4. \(i=\tan ^{-1}\left(\frac{1}{\mu}\right)\)

Answer: 2. Reflected light is polarised with its electric vector perpendicular to the plane of incidence

When reflected light rays and refracted rays are perpendicular, reflected light is polarised with an electric field vector perpendicular to the plane of incidence.

Also Tan i = μ (Brewster’s Law)

Question 44. Two polaroids P₁ and P₂ are placed with their axes perpendicular to each other. Unpolarised light IQ is incident on P₁. A third polaroid P₂ is kept in between P₁ and P₂ such that its axis makes an angle of 45° with that of P₁. The intensity of transmitted light through P₂ is:

1. \(\frac{\mathrm{I}_0}{2}\)
2. \(\frac{I_0}{4}\)
3. \(\frac{\mathrm{I}_0}{8}\)
4. \(\frac{I_0}{16}\)

Answer: 3. \(\frac{\mathrm{I}_0}{8}\)

According to the given question,

From the explained diagram, the intensity transmitted through P3

⇒ \(I_2=\frac{I_0}{2} \cos ^2 45^{\circ}\)

= \(\frac{I_0}{2} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{4}\)

Again intensity is transmitted through \(P_2\)

∴ \(I_3=\frac{I_0}{4} \cos ^2 45^{\circ}=\frac{I_0}{4} \times\left(\frac{1}{\sqrt{2}}\right)=\frac{I_0}{8}\)

Ray Optics and Optical Instruments

Question 1. An object is placed on the principal axis of a concave mirror at a distance of 1.5 f (f is the focal length). The image will be at:

1. -3f
2. 15f
3. -15f
4. 3f

Answer: 1. -3f

From question, object distance,

u=-1.5 f

As we know,

The mirror formula is: \(\frac{1}{v}+\frac{1}{u} =\frac{1}{f} \)

⇒ \(\frac{1}{v}+\frac{1}{-1.5 f} =\frac{1}{-f} \)

⇒ \(\frac{1}{v} =-\frac{1}{f}+\frac{1}{1.5 f} \)

= \(\frac{1}{f}\left[-1+\frac{1}{1.5}\right] \)

= \(\frac{1}{f}\left[-1+\frac{2}{3}\right]\)

⇒ \(\frac{1}{v} =\frac{1}{f}\left(-\frac{1}{3}\right)=-\frac{1}{3 f}\)

v =-3 f.

Question 2. An object is placed at a distance of 40 cm from a concave mirror of a focal length of 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be :

1. 30 cm towards the mirror
2. 36 cm away from the mirror
3. 30 cm away from the mirror
4. 36 cm towards the mirror

Answer: 2. 36 cm away from the mirror

In this problem, we consider two cases,

In Case (1)

When object is placed, \(u_1=-40 \mathrm{~cm}\) and focal length of mirror, f=-15 cm Using the mirror formula,

⇒ \(\frac{1}{f} =\frac{1}{v_1}+\frac{1}{u_1}\)

⇒ –\(\frac{1}{15} =\frac{1}{v_1}+\frac{1}{40}\)

⇒ \(\frac{1}{v_1}=\frac{1}{-15}+\frac{1}{40}=-\frac{5}{120}\)

∴ \(V_1=-24 \mathrm{~cm}\)

Case (2)

When an object is displaced by 20 cm towards the mirror

Then, \(u_2\)=-20

⇒ –\(\frac{1}{f} =\frac{1}{v_2}+\frac{1}{u_2}\)

⇒ –\(\frac{1}{15} =\frac{1}{v_2}-\frac{1}{20}\)

⇒ \(\frac{1}{v_2}=\frac{1}{20}-\frac{1}{15}=-\frac{1}{60}\)

⇒ \(v_2\)=-60 cm

So image shifts away from the mirror by =60-24=36 cm

Read and Learn More NEET Physics MCQs

Question 3. A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle of θ, the spot of the light is found to move through a distance y on the scale. The angle 0 is given by:

1. \(\frac{y}{2 x}\)
2. \(\frac{y}{x}\)
3. \(\frac{x}{2 y}\)
4. \(\frac{x}{y}\)

Answer: 1. \(\frac{y}{2 x}\)

⇒ \(\frac{Y}{x}=2 \theta\)

∴ \(\theta=\frac{Y}{2 x}\)

Question 4. Match the corresponding entries of Column 1 with Column 2. [Where m is the magnification produced by the mirror]

1. A → 1 and 3; B → 1 and 4; C → 2 and 3; D →3 and 4
2. A → 1 and 4; B → 2 and 3; C → 3 and 5; D → 2 and 3
3. A → 3 and 4; B → 2 and 3; C → 2 and 3; D →1 and 4
4. A → 2 and 3; B → 2 and 3; C → 3 and 4; D →3 and 4

Answer: 4. A → 2 and 3; B → 2 and 3; C → 3 and 4; D →3 and 4

⇒ \( m =+v_e\) (virtual image)

m =\(-v_e\) (real image)

∴ \(|m| <1\) (diminished image)

Question 5. Two plane mirrors are inclined at 70°. A ray incident on one mirror at angle θ, after reflection falls on the second mirror and is reflected from there parallel to the first mirror. The value of θ is:

1. 45°
2. 30°
3. 55°
4. 50°

Answer: 4. 50°

The actual situation is shown in the given degree,

⇒ \(\theta+40^{\circ} =90^{\circ}\)

∴ \(\theta =90-40=50^{\circ}\)

Question 6. A concave mirror of focal length/ is placed at a distance of d from a convex lens of focal length l. A beam of light coming from infinity and falling on this convex lens concave mirror combination returns to infinity. The distance d must be equal:

1. \(f_1+f_2\)
2. –\(f_1+f_2\)
3. \(2 f_1+f_2\)
4. –\(2 f_1+f_2\)

Answer: 3. \(2 f_1+f_2\)

\(d=2 f_1+f_2\)

Question 7. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm so that its end is closer to the pole 20 cm away from the mirror. The length of the image is:

1. 10 cm
2. 15 cm
3. 2.5 cm
4. 5 cm

Answer: 4. 5 cm

Image Position Of End A,

⇒ \(\frac{1}{v_A}+\frac{1}{-20} =\frac{1}{-10} \)

⇒ \(v_A \)=-20 cm

Image position of end B

⇒ \(\frac{1}{v_B}+\frac{1}{-30} =\frac{1}{-10}\)

⇒ \(v_B\) =-15 cm

The length of the image is

∴ \(\mathrm{A}^{\prime} \mathrm{B}^{\prime}=\left|v_A\right|-\left|v_B\right|=20-15=5 \mathrm{~cm}\)

Question 8. If two mirrors are kept inclined at 60° to each other and a body is placed in the middle, then the total number of images formed is:

1. six
2. five
3. Four
4. Three

Answer: 2. five

Many images are created when an object is held between two plane mirrors M₁ and M₂ inclined at (<θ) due to multiple reflections of light from the mirrors. The total number of images formed.

n=\(\frac{360^{\circ}}{\theta}-1\)

Here, \(\theta=60^{\circ}\)

n=\(\frac{360^{\circ}}{60^{\circ}}-1=6-1=5\)

Question 9. A light ray falls on a glass surface with a refractive index of √3″, at an angle of 60°. The angle between the refracted and reflected rays would be:

1. 30°
2. 60°
3. 90°
4. 120°

Answer: 1. 30°

By Snell’s law, 1 \(\sin 60^{\circ} =\sqrt{3} \sin r\)

⇒ \(\frac{\sqrt{3}}{2} =\sqrt{3} \sin r\)

⇒ \(\sin r =\frac{1}{2}\)

r =\(30^{\circ}\)

Question 10. An air bubble in a glass slab with a refractive index of 1.5(near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep from the opposite face. The thickness (in cm) of the slab is:

1. 8
2. 10
3. 12
4. 16

Answer: 3. 12

Let, t= thickness of the glass slab.

From question, \(\frac{x}{\mu}+\frac{t-x}{\mu}\) =3+50

⇒ \(\frac{t}{\mu} =8 \mathrm{~cm}\)

t =\(\mu \times 8=8 \times \frac{3}{2}\)=12 cm

Question 11. A glass tube having a curved surface at one face is shown below with a refractive index of 1.5. Its center of the radius of curvature R lies inside the glass tube. If a particle is placed at point P. It forms the real image at point Q. The point O cuts PQ such that OP = 20Q then the value of OP is:

1. 2R
2. 4R
3. 8R
4. R

Answer: 3. 8R

When an object is in the rarer medium then

⇒ \(\frac{\mu_1}{v}=\frac{\mu_2}{u} =\frac{\mu_1-\mu_2}{R}\)

Now v =\(\mathrm{OQ} \)

u =\(\mathrm{OP}=-20 \mathrm{Q} \)

⇒ \(\frac{1.5}{\mathrm{OQ}}+\frac{1}{2 \mathrm{OQ}} =\frac{1.5-1}{R} \)

⇒ \(\frac{3+1}{2 \mathrm{OQ}} =\frac{0.5}{R}\)

⇒ \(\mathrm{OQ} \)=4 R

∴ \(\mathrm{OP} =20 \mathrm{Q}=2 \times 4 R=8 R\)

Question 12. The frequency of a light wave in a material is 2 × 10¹⁴ Hz and the wavelength is 5000 Å. The refractive index of the material will be:

1. 1.50
2. 3.00
3. 1.33
4. 1.40

Answer: 2. 3.00

Refractive index

⇒ \(\mu =\frac{c}{v \lambda} \)

∴ \(\mu =\frac{3 \times 10^8}{2 \times 10^{10} \times 5000 \times 10^{-10}}=3\)

Question 13. A small coin is resting on the bottom of a beaker filled with liquid. A ray of light from the coin travels up to the surface of the liquid and moves along its surface. How fast is the light traveling in the liquid?

1. 2.4 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
2. 3.0 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
3. 1.2 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
4. 1.8 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

Answer: 4. 1.8 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

⇒ \(\frac{1}{\mu}=\sin \theta_c=\frac{3}{5}\)

⇒ \(\mu=\frac{5}{3}\)

And v =\(\frac{c}{\mu}=\frac{3 \times 10^8}{5 / 3} [3 \times 10^8 \mathrm{~m} / \mathrm{s}]\)

= \(\frac{9}{5} \times 10^8=1.8 \times 10^8 \mathrm{~ms}^{-1}\)

Question 14. A beam of light composed of red and green rays is incident obliquely at a point on the face of a rectangular glass slab. When coming out on the opposite parallel face, the red and green-ray emerge from:

1. two points prepared in two different nonparallel directions
2. two points propagating in two different parallel directions
3. one point propagating in two different directions
4. one point propagating in the same directions

Answer: 2. two points propagating in two different parallel directions

Red and green rays emerge from two points propagating in two different parallel directions.

Question 15. Electromagnetic radiation of frequency v, velocity v, and wavelength \(\lambda\), in air, enters a glass slab of refractive index p. The frequency, wavelength, and velocity of light in the glass slab will be respectively:

1. \(\frac{v}{\mu}, \frac{\lambda}{\mu}, v\)
2. v,\( \lambda, \frac{v}{\mu}\)
3. v, \(\frac{\lambda}{\mu}, \frac{v}{\mu}\)
4. \(\frac{v}{\mu}, \frac{\lambda}{\mu}, \frac{v}{\mu}\)

Answer: 3. v, \(\frac{\lambda}{\mu}, \frac{v}{\mu}\)

When an electromagnetic wave travels through another medium, the frequency is unchanged, but the wavelength and velocity are \(\frac{1}{\mu}\) times.

Thus, after passing from air to a glass slab with a refractive index of \(\lambda\), frequency remains v, and wavelength remains \(\lambda^{\prime}=\frac{\lambda}{\mu}\) and velocity of light \(\mathrm{v}^{\prime}=\frac{\mathrm{v}}{\mu}\)

Question 16. Time taken by sunlight to pass through a window of 3 thickness 4 mm Whose refractive index is \(\frac{3}{2}\), is:

1. \(2 \times 10^{-4} \mathrm{~s}\)
2. 2 \(\times 10^8 \mathrm{~s}\)
3. 2 \(\times 10^{-11} \mathrm{~s}\)
4. 2 \(\times 10^{11} \mathrm{~s}\)

Answer: 3. 2 \(\times 10^{-11} \mathrm{~s}\)

Let x be the thickness of the window, v be the velocity of light entering the window, so the time taken by sunlight to pass through the window

t= \(\frac{x}{v}=\frac{x}{\frac{c}{\mu}}=\frac{\mu x}{c} \left[v=\frac{c}{\mu}\right]\)

t= \(\frac{1.5 \times 4 \times 10^{-3}}{3 \times 10^8}=2 \times 10^{-11} \mathrm{~s}\)

Question 17. A beam of monochromatic light is refracted from a vacuum into a medium with a refractive index of 1.5. The wavelength of refracted light will be:

1. dependent on the intensity of refracted light
2. same
3. smaller
4. larger

Answer: 3. smaller

We have, a refractive index,

⇒ \(\mu=\frac{\mathrm{c}}{\mathrm{v}}=\frac{\mathrm{v} \lambda_{\mathrm{v}}}{\mathrm{v} \lambda_{\mathrm{m}}}\)

As,c= velocity of light in vacuum

v = velocity of light in medium

⇒ \(\mathrm{v}=v \lambda\) and

v remains constant during refraction.

⇒ \(\lambda_{\mathrm{m}}=\frac{\lambda_{\mathrm{v}}}{\mu}\)

⇒ \(\lambda_{\mathrm{m}}<\lambda_{\mathrm{v}} ( \mu=1, \text { given })\)

Hence, in the second medium, the wavelength falls.

Question 18. Two transparent media A and B are separated by a plane boundary. The speed of light in those media are \(1.5 \times 10^8 \mathrm{~m} / \mathrm{s}\) and \(2.0 \times 10^8 \mathrm{~m} / \mathrm{s}\) respectively, The critical angle for a ray of light for these two media is:

1. \(\sin ^{-1}(0.500)\)
2. \(\sin ^{-1}(0.750)\)
3. \(\tan ^{-1}(0.500)\)
4. \(\tan ^{-1}(0.750)\)

Answer: 2. \(\sin ^{-1}(0.750)\)

For medium A, Speed of light \(v_{\mathrm{A}}=1.5 \times 10^8 \mathrm{~m} / \mathrm{s} \)

For medium B, Speed of light \(v_B=2.0 \times 10^8 \mathrm{~m} / \mathrm{s} \)

For medium \(\mathrm{A}, n_{\mathrm{A}}=\frac{C}{v_{\mathrm{A}}}=\frac{3 \times 10^8}{1.5 \times 10^8}\)=2

For medium \(\mathrm{B}, n_{\mathrm{B}}=\frac{C}{v_{\mathrm{B}}}=\frac{3 \times 10^8}{2 \times 10^8}\)=1.5

Critical angle, \(\sin c=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}}\)

c =\(\sin ^{-1}\left(\frac{1.5}{2}\right) \)

= \(\sin ^{-1}(0.750)\)

Question 19. If the critical angle for total internal reflection from a medium to vacuum is 45°, then the velocity of light in the medium is:

1. 15 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
2. \(\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\)
3. \(\sqrt{2} \times 10^8 \mathrm{~m} / \mathrm{s}\)
4. 3 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

Answer: 2. \(\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\)

It is given that critical angle \(i_c=45^{\circ}\).

We know \(\mu=\frac{1}{\sin i_{\mathrm{C}}}=\frac{1}{\sin 45^{\circ}}\)

= \(\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2} \)

⇒ \(\frac{\text { Velocity of light in air }}{\text { Velocity of light in medium }}=\sqrt{2}\)

⇒ \(\frac{3 \times 10^8}{v_m}=\sqrt{2}\)

∴ \(v_m=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\) .

Question 20. In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be the angle of refraction?

1. 0°
2. Equal to the angle of incidence
3. 90°
4. 180°

Answer: 3. 90°

At i=\(i_c\), refracted ray grazes with the surface.

So, the angle of refraction is \(90^{\circ}\).

Question 21. Which of the following is not due to total internal reflection?

1. Difference between apparent and real depth of a pond.
2. Mirage on hot summer days.
3. Brilliance of diamond.
4. Working of optical fiber.

Answer: 1. Difference between the apparent and real depth of a pond

The difference between the apparent and real depth of a pond is not due to internal reflection.

Question 22. A ray of light traveling in a transparent medium of refractive index μ falls, on the surface separating the medium from air at an angle of incidence of 45°. For which of the following values of p the ray can undergo total internal reflection?

1. μ = 1.33
2. μ = 1.40
3. μ = 1.50
4. μ = 1.25

Answer: 3. μ = 1.50

We know that, for total internal reflection

⇒ \(\sin i>\sin c\)

⇒ \(\sin 45^{\circ}>\frac{1}{\mu}\)

⇒ \(\mu>\sqrt{2} \)

∴ \(\mu>1.4\)

Question 23. The speed of light in medium M₁ and M₂ is 1.5 x 10⁸ m/s and 2.0 x 10⁸ m/s respectively. A ray of light enters from medium M₁ to M₂ at an incidence angle I, if the ray suffers total internal reflection the value of i is:

1. equal to sin\(1^{-1}\left(\frac{2}{3}\right)\)
2. equal to or less than sin\(\mathrm{n}^{-1}\left(\frac{3}{5}\right)\)
3. equal to or greater than\(\sin ^{-1}\left(\frac{3}{4}\right)\)
4. less than sin\(\mathrm{a}^{-1}\left(\frac{2}{3}\right)\)

Answer: 3. equal to or greater than\(\sin ^{-1}\left(\frac{3}{4}\right)\)

We know that in total internal reflection the angle of incidence i is greater than critical angle C.

i >C

2 \(\mu_1 =2 and \mu_2=\frac{3}{2}\)

⇒ \(2 \sin i \geq \frac{3}{2} \sin 90^{\circ}\)

⇒ \(\sin i \geq \frac{3}{4}\)

i \(\geq \sin ^{-1}\left(\frac{3}{4}\right)\)

Question 24. A boy is trying to start a fire by focusing sunlight on a piece of paper using a biconvex lens with a focal length of 10 cm. The diameter of the sun is \(1.39 \times 10^9\)m and its mean distance from the earth is \(1.5 \times 10^{11}\) m. What is the diameter of the sun’s image on the paper?

1. \(9.2 \times 10^{-4} \mathrm{~m}\)
2. 6.5 \(\times 10^{-4} \mathrm{~m}\)
3. 6.5 \(\times 10^{-5} \mathrm{~m}\)
4. 12.4 \(\times 10^{-4} \mathrm{~m}\)

Answer: 1. \(9.2 \times 10^{-4} \mathrm{~m}\)

From the relation, \(\frac{I}{O}=\frac{v}{u}\)

Here, \(\mathrm{O}=1.39 \times 10^9\)

and v=0.1 \(\mathrm{~m}\)

u=1.5 \(\times 10^{11} \mathrm{~m}\)

I =\(\frac{0.1}{1.5 \times 10^{11}} \times 1.39 \times 10^9\)

= 9.2 \(\times 10^{-4} \mathrm{~m}\)

Question 25. A convex lens and a concave lens, each having the same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is:

1. zero
2. 25
3. 50
4. infinite

Answer: 1. zero

The focal length of the lens

⇒ \(f_1\)=25 cm

focal length of concave lens

⇒ \(f_2\)=25 cm

Power of combination in dioptres

P=\(P_1+P_2\)

P=\(\frac{100}{f_1}+\frac{100}{f_2}\)

P=\(\frac{100}{25}+\frac{100}{-25}\)=0

Question 26. A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is:

1. + 2D
2. + 20 D
3. + 5 D
4. infinity

Answer: 3. + 5 D

Given: u=1.5, \(R_1=+R, R_2=-R_I=-R\)

⇒ \(R_1=+20\) cm and \(R_2=-20\) cm

Lens maker’s formula, \(\frac{1}{f}=(u-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \(\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{(-20)}\right)\)

⇒ \(\frac{1}{f}=0.5\left(\frac{1}{20}+\frac{1}{20}\right) \)

⇒ \(\frac{1}{f}=\frac{1}{20} \)

f = 20 cm

Now Power, P=\(\frac{100}{20}=5 \mathrm{D}\)

Question 27. A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of:

1. 20 cm from the lens, it would be a real image
2. 30 cm from the lens, it would be a real image
3. 30 cm from the plane mirror, it would be a virtual image
4. 20 cm from the plane mirror, it would be a virtual image

Answer: 4. 20 cm from the plane mirror, it would be a virtual image

Using the lens formula for the first refraction from the convex lens,

⇒ \(\frac{1}{v_1}-\frac{1}{u} =\frac{1}{f} \)

⇒ \(v_1\)=?, u =-60 cm, f=30 cm

⇒ \(\frac{1}{v_1}+\frac{1}{60} =\frac{1}{30} \)

⇒ \(v_1\) =60 cm

Here \(I_1\) is the first image by the lens,

The plane mirror will produce an image at a distance of 20 cm to the left of it.

For the second refraction from a convex lens,

u=-20 cm, v=?, f=30 cm

From lens formula, \(\frac{1}{v}-\frac{1}{(-20)} =\frac{1}{30} \)

v =-60 cm

Thus, the final image is virtual and at a distance, of 60 cm – 40 cm=20 cm from the plane mirror.

Question 28. A convex lens ‘A’ with a focal length of 20 cm and a concave lens ‘B’ with a focal length of 5 cm is kept along the same axis with a distance between them. If a parallel beam of light falling on ‘A’ leaves ‘B’ as a parallel beam, then the distance ‘cT in cm will be:

1. 25
2. 15
3. 50
4. 30

Answer: 2. 15

Given, Focal length of A=\(f_{\mathrm{A}}=+20 \mathrm{~cm}\) [convex]

Focal length of B=\(f_{\mathrm{B}}=-5 \mathrm{~cm}\) [concave]

distance between A and B=d

For a parallel beam of light entering A to remain parallel while exiting B.

The net focal length of the combination should be infinite.

⇒ \(\frac{1}{f}=\frac{1}{f_{\mathrm{A}}}+\frac{1}{f_{\mathrm{B}}}-\frac{d}{f_{\mathrm{A}} f_{\mathrm{B}}}\) =0

⇒ \(\frac{1}{20}-\frac{1}{5}+\frac{d}{100}\) =0

⇒ –\(\frac{15}{100}+\frac{d}{100} \)=0

d =15 cm

Question 29. A piano convex lens of unknown material and unknown focal length is given. With the help of a spherometer, we can measure the:

1. focal length of the lens
2. radius of curvature of the curved surface
3. the aperture of the lens
4. the refractive index of the material

Answer: 2. radius of curvature of the curved surface

The first known spherometer was invented by French Optician Robert Aglae Cauchoix in 1810. They were manufactured starting in the nineteenth century primarily for the use of opticians in grinding lenses and curved mirrors.

Question 30. The power of a biconvex lens is 10 D and the radius of curvature of each surface is 10 cm. Then, the refractive index of the material of the lens is:

1. \(\frac{4}{3}\)
2. \(\frac{9}{8}\)
3. \(\frac{5}{3}\)
4. \(\frac{3}{2}\)

Answer: 4. \(\frac{3}{2}\)

From the question power of the biconvex lens,

P =\(10 \mathrm{D}\)

f =\(\frac{1}{P}=\frac{1}{10}=0.1 \mathrm{~m}=10 \mathrm{~cm} \)

⇒ \(R_1 =R_2=10 \mathrm{~cm}\)

Now from the lens maker’s formula

⇒ \(\frac{1}{f} =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)

⇒ \(\frac{1}{10} =(\mu-1)\left(\frac{1}{10}-\frac{1}{-10}\right) \)

⇒ \(\frac{1}{10} =(\mu-1)\left(\frac{2}{10}\right)\)

1 =\((\mu-1) \times 2 \)

⇒ \(\mu-1 =\frac{1}{2} \)

∴ \(\mu =1+\frac{1}{2}=\frac{3}{2}\)

Question 31. Two similar thin equi-convex lenses, of focal length / each, are kept co-axially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerine which has the same refractive index (μ = 15) as that of glass then the equivalent focal length is F₂. The ratio F₁ : F₂ will be:

1. 1:2
2. 2:3
3. 3:4
4. 2:1

Answer: 1. 1:2

First, find the focal length of the lens (1),

⇒ \(\frac{1}{f_1}=\left(\frac{\mu_2}{\mu_3}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\mathrm{L}\)= Lens

S = Surrounding

⇒ \(R_1\)= Radius of first surface

⇒ \(R_2\)= Radius of 2nd surface

⇒ \(\frac{1}{f_1} =\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R}-\frac{1}{-R}\right)\)

=\(\frac{1}{2} \times \frac{2}{R}=\frac{1}{R}\)

Now focal length of (2),

⇒ \(\frac{1}{f_2}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{1}{R}\)

The combination of these two

⇒ \([latex]\frac{1}{F_1}=\frac{1}{f_2}+\frac{1}{f_2}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R} \)

⇒ \(F_1=\frac{R}{2}\)

Now focal length of concave (3)

⇒ \(\frac{1}{f_3} =\left(\frac{\mu_2}{\mu_3}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \(\frac{1}{f_3} =\left(\frac{1.5}{1}-1\right)\left(-\frac{1}{R}-\frac{1}{R}\right) \)

= \(\frac{1}{2} \times \frac{-2}{R} \)

=-\(\frac{1}{R}\)

Focal length, \(\frac{1}{F_2} =\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3} \)

⇒ \(\frac{1}{F_2} =\frac{1}{R}+\frac{1}{R}-\frac{1}{R}=\frac{1}{R} \)

⇒ \(F_2 \)=R

According to the question,

∴ \(\frac{F_1}{F_2}=\frac{R / 2}{R}=\frac{R}{2 \times R}=\frac{1}{2}\)

Question 32. An equi-convex lens has power P. It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be:

1. 0
2. \(\frac{P}{2}\)
3. \(\frac{P}{4}\)
4. P

Answer: 3. \(\frac{P}{4}\)

In both cases (1) and (2) focal length does not change. Hence power does not change. So, the power of one part is also P.

Question 33. A double convex lens has a focal length of 25 cm. The radius of curvature of one of the surfaces is doubled that of the other. Find the radii, if the refractive index of the material of the lens is 1.5:

1. 100 cm. 50 cm
2. 25 cm. 50 cm
3. 18.75 cm. 37.5 cm
4. 50 cm, 100 cm

Answer: 3. 18.75 cm. 37.5 cm

Focal length of lens is, \(\frac{1}{f} =(\mu-1)\)

⇒ \(\frac{1}{25} =(1.5-1)\left(\frac{1}{R}+\frac{1}{2 R}\right)\)

⇒ \(\frac{1}{25} =0.5\left(\frac{3}{2 R}\right)\)

2 R =37.5 \(\mathrm{~cm} \)

R =18.75 \(\mathrm{~cm}\)

Hence radii are 18.75 \(\mathrm{~cm}, 37.5 \mathrm{~cm}\).

Question 34. Two identical glass (\(\mu_g\) = 3/2) equi-convex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (\(\mu_w\) = 4/3). The focal length of the combination is:

1. f/3
2. f
3. \(\frac{4 f}{3}\)
4. \(\frac{3 f}{4}\)

Answer: 4. \(\frac{3 f}{4}\)

The diagram shows the actual situation

Let R = Radius of curvature of the lens

and \(\frac{1}{f_{e q}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}\left\{ f_1=f_3\right\}\)

⇒ \(f_1=f_3=\frac{R}{2\left(\frac{3}{2}-1\right)}=R=f \text { (given) } \)

⇒ \(f_2=\frac{-R}{2\left(\frac{4}{3}-1\right)}=-\frac{3}{2} R=\frac{3}{2} f \)

⇒ \(\frac{1}{f_{\text {eq }}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3} \)

⇒ \(\frac{1}{f_{\text {eq }}}=\frac{1}{f}+\left(-\frac{2}{3 f}\right)+\frac{1}{f}\)

⇒ \(\frac{1}{f_{e q}}=\frac{4}{3 f} \)

∴ \(f_{e q}=\frac{3 f}{4} \)

Question 35. A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be:

1. convex, + 2.25 diopter
2. concave, – 0.25 diopter
3. concave – 0.2 diopoter
4. convex, + 0.15 diopter

Answer: 2. concave, – 0.25 diopter

As we want to correct myopia. So, the far point must go to infinity.

V =-4 \(\mathrm{~m}, u=\infty\), P=?

P =\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

=\(\frac{1}{-4}-\frac{1}{\infty}=-0.25 \mathrm{D}\)

∴ – ve sign indicates that the mirror is concave.

Question 36. Two identical thin plano-convex glass lenses (refractive index 1.5) each having a radius of curvature of 20 cm are placed with their convex surface in contact at the center. The intervening space is filled with oil with a refractive index of 1.7. The focal length of the combination is:

1. – 20 cm
2. – 25 cm
3. – 50 cm
4. 50 cm

Answer: 3. – 50 cm

The radius of curvature of plano-convex lensR:20 cm Refractive index of plano-convex lens μ: 1.5 Using mirror formula

⇒ \(\frac{1}{f}=(\mu-1)\left[\frac{1}{R}\right] \)

⇒ \(\frac{1}{f}=(1.5-1) \cdot \frac{1}{R}\)

⇒ \(\frac{1}{f}=\frac{0.5}{R}\)

focal length, \(\frac{1}{f_{\text {concave }}} =\left(\mu_1-1\right)\left(-\frac{1}{2}-\frac{1}{2}\right)\)

= \((1.7-1)\left[-\frac{1}{R}-\frac{1}{R}\right]\)

⇒ \(\frac{1}{f_{\text {concave }}} =\frac{-0.7 \times 2}{R}=-\frac{1.4}{K}\)

Equivalent focal length will be

⇒ \(\frac{1}{f_{\text {eq }}}=\frac{0.5}{R}-\frac{1.4}{R}+\frac{0.5}{R}=-\frac{0.4}{R}\)

∴ \(f_{e q}=-50 \mathrm{~cm}\)

Question 37. If the focal length of the objective lens is increased, then the magnifying power of:

1. the microscope will increase but that of the telescope decrease
2. microscope and telescope both will increase
3. microscope and telescope both will decrease
4. microscope will decrease but that of the telescope will increase.

Answer: 4. The microscope will decrease but that of the telescope will increase.

Focal length m \(\propto \frac{1}{f_0}\) microscope

⇒ \(\frac{f_0}{f_e}=m \)

m\( \propto f_0 \)telescope

The magnifying power of the microscope decreases and the magnifying power of the telescope increases.

Question 38. A plano-convex lens fits exactly into a plano-concave lens. Their plane surface is parallel to each other. If lenses are made of different materials of refractive indices \(\mu_1\) and \(\mu_2\) and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

1. \(\frac{R}{2\left(\mu_1+\mu_2\right)}\)
2. \(\frac{R}{2\left(\mu_1-\mu_2\right)}\)
3. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)
4. \(\frac{2 R}{\left(\mu_2-\mu_1\right)}\)

Answer: 3. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)

Lens maker formula: \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)  → Equation 1

For Plano-convex lens: \(R_1=\infty, R_2=-R\)

Using (1), \(\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{\infty}-\frac{1}{-R}\right)\)

⇒ \(\frac{1}{f_1}=\frac{\mu_1-1}{R} \)

For Plano-concave lens: \(R_1=-R, R_2=\infty\) Using (2),

⇒ \(\frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)\)

\(\frac{1}{f_2}=\frac{1-\mu_2}{R}\)

Thus focal length of the combination,

⇒ \(\frac{1}{F} =\frac{1}{f_1}+\frac{1}{f_2}=\frac{\mu_1-1}{R}+\frac{1-\mu_2}{R}\)

F =\(\frac{R}{\mu_1-\mu_2}\)

Question 39. For a normal eye, the corneas of the eye provide a converging power of 40 D, and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be:

1. 5 cm
2. 2.5 cm
3. 1.67 cm
4. 1.5 cm

Answer: 3. 1.67 cm

For a normal eye rays coming from the infinite should go to the retina without effort when we look at infinity, the lens offers minimum power and hence combination gives 40D + 20D = 60D.

Distance between the retina and the cornea of the eye must be equal to focal length \(f=\frac{1}{60} \mathrm{~m}=1.67 \mathrm{~cm}\)

Question 40. When a biconvex lens of glass has refractive. index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have a refractive index:

1. equal to that of glass
2. less than one
3. greater than that of glass
4. less than that of glass

Answer: 1. equal to that of glass

According to the question,

and \(\mu_g \)=1.47

f =\(\infty\)

Using the formula,

⇒ \(\frac{1}{f} =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

= \(\left(\frac{\mu_g}{\mu_2}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

⇒ \( \frac{1}{\infty} =\left(\frac{\mu_g}{\mu_2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)

0 =\(\left(\frac{\mu_g}{\mu_2}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)

Since \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \neq 0\)

⇒ \((\frac{\mu_g}{\mu_2}-1)\)=0

⇒ \(\frac{\mu_g-\mu_2}{\mu_2}\)=0

This shows that the liquid must have a refractive index equal to that of glass.

Question 41. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object height of 2 cm placed 20 cm from the lens?

1. Virtual, upright, height = 0.5 cm
2. Real, inverted, height = 4 cm
3. Real, inverted, height = 1 cm
4. Virtual, upright, height = 1 cm

An⇒swer: 2. Real, inverted, height = 4 cm

Here,\(\mu\)=1.5 (considered)

f=20 cm

and Using the formula,

⇒ \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

⇒ \(\frac{1}{20}=\frac{1}{v}+\frac{1}{30}\)

⇒ \(\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{100}{600}\)

v=60 m

Question 42. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is:

1. – 10 cm
2. 20 cm
3. – 30 cm
4. 5cm

Answer: 3. 20 cm

In the question, u=10 \(\mathrm{~cm}\), v=15 \(\mathrm{~cm}\), f= ?

⇒ \(\frac{1}{f} =\frac{1}{v}-\frac{1}{u}\)

= \(\frac{1}{15}-\frac{1}{10}=\frac{10-15}{150}=-\frac{5}{150} \)

⇒ \(\frac{1}{f} =-\frac{1}{30} \)

f=-30

Question 43. A lens having focal length l and an aperture of diameter of d forms an image of intensity I. The aperture of diameter \(\frac{d}{2}\) in the central region of the lens is covered by black paper. The focal length of the lens and intensity of the image now will be respectively:

1. f and \(\frac{I}{4}\)
2. \(\frac{3 f}{4}\) and \(\frac{I}{2}\)
3. f and \(\frac{3 I}{4}\)
4. \(\frac{f}{2} and \frac{I}{2}\)

Answer: 3. f and \(\frac{3 I}{4}\)

We know that,Intensity \(\mathrm{I} \propto \mathrm{A}^2\) then

⇒ \(\frac{I_2}{I_1} =\left[\frac{A_2}{A_1}\right]^2=\frac{\pi r^2-\frac{\pi r^2}{4}}{\pi r^2}=\frac{3}{4} \)

∴ \(I_2 =\frac{3}{4} I_1\)

Question 44. Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial.mtrThe power of the combination is;

1. \(\sqrt{\frac{f_1}{f_2}}\)
2. \(\sqrt{\frac{f_2}{f_1}}\)
3. \(\frac{f_1+f_2}{f_1 f_2}\)
4. \(\frac{f_1-f_2}{f_1 f_2}\)

Answer: 3. \(\frac{f_1+f_2}{f_1 f_2}\)

We can write,

⇒ \(P=P_1+P_2 \)

P=\(\frac{1}{f_1}+\frac{1}{f_2} \)

P=\(\frac{f_1+f_2}{f_1 f_2}\)

Question 45. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:

1. become zero
2. become infinite
3. become small, but non-zero
4. remain unchanged

Answer: 2. become infinite

When the refractive index of the lens is equal to the refractive index of the liquid, the lens behaves like a plane surface with a focal length of infinity.

Question 46. An equiconvex lens is cut into two halves along (1) XOX’ and (2) YOY as shown in the figure. Let f,f’,f” be the focal length of the complete lens, of each half in case (1) and each half in case (2), respectively. Choose the correct statement from the following:

1. f’: f, f”:2f
2. f’ :2f, f” : f
3. f’: f, f”: f
4. f’ :2f, f” :2f

Answer: 1. f’: f, f”:2f

Since the lens is biconvex, the radius of curvature of each half is the same.

Using the lens maker formula,

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Here,\( R_2\)=-R by convection

⇒ \(\frac{1}{f}=(\mu-1) \frac{2}{\mathrm{R}},(\mu-1) \frac{1}{\mathrm{R}}=\frac{1}{2 f}\)

If we cut the lens along X O\( X^{\prime}\) then two halves of the lens will have the same radii of curvature and

⇒ \(R_1=R, R_2=-R\)

⇒ \(\frac{1}{f^{\prime}}=(\mu-1)\left(\frac{2}{R}\right)\)

⇒ \(\frac{1}{f},=\frac{2}{2 f}=\frac{1}{f}\)

From (1) it is clear that\( f^{\prime}\)=f

But when we cut along YOY’ then

⇒ \(R_1=R\) and \(R_2=\infty \)

⇒ \(\frac{1}{f^n} =(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)\)

=\((\mu-1) \frac{1}{R}=\frac{1}{2 f}\)

∴ \(f^{\prime \prime} \)=2 f

Question 47. A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance between parallel walls is then the required focal length of the lens placed in between the walls is:

1. only \(\frac{d}{4}\)
2. only \(\frac{d}{2}\)
3. more than \(\frac{d}{4} but less than \frac{d}{2}\)
4. less than or equal to \(\frac{d}{4}\)

Answer: 4. less than or equal to \(\frac{d}{4}\)

Since , v+u =d

v =d-u

From lens formula, \(\frac{1}{f} =\frac{1}{v}-\frac{1}{u}\)

⇒ \(\frac{1}{f} =\frac{1}{d-u}-\frac{1}{-u} \)

= \(\frac{d}{(d-u) u} \)

f =\(\frac{d u-u^2}{d}\) →  Equation 1

Differentiating eq. (1) we have,

⇒ \(\frac{d f}{d u}=\frac{1}{d}(d-2 u)\)

for maximum value of \(f_1=\frac{d f}{d u}\)=0

⇒ \(\frac{1}{d}(d-2 u)\) =0

d =2 u

u =\(\frac{d}{2}\)

putting these values in eq. (1),

f=\(\frac{d}{4}\)

Question 48. A planoconvex lens is made of a material with a refractive index μ = 1.5. The radius of curvature of the curved surface of the lens is 20 cm. If its plane surface is silvered, the focal length of the silvered lens will be:

1. 10 cm
2. 20 cm
3. 40 cm
4. 80 cm

Answer: 2. 20 cm

Since, v+u =d

v =d-u

From lens formula

⇒ \(\frac{1}{f} =\frac{1}{v}-\frac{1}{u}\)

⇒ \(\frac{1}{f} =\frac{1}{d-u}-\frac{1}{-u} \)

=\(\frac{d}{(d-u) u} \)

f =\(\frac{d u-u^2}{d}\)  → Equation 1

Differentiating eq. (1) we have,

⇒ \(\frac{d f}{d u}=\frac{1}{d}(d-2 u)\)

for maximum value of \(f_1=\frac{d f}{d u}\)=0

⇒ \(\frac{1}{d}(d-2 u)\) =0

d =2 u

u =\(\frac{d}{2}\)

putting these values in eq. (1),

f=\(\frac{d}{4}\)

When a ray strikes the convex surface of a planoconvex lens, it is first refracted and then reflected from the plane surface, then refracted from the convex surface. As a result, there are two refractions and one reflection. Since re&action and reflection occur twice, and reflection occurs at the same time. So, the focal length of a planoconvex lens \(\frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m}\)

If the plane surface is silvered, then

⇒\(\mathrm{f}_{\mathrm{m}}=\frac{\mathrm{R}_2}{2}=\frac{\infty}{\mathrm{f}_{\mathrm{m}}}\)  →     Equation 1

by lens maker formula,

⇒ \(\frac{1}{f_1}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)

= \((\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{(\mu-1)}{R}\)

⇒ \(\frac{1}{F}=\frac{2(\mu-1)}{R}+\frac{1}{\infty}\)

= \(\frac{R}{2(\mu-1)} \)

or F=\(\frac{R}{2(\mu-1)}\)

Given, R=20 \(\mathrm{~cm}, \mu\)=1.5

Hence, F =\(\frac{20}{2(1.5-1)}=\frac{20}{2 \times 0.5} \)

=20 cm

Question 49. A planoconvex lens is made of material with a refractive index of 1.6. The radius of curvature of the curved surface is 60 cm. The focal length of the lens is:

1. 50 cm
2. 100 cm
3. 200 cm
4. 400 cm

Answer: 2. 100 cm

By lens maker formula,\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)  → Equation 1

We have, the radius of curvature of a plane surface for a planoconvex lens is infinite.

i.e., \(R_2=\infty \)

⇒ \(R_1=60 \mathrm{~cm}, \mu=1.6\)

Given,\(\mathrm{R}_1=60 \mathrm{~cm}, \mu\)=1.6

Substituting the given values in Eq. (1), we get

⇒ \(\frac{1}{f} =(1.6-1)\left(\frac{1}{60}-\frac{1}{\infty}\right)=0.6 \times \frac{1}{60}\)

=100 cm

The focal length of plane convex lens is 100 \(\mathrm{~cm}\)

Question 50. A convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined. What will be their resulting power?

1. + 6.5 D
2. -6.5 D
3. + 7.5 D
4. -0.75 D

Answer: 4. -0.75 D

The ability of a lens to converge a beam of light falling on it is known as the power of the lens. It is measured as the reciprocal of the focal length of the lens.

⇒ \(\mathrm{P}=\frac{1}{f_{\text {(in } \mathrm{m})}}\)

When focal length is in \mathrm{cm}, \(\mathrm{P}=\frac{100}{f} \mathrm{D}\)

Here,\(f_1\)=80 \mathrm{~cm}, \(f_2=-50 \mathrm{~cm}\)

we have, the combined power of two lenses

⇒ \(P_{\text {eq }} =P_1+P_2 \)

⇒ \(\mathrm{P}_{\text {eq }} =\frac{1}{f_1(\mathrm{~m})}+\frac{1}{f_2(\mathrm{~m})}\)

∴ \(\mathrm{P} =\frac{100}{80}-\frac{100}{50}=-0.75 \mathrm{D}\)

Question 51. The value of the angle of emergence from the prism, when the refractive index of the glass is \(\sqrt{3}\); is:

1. 60°
2. 35°
3. 45°
4. 90°

Answer: 4. 90°

⇒ \(\mu_{\mathrm{g}}=\sqrt{3}\)

We know, \(\sin i_c =\frac{\mu_a}{\mu_g}=\frac{1}{\mu_g} \)

⇒ \(i_c =\sin ^{-1}\left(\frac{1}{\mu_g}\right)\)

= \(\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Since the angle with which the light hits the second wall of the prism is 30°. Hence, it will suffer TIR and go out of the prism with the same angle it came in. Hence, the emergence angle is 90°.

Question 52. A ray is an incident at an angle of incidence I on one surface of a small angle prism (with the angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is p, then the angle of incidence is nearly equal to :

1. \(\frac{2 \mathrm{~A}}{\mu}\)
2. \(\mu \mathrm{A}\)
3. \(\frac{\mu \mathrm{A}}{2}\)
4. \(\frac{\mathrm{A}}{2 \mu}\)

Answer: 2. \(\mu \mathrm{A}\)

The Ra Diagram Is,

From the ray diagram,

⇒ \(\mathrm{A}+\alpha+90^{\circ} =180^{\circ}\)

⇒ \(\alpha =90^{\circ}-\mathrm{A}\)

Also \(r+\alpha =90^{\circ} \)

r =\(90^{\circ}-\alpha \)

⇒ \(90^{\circ}-\left(90^{\circ}-\mathrm{A}\right)\) =A

r =A

Also, From snell’s law \(\mu_1 \sin i=\mu_2 \sin r\)

⇒ \(\sin i=\mu \sin \mathrm{A}\) Here \(\mu_1=1 \mu_2=\mu)\)

For small, \(\sin \theta \approx \theta\)

i=\(\mu\) A.

Question 53. The refractive index of the material of a prism is V₂ and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is:

1. 30°
2. 45°
3. 60°
4. zero

Answer: 2. 45°

For retracing its path, a light ray should be normally inclined on a silvered face.

Applying Snell’s law at P,

⇒ \(\frac{\sin i}{\sin 30^{\circ}} =\frac{\sqrt{2}}{1}\)

⇒ \(\sin i =\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\)

i =\(45^{\circ}\)

Question 54. A thin prism having a refracting angle of 10° is made of glass with a refractive index of 1.42. This prism is combined with another thin prism of glass with a refractive index of 1.7. This combination produces dispersion without deviation. The refracting angle of the second prism should be:

1. 4°
2. 6°
3. 8°
4. 10°

Answer: 2. 6°

The dispersion without deviation of the prism is,

⇒ \((\mu-1) \mathrm{A}+\left(\mu^{\prime}-1\right) \mathrm{A}^{\prime}\) =0

⇒ \(|(\mu-1) \mathrm{A}| =\left|\left(\mu^{\prime}-1\right) \mathrm{A}^{\prime}\right| \)

⇒ \((1.42-1) =(1.7-1) \mathrm{A}^{\prime} \)

0.42 =0.7 \(\mathrm{~A}^{\prime} \)

∴ \(\mathrm{A}^{\prime} =6^{\circ}\)

Question 55. The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of the prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:

1. \(30^{\circ} ; \sqrt{2}\)
2. \(45^{\circ} ; \sqrt{2}\)
3. \(30^{\circ} ; \frac{1}{\sqrt{2}}\)
4. \(45^{\circ} ; \frac{1}{\sqrt{2}}\)

Answer: 1. \(30^{\circ} ; \sqrt{2}\)

From the figure, it is clear that,

Here angle of incidence i=\(45^{\circ}\)

angle of refraction r=\(r^{\prime}=30^{\circ}\)

angle of emergence e=\(45^{\circ}\)

Minimum deviation suffered by the ray

⇒ \(u_d=i+e-\left(r+r^{\prime}\right)=90^{\circ}-60^{\circ}=30^{\circ}\)

Using the formula,

⇒ \(\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \frac{A}{2}} \)

⇒ \(\mu=\frac{\sin \left(\frac{60^{\prime \prime}+30^{\prime \prime}}{2}\right)}{\sin \frac{60^{\prime \prime}}{2}}\)

⇒ \(\mu=\frac{\sin 45^{\prime \prime}}{\sin 30^{\prime \prime}}\)

∴ \(\mu=\frac{1 / \sqrt{2}}{1 / \mathrm{R}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Question 56. The refracting angle of a prism is A, and the refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is:

1. 180° – 3A
2. 180°-2A
3. 90°-A
4. 180° + 2A

Answer: 2. 180°-2A

We know that (from the prism formula)

⇒ \(\mu =\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)} \)

⇒ \(\cot \frac{A}{2} =\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) …(Given)

⇒ \(\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} =\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

⇒ \(\sin \left(\frac{\pi}{2}-\frac{A}{2}\right) =\sin \left(\frac{A}{2}+\frac{\delta}{2}\right)\)

=\(\pi-2 A\)

Question 57. A beam of light consisting of red, green, and blue colors is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green, and blue wavelength are 1.9, 1.44, and 1.47, respectively. Blue Green Red The prism will:

1. separate the blue color part from the red and green colors
2. separate all the three colors from one another
3. not separate the three colors at all
4. separate the red color part from the green and blue color.

Answer: 4. separate the red color part from the green and blue colors.

We know that, \(\mu =\frac{1}{\sin i_c}=\frac{1}{\sin 45^{\circ}}=\sqrt{2} \)

⇒ \(\ddots \left(\mu_{\text {red }}=1.39\right) <\mu \)

⇒ \(\mu_v >\mu ; \mu_g>\mu\)

only the red color does not suffer total internal reflection.

Question 58. The angle of the prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface return back through the same path after suffering reflection at the silvered surface. The refractive index p, of the prism is:

1. 2 sin A
2. 2 cos A
3. -cos A
4. tan A

Answer: 2. 2 cos A

MOP is a prism and the diagram is given

⇒ \(\angle \mathrm{MON} =90-\mathrm{A}^{\circ} \)

⇒ \(\angle r =90-(90-\mathrm{A}) \)

⇒ \(\angle r =\mathrm{A}^{\circ}\)

From Snell’s law

⇒ \(\frac{\sin i}{\sin s} =\mu \)

⇒ \(\mu =\frac{\sin (2 \mathrm{~A})}{\sin \mathrm{A}} \)

⇒ \(\mu =\frac{2 \sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A}}=2 \cos \mathrm{A}\)

∴ \(\mu =2 \cos \mathrm{A}\)

Question 59. A ray of light is incident at an angle of incidence, I, on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is p the angle of incidence i, is nearly equal to:

1. \(\mu \mathrm{A}\)
2. \(\frac{\mu A}{2}\)
3. \(\frac{A}{\mu}\)
4. \(\frac{A}{2 \mu}\)

Answer: 1. \(\mu \mathrm{A}\)

According to the question,

⇒ \(\mu=\frac{\sin i}{\sin r}=\frac{\sin i}{\sin \mathrm{A}}\)

If the angle is very small, then

⇒ \(\mu=\frac{i}{A}\)

i=\(\mu A\)

Question 60. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

1. lies between \(\sqrt{2}\) and 1
2. lies between 2 and \(\sqrt{2}\)
3. is less than 1
4. is greater than 2

Answer: 2. lies between 2 and \(\sqrt{2}\)

=\(\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}\)

Values of A varies from 0 to \(90^{\circ}\).

⇒ \(\mathrm{A}_{\min } =2 \cos \frac{90^{\circ}}{2}=\sqrt{2} \)

∴ \(\mathrm{~A}_{\max } =2 \cos 0^{\circ}\)=2

Question 61. A thin prism of angle 15° made of glass of refractive index μ₁= 1.5 is combined with another prism of glass of refractive index μ₂ = 1.75. The combination of the prism produces dispersion without deviation. The angle of the second prism should be:

1. 7°
2. 10°
3. 12°
4. 5°

Answer: 2. 10°

The formula used without deviation is,

⇒ \(\frac{A}{A^{\prime}}=\frac{\mu^{\prime}-1}{\mu-1}\)

putting the given value from the question

⇒ \(\frac{15^{\circ}}{\mathrm{A}^{\prime}} =\frac{1.75-1}{1.5-1} \)

⇒ \(\frac{15}{\mathrm{~A}^{\prime}} =\frac{0.75}{0.5} \)

∴ \(\mathrm{A}^{\prime} =\frac{0.5 \times 15}{0.75}=10^{\circ}\)

Question 62. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first face (i.e. incident face) of the prism is:

1. zero
2. 30°
3. 45°
4. 60°

Answer: 2. 45°

Here refracting angle of the prism

A=\(r_1+r_2\)

And for minimum deviation,

⇒ \(r_1 =r_2=r\)

A =2 r

r =\(\frac{A}{2}=\frac{60}{2}=30^{\circ}\)

Question 63. The refractive index of the material of a prism is V₂ and its refracting angle is 30°. One of the refracting surfaces of the prism is made a mirror inward. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is:

1. 45°
2. 60°
3. 0
4. 30°

Answer: 1. 0

Here, \(\angle r=30^{\circ}\)

(Using the law of triangle)

⇒ \(\mu=\frac{\sin i}{\sin r}\)

⇒ \(\sqrt{2} \cdot \sin 30^{\circ}=\sin i \)

⇒ \(\sin i=\frac{1}{\sqrt{2}}\)

i=\(45^{\circ}\)

Question 64. For the given incident ray as shown in the figure, the condition of the total internal reflection of this ray and the minimum refractive index of the prism will be:

1. \(\frac{\sqrt{3}+1}{2}\)
2. \(\frac{\sqrt{2}+1}{2}\)
3. \(\sqrt{\frac{3}{2}}\)
4. \(\sqrt{\frac{7}{6}}\)

Answer: 3. \(\sqrt{\frac{3}{2}}\)

For given conditions and use Snell’s law

1.\(\sin 45^{\circ}=\mu \sin \left(90-\theta_C\right) \)

⇒ \(\frac{1}{\sqrt{2}}=\mu \cos \theta_C=\sqrt{u^2-1}\)

⇒ \(\mu^2=1+\frac{1}{2}=\frac{3}{2} \)

=\(\mu=\sqrt{\frac{3}{2}} \)

Question 65. The angle of deviation (8) by a prism (refractive index = μ and supposing the angle of prism A to be small) can be given by

1. \(\delta=(\mu-1) \mathrm{A}\)
2. \(\delta=(\mu+1) \mathrm{A}\)
3. \(\delta=\frac{\sin \frac{A+\delta}{2}}{\sin \frac{A}{2}}\)
4. \(\delta=\frac{\mu-1}{\mu+1} \mathrm{~A}\)

Answer: 1. \(\delta=(\mu-1) \mathrm{A}\)

The refractive index of a prism is given by, Here,

⇒ \(\mu=\frac{\sin \frac{(\mathrm{A}+\delta)}{2}}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)

For a very thin prism, we can assume that \(\sin \theta \approx \theta[//latex]

⇒ [latex]\mu=\frac{\frac{A+\delta}{2}}{\frac{A}{2}} \)

or \(\mu=\frac{A+\delta}{A} \)

Or, \(\delta=\mu \mathrm{A}-\mathrm{A}=(\mu-1) \cdot \mathrm{A}\)

Question 66. There is a prism with a refractive index equal to \(\sqrt{2}\) and a refracting angle equal to 30°. One of the refracting surfaces of the prism is polished. A beam of monochromatic will retrace its path if its angle of incidence over the refracting surface of the prism is.

1. 0°
2. 30°
3. 45°
4. 60°

Answer: 3. 45°

When the ray QR retraced its path,

⇒ \(\angle A R Q =90^{\circ}, \angle r=30^{\circ}\)

⇒ \(\angle A Q R =90^{\circ}-30^{\circ}=60^{\circ}\)

By snell’s law, \(\sin i \times 1 =\sin \mathrm{r} \times \mu \)

⇒ \(\sin i =\mu \sin r=\sqrt{2} \sin 30^{\circ} [\mu for air =1]\)

Therefore, \(\sin i=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}\)

i=\(45^{\circ}\)

Question 67. Pick the wrong answer in the context of the rainbow :

1. The order of colors is reversed in the secondary rainbow
2. An observer can see a rainbow when his front is toward the sun
3. Rainbow is a combined effect of dispersion and reflection of sunlight
4. When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed.

Answer: 2. An observer can see a rainbow when his front is towards the sun

Rainbows cannot be observed when the observer faces toward the sun.

Question 68. The reddish appearance of the sun at sunrise and sunset is due to:

1. the scattering of light
2. the polarization of light
3. the color of the sun
4. the color of the sky

Answer: 1. the scattering of light

The reddish appearance of the sun at sunrise and sunset is due to the scattering of light.

Question 69. Rainbows are formed by:

1. reflection and diffraction
2. refraction and scattering
3. dispersion and total internal reflection
4. interference only

Answer: 3. dispersion and total internal reflection

When white light from the sun falls on raindrops, a band of different colors appears in the sky in the shape of a circular arc. This phenomenon is called a rainbow. It is due to that the little drops of water act as prisms for the white sunlight, causing refraction, dispersion, and total internal reflection of white light. The rainbow is not visible after every rain, and often when light rays of a specific color deviate minimally after one or two total internal reflections inside minuscule water drops.

Question 70. A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since:

1. a large aperture contributes to the quality and visibility of the images.
2. a large area of the objective ensures better light-gathering power.
3. a large aperture provides a better resolution
4. all of these

Answer: 4. all of these

Aperture ensure proper collection of light as well as better resolution.

Question 71. Assume that, light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is:

1. 183 \(\times 10^{-7} \mathrm{rad}\)
2. 7.32 \(\times 10^{-7} \mathrm{rad}\)
3. 6.00 \(\times 10^{-7} \mathrm{rad}\)
4. 3.66 \(\times 10^{-7} \mathrm{rad}\)

Answer: 4. 3.66 \(\times 10^{-7} \mathrm{rad}\)

Given the wavelength of light

⇒ \(\lambda =600 \mathrm{~nm} \)

= \(600 \times 10^{-9} \mathrm{~m}\)

Diameter of objective d=2 \(\mathrm{~m}\)

Limit of resolution telescope

d \(\theta =\frac{1.22 \times 600 \times 10^{-9}}{2} \)

=3.66 \(\times 10^{-7} \mathrm{rad}\) .

Question 72. An astronomical refracting telescope will have large angular magnification and high angular resolution when it has an objective lens of:

1. large focal length and large diameter
2. large focal length and small diameter
3. small focal length and large diameter
4. small focal length and small diameter

Answer: 1. large focal length and large diameter

For telescope, angular magnification of. So focal M = \(\frac{f_0}{f_e}\) length of the objective lens should be large. And Angular resolution = \(\frac{D}{1.22 \lambda}\)should be large.S

So objective should have a large focal length \(\left(f_0\right)\) and a large diameter (D). The option (a) is Correct. [Means objective lens should have large focal length \(\left(f_0\right)\) and large diameter (0)]

Question 73. The ratio of resolving powers of an optical microscope for two wavelengths λ₁= 4000 Å and λ₂ = 6000 Å is:

1. 8:27
2. 9: 4
3. 3:2
4. 16: 81

Answer: 3. 3:2

Resolving power of microscope,

(R.P) \(\propto \frac{1}{\lambda_{\text {(wavelength) }}}\)

∴ \(\frac{\mathrm{RP}_1}{\mathrm{RP}_2}=\frac{\lambda_2}{\lambda_1}=\frac{6000}{4000}=\frac{3}{2}\)

Question 74. An astronomical telescope has an objective and eyepiece of focal lengths of 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lens must be separated by a distance:

1. 46.0 cm
2. 50.0 cm
3. 54.0 cm
4. 37.3 cm

Answer: 3. 54.0 cm

Using lens formula for the objective lens

⇒ \(\frac{1}{v}-\frac{1}{u} =\frac{1}{f} \)

⇒ \(\frac{1}{v} =\frac{1}{f}+\frac{1}{u}\)

= \(\frac{1}{40}+\frac{1}{-200}\)

= \(\frac{+5-1}{200} \)

v =50 cm

Tube length =l=\(|v|+f_e \)

=50+4

=54 cm

Question 75. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on the inside part of the objective lens, the eye-piece forms a real image of this line. The length of this image is l. Then magnification of the telescope is:

1. \(\frac{L}{l}+1\)
2. \(\frac{L}{l}-1\)
3. \(\frac{L+1}{L-1}\)
4. \(\frac{L}{l}\)

Answer: 4. \(\frac{L}{l}\)

We know that the magnification of a telescope,

M=\(\frac{f_0}{f_0}/[latex]  → Equation 1

And also [latex]\frac{f_e}{f_e+u}=-\frac{l}{L}\)

⇒ \(\frac{f_e}{f_e-\left(f_0+f_e\right)} =-\frac{l}{L}\)

⇒ \(\frac{f_e}{f_0} =\frac{l}{L}\)  → Equation 2

From eq. (1) and (2),

M=\(\frac{L}{l}\)

Question 76. A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and a refractive index of 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

1. 2 cm upward
2. 1 cm upward
3. 4.5 cm downward
4. 1 cm downward

Answer: 2. 1 cm upward

Shifting in microscope = upward shifting in mark

t\(\left(1-\frac{1}{\mu}\right)=3\left(1-\frac{1}{1.5}\right)=1 \mathrm{~cm}\)

Question 77. The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 A is of the order of:

1. \(10^6 \mathrm{rad}\)
2. \(10^{-2} \mathrm{rad}\)
3. \(10^{-4} \mathrm{rad}\)
4. \(10^{-6} \mathrm{rad}\)

Answer: 4. \(10^{-6} \mathrm{rad}\)

Here, R.P. =\(\frac{1}{\Delta \theta}\)

Angular resolution, \(\Delta \theta =\frac{1.22 \lambda}{\mathrm{D}}\)

= \(\frac{1.22 \times 5000 \times 10^{-10}}{0.1}\)

= \(6.10 \times 10^{-6} \)

∴ \(\cong 10^{-6}\)

Question 78. A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 A, is of the order of:

1. 0.5 m
2. 5 m
3. 5 mm
4. 5 cm

Answer: 3. 5 mm

According To Question, \(\theta =\frac{1.22}{a} \lambda \)

⇒ \(\frac{d}{D} =\frac{\lambda}{\mathrm{a}} \)

⇒ \(a^{\prime} =\frac{\lambda D}{a} \)

a =\(\frac{5000 \times 10^{-10} \times 10^3}{10 \times 10^{-2}}=5 \mathrm{~mm}\)

Question 79. The diameter of the human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 m from the eye? The wavelength of light is 5000 Å.

1. 2.32 m
2. 4.28 mm
3. 1.25 cm
4. 12.48 cm

Answer: 3. 1.25 cm

For the aperture, limit of resolution

⇒ \(\frac{y}\geq \frac{\lambda}{d} \)

y \(\geq \frac{\lambda D}{d} \)

y \(\geq \frac{5 \times 10^{-7}}{2 \times 10^{-3}} \times 50 \geq 1.25 \mathrm{~m}\)

Question 80. An astronomical telescope of ten-fold angular magnification has a length of 44 cm. The focal length of the objective is:

1. 440 cm
2. 44 cm
3. 40 cm
4. 4 cm

Answer: 3. 44 cm

For an astronomical telescope,

Therefore, m=\(\frac{f_0}{f_e}\)

where,f_0= focal length of objective lens,

f_e= focal length of eyepiece lens

Length of telescope tube, L=\(f_{\mathrm{o}}+f_{\mathrm{e}}\)

m=10, L=44 \(\mathrm{~cm}\)

⇒ \(\frac{f_0}{f_e}\)=10

and \(f_o+f_e\)=44

⇒ \(f_o=\frac{f_e}{10}\)

⇒ \(f_o+\frac{f_0}{10}\)=44

or –\(\frac{11 f_0}{10}\)=44

or \(f_o=40 \mathrm{~cm}\)

Electromagnetic Waves

Question 1. A capacitor of capacitance ‘C’ is connected across an ac source of voltage V, is given by \(\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}\). The displacement current between the plates of the capacitor would then be given by :

1. \(\mathrm{I}_d=\mathrm{V}_0 \omega C \sin \omega \mathrm{t}\)
2. \(\mathrm{I}_d=\mathrm{V}_0 \omega C \cos \omega \mathrm{t}\)
3. \(\mathrm{I}_d=\frac{\mathrm{V}_0}{\omega c} \cos \omega \mathrm{t}\)
4. \(\mathrm{I}_d=\frac{\mathrm{V}_0}{\omega c} \sin \omega \mathrm{t}\)

Answer: 2. \(\mathrm{I}_d=\mathrm{V}_0 \omega C \cos \omega \mathrm{t}\)

Displacement Current, \(\mathrm{I}_d =\frac{C d V}{d t} \)

⇒ \(\mathrm{I}_d =C \frac{d}{d t}\left(V_0 \sin \omega \mathrm{t}\right) \)

∴ \(\mathrm{I}_d =C V_0 \omega \cos \omega \mathrm{t}\)

Question 2. For a plane electromagnetic wave propagating in the x-direction, which one of the following combinations gives the correct possible directions for the electric field (E) and magnetic field respectively?

1. \(\hat{j}+\hat{k}, \hat{j}+\hat{k}\)
2. –\(\hat{j}+\hat{k},-\hat{j}-\hat{k}\)
3. \(\hat{j}+\hat{k},-\hat{j}-\hat{k}\)
4. –\(\hat{j}+\hat{k},-\hat{j}+\hat{k}\)

Answer: 2. –\(\hat{j}+\hat{k},-\hat{j}-\hat{k}\)

For a plane EM wave the E and B field are mutually perpendicular. Hence (1), (3) and (4) are not possible.

Question 3. The magnetic field in a plane electromagnetic wave is given by: \(\mathrm{B}_y=2 \times 10^{-7} \sin \left(\pi \times 10^3 x+3 \pi \times 10^{11}\right)\). Calculate the wavelength.

1. \(\pi \times 10^3 \mathrm{~m}\)
2. 2 \(\times 10^{-3} \mathrm{~m}\)
3. 2 \(\times 10^3 \mathrm{~m}\)
4. \(\pi \times 10^{-3} \mathrm{~m}\)

Answer: 2. 2 \(\times 10^{-3} \mathrm{~m}\)

The magnetic field in a plane electromagnetic wave is\( \mathrm{B}_y=\mathrm{B}_{\mathrm{o}} \sin (k x-\omega t)\)  → Equation 1

The given magnetic field in a plane electromagnetic wave is,

⇒ \(\mathrm{B}_y=2 \times 10^{-7} \sin \left(\pi \times 10^3 x+3 \pi \times 10^{11} t\right)\)  → Equation 2

comparing eq. (1) and eq. (2) we get,

k= \(\pi \times 10^3\)

⇒ \(\frac{2 \pi}{\lambda} =\pi \times 10^3 \)

⇒  \(\lambda =\frac{2}{10^3} \)

= 2 \(\times 10^{-3} \mathrm{~m}\) .

Read and Learn More NEET Physics MCQs

Question 4. The ratio of the contribution made by the electric field and magnetic field components, to the intensity of an electromagnetic wave, is (where, c = speed of electromagnetic waves):

1. 1: 1
2. 1: c
3. 1: c²
4. c: 1

Answer: 1. 1: 1

We know that, \(\frac{E_0}{B_0}\)=c

Where, \(E_0\)= Peak values of electric field

⇒  \(B_0\)= Peak value of magnetic field

Then \(E_0: B_0\)=1: 1.

Question 5. For a transparent medium relative permeability and permittivity \(\mu_r\) and \(\varepsilon_r\) are 1.0 and 1.44 respectively. The velocity of light in this medium would be :

1. 2.5 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
2. 3 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
3. 2.08 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)
4. 4.32 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

Answer: 1. 2.5 \(\times 10^8 \mathrm{~m} / \mathrm{s}\)

The velocity of light in a medium is,

v= \(\frac{c}{\sqrt{\mu, \varepsilon_r}}\)

v= \(\frac{3 \times 10^8}{\sqrt{1 \times 1.44}} \)

v= 2.5 \(\times 10^8 \mathrm{~ms}^{-1}\)

Question 6. In an electromagnetic wave in free space, the root mean square value of the electric field in Ems = 6 V/m. The peak value of the magnetic field is :

1. \(1.41 \times 10^{-8} \mathrm{~T}\)
2. 2.83 \(\times 10^{-8} \mathrm{~T}\)
3. 0.70 \(\times 10^{-8} \mathrm{~T}\)
4. 4.23 \(\times 10^{-8} \mathrm{~T}\)

Answer: 2. 2.83 \(\times 10^{-8} \mathrm{~T}\)

According to the question, \(E_{\mathrm{rms}}=6 \mathrm{~V} / \mathrm{m}\)

Peak value of electric field

⇒  \(E_0=\sqrt{2} E_{\mathrm{rms}}\)

⇒  \(E_0=\sqrt{2} \times 6 \mathrm{~V} / \mathrm{m}\)

We know that, c =\(\frac{E_0}{B_0}\)

⇒  \(B_0 =\frac{E_0}{c}=\frac{\sqrt{2} \times 6}{3 \times 10^8} \)

= \(\frac{8.48}{3} \times 10^{-8}\)

B =2.83 \(\times 10^{-8} \mathrm{~T}\)

Question 7. Out of the following options which one can be used to produce a propagating electromagnetic wave?

1. a stationary charge
2. a chargeless particle
3. An accelerating charge
4. A charge moving at constant velocity

Answer: 3. An accelerating charge

To generate electromagnetic waves, we need accelerating charged particles.

Question 8. An electromagnetic wave of frequency v = 3.0 MHz passes from a vacuum into a dielectric medium with relative permittivity 8 = 4.0 Then:

1. Wavelength is double and frequency becomes half.
2. Wavelength is halved and frequency remains unchanged.
3. Wavelength and frequency both remain unchanged.
4. Wavelength is doubled and frequency is unchanged.

Answer: 2. Wavelength is halved and frequency remains unchanged.

The velocity of electromagnetic waves in a vacuum.

c=\(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=v_{\text {vacuum }}\)  → Equation 1

The velocity of electromagnetic waves in a medium

⇒  \(v_{\text {med }}=\frac{1}{\sqrt{\mu_0 \mu_0 \varepsilon_0 c_{n r}}}=\frac{c}{\sqrt{\mu_r \varepsilon_r}}\)

⇒  \(\mu_0\) and \(\varepsilon_0\) is in vacuum and \(\mu \mathrm{r} \varepsilon_0\) is in medium.

For dielectric medium r=1

⇒  \(v_{\text {med }}=\frac{c}{\sqrt{\varepsilon_r}} \text { Given } \varepsilon_0\)=4

⇒ \(v_{\text {med }}=\frac{c}{\sqrt{4}}=\frac{c}{2}\)  →  Equation 2

Wavelength of the wave in medium \(\lambda_{\text {med }}=\frac{v_{\text {med }}}{V}\)

= \(\frac{c}{2 V}=\frac{\lambda_{\text {vacuum }}}{2}\)

Question 9. The ratio of the amplitude of the magnetic field to the amplitude of the electric field for an electromagnetic wave propagating in a vacuum is equal to:

1. the speed of light in a vacuum
2. reciprocal of speed of light in vacuum
3. the ratio of magnetic permeability to the electric susceptibility vacuum
4. unity

Answer: 2. reciprocal of speed of light in vacuum

Since c =\(\frac{E_0}{B_0} \)

∴ \(\frac{B_0}{E_0}=\frac{1}{c}\)

Question 10. The electric and the magnetic field associated with an electromagnetic wave, propagating along the + z-axis, can be represented by:

1. \(\left[E=E_0 \hat{k} B=B_0 \hat{i}\right]\)
2. \(\left[E=E_0 \hat{j} B=B_0 \hat{j}\right]\)
3. \(\left[E=E_0 \hat{j} B=B_0 \hat{k}\right]\)
4. \(\left[E=E_0 \hat{i} B=B_0 \hat{j}\right]\)

Answer: 1. \(\left[E=E_0 \hat{k} B=B_0 \hat{i}\right]\)

⇒  \(\vec{v} =\vec{E} \times \vec{B}\)

= \(E_0 \hat{i}+B_0 \hat{j}\)

= \(E_0 B_0 \hat{k}\)

or (direction of propagation of waves is \(\vec{E} \times \vec{B}\) )

Question 11. Which of the following statements is false for the properties of electromagnetic waves?

1. Both electric and magnetic field vectors attain the maxima and minima at the same place and same time
2. The energy in electromagnetic waves is divided equally between electric and magnetic vectors
3. Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave
4. These waves do not require any material medium for propagation

Answer: 3. Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave

Wrong statement: Electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of the wave.

Option (1), (2) and (4) are correct.

Question 12. The electric field of an electromagnetic wave in free space is given by: \(\vec{E}=10 \cos (10 t+k x) \hat{j}\), where t and x are in seconds and meters respectively. It can be inferred that:

(1) the wavelength \(\lambda\)is 188.4 m.

(2) the wave number k is 0.33 rad/m.

(3) the wave amplitude to 10 V/m.

(4) the wave is propagating along + x direction

Which one of the following pairs of statements is correct?

1. (3) and (4)
2. (1) and (2)
3. (2) and (3)
4. (1) and (3)

Answer: 4. (1) and (3)

It is given that the electric field of the electromagnetic wave is:

⇒  \(\vec{E}=10 \cos \left(10^7 t \pm k x\right) j\)

Here, amplitude =10 \(\mathrm{~V} / \mathrm{m}\)

c= \(\frac{\omega}{K}\)

3 \(\times 10^8=\frac{10^7}{K}\)

K= \(\frac{1}{30} \)

or \(\frac{2 \pi}{\lambda}=\frac{1}{30}\)

∴ \(\lambda=188.4 \mathrm{~m}\)

Question 13. The electric field part of an electromagnetic wave in a medium is represented by: \(\mathrm{E}_x=0\)

\(\mathrm{E}_y=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{m}\right) t-\left(\pi 10^{-2} \frac{\mathrm{rad}}{\mathrm{s}}\right) x\right] \)

\(\mathrm{E}_z\)=0 .

The wave is:

1. moving along y direction with frequency 2TC X 106 Hz and wavelength 200 m.
2. moving along x direction with frequency 106 Hz and wavelength 100 m
3. moving along x direction with frequency 106 and wavelength 200 m.
4. moving along – x direction with frequency 106 Hz and wavelength 200 m.

Answer: 3. moving along x direction with frequency 106 and wavelength 200 m.

The standard equation is \(E_y=E_0 \cos (\omega t-k x)\)

And \(E_y=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) t\right]-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{sec}}\right)\)

Compare the above equations we have,

⇒  \(\omega=2 \pi f=2 \pi \times 10^6\) and f=\(10^6 \mathrm{~Hz}\)

we know that, \(\frac{2 \pi}{\lambda}\) =k

= \(\pi \times 10^{-2} \mathrm{~m}^{-1} \)

∴ \(\lambda =200 \mathrm{~m}\)

Question 14. The velocity of electromagnetic radiation in a medium of permittivity \(\mu_0\) and permeability ε₀ is given by

1. \(\sqrt{\frac{\varepsilon_0}{\mu_0}}\)
2. \(\sqrt{\mu_0 \varepsilon_0}\)
3. \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
4. \(\sqrt{\frac{\mu_0}{\varepsilon_0}}\)

Answer: 3. \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)

The velocity of electromagnetic radiation is the velocity of light (c) i.e.

c=\(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)

Where, \(\mu_0\)= permeability and

∴ \(\varepsilon_0\)= permittivity of free space

Question 15. The electric and magnetic fields of an electromagnetic wave are:

1. in opposite phase and perpendicular to each other
2. in opposite phase and parallel to each other
3. in phase and perpendicular to each other
4. in phase and parallel to each other

Answer: 3. in phase and perpendicular to each other

In EM waves \(\vec{E}\) and \(\vec{B}\) are in the same phase and perpendicular to each other.

Question 16. The velocity of the electromagnetic wave is parallel to:

1. \(\vec{B} \times \vec{E}\)
2. \(\vec{E} \times \vec{B}\)
3. \(\vec{E}\)
4. \(\vec{B}\)

Answer: 2. \(\vec{E} \times \vec{B}\)

According to the Pointing theorem, the direction of the wave is given by: S = E x B

or velocity of the electromagnetic wave is parallel to \(\vec{E} \times \vec{B}\)

Question 17. The wavelength of light of frequency 100 Hz is:

1. \(2 \times 10^6 \mathrm{~m}\)
2. \(3 \times 10^6 \mathrm{~m}\)
3. \(4 \times 10^6 \mathrm{~m}\)
4. \(5 \times 10^6 \mathrm{~m}\)

Answer: 2. \(3 \times 10^6 \mathrm{~m}\)

We have, a relation between the velocity of light (c), frequency (v), and wavelength \((\lambda)\) is given by

c=v \(\lambda\)

Thus, wavelength,\(\lambda=\frac{c}{v}\)

Given, \(\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, v=100 \mathrm{~Hz}\)

∴ \(\lambda=\frac{3 \times 10^8}{100}=3 \times 10^6 \mathrm{~m}\)

Question 18. The frequency of the electromagnetic wave, which is best suited to observe a particle of radius 3 x \(10^{-4}\) cm, is of the order of:

1. \(10^{15}\)
2. \(10^{14}\)
3. \(10^{13}\)
4. \(10^{12}\)

Answer: 2. \(10^{14}\)

We have, \(\lambda=\frac{c}{v}\)

Here, \(\lambda=3 \times 10^{-4} \mathrm{~cm}\),

The velocity of light in a vacuum,

c =3 \(\times 10^{10} \mathrm{~cm} / \mathrm{s}\)

3 \(\times 10^{-4} =\frac{3 \times 10^{10}}{v}\)

Frequency of electromagnetic wave \(v=10^{14} \mathrm{~Hz}\)

Question 19. Match List-1 with List-2.

Choose the correct answer from the options given below.

1. (A)-(4), (B)-(3), (C)-(2), (D)-(1)
2. (A)-(3), (B)-(2), (C)-(1), (D)-(4)
3. (A)-(3), (B)-(4), (C)-(2), (D)-(1)
4. (A)-(2), (B)-(3), (C)-(4), (D)-(1)

Answer: 4. (A)-(2), (B)-(3), (C)-(4), (D)-(1)

The wavelength of AM radio waves is 10²m. The wavelength of microwaves ranges between 10^-3m to 10^-1m. Infrared radiation wavelength lines Between 4 x 10^-7m to 7 x 10^-3m. X-ray wavelength ranges from 10^-8 to  10^-12 m.

Question 20. The electromagnetic wave with the shortest wavelength among the following is:

1. UV-rays
2. X-rays
3. ϒ-rays
4. Microwaves

Answer: 3. y-rays

From the above table, it is clear that Gamma rays have the shortest wavelength and have a higher frequency than microwaves, X-rays, and UV rays.

Question 21. Three stars A, B, and C have surface temperatures T_A, T_B, and T_c, respectively. Star A appears bluish, star B appears reddish, and star C is yellowish. Hence;

1. \(T_A>T_S>T_C\)
2. \(T_B>T_C>T_A\)
3. \(T_C>T_B>T_A\)
4. \(T_A>T_C>T_B\)

Answer: 4. \(T_A>T_C>T_B\)

According to Wein’s displacement law:

⇒ \(\lambda \mathrm{T}\) =b

⇒ \(\lambda \propto \frac{1}{\mathrm{~T}}\)  → Equation 1

From VIBGYOR we know,

⇒ \(\lambda_{\text {bluish }}<\lambda_{\text {yellowish }}<\lambda_{\text {reddish }}\)

\(\mathrm{T}_{\mathrm{A}}>\mathrm{T}_{\mathrm{C}}>\mathrm{T}_{\mathrm{B}}\) [using eq (1)]

Question 22. Which color of light has the longest wavelength?

1. Blue
2. Green
3. Violet
4. Red

Answer: 4. Red

∴ \(\lambda_{\text {red }}>\lambda_{\text {green }}>\lambda_{\text {blue }}>\lambda_{\text {red }}>\lambda_{\text {violet }}\)

Question 23. The energy of the EM waves is of the order of 15 keV. To which part of the spectrum does it belong?

1. X-rays
2. Infrared rays
3. Ultraviolet rays
4. y-rays

Answer: 1. X-rays

We know, E =\(\frac{h c}{\lambda} \)

⇒ \(\lambda_i =\frac{h c}{E}=\frac{6.62 \times 10^{-30} \times 3 \times 10^8}{15 \times 10^3 \times 1.0 \times 10^{-19}}\)

= 0.828 \(\times 10^{-10} \mathrm{~m}\)

⇒ \(\lambda\)=0.828 AA

Since the order of wavelength is between 10 \(\mathrm{~nm}\) and 0.001 \(\mathrm{~nm}\), it belongs to X-ray.

Question 24. The condition under which a microwave oven heats up a food item containing water molecules most efficiently is:

1. the frequency of the microwave must match the resonant frequency of the water molecules
2. the frequency of the microwave has no relation with the natural frequency of water molecules
3. microwaves are heat waves, so always produce heating
4. infrared waves produce heating in a microwave oven

Answer: 1. the frequency of the microwave must match the resonant frequency of the water molecules

It is an electromagnetic wave.

Question 25. The decreasing order of wavelength of infrared, microwave, ultraviolet, and gamma rays is:

1. gamma rays, ultraviolet, infrared, microwaves
2. microwaves gamma rays, infrared, ultraviolet
3. infrared, microwaves ultraviolet, gamma rays
4. microwave infrared, ultraviolet, and gamma rays

Answer: 4. microwave infrared, ultraviolet, gamma rays

Decreasing order of wavelength \((\lambda)\)

Microwave > Infrared > Ultraviolet > Gamma ray

Question 26. If \(\lambda_v, \lambda_x\) and \(\lambda_m\) represent the wavelength of visible light, X-rays and microwaves respectively, then:

1. \(\lambda_m>\lambda_x>\lambda_v\)
2. \(\lambda_m>\lambda_v>\lambda_x\)
3. \(\lambda_v>\lambda_x>\lambda_m\)
4. \(\lambda_v>\lambda_m>\lambda_x\)

Answer: 2. \(\lambda_m>\lambda_v>\lambda_x\)

⇒ \(\lambda_{\mathrm{m}}>\lambda_v>\lambda_x\)

In spectrum X-rays has minimum wavelength and microwave has maximum wavelength

Question 27. We consider the radiation emitted by the human body. Which one of the following statements is true?

1. The radiation emitted is in the infrared region
2. The radiation is emitted only during the day
3. The radiation is emitted during the summers and absorbed during the winters
4. The radiation emitted lies in the ultraviolet region and hence is not visible

Answer: 1. The radiation emitted is in the infrared region

Everybody at all times, at all temperatures emits radiation that falls in the infrared region.

Question 28. Which of the following rays are not electromagnetic waves?

1. X-rays
2. y-rays
3. \(\beta\)-rays
4. heat rays

Answer: 3. (3-rays

∴ \(\beta\)-rays are not electromagnetic waves.

Question 29. Which has a minimum wavelength?

1. X-rays
2. Ultraviolet rays
3. y-rays
4. Cosmic rays

Answer: 4. Cosmic rays

Cosmic rays have very short wavelengths.

Wavelengths →

X-ray → 1 Å to 100 Å

Ultraviolet rays →  100 Å to 1 Å

⇒ \(\gamma-\)ray s→ 0.001 Å  to 1 Å

Cosmic rays \(\rightarrow\) upto 4 \(\times 10^{-3} \) Å

Question 30. Which of the following is the cause of the ‘Greenhouse effect’?

1. Infrared rays
2. Ultraviolet rays
3. X-rays
4. Radio waves

Answer: 1. Infrared rays

Question 31. The frequency of y-rays, X-rays, and ultraviolet rays are a, b, and c respectively. Then,

1. a > b > c
2. a < b < c
3. a = b = c
4. a > c > b

Answer: 1. a > b > c

Question 32. The ozone layer blocks the radiations of wavelength:

1. less than 3 \(\times 10^{-7} \mathrm{~m} \)
2. equal to 3 \(\times 10^{-7} \mathrm{~m}\)
3. more than 3 \(\times 10^{-7} \mathrm{~m}\)
4. All of the above

Answer: 1. less than 3 \(\times 10^{-7} \mathrm{~m} \)

In the ozone sphere, the ozone layer extends from 30 km to nearly 50 km above the earth’s surface. This layer absorbs the majority of the sun’s UV radiation and prevents it from reaching the earth’s surface.

Since the range of UV radiations is 100 \(Å-4000 Å\). As a result, it absorbs radiations of wavelength of less than 3 x \(110^{-7}\)( or 3000 \(Å\)).

Question 33. A signal emitted by an antenna from a certain point can be received at another point of the surface in the form of:

1. sky waves
2. ground waves
3. sea wave
4. Both 1 and 2

Answer: 4. Both 1 and 2

The transmitting, receiving, and processing of data over space is referred to as space communication.

The modes of space communication are listed below.

1. Round or surface wave propagation

2. Space wave or tropospheric wave propagation

3. Sky wave propagation

4. Satellite communication

Question 34. The structure of solids is investigated by using:

1. cosmic rays
2. X-rays
3. γ-rays
4. infrared radiations

Answer: 2. X-rays

X-rays are used to investigate the structure of solids because of their strong penetrating capacity. For this purpose, the lane spot method and rotating cylinder method are unsuitable. X-rays strike the solid under investigation, exposing its structure on a photographic plate.

Question 35. Which of the following is the longest wave?

1. X-rays
2. γ-rays
3. Microwaves
4. Radiowaves

Answer: 4. Radiowaves

The following is a list of wavelength ranges of various waves.

Clearly, radiowaves are the longest waves.

Alternating Current

Question 1. The peak voltage of the AC source is equal to:

1. The value of the voltage supplied to the circuit
2. The RMS value of the source
3. \(\sqrt{2}\) times theRMS value of the ac source
4. \(\frac{1}{\sqrt{2}}\)times the RMS value of the ac source

Answer: 3. \(\sqrt{2}\) times the RMS value of the ac source

We know that, \(v_{r m s}=\frac{v_o}{\sqrt{2}}\)

∴ \(v_0=\sqrt{2} v_{r m s}\)

Question 2. The variation of EMF with time for four types of generators is shown in the figures. Which amongst them can be called AC?

1. 1 and 4
2. 1 and 2
3. 1, 2 and 4
4. only 1

Answer:2. 1 and 2

Ac waves are generally of a sinusoidal nature or which is positive for half cycle and negative for the other half cycle which is satisfied by all 4 options.

Question 3. A light bulb and an inductor coil are connected to an AC source through a key as shown in the figure below. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of a light bulb:

1. decreases
2. remains unchanged
3. will fluctuate
4. increases

Answer: 1. decreases

The diagram is shown in the figure.

According to the question the key is closed and when the iron rod is inserted inside the inductor the inductance (L) of the coil increases.

When L increases then inductive reactance \(X_L=\omega\) L also increases.

When \(X_L \)increases then current I decrease which is confirmed by the equation. I=\(\frac{e}{X_L}\)

Since \(X_L\) increases, hence I decrease and the glow of the bulb also decreases.

Read and Learn More NEET Physics MCQs

Question 4. A coil of self-inductance L is connected in series with a bulb B and AC source. The brightness of the bulb decreases when:

1. frequency of the AC source is decreased
2. number of turns in the coil is reduced
3. a capacitance of reactance XQ= XL is included in the same circuit
4. an iron rod is inserted into the coil

Answer: 4. an iron rod is inserted into the coil

We know that and Z=\(\sqrt{R^2+X_L^2}=\sqrt{R^2+(2 \pi f L)^2} \)

I=\(\frac{V}{Z}, P=I^2 R\)

i.e. V \(\uparrow L \uparrow=Z \uparrow\)

Question 5. In The circuit of the figure, the bulb will become suddenly bright, if:

1. contact’s is made or broken
2. contact is made
3. contact is broken
4. None of the above

Answer: 3. contact is broken

Self-induced current flows in the direction of the main current when the contact is suddenly broken. As a result, the bulb B will instantaneously become bright.

Question 6. A 40 μF capacitor is connected to a 200 V, 50 Hz AC supply. The RMS value of the current in the circuit is, nearly:

1. 2.05 A
2. 2.5 A
3. 25.1 A
4. 1.7 A

Answer: 2. 2.5 A

From the question,

C=40 \(\mu \mathrm{F}=40 \times 10^{-6} \mathrm{~F}, V_{r m s}=200 \mathrm{~V}, v=50 \mathrm{~Hz}\)

⇒ \(\mathrm{X}_{\mathrm{C}} =\frac{1}{2 \pi \nu \mathrm{C}}=\frac{1}{2 \pi \times 50 \times 40 \times 10^{-6}} \)

= \(\frac{10^6}{100 \pi \times 40}=\frac{250}{\pi} \Omega \)

∴ \(\mathrm{I}_{r m s} =\frac{\mathrm{V}_{r m s}}{X_C}=\frac{\frac{200}{250}}{\pi}=2.5 \mathrm{~A}\).

Question 7. A 100 Ω resistance and capacitor of 100 Q reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is:

1. 2.2 A
2. 11 A
3. 4.4 A
4. 1 √2 A

Answer: 1. 2.2 A

We know that the impedance of the R-C circuit,

Z= \(\sqrt{R^2+X_C^2}\)

Here R=100 \(\Omega, X_C=100 \Omega \)

Z= \(\sqrt{(100)^2+(100)^2}=100 \sqrt{2} \Omega \)

Peak value of current,

∴ –\(\mathrm{I}_{\max }=\frac{V_{\max }}{2}=\frac{220 \sqrt{2}}{100 \sqrt{2}}=2.2 \mathrm{~A}\)

Question 8. The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance are doubled, then the new reactance will be:

1. X
2. 2 x
3. 4 x
4. \(\frac{X}{4}\)

Answer: 4. \(\frac{X}{4}\)

When an AC signal is coupled to a capacitor with capacitance C, the circuit’s reactance is simply capacitive.

We have, the capacitive reactance,

⇒ \(X_C =\frac{1}{\omega C}=\frac{1}{2 \pi f C} \quad(\omega=2 \pi f) \)

⇒ \(\mathrm{X}_C \propto \frac{1}{f C}\)=0

Considering two different cases of frequency and capacitance, we have

⇒ \(\frac{X}{X}=\frac{f C}{f^{\prime} C}=\frac{f \times C}{2 f \times 2 C}\)

f=2 f and \(C^{\prime}\)=2 C

⇒ \(\frac{X}{X}=\frac{1}{4} \)

∴ \(X^{\prime}=\frac{X}{4}\)

Question 9. A series LCR circuit with inductance 10H capacitance I OpF, resistance 50\(\Omega\) is connected to the AC source of voltage, V = 200 sin (100 t) volt. If the resonant frequency of the LCR circuit is vQ and the frequency of the a.c. the source is v, then :

1. \(v_0=v=50 \mathrm{~Hz}\)
2. \(v_0=v=\frac{50}{\pi} \mathrm{Hz}\)
3. \(v_0=\frac{50}{\pi} \mathrm{Hz}, v=50 \mathrm{~Hz}\)
4. \(v_0=100 \mathrm{~Hz}, v_0=\frac{100}{\pi} \mathrm{Hz}\)

Answer: 2. \(v_0=v=\frac{50}{\pi} \mathrm{Hz}\)

Given, C=10 \(\mu F, R=50 \Omega, L=10 \mathrm{H} \)

⇒ \(\omega=2 \pi \nu\)

v=\(\frac{\omega}{2 \pi}=\frac{100}{2 \pi}=\frac{50}{\pi} \mathrm{Hz} \)

Resonance, \(v_{\mathrm{r}}=v_{\mathrm{o}} \)

= \(\frac{1}{2 \pi \sqrt{L C}}\)

= \(\frac{1}{2 \pi \sqrt{10 \times 10 \times 10^{-6}}} \)

= \(\frac{100}{2 \pi} \)

= \(\frac{50}{\pi} \mathrm{Hz} \)

Question 10. An inductor of inductance L, a capacitor of capacitance C, and a resistor of resistance ‘R’ are connected in series to an AC source of potential difference ‘ V volts as shown in the figure.

The potential difference across L, C, and R is 40 V, 10 V, and 40 V, respectively. The amplitude of the current flowing through the LCR series circuit is 10\(\sqrt{2}\) A. The impedance of the circuits is:

1. 4 \(\sqrt{2} \Omega\)
2. 5 \(\sqrt{2} \Omega\)
3. 4 \(\Omega\)
4. 5 \(\Omega\)

Answer: 4. 5 \(\Omega\)

Given, \(V_L =40 \mathrm{~V} \)

⇒ \(V_C =10 \mathrm{~V} \)

⇒ \(V_R =40 \mathrm{~V} \)

⇒ \(I_0 =10 \sqrt{2}\)

⇒ \(\mathrm{Z}\) =?

We know, \(\mathrm{Z} =\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \)

V=I Z =\(\sqrt{\mathrm{I}^2 \mathrm{R}^2+\left(\mathrm{IX}_{\mathrm{L}}-\mathrm{IX}_{\mathrm{C}}\right)}\)

V =\(\sqrt{1600+900}=50 \mathrm{~V} \)

⇒ \(I_{R M S} =\frac{I_0}{\sqrt{2}}=\frac{10 \sqrt{2}}{\sqrt{2}}=10 \mathrm{~A}\)

Z = \(\frac{V_{R M S}}{I_{R M S}}=\frac{50}{10}=5 \Omega\)

Question 11. A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is:

1. series LR
2. series RC
3. series LC
4. series LCR

Answer: 4. series LCR

⇒ \(I_1=\frac{V}{Z}\)

⇒ \(I_1=\frac{12}{\sqrt{R^2+\left(X_L-X_C\right)^2}}=0.2 \mathrm{~A}\)

In the second case for the DC source, the capacitor would provide infinite resistance but current is present in the circuit, which means the resistor and inductor can be present in the circuit.

As the current with the AC source and DC source are different, the inductor must be present with resistance.

Question 12. The figure shows a circuit that contains three identical resistors with resistance R = 9.0 fi each, two identical inductors with inductance L = 2.0 mH each and an ideal battery with emf E = 18 V. The current I through the battery just after the switch closed is:

1. 2 mA
2. 0.2 A
3. 2 A
4. 0 A

Answer: 3. 2 A

At, t = 0 no current flows through \(R_1\) and \(R_3.\)

i=\(\frac{E}{R_2} =\frac{18}{9}=2 \mathrm{~A}\) [since E=18 \(\mathrm{~V}, R_2=9 \Omega]\)

Question 13. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?

1. R=200 \(\Omega, L=1.5 H, \mathrm{C}=35 \mu \mathrm{F} \)
2. R=25 \(\Omega, L=2.5 H, \mathrm{C}=45 \mu \mathrm{F} \)
3. R=15 \(\Omega, L=3.5 H, \mathrm{C}=30 \mu \mathrm{F} \)
4. R=25 \(\Omega, L=1.5 \mathrm{H}, \mathrm{C}=45 \mu \mathrm{F}\)

Answer: 3. R=15 \(\Omega, L=3.5 H, \mathrm{C}=30 \mu \mathrm{F} \)

Q=\(\frac{\omega_0 L}{R}\)

Q=\(\frac{1}{\sqrt{L C}}\left(\frac{L}{R}\right)=\frac{1}{R} \sqrt{\frac{L}{C}}\)

R and C should be small and L should be high. So, option (3) is correct.

Question 14. A series R-C circuit is connected to an alternating voltage source. Consider two situations:

1. When the capacitor is air-filled.

2. When the capacitor is mica filled.

Current through the resistor is i and voltage across the capacitor is V then:

1. \(V_a<V_b\)
2. \(V_a>V_b\)
3. \(i_a>i_b\)
4. \(V_a=V_b\)

Answer: 2. \(V_a>V_b\)

When the capacitor is filled with mica then capacitance C increases so \(X_c\) decreases. In case (2) \(X_c\) decreases so the voltage across the capacitor decreases.

∴ \(V_a>V_b\)

Question 15. In an electrical “circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is \(\frac{\pi}{3}\). If instead, C is removed from the circuit, the phase difference is again \(\pi / 3\). The power factor of the circuit is:

1. \(\frac{1}{2}\)
2. 1 \(\sqrt{2}\)
3. 1
4. \(\frac{\sqrt{3}}{2}$\)

Answer: 3. 1

According to the question

Phase difference, \(\tan \phi =\frac{X_L-X_C}{R}\)

⇒ \(\tan \frac{\pi}{3} =\frac{X_L-X_C}{R}\)  → Equation 1

When \mathrm{L} is removed then,

⇒ \(\sqrt{3}=\frac{X_C}{R} X_C=\sqrt{3} R\)

When C is removed then

⇒ \(\tan \frac{\mathrm{p}}{3} =\sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{R} \)

⇒ \(X_{\mathrm{L}} =\sqrt{3} \cdot R\)

From eq. (1),

⇒ \(\tan \phi =\frac{\sqrt{3} R-\sqrt{3} R}{R}\)=0

⇒ \(\phi\) =0

Power factor, \(\cos \phi\) =1

Question 16. An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3 Ω the phase difference between the applied voltage and the current in the circuit is:

1. \(\frac{\pi}{4}\)
2. \(\frac{\pi}{2}\)
3. zero
4. \(\frac{\pi}{6}\)

Answer: 1. \(\frac{\pi}{4}\)

We know that, \(\tan \phi=\frac{X_L}{R}=\frac{L \omega}{R}\)

and \(\tan \phi =\frac{3 \Omega}{3 \Omega}\)

⇒ \(\tan \phi\) =1

⇒ \(\phi =45^{\circ} \)

∴ \(\phi =\frac{\pi}{4} \mathrm{rad}\)

Question 17. A coil has a resistance of 30 Ω and an inductive resistance of 20 Ω at 50 Hz frequency. If an AC source of 200 V. 100 Hz. is connected across the coil, the current in the coil will be :

1. 4.0 A
2. 8.0 A
3. \(\frac{20}{\sqrt{13}} \mathrm{~A}\)
4. 2.0A

Answer: 1. 4.0 A

According to the question

and \(\omega =50 \times 2 \pi, \omega L=20 \Omega\)

⇒ \(\omega^{\prime} =100 \times 2 \pi\), then \(\omega^{\prime} L=40 \Omega\)

Then, I =\(\frac{200}{Z}=\frac{200}{\sqrt{\mathrm{R}^2+\left(\omega^{\prime} \mathrm{L}\right)^2}}\)

= \(\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A}\)

Question 18. In the given circuit the reading of voltmeter V_l and V₂ are 300 V each. The reading to the voltmeter V3 and ammeter A are respectively:

1. 150 V, 2.2 A
2. 220 V, 2.2 A
3. 220 V, 2.0 A
4. 100 V, 2.0 A

Answer: 2. 220 V, 2.2 A

For series LCR circuit Voltage,

V=\(\sqrt{V_R^2+\left(V_L-V_C\right)^2}\)

since, \(V_L=V_C\)

Hence,V=\(V_R=220 \mathrm{~V}\)

I=\(\frac{V}{R}=\frac{220}{100}=2.2 \mathrm{~A}\)

Question 19. What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 pF and co = 1000 s^-1?

1. 1 mH
2. cannot be calculated unless R is known
3. 10 mH
4. 100 mH.

Answer: 4. 100 mH.

According to the question current is maximum at resonance

⇒ \(\omega^2 =\frac{1}{L C}\)

L =\(\frac{1}{\omega^2 C} \)

= \(\frac{1}{(1000)^2\left(10 \times 10^{-6}\right)}\)

= 0.1 \(\dot{\mathrm{H}}=100 \mathrm{mH}\)

Question 20. A coil of inductive reactance 31 Ω. has a resistance of 8 Ω. It is placed in series with a condenser of capacitative reactance 25 Ω. The combination is connected to an a.c. source of 110 V. The power factor of the circuit is:

1. 0.33
2. 0.56
3. 0.64
4. 0.80

Answer: 4. 0.80

⇒ \(X_L=31 \Omega, X_C=25 \Omega, R=8 \Omega\)

The impedance of series LCR is

Z= \(\sqrt{R^2+\left(X_L-X_C\right)^2}\)

Z= \(\sqrt{8^2+(31-25)^2} \)

Z= \(\sqrt{64+36}\)

Z= \(\sqrt{100}=10 \Omega\)

Power factor, \(\cos \phi=\frac{R}{Z}=\frac{8}{10}=0.8\)

Question 21. In a circuit l, c, and r are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of c is:

1. \(\frac{1}{\pi f(2 \pi f(L-R)}\)
2. \(\frac{1}{2 \pi f(2 \pi f L-R)}\)
3. \(\frac{1}{\pi f(2 \pi f L+R)}\)
4. \(\frac{1}{2 \pi f(2 \pi f L+R)}\)

Answer: 4. \(\frac{1}{2 \pi f(2 \pi f L+R)}\)

⇒ \(\tan \phi =\frac{X_C-X_L}{R}\)

⇒ \(\tan \left(\frac{\pi}{4}\right) =\frac{\frac{1}{\omega C}-\omega L}{R} \)

R =\(\frac{1}{\omega C}-\omega L \)

⇒ \((R+2 \pi f L) =\frac{2}{2 \pi f C} \)

Or C =\(\frac{1}{2 \pi f(R+2 \pi f L)}\)

Question 22. A coil of 40 Henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2-volt battery. The time constant of the circuit is:

1. 5 seconds
2. 1/5 seconds
3. 40 seconds
4. 20 seconds

Answer: 1. 5 seconds

According to the question

Time constant LR circuit is \(\tau=\frac{L}{R}\)

∴ \(\tau=40 / 8=5 \mathrm{sec}\)

Question 23. For a series LCR circuit, the power loss at resonance is:

1. \(\frac{V^2}{\left(\omega L-\frac{1}{\omega C}\right)}\)
2. \(I^2 L \omega\)
3. \(I^2 R\)
4. \(\frac{V^2}{C \omega}\)

Answer: 3. \(I^2 R\)

Power loss is given by,

P=V I \(\cos \phi\)

At resonance current and voltage are in the same phase, that is \(\phi\)=0.

P=V I\( \cos 0=V I=I^2 R\)

Question 24. A wire of resistance R is connected in series with an inductor of reactance oiL. The quality factor of the RL circuit is:

1. \(\frac{R}{\omega L}\)
2. \(\frac{\omega L}{R}\)
3. \(\frac{R}{\sqrt{R^2+\omega^2 L^2}}\)
4. \(\frac{\omega L}{\sqrt{R^2+\omega^2 L^2}}\)

Answer: 2. \(\frac{\omega L}{R}\)

We have,

Quality factor,Q=2 \(\pi \times \frac{\text { total energy stored in the circuit }}{\text { loss in energy in each cycle }}\)

But the total energy stored in circuit =\(\mathrm{Li}^2{ }_{\mathrm{ms}}\) and the energy loss per second =\(i^2{ }_{\mathrm{ms}}\)

So, energy lose per cycle =\(\frac{i_{m s}^2 R}{f}\)

⇒ \(\mathrm{Q}=2 \pi \times \frac{L i_{m s}^2}{i_{m s}^2 R / f} \)

∴ \(\mathrm{Q}=\frac{2 \pi f L}{R}=\frac{\omega L}{R}\)

Question 25. An LCR series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of:

1. \(\pi\)
2. 0
3. \(\frac{\pi}{4}\)
4. \(\frac{R}{C}\)

Answer: 2. 0

A circuit that allows maximum current by connecting inductance L, capacitance C, and resistance R in series. It is the series resonance circuit that corresponds to a specific AC frequency. An RLC circuit’s impedance (Z) is given by

⇒ \(\mathrm{Z}={\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)}}^2\)

At resonance,\( X_L=X_C\)

i.e. \(\omega L=\frac{1}{\omega C}\) or Z=R

As a result, the circuit behaves as if it just has R. So, phase difference =0.

Question 26. The time constant of the C-R circuit is:

1. \(\frac{1}{C R}\)
2. \(\frac{C}{R}\)
3. \(\mathrm{CR}\)
4. \(\frac{R}{C}\)

Answer: 3. \(\mathrm{CR}\)

We have, R =\(\frac{V}{i} \text { and } C=\frac{q}{V} \)

⇒ \(\mathrm{RC} =\frac{V}{i} \frac{q}{V}=\frac{q}{i}=\frac{i \times t}{i}\)=t

⇒ \({[\mathrm{RC}] } =[t]=[\mathrm{T}]\)

CR is referred to as the CR circuit’s time constant.

Question 27. A series LCR circuit containing a 5.0 H inductor, 80 μF capacitor, and 40 Ω resistor is connected to a 230 V variable frequency AC source. The angular frequencies of the source at which power transferred to the circuit are half the power at the resonant angular frequency are likely to be:

1. 25 rad/s and 75 rad/s
2. 50 rad/s and 25 rad/s
3. 46 rad/s and 54 rad/s
4. 42 rad/s and 58 rad/s

Answer: 3. 46 rad/s and 54 rad/s

Given, L =5 \(\mathrm{H} \)

C =80 \(\mu \mathrm{F} \)

R =40 \(\Omega\)

r =230 \(\mathrm{~V}\)

Let the frequencies be

⇒ \(\omega^{\prime}=\omega_R \pm \Delta \omega\)

where, \(\Delta \omega=\frac{R}{2 L}\)

⇒ \(\Delta \omega=\frac{40}{2 \times 5}=4 \mathrm{rad} / \mathrm{s}\)

And \(\omega_{\mathrm{R}}=\frac{1}{\sqrt{L C}}=50 \mathrm{rad} / \mathrm{s}\)

Frequencies are 50-4 and 50+4

= 46 \(\mathrm{rad} / \mathrm{s}\) and 54 \(\mathrm{rad} / \mathrm{s}\)

Question 28. A series L-C-R circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is \(\frac{\pi}{3}\). If instead C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) between current and voltage. The power factor of the circuit is:

1. 0.5
2. 1.0
3. -10
4. zero

Answer: 2. 1.0

For LCR circuit phase difference,

⇒ \(\phi=\tan ^{-1}\left[\frac{X_L-X_C}{R}\right]\)

When L is removed then \(X_L\)=0

⇒ \(\phi =\tan ^{-1}(\frac{-X_C}{R})\)

⇒ \(\tan \phi =|\frac{X_C}{R}|\)  → Equation 1

When C is removed then \(X_C\)=0

⇒ \(\phi =\tan ^{-1}(\frac{X_L}{R})\)

⇒ \(\tan \phi =(\frac{X_L}{R})\) →  Equation 2

From eq. (1) and eq. (2),

⇒ \(X_L =X_C \)

Impedance \(\mathrm{Z} =\sqrt{R^2+\left(X_L-X_C\right)^2}\)

Power Factor

⇒ \(\cos \phi=\frac{R^2}{Z}\)=1

Power factor \(\cos \phi\)=1.

Question 29. An inductor of 20 mH, a capacitor of 100 μF, and a resistor of 50 Ω are connected in series across a source of emf, V= 10 sin 314 t. The power loss in the circuit is:

1. 2.74 W
2. 43 W
3. 0.79 W
4. 1.13 W

Answer: 3. 0.79 W

Given,Inductance, L=20 \(\mathrm{mH}=20 \times 10^{-3} \mathrm{H}\)

Capacitance, C=100 \(\mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}\)

Resistance,R=50 \(\Omega\)

V=10 \(\sin 314 t\) →  Equation 1

and equation of emf is, V=\(V_0 \sin \omega t\)  → Equation 2

Compare eq. (1) and (2),

⇒ \(V_0=10 \mathrm{~V}, and \omega=314 \mathrm{rad} \mathrm{s}^{-1}\)

Power of the AC circuit is, P =\(V_{r m s} I_{r m s} \cos \phi \)

= \(V_{r m s}\left(\frac{V_{r m s}}{Z}\right)\left(\frac{R}{Z}\right)=\left(\frac{V_{r m s}}{Z}\right)^2 R\)

P =\(\left(\frac{V_0}{\sqrt{2} \cdot Z}\right)^2 \cdot R\) →  Equation 3

Impedance, Z=\(\sqrt{R^2+\left(X_L-X_C\right)^2}\)

= \(\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \)

= \(\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \)

= \(\sqrt{(50)^2+\left(\frac{1}{314 \times 100 \times 10^{-6}}-314 \times 20 \times 10^{-3}\right)^2} \)

= \(\sqrt{2500+(25.56)^2}=56 \Omega\)

Putting the value of Z=56 \(\Omega\) in eq. (2)

P =\(\left(\frac{10}{\sqrt{2} \times 56}\right)^2 \times 50\)

P =\(\frac{100}{2 \times 3136} \times 50 \)

P =0.79 \(\mathrm{~W}\)

Power loss =0.79 \(\mathrm{~W}\)

Question 30. An inductor 20 mH, a capacitor 50 μF, and a resistor 40Ω are connected in series across a source of emf, V = 10 sin 340t. The power loss in the AC circuit is:

1. 0.67 W
2. 0.76 W
3. 0.89 W
4. 0.51 W

Answer: 4. 0.51 W

According to the question

Induction L=20 \(\mathrm{mH}\)

Capacitance C=50 \(\mu \mathrm{F}\)

Resistance R=40 \(\mathrm{~W}\)

emf V=10 \(\sin 340 \mathrm{t}\)

Power loss in \(\mathrm{AC}\) is

P=\(I_v^2 R=\left[\frac{E_y}{2}\right]^2 \cdot R=\left(\frac{10}{\sqrt{2}}\right)^2\) .

= 40\(\left[\frac{1}{(40)^2+\left(340 \times 20 \times 10^3-\frac{1}{340 \times 50 \times 10^{-6}}\right)^2}\right]\)

= \(\frac{100}{2} \times 40 \times \frac{1}{1600+(6.8-58.8)^2} \)

= \(\frac{2000}{1600+2704}=0.51 \mathrm{~W}\)

Question 31. The potential differences across the resistance capacitance and inductance are 80 V, 40 V, and 100 V respectively in an L-C-R circuit. The power factor of this circuit is:

1. 0.4
2. 0.5
3. 0.8
4. 1.0

Answer: 3. 0.8

For this, we know the formula,

⇒ \(\tan \phi =\frac{V_L-V_C}{V_R}=\frac{100-40}{80}=\frac{3}{4}\)

⇒ \(\phi =37^{\circ} \)

Power factor =\(\cos \phi=\cos 37^{\circ}\)

= \(\frac{4}{5}\) or 0.8

Question 32. A resistance ‘R’ draws power ‘F when connected to an AC source. If an inductance is introduced such that the impedance of the circuit becomes ‘Z the power drawn will be:

1. P\(\left(\frac{R}{Z}\right)^2\)
2. P \(\sqrt{\frac{R}{Z}}\)
3. P\(\left(\frac{R}{Z}\right)\)
4. P

Answer: 1. P\(\left(\frac{R}{Z}\right)^2\)

According to the question, the power drawn is

P =\(V_{\mathrm{rms}} \cdot I_{\mathrm{rms}}=V_{\mathrm{rms}} \cdot \frac{V_{\mathrm{rms}}}{R}\)

⇒ \(V_{\mathrm{ms}}^2\) =P R

When the inductor is connected in series with the resistor the power drawn is,

⇒ \(P^{\prime}=V_{\mathrm{rms}} \cdot I_{\mathrm{rms}} \cos \phi\)

Where, \(\phi\)= phase difference

⇒ \(^{\prime}=\frac{V_{\mathrm{ms}}^2}{R} \cdot \frac{R^2}{Z^2}=P \cdot R \cdot \frac{R}{Z^2}\)

∴ \(P^{\prime}=\frac{P \cdot R^2}{Z^2}=P\left(\frac{R}{Z}\right)^2\)

Question 33. The instantaneous value of alternating current and voltage in a circuit are given as:

i=\(\frac{1}{\sqrt{2}} \sin (100 \pi t)\) ampere

e=\(\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)\) volt

The average power in Watts consumed in the circuit is :

1. \(\frac{1}{4}\)
2. \(\frac{\sqrt{3}}{4}\)
3. \(\frac{1}{2}\)
4. \(\frac{1}{8}\)

Answer: 4. \(\frac{1}{8}\)

P =\(V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \)

= \(\frac{1}{2} \times \frac{1}{2} \times \cos \left(\frac{\pi}{3}\right)=\frac{1}{8} \mathrm{~W}\)

Question 34. The power dissipated in an L-C-R series circuit connected to an AC source of emf ε is: 

1. \(\frac{\varepsilon^2 R}{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}\)
2. \(\frac{\varepsilon^2 \sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}{R}\)
3. \(\frac{\varepsilon^2\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}{R}\)
4. \(\frac{\varepsilon^2 R}{\sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}}\)

Answer: 1. \(\frac{\varepsilon^2 R}{\left[R^2+\left(L \omega-\frac{1}{C \omega}\right)^2\right]}\)

The power dissipated in series L-C-R

P=\(I_{\mathrm{rms}}{ }^2 R=\frac{E_{\mathrm{ms}}^2 R}{|Z|^2}\)

P=\(\frac{E^2 R}{\left[R^2+\left(\omega L-\frac{1}{\omega C}\right)^2\right]}\)

Question 35. In an AC circuit, the emf (e) and the current (i) at any instant are given respectively by:

e=\(E_0 \sin \omega t \)

i=\(I_0 \sin (\omega t-\phi T)\)

The average power in the circuit over one cycle of AC is:

1. \(\frac{E_0 I_0}{2}\)
2. \(\frac{E_0 I_0}{2} \sin \phi\)
3. \(\frac{E_0 I_0}{2} \cos \phi\)
4. \(E_0 I_0\)

Answer: 3. \(\frac{E_0 I_0}{2} \cos \phi\)

Power = rate of work done in one complete cycle

P \(\omega=\frac{W}{T}\)

P \(\omega=\frac{\left(\mathrm{E}_0 I_0 \cos \phi\right) T / 2}{T} \)

P \(\omega=\frac{\mathrm{E}_0 I_0 \cos \phi}{2}\)

Where \(\cos \phi\) is called the power factor of an AC circuit.

Question 36. An eclectic Kettle has two heating coils. When one of the coils is connected to a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel the time taken by the same quantity of water to boil will be :

1. 8 minute
2. 4 minute
3. 25 minute
4. 15 minute

Answer: 1. 8 minute

Here, Q =\(\frac{V^2}{R_1} \times t_1=\frac{V^2}{R_2} \times t_2 \)

=\(\frac{V^2}{R} \times t\)

⇒ \(\frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}\)

⇒ \(\frac{Q}{V^2 t} =\frac{Q}{V^2 t_1}+\frac{Q}{V^2 t_2} \)

⇒ \(\frac{1}{t} =\frac{1}{t_1}+\frac{1}{t_4}\)

t =\(\frac{t_1 t_2}{t_1+t_2} \)

= \(\frac{10 \times 40}{10+40}=8 \mathrm{~min}\)

Question 37. A step-down transformer connected to an AC mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

1. 0.2 A
2. 0.4 A
3. 2 A
4. 4 A

Answer: 1. 0.2 A

Given, Step down transformer

⇒ \(V_P =220 \mathrm{~V}\)

⇒ \(V_S =11 \mathrm{~V} \)

⇒ \(P_S =44 \mathrm{~W}\)

⇒ \(I_P\) =?

⇒ \(P_P =P_S \)

⇒ \(V_P I_P =V_S I_S \)

⇒ \(I_S =\frac{P_s}{V_s}=4 \mathrm{~A}\)

∴ \(I_P =\frac{V_s I_s}{V_P} \)

= \(\frac{11 \times 4}{220}=\frac{4}{20} \)

= \(\frac{1}{5}=0.2 \mathrm{~A}\)

Question 38. A transformer having an efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:

1. 300 V, 15 A
2. 450 V, 15 A
3. 450 V, 13.5 A
4. 600 V, 15 A

Answer: 2. 450 V, 15 A

According to the question,

Efficiency, 90 \%=0.9

Input voltage, \(V_P=200 \mathrm{~V}\)

Input power, \(P_P=300 \mathrm{~W}\)

Secondary coil current, \(I_S=6 \mathrm{~A}\)

⇒ \( V_P I_P \)=3000

⇒ \(I_P =\frac{3000}{V_P}=\frac{3000}{200}=15 \mathrm{~A}\)

Now efficiency =\(\frac{\text { Output }}{\text { input }}=\frac{V_S I_S}{V_P I_P}\)

0.9=\(\frac{V_S \times 6}{200 \times 15}\)

\(V_S=450 \mathrm{~V}\)

Secondary coil current =450 \(\mathrm{~V}\).

Question 39. The primary winding of the transformer when connected to a DC battery of 10 Volt draws a current of 1 mA. The number of turns of the primary and secondary windings are 50 and 100 respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively:

1. 20 V and 2.0 mA
2. 10 V and 0.5 mA
3. Zero volts and therefore no current
4. 20 V and 0.5 mA

Answer: 3. Zero volt and therefore no current

The transformer can not work on DC.

Question 40. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is:

1. 3.6 A
2. 2.8 A
3. 2.5 A
4. 5.0 A

Answer: 4. 5.0 A

Efficiency, n %=\(\frac{P_{\text {out }}}{P_{\text {in }}} \times 100 \)

= \(\frac{V_s i_s}{V_p i_p} \times 100\)

80=\(\frac{2 \times 440}{220 \times l_p} \times 100\)

∴ \(i_P=5 \mathrm{~A} \)

Question 41. The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux \(\phi\) linked with the primary coil is given by \(\phi=\phi_v+4\) t, where \(\phi\) is in webers, t is time in seconds and \(\phi_v\) is a constant, the output voltage across the output voltage across the secondary coil is:

1. 120 volts
2. 220 volts
3. 30 volts
4. 90 volts.

Answer: 1. 120 volts

According to the question,

⇒ \(N_p =50, N_s\)=1500

⇒ \(\phi =\phi_0+4 t\)

Voltage across primary coil,

⇒ \(V_p =\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right)\)

= 4 volt

⇒ \(\frac{V_s}{V_p} =\frac{N_s}{N_p}\)

∴ \(V_s =\frac{1500}{50} \times 4=120 \mathrm{~V}\)

Question 42. A transformer is used to light a 100 W and 110 V lamp from a 220 V main. If the main current is 0.5 amp, the efficiency of the transformer is approximately:

1. 50%
2. 90%
3. 10%
4. 30%

Answer: 2. 90%

The efficiency of the transformer

⇒ \(\eta =\frac{V_s I_s}{V_p I_p} \times 100\)

=\(\frac{100}{220 \times 0.5} \times \)100=90 %

Question 43. The core of a transformer is laminated because:

1. ratio of voltage in primary and secondary may be increased
2. energy losses due to eddy currents may be minimized
3. the weight of the transformer may be reduced
4. rusting of the core may be prevented

Answer: 2. energy losses due to eddy currents may be minimized

The core of a transformer is laminated to minimize the energy losses due to eddy currents.

Question 44. A step-up transformer operates on a 230 V line and supplies a current of 2 A to a load. The ratio of the primary and secondary windings is 1:25. The current in the primary coil is

1. 15 A
2. 50 A
3. 25 A
4. 12.5 A

Answer: 2. 50 A

We have, the change in flux of the primary and secondary coils is proportional to the number of turns in the primary and secondary coils, respectively.

So, \(\frac{\phi_P}{N_P} =\frac{\phi_S}{N_S}\)

or \(\frac{1}{N_P} \times \frac{d \phi_P}{d t} =\frac{1}{N_S} \frac{d \phi_P}{d t}\)

∴ \(\frac{V_S}{V_P} =\frac{N_S}{N_P} \quad\left(\text { as } V \propto \frac{d \phi}{d t}\right)\)

Question 45. The primary winding of the transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an AC supply of 20 V-50 Hz. The secondary will have an output of

1. 2 V, 5 Hz
2. 200 V, 500 Hz
3. 2 V, 50 Hz
4. 200 V, 50 Hz

Answer: 4. 200 V, 50 Hz

The transformer converts high-voltage AC into low-voltage AC without changing the frequency. The voltage across the input is divided by the voltage across the output.

We have, the voltage is given by,\(\frac{V_S}{V_P}=\frac{N_S}{N_P}\)

⇒ \(\mathrm{N}_{\mathrm{S}}\)= No. of turns in secondary coil \(\mathrm{N}_{\mathrm{P}}\)= No. of turns in the primary coil

⇒ \(\mathrm{V}_{\mathrm{S}}=\frac{N_S}{N_P} \mathrm{~V}_{\mathrm{P}}\)

Substitute the given values, we get

= \(\frac{5000}{500} \times 20\)=200 V

∴ Thus, the output has a voltage of 200 V and a frequency of 50 Hz.

Electromagnetic Induction

Question 1. A square loop of side 1 m and resistance 1 \(\Omega\) is placed in a magnetic field of 0.5%T. If the plane of the loop is perpendicular to the direction of the magnetic field, the magnetic flux through the loop is:

1. 2 weber
2. 0.5 weber
3. 1 weber
4. Zero weber

Answer: 2. 1 weber

Here, l=1 \(\mathrm{~m}, R=1 \Omega, \mathrm{B}=0.5 \mathrm{~T} \text {, }\)

Angle\(\theta=0^{\circ}\)

Magnetic flux,\(\phi=\mathrm{BA} \cos \theta \)

= 0.5 \(\times 1 \times 1 \times \cos 0^{\circ}\)

= 0.5 \(\mathrm{~Wb}\)

Question 2. A big circular coil of 1000 turns and an average radius of 10 m is rotating about its horizontal diameter at 2 rad s^-1. If the vertical component of the earth’s magnetic field at that place is 2 x 10^-5 T and the electrical resistance of the coil is 12.56 Ω, then the maximum induced current in the coil will be:

1. 0.25 A
2. 1.5 A
3. 1A
4. 2 A

Answer: 3. 1A

Here, the number of turns (N) =1000

Radius (r) =10 \(\mathrm{~m}\)

Magnetic field (B) =2 \(\times 10^{-5} \mathrm{~T}\)

Resistance(R) =12.56 \(\Omega \)

Maximum induced emf, \(\mathrm{E} =N \omega A B\)

= N \(\omega\left(\pi r^2\right) B\)

= 1000 \(\times 2 \times 3.14 \times(10)^2 \times 2 \times 10^{-5}\)

= 12.56 \(\mathrm{~V}\)

Maximum induced current, \(I_{\max }=\frac{E}{R}\)

= \(\frac{12.56}{12.56}=1 \mathrm{~A}\)

Question 3. The magnetic flux linked with a coil (in Wb) is given by the equation \(\phi=5 t^2+3 t+16\). The magnitude of induced emf in the coil at the fourth second will be :

1. 33 V
2. 43 V
3. 108 V
4. 10 V

Answer: 4. 10 V

Given: Magnetic flux, \((\phi)=5 t^2+3 t+16\)

Induced emf, |E| =\(\frac{d \phi}{d t}\)

= \(\frac{d}{d t}\left(5 t^2+3 t+16\right)\)

= 10 t+3

Therefore induced emf, when t=3,

⇒ \(|E_3|=(10 \times 3)+3=33 \mathrm{~V}\)

and induced emf, when, t =4,

⇒ \(\left|E_4\right| =(10 \times 4)+3=43 \mathrm{~V}\)

Therefore emf induced in the fourth second

= \(|\varepsilon_4|-|\varepsilon_3|\)

= 43-33=10\( \mathrm{~V}\)

Read and Learn More NEET Physics MCQs

Question 4. A coil of 800 turns and an effective area of 0.05 m² is kept perpendicular to a magnetic field 5 x 10^-5 T. When the plane of the coil is rotated by 90° around any of its co-planner axis in 0.1 s. The emf induced in the coil will be :

1. 0.2 V
2. 2 x 10^-3 V
3. 0.02 V
4. 2 V

Answer: 3. 0.02 V

Here, N = 800 turn

A =0.05 \(\mathrm{~m}^2=5 \times 10^{-2} \mathrm{~m}^2\)

B =5 \(\times 10^5 \mathrm{~T}\)

⇒ \(\Delta t =0.15 \mathrm{sec}\)

We know that,

⇒ \(e_{\text {induced }} =-\frac{\Delta \phi}{\Delta t}\)

⇒ \(\Delta \phi =\phi_f-\phi_i \)

⇒ \(\phi_f=0, \phi_i =N(\vec{B} \cdot \vec{A})\)

= 800 \(\times 5 \times 10^{-5} \times 5 \times 10^{-2}\)

= 2 \(\times 10^{-3} \)

⇒ \(e_{\text {induced }} =-\frac{\Delta \phi}{\Delta t}=\frac{-\left(0-2 \times 10^{-3}\right)}{0.1}\)

=0.02 \(\mathrm{~V}\)

Question 5. A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate\(\frac{d B}{d t}\). Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of the magnetic field as shown in the figure. Then, the emf generated is:

1. zero in loop 1 and zero in loop 2
2. –\(\frac{d B}{d t} \pi r^2 \)in loop 1 and\( -\frac{d B}{d t} \pi r^2\) in loop 2
3. –\(\frac{d B}{d t} \pi \mathrm{R}^2\) in loop 1 and zero in loop 2
4. –\(\frac{d B}{d} \pi r^2\) in loop 1 and zero in loop 2

Answer: 3. –\(\frac{d B}{d t} \pi \mathrm{R}^2\) in loop 1 and zero in loop 2

We know that induced emf in the region is given by

|e| =\(\frac{d \phi}{d t}\)

⇒ \(\frac{d \phi}{d t} =-\pi r^2 \frac{d B}{d t} \)

⇒ \({\phi=B A=\pi r^2 B e A=\pi r^2}\)

⇒ \(\frac{d \phi}{d t} =-\pi r^2 \frac{d B}{d t}\)

From the question, Rate of change of magnetic flux associated with loop (1)

⇒ \(e_1=-\frac{d \phi_1}{d t}=-\pi r^2 \frac{d B}{d t}\)

and Rate of change of magnetic flux associated with loop (2)

=\(-\frac{d \phi_2}{d t}\)=0 { since } \({\phi_2=0}\)

Question 6. A thin semicircular conducting ring (PQR) of radius V is falling with its plane vertically in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is v is:

1. Zero
2. Bvπ² and P is at higher potential
3. πrBv and R is at higher potential
4. 2rBr and R is at higher potential.

Answer: 4. 2rBr and R is at higher potential.

Rate of decrease of area of semicircular ring = \(\frac{d A}{d t}=(2 r) \mathrm{V}\)

From Faraday’s law of electromagnetic induction = e=\(-\frac{d \phi}{d t}=-B \frac{d A}{d t}=-B(2 r \mathrm{~V})\)

As induced current in the ring produces a magnetic field in an upward direction hence R is at higher potential.

Question 7. A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f is:

1. twice per revolution
2. four times per revolution
3. six times per revolution
4. once per revolution.

Answer: 1. twice per revolution

This is the case of periodic EMI.

From the graph, it is clear that direction is changing once in \(\frac{1}{2}\) cycle.

Question 8. A coil of resistance 400 \(\Omega\) is placed in a magnetic field. If the magnetic flux \(\phi\)(Wb) linked with the coil varies with time t (sec) as \(\phi=50 t^2+4\). The current in the coil at t = 2 s is:

1. 0.5 A
2. 0.1A
3. 2 A
4. 1 A

Answer: 1. 0.5 A

We know that, Induced emf of coil,

E= \(|-\frac{d \phi}{d t}|_t\)

Given that, \(\phi=50 t^2+4\) and R=400 \(\Omega\)

E = \(|\frac{-d \phi}{d t}|_{t=2}[100 t]_{t=2}\)

= 200 \(\mathrm{~V}\)

Current in the coil I=\(\frac{E}{R}=\frac{200}{400}\)

=0.5 \(\mathrm{~A} \text {. }\)

Question 9. A conducting circular loop is placed in a uniform magnetic field. B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mms^-1. The induced emf when the radius is 2 cm, is:

1. 2 \(\pi \mu V\)
2. \(\pi \mu \mathrm{V}\)
3. \(\frac{\mu}{2} \mu \mathrm{V}\)
4. 2 \(\mu \mathrm{V}\)

Answer: 2. \(\pi \mu \mathrm{V}\)

We know that, \(\phi=B A\)

⇒ \(E_{\text {in }} =\frac{-d \phi}{d t} \)

⇒ \(E_{\text {in }} =-\frac{d}{d t}(B A)\)

⇒ \(E_{\text {ir }} =-B \pi 2 r \frac{d r}{d t} \)

E =-B \(\pi 2 r \frac{d r}{d t} \)

-0.025 \(\times \pi \times 2 \times 2 \times 10^{-1} \times 10^{-3}\)

Compare with original formula, induced emf =\(\pi \mu \mathrm{V}\)

Question 10. A conducting circular loop is placed in a uniform magnetic field of 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mms^-1. The induced emf in the loop, when the radius is 2 cm, is:

1. \(3.2 \pi \mu \mathrm{V}\)
2. 4.8 \(\pi \mu \mathrm{V}\)
3. 0.8 \(\pi \mu \mathrm{V}\)
4. 1.6 \(\pi \mu \mathrm{V}\)

Answer: 1. \(3.2 \pi \mu \mathrm{V}\)

According to the question,

Magnetic field, B=0.04 \(\mathrm{~T}\)

and \(-\frac{d r}{d t}=2 \mathrm{~ms}^{-1}\)

Induced emf is equal to the rate of change of magnetic flux linked with the circuit

Induced emf, e =\(-\frac{d \phi}{d t}\)

=-\(\frac{B d A}{d t}=-B \frac{d\left(\pi r^2\right)}{d t}\)

If r=2 \(\mathrm{~cm}\) then,

e =-0.04 \(\times \pi \times 2 \times 2 \times 10^{-2} \times 2 \times 10^{-3} \)

=3.2 \(\pi \mu \mathrm{V}\)

Question 11. A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move:

1. in a circular orbit
2. along a parabolic path
3. along a straight line
4. in an elliptical orbit.

Answer: 1. in a circular orbit

In the magnetic field a charged particle moves in a circular orbit.

Question 12. As a result of a change in the magnetic flux linked to the closed loop as shown in the figure, an e.m.f V volt is induced in the loop. The work done (joules) in taking a charge Q coulomb once along the loop is:

1. QV
2. 2 QV
3. \(\frac{Q V}{2}\)
4. zero

Answer: 1. QV

Work done due to a charge W= QV

Question 13. A rectangular coil of 20 turns and an area of cross-section of 25 sq cm has a resistance of 100 fl If a magnetic field that is perpendicular to the plane of the coil changes at a rate of 1000 T/s, the current in the coil is:

1. 1A
2. 50 A
3. 0.5 A
4. 5 A

Answer: 3. 0.5 A

Total number of turns, N=20

Area of coil, A =25 \(\mathrm{~cm}^2 \)

=25 \(\times 10^{-4} \mathrm{~m}^2\)

Change in magnetic field w.r.t. $t

⇒ \(\frac{d B}{d t}=1000 \mathrm{~T} / \mathrm{s}\)

Resistance of coil, R=100 \(\Omega\)

i = ?

We have, Induced current, i=\(\frac{e}{R}\)

= \(\frac{N A \frac{d B}{d t}}{R} \cdot[e=N A \frac{d B}{d t}]\)

= \(\frac{20 \times 25 \times 10^{-4} \times 1000}{100}\)

= 0.5 \(\mathrm{~A}\)

Question 14. An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of the electron. What will be the direction of current, If any induced in the coil?

1. abcd
2. adcb
3. The current will reverse its direction as the electron goes past the coil
4. No current induced.

Answer: 3. The current will reverse its direction as the electron goes past the coil

When the electron goes in a straight line electric flux first increases and then decreases.

Question 15. A conductor of length 0.4 m is moving with a speed of 7 m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced emf across the conductor is:

1. 1.26 V
2. 2.52 V
3. 5.04 V
4. 25.2 V

Answer: 2. 2.52 V

Given, length of conductor (l)=0.4 \(\mathrm{~m}\),

speed \((v)=7 \mathrm{~m} / \mathrm{s}\)

Magnetic field (B)=0.9 \(\mathrm{~Wb} / \mathrm{m}^2\)

Induced emf, e=B I v \(\sin \theta\)

⇒ \([\theta=90^{\circ}.\) as B is perpendicular to V

=0.9 \(\times 0.4 \times 7 \times \sin 90^{\circ}\)

=2.52 \(\mathrm{~V}\)

Question 16. In the circuit of Fig, the bulb will become suddenly bright if

1. contact is made or broken
2. contact is made
3. contact is broken
4. won’t become bright at all.

Answer: 3. contact is broken

When a circuit is broken, the induced e.m.f. is largest. So the answer is (C).

Question 17. A wheel with 20 metallic spokes, each 1 m long, is rotated with a speed of 120 rpm in a plane perpendicular to a magnetic field of 0.4 G. The induced emf between the axle and rim of the wheel will be (1 G = \(10^{-4}\) T):

1. 2.51 \(\times 10^{-4} \mathrm{~V}\)
2. 2.51 \(\times 10^{-5} \mathrm{~V}\)
3. 40 \(\times 10^{-5} \mathrm{~V}\)
4. 2.51 \(\mathrm{~V}\)

Answer: 1. 2.51 \(\times 10^{-4} \mathrm{~V}\)

In the question, B =0.4 \(\mathrm{G}=0.4 \times 10^{-4} \mathrm{~T}\)

l =1 \(\mathrm{~m} \)

f =120 \(\mathrm{rpm}=\frac{120}{60} \mathrm{rps}=2 \mathrm{~Hz}\)

Induced emf between the axle and rim of the wheel is:

e =\(\frac{1}{2} B \omega l^2 \)

= \(\frac{1}{2} B(2 \pi f) l^2=\pi B f l^2 \)

= 3.14 \(\times 0.4 \times 10^{-4} \times 2 \times 1\)

= 2.51 \(\times 10^{-4} \mathrm{~V}\) .

Question 18. A cycle wheel of radius 0.5 m is rotated with a constant angular velocity of 10 rad/s in a region of the magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its center and the rim is :

1. 0.25 V
2. 0.125 V
3. 0.5 V
4. zero

Answer: 2. 0.125 V

According to the question

e =\(\frac{B I^2 \omega}{2} \)

= \(\frac{1}{2} \times 0.1 \times\left(\frac{1}{2}\right)^2 \times 10=\frac{1}{8} \)

= 0.125 \(\mathrm{~V}\)

Question 19. A conducting square frame of length ‘a’ and a long straight wire carrying current / are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity of V. The emf induced in the frame will be proportional to:

1. \(\frac{1}{x^2}\)
2. \(\frac{1}{(2 x-a)^2}\)
3. \(\frac{1}{(2 x+a)^2}\)
4. \(\frac{1}{(2 x-a)(2 x+a)}\)

Answer: 4. \(\frac{1}{(2 x-a)(2 x+a)}\)

Potential difference across \(\mathrm{AB}\) is,

⇒ \(V_A-V_B=e_1=B_1(a) v\)

Where, \(B_1\)= magnetic field induction at A B

But, \(B_1=\frac{\mu_0}{4 \pi} \frac{2 i}{\left(x-\frac{a}{2}\right)}\)

Putting the value of B_1 in eq. (1),

⇒ \(e_1=\frac{\mu_0 i}{2 \pi\left(x-\frac{a}{2}\right)} a v\)

Potential difference \(\mathrm{CD}\) is given by

⇒ \(V_C-V_D=e_2=B_2(a) v\)

But, \(\mathrm{B}_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 i}{\left(x+\frac{a}{2}\right)}\)

⇒ \(e_2=\frac{\mu_0 i}{2 \pi\left(x+\frac{a}{2}\right)} a v\)

The net potential difference in the loop \(\mu\)

⇒ \(e_1-e_2 =\left(V_A-V_B\right)-\left(V_C-V_D\right) \)

= \(\frac{\mu_0 i^{\prime} a v}{2 \pi}\left(\frac{1}{x-\frac{a}{2}}-\frac{1}{x+\frac{a}{2}}\right) \)

= \(\frac{\mu_0 i^{\prime} a v}{2 \pi}\left(\frac{2 a}{\left(x-\frac{a}{2}\right)\left(x+\frac{a}{2}\right)}\right)\)

Hence clear that it is Proportional to \(\frac{1}{(2 x-a)(2 x+a)}\)

Question 20. In coil of resistance 10 \(\Omega\) the induced current developed by changing magnetic flux through it is shown in the figure as a function of time. The magnitude of change in flux through the coil in Weber is:

1. 8
2. 2
3. 6
4. 4

Answer: 2. 2

Here, I=\(\left|\frac{1}{R} \frac{d \phi}{d t}\right|\)

⇒ \(|d \phi|=|I R d t| \)

⇒ \(d \phi=(\text { Area of triangle }) \times R\)

=\(\left(\frac{1}{2} \times 4 \times 0.1\right) \times 10=2 \mathrm{~Wb}\)

Question 21. The magnetic flux through a circuit of resistance R changes by an amount \(\Delta \phi\) in a time \(\Delta t\). Then the total quantity of electric charge Q that passes any point in the circuit during the time \(\Delta t\) is represented by:

1. Q=\(\frac{1}{R} \cdot \frac{\Delta \phi}{\Delta t}\)
2. Q=\(\frac{\Delta \phi}{R}\)
3. Q=\(\frac{\Delta \phi}{\Delta t}\)
4. Q=R \(\cdot \frac{\Delta \phi}{\Delta t}\)

Answer: 2. Q=\(\frac{\Delta \phi}{R}\)

According to the question on induced emf is:

V =\(\frac{\Delta \phi}{\Delta \lambda}\)

and current, I =\(\frac{Q}{\Delta t}=\frac{\Delta \phi}{\Delta t} \times \frac{1}{R} \)

Q =\(\frac{\Delta \phi}{R}\)

Question 22. In which of the following devices, the eddy current effect is not used?

1. Magnetic braking in train
2. Electromagnet
3. Electric heater
4. Induction furnace

Answer: 3. Electric heater

Eddy currents are loops of electrical current induced within a conductor by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. Electric heaters are not based on the eddy current effect. It is based on Joule’s heating effect

Question 23. Two conducting circular loops of radii \(R_1\) And \(R_2\) are placed in the same plane with their centers coinciding. If \(R_1 \gg R_2\), the mutual inductance between them will be directly proportional to:

1. \(\frac{R_1}{R_2}\)
2. \(\frac{R_2}{R_1}\)
3. \(\frac{R_1^2}{R_2}\)
4. \(\frac{R_2^2}{R_1}\)

Answer: 3.\(\frac{R_1^2}{R_2}\)

Given, Radii of loops, \(R_1\) and \(R_2\),

⇒ \(R_1 >>R_2 \)

M =?

we know, \(M_{12}=\frac{\mu_0 N_1 N_2 \pi R_1^2}{l}\)

⇒ \(M_{21}=\frac{\mu_0 N_1 N_2 \pi R_2^2}{l}\)

∴ \(\frac{M_1}{M_2}=\frac{R_1^2}{R_2^2}\)

Question 24. The variation of EMF with time for four types of generators is shown in the figures. Which amongst them can be called AC?

1. Only (1)
2. (1) and (4)
3. (1), (2), (3) and (4)
4. (1) and (2)

Answer: 3. (1), (2), (3) and (4)

A current that changes its direction periodically is called an alternating current

Question 25. The magnetic potential energy stored in a certain inductor is 25 ml, when the current in the inductor is 60 mA. This inductor is of inductance:

1. 1.389 H
2. 138.88 H
3. 0.138 H
4. 13.89 H

Answer: 4. 13.89 H

Given, U=25 \(\mathrm{~mJ} =25 \times 10^{-3} \mathrm{~J}\)

I =60 \(\mathrm{~mA}=60 \times 10^{-3} \mathrm{~A}\)

Energy stored in inductor, E=\(\frac{1}{2} L \times I^2\)

⇒ \(25 \times 10^{-3} =\frac{1}{2} L \times(60 \times(-3))^2\)

L =\(\frac{25 \times 2 \times 10^6 \times 10^{-3}}{3600}=\frac{500}{36} \)

=13.89 Henry

Question 26. A long solenoid has 1000 turns. When a current 4 A flows through it, the magnetic flux linked with each turn of the solenoid “is 4 x 10^-5 Wb. The self-inductance of the solenoid is:

1. 3 H
2. 2 H
3. 1 H
4. 4 H

Answer: 3. 1 H

From the question,

Number of turns of the solenoid, \(\mathrm{N}\)=1000

Current, I=4 \(\mathrm{~A}\)

And magnetic flux,\( \phi=4 \times 10^{-3} \mathrm{~Wb}\)

self induction, L =\(\frac{\phi . N}{I}\)

= \(\frac{4 \times 10^{-3} \times 1000}{4}=1 \mathrm{H}\)

Question 27. A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is.

1. 2 Wb
2. 0.5 Wb
3. 12.5 Wb
4. Zero

Answer: 3. 12.5 Wb

Given: Current I =2.5

Inductance, L = 5H

Magnetic flux, \(\phi\)=?

We know, \(\phi\) =L I

=5 \(\times 2.5 \mathrm{~Wb}=12.5 \mathrm{~Wb}\)

Question 28. The current (I) in the inductance is varying with time according to the plot shown in the figure.

Which one of the following is the correct variation of voltage with time in the coil?

Answer: 4.

From the diagram, for t=0 to t=T / 2. induced voltage

V =L \(\frac{d I}{d t}\)

= L \(\frac{d}{d t}\left(\frac{2 I_0 t}{T}\right)\)= constant

For t=\(\frac{T}{2}\) to t=T induced

V =L \(\frac{d I}{d t}\)

=L \(\frac{\mathrm{d}}{d t}\left(\frac{-2 I_0 t}{T}\right)\)=- constant

This confirms that option (D) is correct.

Question 29. The current i in a coil varies with time as shown in the figure. The variation induced emf with time would be

Answer: 4.

We know that, e=-L \(\frac{d I}{d t}\)

During 0 to \(\frac{T}{4}, \frac{d I}{d t}\)= constant

e=\(-V_0\)

For \(T_4\) to\( T_2\), \(\frac{d I}{d t}\)=0

e=0

For \(\frac{T}{2} to \frac{3 T}{4}, \frac{d I}{d t}\)= constant

e=+ve

Question 30. Two coils of self-inductance 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

1. 16 mH
2. 10 mH
3. 6 mH
4. 4 mH

Answer: 4. 4 mH

Mutual inductance between coils is,

M =k \(\sqrt{I_1 I_2}\)

M =1 \(\sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}} \) {x=1}

= 4 \(\times 10^{-3}=4 \mathrm{mH} \)

Question 31. In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = At what time emf ‘xs zero?

1. 4s
2. 3s
3. 2s
4. 1s

Answer: 3. 2s

L=2 \(\mathrm{mH}, i=i^2 e^{-t}\)

E=-L \(\frac{d i}{d t}=-L\left[-t^2 e^{-t}+2 t e^{-t}\right]\)

when, E=0,

⇒ \(-e^{-t} t^2+2 t e^{-1}\)=0

i.e. \(t e^{-t}(t-2)\)=0

t \(\neq \infty\) and t \(\neq 0\)

-t \(\neq 2 \mathrm{sec} (t e^{-t}=0\)

or, \(2 t e^{-t}=e^{-1} t^2 \)

t=2 \(\mathrm{sec} \)

Question 32. A varying current in a coil changes from 10 A to zero in 0.5 s. If the average emf induced in the coil is 220 V, the self-inductance of the coil is:

1. 5 H
2. 6 H
3. 11 H
4. 12 H.

Answer: 3. 11 H

Emf induced in the coil of self-inductance, L

e=-L \(\frac{d \phi}{d t}=-\frac{d}{d t}(L i) \text { or } e=-L \frac{d i}{d t}\)

⇒ \((\frac{d i}{d t}\)= rate of flow of current in coil

As, d i\(=i_2-i_1=0-10=-10 \mathrm{~A} \)

d t=0.5 \(\mathrm{~s}\)

e=220 \(\mathrm{~V} \)

220=-L \(\frac{(-10)}{0.5}\)

or , L=\(\frac{220}{20}=11 \mathrm{H} \)

Question 33. What is the self-inductance of a coil that produces 5 V when the current changes from 3 A to 2 A in one millisecond?

1. 5000 H
2. 5 mH
3. 50 H
4. 5 H

Answer: 2. 5 mH

Emf induced in the coil is given by,

e=-\(\frac{d \phi}{d t}\)

If the coil has self-inductance (L) and current i, then induced emf is given by

e =-\(\frac{d}{d t}(\mathrm{Li}) or \mathrm{e}=-\mathrm{L} \frac{d i}{d t} \)

⇒ \(\mathrm{L} =\frac{\frac{|e|}{d i}}{d t}\)

Given, e \(\mid =5 \mathrm{~V}, d i=3-2=1 \mathrm{~A} \)

d t =1 \(\mathrm{~ms}=1 \times 10^{-3} \mathrm{~s} \)

∴ \(\mathrm{~L} =\frac{5 \times 10^{-3}}{1}=5 \mathrm{mH}\)

Question 34. If the number of turns per unit length of a coil of solenoid is doubled, the self-inductance of the solenoid will:

1. remain unchanged
2. halved
3. be doubled
4. become four times

Answer: 4. become four times

A long solenoid is one whose length is significantly greater than its cross-section radius. If N is the total number of turns in the solenoid, A is the area of each turn, and 1 is the length of the solenoid, then the self-inductance of the solenoid is L=\(\frac{\mu_0 N^2 A}{I} \Rightarrow L=\mu_0 n^2 A I\)

L=\(\frac{\mu_0 N^2 A}{I} \Rightarrow L=\mu_0 n^2 A I\)

(If the number of turns per unit length is \(\mathrm{n}.)\)

So,\(\mathrm{L} \alpha \mathrm{n}^2\)

When \(n^2\) is doubled, L becomes 4 times.

Question 35. A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is:

1. once per revolution
2. twice per revolution
3. four times per revolution
4. six times per revolution

Answer: 2. twice per revolution

The flux of the magnetic field through the 100 \(\pi\) is,

⇒ \(\phi=B \pi r^2 \cos \omega t\)

where \(\theta\)= angle the normal makes with the portion of the 100 \(\pi\) induced emf

⇒ \(\varepsilon=\omega B \pi r^2 \sin \omega t\)

These is zero when \(\omega t=n \pi\) i.e.,

when, \(\theta=0, \pi, 2 \pi \ldots\). etc.,

So, the induced emf changes direction every half rotation.

Magnetism And Matter

Question 1. A short bar magnet of magnetic moment 0.4JT^-1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is:

1. – 0.64 J
2. zero
3. – 0.082 J
4. -0.064 J

Answer: 4. -0.064 J

We know that, U=-M B \(\cos \theta\)

for stable equilibrium \(\theta=0^{\circ}\)

U =\(-M B \cos 0^{\circ}\)

= -M B=-(0.4)(0.16)

= -0.064 \(\mathrm{~J}\)

Question 2. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in an equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in the new position is:

1. \(\frac{W}{\sqrt{3}}\)
2. \(\sqrt{3} W\)
3. \(\frac{\sqrt{3} W}{2}\)
4. \(\frac{2 W}{\sqrt{3}}\)

Answer: 2. \(\sqrt{3} W\)

Since work was done by rotating the magnet,

W=M B\(\left(\cos \theta_0-\cos \theta\right)\)

Where, \(\mathrm{M}\)= magnetic moment of the magnet

B= magnetic field

W =M B\(\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \)

=M B\(\left(1-\frac{1}{2}\right)=\frac{M B}{2}\)

M B =2 W  → Equation 1

And Torque,\(\tau=M \times B=M B \sin 0^{\circ}\) putting M B=2 W from (1),

⇒ \(\tau =2 W \sin 60^{\circ}\)

=2 W \(\times \frac{\sqrt{3}}{2}=W \sqrt{3}\)

Read and Learn More NEET Physics MCQs

Question 3. The following figures show the arrangement of bar magnets in different configurations. Each magnet has a magnetic dipole moment of m. Which configuration has the highest net magnetic dipole moment

Answer: 3.

Given, that each magnet has a magnetic dipole moment =m.

Net magnetic moment =\(\sqrt{m^2+m^2+2 m^2 \cos \theta}\)

Here, if \(m_{\text {net }}\) is \(\max\) when \(\cos \phi\) is \(\max\) when \(\cos \phi\) will be \(\max\) when \(\theta\) in minimum.

So, \(\theta=30^{\circ} m_{\text {net }}\) will be maximum.

Question 4. A bar magnet of length l and magnetic dipole moment M is bent in the form of an are as shown in the figure. The new magnetic dipole moment will be:

1. M
2. \(\frac{3}{\pi} M\)
3. \(\frac{2}{\pi} M\)
4. \(\frac{M}{2}\)

Answer: 2. \(\frac{3}{\pi} M\)

The magnetic pole strength is m then the magnetic moment, M=m l

From the given figure, l =\(\frac{3}{\pi} \times r\)

r =\(\frac{31}{\pi}\)

New magnetic moment, \(\mathrm{M}^{\prime} =m v r=m \times \frac{31}{\pi} \)

= \(\frac{3}{\pi} \times m l=\frac{3 M}{\pi}\)

Question 5. A bar magnet of magnetic moment M is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet. The length of the magnet will be:

1. \(\frac{M B}{F}\)
2. \(\frac{B F}{M}\)
3. \(\frac{M F}{B}\)
4. \(\frac{F}{M B}\)

Answer: 1. \(\frac{M B}{F}\)

Here, \(\tau =\vec{M} \times \vec{B}\)

⇒ \(|\tau| =M B \sin 90^{\circ}\)

and we know that, \(\tau=F l\)

Fl =M B

l = \(\frac{M B}{F}\)

Question 6. Two identical bar magnets are fixed with their centers at a distance d apart. A stationary charge Q is placed at P in between the gap of the two magnets at a distance D from the center O as shown in the figure. The force on the charge Q is:

1. zero
2. directed along OP
3. directed along PO
4. directed perpendicular to the plane of the paper.

Answer: 1. zero

The force on charge Q is zero.

Question 7. A bar magnet having a magnetic moment of 2 x 10⁴ JT^-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:

1. 0.6 J
2. 12 J
3. 6 J
4. 2 J

Answer: 3. 6 J

According to the question; When the magnetic dipole is rotated from the initial position \(\theta_1\) to the final position \(\theta_2\) then work is done.

= M B\(\left(\cos \theta_1-\cos \theta_2\right)\)

= M B\(\left(1-\frac{1}{2}\right)=\frac{2 \times 10^4 \times 6 \times 10^{-4}}{2}\)

= 6 \(\mathrm{~J}\)

Question 8. A bar magnet of magnetic moment M is placed in the magnetic field of induction B. The torque exerted on it is

1. M. B
2. -M. B
3. M x B
4. -M x B

Answer: 3. M x B

A magnetic torque \(\tau\) acts on a bar magnet when it is put in an external magnetic field B, which is given by,

∴ \(\tau=M \times B =[M] \times[B] \sin \theta (\theta\) = angle between M and B)

Question 9. Does a bar magnet oscillate in the earth’s magnetic field with a period T. What happens to its period of motion if its mass quadrupled?

1. Motion remains simple harmonic with new period \(\frac{T}{2}\)
2. Motion remains simple harmonic with new period = 2T
3. Motion remains simple harmonic with new period = 4T
4. Motion remains simple harmonic and the period stays nearly constant.

Answer: 2. Motion remains simple harmonic with new period = 2T

We have, the time period of a bar magnet in a magnetic field,\(\mathrm{T}=2 \pi \sqrt{\left(\frac{\mathrm{I}}{M B}\right)}\)

Where I is the moment of inertia of the bar magnet, M is the magnetic moment and B is the magnetic induction.When mass is made 4 times (Since, I=\(m r^2, \mathrm{I} \propto m \)). From the above equation of period T \(\propto \sqrt{I}\). So, when mass is made 4 times, then T becomes twice.

Question 10. At point A on the earth’s surface the angle of dip \(\tan\) = + 25°. At a point B on the earth’s surface the angle of dip, \(\tan\) = – 25°. We can interpret that:

1. A is located in the southern hemisphere.
2. A is located in the northern hemisphere and B is located in the southern hemisphere.
3. A and B are both located in the southern hemisphere.
4. A and B are both located in the northern hemisphere.

Answer: 2. A is located in the northern hemisphere and B is located in the southern hemisphere.

The angle of dip is the angle between the earth’s resultant magnetic field from the horizontal. The dip is zero at the equator and positive in the northern hemisphere.

In the southern hemisphere dip angle is considered negative.

Question 11. The relations amongst the three elements of earth’s magnetic field, namely horizontal component H, vertical component V, and dip \(\tan\) are, (\(B_E\) = total magnetic field):

1. V=\(B_E \tan \delta, H=B_E\)
2. V=\(B_E \sin \delta, H=B_E \cos \delta\)
3. V=\(B_E \cos \delta, H=B_E \sin \delta\)
4. V=\(B_{E^{\prime}} H=B_E \tan \delta\)

Answer: 2. V=\(B_E \sin \delta, H=B_E \cos \delta\)

Here \(B_E\)= net magnetic field and H and V are the horizontal and vertical components of \(B_E\).

⇒  \(\delta \) is the angle between \(\mathrm{B}_{\mathrm{E}}\) and \(\mathrm{H}\).

From figure we have, H =\(B_E \cos \delta \)

V =\(B_E \sin \delta \)

H =\(B_E \cos \delta\)

Concept in \(\triangle\) A O B, \(\cos \delta=\frac{H}{B_E}\)

⇒ \(\mathrm{H}=B_E \cos S\)

And, in\( \triangle \mathrm{OBC}\),

⇒ \(\sin \delta =\frac{V}{B_E}\)

V =\(B_E \sin \delta\) .

Question 12. If \(\theta_1\) and \(\theta_2\) are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip is given by:

1. \(\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2\)
2. \(\tan ^2 \theta=\tan ^2 \theta_1+\tan ^2 \theta_2\)
3. \(\cot ^2 \theta=\cot ^2 \theta_1-\cot ^2 \theta_2\)
4. \(\tan ^2 \theta=\tan ^2 \theta_1+\tan ^2 \theta_2\)

Answer: 1. \(\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2\)

⇒ \(B_H\) and \(B_V\) be the Horizontal and vertical component of magnetic field B

⇒ \(\tan \theta =\frac{B_V}{B_H}\)

⇒ \(\cot \theta=\frac{B_H}{B_V}\)  → Equation 1

Again we have,\( B_1 =B_H \cos \theta^{\prime} \)

⇒ \(B_2 =B_H \sin \theta^{\prime}\)

⇒ \(\tan \theta_1 =\frac{B_V}{B^{\prime}}=\frac{B_V}{B_H \cos \theta^{\prime}} \)

⇒ \(\cot \theta_1 =\frac{B_H \cos \theta^{\prime}}{B_V}\)

Similarly, \(\tan \theta_2=\frac{B_V}{B_H} and \quad \cot \theta_2=\frac{B_H \sin \theta^{\prime}}{B_V}\)  →  Equation 2

From eq. (1) and (2)

⇒ \(\cot ^2 \theta_1+\cot ^2 \theta_2=\frac{B_H^2 \cos ^2 \theta^{\prime}}{B_V^2}+\frac{B_H^2 \sin ^2 \theta^{\prime}}{B_V^2}\)

∴ \(\cot ^2 \theta_1+\cot ^2 \theta_2=\cot ^2 \theta\)

Question 13. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It:

1. will become rigid showing no movement
2. will stay in any position
3. will stay in North-South direction only
4. will stay in the East-West direction only.

Answer: 2. Will stay in any position

A compass needle when allowed to move in a horizontal plane is taken to a geomagnetic pole. It will stay in any position at the geomagnetic north and south pole.

Question 14. A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a period of 2s in Earth’s horizontal magnetic field at 24 μ. When a horizontal field at 18 μ is produced opposite to the earth’s field by placing a current-carrying wire, the new time period of the magnet will be:

1. 1s
2. 2s
3. 3s
4. 4s

Answer: 2. 2s

Period of vibrating magneto-meter is,

⇒ \(\mathrm{T} =2 \pi \sqrt{\frac{I}{M \times B_H}} \)

T \(\propto \sqrt{\frac{1}{B_H}} \)

⇒ \(\frac{T_1}{T_2} =\sqrt{\frac{\left(B_H\right)_1}{\left(B_H\right)_2}} \)

⇒ \(\frac{2}{T_2} =\sqrt{\frac{18}{24}} \)

T =2.35=2 \(\mathrm{~s}\)

Question 15. Due to the earth’s magnetic field, charged cosmic ray particles:

1. can never reach the poles
2. can never reach the equator
3.  require less kinetic energy to reach the equator than the poles
4. require greater kinetic energy to reach the equator than the poles.

Answer: 4. require greater kinetic energy to reach the equator than the poles.

Since the earth’s magnetic field is stronger near the poles, so, charged particles of cosmic rays are deflected away from the equator.

Since, F = B.qv and K.E \(\propto\) v, they must have a higher K.E in order to reach the equator.

Question 16. An iron rod of susceptibility 599 is subjected to a magnetizing field of 1200 A \(\mathrm{m}^{-1}\). The permeability of the material of the rod is (Take, \(\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}\)):

1. \(80 \times 10^{-5} \mathrm{~T} \mathrm{~mA}^{-1}\)
2. \(2.4 \pi \times 10^{-5} \mathrm{~T} \mathrm{~mA}^{-1}\)
3. \(2.4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}\)
4. \(2.4 \pi \times 10^{-4} \mathrm{~T} \mathrm{~mA}^{-1}\)

Answer: 4. \(2.4 \pi \times 10^{-4} \mathrm{~T} \mathrm{~mA}^{-1}\)

Given, \(\chi\)=599

Magnetic field, B=1200 \(\mathrm{Am}^{-1}\)

Permeability of the material of the rod,

⇒ \(\mu_m =\mu_0(1+\chi)\)

= 4 \(\pi \times 10^{-7}(1+599) \)

= 2.4 \(\pi \times 10^{-4} \mathrm{Tm} \mathrm{A}^{-1}\)

Question 17. The magnetic moment of a diamagnetic atom is:

1. much greater than one
2. one
3. between zero and one
4. equal to zero

Answer: 4. equal to zero

The magnetic moment of a diamagnetic atom is Related Theory equal to zero

Question 18. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence, the rod gains gravitational potential energy. The work required to do this comes from:

1. the lattice structure of the material of the rod
2. the magnetic field
3. the current source
4. the induced electric field due to the changing field.

Answer: 3. the current source

The energy of the current source will be converted into the potential energy of the rod.

Question 19. The magnetic susceptibility is negative for:

1. paramagnetic material only
2. ferromagnetic material only
3. paramagnetic and ferromagnetic materials
4. diamagnetic material only.

Answer: 4. diamagnetic material only.

Magnetic susceptibility = \(\chi\)

It is negative for diamagnetic material only

Question 20. There are four lightweight rod samples A, B, C, and D respectively suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted:

(1) A is freely repelled

(2) B is freely attracted

(3) C is strongly attracted

(4) D remains unaffected

Which one of the following is true?

1. C is of diamagnetic material
2. D is of a ferromagnetic material
3. A is of non-magnetic material
4. B is of a paramagnetic material

Answer: 4. B is of a paramagnetic material

(1) Paramagnetic material will be freely attracted,

(2) Diamagnetic material will be freely repelled,

(3) Ferromagnetic material will be strongly attracted

Question 21. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is:

1. repelled by both the poles
2. repelled by the North Pole and attracted by the South Pole
3. attracted by the North Pole and repelled by the South Pole
4. attracted by both poles.

Answer: 1. repelled by both the poles

If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is repelled by both poles.

Question 22. Curie temperatures are the temperature above which:

1. ferromagnetic material becomes paramagnetic material
2. paramagnetic material becomes diamagnetic material
3. paramagnetic material becomes ferromagnetic material
4. ferromagnetic material become diamagnetic material

Answer: 1. ferromagnetic material becomes paramagnetic material

Curie temperature is that temperature above which a ferromagnet becomes paramagnet

Question 23. Nickel shows ferromagnetic properties at room temperature. If the temperature is increased beyond Curie’s temperature, then it will show:

1. anti ferromagnetism
2. no magnetic property
3. diamagnetism
4. paramagnetism.

Answer: 4. paramagnetism.

Above curie’s temperature, ferromagnetic material behaves as paramagnetic material.

Question 24. If the magnetic dipole of an atom of diamagnetic material, and paramagnetic material are denoted by \(\mu_d, \mu_p,\mu_f\) respectively, then:

1. \(\mu_d=0 and \mu_p \neq 0\)
2. \(\mu_d \neq 0 and \mu_p=0\)
3. \(\mu_p=0 and \mu_f \neq 0\)
4. \(\mu_d \neq 0 and \mu_f \neq 0\).

Answer: 1. \(\mu_d=0 and \mu_p \neq 0\)

∴ \(\mu_{\mathrm{P}} \neq \mu_f \neq 0 \text { and } \mu_d\)=0

Question 25. A diamagnetic material in a magnetic field moves:

1. from stronger to the weaker parts of the field
2. from weaker to the stronger parts of the field
3. perpendicular to the field
4. in none of the above directions

Answer: 1. from the stronger to the weaker parts of the field

A diamagnetic material in a magnetic field moves from the stronger to the weaker part of the field.

Question 26. According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to:

1. \(\frac{1}{T}\)
2. T
3. \(\frac{1}{T^2}\)
4. \(T^2\)

Answer: 1. \(\frac{1}{T}\)

From Curie Law,\(\chi_m \propto \frac{1}{T}\)

Question 27. In which type of material the magnetic susceptibility does not depend on temperature?

1. Diamagnetic
2. paramagnetic
3. Ferromagnetic
4. Ferrite

Answer: 1. Diamagnetic

Material’s magnetic susceptibility is a measurement of how easily a specimen of that material may be magnetized in a magnetic field. Magnetic susceptibility \((\chi_m)\) of a diamagnetic substance is temperature independent.

Question 28. For protecting sensitive equipment from the external magnetic field, it should be:

1. placed inside an aluminum can
2. placed inside an iron can
3. wrapped with insulation around it when passing a current through it
4. surrounded with fine copper sheets.

Answer: 2. placed inside an iron can

Iron is a ferromagnetic substance. Since ferromagnetic substances have no magnetic lines of force. Hence, equipment can be protected by placing it within a ferromagnetic substance container. As a result, it’s kept in an iron can.

Question 29. A diamagnetic substance is brought near a strong magnet, then it is:

1. attracted by a magnet
2. repelled by a magnet
3. repelled by North pie and attracted by South pole
4. attracted by the North Pole and repelled by the South Pole.

Answer: 2. repelled by a magnet

When diamagnetic substances are placed in the magnetic field of a powerful magnet, they are either weakly magnetized in the opposite direction of the field or repelled by the powerful magnet.