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Tissues Multiple Choice Questions

Question 1. At maturity, the sieve plates become impregnated with

1. Callose
2. Lignin
3. Suberin
4. Pectin

Answer: 1. Callose

Sieve plates are formed by two adjoining end walls of neighboring sieve elements of the sieve tube of phloem. At maturity, they become impregnated with a callose pad, which may be seasonal callose (for only a limited unfavorable period) or definitive (formed permanently in functionless old sieve tubes).

Question 2. Which of the following is not a part of this? epidermal tissue system?

1. Companion cells
2. Guard cells
3. Root hairs
4. Subsidiary cells

Answer: 1. Companion cells

Companion cells are characteristic elements of phloem tissue associated with the sieve tubes in the angiosperms. Thus, they are the part of vascular tissue system. They are absent in pteridophytes and gymnosperms. Guard cells and subsidiary cells are associated with stomata, found on the surface of the epidermal tissue system.

Question 3. A keratinized dead layer of skin is made of

1. Stratified columnar epithelium
2. Simple cuboidal epithelium
3. Stratified squamous epithelium
4. Simple columnar epithelium

Answer: 3. Stratified squamous epithelium

Stratified squamous epithelium is seen in the adult human body. It may be keratinized or non-keratinized. In keratinized stratified squamous epithelium, the outer few layers contain a hard waterproof protein in their cytoplasm, called keratin. It is found on the epidermis of skin, hair, and nails.

Question 4. Connective tissue is

1. Ectodermal in origin with intercellular spaces
2. Ectodermal in origin without intercellular spaces
3. Mesodermal in origin without intercellular spaces
4. Mesodermal in origin with intercellular spaces

Answer: 4. Mesodermal in origin with intercellular spaces

Connective tissue is mesodermal in origin with intercellular spaces, connective tissue cells, and fibers. The major functions of the connective tissue are binding, support, protection, transport, insulation, fat storage, and body defense.

Question 5. The striated appearance of a myofibril is due to the distribution pattern of

1. Meromyosin
2. Actin and myosin
3. Sarcoplasmic reticulum
4. Troponin and fascicles

Answer: 2. Actin and myosin

The striated appearance of a myofibril is due to the distribution pattern of both actin and myosin proteins, which are arranged as a rod-like structure, parallel to each other and also to a longitudinal axis of myofibrils. A myofibril has an alternative dark and light band. The dark band contains myosin, while the light band contains actin.

Question 6. Simple mechanical tissue devoid of lignin is

1. Chlorenchyma
2. Sclerenchyma
3. Collenchyma
4. Parenchyma

Answer: 3. Collenchyma

Collenchyma is a specialized supporting or simple mechanical permanent tissue of living cells possessing characteristically unevenly distributed thickenings of cellulose, pectin, and hemicellulose. Lignin is found in the cell walls of sclerenchyma.

Question 7. The type of tissue lining of the nasal passage, bronchioles, and Fallopian tubes is

1. Cuboidal epithelium
2. Columnar ciliated epithelium
3. Stratified squamous epithelium
4. Simple squamous epithelium

Answer: 2. Columnar ciliated epithelium

• Ciliated columnar epithelium comprises columnar cells, which have cilia on the free surface.
• This epithelium lines most of the respiratory tract and Fallopian tubes (oviducts). It also lines the ventricles of the brain and the central canal of the spinal cord.

Question 8. Healing of wounds is done in plants by the activity of

1. Lateral meristem
2. Primordial meristem
3. Intercalary meristem
4. Apical meristem

Answer: 1. Lateral meristem

Lateral meristems are present along the side of the organs, for example., vascular cambium in the plants. It is responsible for healing wounds of plants by its meristemic activity.

Question 9. The tissue whose cells are thin-walled and living, more or less isodiametric, and have intercellular spaces, is known as

1. Collenchyma
2. Sclerenchyma
3. Parenchyma
4. Chlorenchyma

Answer: 3. Parenchyma

Parenchymatous cells are isodiametric, spherical, oval, or polygonal with intercellular space or may be closely packed. They are living cells with thin cellulosic cell walls. Their main function is to store food and water. They are found in most of the plant body.

Question 10. Choose the correctly matched pair.

1. The inner lining of salivary ducts- Ciliated epithelium
2. Moist surface of buccal cavity- Glandular epithelium
3. Tubular parts of nephrons- Cuboidal epithelium
4. The inner surface of bronchioles- Squamous epithelium

Answer: 3. Tubular parts of nephrons- Cuboidal epithelium

• Cuboidal epithelium is present in the tubular parts of the nephron. It consists of short, cube-shaped cells with round nuclei located in the centre of cell. These cells often form microvilli to increase the absorptive surface area of a cell.
• Others are incorrectly matched pairs as, the moist surface of the buccal cavity and inner lining of the salivary duct is lined by compound epithelium, while the inner surface of bronchioles is lined by ciliated epithelium.

Question 11. Aerenchyma provides

1. Mechanical strength of plants
2. Extra space for photosynthesis and storage of food
3. Flexibility to plants
4. Buoyancy to hydrophytic plants

Answer: 4. Buoyancy to hydrophytic plants

Aerenchyma is a special tissue found in hydrophytes (for example., Hydrilla, Potamogeton, etc.) and some land plants (for example., petiole of banana, Canna), formed by parenchyma cells. They make a network by leaving wide air spaces (air cavities) for the gaseous exchange and make the aquatic plants light and buoyant so that they can easily afloat.

Question 12. Axons and dendrites are special features of

1. Cardiac muscle
2. Cartilage
3. Specialized epithelium
4. Neuron

Answer: 4. Neuron

Axons and dendrites are parts of neurons or nerve cells. In a neuron, dendrites may be one to several, but an axon is always one. They help in the conduction of impulses.

Question 13. A group of cells similar in structure, function, and origin is known as

1. Organ
2. Organ system
3. Tissue
4. Organism

Answer: 3. Tissue

Question 14. The tissue responsible for an increase in the girth of the stem is

1. Lateral cambium
2. Apical meristem
3. Intercalary meristem
4. None of these

Answer: 1. Lateral cambium

Question 15. Which of the following are living cells?

1. Sclereids
2. Sclerenchyma
3. Vessels
4. Collenchyma

Answer: 4. Collenchyma

Question 16. Aerenchyma tissue is found in

1. Halophytes
2. Hydrophytes
3. Xerophytes
4. Mesophytes

Answer: 2. Hydrophytes

Question 17. Which of the following is an example of permanent tissue?

1. Apical meristem
2. Parenchyma
3. Xylem
4. Both (2) and (3)

Answer: 4. Both (2) and (3)

Question 18. Simple squamous epithelium is found in

1. Liver
2. Fallopian tube
3. Lungs
4. Stomach

Answer: 3. Lungs

Question 19. Blood devoid of blood cells is called as

1. Serum
2. Plasma
3. Lymph
4. None of these

Answer: 2. Plasma

Question 20. The tissues, which join muscles to the bones are called as

1. Ligament
2. Tendons
3. Cartilage
4. Adipose tissue

Answer: 2. Tendons

Question 22. Name the type of tissue found in the iris of the eye, walls of the alimentary canal, etc.

1. Cardiac muscle
2. Smooth muscle
3. Involuntary muscle
4. Both (2) and (3)

Answer: 4. Both (2) and (3)

Question 22. The gap between two nerve cells is known as

1. Node of Ranvier
2. Synapse
3. Matrix
4. None of these

Answer: 2. Synapse

 Class 9 Science Chapter 7  Motion Very Short Answer Type Questions

Question 1. What is the importance of reference points?
Answer:

The reference point is important because it states the position of the object correctly, as motion is relative.

Question 2. Is it possible that the train in which you are sitting appears to move while it is at rest?
Answer:

Yes, if another train is moving in an adjacent line to the train we are sitting on, then it seems that our train is moving in the opposite direction.

Question 3. When is an object in motion considered to be a point object?
Answer:

If the object in motion has a dimension very small as compared to the distance travelled, then the object is considered to be a point object.

The odometer measures the displacement of the vehicle. Correct this statement.

No, the Odometer measures the distance covered by a vehicle.

Question 5. Suppose Ram throws a ball vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?
Answer:

The net displacement is zero. The distance travelled by the ball is twice the distance between positions P and Q.

Read and Learn More Class 9 Science Solutions

Question 6. A particle is moving in a circular path of radius r. What will be the displacement after half a circle?

Answer:

So After half the circle, the particle will reach the diametrically opposite point, i.e. from point A to B. Displacement after half circle = AB = r + r = 2r

Question 7. A particle moves 3m North, 4m East and finally 6m South. Calculate the displacement.
Answer:

According to the question,

Now, by applying Pythagoras theorem in \(\triangle\) ODC,

|O C| =\(\sqrt{O D^2+D C^2}\)

=\(\sqrt{A B^2+(B C-B D)^2}\)

=\(\sqrt{4^2+(6-3)^2}\)

=\(\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5 \mathrm{~m}\)

Question 8. In which condition, will the magnitude of the displacement be equal to the distance travelled by an object?
Answer:

If an object moves in a straight line from one point to another, then the magnitude of displacement and distance will be equal.

Question 9. Give one condition for the motion of a body to be uniform.
Answer:

The body should travel along a straight line path in one direction.

Question 10. Does the speedometer of a car measure its average speed?
Answer:

No, the speedometer of a car does not measure its average speed. It measures only instantaneous speed.

Question 11. What do you mean by an acceleration of 5 m/s2?
Answer:

The velocity of the body increases by 5 m/s after every second.

Question 12. A physical quantity measured is -10 m/s. Is it a speed or a velocity?
Answer:

It is a velocity because velocity can be positive, zero or negative while speed is always positive.

Question 13. A person while travelling with his car found that the magnitude of average velocity was equal to its average speed. Identify the path formed by him.
Answer:

If a person moves in a straight line, then their average velocity will be equal to the average speed.

Question 14. What is the numerical ratio of average velocity to average speed of an object when it is moving along a straight path?
Answer:

The numerical ratio of the average velocity and average speed of an object when it is moving along a straight path is 1: 1.

Question 15. A person standing at A goes to B by following any of the paths 1, 2 and 3. Which path can we use to find the average velocity?

Can a particle be accelerated?

1. Is its speed constant?
2. Is its velocity constant?

Answer: Path-1

1. Yes, a particle moving with constant speed can be accelerated.
2. No

Question 16. When is the acceleration taken as negative?
Answer:

Acceleration is taken as negative if it is in the opposite direction of motion of the object.

Question 17. A body accelerates uniformly so that its velocity changes from u to v. What will be its average velocity?
Answer:

The average velocity is \(\frac{u+v}{2}\).

Question 18. What kind of motion of a body is represented by the graphs given below?

Answer:

1. The object is at rest
2. Non-uniform motion.

Question 19. The distance-time graph for the motion of Ram and Shyam is shown alongside. Which of them has greater acceleration? Justify your answer.

Answer:

Both Ram and Shyam have zero acceleration as they are moving with constant velocity. Since the distance-time graph is a straight line.

Question 20. Figure shows the displacement-time graph of four children A, B, C and D. Which child has the highest velocity?

Answer:

As we know, velocity = slope of the displacement-time graph. Here, the slope of child B is greater than all other children. Hence, child B has the highest velocity.

Question 21. From the given v-t graph (see figure), what can be inferred?

Solution:

From the given v-t graph, it is clear that the velocity of an object does not change with time, i.e. the object is in a uniform motion.

Question 22. How do we measure the magnitude of the displacement from a v-t curve?
Answer:

By measuring the area under the v-t curve along the time axis.

Question 23. Give one example of uniform circular motion.
Answer:

A satellite in a circular orbit around the Earth is an example of uniform circular motion.

Question 24. Is the motion of a body uniform or accelerated, if it goes round the sun with constant speed in a circular orbit?
Answer:

In this case, the motion of a body is accelerated as its velocity changes due to the change in direction.

Question 25. Why is the motion of a circulating fan non-uniform?
Answer:

The motion of a circular fan is non-uniform because the direction of motion changes at every point.

Question 26. Give an expression for the speed of an athlete if he takes time t to go around a circular track, of radius r.
Answer:

v=\(\frac{2 \pi r}{t}\)

Question 27. Find the angular velocity of a satellite which revolves in a circular orbit of radius 35000 km and completes one round in 12 h.
Answer:

Angular velocity =\(\frac{\text { Angle subtended at centre }}{\text { Time taken }}\)

= \(\frac{2 \pi}{12}\)

= \(\frac{\pi}{6} \mathrm{rad} / \mathrm{h}\)

 Class 9 Science Chapter 7  Motion Short Answer Type Questions

Question 1. (1) What is motion?

(2) State the types of motion.

(3) Write the unit of acceleration.

Answer:

(1) If a body changes its position concerning time and its surroundings, then it is said to be in motion.

(2) Generally, there are two types of motion:

(a) Uniform motion (b) Non-uniform motion

(3) The SI unit of acceleration is m/s².

Question 2. A cow and a bird both A 15 m B travelled from point A to point B. The cow travelled in a straight line but the bird travelled along the curved path as shown alongside:

1.  What is the distance travelled by the cow?
2. What is the distance travelled by the bird?
3.  Which one of them has more displacement?

Answer:

1. Distance travelled by the cow = 15 m
2. Distance travelled by the bird = 30 m
3. Displacement of the cow = 15 m

Displacement of the bird = 15 m

So, cows and birds have the same displacement.

Question 3. Give an example of a body which may appear to be moving for one person and stationary for the other.
Answer:

For the passengers in a moving bus, the trees, buildings as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in a forward direction.

But at the same time, each passenger in a moving bus or train observes that his fellow passengers are at rest. Thus, we can tell that motion is relative.

Question 4. How can we describe the location of an object?
Answer:

To describe the location of an object we need to specify a reference point called origin, e.g. suppose that a library in a city is 2 km North of railway station. We have specified the position of the library with respect to the railway station, i.e. in this case, the railway station acts as the reference point.

Question 5. Differentiate between distance and displacement.
Answer:

Differences between distance and displacement are as given below:

Question 6. Express average velocity when the velocity of a body changes at a non-uniform rate and a uniform rate.
Answer:

When the velocity of a body changes at a non-uniform rate, then its average velocity is calculated by dividing the net displacement covered by the total time taken.

Average velocity =\(\frac{\text { Net displacement }}{\text { Total time taken }}\)

In case, the velocity of a body changes at a uniform rate, the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time, i.e.

Average velocity= \(\frac{\text { Initial velocity }+ \text { Final velocity }}{2}\)

Question 7. Give one example each of type of motion when

1. acceleration is in the direction of motion.
2.  acceleration is against the direction of motion.

Answer:

1. A train starts moving from a station.
2. Brakes applied to a moving car.

Question 8. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Answer:

When a stone is thrown vertically upwards, then it has some initial velocity (u). As the stone goes up its velocity decreases (since it is moving against gravity) and at the highest point, i.e. maximum height,) its velocity becomes zero.

Let the stone take time the t second to reach the highest point. After that stone begins to fall (with zero initial velocity) and its velocity goes on increasing (since it is moving with gravity) and it reaches its initial point of projection with the velocity (—u) at the same time (with which it was thrown). So,

Question 9. A biker rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east and finally 100 m north. Draw the path of motion of the biker. What distance did he cover? What was his displacement?
Answer:

(2) Distance travelled by the biker

= OA + AB + BC + CD + DE + EF + FO = 700 + 300 + 400 +600 + 1200 + 300 + 100 = 3600 m

(3) Displacement of the biker = 0.

Question 10. An object P is moving with a constant velocity for 5 min. Another object O is moving with changing velocity for 5 min. Out of these two objects, which one has acceleration? Explain.
Answer:

Acceleration =\(\frac{\text { Change in velocity }}{\text { Time taken }}\)

Since the velocity of object P is not changing or the change in velocity of the object is gro, therefore object-t Phas no acceleration. On the other hand, there is i change in velocity of the object Q, so it has acceleration.

Question 11. A car covers the first 50 km at a uniform velocity of 45 km/h and the next 50 km at 90 km/h. Find the average velocity of the car.
Answer:

Total displacement = 50 + 50 = 100 km

Total time =\(\frac{50}{45}+\frac{50}{90} =50\left(\frac{1}{30}\right)=\frac{5}{3} \mathrm{~h}\)

∴ \({\left[\text { Time }=\frac{\text { Displacement }}{\text { Velocity }}\right] }\)

Average velocity =\(\frac{\text { Total displacement }}{\text { Time taken }}=\frac{100}{5 / 3}\)

= 60 km/h

The average velocity of the car is 60 km/h.

Question 12. A girl moves at a speed of 6 km/h for two hours and with a speed of 4 km/h for the next three hours. Find the average speed of the girl and the total distance moved.
Answer:

Here, \(v_1=6 \mathrm{~km} / \mathrm{h}, t_1=2 \mathrm{~h}, v_2=4 \mathrm{~km} / \mathrm{h}, t_2=3 \mathrm{~h}\)

Total distance =\(v_1 t_1+v_2 t_2=6 \times 2+4 \times 3=24 \mathrm{~km}\)

Average speed =\(\frac{\text { Total distance }}{\text { Total time }}\)

= \(\frac{24}{2+3}=4.8 \mathrm{~km} / \mathrm{h}\)

Question 13. A person travelling in a bus noted the timings and the corresponding distances as indicated on the km stones.

1. Name this type of table.
2. What conclusion do you draw from this data?

Answer:

1.  It is a distance-time table.
2. According to the given table, we conclude that the given motion is uniform motion as in every 15 the body covers 10 km.

Question 14. Represent the given data graphically.

 

Answer:

From the given distance-time table, the graph can be drawn as

Question 15. Two trains A and B start moving at the same time. The distances travelled by them in given internals of time are shown below. State which train has uniform motion and which train has non-uniform motion.
Answer:

Since, train A travels equal distances in equal intervals of time, i.e. every 15 min, the motion of train A is uniform motion.

On the other hand, train B travels unequal distances in equal intervals of time, i.e. every 15 min, so the motion of train B is non-uniform motion.

Question 16. Make a velocity-time graph from the following displacement-time graph:

Answer:

From the graph,

velocity after 2 \(\mathrm{~s}=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}\)

Velocity after 4 \(\mathrm{~s}=\frac{20}{4}=5 \mathrm{~m} / \mathrm{s}\) and velocity after 6 \(\mathrm{~s}=\frac{30}{6}=5 \mathrm{~m} / \mathrm{s}\)

For Example. Velocity is constant, acceleration = 0

So, the velocity-time graph will be a straight line parallel to the time axis as shown below:

Question 17. Find the displacement of a body whose velocity-time graph is shown as:

Answer: 

Displacement = Area under v-t curve (i.e. the area of trapezium ABCD )

= \(\frac{1}{2}(\text { sum of parallel sides }) \times \text { height }\)

= \(\frac{1}{2}(A D+B C) \times C F=\frac{1}{2}(10+6) \times 10=80 \mathrm{~m}\)

Question 18. The graph given below is the distance-time graph of an object.

1. Find the speed of the object during the first four seconds of its journey and how long was it stationary.
2. Does it represent a real situation? Justify your answer.

Answer:

(1) Speed during first 4s = Slope of OP

⇒ \(\tan \theta=\frac{75}{4}=18.75 \mathrm{~m} / \mathrm{s}\)

From 4 s to 14 s, the distance does not change with time, so it is stationary for 10 s.

(2) No, it is not a real situation because distance travelled cannot decrease with time. Moreover, time cannot flow backwards as shown by the QR part of the graph.

Question 19. The velocity-time graph (see figure) shows the motion of a cyclist. Find

1. its acceleration,
2. its velocity and the distance covered by the cyclist in 15 seconds.

Answer:

1. From the graph, it is clear that velocity does not change with time, i.e. acceleration is zero.
2. Again from the graph, we can see that there is no change in the velocity with time, so velocity after 15 seconds will remain the same as 20 m/s.

Distance covered in 15 s = Velocity x Time = 20×15 = 300 m

Question 20. Find the total displacement of the body from the following graph

Answer:

Total displacement = Sum of areas of (\(\triangle\) A B F+ Rectangle BCGF + \(\triangle\) CDE +{ Rectangle CEIG)

=\([\frac{1}{2}(A F \times F B)+(B C \times C G)\).

⇒ \(\left.\quad+\frac{1}{2}(C E \times D J)+(C E \times E I)\right]\)

= \(\left[\frac{1}{2}(2 \times 5)+(4 \times 5)+\frac{1}{2}(4 \times 5)+(5 \times 4)\right]\)

=55 m

Question 21. The speed-time graphs of the two cars are represented by P and Q as shown below:

1. Find the difference in the distance travelled by the two cars (in m) after 4 s.
2.  Do they ever move at the same speed? If, so when?

Answer:

(1) Total distance travelled by P (area of A POR)

= 1/2 x4x 6²

= 12 m

Total distance travelled by Q (area of RQSO)

= 4×3 = 12m

Difference in the distance = 12-12=0

(2) Yes, they move with the same velocity equal to 3 m/s at time 2 s. It can be determined, by viewing the intersection point of two v-t graphs.

Question 22. Study the v-t table and answer the following:

1. What is the value of an in 0 to 15 s?
2. Predict the nature of acceleration in different parts of the graph.

Answer:

(1) a =\(\frac{\text { Change in velocity }}{\text { Time interval }}\)

=\(\frac{20-0}{15-0}=\frac{20}{15}=\frac{4}{3}\)

=1.33 \(\mathrm{~m} / \mathrm{s}^2\)

(2) The body is in uniform motion from 0 to 15 s, then non-uniform related motion from 15 s to 30 s. Here, OA = uniformly accelerated AB, BC, CD = uniformly related.

Question 23. What is the difference between uniform motion in a straight line and circular motion?
Answer:

Question 24. Study the following distance-time graph and compare the velocity of the two objects

Answer:

Since A and B have parallel slopes for distance-time graphs, their velocities are equal. However, A starts from the 10 m mark and B starts from the origin, (i.e. A is ahead of B by 10 m)

Question 25. What do the graphs (1) and (2) shown here indicate?

Answer:

Graph (1) indicates that a body has a certain initial speed and then its speed decreases to zero.

Graph (2) shows that a body first retards and then accelerates non-uniformly.

Question 26. Can an object be accelerated, if it is moving with constant speed? Justify your answer with an example.

Or Can a body have a constant speed and still be accelerating? Give an example.

Or Explain, how is it possible for an object to move with a constant speed but with uniform acceleration.

Answer:

An object moving with a constant speed can be accelerated, if its direction of motion changes, For Example. An object moving with a constant speed in a circular path has acceleration because its direction of motion changes continuously. Its acceleration is always directed toward the centre of the circular path.

Question 27. A bridge is 500 m long. A 100 m long train crosses the bridge at a speed of 30 m/s. Find the time taken by train to cross it.
Answer:

Total length of path covered by train = 500 + 100 =600 m

Speed of train = 30 m/s

Time taken by train to cover the bridge

Time taken by train to cover the bridge

= \(\frac{\text { Distance }}{\text { Speed }}=\frac{600}{30}=20 \mathrm{~s}\)

Question 28. When two bodies move uniformly towards each other, then the distance between them decreases by 8 m/s. If both the bodies move in the same direction at the same speeds, then the distance between them increases by 4 m/s. What are the speeds of two bodies?
Answer:

Let u and v be the speeds of two bodies. According to the question, when two bodies move towards each other the distance between them decreases by u + v = 8 and when they are moving in the same direction,

then the distance between them increased by u – v = 4

Solving the above equations, we get 2u=12 ⇒ u — G m/s

and v =8 —u =8 -6 = 2 m/s The speed of two bodies are 6 m/s and 2 m/s.

Question 29. A fan rotates at 100 rpm. Find its angular velocity and linear velocity, if the tip of its blades is 0.20 m away from the axis of rotation.
Answer:

Given, angular velocity, \(\omega\)=100 rpm=100 \(\times \frac{2 \pi}{60}\)

= \(\frac{10 \pi}{3} \mathrm{rad} / \mathrm{s}\)

Radius, r=0.20 \(\mathrm{~m}\)

Linear velocity =\(\text { Radius } \times \text { Angular velocity }\)

v=r \(\omega\)

Linear velocity =0.20 \(\times \frac{10 \pi}{3}\)

= \(\frac{2 \pi}{3} \mathrm{~m} / \mathrm{s}\)

Question 30. An object is moving with a uniform speed in a circle of radius r. Calculate the distance and displacement,

1. when it completes half the circle,
2. when it completes full circle,
3. what type of motion does the object possess?

Answer:

(1) When an object completes half the circle, then the distance travelled by an object

= \(\frac{1}{2} \times \text { circumference of a circle }\)

= \(\frac{1}{2} \times 2 \pi r=\pi r\)

(2) When an object completes a full circle, then the distance travelled by an object = \(2 \pi\) r

Displacement travelled by an object = 0 [since initial and final positions are the same]

(3) The direction of motion of an object changes continuously in the circle, hence its velocity changes and its motion is accelerated motion.

Question 31. An athlete runs on a circular track of a radius of 50 m from the starting point A with a constant speed. It takes 50 seconds for him to reach point B which is at the other end of the diameter drawn from A. Calculate the distance covered, displacement and speed.
Answer:

Distance covered by the athlete =\(\pi r\)

= \(\frac{22}{7} \times 50=157.14 \mathrm{~m}\)

Displacement of the athlete

=2 r=2 \(\times 50=100 \mathrm{~m}\)

The speed of the athlete is given by

= \(\frac{\text { Distance }}{\text { Time }}\)

= \(\frac{\pi r}{50}\)

= \(\frac{22}{7} \times \frac{50}{50}=3.14 \mathrm{~m} / \mathrm{s}\)

 Class 9 Science Chapter 7  Motion Long Answer Type Questions

Question 1. Give one example of each of the following situations:

1. Uniformly accelerated motion.
2. Motion with uniform retardation.
3. Accelerated motion with a uniform magnitude of velocity.
4. Motion in a direction with acceleration in a perpendicular direction.
5. Motion in which the v-t graph is a horizontal line parallel to the X-axis.

Answer:

1. An object dropped down from a height towards the surface of the earth.
2. The object thrown up with a velocity retards uniformly.
3. Uniform circular motion, say planetary motion.
4. A bullet fired horizontally from a rifle, has acceleration in a downward direction (due to gravity).
5. A car moving with uniform velocity along a straight line

Question 2. The position-time graphs of two objects A and B in three different situations for a particular duration are shown below:

1. In which situation they are moving in opposite directions?
2. Is the velocity of object A positive or negative in the situation (2)?
3. Are they crossing each other in any situation (s)? If so, why?In which situation the distance between them will remain the same?

Answer:

1. In graph (3), the distance between them will be the same.
2. In the graph (1), they are moving in opposite directions.
3. The velocity of object A is positive in situation (1) because the slope of the s-t graph is positive.
4. Yes, in situations (1) and (2) they are crossing each other because in these two cases, the s-t graph of A and B intersect each other.

Question 3. On a 100 km track, a train travels the first 30 km at a uniform speed of 30 km/h. How fast must the train travel the next 70 km, so as to average 40 km/h for the entire trip?
Answer: 

Total time, t=\(t_1+t_2=\frac{30 \mathrm{~km}}{30 \mathrm{~km} / \mathrm{h}}+\frac{70 \mathrm{~km}}{x(\text { say })}\)

t=1+\(\frac{70}{x}=\left(\frac{x+70}{x}\right) \mathrm{h}\)

Now, average speed, v=\(\frac{s}{t}=40 \mathrm{~km} / \mathrm{h}\)

According to question, \(\frac{100}{\underline{x+70}}=40\)

10=4\(\left(\frac{x+70}{x}\right)^x\)

10 x=4 x+280

6 x=280

x=46.67 \(\mathrm{~km} / \mathrm{h}\)

The speed for the next 70 km to maintain an average of 40 km/h is 46.67 km/h.

Question 4. Study the speed-time graph of a body given here and answer the following questions:

1. What type of motion is represented by OA?
2. What type of motion is represented by AB?
3. What type of motion is represented by BC?
4. Find out the acceleration of the body.
5. Calculate the retardation of the body.
6. Find out the distance travelled by the body from A to B.

Answer:

1. OA is a straight line graph between speed and time and it is sloping upward from O to A. Here, OA represents uniform acceleration.

2. AB is a straight line graph between speed and time which is parallel to the time axis (X-axis). So, AB represents uniform speed. There is no acceleration from A to B.

3. BC is a straight line graph between speed and time which is sloping downwards from B to C. Therefore, BC represents uniform retardation or negative acceleration.

4. Acceleration of the body as we see from graph line OA represents it.

So, the slope of the velocity-time graph OA will give the acceleration of the body.

Thus, acceleration = slope of line OA=\(\frac{A D}{O D}\)

We have, AD=6 \(\mathrm{~m} / \mathrm{s} and OD=4 \mathrm{~s}\)

So, acceleration =\(\frac{6 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}=1.5 \mathrm{~m} / \mathrm{s}^2\)

5. The slope of line graph BC represents the retardation of the body.

So, retardation = slope of line BC = \(\frac{B E}{E C}\)

We have, BE= 6 \(\mathrm{~m} / \mathrm{s}\),

E C=16-10=6 \(\mathrm{~s}\)

Retardation =\(\frac{6 \mathrm{~m} / \mathrm{s}}{6 \mathrm{~s}}=1 \mathrm{~m} / \mathrm{s}^2\)

6. The distance travelled by the body is equal to the area enclosed between the speed-time graph and the time axis.

Distance travelled from A area under the line AB and the time axis = area of rectangle DABE — DA x DE.

Here, DA = 6 m/s and DE = 10 — 4 = 6 s Distance travelled from A to B=6×6 = 36 m

Question 5. The v-t graph of an object is shown below:

1. Identify the type of motion by lines OA and BC.
2. Velocity at t = 8s.
3. Calculate acceleration
   1.  between 3rd and 10th second.
   2.  last 2s.

Answer:

1. OA represents uniformly accelerated motion, and BC represents uniformly retarted motion.

2.  At t =8 s, the velocity of the object is 4.6 m/s.

3. (1) Between 3rd and 10th second, a – 0.

(2) During last 2s,

Acceleration, a=\(\frac{\text { change in velocity }}{\text { time interval }}\)

=\(\frac{0-4.6}{12-10}=\frac{-4.6}{2}=-\frac{4.6}{2}=-2.3 \mathrm{~m} / \mathrm{s}^2\)

Question 6. An insect moves along a circular path radius of 10 cm with a constant speed. It takes 1 minute to move from a point on the path to the diametrically opposite point, find

1. the distance covered,
2. the speed,
3.  the displacement and
4. the average velocity. Suppose the insect was at A initially and it moves along ACB to reach the diametrically opposite point B in 1 min.

Answer:

1. The distance moved in 1 min = nr = 3.14×10 =31.4 cm
2. Speed =\(\frac{\text { Distance }}{\text { Time }}=\frac{31.4}{1}=31.4 \mathrm{~cm} / \mathrm{min}\)
3.  Displacement, AB = 2r = 2 x 10 = 20cm
4.  Average velocity,

• • \(v_{\mathrm{av}} =\frac{\text { Displacement }}{\text { Time }}\)
  • =\(\frac{20 \mathrm{~cm}}{1 \mathrm{~min}}=20 \mathrm{~cm} / \mathrm{min}\)

Question 7. The driver of train A travelling at a speed of 54 km/h applies brakes and retards the train uniformly. The train stops in 5 s. Another train B is travelling on the parallel with a speed of 36 km/h.

Its driver applies the brakes and the train retards uniformly, train B stops in 10 s. Plot speed-time graphs for both trains on the same axis. Which of the trains travelled farther after the brakes were applied?

Answer:

For train A, the initial velocity,

u=54 \(\mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)

Final velocity, v=0 and time, t=5 s

For train B, u=36 \(\mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}\)

⇒ \(\nu\)=0, t=10 \(\mathrm{~s}\)

Speed-time graphs for trains A and B are shown

Distance travelled by train A= Area under straight line graph

RS = area of \(\triangle\) ORS

= \(\frac{1}{2} \times O R \times O S\)

= \(\frac{1}{2} \times 15 \times\) 5=37.5 m

Distance travelled by train B Area under PQ

= \(\frac{1}{2} \times O P \times OQ \)

= \(\frac{1}{2} \times 10 \times 10=50 \mathrm{~m}\)

Distance travelled by train B is more than train A.

Motion Multiple Choice Questions

Question 1. The minute hand of a wall clock is 14 m long. The distance and displacement covered by the tip of the minute hand from 9: 00 am to 9: 30 am, are

1. 88 cm, 28 cm
2. 28 cm, 44 cm
3. 28 cm, 88 cm
4. 44 cm, 28 cm

Answer: 4. 44 cm, 28 cm

Distance =\(\pi r=\frac{22}{7} \times 14=44 \mathrm{~cm}\)

Displacement = 2r = 2×14 = 28 cm

Question 2. If the velocity of an object decreases with time, then it is called

1. retardation
2. de-acceleration
3. negative acceleration
4. All of the above

Answer: 4. All of the above

When the velocity of an object decreases with time, then it is said to be retardation, de-acceleration or negative acceleration.

Question 3. A speedometer is a device which is used to measure

1. average speed
2. average acceleration
3. instantaneous speed
4. instantaneous acceleration

Answer: 3. instantaneous speed

A speedometer is used to measure instantaneous speed.

Question 4. An object travels 24 m with a speed of 4 m/s and then another 32 m with a speed of 8 m/s. The average speed of the object is

1. 6 m/s
2. 4 m/s
3. 5.6 m/s
4. 6.5 m/s

Answer: 3. 5.6 m/s

Time travel to cover 24 m with a speed of 4 m/s,

Time Travel, \(t_1=\frac{24}{4}=6 \mathrm{~s}\)

Time travel to cover 32 \(\mathrm{~m}\) with speed 8 \(\mathrm{~m} / \mathrm{s}\),

⇒ \(t_2=\frac{32}{8}=4 \mathrm{~s}\)

⇒ \(\text { Average } \text { speed }=\frac{\text { Total distance }}{\text { Total time }}=\frac{24+32}{t_1+t_2}\)

= \(\frac{56}{6+4}=5.6 \mathrm{~m} / \mathrm{s}\)

Question 5. A body moves on a circular path with a uniform speed of 6 m/s. If the body completes one revolution in 22 seconds, then the radius of the circular path is

1. 21 m
2. 7 m
3. 14 m
4. 28 m

Answer: 1. 21 m

Given, speed, v=6 m/s, time period, T=22 s

v=\(\frac{2 \pi r}{T}\)

r =\(\frac{v T}{2 \pi}=\frac{6 \times 22 \times 7}{2 \times 22}=21 \mathrm{~m}\)

Question 6. A bridge is 400 m long. A 150 m long train crosses the bridge at a speed of 50 m/s. Time taken by the train to cross it.

1. 5 s
2. 8 s
3. 6 s
4. 11 s

Answer: 4. 11 s

Total length of path covered by train = 400 + 150 = 550m

Time taken by train to cross the bridge

= \(\frac{\text { Total distance }}{\text { Speed }}=\frac{550}{50}\)=11 s

Question 7. If a body is moving on a circular path of radius 21 cm with a velocity of 2 m/s, then the time taken by the body to complete half revolution is

1. 11s
2. 22 s
3. 44 s
4. 33 s

Answer: 4. 33 s

Given, radius, r=21 cm and velocity, v=2 m/s

Time period (T)=\(\frac{2 \pi r}{v}\)

Half-time period (T)=\(\frac{\pi r}{v}=\frac{\frac{22}{7} \times 21}{2}=33 \mathrm{~s}\)

Question 8. A velocity-time graph for a moving object is found to be a curved line, then its acceleration is

1. constant
2. variable
3. zero
4. None of these

Answer: 2. variable

Since the velocity-time graph for a moving object is a curved line, this means that the object is moving with non-uniform accelerated motion, therefore the body is moving with variable acceleration.

Question 9. In the following figure of the velocity-time graph for the motion of the body, the total distance covered by the body from 3 s to 7 s is

1. 28 m
2. 56 m
3. 14 m
4. 35 m

Answer: 1. 28 m

Total distance moved by the body from 3 s to 7 s

= Area of shape ABCD =\(\frac{(A B+D C) \times A D}{2}\)

=\(\frac{(4+10)(7-3)}{2}=\frac{14 \times 4}{2}=28 \mathrm{~m}\)

Question 10. The speed-time graph in the following figure indicates the body starts from a certain initial velocity and accelerates

1. the body has a certain initial speed and then its speed decreases to zero
2. the body has constant acceleration
3. None of the above

Answer: 2. The body has a certain initial speed and then its speed decreases to zero.

Question 12. In a uniform circular motion, angular displacement is measured in

1. radian
2. steradian
3. meter
4. radian/second

Answer: 1. radian

Angular displacement is measured in radians.

Question 13. When two bodies move uniformly towards each other, then they cross each other at the speed of 10 m/s. If both the bodies move in the same direction, then they cross each other at the speed of 6 m/s. The speed of both bodies are

1. 8 m/s, 2 m/s
2. 8 m/s, 4 m/s
3. 6 m/s, 2 m/s
4. 6 m/s, 4 m/s

Answer: 1. 8 m/s, 2 m/s

Let \(v_1\) and \(u_2\) be the speeds of two bodies, then

⇒ \(u_1+u_2\)=10

(when they are moving in the opposite direction)  →  Equation 1

⇒ \(u_1-u_2\)=6

(when they are moving in the same direction)  →  Equation 2

Solving Eqs. (1) and (2), we get

∴  \(u_1=8 \mathrm{~m} / \mathrm{s} \text { and } u_2=2 \mathrm{~m} / \mathrm{s}\)

Question 14. A cyclist appears to be stationary with respect to another moving cyclist, then both cyclists are moving

1. In the opposite direction with the same velocity
2. In the same direction with the same velocity
3. In the perpendicular direction with any velocity
4. In the perpendicular direction with the same velocity

Answer: 2. In the same direction with the same velocity

When both cyclists are moving in the same direction with the same velocity, then they appear to be stationary.

Question 15. The ratio of displacement and distance is

1. <1
2. >1
3. >1
4. None of these

Answer: 2. >1

Question 16. A body moves in a circular path radius of 10 cm. If it completes two revolutions along the circular path, then the displacement of a body is

1. zero
2. 40 cm
3. 20 cm
4. None of these

Answer: 1. zero

Question 17. If a car moves along a straight line path, that covers an equal distance in equal intervals of time, it is said to be

1. In non-uniform motion
2. In uniform motion
3. Both (1) and (2)
4. None of the above

Answer: 2. In uniform motion

Question 18. The SI unit of acceleration is

1. m/s
2. m²/s
3. m/s²
4. s/m

Where m is the meter and s is the second.

Answer: 3. m/s²

Question 19. Which of the following is an example of a uniform acceleration?

1. The motion of a free-falling body
2. The motion of a ball rolling down on an inclined plane
3. Both (1) and (2)
4. None of the above

Answer: 3. Both (1) and (2)

Question 20. Which of the following statements is correct?

1. The average speed of an object can never be zero
2. The average velocity of a moving object can be zero
3. The motion of the moon and the earth is a uniform circular motion
4. All of the above

Answer: 4. All of the above

Question 21. Uniform circular motion

1. Is an accelerated motion
2. Is an unaccelerated motion
3. Have uniform velocity
4. None of the above

Answer: 1. Is an accelerated motion

Question 22. A wheel completes 2000 revolutions to cover the 9.5 km distance, then the diameter of the wheel is

1. 1.5 m
2. 1.5 cm
3. 7.5 cm
4. 7.5 m

Answer: 1. 1.5 m

Class 9 Science Chapter 8 Force And Laws Of Motion Very Short Answer Type Questions

Question 1. Give two examples of the effects of force.
Answer:

A toy car starts moving when pushed.

The shape of the dough ball changes when rolled.

Question 2. Apart from changing the magnitude of velocity of an object or changing the direction of motion of an object, what other changes can force bring on an object?
Answer:

Force can change the state of rest to the state of motion and vice-versa. It can also change the shape and position of the body.

Question 3. Why friction does not pull a ball backward if no force is pushing or pulling it?
Answer:

Friction comes into action only in the presence of an applied force.

Question 4. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the frictional force that will be exerted on the cabinet?
Answer:

The frictional force is 200 N but in the opposite direction.

Question 5. What do you mean by resultant force?
Answer:

When two or more forces act on a body simultaneously then the single force which produces the same effect as produced by all the other forces acting together is known as the resultant force.

Question 6. While riding on the bicycle, if we stop peddling, why does the bicycle begin to slow down?
Answer:

The bicycle begins to slow down because of the force of friction acting in the opposite direction.

Question 7. While performing a practical, student A kept the wooden block on a polished wooden surface and measured the minimum force required to pull it as F₁ while student B kept the wooden block on a rough surface, covered with sand, and measured the minimum force required to pull it as F₂. They repeated the experiment five times. Which student applied more force to pull the block?
Answer:

Student B applied more force to pull the block because the friction force acting on the block placed on a rough surface was of large magnitude and opposite to the direction of force applied.

Read and Learn More Class 9 Science Solutions

Question 8. What did Galileo conclude on the basis of his experiments on the motion of objects?
Solution:

A body continues to move with the same velocity if no unbalanced force acts on it.

Question 9. Velocity-time graph of a moving particle of mass 1 kg

Is any force acting on the body? Justify your answer.

Answer:

The velocity of the body is uniform, thus acceleration is zero. Hence, no unbalanced force acts on the body.

Question 10. Why should we wear safety seat belts in a car?
Answer:

Due to our inertia, we may get hurt when the car suddenly starts or stops. Safety belts protect us by slowing down the motion in such situations.

Question 11. A water tanker filled up to — of its height is moving with a uniform speed. On sudden application of the brake, in which direction the water in the tanker would flow?
Answer:

On the sudden application of brake, the tanker will come in the state of rest but the water remains in the state of motion, so the water will move forward.

Question 12. Why do we tend to fall outwards when a car takes a steep turn?
Answer:

We tend to fall outwards when a car takes a steep turn due to inertia of direction.

Question 13. Name the agency which when applied to a body is directly proportional to the rate of change of momentum which it produces in the body.
Answer:

The force applied to a body is directly proportional to the rate of change of momentum.

Question 14. Two similar vehicles are moving with the same velocity on the roads such that one of them is loaded and the other is empty. Which of the vehicles will require a larger force to stop it? Give reasons.
Answer:

Both have the same velocity but one of them is loaded, so its momentum is greater than the other. Thus, for the loaded vehicle greater change of momentum, so larger force will be required to stop it.

Question 15. Athletes in pole jump events fall on a cushioned surface and not on the floor. Why?
Answer:

The change in momentum occurs over a long duration of time which reduces the force acting on the athlete and he does not hurt.

Question 16. A passenger in a moving train tosses a coin which falls behind him. From this incident, what can you predict about the motion of the train?
Answer:

If the coin falls behind the passenger that means the train is accelerated. When the coin is tossed, then it has the same velocity as that of the train but during the time it is in the air its velocity becomes less than that of the train (because the train is accelerated), so it falls behind the passenger.

Question 17. An athlete always runs for some distance before taking a jump. Why?
Answer:

When an athlete runs for some distance, then he increases his inertia of motion. At the jumping time, when he takes a jump, his increased inertia of motion carries him over a long distance or height.

Question 18. Name the physical quantity which corresponds to the rate of change of momentum.
Answer:

Force (according to Newton’s second law rate of change of momentum is known as force).

Question 19. A body accelerates when a balanced force acts on it. Is it true or false? Why?
Answer:

It is false because according to Newton’s second law, the momentum of a body changes only when an unbalanced force acts on it.

Question 20. A goalkeeper in a game of football pulls his hands backward after holding the ball shot at the goal. Explain.
Answer:

The goalkeeper pulls his hands backward after holding the ball to decrease the rate of change of momentum by increasing the time.

By doing this, less force is exerted on his hands. Since force is directly proportional to the rate of change of momentum.

Question 21. If a man jumps out from a boat to the bank of the river, the boat moves backward. Identify the action and reaction in this situation.
Answer:

The push of the man on the boat will be action while the boat exerts an equal force on the man, which will be a reaction.

Question 22. A runner presses the ground with his feet before he starts his run. Identify action and reaction in this situation.
Answer:

When a runner presses the ground, action is done by his feet and a reaction is produced by the ground, as a result runner moves forward.

Question 23. According to the third law of motion, do action and reaction forces act on the same body or different bodies?
Answer:

According to the third law of motion, action and reaction force always act on different bodies in opposite directions.

Question 24. Does Newton’s third law apply to a system where bodies do not touch each other?
Answer:

Yes. For example, in the case of attraction or repulsion between two magnets, etc. Newton’s third law is applicable.

Question 25. A balloon is inflated and released. Why does it fly forward as air escapes out of it?
Answer:

The reaction force of air coming out of the balloon pushes the balloon in the forward direction.

Question 26. Shyam throws a heavy stone out of his boat. As a result, the boat moves in the opposite direction, why?
Answer:

According to Newton’s third law of motion, action and reaction are always equal and opposite and act on two different bodies.

As the heavy stone is thrown in the forward direction, the boat moves backward.

Question 27.  If F and F’ are balanced forces, then what will be the magnitude of \(F_2\)?
Answer:

As per the question, F and F’ are balanced forces.

So, F=\(F^{\prime}\)

15+F_2=25 N

⇒ \(F_2=10 \mathrm{~N}\)

Class 9 Science Chapter 8 Force And Laws OF Motion Short Answer Type Questions

Question 1. A car is moving horizontally on a surface along a straight line at a constant speed. Is any force acting on it? If not, why is the fuel required to run it?
Answer:

The car moves with constant velocity. So, no unbalanced force acts on it.

The fuel is used to move the car with a force that balances with the force of friction exerted by the surface due to the motion of the car.

Question 2. Make a pile of similar carom coins on a table. Then, remove the lower coin without touching the other coins. With your fingers, you may give a sharp horizontal hit at the bottom of the pile using a striker.

1.  What will happen, if the hit is strong enough?
2. What will happen to the remaining coins once the lower coin is removed?

Answer:

1.  In case, the hit is strong enough, the bottom coin moves quickly, while the pile above remains at rest due to inertia.
2.  Once the lower coin is removed, the inertia of the remaining coins makes them fall vertically on the table.

Question  3. Why does the pillion rider fall forward when brakes are applied?
Answer:

During the ride, the pillion rider and driver are in a state of motion. But when the driver applies brakes, then the body of the pillion rider continues moving forward on account of the inertia of motion. Therefore, the pillion rider falls forward.

Question  4. There are three solids made up of aluminum, steel, and wood of the same shape and volume. Which of them would have the highest inertia?
Answer:

So Steel. As the mass is a measure of inertia, the ball having more mass than other balls will have the highest inertia. Since steel has the greatest density among these. So, it has the highest mass and inertia.

Question 5. A bullet fired against a glass window pane makes a hole in it and the glass pane is not cracked. But on the other hand, when a stone strikes the same glass pane, then it gets smashed. Why is it so?
Answer:

When the bullet strikes the glass pane, then the part of the glass pane that comes in contact with the bullet immediately shares the large velocity of the bullet and makes a hole, while the remaining part of the glass remains at rest and is therefore not smashed due to inertia of rest.

But when a slow-moving stone strikes the same glass pane, then the various parts of the glass pane get enough time to share the velocity of the stone, and the glass is smashed.

Question 6. Why can a small mass such as a bullet kill a person when fired from a gun?
Answer:

It is so because even, if the mass of the bullet is small, it moves out of the gun with a very high velocity, due to which its momentum is high (p =mv). This high momentum of the bullet exerts a large force and kills a person.

Question 7. Why does a boat tend to leave the shore, when passengers are alighting from it?
Answer:

When the passengers are alight from the boat, the boat moves in a backward direction. This is an example of Newton’s third law. This difficulty is usually overcome by the boatman by binding the boat to some rigid support.

Question 8. How are action and reaction forces related to magnitude and direction?
Answer:

Action and reaction forces have the same magnitude but opposite directions.

Question 9. Describe our walking in terms of Newton’s third law of motion.
Answer:

When we walk on the ground or road, our foot pushes the ground backward (action) and the ground pushes our foot forward (reaction). Thus, the forward reaction force exerted by the ground on our foot makes us walk forward.

Question 10. Two identical bullets are fired one by one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:

Both the bullets are identical and are fired with the same force. So, according to Newton’s third law of motion, the same force will be applied to both rifles. With the application of the same force, the light rifle will move more quickly in the forward direction, so it will hurt more to the shoulder.

Question 11. Which Newton’s law of motion is applied in the flight of a bird?
Answer:

For flight, a bird applies force on air in a backward direction by flapping its wings (action). Air exerts (a force) on the bird in the forward direction. Therefore Newton’s third law of motion is applied in the flight of a bird.

Question 12. (1) A heavy and a light object has the same momentum. Which of these is traveling faster?

(2) State the law of conservation of momentum. Give examples.

Answer:

1. If two objects are moving with the same momentum, then the object having lightweight moves faster.
2. According to the law of conservation of momentum, if an external force is not present, then the total momentum before the collision between the two bodies is equal to the total momentum after the collision.

Question 13. Two children stand on wheel carts facing each other. One student throws a heavy mass towards the other who catches it. What will be the direction of motion of both the children? Give a reason for your answer.
Answer:

The child who throws the mass towards the other and the one who catches it, both move backwards.

Reason: The Thrower moves due to a backward reaction of mass pushed forward.

The catcher moves due to the forward reaction of mass pushed and held backward.

Question 14. Take a big rubber balloon and inflate it fully. Tie its neck using a thread. Also, using adhesive tape, fix a straw on the surface of this balloon. Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.

Ask your friend to hold the other end of the thread or fix it on a wall at some distance.

This arrangement is shown in the figure. Now, remove the thread tied to the neck of the balloon. Let the air escape from the mouth of the balloon. Observe the direction in which the straw moves. Give reason.

Answer:

The straw moves in the opposite direction of air because when the air is released out of the balloon, it imparts momentum to the straw from Newton’s third law of motion.

Question 15. Two balls of the same size but of different materials, rubber, and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If, so in which direction? Will they move at the same speed? Give reasons for your answer.
Answer:

When the train is stopped suddenly, then it comes in the state of rest but the balls remain in the state of motion. So, due to the inertia of motion, the balls move in the forward direction.

As the balls are of the same size but of different materials that means their masses will be different. So, both the balls will move at different speeds.

Question 16. Suppose a ball of mass m is thrown vertically upward with an initial speed of v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attain the speed v again before striking the ground. It implies that the magnitude of the initial and final momentum of the ball are the same. Yet, it is not an example of conservation of momentum. Explain, why.
Answer:

The momentum of a system remains conserved if no external force acts on the system. In the given example, there is a gravitational force acting on the ball, so it is not an example of conservation of momentum.

Question 17. Two friends on roller skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other, who catches it? How will this activity affect the position of the two? Explain your answer.
Answer:

Separation between them will increase. Initially, the momentum of both of them is zero as they are at rest. In order to conserve the momentum, the one who throws the ball would move backward.

The second will experience a net force after catching the ball and therefore will move backwards that is in the direction of the force.

Question 18. On what factors do the following physical quantities depend?

1. Inertia
2. Momentum
3. Force
4. Impulse

Answer:

1. Inertia depends on the mass of a body.
2. Momentum depends on the mass and velocity of the body.
3.  The force depends on the mass and acceleration of the body.
4.  Impulse depends on force and the short time in which force acts.

Question 19. Give a reason for the following questions:

1. Road accidents occurring due to high speeds are much worse than accidents due to low speeds of vehicles.
2. When a motorcar makes a sharp turn at a high speed passengers tend to get thrown to one side.

Answer:

1. Road accidents occurring due to high speeds are much worse than accidents due to low speeds of vehicles. This is because the momentum of high-speed vehicles is more than that of the low speeds of vehicles.
2. When a motorcar makes a sharp turn left or right at a high speed. The lower portion of their passengers turns suddenly along with the motorcar but the upper portion does not change its direction due to inertia.

• • So, this portion of a passenger moves forward and the passenger tends to get thrown to one side or another side.

Question 20. Glasswares are wrapped in straw during their transportation. Justify giving a reason.
Answer:

During transportation, the glasswares may break, if they collide with each other in the event of jerks. When they are wrapped in straw, then the force of the jerk is transmitted to them through the pieces of straw for a longer period of time.

Thus, the change in momentum of the glasswares takes place over a longer period of time. Therefore, a very small force is experienced by them in the event of jerks and hence they do not break.

Question 21. What would happen if a fielder stops the fast-moving ball suddenly? Justify your answer.
Answer:

The high velocity of the ball decreases to zero in a short time. It means that in a short time, there is a large change in the momentum of the ball.

Therefore, to stop the fast-moving ball, the fielder will have to apply a large force and in the process, he may hurt his palm.

Question 22. Give a reason for the following:

1. The water sprinkler used for grass lawns begins to rotate as soon as water is supplied.
2. Water drops are removed from wet clothes by giving a tight jerk to the cloth.

Answer:

1. When water is pushed out of the sprinkler with a force, it exerts reaction force on the sprinkler causing it to rotate.
2. Water drops have the inertia of rest. They do not move with the cloth as it is jerked and comes out.

Question 23. What do you understand by momentum? A vehicle is moving with a velocity of 5 m/s. If the momentum of the vehicle is 5000 kg-m/s, then what is its mass?
Answer:

Momentum is the physical quantity which is the product of mass and velocity of an object. Momentum has both masses as well as the direction of motion. It is denoted by p. Its SI unit is kg-m/s.

Momentum, p = mv

Given, p – 5000 kg-m/s, v = 5 m/s, m -?

5000 = m x 5 => m = \(\frac{5000}{5}\) = 1000 kg

Question 24. A man Throws a ball weighing 500 g vertically upwards with a speed of 20 m/s.

1.  What will be its initial momentum?
2. What would be its momentum at the highest point of its flight?

Answer:

Given, initial velocity, u = 20 m/s

Mass, m=500 \(\mathrm{~g}=\frac{1}{2} \mathrm{~kg}\)

Initial momentum, p_1=m u

⇒ \(p_1=\frac{1}{2} \times 20=10 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

At the highest point, v=0

Final momentum,\(p_2\)=m v

∴ \(p_2=\frac{1}{2} \times 0=0\)

Question 25. A force of 30 N produces an acceleration of 2 m/s² in a body of mass m. Find the mass of the body. Also, find the acceleration, if force is doubled in magnitude along the same direction.
Answer:

Acceleration, a=2 \(\mathrm{~m} / \mathrm{s}^2\)

Mass of a body, m=\(\frac{F}{a}=\frac{30}{2}=15 \mathrm{~kg}\) [F=m a]

If F is doubled, a is also doubled as

F \(\propto\) a [since, m is constant]

So, a=4 \(\mathrm{~m} / \mathrm{s}^2\)

Thus, acceleration when force is doubled is 4 m/s.

Question 26. For how much time should a force of 200 N act on an object having a mass 5 kg, so as to increase its velocity from 50 m/s to 100 m/s? Given, force, F = 200 N Mass of an object, m = 5 kg Initial velocity, u – 50 m/s Final velocity, v = 100 m/s
Answer:

Given, force, F = 200 N

Mass of an object, m =5l.g

Initial velocity u -50 m/s

Final velociry u -100 m/s

From Newton’s second law of motion, we get

F=m a = \(m(\frac{v-u}{t}) [a=\frac{v-u}{t}]\)

200 =5\(\left(\frac{100-50}{t}\right)\)

t =5\(\left(\frac{100-50}{200}\right)\)

= 1.25 s

Thus, the time taken is 1.25 s.

Question 27. A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force exerted by the player to stop it?
Answer:

Given, the mass of the cricket ball, m = 70 g = 0.07 kg

Initial velocity, u =0.5 m/s, final velocity, v = 0 Time, t =0.5 s

From Newton’s second law of motion, we get

F =m a=m\((\frac{v-u}{t}) [ a=\frac{v-u}{t}]\)

=0.07 \(\times\left(\frac{0-0.5}{0.5}\right)=-0.07 \mathrm{~N}\)

A negative sign shows that force is in the opposite direction to the motion of the ball. So, the magnitude of the force is 0.07 N.

Question 28. The velocity-time graph of a ball of mass 30 g moving along a straight line on a long table is given in the figure below.

How much force does the table exert on the ball to bring it to rest?

Answer:

Given, the mass of the ball, m= 30 g = 0.03 kg

From the given graph, we note that u =20 cm/s = 0.20 m/s

v =0, t =10 -0 =10 s

As we know, the force exerted on the ball is given by

F = \(m\left(\frac{v-u}{t}\right)=0.03\left(\frac{0-0.20}{10}\right)\)

=-0.0006 N

Thus, the force exerted by the table on the ball is -0.0006 N.

Question  29. Two billiard balls each of mass 0.05 kg moving in opposite directions each with a speed of 6 m/s collide and rebound with the same speed. Find the impulse imported to each ball due to the other.
Answer:

The initial momentum of the ball

A = 0.05 x 6 = 03kg m/s and final momentum of the same ball

A = 0.05 x (-6) = -03 kg m/s /.

Impulse imported to ball A = Change in momentum = -03 -03 = -0.6kg m/s

An equal impulse is imported to ball B by ball A.

Question  30. A bullet of mass 20 g is fired from a pistol of mass 2 kg with a horizontal velocity of 150 m/s. Calculate the recoil velocity of the pistol.
Answer:

Given, mass of bullet, m = 20 g = 0.02 kg

Mass of pistol, M = 2 kg

Velocity of bullet, =150m/s

From the law of conservation of momentum, momentum of pistol = momentum of bullet MV = -mv

[Here, a negative sign shows that the direction of motion of the bullet and pistol are opposite to each other]

Velocity of pistol, V =\(-\frac{m v}{M}\)

= \(\frac{-0.02 \times 150}{2}\)

V =-1.5 m/s

Thus, the recoil velocity of the pistol is 1.5 m/s.

Velocity of pistol, V =-\(\frac{m v}{M}\)

= \(\frac{-0.02 \times 150}{2}\)

V =-1.5 m/s

Question 31. Find the acceleration and force acting on a body of mass 4 kg using the following v-t graph.

Answer:

Acceleration, a= Slope of the v-t graph

=\(\frac{Y \text {-intercept }}{X \text {-intercept }}=\frac{10 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}=2.5 \mathrm{~m} / \mathrm{s}^2\)

From Newton’s second law of motion, we get

F=m a=4 \(\times\) 2.5=10 \(\mathrm{~N}\)

Thus, acceleration and force on the body are 2.5 \(\mathrm{~m} / \mathrm{s}^2\) and 10 N, respectively.

Question 32. The force of ION applied to a mass mx produces an acceleration of 5 m/s² and when applied to mass produces an acceleration of 15 m/s² in the mass. How much acceleration will the same force produce, if the two masses are tied together?
Answer:

We know that, F = ma

10 = \(m_1 \times 5 or m_1=\frac{10}{5}=2 \mathrm{~kg}\)

Also, \(m_2=\frac{10}{15}=\frac{2}{3} \mathrm{~kg}\)

Required acceleration produced when the masses are tied together (i.e. m = \(m_1+m_2\) ),

a = \(\frac{F}{m}=\frac{10}{m_1+m_2}=\frac{10}{2+\frac{2}{3}}\)

=\(\frac{30}{8}=3.75 \mathrm{~m} / \mathrm{s}^2\)

Question 33. A bullet of mass 4 g when fired with a velocity of 50 m/s can enter a wall up to a depth of 10 cm. How much will be the average resistance offered by the wall?
Answer:

The hindrance offered by the wall to the motion of a bullet is called the resistance offered by the wall. Given, Mass of the bullet, m = 4g = 4 x 10^-3 kg

Initial velocity, u=50 \(\mathrm{~m} / \mathrm{s}\),

Depth, s=10 \(\mathrm{~cm}=\frac{1}{10} \mathrm{~m}\)

Final velocity, v=0, Force, F=?

Using the second equation of motion,

⇒ \(v^2 = u^2+2 a s\)

⇒ \(v^2-u^2 =2 a s \)

⇒ \(0-(50)^2  =2 a \times \frac{1}{10}\)

⇒ \(-2500 =\frac{a}{5} \Rightarrow a=-12500 \mathrm{~m} / \mathrm{s}^2\)

Force, \(F=m a=4 \times 10^{-3} \times(-12500)=-50 \mathrm{~N}\)

Thus, the average resistance offered is 50 N.

The negative sign indicates that the force is acting opposite to the motion.

Question 34. A heavy car A of mass 2000 kg traveling at 30 m/s has a head-on collision with a sport car B of mass 1000 kg. If both cars stop dead on colliding, then what was the velocity of car B?
Answer:

Given, mass of car  A, \(m_1 =2000 \mathrm{~kg}\)

Mass of car B, \(m_2\) =1000 kg

Velocity of car A, \(v_1 =30 \mathrm{~m} / \mathrm{s}\)

and velocity of car B, \(v_2\)=?

Now, according to the law of conservation of momentum, \(m_1 v_1=m_2 v_2\)

2000 \(\times\)30 =1000 \(\times v_2\)

⇒ \(v_2 =2 \times 30\)

=60 \(\mathrm{~m} / \mathrm{s}\)

Question 35. The motion of a body of mass 5 kg is shown in the velocity-time graph.

Find from the graph.

1.  Its acceleration.
2. The force acting on the body.
3. The change in momentum of a body in 2 seconds after the start.

Answer:

(1) Acceleration = Slope of the line of the velocity-time graph,

a =\(\frac{v_2-v_1}{t-t_1}=\frac{5-0}{2-0}=\frac{5}{2}=\frac{10}{4}\)

= \(\frac{15}{6}=2.5 \mathrm{~m} / \mathrm{s}^2\)

(2) The force acting on the body is given by

F=m a=5 \(\times\) 2.5=12.5 N

Change in momentum =m v-m u  [ u=0 and v=5 m/s]

=5 \(\times\) 5-5 \(\times\) 0=25 kg- m/s

Question 36. The velocity-time graph of a ball moving on the surface of the floor is shown in the figure. Calculate the force acting on the ball, if the mass of the ball is 100 g.

Answer:

The velocity-time graph shows that the velocity of the ball at =0 is zero.

So, the initial velocity of the ball, u = 0.

Velocity of the ball at t = 4 s is 20 m/s

i.e. final velocity, v =20 m/s; Time, t = 4 s

Acceleration of the ball, a =\(\frac{v-u}{t}=\frac{20 \mathrm{~m} / \mathrm{s}-0}{4 \mathrm{~s}}\)

=5 \(\mathrm{~m} / \mathrm{s}^2\)

Also, the mass of the ball,

m=100 \(\mathrm{~g}=\frac{100}{1000} \mathrm{~kg}=\frac{1}{10} \mathrm{~kg}\)

Force acting on the ball, F=m a=\(\frac{1}{10} \mathrm{~kg} \times 5 \mathrm{~m} / \mathrm{s}^2\)

=0.5 \(\mathrm{~kg}-\mathrm{m} / \mathrm{s}^2=0.5 \mathrm{~N}[1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}^2=1 \mathrm{~N}]\)

Question 37. The speed-time graph of a car is given. The car weighs 1000 kg.

1. What is the distance traveled by the car in the first 2s?
2.  What is the braking force applied at the end of 5 s to bring the car to stop within one second?

Answer:

(1) Distance traveled by the car in first 2 s= Area of \(\triangle\) O A D=\(\frac{1}{2} \times 2 \times 15=15 \mathrm{~m}\)

(2) Braking force, F=m \(\times\) a

Given, mass of the car, m=1000 kg, initial velocity, u=15 m/s,

final velocity, v=0, time, t = 1s

On applying, a=\(\frac{v-u}{t} \Rightarrow a=\frac{0-15}{1}=-15 \mathrm{~m} / \mathrm{s}^2\)

F = ma =1000 \(\times(-15)\)=-15000 N

Question 38. A constant force of friction of 50 N is acting on a body of mass 200 kg moving initially with a speed of 15 m/s. How long does the body take to stop? What distance will it cover before coming to rest?
Answer:

Given, F = 50 N, m = 200 kg, u= 15 m/s

The acceleration of the body is obtained from F = ma

or a=\(\frac{F}{m}=\frac{50}{200}=-0.25 \mathrm{~m} / \mathrm{s}^2\)

Now, from the first equation of motion

v=u+a t

t=\(\left(\frac{v-u}{a}\right)=\left(\frac{0-15}{-0.25}\right)=60 \mathrm{~s}\)

Also, the distance traveled is obtained from

s =u t+\(\frac{1}{2} a t^2\)

=50 \(\times 60+1 / 2 \times(-0.25) \times(60)^2=450 \mathrm{~m}\)

Question 39. Look at the diagram below and answer the following questions:

1.  When a force is applied through the free end of the spring balance A, then the reading on the spring balance A is 15 g-wt. What will be the measure of the reading shown by spring balance B?
2.  Write reasons for your answer.
3.  Name the force that balance A exerts on balance B and the force of balance B on balance A.

Answer:

1. 15 g-wt.
2. From Newton’s third law, the force exerted by B on A and the force exerted by A on B is equal.
3.  Force of reaction balance A exerts on balance B and force of action balance B exerts on balance A.

Question 40. An iron sphere of mass 1 kg is dropped from a height of 10 m. If the acceleration of the sphere is 9.8 m/s², then calculate the momentum transferred to the ground by the ball.
Answer:

Here, initial velocity of sphere, u = 0 Distance travelled, s = 10 m

Acceleration of sphere, a = 9.8 m/s²

Final velocity, v =?

The final velocity of the sphere when it just reaches the ground can be calculated using the,

v² – u² = 2 as

⇒ v² -0 = 2 x 9.8 x 10 =196 m/s

v = √196 m/s = 14 m/s

Momentum of the sphere just before it touches the ground = mv

= 1 kg x 14 m/s =14 kg-m/s

On reaching the ground, the iron sphere comes to rest, so its final momentum is equal to zero.

According to the law of conservation of momentum, momentum transferred to the ground = momentum of the sphere just before it comes to rest = 14 kg-m/s

Class 9 Science Chapter 8 Long Answer Type Questions

Question 1. What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically:

1. Momentum versus velocity when mass is fixed.
2. Momentum versus mass when velocity is constant.

Answer:

Momentum The quantity of motion possessed by a moving body is known as the momentum of the body.

It is the product of the mass and velocity of the body. Momentum, p = mu. Its SI unit is kg-m/s

Force applied on an object of mass m moving with acceleration a.

F=m a=m \(\frac{\Delta v}{\Delta t}\)

[acceleration = rate of change of velocity =\(\frac{\Delta v}{\Delta t}]\)}

=\(\frac{\Delta p}{\Delta t}\)

Force applied on an object is equal to the rate of change of momentum of the object.

(1) Momentum versus velocity graph when mass is fixed, p=m v. If m is fixed, then p \(\propto v\)

The momentum versus velocity graph will be a straight line passing through the origin (if v=0, then p=0 ).

(2) Momentum versus mass graph when velocity is constant, p — mv.

If velocity is constant, then p m.

So, the momentum versus mass graph will be a straight line passing through the origin (if m = 0, then p = 0).

Question 2. (1) When a carpet is beaten with a stick dust comes out of it. Explain.

(2) Calculate the force required to impart a car with a velocity of 30 m/s in 10 s starting from rest. The mass of the car is 15000 kg.

Answer:

(1) When we beat the carpet with a strick, then it comes into motion. However, the dust particles continue to be at rest due to inertia and get detached from the carpet.

(2) Given, initial velocity, u = 30 m/s

Time, t = 10 s, final velocity, v = 0

Mass, m = 15000 kg

From first equation of motion, a=\(\frac{v-u}{t}=\frac{-30}{10}=-3 \mathrm{~m} / \mathrm{s}^2\)

So, retarding force required = ma

= 15000 x (-3) = – 45000 N

Thus, the force required by the car is 45000 N.

Question 3. An object of mass 5 kg is accelerated uniformly from a velocity of 4 m/s to 8 m/s in 8 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Given, mass, m=5 kg, initial velocity, u=4 m/s

Final velocity, v=8 m/s, time, t=8 s

Initial momentum, \(p_1\)=m u=5 \(\times\) 4=20 kg- m/s and final momentum, \(p_2=m v=5 \times 8\)

=40 kg – m/s

Now, force =\(\frac{\Delta p}{t}=\frac{p_2-p_1}{t}=\frac{40-20}{8}=\frac{20}{8} N\) = 2.5 N

Question 4. Using the second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sandbag at a speed of \(10^3 \mathrm{~m} / \mathrm{s}\) and gets embedded after traveling 5 cm. Calculate

1.  The resistive force exerted by sand on the bullet.
2.  The time is taken by a bullet to come to rest.

Answer:

Given, m=10 g=0.01kg, u=\(10^3 \mathrm{~m} / \mathrm{s}\), v=0,

s=5 cm=0.05 m, t=?, a=?, F=?

(1) From the third equation of motion,

a=\(\frac{v^2-u^2}{2 s}=\frac{0-\left(10^3\right)^2}{2 \times 0.05}=-10^7 \mathrm{~m} / \mathrm{s}^2\)

The force applied by the bullet,

F=m a=0.01 \(\times\left(-10^7\right)=-10^5 \mathrm{~N}\)

[negative sign shows that force is against the direction of motion]

The resistive force exerted by the sand on the bullet is \(10^5\) N.

(2) Time taken by a bullet to come to rest,

t=\(\frac{v-u}{a}=\frac{0-10^3}{-10^7}=10^{-4} \mathrm{~s}\)

Question 5. Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 \(m/s^2\) on a mass \(m_1\) and an acceleration of 24 \(\mathrm{~m} / \mathrm{s}^2\) on a mass \(m_2\). What acceleration would the same provide, if both the masses are tied together?
Answer:

⇒ \(F_1=5 \mathrm{~N}, a_1=8 \mathrm{~m} / \mathrm{s}^2, m_1\)= ?

⇒ \(F_2=5 \mathrm{~N}, a_2=24 \mathrm{~m} / \mathrm{s}^2, m_2\)=?

⇒ From , F=m a, 5 = \(m_1 \times 8\)

⇒ \(m_1=\frac{5}{8} \mathrm{~kg}\)

Similarly, \(5=m_2 \times 24\)

⇒ \(m_2=\frac{5}{24} \mathrm{~kg}\)

⇒ \(m_1+m_2=\frac{5}{8}+\frac{5}{24}\)

= \(\frac{15+5}{24}=\frac{20}{24}=\frac{5}{6}\)

Given that acceleration produced by the same force provided, if both the masses are tied together is

a =\(\frac{F}{\left(m_1+m_2\right)}\)

= \(\frac{5}{5 / 6}=6 \mathrm{~m} / \mathrm{s}^2\)

Question 6. If the engine of a car provides an acceleration of 2 m/s² to start it from rest, then assuming the mass to be roughly 1000 kg. Calculate

1.  force provided by the engine.
2.  velocity after 10 s.
3.  time after which the car comes to rest, if the engine is turned off after 15 s. (take, frictional force =15 N)

Answer:

Given, initial velocity, u=0 m/s

Acceleration, a=2 \(m / s^2\),

Mass, m=1000 kg

(1) From Newton’s second law of motion, force = mass \(\times\) acceleration

F=m a=1000 \(\times\) 2=2000 N

(2) Velocity after 10 s,

from the first equation of motion,

v=u+a t

v=0+2 \(\times\) 10=20 m/s

(3) Final velocity of the car when the engine is on after 15 s.

v=0+2 \(\times\) 15=30 m/s [t=15 s]

So, p=m v

=1000 \(\times\) 20=20000 \(\mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

After 15 s, net force = force exerted by engine + friction force

=2000-15=1985 N [friction force=15 N]

[Here, friction force is subtracted because it opposes the motion]

Acceleration =\(\frac{\text { Net force }}{\text { Mass }} \Rightarrow a=\frac{1985}{1000}\)

a=1.985 \(\mathrm{~m} / \mathrm{s}^2 \approx 2 \mathrm{~m} / \mathrm{s}^2\)

Time taken by the car (comes to rest)

v=u-a t [first equation of motion][final velocity, v=0]}

0=30-2 \(\times\) t

2 t=30 [from Eq. (1)]

t=15 s

Force And Laws Of Motion Multiple Choice Questions

Question 1. The resultant force acting on a body is zero, then a

1. The body is in unequilibrium
2. The body is in equilibrium
3. The body moves with constant acceleration
4. Body moves with retardation

Answer: 2. Body is in equilibrium

When the resultant force acting on a body is zero, then the body will be in equilibrium.

Question 2. If a ball thrown up in a moving car comes back to the boy’s hands, then it explains

1. Newton’s first law of motion
2. Newton’s second law of motion
3. Newton’s third law of motion
4. law of conservation of momentum

Answer: 1. The balls come back to the boy’s hand due to the law of inertia (Newton’s first law of motion). During the period, when the ball is in the air, the ball covers the same horizontal distance as the train and so it comes back to the boy’s hand.

Question 3. A thrust of 200 N is applied on the surface of the wall, then the normal reaction on the wall is

1. 200 N
2. 100 N
3. 400 N
4. 300 N

Answer: 1. 200 N

Every action has an equal and opposite reaction, therefore the normal reaction is 200 N.

Question 4. A bus of mass 500 kg is moving with a velocity of 5 m/s and is acted upon by a forward force of 500 N due to the engine and retarding force of 200 N due to friction velocity of the bus after 20 s will be

1. 15 m/s
2. 17 m/s
3. 19 m/s
4. 21 m/s

Answer: 2. 17 m/s

Given, m = 500 kg,

u=5 m/s and t=20 s

Resultant force, F=500-200=300 N

Acceleration, a=\(\frac{F}{m}=\frac{300}{500}=0.6 \mathrm{~m} / \mathrm{s}^2\)

By equation of motion,

v=u+a t=5+0.6 \(\times 20=17 \mathrm{~m} / \mathrm{s}\)

Question 5. 1 dyne is equal to

1. \(10^5 \mathrm{~N}\)
2. \(10^7 \mathrm{~N}\)
3. \(10^{-5} \mathrm{~N}\)
4. \(10^{-7} \mathrm{~N}\)

Answer: 3. \(10^{-5} \mathrm{~N}\)

Dyne is the unit of force in the CGS system.

1 N = \(\frac{1 \mathrm{~kg}-\mathrm{m}}{\mathrm{s}^2}=\frac{1000 \mathrm{~g} \times 100 \mathrm{~cm}}{1 \mathrm{~s}^2}\)

1 dyne =\(\frac{1}{10^5} \mathrm{~N}=10^{-5} \mathrm{~N}\)

Question 6.  1 kg-wt is equal to

1. 9.8 kg
2. 1 kg
3. 9.8 N
4. 98 N

Answer: 3. 9.8 N

1 kg-wt = mg N = 1 x 9.8 N = 9.8 N

Question 7. A truck moving with a speed of 54 km/h. The truck driver applied brakes suddenly and brings the truck to rest in 5 s, then the average retarding force on the truck, if the mass of the truck and driver is 400 kg, will be

1. 1200 N
2. 600 N
3. 800 N
4. 500 N

Answer: 1. 1200 N

Given, u=54 \(\mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\)

=15 m/s, v=0 and t=5 s

Retardation, a=\(\frac{v-u}{t}=\frac{0-15}{5}=-3 \mathrm{~m} / \mathrm{s}^2\)

Average retarding force, F=m a=400 \(\times\) 3=1200 N

Question 8. Two forces acting on a body in different directions, then acceleration produced in the body is due to

1. Resultant of both forces
2. The sum of both forces
3. Difference between both forces
4. None of the above

Answer: 1. Resultant of both forces

According to Newton’s second law of motion, the resultant force on a body is responsible for its acceleration.

Question 9. Two forces F₁ and F2 are acting on a body as shown in the figure, then acceleration in the body is

1. 23 \(\mathrm{~m} / \mathrm{s}^2 \)
2. 3 \(\mathrm{~m} / \mathrm{s}^2 \)
3. 2 \(\mathrm{~m} / \mathrm{s}^2 \)
4. 22 \(\mathrm{~m} / \mathrm{s}^2 \)

Answer: 2. 3 \(\mathrm{~m} / \mathrm{s}^2 \)

The resultant force on a body,

F=\(F_2-F_1=65-50=15 \mathrm{~N}\)

By Newton’s second law of motion,

F=m a \(\Rightarrow\) 15=5 a

a=3 \(\mathrm{~m} / \mathrm{s}^2\)

Question 10. The dirty Blanket is beaten with a stick to remove dust particles. Which law holds good for this?

1. Law of conservation of momentum
2. Law of inertia
3. Law of impulse
4. Law of conservation of energy

Answer: 2. Law of inertia

When we beat a blanket with the help of a stick, then it comes into motion. However, the dust particles continue to rest due to inertia and get detached from the blanket.

Question 11. A bullet of 20 g strikes a sandbag at a speed of 200 m/s and gets embedded after traveling 2 cm, then the resistive force exerted by sand on the bullet is

1. 2 \(\times 10^3 \mathrm{~N}\)
2. 2 \(\times 10^4 \mathrm{~N}\)
3. 2 \(\times 10^6 \mathrm{~N}\)
4. 2 \(\times 10^5 \mathrm{~N}\)

Answer: 2. 2 \(\times 10^4 \mathrm{~N}\)

Given, u=200 \(\mathrm{~m} / \mathrm{s}\), v=0, s=0.02 m

and m=20 \(\mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\)

From the third equation of motion,

⇒ \(v^2 =u^2+2 a s\)

a =\(\frac{v^2-u^2}{2 s}\)

=\(\frac{(0)^2-(200)^2}{2 \times 0.02}\)

=-\(10^6 \mathrm{~m} / \mathrm{s}^2\)

Negative sign indicates retardation.

Resistive force, F =m a=20 \(\times 10^{-3} \times 10^6\)

=2 \(\times 10^4 \mathrm{~N}\)

Question 12. Impulse is equal to

1. Rate of change in momentum
2. Rate of change in force
3. Change in reaction force
4. Change in momentum

Answer: 4. Change in momentum

Impulse, I =F \(\times \Delta t\)

= \(\frac{\Delta p}{\Delta t} \cdot \Delta t=\Delta p\) = Change in momentum.

Question 13. Jetplanes and rockets work on the principle of conservation of

1. Energy
2. Momentum
3. Mass
4. Heat

Solution: 2. Momentum

Initially, the total momentum of the rocket and its fuel is zero, when the fuel is exploded. The burnt gases are allowed to escape through a nozzle with very high downward velocity and with large momentum in a downward direction.

To conserve this momentum, the rocket acquires an equal momentum in the upward direction and hence starts moving upwards.

Question 14. A Bullet Of Mass 30g is fired horizontally with a velocity of 120 m/s from a pistol of mass 2 kg, the recoil velocity of the pistol is

1. 1.6 m/s
2. 2.8 m/s
3. 2.4 m/s
4. 1.8 m/s

Answer: 4. 1.8 m/s

By the law of conservation of momentum,

Initial momentum = Final momentum 0=0.03×120 + 2xv

where v is recoil velocity.

0 = 3.6 + 2v

v=\(\frac{-3.6}{2}=-1.8 \mathrm{~m} / \mathrm{s}\)

Question 15. An object of mass 4 kg moves with a velocity of 4 m/s, then its momentum will be

1. 16 m/s
2. 4 m/s
3. 16 kg-m/s
4. 4 kg-m/s

Solution: 3. 16 kg-m/s

Momentum,p = mv = 4x 4 = 16kg-m/s

Question 16. A force acts on a body of mass 5 kg and changes its velocity from 8 m/s to 12 m/s in 4s, then the magnitude of the force is

1. 8 N
2. 4 N
3. 5 N
4. 6 N

Answer: 3. 5 N

According to Newton’s second law of motion, Force = Rate of change in momentum

F=\(\frac{m(v-u)}{t}=\frac{5(12-8)}{4}=5 \mathrm{~N}\)

Question 17. When a person jumps down from a tower into a stretched tarpaulin, then receives

1. Greater injury
2. Less injury
3. No injury
4. None of the above

Answer:

When a person jumps, the tarpaulin gets depressed at the place of impact, therefore impact time interval increases, As a result, a person experiences a very small force, and hence he receives no injury.

Question 18. A cricket ball of mass 0.25 kg moving with a velocity of 10 m/s is brought to rest by a player in 0.1 seconds, and then the impulse exerted by the player is

1. 2.5 N-s
2. 1.5 N-s
3. 25 N-s
4. 30 N-s

Answer: 1. 2.5 N-s

Impulse = Change in momentum -m{v -u)

= 0.23 (0-10) = -2.3 N-s

A negative sign indicates the direction of the impulse is opposite to the motion of the ball.

Gravitation Multiple-Choice Questions

Question 1. Newton’s law of gravitation is universal because it

1. acts on all bodies and particles in the universe
2. it acts on all the masses at all distances and is not affected by the medium
3. is always attractive
4. None of the above

Answer: 2. it acts on all the masses at all distances and is not affected by the medium

According to Newton’s law of gravitation, the force between two masses in the universe, F=\(\frac{G m_1 m_2}{r^2}\)

Which does not depend on medium and G is the same in the whole universe.

Question 2. The weakest force in the following is

1. magnetic force
2. nuclear force
3. gravitational force
4. electric force

Answer: 3. gravitational force

Gravitational force is the weakest force among given forces.

Question 3. The atmosphere is held to the earth by

1. earth’s magnetic field
2. earth’s rotation
3. gravity
4. earth’s electric field

Answer: 3. gravity

The atmosphere is held to the earth by gravity.

Question 4. Between the two gravitational constants G and g, which is called the universal gravitational constant?

1. g
2. G
3. Both (1) and (2)
4. None of these

Answer: 2. G

G is called the universal gravitational constant because its value is the same in the whole universe.

G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Question 5. The value of gravitational acceleration (g) is

1. highest at poles
2. highest at equator
3. lowest at poles
4. lowest at equator

Answer: 1. highest at poles

Gravitational acceleration g is given by g = \(\frac{G m_e}{R_e^2}\)

Since the radius of the earth at the equator is greater than the poles, the value of g at the poles is greater than the equator

Question 6. The law of gravitation gives the gravitational force between

1. The Earth and other planets only
2. the earth and sun only
3. two charged bodies only
4. any two bodies having some mass

Answer: 4. any two bodies having some mass

The law of gravitation is applicable to all bodies having some mass.

Question 7. The gravitational force between two objects is 10 N. If the masses of both objects are doubled without changing the distance between them, then the gravitational force would become

1. 2.5 N
2. 20 N
3. 40 N
4. 10 N

Answer: 3. 40 N

Gravitational force, \(F_1=\frac{G m_1 m_2}{r^2}\)

When masses are doubled, then

⇒ \(F_2=\frac{G 2 m_1 2 m_2}{r^2}=4 \frac{G m_1 m_2}{r^2}\)

⇒ \(F_2=4 F_1=4 \times 10=40 \mathrm{~N}\)

Question 8. The magnitude of the weight of a body at the centre of the earth is

1. zero
2. equal to the mass of the body
3. greater than g
4. less than g

Answer: 1. zero

The value of gravitational acceleration (g) at the centre is zero, hence weight, w =mg = mx 0 = 0.

Question 9. When a ball is fully immersed in a liquid, its weight decreases. It happens due to

1. gravitational force
2. magnetic force
3. buoyant force
4. friction force

Answer: 3. buoyant force

When a ball is fully immersed in a liquid, its weight decreases due to buoyant force.

Question 10. Why the dam of the water reservoir is thick at the bottom?

1. The quantity of water increases with the depth
2. The density of water increases with the depth
3. The pressure of water increases with the depth
4. Temperature of water increases with depth

Answer: 3. Pressure of water increases with depth

The dam of water is made thick at the bottom, because due to maximum depth, the pressure of water is maximum at the bottom and to bear this maximum pressure dam should be thick at the bottom.

Question 11. Three spheres have radii of 1 cm, 2 cm and 3 cm, respectively. Which sphere exerts maximum pressure on earth?

1. First
2. Second
3. Third
4. All Equal

Answer: 1. First

We know that, pressure =\(\frac{\text { force }}{\text { area }}=\frac{F}{A}\)

or p=\(\frac{F}{A} \Rightarrow \) p \(\propto \frac{1}{A}\)

p \(\propto \frac{1}{r^2}\)

Therefore, the sphere which has a lesser area will exert maximum pressure. Since the first sphere has a small radius, its area will be small. Hence, it will exert maximum pressure.

Question 12. A petrol cane of mass 5 kg has a volume of 2880. What is the density of the cane?

1. 17.3 \(\mathrm{~g} / \mathrm{cm}^3\)
2. 16.5 \(\mathrm{~g} / \mathrm{cm}^3\)
3. 1.65 \(\mathrm{~g} / \mathrm{cm}^3\)
4. 1.73 \(\mathrm{~g} / \mathrm{cm}^3\)

Answer: 4. 1.73 \(\mathrm{~g} / \mathrm{cm}^3\)

Given, mass of cane, m=5 kg =5000 g, volume of cane, V=2880 \(\mathrm{~cm}^3\)

Density of cane, \(\rho=\frac{m}{V}\)

⇒ \(\rho=\frac{5000}{2880}=1.73 \mathrm{~g} / \mathrm{cm}^3\)

Question 13. The pressure exerted by the weight of a cubical box of side 2 cm on the surface of table is 4 Pa. What is the weight of the box?

1. 1.6 \(\times 10^{-4} \mathrm{~N}\)
2. 1.6 \(\times 10^{-5} \mathrm{~N}\)
3. 16 \(\times 10^{-4} \mathrm{~N}\)
4. 15 \(\times 10^{-4} \mathrm{~N}\)

Answer: 3. 16 \(\times 10^{-4} \mathrm{~N}\)

Given, pressure, p = 4 Pa and side of cube,

a = 2 cm = 0.02 m

Area of cubical box, A =\(a^2=(0.02)^2\)

=4 \(\times 10^{-4} \mathrm{~m}^2\)

Since, pressure, p =\(\frac{F}{A}\)

F =p \(\times A=4 \times 4 \times 10^{-4}\)

=16 \(\times 10^{-4} \mathrm{~N}\)

So, the weight of the block is \(16 \times 10^{-4} \mathrm{~N}\).

Question 14. A large ship can float, but a steel needle sinks, because of

1. mass
2. volume
3. density
4. None of these

Answer: 4. None of these

This concept is explained by Archimedes’ principle which states that when a body is fully or partly submerged in a fluid, a buoyant force from the surrounding fluid, acts on the body.

Question 15. A wooden object of mass 700 g is thrown into the water tub. What is the buoyant force acting on the object, if it displaces 700 g of water?

1. 5 N
2. 8 N
3. 7 N
4. 10 N

Answer: 3. 7 N

Given, the mass of the object (m) = 700 g = 0.7 kg

Acceleration due to gravity (g) =10 m/s²

According to Archimedes’ principle, the buoyant force is equal to the weight of displaced water,

i.e. weight of displaced water = mg = 0.7 x 10 = 7 N So, buoyant force is 7 N.

Question 16. A ball weighs 90 g in air, 60 g in water and 40 g in a liquid. If the density of water is 1 g cm-3, then what is the density of this liquid?

1. 3 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
2. 1 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
3. 1.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)
4. 2.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

Answer: 3. 1.66 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

1. As given, when immersed in water, the ball displaces 90 – 60 = 30 g of water.
2. So, it’s volume = volume of water displaced = \(\frac{30}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=30 \mathrm{~cm}^3\)
3. Density of ball = \(\frac{90}{30}=3 \mathrm{~g} \mathrm{~cm}^{-3}\)
4. So, the ball displaces (90-40)=50 \(\mathrm{~g} \) of liquid with density =  \(\frac{50 \mathrm{~g}}{30 \mathrm{~cm}^3}=1.66 \mathrm{~g} \mathrm{~cm}^{-3} \)

Question 17. If two forces in the ratio 3 :8 act on two surface areas in ratio 2 :5. What is the ratio of pressure exerted?

1. 3: 8
2. 6: 40
3. 15: 16
4. 4: 8

Answer: 3. 15: 16

Ratio of areas =2: 5, i.e. \(\frac{A_1}{A_2}=\frac{2}{5}\)

Ratio of forces =3: 8, i.e. \(\frac{F_1}{F_2}=\frac{3}{8}\)

Ratio of pressure =\(\frac{p_1}{p_2}=\frac{F_1}{F_2} \times \frac{A_2}{A_1}\)

=\(\frac{3}{8} \times \frac{5}{2}=\frac{15}{16}\)

Class 9 Science Chapter 9 Very Short Answer Type Questions

Question 1. When do we use the term force of gravity rather than force of gravitation?
Answer:

We use the term force of gravity rather than force of gravitation for the force of attraction between two bodies in which one body has an infinitely large mass.

Question 2. Name the scientist who determined the value of the universal gravitational constant.
Answer:

Henry Cavendish determined the value of the universal gravitational constant.

Question 3. Which force brings tides into the ocean?
Answer:

The gravitational force of the moon brings tides into the ocean.

Question 4. Is it possible to shield a body from gravitational effects?
Answer:

No, as gravitational interaction does not depend on the nature of the intervening medium.

Question 5. Which force keeps the moon in a uniform circular motion around the Earth?
Answer:

The gravitational force between the moon and the Earth keeps the moon in a uniform circular motion around the Earth.

Read and Learn More Class 9 Science Solutions

Question 6. Suppose gravity of the earth suddenly becomes zero, then which direction will the moon begin to move, if no other celestial body affects it?
Answer:

The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of the moon is due to the centripetal force provided by the gravitational force of the earth.

Question 7. A stone and the earth attract each other with an equal and opposite force. Why, then we see only the stone falling towards the earth but not the earth rising towards the stone?
Answer:

As the mass of the earth is very large, the acceleration produced in the earth is too small, hence we see only the stone falling towards the earth but not the earth rising towards the stone.

Question 8. Two objects kept in a room do not move towards each other as per the universal law of gravitation. Why?
Answer:

The size of the bodies is very small. Therefore the force of attraction between them is very small. So, both objects do not move towards each other.

Question 9. The earth is acted upon by the gravitation of the sun even, then it does not fall into the sun. Why?
Answer:

The earth does not fall into the sun because the earth remains in its circular orbit due to the gravitational force acting on it.

Question 10. Write the direction of acceleration due to gravity.
Answer:

The direction of acceleration due to gravity is downwards i.e. towards the centre of the earth.

Question 11. Two objects of masses m₁ and m₂ are dropped in a vacuum from a height above the surface of the earth (m₁ is greater than m₂). Which one will reach the ground first and why?
Answer:

Both will reach the ground at the same time because the acceleration due to gravity is independent of the masses of freely falling bodies.

Question 12. When a body is thrown upwards, its velocity becomes zero at the highest point. What will be its acceleration at this point?
Answer:

The acceleration at this point is equal to the value of the downward direction.

Question 13. Does the velocity of a body during free fall remain constant? Why/Why not?
Answer:

No, the velocity of a body increases at every point of its motion during free fall as acceleration due to gravity acts on it.

Question 14. At which place on the earth, the acceleration due to gravity is zero?
Answer:

At the centre of the earth, the acceleration due to gravity is zero.

Question 15. Name the place on the earth’s surface, where the weight of a body is maximum and minimum.
Answer:

Weight is maximum at the poles and minimum at the equator.

Question 16. Anu buys 300 g of gold at the poles. What will be the weight of gold at the equator?
Answer:

The value of acceleration due to gravity (g) is less at equators than at poles, so the weight of 300 g gold will be less at the equator

Question 17. The astronauts in space feel weightless. Why?
Answer:

They do not exert any force/weight on their spaceship in the absence of gravity in space.

Question 18. What is the final velocity when a body is dropped from a height?
Answer:

The final velocity is zero when a body is dropped from a height.

Question 19. Which of the two will double the pressure? Doubling the area or making the area half.
Answer:

Making the area half \(\left[ p=\frac{F}{A}\right]\)

Question 20. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways i.e. when length and breadth form the base, when breadth and width form the base, and when width and length form the base. In which condition, the pressure exerted by the box will be maximum?
Answer:

Pressure =\(\frac{\text { Thrust }}{\text { Area }}\)

The pressure exerted by the box will be maximum when the area is small.

The area will be minimal when breadth and width form the base.

Question 21. When a floating body is pressed down a little, which force will increase at that time?
Answer:

Upthrust.

Question 22. A hydrometer is made heavy near the bottom.
Answer:

A hydrometer is made heavy near its bottom so that it can float with the stem in the vertical position.

Question 23. A bucket of water weighs less inside the well water. Why?
Answer :

Due to the upthrust exerted by the well water on the bottom of the bucket in an upward direction.

Question 24. If the density of a body is 800 kg m^-3. Will it sink or float when dipped in a bucket of water? (Take, density of water = 1000 kg^-3)
Answer:

Since its density is less than that of water, hence it will float.

Question 25. If 100 cc of water is heated to three different temperatures i.e. 4°C, 20°C and 45°C. At what temperature, its density will be maximum?
Answer:

4° C; because density increases with a decrease in temperature.

Question 26. The density of glass is 3.5 g cm^-3. What does it mean?
Answer:

3.5 g cm^-3 density of glass means that the volume of 3.5 glass is 1 cm³.

Question 27. An object is dropped one by one in three liquids having different densities. The object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of their volumes outside the liquid surface, in liquids of densities \(d_1, d_2\) and \(d_3\), respectively. Arrange them in increasing order.
Answer:

In a liquid of higher density, more part of the object remains outside the liquid. Since the order of part of their volume outside the liquid is \(\frac{1}{9}<\frac{2}{11}<\frac{3}{7}\)

Thus, the order of densities is \(d_1<d_2<d_3\).

Question 28. Name the instrument which is used to determine the density of the liquid.
Answer:

The instrument which is used to determine the density of liquid is a “hydrometer.”

Question 29. Arrange the following in the increasing order of their relative densities.

Iron, air and water

Answer:

The increasing order of the relative densities is as follows: Air < water < iron

Question 30. A body projected horizontally moves with some horizontal velocity although it is under the action of the force of gravity, why?
Answer:

The force of gravity has no effect on the horizontal velocity because it acts in a vertically downward direction. So the body moves with uniform horizontal velocity.

Question 31. A stone dropped from a tree takes 2 seconds to reach the ground. Find its velocity on striking the ground.
Answer:

Given, u =0, t =2 s, a = g =9.8 m/s², v = ?

From the first equation of motion, v =u + at

= 0 + 9.8x 2 = 19.6 m/s

Question 32. Find pressure, when a thrust of 20 N is applied on a surface area of 10 cm².
Answer:

Area, A=10 \(\mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2[1 \mathrm{~m}=100 \mathrm{~cm}]\)

Pressure(p) =\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{20}{10 \times 10^{-4}}\)

=2 \(\times 10^4 \mathrm{~Pa}\)

Pressure is 2 \(\times 10^4 \mathrm{~Pa}\).

Question 33. A metal cuboid of mass M kg rests on a table. A surface area of 40 cm² is in contact with the table. The pressure exerted by the cuboid on the table surface is 10000 Pa. Find the value of M is (given that, g=10 \(\mathrm{~ms}^{-2}\)  )
Answer:

Given, A=40 \(\mathrm{~cm}^2=0.004 \mathrm{~m}^2, p=10000 \mathrm{~Pa}\)

p=\(\frac{F}{A}=\frac{M g}{A}\) [e F=M g]

M=\(\frac{p \times A}{g}=\frac{10000 \times 0.004}{10}=4 \mathrm{~kg}\)

Question 34. Find the mass of 5 \(\mathrm{~m}^3\) of cement of density 3000 kg \(\mathrm{~m}^{-3}\).
Answer:

Given, volume, V=5 \(\mathrm{~m}^3\)

Density, \(\rho=3000 \mathrm{~kg} \mathrm{~m}^{-3}\), mass, m= ?

As, \(\rho=\frac{m}{V}\)  or m=\(\rho\) V

= 3000×5 =15000 kg

Question 35. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. What will be the weight of the liquid displaced by the object?
Answer:

Weight of liquid displaced, F =10 – 8 = 2 N

Class 9 Science Chapter 9 Gravitation Short Answer Type Questions

Question 1. “All the objects in the universe attract each other.”

1. What is the force of attraction called?
2. Name any two factors on which this force of attraction depends.

Answer:

1. This force of attraction is called as force of gravitation.
2.  The force of gravitation depends on two factors:
   1.  Directly proportional to the product of their masses.
   2. Inversely proportional to the square of the distance between them.

Question 2. What is the source of centripetal force that a planet requires to revolve around the sun? On what factors does that force depend?
Answer:

The motion of the planet around the sun is due to the centripetal force. This centripetal force is provided by the gravitational force between the planet and the sun.

This force depends on the mass of the sun and mass of the planet and the distance between the sun and the planet.

Question 3. State the source of centripetal force that a planet requires to revolve around the sun. On what factors does the force depend? Suppose this force suddenly becomes zero, then in which direction will the planet begin to move, if no other celestial body affects it?

Answer:

The source of centripetal force is the gravitation force. It depends upon the following factors:

1. Mass of the planet and the sun.
2.  Distance between the planet and the sun.

If this force suddenly becomes zero, then the planet will begin to move in a straight line in the direction in which it was moving at that instant.

Question 4. (1) Seema buys a few grains of gold at the poles as per the instruction of one of her friends. She hands over the same when she meets her at the equator. Will the friend agree with the weight of gold bought? If not, why?

(2) If the moon attracts the Earth, then why does the Earth not move towards the moon?

Answer:

1.  No, her friend will not agree with the weight of the gold bought because the weight at the poles is greater than the weight at the equator.
2. We know that the gravitational force is always attractive, still, the moon does not fall on the Earth because the gravitational force between the Earth and the moon works as the necessary centripetal force for the moon to make it revolve around the Earth.

Question 5. Give two reasons for the variation of g at the equator and at the poles.

Answer:

The variation of g at the equator and at the poles are

1. due to the difference in the radius and
2. due to the rotation of the earth.

Question 6. Two solid objects of masses 1 kg and 2 kg are dropped from a helicopter at the same time. Which one will reach the ground earlier? Justify your answer with a suitable reason.

Answer:

Both will reach the ground at the same time as we know that an object experiences acceleration during free fall. This acceleration experienced by an object is independent of its mass because g=\(\frac{G M}{R^2}\)

As they are dropped at the same time, they will reach the ground at the same time.

Question 7. Give reasons.

1.  The moon does not have an atmosphere.
2.  If you jump on the moon, you will rise much higher than, if you jump on the earth.

Answer:

Moon does not have strong gravity to hold atmospheric gases.

Acceleration due to gravity g is much less on the moon’s surface than that of the earth’s surface. v² -u²

Hence, h=\(\frac{v^2-u^2}{2 g}\) is larger.

Question 8. Give one example of each where the same force acting on

1. a smaller area exerts a larger pressure.
2. a larger area exerts a smaller pressure.

Answer:

1.  Needles have sharp tips having smaller areas to exert a larger pressure.
2. School bags have broad bases and wide straps having larger areas to exert less pressure.

Question 9. Explain, why a camel walks easily on a sandy surface than a man in spite of the fact that a camel is much heavier than a man.

Answer:

The feet of the camel are larger and so cover a larger area, which results in low pressure on the sand bed as compared to human beings and thus, enables them to walk easily on sand without sinking.

Question 10. Give a brief description of why the bottom of dams is broad.

Answer:

As we know, pressure is the force acting per unit area. Dams have large water storage. If the bottom of the dams is not made broad, the large hydraulic pressure may cause dams to sink into the Earth’s basin. So, an increase in the base area decreases the pressure exerted by large water storage.

Question 11. Why pressure exerted by the solid and fluid are different?

Answer:

1. Solids exert pressure because of their weight, i.e. pull of mass by the centre of the Earth with an acceleration of g = 9.8 ms^-2.
2. Similarly, fluids, i.e. gases and liquids both exert equal pressure in all directions over the inner walls of the container in which it is kept.

Question 12. Name two forces which act on a body immersed in a liquid. Give the directions in which they act.

Answer:

The two forces are

1. Weight of the body acting downwards.
2. The buoyant force acts upwards.

Question 13. State the condition under which an object floats on the surface of a liquid. What is the volume of the liquid displaced by the object?

Answer:

1. When the upthrust on the body in a liquid is greater than the weight of the body, then the body floats on the surface of the liquid.
2. The volume of the body = volume of liquid displaced.

Question 14. (1) Name the SI units of thrust and pressure.

(2) In which situation, do we exert more pressure on the ground when we stand on one foot or on both feet? Justify your answer.

Answer:

1. The SI unit of thrust is Newton (N). The SI unit of pressure is Nm^-2or pascal (Pa).
2.  We exert more pressure on the ground when we stand on one foot than both feet, as the area of one foot is half that of two feet as p \(\propto \frac{1}{A}\).

Question 15. (1) Explain, why a completely immersed bottle in water bounces back on the surface.

(2) Why does a bucket of water weigh less inside the well water?

Answer:

1. Since it is known that a body can sink in water only when its weight is greater than the upthrust acted on it by the water. But in this case, the upthrust act on the bottle is greater than its weight, that’s why, it bounces back on the water surface.
2. A bucket of water weighs less inside the well water, it is because when the bucket immersed in the water fully, upthrust “acts on it by water which reduces its actual weight.

Question 16. When a fresh egg is put into a beaker filled with water, it sinks in water. But when a lot of salt is dissolved in the water, the egg begins to rise and then floats. Why?

Answer:

On dissolving a lot of salt in water, the density of salt solution becomes higher than that of pure water. Due to its much higher density, the salt solution exerts much more upward buoyant force on the egg, making the egg rise and then float.

Question 17.  Two different bodies are completely immersed in water and undergo the same loss in weight. Is it necessary that their weights in the air should also be the same? Justify your answer.

Answer:

No, their weights in air don’t need to also be the same. This is because the two bodies have undergone the same loss in weight on completely immersing in water due to their equal volumes, not due to their equal weights. So, they may have different weights in the air.

Question 18. (1) Why does a bucket of water feel heavier when taken out of water?

(2) Lead has greater density than iron and both are denser than water. Is the buoyant force on a lead object greater than, less than or equal to the buoyant force on an iron object of the same volume?

Answer:

1.  A bucket of water feels heavier when taken out of water because when immersed in water, an upward force, i.e. buoyant force acts on it which is equal to the weight of water displaced by the bucket.
2. The buoyant force on a lead object is lesser than the buoyant force on an iron object because lead has greater density, so it displaces a lesser amount of water consequently lesser amount of buoyant force acts on it.

Question 19. Verify Archimedes’ principle of buoyancy with an activity. For the activity, you are provided with a piece of stone, a rubber string and a container filled with water.

Answer:

1. First of all, tie up the stone with the rubber string and hold it against a scale fixed on a wall. Put a mark on the elongated rubber string when the stone is tied.
2. Repeat this experiment but this time suspend the stone in a beaker, filled with water. Now, compare the markings.
3. Explanation: As the buoyant force is acting on the stone in an upward direction, this, gravitational pull decreases and in turn, the stretch of the rubber is lost.

Question 20. State Archimedes’ principle. Write its two applications.

Answer:

Archimedes’ principle states that “when a body is immersed fully or partially in a liquid, it experiences an upward force that is equal to the weight of the liquid displaced by it.”

Applications

1. It is used in designing ships and submarines.
2. It is used in making lactometers, which are used to determine the purity of milk.

Question 21. When a metallic block is immersed below the surface of a liquid, state and define the upward force acting on it.

Answer:

When a metallic block is immersed below the surface of the liquid, it experiences an upward force known as buoyant force.

Buoyant force acting on the block =Weight of liquid displaced by object = mg=(V \(\rho\)) g

where, m – mass of the metallic block

⇒ \(\rho\) = density of metallic block V = volume

g = acceleration due to gravity.

Question 22. If a body is compressed to half its previous volume, what will be the effect on its density and why?

Answer:

1. Since, density \((\rho)=\frac{\text { mass }(m)}{\text { volume }(V)}\)
2. Therefore, if the volume of a body is compressed to half of its previous volume, then the density will be doubled.

Question 23. A block of ice is floating in a bucket of water full up to the brim. Some portion of the ice is visible above the water level. As it melts completely do you expect water to spill out? Give a reason for your answer.

Answer:

We know that the volume of ice is greater than that of water. So in this case when the ice melts, it occupies a volume less than that occupied by the ice dipped in it, so the water level will not spill out.

Question 24. Identical packets are dropped from two aeroplanes, one above the equator and the other above the North Pole, both at height h. Assuming all conditions are identical, will those packets take the same time to reach the surface of the earth? Justify your answer.

Answer:

1. No, those packets do not take the same time to reach the surface of the earth.
2. As the value of g is less at the equator than poles, so packets dropped at the poles reach the surface of the earth first.

Question 25. How will the weight of a body of mass 100 g change, if it is taken from the equator to the poles? Give reasons.

Answer:

Since, the acceleration due to gravity increases from the equator to the poles, so its weight increases because the radius of the earth is less at the poles than at the equator.

Question 26. When you immerse an empty plastic bottle in a bucket of water, it comes above the surface of the water. Why does this happen? How can it remain immersed in water and why?

Answer:

1. When we immerse an empty plastic bottle in a bucket of water, the upward force (upthrust or buoyant force) exerted by water on the bottle is greater than its own weight, therefore it comes above the surface of the water.
2. To keep the bottle completely immersed, an external force which is equal to the difference between the upward force and the weight of the bottle, must be applied to the bottle in a downward direction. This is because the upthrust on the bottle due to water must be balanced.

Question 27. What happens to the magnitude of the force of gravitation between two objects, if (1) distance between the objects is tripled? (2) mass of both objects is doubled?

Answer:

As we know,F=\(\frac{G m_1 m_2}{r^2}\) [symbols have their usual meanings]

(1) \(r^{\prime}=3 r \Rightarrow F^{\prime}=\frac{G m_1 m_2}{9 r^2}=\frac{F}{9}\) [force decreases by 9 times]

(2) \(m_1^{\prime}=2 m_1 and m_2^{\prime}=2 m_2\)

⇒ \(F^{\prime}=\frac{4 G m_1 m_2}{r^2}=4 F\) [force increases by 4 times]

To maintain the same force one of the mass is to be increased by 36 times.

Question  29. A body weighs 25 kg on the surface of the earth. If the mass of the earth is \(6 \times 10^{24}\) kg then the radius of the earth is 6.4 \(\times 10^{6}\) and the gravitational constant 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\).

Calculate

1. the mutual force of attraction between the body and the earth.
2. the acceleration produced in the body.

Answer:

Given, \(M_e=6 \times 10^{24} \mathrm{~kg}\),m=25 kg

⇒ \(R_e=6.4 \times 10^6 \mathrm{~m}\) and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

(1) Mutual force, F=G \(\frac{M_e}{R_e^2} m \)

= \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 25}{\left(6.4 \times 10^6\right)^2}=244 \mathrm{~N}\)

(2) Acceleration produced in the body,

a=\(\frac{F}{m}=\frac{244}{25}=9.8 \mathrm{~m} / \mathrm{s}^2\)

Question 30. Two bodies of masses 3 kg and 12 kg are placed at a distance of 12 m. A third body of mass 0.5 kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point, for Given, mx =3 kg and m² =12 kg

Answer:

Given, \(m_1\)=3 kg and m_2=12 kg

Let the mass, \(m_3\)=0.5 kg be placed at a distance of x from m_1 as shown

Then, the force acting on \(m_3\) due to \(m_1\) is equal and opposite to the force acting on \(m_3\) due to \(m_2\).

⇒ \(F_{31}=F_{32}\)

⇒ \(\frac{G m_1 m_3}{x^2}=\frac{G m_3 m_2}{(12-x)^2} \Rightarrow \frac{3}{x^2}=\frac{12}{(12-x)^2}\)

⇒ \(\left(\frac{12-x}{x}\right)^2=\frac{12}{3}\) =4

⇒ \(\frac{12-x}{x}\) =2

12-x=2 x  \(\Rightarrow\) 12=3 x \(\Rightarrow x=4 \mathrm{~m}\)

The required point is at a distance of 4 m from the mass of 3 kg.

Question 31. Calculate the acceleration due to gravity on the surface of satellite having mass 7.4 x \(10^22\) kg and radius 1.74 x \(10^6\) cm. (Take, G = 6.7 x \(10^{-11} \mathrm{~N}-\mathrm{m} / \mathrm{kg}^2\))

Answer:

As we know, acceleration due to gravity, g=\(\frac{G M}{R^2}\).

For the satellite, R =1.7 .4 \(\times 10^6 \mathrm{~cm}=\frac{1.74 \times 10^6}{100}=1.74 \times 10^4 \mathrm{~m}\)

M =7.4 \(\times 10^{22} \mathrm{~kg}\)

g =\(\frac{6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{1.74 \times 10^4 \times 1.74 \times 10^4}\)

=\(\frac{6.67 \times 7.4}{1.74 \times 1.74} \times 10^3 \mathrm{~g}=16.30 \times 10^3 \mathrm{~m} / \mathrm{s}^2\)

Question 32. What height above the surface of the earth, the value of g becomes 64% of its value at the surface of the earth? Take, the radius of the earth = 6400 km. r? Let g – acceleration due to gravity at the earth’s surface.

Answer:

g, = acceleration due to gravity at height

⇒ \(g_h\) = acceleration due to gravity at height

h =\(\frac{64}{100} \times g=0.64 g\)

g =\(\frac{G M}{R_e^2}\)

Similarly, \(g_h=\frac{G M}{\left(R_e+h\right)^2}\)

From Eqs. (1) and (2), we get

⇒ \(g_h=\frac{g R_e{ }^2}{\left(R_e+h\right)^2}\)

0.64 g=\(\frac{g R_e{ }^2}{\left(R_e+b\right)^2}\)

0.64\(\left(R_e+b\right)^2=R_e^2\)

0.8\(\left(R_e+h\right)=R_e\)

0.8 h=\(R_e-0.8\)

⇒ \(R_e=0.2 R_e\)

h=\(\frac{2 \times 6400}{8}=1600 \mathrm{~km} \left[ R_e=6400 \mathrm{~km}\right]\)

Thus, the height above the surface of the earth is 1600 km.

Question 33. Estimate the gravitational force between two protons (1.6x \(10^{-27}\) kg) separated by a distance of I A.

Answer:

Since, gravitational force, F=\(\frac{G M_p M_p}{r^2}\)

Given, mass of proton, \(M_p=1.6 \times 10^{-27} \mathrm{~kg}\)

Distance, r=1 \(Å=10^{-10} \mathrm{~m}\)

F =\(\frac{6.67 \times 10^{-11} \times 1.6 \times 10^{-27} \times 1.6 \times 10^{-27}}{\left(10^{-10}\right)^2}\)

=17.1 \(\times 10^{-45} \mathrm{~N}\)

Question 34. A ball is dropped from a height half of the earth’s radius. Find the value of g at this point.

Solution:

According to the question, the distance of the ball from the earth’s centre,

⇒ \((R+h)=R+\frac{R}{2}=\frac{3 R}{2} [height, h=R / 2]\)

From the formula of acceleration due to gravity,

g =\(\frac{G M}{(R+h)^2}=\frac{G M}{(3 R / 2)^2}=\frac{4 G M}{9 R^2}\)

= \(\frac{4}{9} \times 10=4.44 \mathrm{~m} / \mathrm{s}^2 \left[ \frac{G M}{R^2}=10 \mathrm{~m} / \mathrm{s}^2\right]\)

Thus, the value of g is 4.44 \(\mathrm{~m} / \mathrm{s}^2\).

Question 35. A body is dropped from a height of 100 m. What is its height above the ground after 2 seconds of its fall? (Take, g=10m/s² )

Solution:

Given, initial velocity, u = 0 Time taken, t – 2 s

Acceleration due to gravity, a = g

From the second equation of motion,

⇒ \(s=u t+\frac{1}{2} g t^2=0+\frac{1}{2} \times 10 \times(2)^2=20 \mathrm{~m}\)

The height of the body above the ground after 2 s of its fall, h =100 – 20 =80 m

Question 36.  A ball is thrown up with a velocity of 19.6 m/s.

1. How long will it take to reach the maximum height?
2. How high will it go?

Answer:

Given, initial velocity, u = 19.6 m/s

(1) Final velocity, v = 0 at maximum height.

So, t=\(\frac{u}{g}=\frac{19.6}{9.8}=2 \mathrm{~s}\) [ v=u-g t and v=0]

From the third equation of motion,

⇒ \(v^2=u^2-2 g h\) [negative sign is taken due to upward motion]

⇒ \(u^2=2 g h\) {[ v=0]}

h=\(\frac{u^2}{2 g}=\frac{(19.6)^2}{2 \times 9.8}=19.6 \mathrm{~m}\)

Question 37. Your mass on the earth is 50 kg. Planet M has two times the force of gravity of that of the Earth. What will be your mass and weight on planet M? (Take, g = 9.8 m/s²)

Answer:

As the mass remains constant, so on the planet M your mass will be 50 kg.

Given, that acceleration due to gravity on planet M is two times that of the Earth.

So, weight on planet M will be 50 x 2 x g = 50x2x9.8 =980

Question 38. Suppose that the radius of the earth becomes twice its original radius without any change in its mass. Then, what will happen to your weight?

Answer:

The weight of a body is the force with which a body is attracted towards the earth, w=\(\frac{G M m}{R^2}\)

If the radius of the earth becomes twice its original radius, then w=G \(\frac{M m}{(2 R)^2}=\frac{G M m}{4 R^2}=\frac{w}{4}\)

Thus, the weight will be reduced to one-fourth of the original

Question 39. 39 Find the weight of an 80 kg man on the surface of the moon. What should be his mass on the earth and on the moon?

(Take, \(g_e=9.8 \mathrm{~m} / \mathrm{s}^2, g_m=1.63 \mathrm{~m} / \mathrm{s}^2\) )

Answer:

Given, mass on the earth = mass on the moon =80 kg

Man’s weight on the earth,

⇒ \(w_e=9.8 \times\) 80=784 N

Man’s weight on the moon,

⇒ \(w_m=1.63 \times\) 80=130.4 N

Thus, the weight of the moon is 130.4 N.

Question 40. 40 The weight of any person on the moon is about l/6th of that on the earth. He can lift a mass of 15 kg on the earth. What will be the maximum mass which can be lifted by the same force applied by the person on the moon?

Answer:

Maximum weight which can be lifted = mg

=15 \(\mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2=147 \mathrm{~N}\)

Mass which can be lifted on the moon

=147 \(\mathrm{~N} \times \frac{1}{g_{\text {earth }} / 6}\)

[at moon, acceleration due to gravity is 1/6 th to that of the earth]

= \(\frac{147 \times 6}{9.8}=\frac{147}{1.63}=90 \mathrm{~kg}\)

Thus, the maximum mass which can be lifted is 90 kg.

Question 41. Calculate the average density of the earth in terms of g, G and R.
Answer:

According to the formula, density =\(\frac{\text { mass }}{\text { volume }}\).

If the radius of the earth is R, then its volume =\(\frac{4}{3} \pi R^3\). [The shape of the earth is spherical]

Mass of the earth, M=\(\frac{g R^2}{G} [ g=\frac{G M}{R^2}]\)

Density =\(\frac{g R^2}{G} / \frac{4}{3} \pi R^3=\frac{3 g}{4 \pi R G}\)

Thus, the average density of the earth is \(\frac{3 g}{4 \pi R G}\)

Question 42. Which will exert more pressure, 100 kg mass on 10 m² or 50 kg mass on 4 m²? Give reason. (Take, g = 10 ms^-2)

Answer:

For 100 kg of mass,

Pressure, \(p_1=\frac{\text { Force }}{\text { Area }}=\frac{m_1 g}{A_1}=\frac{100 \times 10}{10}=100 \mathrm{~Pa}\)

For 50 kg of mass,

Pressure, \(p_2=\frac{m_2 g}{A_2}=\frac{50 \times 10}{4}=125 \mathrm{~Pa}\)

⇒ \(p_2>p_1\)

Hence, a 50 kg mass will exert more pressure.

Question 43. A block of wood of mass 5 kg and dimensions 40 cm x 20 cm x 10 cm is placed on a tabletop. Find the pressure exerted, if the block lies on the tabletop with sides of dimension

1. 40 cm x 20 cm
2. 40 cm x 10 cm (Take, g = 10 \(\mathrm{~ms}^{-2}\))

Answer:

Given, mass, m =5 kg, acceleration due to gravity, g =10 \(\mathrm{~ms}^{-2}\)

g=10 \(\mathrm{~ms}^{-2}\)

Weight =m g=5 \(\times 10=50 \mathrm{~N}\)

(1) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 20 \times 10^{-4}} \mathrm{~Pa}\)

=6.25 \(\times 10^2 \mathrm{~Pa}\)

(2) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 10 \times 10^{-4}} \mathrm{~Pa}\)

=1.25 \(\times 10^3 \mathrm{~Pa}\)

Question 44. If two forces in the ratio 5:9 act on two areas in the ratio 10:3, find the ratio of pressure exerted.

Answer:

Given, ratio of areas =10: 3$, i.e. \(A_1 / A_2\)=10 / 3

Ratio of forces =5: 9, i.e. \(F_1 / F_2\)=5 / 9

Ratio of pressures =\(\frac{p_1}{p_2}=\frac{F_1}{A_1} \times \frac{A_2}{F_2}\)

= \(\frac{F_1}{F_2} \times \frac{A_2}{A_1}=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}\)

Therefore, the ratio of the pressures is 1:6.

Question 45. A ball filled with air has a volume of 500 cm3. Calculate the minimum force applied by a child to put it completely inside the water. (Take, g = 10 \(\mathrm{~ms}^{-2}\))

Answer:

Given that,

Volume, V=500 cm³ =500 x 10^-6 m³,

g = 10 ms^-2, F =?

The force required to put the ball inside the water = Buoyant force

= Weight of water displaced – mg … (1)

Now, we know that,

Mass of water = Density of water x Volume m =pV

On substituting this value in Eq. (1), we get Force = p Vg

= (1000 kg m^-3) X (500 X 10^-6 m³) X (10 ms^-2)

= 1000 x 500 x 10~6 x 10 N =5 N

The minimum force applied by a child to put the ball completely inside the water is 5 N.

Question 46. A ball weighs 80 g in air, 60 g in water and 50 g in liquid. If the density of water is 1 g cm^-3, find the density of this liquid.

Answer:

As given, when immersed in water, the ball displaces 80 -60 = 20 g of water.

So, its volume = volume of water displaced

=\(\frac{20 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=20 \mathrm{~cm}^3\)

Density of ball =\(\frac{80 \mathrm{~g}}{20 \mathrm{~cm}^3}=4 \mathrm{~g} \mathrm{~cm}^{-3}\)

So, the ball displaces (80-50) \(\mathrm{g}=30 \mathrm{~g}\) of liquid with density =\(\frac{30 \mathrm{~g}}{20 \mathrm{~cm}^3}=1.5 \mathrm{~g} \mathrm{~cm}^{-3}\).

Question 47. Prove that, if a body is thrown vertically upwards, then the time of ascent is equal to the time of descent.

Answer:

For the upward motion,

v=u-g \(t_1\),  0=u-g \(t_1\),  \(t_1=\frac{u}{g}\)

and the downward motion,

v=u+g \(t_2\), v=0+g \(t_2\)

The body falls back to the earth at the same speed as it was thrown vertically upwards.

v=u,  u=0+g \(t_2 \Rightarrow t_2=\frac{u}{g}\)

From Eqs. (1) and (2), we get \(t_1=t_2 \Rightarrow\) Time of ascent = Time of descent

Question 48. A ball is dropped from the edge of a roof. It takes 0.1s to cross a window of height 2.0 m. Find the height of the roof above the top of the window.

Answer:

Let AB be the window and suppose the roof T is at a height y above A. Also, suppose it takes a time TV for the ball to reach A. The velocity of the ball at A is

⇒ \(v_1=0+g t_1=9.8 t_1\) (1)

Now, consider the motion of the ball from A to B.

Here, the initial velocity is v1 the distance covered is 2 m and the time taken is 0.1 s.

From second equation of motion, s=u t+\(\frac{1}{2} g t^2\)

2.0=\(v_1(0.1)+\frac{1}{2} \times 9.8 \times(0.1)^2 =9.8 t_1(0.1)+0.049\)

⇒ \(t_1 =1.99 \approx 2 \mathrm{~s}\)

The height y is y=\(\frac{1}{2} g t_1^2\)

= \(\frac{1}{2} \times 9.8 \times(2)^2=19.6 \mathrm{~m}\)

The roof is at a height of 19.6 m above the top of the window.

Question 49. On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?

Answer:

For both the stones, initial velocity, u- 0

Acceleration in a downward direction = g

Now, from the second equation of motion,

h=u t+\(\frac{1}{2} g t^2 \Rightarrow  h=0+\frac{1}{2} g t^2\)

h=\(\frac{1}{2} g t^2 \quad \Rightarrow t=\sqrt{\frac{2 h}{g}}\)

Both stones will take the same time to reach the ground because the two stones fall from the same height.

Question 50. A ball is thrown with some speed u m/s. Show that under the free fall, it will fall on the ground at the same speed.

Answer:

When the ball is thrown upwards, then it will reach a certain height h and start falling. At maximum height h, the final velocity will be v = 0.

The maximum height reached by the ball,

⇒ \(v^2-u^2=2 g h\)

⇒ \(0-u^2=-2 g h\) [acceleration =-g]

h=\(\frac{u^2}{2 g}\)  [using equation]

In the second case, when the ball starts to fall, then the initial velocity u=0. It will accelerate due to gravity, i.e. a=g and reach ground with speed (say \(v_2\) ).

Using equation, \(v_2^2-u^2\) =2 g h

⇒ \(v_2^2-0 \)=2 g h

⇒ \(v_2^2 =2 g\left(\frac{u^2}{2 g}\right)=u^2\) [{ from Eq. (1) })

⇒ \(v_2\) =u

Thus, the ball reaches the ground at the same speed.

Class 9 Science Chapter 9 Gravitation Long Question And Answers

Question 1. (1) Write the formula to find the magnitude of the gravitational force between the earth and an object on the earth’s surface.

(2) Derive how does the value of gravitational force F between two objects change when

1.  distance between them is reduced to half and
2. mass of an object is increased four times.

Answer:

(1) Formula to find the magnitude of gravitational Force, F=\(\frac{G M m}{R^2}\)

M=mass of the earth

m = mass of the object

R = radius of the earth

and universal gravitational constant, G=\(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

(2) (1) Let the gravitational force be F when the distance between them is R,

F=\(\frac{G M m}{R^2}\)  → Equation 1

Now, when the distance is reduced to half,

⇒ \(F^{\prime}=\frac{G M m}{\left(\frac{R}{2}\right)^2}=\frac{4 G M m}{R^2}\)  → Equation 2

On dividing Eq. (1) by Eq. (2), we get

⇒ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m}\)

⇒ \(F^{\prime}\) =4 F

(2) When the mass becomes 4 times,

∴ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m} \Rightarrow F^{\prime}=4 F\)

Question 2.  (1) Prove that, if the earth attracts two bodies placed at the same distance from the centre of the earth with equal force, then their masses will be the same.

(2) Mathematically express the acceleration due to gravity in terms of mass of the earth and radius of the earth.

(3) Why is G called a universal constant?

Answer:

Let the two bodies have masses m₁ and m₂ and they are placed at the same distance R from the centre of the earth. According to the question, if the same force acts on both of them, then

⇒ \(F_1=\frac{G M m_1}{R^2}\)

and\(F_2=\frac{G M m_2}{R^2}\)

As, \(F_1=F_2\)

Hence, \(\frac{G M m_1}{R^2}=\frac{G M m_2}{R^2}\)

So, \(m_1=m_2\), their masses will be the same.

(2) Mathematically, g=\(\frac{G M}{R^2}\).

where, g = acceleration due to gravity

G = universal gravitational constant M mass of the earth and R = radius of the earth

(3) G is known as the universal gravitational constant because its value remains the same all the time everywhere in the universe, applicable to all bodies whether celestial or terrestrial.

Question 3. (1) A person weighs 110.84 N on the moon, whose acceleration due to gravity is 1/6 of that of the earth. If the value of g on the earth is 9.8 m/s², then calculate

1.  g on the moon
2.  mass of people on the moon
3. weight of the person on the earth

(2) How does the value of g on the earth is related to the mass of the earth and its radius? Derive it.

Answer:

(1) g on the moon is given by

\(g^{\prime}=\frac{g}{6}=\frac{9.8}{6}=1.63 \mathrm{~m} / \mathrm{s}^2\)

(2) Mass of the person on the moon

=\(\frac{110.84}{1.63}=68 \mathrm{~kg}\)

(3) Weight of person on the earth = mg

=68 \(\times\) 9.8

=666.4 \(\mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Weight of a person on the earth will be

w=\(\frac{G M m}{R^2}\)

where, M=mass of the earth R = radius of the earth m = mass of person and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)

Question 4. Two objects of masses mx and having the same size are dropped simultaneously from heights ha and h₂, respectively. Find out the ratio of time they would take to reach the ground. Will this ratio remain the same, if

1.  one of the objects is hollow and the other one is solid and
2. both of them are hollow, size remains the same in each case. Give reason.

Answer:

Height of object A, \(h_1=\frac{1}{2} g t_1^2\)

Height of object B, \(h_2=\frac{1}{2} g t_2^2\)

⇒ \(h_1: h_2=t_1^2: t_2^2\)

or \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)

(1) Acceleration due to gravity is independent of the mass of the falling body. So, the ratio remains the same.

(2) If bodies are hollow, then also ratio remains the same, \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)

Question 5. A stone is dropped from the edge of a roof.

1. How long does it take to fall 4: 9 am
2.  How fast does it move at the end of that fall?
3. How fast does it move at the end of 7.9 m?
4.  What is its acceleration after Is and after 2s?

Answer:

Given, initial velocity u=0

Acceleration g=9.8 \(\mathrm{~m} / \mathrm{s}^2\)

(1) We have, s =u t+\(\frac{1}{2} g t^2\)

4.9 =0 \(\times t+\frac{1}{2} \times 9.8 \times t^2\)

⇒ \(t^2 =\frac{9.8}{9.8}=1 \Rightarrow t=1 s\)

The stone takes 1 s to fall 4.9 m

(2) We have, \(v^2-u^2=2\) as

⇒ \(v^2-0^2 =2 \times 9.8 \times\) 4.9

v =\(\sqrt{96.04}=9.8 \mathrm{~m} / \mathrm{s}\)

At the end of 4.9 m, the stone will be moving at a speed of 9.8 m/s.

We have, \(v^2-u^2\)=2 as

⇒ \(v^2-0^2 =2 \times 9.8 \times 7.9\)

v =12.44 \(\mathrm{~m} / \mathrm{s}\)

=12.44 m/s

The stone will be moving with a speed of 12.44 m/s at the end of 7.9 m

(4) During the free fall the acceleration produced in a body remains constant.

So, acceleration after 1 s = 9.8 m/s² Acceleration after 2 s = 9.8 m/s²

Question 6. (1) A steel needle sinks in water but a steel ship floats. Explain, how.

(2) Why do you prefer a broad and thick handle for your suitcase?

Answer:

(1) The ship displaces more water than the needle as the volume of the ship is more than that of the needle. Since upthrust depends on the volume of the object (U = Vdg), so more the volume of the object, the more upthrust acts on it and the object floats.

(2) Since, pressure act on the body is inversely proportional to the surface area of contact, i.e. p \(\propto \frac{1}{A}\)

It means that the more the area of contact, the less pressure will act on the body. As the broad and thick handle of our suitcase has a large area, due to which less pressure acts on our hand and it is very easy to take from one place to another

Question 7. The radius of the earth at the poles is 6357 km and the radius at the equator is 6378 km. Calculate the percentage change in the weight of a body when it is taken from the equator to the poles.

Answer:

Let acceleration due to gravity at the equator,

⇒ \(g_e=\frac{G M_e}{R_e^2}\)

and acceleration due to gravity at poles,

⇒ \(g_p=\frac{G M_e}{R_p^2}\)

The variation of acceleration due to gravity,

⇒ \(\Delta g=g_p-g_e=G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)\)

e variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)

Percentage variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)

= \(\frac{R_e^2-R_p^2}{R_e^2 R_p^2} \times 100 \times R_e^2=\frac{R_e^2-R_p^2}{R_p^2} \times 100\)

= \(\frac{(6378)^2-(6357)^2}{(6357)^2} \times 100 \approx 0.7 \%\)

% variation in the weight of a body = % Change in  g=0.7 \%

Work, Energy And Power Multiple Choice Question And Answers

Question 1. When a horse-pulls a cart, who does the work ?

1. Cart
2. Wheels
3. Road
4. Horse

Answer: 4. Horse

When a horse-pulls a cart, work is being done by horse.

Question 2. If 10 N of force is applied to an object, but the object does not move, then how much work being done by the force ?

1. Zero
2. 10 J
3. 10N
4. 20 J

Answer: 1. Zero

If there is no displacement due to application of force, then net work done will be zero.

Question 3. A 4 N of force displaces a body by 2m, the work done will be

1. 5 J
2. 8 J
3. 8 N
4. 5 N

Answer: 2. 8 J

Work done by given force =force x displacement = 4 x 2 =8J

Question 4. A car is moving uniformly in a circular racing track of radius 1 km. If the car makes two complete round. How much work is done by the car, if car has a force of 20 kN?

1. Zero
2. 20 kJ
3. 200 kJ
4. 2000 kJ

Answer: 1. Zero

During a circular motion, force is perpendicular to the displacement, so net work done is zero.

Question 5. What kind of energy is possessed by a running horse?

1. Kinetic energy
2. Potential energy
3. No energy
4. Heat energy

Answer: 1. Kinetic energy

A running horse has kinetic energy.

Question 6. If two stones A and B are dropped from a tower, then which one has maximum kinetic energy ?

1. Lighter stone
2. Heavier stone
3. Both have equal
4. None of the above

Answer: 2. Heavier stone

If two stones are dropped from some height, then the heavier stone has greater kinetic energy w.r.t. lighter one.

Question 7. A monkey weighing 50 kg climbs up a vertical tree of height 300 m. How much potential energy does it gain? [Take, g=9.8 m / \(s^2\) ]

1. 147 \(\times 10^4 \mathrm{~J}\)
2. 14.7 \(\times 10^3 \mathrm{~J}\)
3. 14.7 \(\times 10^2 \mathrm{~J}\)
4. 14.7 \(\times 10^4 \mathrm{~J}\)

Answer: 4. 14.7 \(\times 10^4 \mathrm{~J}\)

Given, m=50 kg, g=9.8 \(ms^{-2}\), b=300 m

Work done by the body = mgh

=50 \(\times 9.8300=14.7 \times 10^4 \mathrm{~J}\)

Gain in PE = Work done =14.7 \(\times 10^4\) J

Question 8. A scooter of mass 150 kg is travelling at 10 m/s. If its speed increases to 15 m/s, then by how much amount does its kinetic energy increases ?

1. 9500 J
2. 9425 J
3. 9375 J
4. 9753 J

Answer: 3. 9375 J

Given, m=150 kg, \(v_1\)=10 m / s and \(v_2\)=15 m/ s

The change in kinetic energy

=\(\mathrm{KE}_2-\mathrm{KE}_1=\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2\)

=\(\frac{1}{2} \times 150 \times 15^2-\frac{1}{2} \times 150 \times 10^2\)

= 16875-7500= 9375 J

Question 9. A ball is allowed to fall freely from a tower. Which energy is possessed at the middle point during the fall ?

1. Kinetic only
2. Potential only
3. Both (1) and (2)
4. Heat only

Answer: 3. Both (1) and (2)

At middle point of fall, the ball has both kinetic and potential energy.

Question 10. In head phone, the electrical energy is converted to what ?

1. Light energy
2. Sound energy
3. Chemical energy
4. Heat energy

Answer: 2. Sound energy

Electrical energy changes into sound energy.

Question 11. Which mathematical relation of energy of a stone of mass m fading freely from height h remains conserved at every point in its downward motion?

1. E = mgh
2. E=\(\frac{1}{2} m v^2\)
3. E=\(m g h+\frac{1}{2} m v^2\)
4. None of these

Answer: 3. E=\(m g h+\frac{1}{2} m v^2\)

Total energy ot the free hilling stone is conserved.

So, total energy = kinetic energy + potential energy.

Or E=\(m g+\frac{1}{2} m v^2[latex]

Question 12. If a stone of mass 5 kg drops from height 20 m, then what will be the velocity at the surface of earth ? (hike, g = 10 m/s²)

1. 20 m/s
2. 30 m/s
3. 35 m/s
4. 40 m/s

Answer: 1. 20 m/s

Given, mass of stone, m = 5 kg

Height, h = 20 m and g =10 m/s²

So, potendal energy at highest point = mgh

= 5 x 10 x 20 = 1000 J

Now, according to conservation of energy,

kinetic energy at surface of the earth = potential energy at height h

So, 1/2 m [latex]v^2\) =\(m g h^h\)

⇒ \(v^2 =\frac{2 m g^b}{m}\)

v =\(\sqrt{2 g^b}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}\)

Question 13. Mukesh drops a ball of 200 g from a tower of height 20 m. What will be its kinetic energy at the height of 5 m ?

1. 30 J
2. 50 J
3. 60 J
4. 65 J

Answer: 1. 30 J

Given, m = 200 g = 0.2 kg, height, h = 20 m

When ball reaches at height of 5 m, its velocity is given by

⇒ \(v^2=v^2+2 g\)

⇒ \(v^2=0^2+2 \times 10 \times 15\)

v=\(\sqrt{300}=\sqrt{100 \times 3}=10 \sqrt{3}\)

So, kinetic energy \(\left.=\frac{1}{2} m v^2=\frac{1}{2} \times 0.2 \times 300=30\right]\)

Question 14. An object of mass 50 kg is raised to a height of 7 m above the ground. What is its potential energy? If object is allowed to fall, then its kinetic energy when it is half way down.

1. 3500 J, 1650 J
2. 3800 J, 1750 J
3. 3800 J, 1650 J
4. 3500 J, 1750 J

Answer: 4. 3500 J, 1750 J

Given, mass, m = 50 kg, height, h = 7 m

As, potential energy is given by PR = mgh =50x 10x 7 = 3500J

When it is allowed to fall its potential energy gets converted into kinetic energy.

So, when it reaches the half-way, half of its potential energy gets converted to kinetic energy.

Question 15. A horse does 5000 J of work in 100 s. What is its power ?

1. 50 W
2. 50 J
3. 10W
4. 10J

Answer: 1. 50 W

Given, work done (W) = 5000 J

Time taken (r) =100 s

We know that,

Power,P=\(\frac{W}{t}=\frac{5000}{100}=50 \mathrm{~J} / \mathrm{s}\)

Thus, power, P=50 W

Question 16. The power uses of household water pump is 200 W. What is the mass of water, the pump can lift per minute from a 7 m deep tank? [Take, g =10 m/s²]

1. 172 kg
2. 272 kg
3. 150 kg
4. 250 kg

Answer: 1. 172 kg

Given,power,.P = 200 W^ = 60 s,height,h = 7m

Energy supplied to the pump = Power x Time = 200 Wx 60s =12000J

So, energy =mgh

12000 = m x 10 x 7

m = 172 kg

So, mass of the water is 172 kg.

Question 17. A certain household has consumed 200 units of energy during a month. How much energy is then in joules?

1. 72 \(\times 10^5 \mathrm{~J}\)
2. 72 \(\times 10^7 \mathrm{~J}\)
3. 7.2 \(\times 10^8 \mathrm{~J}\)
4. 72 \(\times 10^{\mathrm{B}} \mathrm{J}\)

Answer: 3. 7.2 \(\times 10^8 \mathrm{~J}\)

Given, energy consumed = 200 units

= 200 kWh = 200x 1000×3600

= 7.2 x 108 J

Question 18. Ankit does 500 J of work in 15 min and Bablu does 700 J of work in 30 min. Who expends more power, Ankit or Bablu ?

1. Ankit
2. Bablu
3. Both same
4. Insufficient information

Answer: 1. Ankit

Power expended by Ankit =\(\frac{W}{t}=\frac{500}{15 \times 60}\)=0.55J

and power expended by Bablu =\(\frac{700}{30 \times 60}\)=0.38 J

Ankit expends more power.

Class 9 Science Chapter 10 Work, Energy And Power Very Short Answer Type Questions

Question 1. What is the work done when you apply an ION force on a wall?
Answer:

Zero work because there is no displacement of the wall.

Question 2. Why do we say work done against gravity is negative?
Answer:

It is because force and displacement are in opposite directions to each other.

Question 3. A man is holding a suitcase in his hand at rest. What is the work done by him?
Answer:

Zero, as displacement is zero.

Question 4. A boy is running along a circular path at a uniform speed. What is the work done by the boy?
Answer:

Work done by the boy is zero because the angle between the force (centripetal) and displacement in case of circular motion is 90°.

Question 5. Think of two situations for each of the following given conditions.

1. Large mass but low kinetic energy.
2. Small mass but high kinetic energy.

Answer:

1. A shot put thrown by an athlete.
2. A hockey ball which has been hit hard by a player and is travelling fast.

Read and Learn More Class 9 Science Solutions

Question 6. What will cause greater change in kinetic energy of a body? Changing its mass or changing its velocity.
Answer:

Change in velocity will cause greater change in kinetic energy because KE = 1/2 mv2

Question 7. Name the type of energy stored in spring of a watch.
Answer:

Elastic potential energy is stored in spring of a watch.

Question 8. In which situation, the potential energy of a spring be minimum?
Answer:

When a spring is at its natural length (i.e. neither stretched nor compressed), the potential energy of a spring will be minimum.

Question 9. If a body is thrown vertically upwards, its velocity goes on decreasing. What happens to its kinetic energy when it stops at the top and its velocity becomes zero?
Answer:

The whole of its kinetic energy gets converted into potential energy (mgh), where m is the mass of body and h is the height and g is the acceleration due to gravity.

Question 10. Can any object have mechanical energy even, if its momentum is zero?
Answer:

1. Since, mechanical energy is the sum of kinetic energy and potential energy. And as given that, momentum of the body is zero, it means velocity of the body is zero, so it has kinetic energy equals to zero.
2. But it may have potential energy. So, even if the momentum of the body is zero, it may have mechanical energy.

Question 11. The head of a nail becomes hot when it is hammered into a plank of wood. What energy transformation takes place in this case?
Answer:

Kinetic energy of falling hammer is converted into heat energy when it strikes the head of a nail. Therefore, the head of a nail becomes hot.

Question 12. A car and a truck have the same speed of 60 ms”1. If their masses are in the ratio 1:4. Find the ratio of their KE.
Answer:

As, \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{1 / 2 m_2 v^2}{1 / 2 m_2 v^2}=\frac{m_1}{m_2}=\frac{1}{4}\)=1: 4

Question 13. If a speed of a particle is doubled, what will be the ratio of its kinetic energy to its momentum?
Answer:

Kinetic energy, \(\mathrm{KE}^{\prime} =\frac{1}{2} m\left(v^{\prime}\right)^2 [v^{\prime}=2 v]\)

= \(\frac{1}{2} m(2 v)^2=4 \mathrm{KE}\)

and momentum, \(p^{\prime =m v^{\prime}}\)

= \(m(2 v)=2 p [ v^{\prime}=2 v]\)

⇒ \(\frac{\mathrm{KE}^{\prime}}{p^{\prime}} =\frac{4 \mathrm{KE}}{2 p}=2 \times\left(\frac{\mathrm{KE}}{p}\right)\)

Hence, ratio gets doubled.

Question 14. If a 5 kg mass is raised to a height of 2 m, calculate the work done against the force of gravity. (Given, g = 9.8 ms^-2)
Answer:

Given, mass, m= 5 kg,

height, h = 2m, W =?

We know that work, W = mgh = 5 x 9.8 x 2 = 98 J

Question 15. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (Take, g=10m^-2)
Answer:

If the energy of the ball reduces by 40% after striking the ground, then remaining energy of the ball will be 60% of initial energy.

Let initial energy of the body of mass (m) at height (h) is (mgh).

According to the question,

mgh’ = 60% of mgh [given, h =10 m and g =10 ms^-2]

∴ \(h^{\prime}=60 \% \times h=\frac{60}{100} \times 10=6 \mathrm{~m}\)

Question 16. An object of mass 2 kg is dropped from a height of lm. What will be its kinetic energy as it reaches the ground? (Take, g =9.8 ms^-2)
Answer:

Given, mass, m = 2 kg and height, h =1 m.

On reaching the ground, the kinetic energy of object is converted into its potential energy such that KE= YE-mgh = 2 x 9.8 x 1 = 19.6 J

Question 17. A girl weighing 50 kg runs up a hill raising herself vertically 10 m in 20 s. What is the power expended by girl?
Answer:

Power expended by girl =\(\frac{\text { Work doac by girl }}{\text { Time raken }}\)

= \(\frac{F \times s}{t}=\frac{m g \times s}{t}\)

= \(\frac{50 \times 9.8 \times 10}{20}\)

= 245 W[ F=m g]

Class 9 Science Chapter 10 Work, Energy And Power Short Answer Type Questions

Question 1. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force? Explain it with an example.
Answer:

Yes, when a force acts in a direction perpendicular to the direction of displacement e.g. Earth revolves around the Sun under the gravitational force of Sun on Earth, but no work is done by the Sun, though Earth has a centripetal acceleration

Question 2. At the bottom of the waterfall, the water is warmer than at the top. Give reason.

Answer: 

When water falls on the ground, its mechanical energy (KE + PE) is converted into heat energy, due to which the temperature of water at the bottom of the waterfall increases.

Question 3. What is meant by the transformation of energy? Explain with the help of two suitable examples.
Answer:

One form of energy can be converted into other forms of energy, this phenomenon is called transformation of energy.

1.  When we throw a ball, muscular energy which is stored in our body, gets converted into the kinetic energy of the ball.
2. The wound spring in the toy car possesses potential energy. As the spring is released, its potential energy changes into kinetic energy due to which, the toy car moves.

Question 4. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
Answer:

Suppose m₁ and m₂ are masses of light and heavy objects, respectively. As we know,

kinetic energy, K=\(\frac{1}{2} m v^2\) [where, y= velocity of objects.] …(1)

and momentum, p = mv

On multiplying and dividing with m in Eq. (1), we get

So, K=\(\frac{1}{2} \frac{m v^2 \times m}{m}\)

K=\(\frac{1}{2} \frac{(m v)^2}{m}\) as from Eq. (2) [ p=m v]

So, K=\(\frac{p^2}{2 m}\)

We have, kinetic energy, K=\(\frac{p^2}{2 m}\)

Momentum is the same for light and heavy bodies.

So, kinetic enetgy, K \(\propto \frac{1}{m}\)

Thus, kinetic energy is inversely proportional to the mass.

So, a lighter body has larger kinetic energy.

Question 5. When a force retards the motion of a body, what is the nature of work done by force? State reason. List two examples of such a situation.
Answer:

The nature of work done in the case of retarding motion is negative.

Suppose a force F brings a body moving with velocity v to rest (retards), then work done

= change in KE

=\(\mathrm{KE}_f-\mathrm{KE}_i\)

W =\(0-\frac{1}{2} m x^2=-\frac{1}{2} m s^2\)

For Example

When we apply the brakes of a car, the work done is negative.

Work done by frictional force is negative.

Question 6. A car is moving on a levelled road and gets its velocity doubled. In this process,

1. how would the potential energy of the car change?
2. how would the kinetic energy of the car change?

Answer:

1. The potential energy of the car remains the same, since PE (= mgh) is independent of velocity.
2. The kinetic energy of the car becomes four times since KE \(\left(=\frac{1}{2} m \nu^2\right)\) is proportional to the square of velocity.

Question 7. A girl sits and stands repeatedly for 6 min. Draw a graph to show the variation of the potential energy of her body with time.

Answer:

1. From the graph shown above, we can take the sitting position of the girl as the position of zero potential energy.
2. Let m be the mass of the girl and h be the position of the centre of gravity while standing above the sitting position.
3. The PE while standing is + mgh and while sitting is zero. We can assume that there is no acceleration or deceleration while standing and sitting, this is repeated after every minute.

Question 8. (1) The potential energy of a freely falling object decreases progressively. What happens to its

1.  kinetic energy,
2. total mechanical energy?

State the law on which your answer is based.

(2) A household consumes 1 kWh of energy per day How much energy is this in joules?

Answer:

(1) (1) As potential energy decreases with decreasing height, the speed of the object will increase and hence its kinetic energy will increase.

(2) Total mechanical energy will remain constant. It is based on the law of conservation of energy which states that energy can neither be created nor be destroyed. It can only be transformed from one form to another.

(2) 1 kWh = 3.6xl0 6J

Question 9. If a body falls from a height bounces from the ground and again goes upwards with the loss of a part of its energy.

1. How will its potential energy change?
2. What are various energy conversions taking place?

Answer:

1. When it strikes the ground, its PE is zero and after bouncing, its potential energy increases gradually.
2. At the time it strikes the ground, it has maximum KE and after it bounces, its KE starts changing into potential energy.

Question 10. Why do we say that a lift delivers more power in taking a man up than that delivered by a man climbing stairs through the same height?

Answer:

It is true to say that a lift delivers more power in taking a man up than that delivered by him climbing the stairs at the same height. It is because, a lift takes less time in taking the man up, so it delivers more power.

Question 11. When an arrow is shot, from where does the arrow acquire its kinetic energy?
Answer:

A stretched bow has potential energy due to the change in its shape. To shoot the arrow, the bow has to be released.

Therefore, its potential energy is converted into the kinetic energy of the arrow.

Question 12. Seema tried to push a heavy rock of 100 kg for 200 seconds but could not move it. Find the work done by Seema at the end of 200 s.

Answer:

The work done by Seema in pushing a rock for 200 s is zero. Because there is no displacement in this case, i.e. W = Fx s =0 [v s =0]

Question 13. A coolie lifts a box of 15 kg from the ground to a height of 2 m. Calculate the work done by coolie on the box. (Given, g = 9.8 ms^-2)

Answer:

Given, mass, m = 15 kg, distance, s =2m, g =9.8 ms-2, W=>

Now, force applied,

F=mg= 15 x 9.8 = 147 N

Work done, W = F x s = 147 x 2 = 294 N

Question 14. How is work done by a force measured? A porter lifts luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (Take, g = 10 ms^-2 )

Answer:

Work done by a force is measured by the product of force and displacement when they are in the same direction or opposite direction.

Given, m = 20 kg

s =1.7 m [where, m = mass of luggage, g = acceleration due to gravity.]

W = Fs = mgs [ Fand s arc in the same direction.]

= 20x 10x 1.7= 340 J

Question 15. A boy is moving on a straight road against a frictional force of 5 N.

After travelling a distance of 1.5 km, he forgot the correct path at a roundabout of a radius of 100 m as shown in the figure.

However, he moves on the circular path for one and a half cycles and then he moves forward up to 2 km. Calculate the work done by him.

Answer:

Given, the force applied by the boy against friction = 5 N

Displacement on the circular path

= One cycle + Half cycle

= 0 + Half cycle

= 0 + Diameter of circular path [ Displacement depends on initial and final point]

= 0 + 2 r =0 + 2 x 100 [given, r =100 m]

= 0 + 200 =200 m

Total displacement = 1.5 km+ 200 m+ 2 km

= 1.5 x 1000 +200+ 2x 1000 m [1km = 1000 m]

= 3700 m

Work done by boy = Fs

= 5 x 3700 =18500J

Question 16. A bus and a car have the same KE. Which of the two is moving fast? Explain.

Answer:

The KE is given by KE=\(\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 \mathrm{KE}}{m}}\) As KE of both bus and car is same, so v \(\propto \frac{1}{\sqrt{m}}\)

Since the mass of the car is less than that of the bus, the car moves faster than the bus.

Question 17. Compute the speed of a 2 kg ball having a kinetic energy of 4 J.

Answer:

Given, the mass of the ball, m = 2 kg

The kinetic energy of the ball, KE = 4 J

As, KE =\(\frac{1}{2} m v^2\)

⇒ \(v^2=\frac{2 \mathrm{KE}}{m}=\frac{2 \times 4}{2}=4\)

v=2 \(\mathrm{~ms}^{-1}\)

Question 18. A car and a truck have the same speed of 30 ms^-1. If their masses are in the ratio of 1:3, find the ratio of their kinetic energies.

Answer:

Given, \(v_1=v_2=30 \mathrm{~ms}^{-1}\)

According to question, \(\frac{m_1}{m_2}=\frac{1}{3}\)

⇒ \(\frac{\mathrm{KE} \text { of car }}{\mathrm{KE} \text { of truck }}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2}=\left(\frac{m_1}{m_2}\right)=\frac{1}{3}\)

Question 19. If two bodies have masses in the ratio 1: 8 and have their speed in the ratio 4:5, find the ratio of their KE.

Answer:

Given, \(m_1 / m_2=1: 8, and v_1 / v_2=4: 5\)

⇒ \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2} =\frac{1 / 2 m_1 v_2^2}{1 / 2 m_2v_2^2}=\frac{m_2}{m_2}\left(\frac{v_1}{v_2}\right)^2\)

= \(\frac{1}{8} \times\left(\frac{4}{5}\right)^2\)

= \(\frac{1}{8} \times \frac{16}{25}=\frac{2}{25}=2: 25\)

Question 20. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of the two kinetic energies?

Answer:

Given, \(v_1=v\) and \(v_2=3 v\)

Kinetic energy of rocket, K=\(\frac{1}{2} m v^2\)

The ratio of two kinetic energies,

⇒ \(\frac{K_1}{K_2}=\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}\)

⇒ \(\frac{K_1}{K_2}=\frac{v_1^2}{v_2^2} \quad\left[\text { put } v_2=3 v \text { and } v_1=v\right ]\)

We get =\(\frac{v^2}{(3 v)^2}=\frac{v^2}{9 v^2}=\frac{1}{9} \Rightarrow \frac{K_1}{K_2}=\frac{1}{9}\)
Thus, the ratio of two kinetic energies \(K_1: K_2=1: 9\)

Question 21. A child drops a stone of 1 kg from the top of a tower. Find its kinetic energy, 5 s after it starts falling. (Take, g = 10 ms^-2)

Answer:

Given, the mass of the stone, m = 1 kg

Initial velocity, u =0, time, t = 5 s Acceleration due to gravity, g = 10 ms^-2

As from the equation of motion.

r=\(N+s^t\) [for downward motion]

r=0+10 x 5

r=50 \(\mathrm{~ms}^{-1}\) [for downward motion]

The kinetic energy of the stone is given by

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} \times 1 \times(50)^2=1250 \mathrm{~J}\)

Question 22. A mass of 20 kg is dropped from a height of 0.5 m. Find its (1) velocity and

(2) KE as it just reaches the ground.

Answer:

Given, height, h=0.5 m, mass, m=20 kg

(1) \(p^2=2 g h=2 \times 10 \times\) 0.5=10

⇒ \({\left[ g=10 \mathrm{~ms}^{-2}\right]}\)

∴ \(\nu=\sqrt{10}=3.16 \mathrm{~ms}^{-1}\)

(2) We know that, KE =\(\frac{1}{2} m \nu^2\)

= \(\frac{1}{2} \times 20 \times 10=100 \mathrm{~J}\)

Question 23. What is the amount of work done in the following cases? Justify your answer by giving the appropriate reason.

1. By an electron revolving in a circular orbit of radius r around a nucleus.
2.  By the force of gravity, when a stone of mass m is dropped from the top of a multi-storeyed building of height h.

Answer:

(1) Work done is zero as shown in the figure below.

When an electron revolves around the nucleus, a centripetal force F acts along the radius towards the centre O.

The displacement (s) acts tangentially, therefore the angle between the force and the displacement is 90°. Therefore, W= 0

(2) We know that, W = mgh

As the stone is dropped, its PE starts to convert into KE.

Let its speed be v, then from \(v^2\)=2 g h (when stone reaches the ground)

KE =\(\frac{1}{2} m v^2\)

= \(\frac{1}{2} \times m \times 2 g h=m g h \Rightarrow W\) = mgh

Question 24. A boy lifts a suitcase of 20 kg from the ground to a height of 1.2 m. Calculate the work done by him on the suitcase. (Given, g=10ms^-2)

Answer:

Given, mass, m = 20 kg, height, h = 1.2 m, g =10 ms^-2

Work done, W = mgh = 20 x 10 x 1.2 = 240 J

Question 25. A shotput player throws a shotput of mass 3 kg. If it crosses the top of the wall 2 m high at a speed of 4 ms^-1. Compute the total mechanical energy gained by the shotput when it crosses the wall.(Given, g =9.8 \(\mathrm{~ms}^{-2}\))

Answer:

Given, m=3 \(\mathrm{~kg}, h=2 \mathrm{~m}, y=4 \mathrm{~ms}^{-1}, g=9.8 \mathrm{~ms}^{-2}\)

Total mechanical energy =\(\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v^2+m g h\)

= \(\frac{1}{2} \times 3 \times 16+3 \times 9.8 \times 2\)

= 24+58.8=82.8 J

Question 26. Avinash can run with a speed of 8 ms^-1 against the frictional force of 10 N and Kapil can move with a speed of 3 ms^-1 against the frictional force of 25 N. Who is more powerful and why?

Answer:

Given, the force applied by Avinash = 10 N

Speed of Avinash = 8 m s^-1

Power of Avinash =F. v =10 x 8 =80 W

Now, force applied by Kapil = 25 N

Speed of Kapil = 3 m s^-1

Power of Kapil = F . v

= 25 x 3 = 75 W

Since, Avinash has more power (80 – 75), i.e. 5 W than Kapil. So, Avinash is more powerful.

Question 27. The human heart does 1.5 J of work in every beat. How many times per minute does it beat, if its power is 2 W?

Answer:

Given, power, P=2 W, time, t=1 min =60 s

Total work, W=\(P_t \quad\left[ P=\frac{W}{t}\right]\)

=2 x 60 s=120J

1.5 J work is done in 1 beat.

120 J work will be done in \(\frac{1 \times 120}{15}\)=80 beats

Therefore, number of beats per min = 80

Question 28. For an experiment to measure his power, a student records the time taken by him to run up a flight of steps on a staircase. Use the following data to calculate the power of the student.

Number of steps =28

Height of each step =20 cm

Time taken =5.4 s

Mass of student =55 kg

Acceleration due to gravity =9.8 \(\mathrm{~ms}^{-2}\)

Answer:

Given. n=28, b =20 cm =0.2 m, t=5.4 s

m =55 kg, g=9.8 \(\mathrm{~ms}^{-2}\)

We know that the power of student is given by

P =n \(\times \frac{W}{r}=n \times \frac{m g b}{t}\)

= \(\frac{28 \times 55 \times 9.8 \times 0.2}{5.4}\) [ W = mgh]

=559 W

Question 29. A boy X can run with a speed of 8 ms^-1 against the frictional force of 10 N and another can move with a speed of 3 ms^-1 against the frictional force of 20 N. Find the ratio of powers of X and Y.

Answer:

Given, the distance travelled by the boy X in 1 s = 8 m Distance travelled by the boy Fin ls=3m

As we know, work done by the boy X to run against the frictional force of 10 N = 10 Nx 8 m= 80 J

So, power of 80 J of work done by X=\(\frac{W}{t}=\frac{80 \mathrm{~J}}{1 \mathrm{~s}}\)

=80 W

Similarly, work done by the boy Y to run against the frictional force of 20 \(\mathrm{~N}=20 \mathrm{~N} \times 3 \mathrm{~m}=60 \mathrm{~J}\)

Power of Y=\(\frac{60 \mathrm{~J}}{1 \mathrm{~s}}=60 \mathrm{~W}\)

So, the ratio of two values of powers is given by

∴ \(\frac{\text { Power of } X}{\text { Power of } Y}=\frac{80}{60}=\frac{4}{3}=4: 3\)

Question 30. The power of a motor pump is 60 W. Find the mass of water, the pump can lift per minute from a 10 m deep well. (Take, g =10 ms^-2)

Answer:

Given, power, P = 60 W, time, t = 60 s Height, h =10m

The energy supplied to the pump = Power x Time

= 60 W x 1 min [vlmin=60s]

= 60 Wx 60s

= 3600 Ws= 3600J

So, energy, E = mgb

3600 = mx 10×10

m = 36 kg

So, the mass of the water is 36 kg.

Question 31.

1.  What is meant by the potential energy of a body?
2. A body of mass m is raised to a vertical height of h through two different paths A and B.

What will be the potential energy of the body in the two cases? Give a reason for your answer.

Answer:

1. The energy possessed due to the position of a body is called potential energy.
2. The work done against gravity in both cases is mgh. It is independent of the path along which the body is moved and it depends only on the initial and final positions of the body.

Question 32. At a height of 20 m above the ground, an object of mass 4 kg is released from rest. It is travelling at a speed of 20 ms^-1 when it hits the ground. The object does not rebound and the gravitational field strength is 10Nkg^-1.

How much energy is converted into heat and sound on impact?

Answer:

Given, height above the ground, b = 20 m Mass of the ball, m = 4 kg

Speed of the ball while striking the ground, v =20 ms^-1

Acceleration due to gravity,

g =10 Nkg^-1

According to the law of conservation of energy, all the energy of the ball will be converted into sound and heat energy because the ball does not rebound.

Energy of the ball =\(\frac{1}{2} m v^2\)

=\(\frac{1}{2} \times 4 \times(20)^2=800 \mathrm{~J}\)

Hence, 800 J of energy will be converted into heat and sound.

Question 33. A labourer whose own mass is 50 kg carries a load of an additional 60 kg on his head to the top of a building 15 m high. Find the total work done by him. Also, find the work done by him, if he carries another additional block of mass 10 kg to the same height. (Take, g =10 ms^-1)

Answer:

Given, total mass, m =50 + 60 =110 kg

Displacement, s =15 m

Work done by him is given by W =0 [since force and displacement are perpendicular to each other.]

If an additional block of mass 10 kg is carried by him to the same height, then also work done by him remains zero because force and displacement are perpendicular to each other.

Question 34. A girl with a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms^-1 by applying a force.

The trolley comes to rest after traversing a distance of 16 m.

1.  How much work is done on the trolley?
2. How much work is done by the girl?

To find the value of acceleration (a) by using the third equation of motion, i.e. v² = u² + 2as and then putting this value in W =mass to get the required work.

Answer:

Given, w = 4 ms^-1,v =0 and 5 =16 m From the third equation of motion,[for retardation, the acceleration is negative, i.e. a=-a]

⇒ \(v^2=u^2-2 a \Rightarrow(0)^2=(4)^2-2 a \times 16\)

0=16-32 a \(\Rightarrow a=\frac{16}{32}=0.5 \mathrm{~ms}^{-2}\)

where, u = initial velocity, v = final velocity, a = acceleration and s = displacement.

(1) Total mass = 35 + 5 = 40 kg

Work is done on the trolley,

W = F-d =ma s [v F =ma]

= 40×0.5×16 =320 J

(2) Given, the mass of the girl, m =35 kg

Work done by the girl,

W =F-d =ma r = 35×0.5×16 = 280]

Question 35. Calculate the kinetic energy of a car of mass 750 kg moving with a velocity of 54 km. Find the new kinetic energy of the car, if a passenger of mass 50 kg sits in the car.

Answer:

Given, mass, m=750 kg, velocity, v=54 \(\mathrm{kmh}^{-1}\)

=54 \(\times \frac{5}{18} \mathrm{~ms}^{-1}=15 \mathrm{~ms}^{-1}\)

Kinetic energy, \(\mathrm{KE}=\frac{1}{2} m v^2\)

KE =\(\frac{1}{2} \times 750 \times(15)^2=84375 \mathrm{~J}\)

If a passenger of mass 50 kg sits in the car, then total mass becomes (750+50) kg, i.e. 800 kg

New KE =\(\frac{1}{2} \times 800 \times(15)^2=90000 J\)

Question 36. On a level road, a scooterist applies brakes to slow down from a speed of 54 kmh^-1 to 36 kmh-1. What amount of work is done by the brakes? (Assuming the mass of the empty scooter is 86 kg and that of the scooterist and petrol is 64 kg.)

Answer:

Here, total mass, w = 86 + 64 = 150 kg

Initial velocity, u=54 \(\mathrm{kmh}^{-1}\)

=\(\frac{54 \times 1000}{3600}=15 \mathrm{~ms}^{-1}\)

Final velocity, v=36 \(\mathrm{kmh}^{-1}\)

=36 \(\times \frac{1000}{3600}=10 \mathrm{~ms}^{-1}\)

Work done by brakes = KE lost by the scooter

= Final KE – Initial KE

=\(\frac{1}{2} m v^2-\frac{1}{2} m u^2=\frac{1}{2} m\left(v^2-u^2\right)\)

=\(\frac{1}{2} \times 150\left[(10)^2-(15)^2\right]=\frac{1}{2} \times 150(100-225)\)

= -75×125 = -9375 J

Question 37. (1) Define potential energy.

(2) Give an example where potential energy is acquired by a body due to a 75 m change in its shape

(3) A skier of mass 50 kg stands at A, at the top of a ski jump. He takes off from A for his jump to B. Calculate the change in his gravitational potential energy between A and B.

Answer:

(1) Potential Energy It is the energy possessed by a body by virtue of its position or shape.

(2) In a toy car, the wound spring possesses potential energy. When spring is released, its potential energy changes into kinetic energy due to which the toy car moves.

(3) Given, m=50 \(\mathrm{~kg} . h=75 \mathrm{~m}_1 h_2=60 \mathrm{~m}\)

At point A, \(\mathrm{PE}_1=m g h_h=50 \times 10 \times 75\)

=37500 J

At point B, \(\mathrm{PE}_2=m g h_2=50 \times 10 \times 60\)

=30000 J

Change in PE =\(\mathrm{PE}_1-\mathrm{PE}_2\)=37500-30000 =7500 J

Question 38. A body of mass 5 kg is thrown vertically upwards with an initial velocity of 50 ms^-1. What will be its potential energy at the end of 5 seconds?

Answer:

Given, m=5 \(\mathrm{~kg}, w=50 \mathrm{~ms}^{-1}, t=5 \mathrm{~h} g=-10 \mathrm{~ms}^{-2}\)

Height covered by the body in 5 s is h=w t+\(\frac{1}{2} g t^2\)

50 \(\times 5-\frac{1}{2} \times 10 \times(5)^2\)

=250-125=125 m

Therefore, PE of the body after 5 s

= mgh =5 x 10 x 125 = 6250 J

Question 39. A body of mass 1.5 kg is thrown vertically upwards with an initial velocity of 15 ms^-1. What will be its potential energy at the end of 2 s? (Take, g =10 ms^-2)

Answer:

Given, mass, m = 1.5 kg

Initial velocity, u =15 ms^-1

Time, t = 2s

From Newton’s second equation of motion,

t=w t+\(\frac{1}{2} g t^2\)

s=15 \(\times 2+\frac{1}{2} \times 10 \times 4=50 \mathrm{~m}\)

Potential energy, PE = mgh

=1.5 \(\times 10 \times 50\) [ b=t]

=750 J

Question 40. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.

1. How much work is done by the man in lifting the box?
2. How much work do they do in just holding it?
3.  Why do they get tired while holding it? (Given, g = 10 ms^-2)

Answer:

1. Given, mass, m = 250 kg, height, A = lm and acceleration due to gravity, g =10 m s^-2
2. Work done by the man in lifting the box W = Potential energy of box W = mgh = 250 x 10 x 1 = 2500 J (it) Work done is zero in holding a box because displacement is zero.
3. In holding the box, the energy of each man is lost. Due to loss of energy, they felt tired.

Question 41. The weight of a person on planet A is about half that on the Earth. He can jump up to 0.4 m height on the surface of the Earth. How high can he jump on planet A?

Answer:

It is given that the weight of a person on the Earth = w (i .e.w =mg)

And as he can jump up to height (hy) = 0.4 m So, potential energy at this point

= mgh = mgh x 0.4  → Equation .(1)

And it is given that

weight of the person on the other planet = \(\frac{w}{2}\)  → Equation 2

And if he could jump to height \(\left(b_2\right)\) its potential energy would be \(\frac{w}{2} h_2=\frac{m g}{2} b_2\)

Since he applied the same amount of effort in both cases to lift his body, its potential energy will be the same.

From Eqs. (1) and (2), we get

m g \(\times 0.4=\frac{m g}{2} b_2 \Rightarrow b_2=0.4 \times 2=0.8 \mathrm{~m}\)

Question 42. 300 J of work is to be done in lifting a bag of mass 5 kg in weight up to a height of 4 m from the ground. What will be the acceleration with which the bag was revised? (Take, g = 10 ms^-2)

Answer:

Given, mass, m=5 kg, work, W=300 J

h=4 \(\mathrm{~m}, g=10 \mathrm{~ms}{ }^{-2}\)

Work done, W=m g h+m a b=m(g+a) b

300=5 \(\times(10+a) \times 4\)

300=20(10+a)

10+a=\(\frac{300}{20}\)=15

a=15-10=5 \(\mathrm{~ms}^{-2}\)

Question 43. Shyam drops a ball of 100 g from a building of a height of 10 m. What will be its kinetic energy at the height of 4 m? What will happen to its total mechanical energy? Give reasons to justify your answer.

Answer:

Given, mass, m = 100 g = 0.1 kg, height, A = 10m

Potential energy, PE = mgh = 0.1x10x10 = 10 J [v g =10 ms^-2]

When the body reaches a height of 6 m, its velocity is given by

⇒ \(r^2=s^2+2 s^h\)

⇒ \(r^2=0^2+2 \times 10 \times 6\) [h=10-4 or h=6]

v=\(\sqrt{2 \times 10 \times 6}=2 \sqrt{30}\)

Now, its kinetic energy at this point is given by

=\(\frac{1}{2} m r^2=\frac{1}{2} \times 0.1 \times 4 \times 30=6 \mathrm{~J}\)

As, mechanical energy = KE + PE [at highest point] Mechanical energy = (0 + 10)J=10J Mechanical energy at 4 m height

= KE+PE = 6+0.1 x 10x 4= 10]

Hence, it is seen that total mechanical energy always remains constant during the motion.

Question 44. A mass of 10 kg is dropped from a height of 50 cm. Find its

1. potential energy just before dropping.
2. kinetic energy just touching the ground.
3. the velocity with which it hits the ground (Take, g = 10 ms^-2)

Answer:

Given, mass of the object, m= 10 kg Height, h =50 cm= 0.3 m

(1) As potential energy is given by PE = mgh =10 x 10 x 0.5 = 50]

(2) From the law of conservation of energy,

the total energy of the ball just before dropping =total energy of the ball just on touching the ground KE + PE of the ball just before dropping

= KE of the ball just on touching the ground => KE = 50J

(3) As we know, KE = 50 J

So, \(\frac{1}{2} m v^2=50 \Rightarrow v^2=\frac{50 \times 2}{10}=10\)

So, the velocity with which it hits the ground,

∴ \(\mathrm{v}=\sqrt{10}=3.16 \mathrm{~ms}^{-1}\)

Question 45. A small child tends to mimic his father by lifting a mass of 10 kg on his head. As soon as he succeeds in lifting it, he loses the object and falls back to the ground. If the child has a height of 90 cm, find the kinetic energy

1. At half the height of the child,
2. With which the object strikes the ground.

Answer:

Given, the mass of the object, m =10 kg Height of the child, h- 90 cm = 0.9 m Total energy of the object at the head of the child is given by

= PE+ KE = mgh + 0 [v KE=0, asrr =0] = mgh= lOx 9.8x 0.9= 88.2 J

(1) At half the height of the child, i.e. at 0.45 m

Total energy, TK = PK + KE = mgh + KE

= 10 x 9.8 x 0.45+ KE

Now, from the law of conservation of energy,

KE+ 10x 9.8x 0.45= 88.2 or KE+ 44.1 = 88.2

KE= 88.2-44.1 =44.1 J

(2) When the object strikes the ground,

Total energy = PE + KE = 0 + KE

= KE [PE = 0, as h =0]

From the law of conservation of energy,

KE= 88.2 J [v TE =88.2 J]

Question 46. (1) State and define the SI unit of power.

(2) A person carrying 10 bricks each of mass 2.5 kg on his head moves to a height 20 m in 50 s. Calculate the power spent in carrying the bricks of the person. (Given, g =10 ms^-2)

Answer:

(1) The SI unit of power is watt.

1 watt is the power of a body which does work at the rate of 1 joule per second.

wart =\(\frac{1 \text { Joule }}{1 \text { Second }}\)

(2) Given, the mass of one brick = 2.5 kg

Mass of 10 bricks = 2.5 x 10 = 25 kg

Height, h = 20 m, time, t = 50 s, power, P = ?

Power P =\(\frac{m g h}{t}\)

= \(\frac{25 \times 10 \times 20}{50}=100 \mathrm{Js}^{-1}\)

Question 47. Mohan lifts his cell phone to a height of 1 m. He takes Is to do this. After lowering the phone, he then lifts it 2 m in 2 s. Has he generated more power in doing the second task? Give a reason to justify your answer.

Answer:

Let, the mass of cellphone be m kg

Work done (W) by Mohan = mgh

where, h = height at which the cellphone is raised and

g = acceleration due to gravity

W = mg x 1 = mg

Power delivered (R)=\(\frac{\text { Work done }(W)}{\text { Time }(T)}\)

=\(\frac{m g}{1}=m g\)

Work done to lift cell phone by 2 m

= mg x 2 = 2 mg

Power delivered \(\left(R_2\right)=\frac{\text { Work done (IV) }}{\text { Tine (T) }}\)

=\(\frac{2 m g}{2}\)=m g

So in both cases, the power delivered by Mohan is the same.

Question 48. A force applied on a body of mass 4 kg for 5 s changes its velocity from 10 \(\mathrm{~ms}^{-1}\) to 20 \(\mathrm{ms}^{-1}\). Find the power required.

Answer:

Given, m=4 kg, t=5 s, w=10 \(ms^{-1}\),

v=\(20 \mathrm{~ms}^{-1}\), P=?

Power, \(P=\frac{W}{t}=\frac{\text { Change in } \mathrm{KE}}{\text { Time taken }}\)

P =\(\frac{1}{2} \times \frac{m\left(v^2-u^2\right)}{t}=\frac{1}{2} \times \frac{4\left\{(20)^2-(10)^2\right]}{5}\)

=\(\frac{1}{2} \times \frac{4 \times 300}{5}=120 \mathrm{Js}{ }^{-1}\)

Question 49. A boy of mass 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step of the staircase is 15 cm, find the power of the boy. (Given, g=10 \(\mathrm{~ms}^{-2}\) )

Answer:

Given, the mass of the body, m=50 kg

b=45 \(\times 15=675 \mathrm{~cm}=6.75 \mathrm{~m}\)

t=9 \(\mathrm{~s}, g=10 \mathrm{~ms}^{-2}\)

PE =m g b=50 \(\times 10 \times 6.75=3375 \mathrm{~J}\)

So, the energy of the boy is 3375 J.

Power of the boy,

y, P =\(\frac{\mathrm{PE}}{\text { Time }}=\frac{\text { Energy }}{\text { Time }}\)

=\(\frac{3375}{9}=375 \mathrm{~W}\)

Question 50. A machine gun takes 10 s to fire 30 bullets with a velocity of 500 \(\mathrm{~ms}^{-1}\). Find the power developed by the gun when each bullet has a mass of 100g.

Answer:

Given, mass of 30 bullets =30 x 100=3 kg

Velocity, v=500 \(\mathrm{~ms}^{-1}\), time, t=10s

Power developed by the gun,\(P=\frac{W}{t}\)

(Here, work done by a gun will be equal to the kinetic energy of all the bullets.]

P =\(\frac{K E}{t}=\frac{\frac{1}{2} m v^2}{t}=\frac{m v^2}{2 t}\)

= \(\frac{3 \times(500)^2}{2 \times 10}=\frac{3 \times 500 \times 500}{20}=37500 W\)

Class 9 Science Chapter 10 Work, Energy And Power Long Answer Type Questions

Question 1. The velocity of a body moving in a straight line is increased by applying a constant force F for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Answer:

Consider an object of mass m moving with a uniform velocity. Let, it now be displaced through a distance s, when a constant force F acts on it in the direction of its displacement.

From the third equation of motion, \(v^2=u^2+2 a t\)

⇒ \(v^2-\mu^2=2 a t \Rightarrow r=\frac{p^2-u^2}{2 a}\)

We know that the work done by F is

W=F r \(\cos \theta\) (since, force and displacement are in some direction, so \(\theta=0^{\circ} )\)

W =m a \(\times s\) [ F=m a]

= ma \(x\left(\frac{v^2-u^2}{2 a}\right) {[\text { from Eq. (1) }]}\)

W =\(\frac{1}{2} m\left(v^2-u^2\right)\)

If the object is starting from its stationary position, i.e. w=0, from Eq. (2), we get

W=\(\frac{1}{2} m v^2\)

It is clear that the work done is equal to the increase in the kinetic energy of an object.

Question 2. (1) Name two forms of mechanical energy. Define the SI unit of energy.

(2) A man of mass 50 kg jumps from a height of 0.5 m. If gr = 10 ms^-2, what will be his energy at the highest point?

(3) Calculate the energy of a body of mass 20 kg moving with a velocity of 0.1 ms^-1.

Answer:

(1) Two forms of mechanical energy are

•  Kinetic energy and
• Potential energy

The SI unit of energy is joule (J). 1 J is the amount of work done on an object when a force of IN displaces it by lm along the line of action of force.

(2) Given, w = 50kg, h = 0.5 m, g=10 ms^-2

At the highest point, kinetic energy is converted into potential energy.

PE =\(m g^b=50 \times 10 \times 0.5\)=250

(3) Given, m=20 kg, v=0.1 \(\mathrm{~mm}^{-1}\)

As we know, a moving body has kinetic energy.

KE =\(\frac{1}{2} m v^2=\frac{1}{2} \times 20 \times(0.1)^2\)

= \(\frac{1}{2} \times 20 \times 0.1 \times 0.1=0.1 \mathrm{~J}\)

Question 3. (1) Define the kinetic energy of an object. Can the kinetic energy of an object be negative? Give reason.

(2) A car weighing 1200 kg is uniformly accelerated from rest and covers a distance of 40 m in 5 s. Calculate the work, the car engine had to do during this time.

Answer:

(1) The energy possessed by a body by virtue of its motion is called its kinetic energy.

No, the kinetic energy of an object cannot be negative because both m and v2 are always positive and KE =\(\frac{1}{2} m v^2\)

(2) Given, m =1200 kg, r = 40 m, t= 5 s, u= 0, W = ?.

We know that, W=F s = mas [ F=m a]

or \(s=u t+\frac{1}{2} a t^2\)

40=0 \(\times t+\frac{1}{2} \times d \times(5)^2\)

⇒ \(a=\frac{40 \times 2}{25}=3.2 \mathrm{~ms}^{-2}\)

⇒ \(W=F_5= matr\)

=1200 \(\times 3.2 \times 40=153600 \mathrm{~J}\)

Question 4. A vehicle of 1 tonne travelling with a speed of 60 ms^-1 notices a cow on the road 9 m ahead and applies brakes. It stops just in front of the cow.

1.  Find out the KE of the vehicle before applying brakes.
2. Calculate the retarding force provided by the brakes.
3.  How much time did it take to stop after the brakes were applied?
4.  What is the work done by the braking force?

Answer:

Given, the mass of the vehicle, m= tonne = 1000 kg

Initial speed, u = 60 ms^-1

Distance between vehicle and the cow, r=9 m

Final velocity, v=0

(1) KE of the vehicle before applying brakes is given by

=\(\frac{1}{2} m u^2=\frac{1}{2} \times 1000 \times 60 \times 60=1800000 \mathrm{~J} \text { (1W) }\)

(2) From the third equation of motion,

⇒ \(v^2-N^2=2 a\)

⇒ \((0)^2-(60)^2=2 \times a \times 9 \Rightarrow a=-200 \mathrm{~ms}^{-2}\)

So, retarding force provided by the brakes

∴ \(=m a d=1000 \mathrm{~kg} \times(-200) \mathrm{ms}^{-2}=-200000 \mathrm{~N}\)

(3) Now; again from the second equation of motion,

\(t=u t+\frac{1}{2} a t^2 \Rightarrow 9=60 t+\frac{1}{2} \times(-200) t^2\)

or 9=60 t-100\( t^2\)

or \(100 t^2-60 t+9=0 \Rightarrow(10 t-3)^2=0\)

or \(10 t-3=0 \Rightarrow t=\frac{3}{10}=0.3 \mathrm{~s}\)

(5) So, work done by the braking force is given by

=\(F_s=-200000 \mathrm{~N} \times 9 \mathrm{~m}=-1800000 \mathrm{~J}\)

Question 5. (1) A battery lights a bulb. Describe the energy changes involved in the process,

(2) Calculate the amount of work needed to stop a car of 500 kg moving at a speed of 36 kmh^-1.

Answer:

(1) When a battery lights a bulb, its chemical energy changes into light and heat energy.

(2) Here, m = 500 kg, v = 0 and

⇒ \(\mu=36 \mathrm{kmh}^{-1}=\frac{36 \times 1000}{3600}=10 \mathrm{~ms}^{-1}\)

Work done = Change in kinetic energy

=\(\frac{1}{2} m\left(v^2-v^2\right)=\frac{1}{2} \times 500\left(0-10^2\right)\)

=-\(\frac{1}{2} \times 500 \times 100=-25000 \mathrm{Js}^{-1}\)

So, it is negative because work is done to stop the car.

Question 6. A body of mass 20 kg is raised to the top of a building 15 m high and then dropped freely under gravity.

1. Find out the work done in raising the body to the top of the building.
2. What will be the value of gravitational potential energy at the top of the building?
3. By what factor will the gravitational potential energy of the same body increases, if it is raised to the top of a multi-storey building 45 m high?
4.  When will the kinetic energy of the body be maximum?

Answer:

Given, m=20 kg, h=15 m, g=10 \(ms^{-2}\)

(1) Work done =m g h=20 x 10 x 15=30001

(2) Gravitational PE = Work done=3000 J

(3) Gravitational PE at a height of 45m

= \(mgh^{\prime}=20 \times 10 \times 45=9000 \mathrm{~J}\)

Now,\(\frac{\text { Potential energy at height } 45 \mathrm{~m}}{\text { Potential energy at height } 15 \mathrm{~m}}=\frac{9000}{3000}\)=3

Therefore, PE increases by 3 times.

(4) KE will be maximum just before the body strikes the ground.

Question 7. An automobile engine propels a 1000 kg car A along a lavelled road at a speed of 36 kmh^-1. Find the power, if the opposing frictional force is 100 N.

Now, suppose after travelling a distance of 200 m, this car collides with another stationary car B of the same mass and comes to rest. Let its engine also stop at the same time.

Now, car B starts moving on the same level road without getting its engine started. Find the speed of the car B just after the collision.

Answer:

Given, the mass of the car = 1000kg

Mass of car B =1000 kg

The force applied by car A = 100 N

Speed of car \(A,\left(v_A\right)=36 \mathrm{~km} \mathrm{~h}^{-1}\)

=36 \(\times \frac{5}{18}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ [\(1 \mathrm{kmh}^{-1}=\frac{5}{18} \mathrm{~ms}^{-1}]\)

Power of car A, \(P_A=F \cdot v_A=100 \times 10=1000 \mathrm{~W}\) [F=force exerted by the car A against friction]

Again, for car A Newton’s second law, F = ma

100 =1000 x a

a =\(\frac{100}{1000}\)

a =\(\frac{1}{10} \mathrm{~m} \mathrm{~s}^{-2}\)

The velocity of car A after travelling 200 m is given by from the third equation of motion, \(v^2=u^2+2 an s\)

[for car A here, u=10 \(\mathrm{~ms}^{-1}, s=200 \mathrm{~m}]\)

⇒ \(v^2 =(10)^2+2 \times \frac{1}{10} \times 200\)

⇒ \(v^2 \)=100+40=140

v =\(\sqrt{140}=11.8 \mathrm{~ms}^{-1}\)

According to the question, after moving 200 m, the speed of car A, m =11.8 \(ms^{-1}\)

Just after the collision, the final speed of car A, \(v_1\)=0 before the collision, the initial speed of car B, \(w_2\)=0

From the conservation of linear momentum,

⇒ \(m_1 \mathrm{H}_1+m_2 \mathrm{H}_2=m_1 v_1+m_2 \mathrm{~N}_2\)

[Let just after the collision, the speed of the car B is \(v_2\)]

⇒ \(m_2 \times 11.8+m_2 \times 0=m_4 \times 0+m_2 \times v_2\)

⇒ \(11.8 m_2=m_2 v_2\)

⇒ \(118 m_2=m_1 v_2 \quad\left\{ m_1=m_2\right\}\)

∴ \(v_2=11.8 \mathrm{~m} / \mathrm{s}\)

Question 8. A car is moving with uniform velocities; 18 kmh^-1, 36 kmh^-1, 54 \(kmh{-1}\) and 72 \(kmh{-1}\) at some intervals. Find the KE of the boy of 40 kg sitting in the car at these velocities. Draw a graph between the KE and the velocities. Also, find the nature of the curve.

Answer:

At \(v_1=18 \mathrm{kmh}^{-1}=\frac{18 \times 1000}{3600}=5 \mathrm{~ms}^{-1}\),

⇒ \(KE_1=\frac{1}{2} m \mathrm{~m}^2=\frac{1}{2} \times 40 \times(5)^2=500 \mathrm{~J}\)

At \(v_2 =36 \mathrm{kmh}^{-1}=10 \mathrm{~ms}^{-1}\),

⇒ \(\mathrm{KE}_2 =\frac{1}{2} m v_2^2=\frac{1}{2} \times 40 \times(10)^2=2000 \mathrm{~J}\)

At \(v_3 =54 \mathrm{kmh}^{-1}=15 \mathrm{~ms}^{-1}\)

⇒ \(\mathrm{KE}_3 =\frac{1}{2} m v_3^2=\frac{1}{2} \times 40 \times(15)^2=4500 \mathrm{~J}\)

At \(v_4 =72 \mathrm{kmh}^{-1}=20 \mathrm{~ms}^{-1}\)

⇒ \(\mathrm{KE}_4 =\frac{1}{2} m v_4^2=\frac{1}{2} \times 40 \times(20)^2=8000 \mathrm{~J}\)

The graph between the KE and the velocities is shown below.

The graph is a parabolic curve because \(\mathrm{KE} \propto v^2\).

Class 9 Science Chapter 11  Sound Very Short Answer Type Questions

Question 1. What is the name of the strings which vibrate in our voice box when we talk?
Answer:

Vocal cord.

Question 2. How would you communicate in an atmos-phereless region?
Answer:

We can communicate through wireless sets using radio waves because radio waves can travel without a medium.

Question 3. If a freely suspended vertical spring is pulled in a downward direction and then released, which type of waves are produced in the spring?
Answer:

In this situation, longitudinal waves will be produced in spring.

Question 4. If a stone is dropped on the surface of water in a pond. Which type of waves are produced in water?
Answer:

In this situation, transverse waves will be produced in water.

Question 5. Give one example of a transverse and longitudinal wave.
Answer:

Transverse wave – Light Longitudinal wave – Sound

Question 6. A baby recognises her mother by her voice. Name the characteristic of sound involved.
Answer:

Quality of the sound is that characteristic of sound due to which distinction can be made between two sounds.

Question 7. On what factor does the quality of the sound depend?
Answer:

The quality of the sound depends on the shape of the sound wave.

Question 8. Why do we hear the loud sound of the horn of an approaching car before the car reaches us?
Answer:

This is because the velocity of the soul is much greater than that of the car.

Read and Learn More Class 9 Science Solutions

Question 9. Among air, water and steel, in which medium, the sound wave will travel faster?
Answer:

The sound wave will travel faster in steel because the speed of sound is the fastest in the solids.

Question 10. How does temperature affect the speed of sound?
Answer:

The speed of sound increases with the increase in temperature.

Question 11. Give the correct picture of the reflection of sound marked with the angle of incidence, angle of reflection and the normal.
Answer:

1. \(\angle i\)= angle of incidence,
2. \(\angle r\)=angle of reflection

Question 12. From which type of surface, the reflection of sound is better?
Answer:

The hard surface is a better reflector of sound.

Question 13. Give examples of multiple echoes.
Answer:

1.  Rolling of thunder,
2. Whispering galleries.

Question 14. At 20°C, what should be the minimum distance of a person from a sound-reflecting surface to hear an echo?
Answer:

The minimum distance from the sound-reflecting surface to hear an echo should be at least 17.2 m.

Question 15. Why are roofs and walls of an auditorium/hall generally covered with sound-absorbent materials?
Answer:

To reduce reverberation, the roofs and walls of an auditorium or hall are generally covered with sound-absorbent materials.

Question 16. Name the phenomenon responsible for the rolling of thunder.
Answer:

It is due to multiple reflections of sound from objects like clouds, land, etc.

Question 17. Why should the curved soundboard be placed behind the stage?
Answer:

It is because, after reflection from the curved sound hoard, sound waves spread evenly apart the width of the hall.

Question 18. What kind of waves are produced in an earthquake before the main shock wave begins?
Answer:

Infrasonic waves are produced in an earthquake before the main shock wave begins.

Question 19. State two important uses of ultrasounds for medical purposes.
Answer:

1. Diagnosing the tumours in the human body.
2. To analyse the development of the foetus.

Question 20. What is the frequency of the wave with a period of 0.025 s?
Answer:

Given, T=0.025 s

v=\(\frac{1}{T}=\frac{1}{0.025}\)=40 Hz

Question 21. A human heart beats 72 times in a minute. Calculate its frequency.
Answer:

Frequency =\(\frac{\text { Number of beats }}{\text { Total time }(\text { in second) }}\)

= \(\frac{72}{60}=1.2 \mathrm{~Hz}[1 \mathrm{~min}=60 \mathrm{~s}]\)

Class 9 Science Chapter 11  Sound Short Answer Type Questions

Question 1. Why do astronauts talk to each other through radiotelephone?
Answer:

As sound waves cannot travel through a vacuum in space. So, they use electromagnetic waves such as radio waves to communicate as it does not require a medium for propagation.

Question 2. Why are the longitudinal waves also called pressure waves?
Answer:

It is because longitudinal waves travel in a medium as a series of alternate compressions and rarefactions, i.e. they travel as variations in pressure. Therefore, they are called pressure waves.

Question 3. When the wire of a guitar is plucked, what types of waves are produced in

1. air
2. wire

Give reasons in support of your answer.

Answer:

1. When the wire of a guitar is plucked, a longitudinal wave is produced in the air due to the to-and-fro motion of the string of the guitar.
2. In the wire of a guitar, the transverse wave is produced as the particle vibrates perpendicular to the direction of motion.

Question 4. Through what type of medium, can

1. the transverse waves and
2. the longitudinal waves be transmitted? Explain in brief.

Answer:

(1) Since transverse waves travel in the form of crests and troughs, the)’ involves changes in the shape of the medium.

So, they can be transmitted through a medium having elasticity of shape. As solids and liquids have elasticity of shape, hence transverse waves can be transmitted through solids and liquids.

(2) Since longitudinal waves travel in the form of compressions and rarefactions, they involve changes in the volume and density of the medium.

All media, i.e. solids, liquids and gases have elasticity of volume. Hence, these waves (longitudinal waves) can be transmitted through all three types of media.

Question 5. Name the types of waves and two examples associated with

1. compressions and rarefactions
2. crests and troughs.

Answer:

1. The waves which travel along a slinky when it is pushed and pulled at one end, are the longitudinal waves. The following figure shows longitudinal waves.
2. The waves produced by moving one end of a long rope (or spring) up and down rapidly whose other end is fixed, are transverse waves as shown

Question 6. When the wire of a sitar is plucked, what kind of waves are produced in

1.  the wire and
2.  air

Answer:

1. Transverse waves are produced in the wire.
2. Longitudinal waves are produced in the air.

Question 7. Why the waves produced by a motorboat sailing in the sea, are both longitudinal and transverse?
Answer:

Longitudinal waves are produced due to vibrations of the rudder inside the sea water but on the surface of the water, transverse waves are generated. Hence, waves produced by motorboats are both longitudinal and transverse

Question 8. At what frequency, a source produces 500 sound waves per second?
Answer:

The frequency of sound waves is called the number of waves, produced in 1 s.

Here, several waves are produced in 1 s = 500 So, the frequency of this sound wave is 500 Hz.

Question 9. What happens to the wavelength, velocity and frequency of sound waves, when it travels from one medium to another?
Answer:

When a sound wave travels from one medium to another medium, its wavelength as well as velocity may change. But frequency does not change.

Question 10. Some animals get disturbed before the earthquakes. Give reason.
Answer:

During earthquakes, some animals get disturbed because earthquake produces low-frequency infrasound before the main shock waves, which is audible to some animals.

Question 11. Which of the two graphs (1) and (2) representing the human voice is likely to be the male voice? Give a reason for your answer

Answer:

Graph (1) represents the male voice. Since the pitch and frequency of the male voice are lower than the pitch of the female voice vibration of the graph (2) represents a higher frequency and higher pitch.

Question 12. Waves of frequency 100 Hz are produced in a string as shown. Give its

1. Amplitude
2. Wavelength
3. Velocity
4. Nature

Answer:

1. Amplitude = 5 cm
2. Wavelength – 20 cm
3. Velocity, v=\(v \lambda=100 \times 20 \times 10^{-2}=20 \mathrm{~ms}^{-1}\)
4. Nature It is a transverse wave.

Question 13. What is wave motion? Write any four characteristics of wave motion.
Answer:

• Wave motion is nothing but a mode of transfer of energy from place to place periodically without material transport.
• Four characteristics of wave motion are:
• It is the disturbance which travels forward through the medium but not the particles of the medium.
• Each particle receives vibrations a little later than its preceding particle.
• The wave velocity is different from the velocity of the particles with which they vibrate about their mean positions.
• The wave velocity remains constant in a given medium, whereas particle velocity changes continuously during its vibrations about the mean position.

Question 14. The graph shows a trace of a sound wave which is produced by a particular tuning fork.

1. Draw a trace of the sound wave which has a higher frequency than that shown
2. Draw a trace of the sound wave which has a larger amplitude than that shown

Answer:

1. The property which leads to the formation of echoes is the reflection of sound. When a sound is repeatedly reflected and if the minimum distance between the sound source and the reflecting surface is more than 17.2 m, then we hear echoes.
2. If the reverberation time in a big hall is too long, then the sound becomes distorted, blurred and confusing.

Question 15. Give the basic difference between wave velocity and particle velocity.

Answer:

The wave velocity is constant for a given medium and is expressed by v = vA, while the particle velocity changes with time. It is maximum at the mean position and zero at the extreme position.

Question 16 (1) Which property of sound leads to the formation of echoes? Briefly explain.

(2) What will happen, if the reverberation time in a big hall is too long?

Answer:

The property which leads to the formation of echoes is the reflection of sound. When a sound is repeatedly reflected and if the minimum distance between the sound source and reflecting surface is more than 17.2 m, then we hear echoes.

If the reverberation time in a big hall is too long, then the sound becomes distorted, blurred and confusing.

Question 17. What is meant by the loudness of sound? State the factor on which it depends. Draw figures to illustrate

1. soft sound and
2. loud sounds.

Answer:

1. Loudness is the measure of sound energy reaching the ear per second. The greater the sound energy reaching our ears per second, the louder the sound will appear to be.
2. It depends on the amplitude of the sound wave, which depends upon the force with which an object is made to vibrate.

Question 18. A nail was gently touched by the hammer and then was hit harder.

1. When will be the sound created louder?
2. Which characteristic of sound here is responsible for the change in sound?

Answer:

1. Sound will be produced when we beat hard on the nails.
2. The amplitude of the vibrating body is responsible for the change in sound.

Question 19. State the conditions required to hear an echo.

Answer:

The conditions to hear an echo are

1. The time interval between source sound and reflected sound must be at least 0.1s.
2. The minimum distance between the obstacle and the source of sound should be at least 17.2 m.

Question 20. When can we distinctly hear the echo of a sharp sound? Why cannot we hear an echo in a small hall?

Answer:

1. As we know, to listen to echo for a sound wave whose speed in air is 344 ms¹ and persistence of sound is 0.1 s, a minimum distance of 17.2 m between the observer and reflecting surface is required.
2. The condition of the minimum distance between the die source and the reflecting surface is not satisfied, so in a small hall, we do not hear echoes.

Question 21. Explain the rolling action of thunder.

Answer:

The rolling of thunder is due to multiple reflections of the sound of thunder from a number of reflecting surfaces such as clouds and land as several echoes may be heard.

Question 22. Distinguish between the terms

1. music and noise
2. tone and note.

Answer:

(1) Music The sound which is pleasant to the ears is called music. It is produced by regular periodic vibrations. There is no sudden change in loudness, e.g. Sound produced from a tabla.

Noise The sound which is unpleasant to the cars is called noise. It is produced at irregular intervals. There is a sudden change in its loudness, For Example. The sound produced in a market and the sound produced by an explosion.

(2) Tone The sound of a single frequency is called a tone.

Note The sound which is a mixture of several frequencies is called a note.

Question 23. When a workman hammers to one end of the long iron pipeline, an observer places his ear on the other end of the pipeline. How he can distinctly hear two sounds? Justify your answer.

Answer:

Due to the propagation of sound through solids such as iron, the sound of hammering to one end will be heard by the observer on the other end of the pipeline.

Also, the sound of hammering will be propagated through air to reach the observer.

As we know, sound travels faster in iron than in air. So, the observer hears two sounds. The first one, travelling through the iron pipeline and the second travelling through air.

Question 24. (1) Define

1.  infrasonic wave
2. ultrasonic wave.

(2) Name two species of animals which can produce and detect

1.  infrasonic waves
2. ultrasonic waves.

Answer: 

(1) (1) Infrasonic wave The sound of frequency lower than 20 Hz is known ai infrasonic sound (or wave).

(2) Ultrasonic wave The sound of frequency higher than 20,000 Hz arc is called ultrasonic wave.

(2) (A) Animals like whales, elephants and rhinoceroses produce infrasonic sound of frequency 5 Hz,

(B) Dogs can hear and detect ultrasonic waves of frequency up to 50 Hz.

Question 25. (1) Which has a shorter wavelength infrasonic or ultrasonic?

(2) Can dolphins detect ultrasonic waves?

(3) Name any one living organism which can detect infrasonic waves.

Answer:

1. The ultrasonic wave has shorter wavelengths.
2. Yes, dolphins can detect ultrasonic waves.
3. Rhinoceros can detect infrasonic waves.

Question 26. A key of a piano is struck gently and then struck again but much harder this time. What will happen in the second case?

Answer:

1. In the second case, the key is struck harder, so the amplitude of vibration of the string increases and hence loudness increases.
2. Also, frequency and hence pitch increase with an increase in force or tension in the string.

Question 27. A student went to a hill station early in the morning, he could hear the echo of his clap after 0.1 s. When he went to the same place in the afternoon he could not hear the echo at all. Explain the reason for his changed observation.

Answer:

There is a rise in temperature in the afternoon, so the speed of sound increases with an increase in temperature, As speed increases, the time taken by reflected sound will be less, which may be less than 0.1 s in the afternoon.

That is why the student could not hear the echo at all in the afternoon.

Question 28. A girl is sitting in the middle of a park of dimension 12 m x 12 m. On the left side of it, there is a building adjoining the park and on the right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.

Answer:

No, the girl can’t hear the echo of this sound because, the distance between the girl and the obstacle (building) is only 6 m approx but an echo is heard only, if the minimum distance between the observer at the source of sound and the obstacle is 11.3 m

Question 29. The wavelength of a sound wave was measured by Rohit as 8 m. The frequency of a sound wave is given as 40 Hz. What is the speed of sound as calculated by Rohit?

Answer:

Given, \(\lambda\)=8 m and v=40 Hz

As speed of sound, v= v \(\lambda=40 \times 8=320 \mathrm{~ms}^{-1}\)

Question 30. How are ultrasonic waves different from ordinary sound waves? State two applications of ultrasound.

Answer:

Ultrasonic waves have greater frequency (more than 20,000 Hz). Ordinary sound has a lower frequency than ultrasonic waves.

Due to their high frequencies,

1.  they have high power.
2. they can penetrate anywhere to a large extent.
3. they are able to travel along well-defined straight paths, even in the presence of obstacles.

Applications of ultrasound are

1. To deter the flaw or defect in metal.
2. In the diagnosis of diseases.

Question 31. An echo is heard on a day when the temperature is about 22°C. Will echo be heard sooner or later, if the temperature increases to 40°C?

Answer:

An echo will be heard sooner than the echo heard when the temperature is 22°C because the speed of sound increases with an increase in temperature.

Also, speed of sound in air = \(\frac{\text { distance }}{\text { time }}\)

i.e. If the speed of sound increases, then the time after which the echo will be heard decreases.

Question 32. (1) Write two main properties of ultrasound.

(2) Mention one application of ultrasound in (A) industries (B) the medical field.

Answer:

(1) The two main properties of ultrasound are

•  High frequency
• These travel along well-defined paths even in the presence of obstacles.

(2) (A) In industries, ultrasound is used to detect cracks and flaws in metal blocks.

(B) In the medical field, ultrasound is used to break stones in the gall bladder and kidney.

Question 33. How is it Mint, hots or able to fly at night without colliding with other objects?

Answer:

1. Bats search out prey and fly at night by emitting and detecting reflections of ultrasonic waves.
2. Bats emit high-frequency ultrasonic squeaks while flying and listen to the echoes produced by the reflection of their squeaks from the obstacle i.e. prey or object in their path.
3. From the time taken by the echo to be heard, bats can determine the distance of the object and can avoid the object by changing the direction without colliding with it.

Question 34. Find the wavelength of sound for frequencies up to 120 kHz at which a bat can hear it. Take the speed of sound in air as 344 99 \(\mathrm{~ms}^{-1}\).

Answer:

Given, speed of sound in air, v=340 \(\mathrm{~ms}^{-1}\)

Time taken to hear echo, t=3 s

As we know that, distance, s= speed (v) \(\times time (t)\)

=340 \(\times\) 3=1020 m

As in 3s, sound has to travel twice the distance between man and the cliff. So, the distance between man and the cliff is

=\(\frac{s}{2}=\frac{1020}{2}\)=510 m

Question 35. Find the wavelength of the tone produced by a body vibrating with a frequency of 4 kHz. [Given the speed of sound in air is 344 m/s]

Answer:

Given,v=4 \(\mathrm{kHz}=4000 \mathrm{~Hz}\) and v=344 \(\mathrm{~m} / \mathrm{s}\)

As, v=v \(\lambda \Rightarrow \lambda=\frac{v}{v}=\frac{344}{4000}=0.086 \mathrm{~m}\)

Question 36. A man fires a rifle in front of a cliff and hears the echo after 3 seconds. Calculate the distance of man from the cliff, if the velocity of sound in air is 340 ms-^-1.

Answer:

Given, speed of sound in air, v = 340 ms^-1 Time taken to hear echo, t = 3 s As we know that, distance, s=speed (v) x time (t)

= 340 x 3 =1020m

In 3 s, sound has to travel twice the distance between man and the cliff. So, the distance between man and the cliff is

=\(\frac{s}{2}=\frac{1020}{2}=510 \mathrm{~m}\)

Question 37. A sound wave has a frequency of 3 kHz and a wavelength of 45 cm. How long will it take to travel 1.8 km?

Answer:

Given, frequency,v=3 kHz=3 \(\times 10^3 \)Hz

Wavelength, \(\lambda=45 \mathrm{~cm}\)=45 \(\times 10^{-2} \mathrm{~m}\)

From the relation, speed of wave, v=v \(\lambda\)

=3 \(\times 10^3 \times 45 \times 10^{-2}=1350 \mathrm{~ms}^{-1}\)

Distance to be covered =\(1.8 \mathrm{~km}=1.8 \times 1000\)

=1800 m

Time taken by the wave to cover the distance of 1.8 km

= \(\frac{\text { Distance }}{\text { Speed }}=\frac{1800}{1350}=1.33 \mathrm{~s}\)

Question 38. Ocean waves of time period of 0.01 s have a speed of 15 ms-1. Calculate the wavelength of these waves. Find the distance between a crest and an adjoining trough.

Answer:

Given, time period of the waves, r=0.01 s

Speed of the waves, v=15 \(\mathrm{~ms}^{-1}\)

Wavelength of these waves, \(\lambda=\frac{v}{v} [ v=v \lambda]\)

or \(\lambda=\nu \times T=15 \times 0.01=0.15 \mathrm{~m}\)

The distance between the crest and adjoining trough

= \(\lambda / 2=\frac{0.15}{2}\)

= 0.075 m

Question 39. In a ripple tank, 14 full ripples are produced in 1 s. If the distance between a crest and the next trough is 12 cm, calculate (1) wavelength and (2) velocity of the wave.

Answer:

(1) Given, the distance between a crest and the next trough

=\(\frac{\lambda}{2}=12 \mathrm{~cm}\)

The wavelength of the wave,

⇒ \(\lambda=12 \times 2=24 \mathrm{~cm}\)=0.24 m

(2) Number of ripples produced in 1 s=14 Hz

Frequency of wave, v=14 Hz

As velocity, \(\nu=\mathrm{v} \lambda\)

v=14 \(\times 0.24=3.36 \mathrm{~ms}^{-1}\)

Question 40. The sound produced by a thunderstorm is heard 10 seconds after the lightning is seen. Calculate the approximate distance of the thundercloud. (Given, the speed of sound = 340 ms^-1)

Answer:

Given, time, t =10 s and speed, v =340 ms^-1

We know that, distance = speed x time

= 340 x 10 = 3400 m = \(\frac{3400}{1000} \mathrm{~km}\)

= 3.4 km

Question 41. Compare the frequencies of notes A and C played during a musical concert when the speed of sound is 340 ms^-1 and the wavelength of A and C are 1.5 m and 1.33 m, respectively.

Answer:

As we know that, from the relation, v =v\(\lambda\)

Frequency of the sound wave (v)

Speed of sound (y)

Wavelength of the sound wave (k)

=\(\frac{\text { Speed of sound }(v)}{\text { Wavelength of the sound wave }(\lambda)}\)

So, frequency of note A, \(v_A=\frac{340}{1.5}=226.66 \mathrm{~Hz}\)

Again, frequency of note C, \(v_C=\frac{340}{1.33}\)=255.63 Hz

So, the frequency of note C is more than the frequency of note A.

Question 42. Aditi clapped her hands near a cliff and heard the echo after 4 seconds. What is the distance of the cliff from her, if the speed of sound is taken as 346 ms^-1?

Answer:

Given, the speed of sound, v =346 ms

Time taken to hear the echo,

t = 4 s

Distance travelled by the sound = v x t

= 346 x 4 = 1384m

In 4 s, sound has to travel twice the distance between the cliff and Aditi. Therefore, the distance between the cliff and Aditi is \(\frac{1384}{2}\)=692 m

Question 43. A sound wave travels at a speed of 340 ms¹. If its wavelength is 1.5 cm. What is the frequency of the wave? Will it be audible?

Answer:

Given that, speed of sound, v=340 \mathrm{~ms}^{-1}

Wavelength, \(\lambda=1.5 \mathrm{~cm}=\frac{1.5}{100} \mathrm{~m}\). Frequency, v=?

Using the formula, v=\(\mathrm{v} \lambda\)

= \(\frac{\nu}{\lambda}=\frac{340}{\left(\frac{1.5}{100}\right)} \mathrm{ms}^{-1}\)

= \(\frac{340 \times 100}{1.5}=22666.6 \mathrm{~Hz}\)

We know that the audible range of frequencies is 20 Hz to 20000 Hz.

So, the given frequency (i.e. 22666.6 Hz) is not audible.

Question 44. If the velocity of sound in air is 330 ms-1, then express the audible range of frequencies in terms of time period.

Answer:

The audible range of frequencies is V1 =20 Hz,

v₂ = 20000 Hz and the velocity, v = 330 ms^-1

So, time period will become,

⇒ \(T_1=\frac{1}{\mathrm{v}_1}=\frac{1}{20}=5 \times 10^{-2} \mathrm{~s}\)

and time period will become,

⇒ \(T_2=\frac{1}{v_2}=\frac{1}{20000}=5 \times 10^{-5} \mathrm{~s}\)

Thus, the audible range in terms of time period is from 5 \(\times 10^{-2} \mathrm{~s}\) to 5 \(\times 10^{-5} \mathrm{~s}\).

Question 45. A ship sends out an ultrasound that returns from the sea bed and is detected after 1.71 seconds. If the speed of ultrasound through seawater is 1531 ms^-1, what is the distance of the sea bed from the ship?

Answer:

Given, the time between transmission and detection, t =1.71 s

Speed of ultrasound through seawater, v =1531 ms^-1

Distance travelled by the ultrasound

= 2 x Depth of the sea = 2d

or 2d = 1 x r =1531 x 1.71 = 2618.01m

d=\(\frac{2618.01}{2}=1309 \mathrm{~m}\)

Hence, the distance of the sea bed from the ship is 1309 m.

Question 46. For hearing the loudest ticking sound heard by the ear, find the angle x in the given figure.

Answer:

We know that in the laws of reflection, the angle of incidence (x) is always equal to the angle of reflection (x). Since AOB is a straight line.

⇒ \(\angle A O B\) =\(180^{\circ}\)

⇒ \(50^{\circ}+x+x+50^{\circ}=180^{\circ}\)

[sum of all angles lies on the same side of a line is \(180^{\circ}\) ]

⇒ \(2 x+100^{\circ}=180^{\circ}\)

⇒ \(2 x=180^{\circ}-100^{\circ} \Rightarrow 2 x=80^{\circ}\)

x=\(\frac{80^{\circ}}{2} \Rightarrow x=40^{\circ}\)

Hence, the value of x is 40°.

Question 47. A boat moving with a velocity of 20 ms¹ in a sea is rocked by waves. If its crests are 80 m apart, then at what time does the boat bounce up?

Answer:

Given that, wavelength, \(\lambda\) = 80 m

(Distance between two consecutive crests and troughs is equal to the wavelength.)

Velocity, v = 20 ms^-1, time, T = ?

As we know that, T=\(\frac{\lambda}{v}=\frac{80}{20}=4 \mathrm{~s}\)

Question 48. The given graph shows the displacement versus time relation for a disturbance travelling with a velocity of 1500 ms^-1. Calculate the wavelength of the disturbance.

Answer:

Given, velocity, v=1500 \(\mathrm{~ms}^{-1}\)

Time taken in one complete cycle is 2 \(\mu \mathrm{s}\)

Time, T=2 \(\mu \mathrm{s}=2 \times 10^{-6} \mathrm{~s}\left[ 1 \mu \mathrm{s}=10^{-6} \mathrm{~s}\right]\)

We know that,v=\(\mathrm{v} \lambda\left[\mathrm{v}=\frac{1}{T}\right]\)

So, v=\(\frac{\lambda}{T} \Rightarrow \lambda\)=v T

where, \(\lambda\)= wavelength, v= frequency and T= time period

So,\(\lambda =1500 \times 2 \times 10^{-6}=3000 \times 10^{-6}\)

=3 \(\times 10^{+3} \times 10^{-6}=3 \times 10^{-3} \mathrm{~m}\)

Question 49. A construction worker’s helmet slips and falls when he is 78.4 m above the ground. He hears the sound of the helmet hitting the ground 4.23 s after it slipped. Find the speed of sound in the air.

Answer:

Here, s = distance travelled by the helmet to reach the ground =78.4 m

t = total time taken to hear the sound of the helmet hitting the ground =4.23 s

Let it be the time taken by the helmet to reach the ground.

So, we know that, according to the equation of motion, where, u = initial velocity g = acceleration due to gravity.

s=u t+\(\frac{1}{2} g t^2\left[\begin{array}{l}
\text { where, } u=\text { initial velocity } \\
g=\text { acceleration due to gravity. }\end{array}\right]\)

78.4=0 \(\times t+\frac{1}{2} \times 9.8 \times t^2 \Rightarrow 78.4=\frac{1}{2} \times 9.8 \times t^2\)

⇒ \(t^2=16 \Rightarrow\) t=4 s

So, time is taken by the sound wave to travel distance.

78.4 \(\mathrm{~m}=4.23 \mathrm{~s}-4 \mathrm{~s}=0.23 \mathrm{~s}\)

So, speed of sound in air =\(\frac{78.4 \mathrm{~m}}{0.23 \mathrm{~s}}\)=340.86 \(\mathrm{~ms}^{-1}\)

Question 50. A sound wave has a frequency of 2 kHz and a wavelength of 45 cm. It takes 4 s to travel. Calculate the distance it travels.

Answer:

Here, the frequency of sound waves,

v = 2 kHz = 2x 10 3 Hz

Wavelength of the wave, \(lambda\) = 45 cm =0.45 m

Velocity of the wave, v = v \(\lambda\)

= 2 x 103 x 0.45 =900 ms^-1

Time to travel = 4 s

Distance travelled by the wave

= vt =900 x 4 =3600 m

Class 9 Science Chapter 11  Sound Long Answer Type Questions

Question 1. (1) What is meant by frequency of sound
waves?

(2) Give the range of frequencies of sound waves that an average human ear can detect.

(3) A source of wave produces 20 crests and 20 troughs in 0.2 s. The distance between a crest and next trough is 50 cm. Find the

• wavelength
•  frequency
• period of the wave.

Answer:

(1) Frequency The number of waves produced per second is called the frequency of the wave.

(2) 20 Hz to 20 kHz

(3)(A) Since the distance between a crest and the next trough is \(\frac{\lambda}{2}\).

Therefore, \(\frac{\lambda}{2}\)=50 cm (given)

⇒ \(\lambda\)=100 cm or 1 m

(B) Distance covered in 20 crests =20 \(\times\) 1=20 m

Velocity of the wave, \(\nu=v \lambda=\frac{\lambda}{t}\)

v \(\times 1=\frac{20}{0.2}\)[ t=0.2 s]

v=100 Hz or

1 \(\text { crest and } 1 \text { trough }=1 \text { wave }\)

20 \(\text { crest and } 20 \text { trough }=20 \text { waves }\)

Frequency =\(\frac{\text { Number of waves }}{\text { Time }}=\frac{20}{0.2}=100 \mathrm{~Hz}\)

1 crest and 1 trough =1 wave

20 crest and 20 troughs =20 waves

Frequency =\(\frac{\text { Number of waves }}{\text { Time }}=\frac{20}{0.2}=100 \mathrm{~Hz}\)

(C)Time period, T=\(\frac{1}{\mathrm{v}}=\frac{1}{100}\)=0.01 s

Question 2. Define frequency and wavelength with reference to sound. Explain, what is echo. Give two applications of ultrasound.

Answer:

Frequency The number of complete sound waves (or oscillations) produced in one second is called the frequency of the sound wave.

Wavelength The minimum distance in which a sound wave repeats itself.

Echo The repeddon of sound caused by the reflection of sound waves is called an echo.

Two applications of ultrasound are

1. Ultrasound is used in industries to detect flaws in metal blocks without damaging them.
2. Ultrasound is used to investigate internal organs of the human body such as the liver, gall bladder, pancreas, kidneys, etc.

Question 3. Establish the relationship between the speed of sound, its wavelength and frequency. If the velocity of sound in air is 340 ms^-1. Calculate

1. wavelength when the frequency is 256 Hz.
2. frequency when the wavelength is 0.85 m.

Answer:

The speed of sound is defined as the distance at which a point on a wave, such as a compression or a rarefaction, travels per unit of time.

We know that, speed, v=\(\frac{\text { distance }}{\text { time }}=\frac{\lambda}{T}\)

Here, \(\lambda\) is the wavelength of the sound wave. It is the distance travelled by the sound wave in one time period (T) of the wave.

⇒ \(\nu=\frac{\lambda}{T}=\lambda \times \frac{1}{T}\)

We know that, v=\(\frac{1}{T}\) [where, v= frequency]

⇒ \(\nu=\lambda v\)

i.e. Speed = Wavelength x Frequency

Given, speed of sound in air, v=340 \(\mathrm{~ms}^{-1}\) and frequency, v=256 Hz

(1) Speed = Wavelength x Frequency

⇒ \(\Rightarrow 340=\lambda \times 256 \Rightarrow \lambda=\frac{340}{256}=133 \mathrm{~m}\)

(2) Again, given wavelength, \(\lambda\)=0.85 m

Then, frequency of sound in air, v=\(\frac{\text { Speed }}{\text { Wavelength }}\)

⇒ {Speed= Wavelength x Frequency}

v=\( \frac{340}{0.85}=\frac{340 \times 100}{85}=400 \mathrm{~Hz}\)

Thus, the frequency of sound is 400 Hz.

Question 4. Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also, define wavelengths and time periods using this curve.

Answer:

We have a curve showing density or pressure variations with respect to distance for a disturbance produced by sound.

Wavelength can be defined as the distance between two successive compressions or rarefactions. It is denoted by λ.

The time taken by the waves to complete one full cycle so that its particles are in the same phase is called time period. It is denoted by T.