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Class 9 Science Notes For Chapter 7 Motion

• In everyday life, we observe several types of motions like vehicles moving on a road, flying birds, movement of needles of a watch, movement of blades of a fan, circulation of blood through veins and arteries, etc.
• Atom molecules, planets, stars and galaxies are all in motion. If a body changes its position concerning time and its surroundings, it is said to be in motion. Motion and rest are always relative but never absolute.

Various Terms Related To Motion

Position

The location of an object concerning a particular point is known as the position of the object. The particular point about which the position of the object is defined, is called reference point or origin.

Scalar and Vector Quantities

• Physical quantities with which we can associate only magnitude, i.e. numbers are called scalar quantities, For Example, mass, time, distance, speed, etc.
• Physical quantities with which we can associate magnitude, i.e. numbers as well as direction are called vector quantities, For Example, weight, displacement, velocity, etc.

Read and Learn  More Class 9 Science Notes

Distance

• The distance travelled by a body is the actual length of the path covered by it, irrespective of the direction in which the body travels. It is a scalar quantity. Its SI unit is metre, For Example, Consider the motion of an object moving along a straight path.
• Let the object start its motion from point O and move through points A, B, and C and reach upto point D.
• Then, the total distance covered by the object = actual length upto D of the path travelled

= OA + AB + BC + CD = 5m + 10m + 8m+20m = 43m

Displacement

• The displacement of an object is the change in the position of the object when it moves from a given position to another position. It is equal to the length of the shortest path measured in the direction from the initial position to the final position of the object. It is a vector quantity. Its SI unit is metre (m).
• Suppose an object starts to move from point O and reaches point B, passing through A.
• Here, the initial position of the object is O and the final position is B. So, displacement of the object = length of the shortest path between the initial position (O) and the final position (B).

= OB = \(\sqrt{O A^2+A B^2}\) [from Pythagoras theorem] }

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

= \(\sqrt{9+16}=\sqrt{25}\)

= 5 m

But, distance in this case = length of actual path = OA + AB = 3 m +4 m = 7 m

Distance versus Displacement

1. Displacement of a moving object can never be greater than the distance travelled by it.

Displacement \(\leq\) Distance

∴ \(\frac{\text { Displacement }}{\text { Distance }} \leq 1\)

i.e. Ratio of displacement and distance is always less than or equal to 1.

2. If a body moves along a straight line (only in one direction), then distance and displacement will be equal.

3. Displacement of the object can be positive, negative or zero but distance can never be negative or zero.

Example 1. A Jogger jogs along one length and breadth of a rectangular park. If the dimensions of park are 150 m x 120 m, then find the distance travelled and displacement of the Jogger.
Answer:

According to question,

Length, l = 150 m, Breadth, b =120 m Distance travelled, s =?

Displacement of the Jogger = ?

Suppose, the Jogger starts from point A and after covering one breadth and one length reaches at point C.

Distance, 5 = Length of the total path covered = AB + BC s = 150 + 120 = 270 m

Displacement = Minimum distance between initial and final position = AC

From Pythagoras theorem, A C=\(\sqrt{(A B)^2+(B C)^2}\)

i.e. Displacement =\(\sqrt{(150)^2+(120)^2}\)=30 \(\sqrt{41} \mathrm{~m}\)

Thus, the Jogger travels a distance of 270 m and his displacement is 30 \(\sqrt{41} \mathrm{~m}\).

Example 2, A body moves in a circular path of radius 20 cm. If it completes two and half revolution along the circular path, then find the distance and displacement of the body.
Answer:

According to question, body moves in a circular path of radius 20 cm. So, during one complete revolution the distance (d) moved by the body is equal to the circumference of the circular path, i.e. 2\(\pi r\). where, r = radius of circular path

So, for two and half revolution, body covers two complete revolutions, i.e. 2×2\(\pi r\) distance and half revolution, i.e. \(\frac{2 \pi r}{2}=\pi r\) distance.

Total distance (s) covered by the body = (2 x 2\(\pi r\)) + \(\pi\)r = 5\(\pi\)r

= 5 x 3.14 x 20

= 314 cm

Now, as we know displacement (d) is the straight line distance between the initial and final position of the body. So, after two and half revolutions, total displacement (d) of the body will be 2r, i.e. displacement (d) = 2×20 = 40 cm.

Uniform And Non-Uniform Motion

Uniform Motion

1. A body is said to have a uniform motion, if it travels equal distances in equal intervals of time, no matter how small these intervals may be. The distance travelled by an object in uniform motion increases linearly.
2. For Example. If a car moving along a straight line path, covers equal distances in equal intervals of time, it is said to be in uniform motion.

Non-uniform Motion

1. A body is said to have a non-uniform motion, if it travels unequal distances in equal intervals of time, no matter how small these intervals may be.
2. For Example, A car moving through a crowded market has non-uniform motion.

Rate of Motion

The ratio of distance travelled by an object to the time taken is called rate of motion. The various terms required to measure the rate of motion are as given below:

Speed

• Speed of an object is defined as the distance travelled by it per unit time.
• Speed or an object (v} = \(\frac{\text { Distance }(s)}{\text { Time }(t)}\)
• Speed is a scalar quantity. The SI unit of speed is metre per second (m/s). The distance travelled by an object is either positive or zero, so the speed may be positive or zero but never negative.

Speed can be classified as:

1. Uniform speed If a moving body covers equal distances in equal intervals of time, then the speed of the body Is said to be uniform, i.e. constant speed.
2. Non-uniform speed If a moving body covers unequal distances in equal intervals of time, then the speed of the body is said to be non-uniform, i.e. variable speed.
3. Average speed It is defined as the ratio of the total distance travelled by a body to the total time taken. It is expressed as the Total distance travelled

• • Average speed =\(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
  • =\(\frac{s_1+s_2+s_3+\ldots}{t_1+t_2+t_3+\ldots}\)

1. Instantaneous speed: The speed of an object at any particular instant of time or a particular point of its path is called the instantaneous speed of the object.

Example 3. The odometer of a bike reads 1600 km at the start of the trip and 2000 km at the end of the trip. If the bike took 16 h, calculate the average speed of the bike in km/h and m/s.
Answer:

Distance covered by bike, s = 2000 km -1600 km = 400 km

Time taken, t =16 h

Average speed \(v_{\mathrm{av}} =\frac{s}{t}\)

=\(\frac{400}{16}=25 \mathrm{~km} / \mathrm{h}\)

= 6.9 m/s

Therefore, the average speed of the bike is 6.9 m/s.

Example 4.  An object travels 14 m and then another 16m in 2s. What is the average speed of an object?
Answer:

According to the question,

First distance, sx =14 m

According to the question,

First distance, \(s_1=14 \mathrm{~m}\)

Second distance, \(s_2=16 \mathrm{~m}\), Times, \(t_1=4 \mathrm{~s}, t_2=2 \mathrm{~s}\)

Average speed =\(\frac{\text { Total distance }}{\text { Total time }}=\frac{s_1+s_2}{t_1+t_2}\)

=\(\frac{14+16}{4+2}=\frac{30}{6}=5 \mathrm{~m} / \mathrm{s}\)

Therefore, the average speed of an object is 3 m/s.

Speed with Direction: Velocity

1. Velocity of an object is defined as the displacement of the body per unit time. i.e. velocity is the speed of an object moving in a definite direction.
2. It is expressed as \(\text { Velocity of an object }(v)=\frac{\text { Displacement }(d)}{{Time}(t)}\)
3. Velocity is a vector quantity. The SI unit of velocity is metres per second (m/s). The velocity of an object can be positive, zero or negative. The velocity of an object can be changed by changing the object’s speed, the direction of motion or both.

Velocity can be classified as:

1. Uniform velocity If an object covers equal displacements in equal intervals of time without changing direction, then its velocity is known as uniform velocity, i.e. constant velocity.

2. Non-uniform velocity If an object covers unequal displacements in equal intervals of time, then its velocity is known as non-uniform velocity, i.e. variable velocity.

3. Average velocity It is defined as the ratio of the total displacement of the object to the total time taken.

It is expressed as  \(v_{\text {av }}=\frac{\text { Total displacement }}{\text { Total time taken }}\)

If the velocity of an object changes at a uniform rate, then the average velocity,

⇒ \(\left(v_{\text {av }}\right)=\frac{{Initial} \text { velocity }(u)+\text { Final velocity }(v)}{2}\)

4. Instantaneous velocity The velocity of an object at a particular instant of time or at a particular point of its path is called its instantaneous velocity.

Note : (1) If a body is moving in a straight line, then the magnitude of its speed and velocity will be equal.

(2) Average speed of an object can never be zero but the average velocity of a moving object can be zero.

Example 5. Rajeev went from Delhi to Chandigarh and returned to Delhi on his motorbike. The odometer of that read 4200 km at the start of trip and 4460 km at the end of his trip. If Rajeev took 4 h 20 min to complete his trip, then find the average speed and average velocity in km/h as well as in m/s.
Answer:

As we know that, the total distance covered,

s = final reading of odometer – initial reading of odometer

= (4460 -4200) km = 260 km

Total time taken, t = 4 h 20 min = 4.33 h

Average speed =\(\frac{\text { Total distance covered }(s)}{\text { Total time taken }(t)}\)

=\(\frac{260 \mathrm{~km}}{4.33 \mathrm{~h}}=60 \mathrm{~km} / \mathrm{h} or \frac{60 \times 5}{18}=16.67 \mathrm{~m} / \mathrm{s}\)

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

= \(\frac{0}{4.33}=0 \mathrm{~m} / \mathrm{s} \text { or } 0 \mathrm{~km} / \mathrm{h}\)

Average speed of Rajeev is 16.67 m/s and average velocity is 0.

Rate of Change of Velocity: Acceleration

Acceleration is defined as the rate of change of velocity with respect to time.

Mathematically, it is expressed as \(\text { Acceleration }(a)=\frac{\text { Change in velocity }(\Delta v)}{\text { Change in time }(\Delta t)}\)

If in a given time interval t, the velocity of a body changes from u to v, then acceleration a is expressed as

a=\(\frac{{Final} \text { velocity }(v)-{Initial}{velocity}(u)}{\text { time interval }(t)}\)

a=\(\frac{v-u}{t}\)

This kind of motion is known as accelerated motion. The SI unit of acceleration is \(\mathrm{m} / \mathrm{s}^2\). It is a vector quantity.

The acceleration is taken to be positive, if it is in the direction of velocity, negative if it is opposite to the direction of velocity and zero when it is moving with a constant velocity.

Note: If velocity of an object decreases with time, then it is said to have negative acceleration. Negative acceleration is also called deceleration or retardation.

Acceleration can be classified as:

(1) Uniform acceleration If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the object is said to be in a uniform acceleration.

For Example:

1. The motion of a freely falling body.
2. The motion of a ball rolling down on an inclined plane.

(2) Non-uniform acceleration If the velocity of an object increases or decreases by unequal amounts in equal intervals of time, then the object is said to be in a non-uniform acceleration.

For Example:

1. The movement of a car on a crowded city road.
2. The motion of the train leaving or entering the platform.

Example 6. Starting from a stationary position, a car attains a velocity of 5 m/s in 20 s. Then, the driver of the car applies a brake such that the velocity of the car comes down to 3 m/s in the next 6 s. Calculate the acceleration of the car in both the cases.
Answer:

Case 1 Initial velocity of the car, u = 0  it starts from stationary position]

Final velocity, v = 5 m/s, Time taken, t = 20 s .

a=\(\frac{v-u}{t}=\frac{5-0}{20}=0.25 \mathrm{~m} / \mathrm{s}^2\)

Case 2 Initial velocity, u=5 m/s

Final velocity, v=3m/s, Time taken, t=6 s

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

= \(\frac{\text { Final velocity }- \text { Initial velocity }}{\text { Time }}=\frac{v-u}{t}\)

a =\(\frac{3-5}{6}=-0.33 \mathrm{~m} / \mathrm{s}^2\)

Thus, the acceleration in both cases are 0.25 \(\mathrm{~m} / \mathrm{s}^2\) and -0.33 \(\mathrm{~m} / \mathrm{s}^2\).

Graphical Representation Of Motion

To describe the motion of an object, we can use line graphs.

In this case, line graphs show the dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.

Note: For a graphical description of a motion, it is convenient to take time along the A-axis, whereas distance, speed or velocity is taken along the Y-axis.

Types of Graph

There are two main types of graphs which we will be studying as given below:

1. Distance-Time Graph

The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice.

To draw a distance-time graph, time is plotted along the X-axis distance of the body is plotted along the T-axis.

In this case, the slope of the distance-time graph is equal to the speed of the object.

Distance-time graphs under various conditions are explained below:

(1) Distance-Time Graph for Uniform Motion

If an object travels equal distances in equal intervals of time, then it moves with uniform speed.

For uniform speed, a graph of distance travelled against time is a straight line as shown:

Distance-time graph for uniform motion

Interpretation

From the graph, it is clear that in equal intervals of time, i.e. 2s, the object covers an equal distance of 4m, so the motion is uniform and the graph is a straight line.

Calculation of speed To calculate the speed of the object from a distance-time graph, choose any two points say A and B on a straight line. From points A and B, draw perpendiculars AE and BC respectively, on the time axis. Now, draw a perpendicular AD on BC. The distance travelled by the object from point A to B is given by

⇒ \(\Delta x=B C-C D=s_2-s_1\)

Time taken by the object to cover this distance

= \(\Delta t=t_2-t_1\) .

Speed, v=\(\Delta x / \Delta t=\left(s_2-s_1\right) /\left(t_2-t_1\right)\)

i.e. \(\Delta x / \Delta t\)= Slope of distance-time graph.

(2) Distance-Time Graph for Non-uniform Motion

• If a body travels unequal distances in equal intervals of time, then the motion of the body is known as non-uniform motion. Non-uniform motion is of two types such as:
• When the speed of the body increases with the passage of time, then the distance-time graph will be a curve with a positive slope as shown alongside:

Interpretation

1. From the graph, it is clear that in equal intervals of time of two seconds, the body is covering unequal distances and this distance goes on increasing. That means, with the passage of time, the body is covering more and more distance in equal time, i.e. the speed of the body is increasing.
2. When the speed of the body decreases with the passage of time, then the distance-time graph will be a curve with a negative slope

Interpretation

From the graph, it is clear that in equal intervals of time of one second, the body is covering unequal distances and this distance goes on decreasing. That means, with the passage of time, the body is covering lesser and lesser distance in equal time, i.e. the speed of the body is decreasing.

The graph below shows the positions of a body at different times. Calculate the speed of the body as it moves from (1) A to B, (2) B to C and (3) C to D.

(1) For the motion from points to B, time taken, t = 5 – 2 = 3 s Distance covered = 3 — 0 = 3 m

Speed during the motion from point to B

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

= \(\frac{3}{3}=1 \mathrm{~m} / \mathrm{s}\)

(2) For the motion from point B to C, time is taken, t-7 — 5 = 2 s

Distance covered = 3 – 3 = 0 m [Body does not change its position from point B to C]

Speed during the motion from point B to C _ Distance Time

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

=\(\frac{0}{2}\)=0

(3) For the motion from point C to D,

time taken, r=10-7=3s

Distance covered = 8 – 3 = 3 m

Speed during the motion from point C to D

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

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

Example 7. The following table gives the data about motion of a car:

Plot the graph and

1. find the speed of the car between 12:00 h and 12:30 h,
2. what is the average speed of the car?
3. Is the car’s motion an example of uniform motion? Justify.

Answer:

(1) Speed of the car between 12: 00h and 12: 30 h is given by

v=\(\frac{65-30}{\frac{30}{60}}=\frac{35}{\left(\frac{1}{2}\right)}=70 \mathrm{~km} / \mathrm{h}\)

(2) Average speed, \(v_{\mathrm{av}}=\frac{100}{2}=50 \mathrm{~km} / \mathrm{h}\)

(3) No, because the car covered unequal distances in equal intervals of time.

2. Velocity-Time Graph

The velocity-time graph shows how the velocity of a body changes with the passage of time. To draw a velocity-time graph, the velocity of the body is plotted along the T-axis and the time taken by the body is plotted along X-axis.

The area under the velocity-time graph gives the displacement. Velocity-time graphs under various conditions are explained as below:

(1) Velocity-Time Graph for a Body Moving with Constant Velocity

When a body moves with constant velocity, i.e. its motion is uniform, then its velocity does not change with time.

The graph will be a straight line parallel to the time axis.

Interpretation

From the graph, it is clear that with time, there is no change in the velocity, i.e. the body is moving with constant velocity.

Calculation of distance or magnitude of displacement

Let us calculate the distance or magnitude of the displacement of a body between time tx and time t₂ • Draw perpendiculars A Cand BD from the points corresponding to time tx and time t₂ on the graph.

Now, AD – BC = velocity of the body

CD =(t₂ -t₁) = time interval

Thus, AD x CD = area of rectangle ABCD

Also, AD x CD = velocity x time

= distance or magnitude of displacement Thus, the magnitude of displacement

= area under velocity-time graph

(2) Velocity-Time Graph for Uniform Accelerated Motion

In uniform accelerated motion, the velocity changes be equal amounts in equal intervals of time. In this case, the velocity-time graph is a straight line passing through the origin.

Interpretation

The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. In equal intervals of time, i. e. 10 s, the change in velocity is 18 m/s, which remains the same, which means the acceleration of the body is constant. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

Calculation of distance or magnitude of displacement To determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. Therefore, S=area of ABCDE

=Area of the rectangle ABCD area of triangle ADE

∴ \(A B \times B C+\frac{1}{2}(A D \times D E)\)

(3) Velocity-Time Graph for Non-uniform Accelerated Motion

A velocity-time graph for non-uniform accelerated motion is given below:

It shows that the velocity of a body (or object) varies non-uniformly with time.

Example 8. The velocity-time graph of an ascending passenger lift is shown in the figure below. What is the acceleration of the lift?

1. During the first two seconds?
2. Between 2nd and 10th second?
3. During the last two seconds?

Answer:

Case 1. \(\Delta v=4-0=4 \mathrm{~m} / \mathrm{s}, \Delta t=2-0=2 \mathrm{~s}, a_1\)=?

⇒ \(a_1=\frac{\Delta v}{\Delta t}=\frac{4}{2}=2 \mathrm{~m} / \mathrm{s}^2\)

Case 2. \(\Delta v=4.6-4=0.6 \mathrm{~m} / \mathrm{s}, \Delta t=10-2=8 \mathrm{~s}, a_2\)= ?

⇒ \(a_2=\frac{\Delta v}{\Delta t}=\frac{0.6}{8}=0.075 \mathrm{~m} / \mathrm{s}^2\)

Case 3.  \(\Delta v=0-4.6=-4.6 \mathrm{~m} / \mathrm{s}, \Delta t=12-10=2 \mathrm{~s}, a_3\)=?

⇒ \(a_3=\frac{\Delta v}{\Delta t}=\frac{-4.6}{2}=-2.3 \mathrm{~m} / \mathrm{s}^2\)

A negative sign shows retardation.

Example 9. A body moves with a velocity of 2 m/s for 5 s, then its velocity increases uniformly to 10 m/s in the next 5 s. Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after 5 seconds.

1. Plot a velocity-time graph for the motion of the body.
2. From the graph, find the total distance covered by the body after 2 s and 12 s.

Answer:

As we know, the distance moved by the body after 2 s = area OAB’C’ = 2 m/s x 2 s = 4 m.

Again distance covered by the body after 12 s = Area OAED + Area of ABEF

=2 \(\mathrm{~m} / \mathrm{s} \times 10 \mathrm{~s}+\frac{1}{2} \times 5 \mathrm{~s} \times 8 \mathrm{~m} / \mathrm{s}+6 \mathrm{~m} / \mathrm{s} \times 2 \mathrm{~s} +\frac{1}{2} \times 2 \mathrm{~s} \times 4 \mathrm{~m} / \mathrm{s}\)

= 20m + 20m+12m + 4m = 56m

Uniform Circular Motion

If an object moves in a circular path with uniform speed, then its motion is called uniform circular motion.

When an object moves along a circular path, its direction of motion keeps changing continuously. The velocity changes due to continuous change in direction and thus motion along a circular path is said to be accelerated.

When a body takes one round of a circular path, then it travels a distance equal to its circumference which is \(2 \pi\) r, where r is the die radius of the circular path.

Then, the speed of the body moving in a circular path, v =2 \(\pi r / t\), where t is the time taken for one round of circular path and K is constant having a value of 22/7.

Some of the examples of uniform circular motion are as follows:

A piece of stone tied to a thread and rotated in a circle with a uniform speed.

The motion of blades of an electric fan around the axle.

The motion of the moon and the earth.

A satellite in a circular orbit around the earth.

A car is moving on a circular path with constant speed.

Example 10. The minute hand of a wall clock is 10 cm long. Find its displacement and the distance covered from 10:00 am to 10:30 am.
Answer:

Given, the length of the minute hand, l = 10 cm

Displacement from 10:00 am to 10:30 am is given by diameter AOB =21 = 20 cm

Total distance covered by the minute hand

ACB = \(\pi l=\frac{22}{7} \times 10 \mathrm{~cm}=\frac{220}{7}=31.43 \mathrm{~cm}\)

The displacement of the minute hand is 20 cm and the distance is 31.43 cm.

Activity 1

Objective: To measure the distance covered and magnitude of displacement

Materials Required: A metre scale and a long rope.

Procedure

1. Take a metre scale and a long rope.

2. Walk from one corner of a basketball court to its opposite corner along its sides.

3. Measure the distance covered by you and the magnitude of the displacement.

4. Note the difference between the two in this case.

Discussion/Conclusion

The distance covered while walking from one comer of the basketball court to its opposite comer along its sides is more, i.e. A to B and B to C is 2x.

The magnitude of the displacement is from A to Q which is less than the distance travelled.

Question 1. What do you understand by the displacement of an object?
Answer:

The displacement of an object is the shortest distance between the initial and final positions of the moving object.

Question 2. Which is least in this activity distance or displacement?
Answer:

Displacement is the least in this activity.

Question 3. Can displacement be negative?
Answer:

Yes, displacement can be negative.

Activity 2

Objective: To find the magnitude of the displacement using a road map.

Materials Required: Road map of India.

Procedure

1. A car fitted with an odometer is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km.

2. Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the road map of India.

Discussion Or Conclusion

We find the aerial distance from the atlas. Then, we measure the straight distance between Bhubaneshwar to New Delhi and convert the. measurement as given in the map.

Question 1. Name the device that is used to measure distance in a vehicle.
Answer:

An odometer is a device used to measure distance in a vehicle.

Question 2. What is the difference between the final reading and the initial reading of the odometer?
Answer:

The difference between the final reading and the initial reading of the odometer is called displacement.

Question 3. How do we find the displacement using a road map?
Answer:

We find the displacement by calculating aerial distance using an atlas map.

Activity 3

Objective: State whether the motion of the objects is uniform or non-uniform.

Materials Required: Given data table.

Procedure

1. The data regarding the motion of two different objects A and B are given in the table.
2.  Examine them carefully and state whether the motion of the objects is uniform or non-uniform.
3. Distances travelled by objects A and B at different times are as given below:
4. Time Distance travelled by object A (in m) Distance travelled by object B (in m)

Discussion Or Conclusion

1. Object A travels equal distances in equal intervals of time. Hence, object A shows uniform motion.
2. Object B travels unequal distances in equal intervals of time. Hence, object B is showing non-uniform motion.

Question 1. How will you identify the non-uniform motion of an object?
Answer:

An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time.

Question 2. What is the path of uniform motion?
Answer:

The path of uniform motion is a straight line.

Question 3. Give two examples of non-uniform motion.
Answer:

1. A racing horse
2. Throwing a softball

Activity 4

Objective: To show that light travels faster than sound.

Materials Required: Digital wristwatch or a stopwatch.

Data Required: Speed of sound in air = 346 m/s Speed of light = 3 x 108 m/s

Procedure

1. At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.
2. Can you answer why this happens?
3. Measure this time interval using a digital wristwatch or a stopwatch.
4.  Calculate the distance of the nearest point of lightning.

Discussion Or Conclusion

1. Lightening travels with the speed of light, i.e. 3 x 10⁸ m/s and thunder travels with the speed of sound in air, i.e. 346 m/s. That is why the sound of thunder reaches us some time later than we see the lightning. Using a digital wristwatch or a stopwatch, we measure this time interval.
2. Let us consider 2 s. Thus, distance = Speed x Time = 346 x 2 =692 m Since the speed of light is almost infinite and takes almost no time to reach us. The measured time interval is the time taken by the sound of thunder to reach us. Thus, light travels much faster than sound.

Question 1. Name the instrument used to measure time intervals.
Answer:

A digital wristwatch/stopwatch is used to measure time intervals.

Question 2. At what speed does the lightning travel?
Answer:

Lightening travels with the speed of light, i.e. 3 x10⁸ m/s.

Question 3. Why do we see lightning first during thunderstorms?
Answer:

We see lightning first because light travels faster than sound.

Activity 5

Objective: To identify a range of motions.

Procedure

In your everyday life, you come across a range of motions in which

1. acceleration is in the direction of motion.
2.  acceleration is against the direction of motion.
3. acceleration is uniform.
4. acceleration is non-uniform.

Can you identify one example of each of the above types of motion?

Discussion Or Conclusion

1. When the speed of a car on the road is increasing, then the acceleration of the car is in the direction of motion.
2.  When we apply brakes, the speed of the car is decreasing. The acceleration produced in the car is against the direction of motion.
3.  When the body is falling freely under the action of gravity, it has a uniform acceleration, i.e. 9.8 m/s (speed of the falling body is increasing at a constant rate.)
4. When a car is passing through city limits on a highway, its acceleration/retardation is non-uniform depending on traffic congestion.

Question 1. Name the instrument used to measure time intervals.
Answer:

A digital wristwatch/stopwatch is used to measure time intervals.

Question 2. At what speed does the lightning travel?
Answer:

Lightening travels with the speed of light, i.e. 3 x10⁸ m/s.

Question 3. Why do we see lightning first during thunderstorms?
Answer:

We see lightning first because light travels faster than sound.

Activity 5

Objective: To identify a range of motions.

Procedure:

In your everyday life, you come across a range of motions in which

1. Acceleration is in the direction of motion.
2.  Acceleration is against the direction of motion.
3. Acceleration is uniform.
4. Acceleration is non-uniform.

Discussion Or Conclusion

1.  When the speed of a car on the road is increasing, then the acceleration of the car is in the direction of motion.
2.  When we apply brakes, the speed of the car is decreasing. The acceleration produced in the car is against the direction of motion.
3. When the body is falling freely under the action of gravity, it has a uniform acceleration, i.e. 9.8 m/s (speed of the falling body is increasing at a constant rate.)
4.  When a car is passing through city limits on a highway, its acceleration/retardation is non-uniform depending on traffic congestion.

Activity 6

Objective: To plot and interpret the distance-time graph.

Materials Required: (1) Graph paper (2) Data given in the table

Procedure

1. Study the times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A as given in the table.
2. Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.Station Distance from A (km) Time of arrival (h) Time of departure (h)

Discussion Or Conclusion

1. The distance-time graph for the train is shown below. From 8:00 am to 8:15 am, the train is standing on station A. From 8:15 am to 11:15 am, the train moves with uniform speed and reaches station B at a distance of 120 km from A.
2. From 11:15 am to 11:30 am, the train is at rest at station B. From 11:30 am to 13:00 pm, the train moves uniformly from station B to station C covering a distance of 60 km (180 -120) and finally halts at point C.
3. Thus, when motion is uniform, the distance-time graph is a straight line and the slope of this straight line represents the velocity of the body.

Question 1. While plotting a distance-time graph, why do we plot time on the x-axis?
Answer:

While plotting a distance-time graph, we should plot time on the x-axis as all the independent quantities are plotted on the x-axis.

Question 2. What does the slope of the distance-time graph represent?
Answer:

The slope of the distance-time graph represents the velocity of the body.

Question 3. What is the geometrical shape of the distance-time graph?
Answer:

The geometrical shape of the distance-time graph is a straight line.

Activity 7

Objective: To plot distance-time graph on the same scale and interpret.

Materials Required: (1) Graph paper (2) Data given in the table

Procedure

1. Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. The table shows the distance travelled by them at different times.
2. Plot the distance-time graph for their motions on the same scale and interpret.

Discussion Or Conclusion

1. The distance-time graph of both Feroz and Sania shows we have taken time on the x-axis and distance along the y-axis choosing a suitable scale.
2. The distance-time graph of Feroz lies above the distance-time graph of Sania.
3. This shows that Feroz is moving faster than Sania. However, speed is non-uniform.

Question 1. Which one of them is moving faster?
Answer:

Feroz is moving faster than Sania.

Question 2. What is the nature of motion?
Answer:

The nature of motion is non-uniform.

Question 3. While plotting a distance-time graph, what will you choose on the X-axis?
Answer:

We will choose distance on the y-axis.

Activity 8

Objective: To describe a circular path with constant speed.

Materials Required: (1) Piece of thread (2) Stone

Procedure:

1. Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end.
2.  Now, let the stone go by releasing the thread.
3. Can you tell the direction in which the stone moves after it is released?
4. By repeating the activity a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.

Discussion Or Conclusion

1. As soon as the stone is released, it moves along the tangent to the circular path at that moment.
2. By releasing the stone at different positions of the circular path, we will find the direction in which the stone moves is always different, but it is along the tangent to the circular path at that instant.
3. The stone is moving along a circular path due to the centripetal force provided by the pull of our hand on the stone.
4. A stone describing a circular path with a . velocity of constant magnitude

Question 1. Which force is required to change the direction in a circular motion?
Answer:

Centripetal force is required to change the direction.

Question 2. What is the direction of the stone which is released?
Answer:

The direction is tangential to the circular path.

Question 3. What is uniform circular motion known as?
Answer:

Uniform circular motion is known as accelerated motion.

Motion Question And Answers

Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:

Yes, the displacement can be zero even, if the object has moved through a distance, e.g. a boy starts from his home to market and comes back. He has covered a distance but his displacement is zero.

Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 min 20 s from his initial position?
Answer:

The farmer takes 40 s to move along the boundary of the square field, i.e. after 40 the farmer is again at his initial position, so his displacement is zero.

Time is given = 2 min 20 s = (2 x 60 + 20) s = 140 s Displacement of farmer after 2 min 20 s, i.e. after 140 s = Displacement after (3 x 40 + 20) s = 0 + displacement after 20 s [v after each 40 s displacement is zero]

The farmer completes one round in 40 s, so he will complete 1/2 round in 20 s, i.e. after the 20s the final position of a farmer is C.

Displacement Of Farmer,

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

= \(\sqrt{10^2+10^2}\)

=10 \(\sqrt{2}=10 \times\) 1.414

=14.14 m

Question 3. Which of the following is true for displacement?

1. It cannot be zero.
2. Its magnitude is greater than the distance travelled by the object.

Answer:

1. The displacement can be zero, so the first statement is false.
2. The magnitude of displacement can never be greater than the distance travelled by the object. So, the second statement is also false.

Question 4. Distinguish between speed and velocity.
Answer:

Differences between speed and velocity are as given below:

Question 5. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Answer:

When the object travels in one direction along a straight line path then its average velocity will be equal to the average speed.

Question 6. What does the odometer of an automobile measure?
Answer:

It measures the distance travelled by an automobile.

Question 7. What does the path of an object look like when it is in a uniform motion?
Answer:

An object having uniform motion has a straight line path.

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, i.e. 3 x 108 m/s.
Answer:

Given, speed of signal = 3 x 10⁸ m/s

Time taken by the signal in reaching the earth = 5 min = 5 x 60 = 300s

Distance of spaceship from ground station

= Distance travelled by the signal in 5 min

= Speed x Time = 3 x 10⁸ x 300

= 9 x 10¹⁰m

Question 9. When will you say a body is in (1) uniform acceleration? (2) non-uniform acceleration?
Answer:

• A body is in uniform acceleration if it travels in a straight path when its velocity increases or decreases by an equal amount in equal time intervals.
•  A body is in non-uniform acceleration, if it travels in a straight path when its velocity increases or decreases by unequal amount in equal time intervals.

Question 10. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Answer:

Given, the initial speed of the bus,

u=80 \(\mathrm{~km} / \mathrm{h}=80 \times \frac{5}{18}=22.22 \mathrm{~m} / \mathrm{s}\)

The final speed of the bus,

v=60 \(\mathrm{~km} / \mathrm{h}=60 \times \frac{5}{18}=16.67 \mathrm{~m} / \mathrm{s}\)

Time taken to decrease the speed, t=5 s

Acceleration of the bus,

a =\(\frac{v-u}{t}\)

= \(\frac{16.67-22.22}{5}\)

=-1.11 \(\mathrm{~m} / \mathrm{s}^2\)

The negative sign of acceleration indicates that the velocity of the bus is decreasing, i.e. the bus retards.

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 min. Find its acceleration.
Answer:

The initial velocity of the train, u = 0

Final velocity of train, v=40 \(\mathrm{~km} / \mathrm{h}=40 \times \frac{5}{18}\)

=11.11 m/s

Time taken, t=10 \(\mathrm{~min}=10 \times 60=600 \mathrm{~s}\)

Acceleration, a=\(\frac{v-u}{t}\)

= \(\frac{11.11-0}{600}\)

=0.0185 \(\mathrm{~m} / \mathrm{s}^2\)

=1.85 \(\times 10^{-2} \mathrm{~m} / \mathrm{s}^2\)

Hence, acceleration of the train is =1.85 \(\times 10^{-2} \mathrm{~m} / \mathrm{s}^2\).

Question 12. What is the nature of distance-time graphs for uniform and non-uniform motion of an object?
Answer:

The distance-time graph for the uniform motion of an object is a straight line.

The distance-time graph for the non-uniform motion of an object is a curved line.

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:

When an object is at rest, then its distance-time graph is a straight line parallel to the 8-time axis. Thus, it indicates 3 that with a change in time, 5 there is no change in the position of the object, i.e. the Time — object is at rest.

Question 14. What can you say about the motion of an object, if its speed-time graph is a straight line parallel to the time axis?
Answer:

A straight line parallel to the time axis in a speed-time graph indicates that a change in time, there is no change in the speed of the object. This indicates the speed of the object is constant.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:

The displacement is measured by the area occupied below the velocity-time graph.

Exercises

Question 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s?
Answer:

According to the question,

diameter of the circular track = 200 m

Hence, radius =100 m

In 40 s, an athlete completes 1 round.

In 140 s, an athlete completes \(\frac{140}{40}=3 \frac{1}{2}\) rounds.

Hence, the distance covered by the athlete

= 3 \(\times 2 \pi r+\frac{1}{2} \times 2 \pi r\)

= 3 \(\times 2 \times \frac{22}{7} \times 100+\frac{1}{2} \times 2 \times \frac{22}{7} \times 100\)

= 2200 m

Displacement = Shortest path between the initial position and the final position = 2 r = 2 x 100 = 200 m

Question 2. Joseph jogs from one end A to the other end B  of a straight 300 m road in 2 min 30 s and then turns around and jogs 100 m back to point C in another 1 min. What are Joseph’s average speeds and velocities in jogging (1) from A to B and (2) from A to C?
Answer:

1. From A to B, Joseph covers distance = 300 m

Time = 2 min 30 s = 2 x 60 + 30

Hence, average speed =\(\frac{\text { total distance }}{\text { total time }}\)

=\(\frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)

Average velocity =\(\frac{\text { Total displacement }}{\text { Total time }}=\frac{300}{150}\)

= 2 m/s

Au in both cases, the distance covered and direction are the same.

2. From A to C, Joseph covers distance = 400 m

Time = 150 + 60 – 210 s

Hence, average speed =\(\frac{400}{210}=1.9 \mathrm{~m} / \mathrm{s}\)

Displacement =200 m

Time =150+60=210 s

Hence, average velocity = \(\frac{200}{210}=0.952 \mathrm{~m} / \mathrm{s}\)

Question 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Answer:

Let the distance covered by Abdul while driving to school =x

While going \(v_1=20=\frac{x}{t_1}\) [where, \(t_1\)= time taken to cover distance x ]

While returning \(v_2=30=\frac{x}{t_2}\) [where, \(t_2\)= time taken to cover x while returning]

Hence, \(t_1=\frac{x}{20}, t_2=\frac{x}{30}\)

Average speed  =\(\frac{\text { Total distance }}{\text { Total time }}=\frac{2 x}{\frac{x}{20}+\frac{x}{30}}=\frac{2 x}{\frac{5 x}{60}}\)

=\(\frac{2 x \times 60}{5 x}=24 \mathrm{~km} / \mathrm{h}\)

Question 4. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer:

Initial speed of car A =52 \(\mathrm{~km} / \mathrm{h}=52 \times \frac{5}{18} \)

=14.44 m/s

The car stops in 5 s, i.e. final speed of the car, v=0, time, t=5 s. For a speed-time graph of car A,

Initial speed of car B=3 km/h=3 \(\times \frac{5}{18}=0.83 \mathrm{~m} / \mathrm{s}\)

The car stops in 10s, i.e. final speed of the car, v=0, time, t=10s.

For a speed-time graph of car B,

The speed-time graph of both cars A and B is shown below:

Distance travelled by car A= Area of \(\triangle\) OCD

=\(\frac{1}{2} O C \times O D=\frac{1}{2} \times 14.44 \times\) 5=36.1 m

Distance travelled by car B= Area of \(\triangle\) OEF

= \(\frac{1}{2} O E \times E F \)

= \(\frac{1}{2} \times 0.83 \times 10\)=4.15 m

Thus, car A travelled farther than car B after the brakes were applied.

Question 5. The figure shows the distance-time graph of three objects A B and C. Study the graph and answer the following questions:

1. Which of the three is travelling the fastest?
2. Are all three ever at the same point on the road?
3. How far has C travelled when B passes A?
4. How far has B travelled by the time it passes C?

Answer:

1. The object for which the slope of the distance-time graph is maximum will have maximum speed, i.e. will travel the fastest. Here, for object B, the slope is maximum, so it is travelling the fastest.
2. All three objects will be at the same point on the road if the speed-time graph intersects each other at any point. Here, all three graphs do not intersect each other, so these three will never be at the same point on the road.
3. When B passes A, then the distance travelled by C =9.6 – 2 = 7.6 km
4. Distance travelled by B when it passes C =6 km.

Question 6. The speed-time graph for a car is shown in the below figure:

1. Find how far the car travels in the first 4 s. Shade the area on the graph that represents the distance travelled by the car during the period.
2. Which part of the graph represents the uniform motion of the car?

Answer:

The area under the slope of the speed-time graph gives the distance travelled by an object.

(1) We will calculate the distance represented by 1 square of the graph.

This can be done as follows. If 5 square on X-axis = 2 s,

1 square on X-axis = 2/5 s,

3 square on Y-axis = 2 m/s,

1 square on Y-axis = 2/3 m/s

So, area of 1 square on graph =\(\frac{2}{5} \times \frac{2}{3}\)

= \(\frac{4}{15} \mathrm{~m}\)

distance =\(\frac{4}{15} \mathrm{~m}\)

1 square represents distance =\(\frac{4}{15} \mathrm{~m}\)

Since approximately 62 squares come under the area of slope for the time of 4 s.

So, distance travelled in 4 s=\(\frac{4}{15} \times 62=16.53 \mathrm{~m}\)

(2) For uniform motion, the speed-time graph is a straight line parallel to the time axis. So, the straight part of the curve parallel to the time axis represents the uniform motion of the car.

Question 7. State which of the following situations are possible and give an example for each.

1.  An object with a constant acceleration but with zero velocity.
2. An object is moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

1. When an object is thrown vertically upward, then at the highest point its velocity is zero but it has a constant acceleration of 9.8 m/s² (acceleration due to gravity).
2. An aeroplane flies in a horizontal direction but the acceleration due to gravity acts on it in a vertically downward direction, i.e. along the direction perpendicular to the direction of motion.

Question 8. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed, if it takes 24 hours to revolve around the earth.
Answer:

Given, the radius of the orbit, r=42250 km

Distance covered in one revolution,

d =2 \(\pi r\) [circumference of the orbit]

=2 \(\times \frac{22}{7} \times 42250=265571.43 \mathrm{~km}\)

Time taken in one revolution, t=24 h

Speed of satellite =\(\frac{\text { Distance }}{\text { Time }}=\frac{265571.43}{24}\)

=11065.48 \(\mathrm{~km} / \mathrm{h}\)

Summary

1. An object is said to be in motion if its position changes with time.
2. The distance travelled by a body is the actual length of the path covered by it, irrespective of the direction in which the body travels.
3. It is a scalar quantity. Its SI unit is metre.
4. The shortest distance between the initial and final position of the moving object is called the displacement of the object. It is a vector quantity. Its SI unit is metre.
5. The ratio of distance travelled by an object to the time taken is called the rate of motion.
6. The speed of an object is defined as the distance travelled by it in unit time.
7. Speed of an object (v) = \(\frac{\text { Distance }(s)}{\text { Time }(t)}\)
8. It is a scalar quantity. Its SI unit is metres per second (m/s).
9. The velocity of an object is defined as the displacement of the body per unit time.
10. Velocity of an object (v)=\(\frac{\text { Displacement }(d)}{\text { Time }(t)}\)
11. It is a vector quantity. Its SI unit is metres per second (m/s).
12. Acceleration is defined as the rate of change of velocity with respect to time.
13. Acceleration (a)=\(\frac{\text { Change in velocity }(\Delta v)}{\text { Change in time }(\Delta t)}\)
14. It is a vector quantity. Its SI unit is m/s².

Types of acceleration

1. Uniform acceleration If the velocity of an object changes by an equal amount in equal intervals of time, then the acceleration of the object is known as uniform acceleration.
2. Non-uniform acceleration If the velocity of an object changes by an unequal amount in equal intervals of time, then the acceleration of the object is known as non-uniform acceleration.
   1. When an object moves in a circular path with a uniform
   2. speed, its motion is called uniform circular motion.

Class 9 Science Notes For Chapter 8 Force And Laws Of Motion

In our everyday life, we observe that some efforts are required to put a stationary object into motion or to stop a moving object. We ordinarily experience this as a muscular effort and say that we must push, hit, or pull on an object to change its state of motion. The concept of force is based on this push, hit, or pull.

Force

Any action which causes pull, hit, or push on a body is called force. Force cannot be seen but it can be judged only by the effects which it produces in various bodies around us. Many effects of force are given below:

1. A force can move a stationary body.
2. A force can stop a moving body.
3. A force can change the direction and speed of a moving body.
4. A force can change the shape and size of a body.
5. Balanced and Unbalanced Forces
6. Forces are of two types such as balanced forces and unbalanced forces which are as given below:

Balanced Forces

When the net effect produced by several forces acting on a body is zero, then the forces are said to be balanced forces. Balanced forces can only bring a change in the shape of the body. A block of wood is placed on a horizontal surface and two strings A and B are connected to it. The block is in a state of rest.

Read and Learn  More Class 9 Science Notes

If we pull A and B strings with equal magnitude of forces, then the block does not change its state of rest. Such types of forces are known as balanced forces.

Unbalanced Forces

When the net effect produced by several forces acting on a body is non-zero, then the forces are said to be unbalanced forces. A boy wants to relocate the refrigerator in his house as shown

He pushes the refrigerator with a small force, the refrigerator does not move due to frictional force acting in a direction opposite to the push. If he pushes the refrigerator harder, then the pushing force becomes more than the friction, and due to this, the refrigerator starts moving in the direction of push as shown in the figure.

In the above example, there is an unbalanced force that causes motion in the refrigerator. The unbalanced forces cause a change in the state of rest or of uniform motion of a body.

For Example, In a tug of war when one of the teams suddenly releases the rope, then an unbalanced force acts on the other team due to which it falls backward.

If an unbalanced force is applied to a moving object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required

Newton’s Laws Of Motion

Newton studied the ideas of Galileo regarding the motion of an object. He formulated three fundamental laws that govern the motion of objects. These three laws are known as Newton’s laws of motion, which are as given below:

Newton’s First Law of Motion

It states that an object will continue to remain in its state of rest or of uniform motion along a straight line path unless an external force acts on it. This means all objects resist change in their state.

The state of any object can be changed by applying external forces only.

(1) A person standing on a bus falls backward when the bus starts moving suddenly. This happens because the person and bus both are at rest while the person is not moving as the bus starts moving. The legs of the person start moving along with the bus, but the rest portion of his body tends to remain at rest. Because of this, a person falls backward if he is not alert.

(2) A person standing in a moving bus falls forward if the driver applies brakes suddenly.

Inertia

The unwillingness (or inability) of an object to change its state of rest or uniform motion along a straight line is called the inertia of the object.

It is the inherent property of all the objects. Newton’s first law of motion is also known as Galileo’s law of inertia.

The inertia of an object is measured by its mass. Inertia is directly proportional to the mass. It means that inertia increases with an increase in mass and decreases with a decrease in mass. A heavy object will have more inertia than a lighter one.

Types of Inertia

Inertia is divided into three types as given below:

1. Inertia of rest The tendency of a body to oppose any change in its state of rest is known as inertia of rest. For Example.

• When a bus suddenly starts moving forward, then the passengers in the bus fall backward.
• The carpet is beaten with a stick to remove the dust particles.
•  When a tree is vigorously shaken, then some of the leaves fall from the tree.

2. Inertia of motion The tendency of a body to oppose any change in its state of uniform motion is known as inertia of motion, For Example.

• The passengers fall forward when a fast-moving bus stops suddenly.
• A person falls forward while getting down from a moving bus or train.
•  Luggage is usually tied with a rope on the roof of a bus.

3. Inertia of direction The tendency of a body to oppose any change in its direction of motion is known as inertia of direction, For Example.

• When a fast-moving bus negotiates a curve on the road, then passengers fall away from the center of the curved road.
• The sparks produced during the sharpening of a knife against a grinding wheel leave tangentially to its rim.
• A stone tied to a string is whirling in a horizontal circle. If the string breaks, then the stone flies away tangentially.

Momentum

Momentum measures the quantity of motion possessed by a body. It is defined as the product of the mass and velocity of the body. Besides magnitude, momentum also has a direction.

At any instant, its direction is the same as the direction of the velocity. If a body of mass m moves with a velocity v, then momentum p is given by p=mv

The SI unit of momentum is kg-m/s.

Example 1. A car of mass 1000 kg is moving with a velocity of 72 km/h. Find its momentum.
Answer:

Given, mass, m = 1000 kg

Velocity, v=72 km/h =72 \(\times \frac{5}{18}\)=20 m /s and Momentum,  p= ?

Momentum = Mass x Velocity => p – mv p =1000 x 20 = 20000 kg-m/s Thus, the momentum of the car is 20000 kg-m/s.

Newton’s Second Law of Motion

The second law of motion states that the rate of change of momentum of an object is directly proportional to the applied external force and takes place in the direction in which external force acts.

Mathematical Formulation of the Second Law of Motion

If a body of mass m moving at initial velocity u accelerates uniformly with an acceleration a for time t, so that its final velocity changes to v, then

Initial momentum,  \(p_1\)= mu

Final momentum, \(p_2\)= mv

Change in momentum = \(p_2-p_1\) = mv – mu  = m(v-u)

According to the second law of motion,

Force, F \(\propto \frac{\text { change in momentum }}{\text { time }}\)

⇒ \(F \propto \frac{p_2-p_1}{t} \Rightarrow F \propto \frac{m(v-u)}{t}\)

⇒ \(F \propto m a [\frac{v-u}{t}=a]\)

F = k ma

The quantity k is a constant of proportionality.

One unit of force is defined as the amount that produces an acceleration of 1 \(\mathrm{~m} / \mathrm{s}^2\) on an object of 1 kg mass.

i.e. 1 unit of force =k \(\times 1 \mathrm{~kg} \times 1 \mathrm{~m} / \mathrm{s}^2 \Rightarrow k=1\)

Thus, the force can be written as F =ma

The SI unit of force is Newton, which is denoted by the symbol N and it is equivalent to kg-m/s2.

When the applied force F is zero, then the acceleration a is also zero and the body remains in its state of rest or of a uniform motion.

Applications of Newton’s Second Law of Motion

The following applications are based on Newton’s second law of motion:

1. A cricket player (or fielder) moves his hands backward while catching a fast cricket ball.
2. During athletics meets, athletes doing the high jump and long jump land on foam or a heap of sand to decrease the force on the body, and the landing is comfortable.

Newton’s First Law from Mathematical Expression of Second Law

The first law of motion can be mathematically stated from the mathematical expression of the second law of motion.

As we know, F = ma

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

Ft = mv – mu

From this equation, if F = 0, then v = u for any value of time. This means that, in the absence of an external force, the object will continue moving with uniform velocity u throughout the time t and if u is zero, then v will also be zero, i.e. the object will remain at rest.

Example 2. The force acts on an object of mass 4 kg and changes its velocity from 10 m/s to 20 m/s in 5 s. Find the magnitude of force.
Answer:

Given, the mass of an object, m = 4 kg

Initial velocity, u = 10 m/s; Final velocity, v = 20 m/s Time taken, t = 5 s

We know that, Newton’s first law of motion,

v = u + at or a=\(\frac{v-u}{t}\)

a=\(\frac{20-10}{5}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}^2\)

Magnitude of force, F = Mass x Acceleration

F =ma = 4kgx2 m/s² = 8 N Hence, the magnitude of the force is 8 N.

Example 3. A force of 50 N acts on a stationary body of mass 10 kg for 2 s. Find the acceleration produced in the body and the velocity attained by it.
Answer:

Initial velocity, u- 0 [since the body is stationary]

Force, F – 50 N, mass, m —10 kg, time, t = 2 s

Final velocity, v = ?, acceleration, a = ? v -u

Acceleration, a=\(\frac{v-u}{t}\)

a=\(\frac{10-0}{2}=5 \mathrm{~m} / \mathrm{s}^2\)

Final velocity,v = \(\frac{F t}{m}+u {[F t=m(v-u)]}\)

v=\(\frac{50 \times 2}{10}+0=10 \mathrm{~m} / \mathrm{s}\)

Thus, the acceleration produced in the body is 5 \(\mathrm{~m} / \mathrm{s}^2\), and the velocity attained by it is 10 m/s.

Example 4 A bullet train is moving with a velocity of 180 km/h and it takes 5 s to stop after the brakes are applied. Find the force exerted by the brakes on the wheel of the train if its mass with the wagon is 2000 kg.
Answer:

Given, initial velocity u=180 km/h =180 \(\times \frac{5}{18}\)=50 m/s

Final velocity v=0,

time t=5 s , mass m = 2000 kg

From the first equation of motion,

v=u+a t

u=-a t

a=-\(\frac{u}{t}=\frac{-50}{5}=-10 \mathrm{~m} / \mathrm{s}\)

Now, the force exerted by the brakes on the wheel is given by Newton’s second law F – ma – 2000 x -10 = -20000 N Negative sign shows that the direction of force is opposite to the motion of the body.

Example 5. A force of 6N gives a mass mv an acceleration of 18 m/s² and a mass m², an acceleration of 24 m/s². What acceleration would it give if both the masses were tied together?
Answer:

Given, force F =6N

Acceleration of mass mx, ax =18 m/s² Acceleration of mass w², a² = 24 m/s²

Acceleration is produced when both masses are tied together a -?

From Newton’s second law of motion Reaction Ground ;

⇒ \(m_1=\frac{F}{a_1}=\frac{6}{18}=\frac{1}{3} \mathrm{~kg}\)

⇒ \(m_2=\frac{F}{a_2}=\frac{6}{24}=\frac{1}{4} \mathrm{~kg}\)

Total mass m=\(m_1+m_2=\frac{1}{3}+\frac{1}{4}=\frac{4+3}{12}=\frac{7}{12} \mathrm{~kg}\)

Acceleration produced in combined mass a=\(\frac{F}{m}\)

a=\(\frac{6}{7 / 12}=\frac{12 \times 6}{7}=10.28 \mathrm{~m} / \mathrm{s}^2\)

Example 6. The velocity-time graph of a ball of mass 30 g moving along a straight line on a long table is given figure. How much force does the table exert on the ball to bring it to rest?

Answer:

The initial velocity of the ball =30 cm/s

Final velocity v=0 and Time t=6 s

Acceleration a=\(\frac{v-u}{t}=\frac{(0-30)}{6}=-5 \mathrm{~cm} / \mathrm{s}^2=-0.05 \mathrm{~m} / \mathrm{s}^2\)

The force exerted on the ball, F = ma

F = ma =\(\left(\frac{30}{1000}\right) \times(-0.05)=-1.5 \times 10^{-3} \mathrm{~N}\)

The negative sign shows that the frictional force exerted by the table is opposite to the direction of motion of the ball.

Impulse

It is termed as the total impact of force. This is equal to the change in momentum of the body. In other words, impulse is defined as the product of force and a small time in which force acts.

According to Newton’s second law, F =ma

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

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

Ft = mv – mu

Impulse, I=F t=\(p_2-p_1\)

or Impulse = Change in momentum

The SI unit of impulse is N-s or kg – m/s

Example 7. If a force of 1000 N is applied over a vehicle of 500 kg, then for how much time the velocity of the vehicle will increase from 2 m/s to 10 m/s? Also, find the impulse.
Answer:

Given, F = 1000 N, mass m = 500 kg

Final velocity =10 m/s

Initial velocity u = 2 m/s

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

t=\(\frac{m(v-u)}{F}=\frac{500 \times(10-2)}{1000}=4 \mathrm{~s}\)

Impulse I = Ft =1000 x 4 = 4000 N-s

Thus, the time required by the vehicle is 4s, and its impulse is 4000 N-s.

Newton’s Third Law of Motion

The third law of motion states that, whenever one object exerts a force on another object, then the second object exerts an equal and opposite force on the first object.

Spring balance action and reaction forces are equal and opposite

Thus, action and reaction forces are equal in magnitude and opposite in direction. They still do not cancel each other’s effect because they act on different objects.

Applications of Newton’s Third Law of Motion

1. Collision of two persons If two persons walking or running in opposite directions collide with each other, then both feel hurt because they apply force to each other. Two opposing forces are in action and reaction pairs.
2.  Walking of a person A person is able to walk because of Newton’s third law of motion. During walking, a person pushes the ground in a backward direction, and in the reaction, the ground also pushes the person with an equal magnitude of force but in the opposite direction. This enables him to move in the forward direction against the push.
3. The recoil of a gun When a bullet is fired from a gun, then the bullet also pushes the gun in the opposite direction with an equal magnitude of force. Since the gun has a greater mass than a bullet, the acceleration of the gun is much less than the acceleration of the bullet.
4. Propulsion of a boat in a forward direction Sailor pushes water with oar in a backward direction resulting in water pushing the oar in the forward direction. Consequently, the boat is pushed in the forward direction. The force applied by oar and water are of equal magnitude but in opposite directions.
5. Rocket propulsion The propulsion of a rocket is based on the principle of action and reaction. The rapid burning of fuel produces hot gases which rush out from the nozzle at the rear end at a very high speed. The equal and opposite reaction force moves the rocket upward at a great speed.

Activity 1

Objective: To demonstrate the property of inertia of rest using carom board and coin.

Materials Required:

1. Carom board
2.  Carom coins

Procedure

1. Make a pile of similar carom coins on a carom board as shown.
2. Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker.
3. If the hit is strong enough, the bottom coin moves out quickly.
4. Once the lowest coin is removed, the inertia of the other coins makes them fall vertically on the table.

Discussion/Conclusion

1. If the hit is strong enough, the bottom coin moves quickly on the carom table without disturbing the upper carom coins. Once the lowest coin is removed, the upper carom coins fall vertically on the table due to the inertia of rest.
2. If the hit is weak, the disturbance may be communicated to upper carom coins, which may fall randomly with or without the actual movement of the coin.

Question 1. What does the Inertia of an object tend to cause?
Answer:

The inertia of an object tends to resist any change in its state of rest or motion.

Question 2. What happens if the hit is weak?
Answer:

If the hit is weak, the upper carom coins may fall randomly with or without actual movement of the coin.

Question 3. On what factor does the inertia of a body depend?
Answer:

The inertia of the body depends on the mass of the body.

Activity 2

Objective: To demonstrate the property of inertia of rest using glass and coins.

Materials Required:

1. Glass tumbler
2. Stiff card
3. Five-rupees coin
4. Table

Procedure

1. Set a five-rupee coin on a stiff card covering an empty glass tumbler, standing on a table.
2. Give a sharp horizontal flick to the card with a finger. If we do it fast, then the card shoots away allowing the coin to fall vertically into the glass tumbler due to its inertia.
3. The inertia of the coin tries to maintain its state of rest even when the card flows off.

Discussion/Conclusion

When the card is given a sharp horizontal flick with a finger, the card underneath the five-rupee coin shoots away. The coin falls vertically into the glass tumbler due to the inertia of the rest.

The basic thing is that force is applied to the card. This is why the coin tends to remain at rest.

Question 1. Which law is associated with this activity?
Answer:

Newton’s first law of motion is associated with it.

Question 2. Name the concept involved in this activity.
Answer:

The concept of inertia is involved in this activity.

Question 3. There are two bodies of mass m₂ >m₂) present on a rough surface, which of them has more inertia and why?
Answer:

Since, m₁ > m₂, so mass m₁ has more inertia than m₂.

Activity 3

Objective: To demonstrate the forces of action and reaction are equal and opposite by throwing the bag full of sand.

Materials Required

1. Two students
2. Two carts
3. A bag full of sand
4. White paint

Procedure

1. Request two children to stand on two separate carts
2. Give them a bag full of sand. Ask them to play a game of catching the bag.
3. Does each of them receive an instantaneous reaction as a result of throwing the sandbag (action)?
4. You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Discussion/Conclusion

Yes, in this case, each of them receives an instantaneous reaction as a result of throwing the sandbag. When the boy throws the bag towards the girl, the cart of the boy moves back. Similarly, when the girl throws the bag towards the boy, the cart of the girl moves back.

Thus, it explains Newton’s third law of motion, i.e. the force is exerted forward in throwing the bag, and the person who is throwing it gets pushed backward. Action and reaction are taking place simultaneously on two different bodies.

Question 1. What happens to the cart when the two children are playing the game of catching the bag?
Answer:

The cart of each child moves outwards when both the children are playing the game of catching the bag.

Question 2. Name the type of reaction involved in throwing the bag.
Answer:

An instantaneous type of reaction is involved in throwing the bag.

Question 3. Are the forces of action and reaction equal?
Answer:

Yes, the forces of action and reaction are equal in magnitude but opposite in direction.

Force And Laws Of Motion Question And Answers

Question 1. Which of the following has more inertia?

1. A rubber ball and a stone of the same size
2. A bicycle and a train
3. A five-rupees coin and a one-rupee coin
4. Give reasons for your answer.

Answer:

1. The inertia of an object is proportional to its mass.
2. A stone of the same size as a of rubber ball will have greater mass, so the stone will have more inertia.
3. A train has much greater mass than of a bicycle, so the train will have more inertia.
4. A five-rupee coin has more mass than a one-rupee coin, so a five-rupee coin will have greater inertia.

Question 2. In the following example, try to identify the number of times the velocity of the ball changes.

A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the ball and kicks it towards a player of his team. Also, identify the agent supplying the force in each case.

Answer:

There are several times at which the direction and magnitude of velocity of the ball changes, which are as given below:

Whenever a force is applied to the ball, then the velocity of the ball will change.

1. When the first player kicks the ball toward another player of his team, then the velocity of the ball will change because the first player applies some force on the ball.
2. When another player kicks the ball toward the goal, then the velocity of the ball will change, here again, the force is applied to the ball.
3. When the goalkeeper of the opposite team collects the ball, then the velocity of the ball will be changed, it becomes zero. Here, the goalkeeper applies some force on the ball to stop.
4. When the goalkeeper kicks the ball towards his own team, then the velocity of the ball changes because the goalkeeper applies some force on the ball.

Question 3. Explain, why some of the leaves may get detached from a tree, if we vigorously shake its branch.

Answer:

Leaves have the inertia of rest. When the branch is shaken, it tends to remain in the same state and gets detached when the position of the branch changes.

Question 4. Why do you fall in the forward direction when a moving bus breaks to a stop and fall backward when it accelerates from rest?

Answer:

When the moving bus brakes to a stop, then the passengers who had inertia of motion, oppose a change in their state. However, the lower portion of their body in contact with the bus comes to rest.

So, they fall forward. When the bus accelerates from rest, then the passengers who had inertia of rest, oppose a change in their state. However, the lower portion of their body starts moving with the bus. Hence, they fall backward.

Question 5. If action is always equal to the reaction, then explain how a horse can pull a cart.

Answer:

The horse pushes the ground in a backward direction. The ground exerts a reaction force on the horse and cart system to push them forward.

When the reaction force exceeds the force of friction between the wheels of the cart and the ground, then the cart is pushed forward.

Question 6. Explain, why is it difficult for a fireman to hold a hose pipe, which ejects a large amount of water with a high velocity.

Answer:

A fireman finds it difficult to hold a hose-pipe which is ejecting a large amount of water at high velocity. Because the stream of water rushing out of the pipe in the forward direction exerts a large force on the pipe.

Due to the reaction of the forward force, a force is applied to the pipe in the backward direction. Therefore, the fireman struggles to keep the hose pipe at rest.

Exercises

Question 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

As per Newton’s first law of motion, no force is needed to move an object that is already moving with a constant (non-zero) velocity. So, when an object experiences a net zero external unbalanced force, then it can move with a non-zero velocity. When an external force is zero then the velocity of the object remains the same both in magnitude and direction.

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

Answer:

When a carpet is beaten with a stick, then the fibers of the carpet attain the state of motion while the dust particles remain at rest due to the inertia of rest, and hence dust particles get detached.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

• When the bus stops suddenly, the bus comes to a state of rest but the luggage remains in a state of motion. So, due to inertia of motion, the luggage moves forward and may fall down from the roof of the bus.
• If the bus starts suddenly, then the bus comes in a state of motion but the luggage remains in the state of rest. Due to the inertia of rest, the luggage does not move in the forward direction and may fall down.

Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. Why does the ball slow down to stop?

1. The batsman did not hit the ball hard enough
2. Velocity is proportional to the force exerted on the ball
3. There is a force on the ball opposing the motion
4. There is no unbalanced force on the ball, so the ball would want to come to rest.

Answer:

The ball slows down to stop because the force of friction between the ground and the ball acts as an external force that opposes the motion of the ball.

Question 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 seconds. Find its acceleration. Find the force acting on it, if its mass is 7 tonne.

(Hint: 1 tonne = 1000 kg)

Answer:

The truck starts from rest, so initial velocity, u = 0,

Distance, s = 400 m , Time, t = 20 s Mass, m – 1 tonne = 7 x 1000 = 7000 kg

From Newton’s second law of motion, s=u t+\(\frac{1}{2} a t^2\)

400 =0 \(\times 20+\frac{1}{2} \times a \times(20)^2\)

400 =200 a

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

From Newton’s second law of motion, the force acting on the truck,

F=m a=7000 \(\times 2 =14000 \mathrm{~N}\)

=1.4 \(\times 10^4 \mathrm{~N}\)

Thus, the acceleration of the truck is 2 m \(s^2\), and the force acting on it is 1.4 \times 10^4 N.

Question 6. A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer:

Given, the mass of the stone, m-1 kg,

Initial velocity, u = 20 m/s

Final velocity, v = 0 [since stone comes to rest]

Distance covered, s = 50 m

From Newton’s third law of motion, v² = u² + 2as

(0)² =(20)² + la (50)

⇒ 100 a =-400

⇒ a = – 4 m/s²

Here, a negative sign shows that there is a retardation in the motion of the stone.

Force of friction between stone and ice = Force required to stop the stone

= ma – 1 x (-4) = – 4 N

Question 7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate

1. The net accelerating force.
2. The acceleration of the train and
3. The force of wagon 1 on wagon 2.

Answer:

(1) Net accelerating force = Force exerted by the engine -Friction force

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

= 40000 -5000 =35000 = 3.5 x 104 N

(2) From Newton’s second law of motion,

Accelerating force = mass of the train x acceleration of train F

a =\(\frac{F}{m}\)

Mass of train =5 x Mass of one wagon

=5 x 2000=10000 kg

Acceleration =\(\frac{35000}{10000}=3.5 \mathrm{~m} / \mathrm{s}^2\)

(3) Force of wagon 1 on wagon 2

=(35000-3.5×2000) N = 28000 N

Question 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road, if the vehicle is to be stopped with a negative acceleration of 1.7 m/s²?

Answer:

Given, mass, m~ 1500 kg, acceleration, = – 1.7 m/s²

From Newton’s second law of motion,

F = ma – 1500 x (-1.7) = -2550N

Question 9. What is the momentum of an object of mass m moving with a velocity v?

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

Answer: 4. mv

The momentum of an object of mass m moving with a velocity v is given by p = mv.

Question 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer:

The cabinet will move across the floor with constant velocity if there is no net external force applied to it. Here, a horizontal force of 200 N is applied to the cabinet, so for the net force to be zero, an external force of 200 N should be applied to the cabinet in the opposite direction.

Thus, the frictional force = 200 N [frictional force always acts in the direction opposite to the direction of motion]

Question 11. According to the third law of motion when we push an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Answer:

1. The logic given by the student is not correct because two equal and opposite forces cancel each other only in the case that they act on the same body. Action and reaction force always act on two different bodies, so they cannot cancel each other.
2. When a massive truck is pushed, then the truck may not move because the force applied is not sufficient to overcome the force produced by friction acting opposite to the applied force to move the truck.

Question 12. A bullet of mass 10 g traveling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer:

Given, the mass of the bullet,

m =10 g

= \(\frac{10}{1000} \mathrm{~kg}=0.01 \mathrm{~kg}\)

Initial velocity, u= 150m/s

Final velocity, v = 0 [since bullet comes to rest]

Time, t = 0.03 s

From Newton’s first law of motion, v = u + at 0 =150+ a x 0.03

a=\(\frac{-150}{0.03}=-5000 \mathrm{~m} / \mathrm{s}^2\)

Distance covered by the bullet before coming to rest is given by \(v^2=u^2+2 a s\)

0 =\((150)^2+2(-5000) \mathrm{s}\)

s \(\frac{(150)^2}{10000}=2.25 \mathrm{~m}\)

The magnitude of the force applied by the bullet on the block,

F = ma= 0.01 x -5000 = -50 N

Question 13. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer:

Given, Mass of the object, m = 100 kg Initial velocity, u = 5 m/s Final velocity, v = 8 m/s,

Time, t = 6 s

(1) Initial momentum = mu

= 100 x 5 = 500 kg – m / s and final momentum = mv

= 100 x 8 = 800 kg – m / s

(2) From Newton’s second law, the force exerted on the object = rate of change of momentum = Change in momentum

=\(\frac{\text { Change in momentum }}{\text { Time }}\)

=\(\frac{\text { Final momentum }- \text { Initial momentum }}{\text { Time }}\)

=\(\frac{800-500}{6}=\frac{300}{6}=50 \mathrm{~N}\)

Question 14. Akhtar, Kiran, and Rahul were riding on a motorcar that was moving with a high velocity on an expressway when an insect hit the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar.

Akhtar said that, since the motorcar was moving with a larger velocity, it exerted a larger force on the insect and as a result insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and change in their momentum. Comment on these suggestions.

Answer:

1. According to the law of conservation of momentum, the momentum of the car and insect system before collision = momentum of the car and insect system after collision
2. Hence, the change in momentum of the car and insect system is zero. The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes by a great amount.
3. On the other hand, the car continues moving with a constant velocity. Kiran’s suggestion is wrong.
4. Akhtar made a correct conclusion because the mass of the car is very large compared to the mass of the insect. Rahul gave a correct explanation.

Question 15.  How much momentum will a dumbbell of mass 10 kg transfer to the floor, if it falls from a height of 80 cm? Take, its downward acceleration to be 10 m/s².

Answer:

Given the mass of the dumb-bell, m = 10 kg

Initial velocity, u = 0 [because it falls from rest]

Distance covered, s = 80 cm = 0.8 m

Acceleration, a = 10 m/s²

From Newton’s third law of motion,

v² =u² + 2 as

v²= 0 + 2 x 10 x  0.8 =16 ⇒ = -v 16 = 4 m/s

The momentum of the dumbbell just before it touches the floor is given by p = mv = 10 x 4 = 40 kg-m/s When the dumbbell touches the floor, then its velocity becomes zero and hence the momentum. Thus, the total momentum of the dumbbell is transferred to the floor.

So, the momentum transferred to the floor is 40 kg- m/s

Question 16. The following is the distance-time table of an object in motion:

1. What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero?
2. What do you infer about the force acting on the object?

Answer:

(1) Here, initial velocity, u=0

Using Newton’s second law of motion,

s=u t+\(\frac{1}{2} a t^2=\frac{1}{2} a t^2 [ u=0]\)

We get, a=\(\frac{2 s}{t^2}\)

Thus, acceleration is increasing.

(2) Since, acceleration is increasing, so net unbalanced force is acting on the object.

Question 17.  Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer: 

Given, the mass of the motorcar, m =1200 kg

Acceleration produced, a = 0.2 m/s²

Force applied on the car by three persons,

F = ma = 1200 x 0.2 = 240 N

Force applied on the car by one person = \(\frac{240}{3}\) = 80 N

Each person pushes the motorcar with a force of 80 N.

Question 18. A hammer of mass 500 g moving at 50 m/s strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer:

Given, Mass of the hammer = 300 g= 0.5 kg

The initial velocity of the hammer, u = 50 m/s

The final velocity of the hammer, v = 0 [because the hammer stops]

Time, t = 0.01 s

According to Newton’s second law of motion, the force of the nail on the hammer = rate of change of momentum of the hammer

= \(\frac{m v-m u}{t}=\frac{0.5 \times 0-0.5 \times 50}{0.01}\)

= \(-\frac{25}{0.01}=-2500 \mathrm{~N}\)

The force of the nail on the hammer is equal and opposite to that of the hammer on the nail.

Question 19. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. its velocity is slowed down to 18 km/h in 4 s by an unbalanced: external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required

Answer:

Given, mass, m =1200 kg

Initial velocity, u=90 \(\mathrm{~km} / \mathrm{h}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)

Final velocity, v=18 \(\mathrm{~km} / \mathrm{h}=18 \times \frac{5}{18}=5 \mathrm{~m} / \mathrm{s}\)

Time, t=4 s

(1) Acceleration, a=\(\frac{v-u}{t}=\frac{5-25}{4}\)

=-\(\frac{20}{4}=-5 \mathrm{~m} / \mathrm{s}^2\) [here, the negative sign indicates that the velocity decreases]

(2) Change in momentum

= Final momentum – Initial momentum

= m v-m u=m(v-u)

= 1200(5-25)

= 1200 \(\times(-20)=-24000 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)

(3)  Magnitude of the force required

= Rate of change of momentum

=\(\frac{\text { Change in momentum }}{\text { Time }}\)

=\(\frac{-24000}{4}=-6000 \mathrm{~N}\)

Summary

Any action which causes pull, hit, or push on a body is called force. Types of forces

1. Balanced forces When the net effect produced by many forces acting on a body is zero, then the forces are said to be balanced forces.
2. Unbalanced forces When the net effect produced by many forces acting on a body is non-zero, then the forces are said to be unbalanced forces.
3. Newton’s first law of motion States that an object will continue to remain in its state of rest or in a uniform motion along a straight line or path unless an external force acts on it.
4. The unwillingness (or inability) of an object to change its state of rest or of uniform motion along a straight line is called the inertia of the object.

Types of inertia

1. The inertia of Rest The tendency of a body to oppose any change in its state of rest is known as the inertia of rest.
2. The inertia of Motion The tendency of a body to oppose any change in its state of uniform motion is known as the inertia of motion.
3.  The inertia of Direction The tendency of a body to oppose any change in its direction of motion is known as the inertia of direction

Momentum Measures the quantity of motion possessed by a body. It is defined as the product of the mass and velocity of the body.

Momentum, p = mv

The SI unit of momentum is kg-m/s.

Newton’s second law of motion States that the rate of change of momentum of an object is proportional to the applied external force and takes place in the direction in which external force acts.

Force applied on an object is equal to the product of the mass of the object and acceleration produced in it. Force, F = ma

The SI unit of force is Newton (N).

Impulse is defined as the product of force and the small time in which the force acts.

Impulse, I = Ft – p₂ – p₁ or Impulse = Change in momentum The SI unit of impulse is N-s or kg-m/s.

Newton’s third law of motion States that whenever one object exerts a force on another object, then the second object exerts an equal and opposite force on the first object.

Class 9 Science Notes For Chapter 9 Gravitation

It has been observed that an object dropped from a height falls towards the earth. Newton generalised this idea and said that not only the earth but every object in the universe attracts every other object.

This force of attraction between two objects is called the force of gravitation or gravitational force. In this chapter, we shall learn about gravitation and the universal law of gravitation.

Gravitation

Gravitation is defined as the force of attraction between any two bodies in the universe. The earth attracts (or pulls) all objects lying on or near its surface towards its centre. The force with which the earth pulls the objects towards its centre is called the gravitational force of the earth or gravity of the earth.

Universal Law of Gravitation

The universal law of gravitation was given by Isaac Newton. According to this law, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The direction of the force is along the line joining the centres of two objects. Consider two bodies A and B having masses m₁ and m_2, whose centres are at a distance d from each other.

Read and Learn  More Class 9 Science Notes

The gravitational force between two objects is directed along the line joining their centres Then, the force between two bodies is directly proportional to the product of their masses,

⇒ \(F \propto m_1 m_2\)  →  Equation 1

and the force between two bodies is inversely proportional to the square of the distance between them, i.e.

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

F \(\propto \frac{m_1 m_2}{d^2}\) or F=G\( \frac{m_1 m_2}{d^2}\)

where, G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\) is called the universal

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

where G= 6.67 x 10_11N-m₂/kg² is called the universal gravitational constant.

Its value does not depend on the medium between the two bodies and the masses of the bodies or the distance between them. Suppose the masses of two bodies are 1 kg each and the distance d between them is 1 m, then

F = G [m₁ = m₂ = 1 kg and d = m]

Hence, the universal gravitational constant is defined as the gravitational force between two objects of unit masses separated by a unit distance from each other placed anywhere in space. The SI unit of G is N-m₂ kg. The value of G was found out by Henry Cavendish (173T1810) by using a sensitive balance.

Importance of Universal Law of Gravitation

1. The universal law of gravitation successfully explained several phenomena given below:
2. The force that binds us to the earth.
3. The motion of the moon around the earth.
4. The motion of planets around the sun.
5. The occurrence of tides is due to the gravitational force of attraction of the moon.
6. The flow of water in rivers is also due to the gravitational force of the earth on water.

Motion of Moon around Earth and Centripetal Force

The force that keeps a body moving along the circular path acting towards the centre is called centripetal (centre-seeking) force. The motion of the moon around the Earth is due to the centripetal force. The centripetal force is provided by the gravitational force of attraction of the earth. If there were no such force, then the moon would pursue a uniform straight-line motion.

Example 1. Find the gravitational force between the earth and an object of 2 kg mass placed on its surface. (Givenr mass of the earth = 6 x \(10^24\) kg and radius of the earth = 6.4 x \(10^6\) m)

Answer:

Given, mass of the earth, \(m_e=6 \times 10^{24} \mathrm{~kg}\)

Mass of an object, \(m_o=2 \mathrm{~kg}\)

Radius of the earth, R=6.4 \(\times 10^6 \mathrm{~m}\)

Gravitational force, F=?

F=G \(\frac{m_e m_o}{d^2}\) [by universal law of gravitation]

F=\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2}{\left(6.4 \times 10^6\right)^2}=19.5 \mathrm{~N}\)

Example 2. The mass of Mars is 639 x 10²³ kg and that of Jupiter is 189 x 10²⁷ kg. If the distance between Mars and Jupiter is 749 x 1²³ m. Calculate the force exerted by the Jupiter on the mars. (G =6.7 x 10~n Nm² kg²)

Answer:

Given, the mass of the mars \(M_m=639 \times 10^{23} \mathrm{~kg}\)

The mass of the Jupiter \(M_j=1.89 \times 10^{27} \mathrm{~kg}\)

The distance between Mars and Jupiter

d=7.49 \(\times 10^5 \mathrm{~m}\)

G=6.7 \(\times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\)

The force exerted by Jupiter on Mars

F =G \(\frac{M_m \times M_j}{d^2}\)

= \(\frac{6.7 \times 10^{-11} \times 639 \times 10^{23} \times 1.89 \times 10^{27}}{\left(7.49 \times 10^5\right)^2}\)

=1.44 \(\times 10^{29} \mathrm{~N}\)

Thus, the force exerted by the Jupiter on the mars is 1.44 \(\times 10^{29} N\).

Free Fall

When objects fall towards the earth under the influence of the earth’s gravitational force alone, then these are called freely falling objects and such a motion is called free fall.

Acceleration due to Gravity [g]

Whenever an object falls towards the earth, acceleration is involved. This acceleration is due to the earth’s gravitational pull and is called acceleration due to gravity. It is denoted by g.

The SI unit of is the same as that of acceleration, i.e. m/s². Let the mass of the earth be M and an object falling freely towards it be m. The distance between the centres of the earth and the object is R.

From Newton’s law of gravitation,

Also, from the second law of motion, the force exerted on an object, F =ma

Since, a = g (i.e. acceleration due to gravity)

F = mg Equation  (2)

Equating RHS of Equation. (1) and (2), we get GMm

m g=\(\frac{G M m}{R^2}\) or g=\(\frac{G M}{R^2}\)

From the formula, it is clear that acceleration due to gravity does not depend on the mass of a falling object. It depends only on the mass of the earth or celestial bodies.

Equations of Motion for Free Fall

The three equations of motion which we have derived earlier are for bodies under uniform acceleration. In the case of motion of bodies under free fall, there is a uniform acceleration, i.e. acceleration due to gravity (g) acting downward.

So, the previous three equations of motion can be applied to the motion of bodies under free fall as follows:

General equations Equations for body under free fall of motion

where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.

In solving numerical problems, we should remember the following points:

1. If an object falls vertically downwards, then acceleration due to gravity is taken as positive, since its velocity increases while falling.
2.  If an object is thrown vertically upwards, then acceleration due to gravity is taken as negative, since its velocity decreases as it moves upward.

Example 3. A car falls off a ledge and drops to the ground in 0.6 seconds. The value of g is 10 m/s2 (for simplifying the calculation).

1.  What is its speed on striking the ground?
2. What is its average speed during the 0.6 s?
3. How high is the ledge from the ground?

Answer:

Initial velocity u = 0

Acceleration due to gravity g=10 \(\mathrm{~m} / \mathrm{s}^2\)

(1) Speed v=u+g t

v=0+10 \(\times\) 0.6

v=6 \(\mathrm{~m} / \mathrm{s}\)

(2) Average speed =\(\frac{u+v}{2}=\left(\frac{0+6}{2}\right)=3.0 \mathrm{~m} / \mathrm{s}\)

(3) Distance travelled h=u t+\(\frac{1}{2} g t^2\)

h=0+\(\frac{1}{2} g t^2\)

h=\(\frac{1}{2} \times 10 \times(0.6)^2=5 \times 0.36=1.80 \mathrm{~m}\)

Example 4. An object is thrown vertically upwards and rises to a height of 13.07 m. Calculate

1. The velocity with, which the object was thrown upwards
2. The Time taken by the object to reach the highest point.

Answer:

Distance travelled, h =13.07 m

Final velocity v = 0

Acceleration due to gravity g=-9.8 \(\mathrm{~m} / \mathrm{s}^2\) (upward motion)

(1) \(v^2=u^2+2 g h\)

0=\(u^2+2 \times(-9.8) \times 13.07\)

⇒ \(u^2=256 \Rightarrow u=16 \mathrm{~m} / \mathrm{s}\)

(2) v=u+a t

0=16-9.8 \(\times t \Rightarrow t=1.63 \mathrm{~s}\)

Example 5. A ball is thrown vertically upwards with a velocity of 25 m/s. If g is 10 m/s², then calculate

1. the height it reaches.
2. time taken to return back.

Answer:

Given, initial velocity, u = 25 m/s, final velocity, v =0

If a body is thrown upwards, then its velocity becomes zero at the highest point, where it reaches, acceleration due to gravity,

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

(1) Height, h =\(\frac{v^2-u^2}{2 g}\)

=\(\frac{0-(25)^2}{2(-10)}=\frac{-625}{-20}=31.25 \mathrm{~m}\)

Height, h=\(\frac{v^2-u^2}{2 g}\)

(2) Time, t=\(\frac{v-u}{g}=\frac{0-25}{-10}=2.5 \mathrm{~s}\)

Time taken to return back,

T = time of ascent + time of descent =21 Time taken to return back, T = 2×2.5 = 5s

Weight

The weight of an object is the force with which it is attracted towards the earth.

Weight of an object, w = mg

where, m = mass, g = acceleration due to gravity or w=\(\frac{G M m}{R^2}\)

Here, M = mass of the earth and R = radius of the earth

Important points regarding weight are as follows:

1. Weight is a vector quantity, it acts in a vertically downward direction, and its SI unit is newton (N). Weight of 1 kg mass is 9.8 N. (i.e. 1 kg-wt =9.8 N)
2. The weight of an object is not constant, it changes from place to place.
3.  In the space, where g = 0, the weight of an object is zero.
4.  At the centre of the earth, weight becomes zero. This is due to the fact that on going down to the earth, the value of g decreases and at the centre of the earth, g = 0.

Weight of an Object on the Moon

Let the mass of an object be m and its weight on the moon be wm. Suppose the mass of the moon is Mm and its radius is Rm. According to the universal law of gravitation, the weight of an object on the moon will be

⇒ \(w_m=G \frac{M_m \times m}{R_m^2}\)

Let the weight of the same object on the earth be we. Let the mass of the earth be Me and the radius of the earth be Re.

⇒ \(\frac{w_m}{w_e} =\frac{\frac{G M_m m}{R_m^2}}{\frac{G M_e m}{R_e^2}}=\frac{M_m}{M_e} \times \frac{R_e^2}{R_m^2}\)

Now, \(M_m =5.98 \times 10^{24} \mathrm{~kg} ; M_e=736 \times 10^{22} \mathrm{~kg}\)

⇒ \(R_m =1.74 \times 10^6 ; R_e=6.37 \times 10^6 \)

⇒ \(\frac{w_m}{w_e} =\frac{5.98 \times 10^{24}}{736 \times 10^{22}} \times \frac{\left(637 \times 10^6\right)^2}{\left(1.74 \times 10^6\right)^2} \approx \frac{1}{6}\)

Thus, the weight of an object on the moon is one-sixth of its weight on the Earth.

Example 6. The mass of an object is 12 kg. Calculate

1. Its weight on the earth.
2. Its weight on the moon.

Answer:

Given, m= 12 kg

(1) Acceleration due to gravity on earth, ge =9.8 m/s² weight earth we = mg =12 x 9.8 = 117.6 N

(2) Acceleration due to gravity on the moon \(g_m=\frac{g_e}{6}\)

⇒ \(g_m=\frac{9.8}{6} \mathrm{~m} / \mathrm{s}^2\)

Weight on moon \(w_m=m g_m=12 \times \frac{9.8}{6}=9.8 \times 2=19.6 \mathrm{~N}\)

Example 7. A man weighs 600 N on the earth. What is his mass, if g is 10 m/s²? On the moon, his weight would be 100 N. What is the acceleration due to gravity on the moon?

Answer:

Given, the weight of a man on the earth, \(w_e=600 \mathrm{~N}\)

Acceleration due to gravity on the earth, \(g_e=10 \mathrm{~m} / \mathrm{s}^2\)

Weight of man on the moon, \(w_m=100 \mathrm{~N}\)

Acceleration due to gravity on the moon, \(g_m\)=?

As we know, mass of the man, m=\(\frac{w_e}{g_e}=\frac{600}{10}=60 \mathrm{~kg}\)

⇒ \(g_e=\frac{w_e}{m}\)

Similarly, for the moon \(g_m=\frac{w_m}{m}\)

⇒ \(g_m=\frac{100}{60}=1.66 \mathrm{~m} / \mathrm{s}^2\)

Thus, acceleration due to gravity on the moon is 1.66 \(\mathrm{~m} / \mathrm{s}^2\), i.e. \(g_m=\frac{g_e}{6}\).

Example 8. A particle weighs 120 N on the surface of the earth. At what height above the earth’s surface will its weight be 30 N? Radius of the earth = 6400 km.

Answer:

The weight of a particle on the surface of the earth is

w=m g=\(\frac{m M G}{R^2} [g=\frac{G M}{R^2}]\)

Let \(w_1\) be the weight of a particle at height h above the earth’s surface.

So, \(\frac{w}{w_1} =\frac{G \frac{M}{R^2}}{G \frac{M}{(R+h)^2}}=\frac{(R+h)^2}{R^2}\)

⇒ \(\frac{120}{30} =\left(\frac{R+h}{R}\right)^2\)

4=\(\left(\frac{R+h}{R}\right)^2 \Rightarrow \quad 2=\frac{R+h}{R}\)

2 R=R+h \(\Rightarrow\) R=h

Height of the particle, h= Radius of the earth,

b=6400 km.

Thrust and Pressure

1. Thrust is the force acting on an object perpendicular to its surface. The effect of thrust depends on the area on which it acts.
2. The unit of thrust is the same as that of force, i.e. the SI unit of thrust is Newton (N). It is a vector quantity.
3. Pressure is the force acting perpendicularly on a unit area of an object.
4. Pressure(p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
5. The SI unit of pressure is Nm -2, which is also called Pascal (Pa) named after the scientist Blaise Pascal. It is a scalar quantity.
6. \(1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}\)
7. From the formula of pressure, it is clear that the same force can produce different pressures depending on the area over which it acts. A force acting on a smaller area exerts a large pressure while the same force acting on a larger area exerts small pressure.

Example 9. A Force of 200 N is applied to an object of an area of 4 m₂. Find the pressure.

Answer:

Given, force, F=200 \(\mathrm{~N}\), area, A=4 \(\mathrm{~m}^2\)

Now, Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{200}{4}=50 \mathrm{Nm}^{-2}\)

Example 10. A woman is wearing sharp-heeled sandals (stilettos). If the mass of a woman is 60 kg and the area of one heel is 1 cm², find out the pressure exerted on the ground, when the woman stands on just one heel. (Given, g = 10ms^-2)

Answer:

In the given case, the force will be the weight of the woman which is given by m x g [where m is the mass of the woman and g is the acceleration due to gravity].

Force, F = mx g [weight of woman]

= 60 x 10 N [given, m = 60 kg, A = 1 cm²]

= 600 N

And area, A=1 \(\mathrm{~cm}^2=\frac{1}{10000} \mathrm{~m}^2\)

Pressure, p=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{600 \times 10000}{1}\)

=6000000 \(\mathrm{Nm}^{-2}\) (or } 6000000Pa

Thus, the pressure exerted by the woman (of 60 kg) standing on only one heel of area 1 cm² is 6000000 Nm-2.

Example 11 A block of wood is kept on a tabletop. The mass of the wooden block is 6 kg and its dimensions are 50 cm x 30 cm x 20 cm.

Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its sides of dimensions

1. 30 cm x 20 cm
2. 50 cm x 30 cm

Answer:

The mass of the wooden block = 6 kg

The dimensions = 50 cm x 30 cm x 20 cm Thrust F = mg = 6 x 9.8 = 58.8 N

(1) Area of a side = Length x Breadth

=30 \(\times 20=600 \mathrm{~cm}^2=0.06 \mathrm{~m}^2\)

Pressure \(p_1 =\frac{F}{A}=\frac{58.8}{0.06}\)

=980 \(\mathrm{~N} / \mathrm{m}^2\)

(2) When the block lies on its side of dimensions 50 \(\mathrm{~cm} \times 30 \mathrm{~cm}\), it exerts the same thrust

Area = Length x Breadth

=50 \(\times\) 30=1500 \(\mathrm{~cm}^2=0.15 \mathrm{~m}^2\)

Pressure \(p_2=\frac{F}{A}=\frac{58.8}{0.15}=392 \mathrm{~N} / \mathrm{m}^2\)

Some Daily Life Applications of Pressure

1. The handles of bags, suitcases, etc., are made broad so that less pressure is exerted on the hand.
2. Buildings are provided with broad foundations so that the pressure exerted on the ground becomes less.
3. Railway tracks are laid on cement or iron sleepers so that the pressure exerted by the train could spread over a larger area and thus pressure decreases.
4. Pins, needles and nails are provided with sharp pointed ends to reduce the area and hence to increase the pressure.
5. Cutting tools have sharp edges to reduce the area so that with lesser force, more pressure can be exerted.
6. Pressure on the ground is more when a man is walking than when he is standing because, in the case of walking, the effective area is less.
7. Depression is much more common when a man stands on the cushion than when he lies down on it because in the standing case, the area is lesser than in the case of lying.
8. The tractors have broad tyres, to create less pressure on the ground so that tyres do not sink into comparatively soft ground in the field.

Pressure in Fluids

All liquids and gases are together called fluids. Water and air are the two most common fluids. Solids exert pressure on a surface due to their weight.

Fluids also have weight, therefore fluids also exert pressure on the base and walls of the container in which they are enclosed. Fluids exert pressure in all directions.

Buoyancy

1. The tendency of a liquid to exert an upward force on an object immersed in it is called buoyancy. Gases also exhibit this property of buoyancy.
2. Buoyant Force is an upward force which acts on an object when it is immersed in a liquid. It is also called upthrust.
3. It is the buoyant force due to which a heavy object seems to be lighter in water. As we lower an object into a liquid, the liquid underneath it provides an upward force.
4. For Example, A piece of cork is held below the surface of water. When we apply a force by our thumb, the cork immediately rises to the surface. This is due to the fact that every liquid exerts an upward force on the objects immersed in it.

Factors Affecting Buoyant Force

The magnitude of buoyant force depends on the following factors:

(1) Density of the Fluid

The liquid having higher density exerts more upward buoyant force on an object than another liquid of lower density. This is the reason why it is easier to swim in seawater in comparison to normal water.

The Sea water has higher density and hence exerts a greater buoyant force on the swimmer than the freshwater having lower density.

(2) Volume of Object Immersed in the Liquid

As the volume of a solid object immersed inside the liquid increases, the upward buoyant force also increases. The magnitude of buoyant force acting on a solid object does not. depend on the nature of the solid object. It depends only on its volume.

For Example, When two balls made of different metals having different weights but equal volume are fully immersed in a liquid, they will experience an equal upward buoyant force as both the balls displace an equal amount of the liquid due to their equal volumes.

Floating or Sinking of Objects in Liquid

When an object is immersed in a liquid, the following two forces act on it:

Weight of the object which acts in a downward direction, i.e. it tends to pull down the object.

Buoyant force (upthrust) which acts in an upward direction, i.e. it tends to push up the object.

Whether an object will float or sink in a liquid, depends on the relative magnitudes of these two forces which act on the object in opposite directions. There are three conditions of floating and sinking of objects. These are:

1. If the buoyant force or upthrust exerted by the liquid is less than the weight of the object, the object will sink into the liquid.
2.  If the buoyant force is equal to the weight of the object, the object will float in the liquid.
3. If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.

Density

1. The density of a substance is defined as mass per unit volume.
2. Density =\(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\) or \(\rho=\frac{m}{V}\)
3. The SI unit of density is kilogram per metre cube (kg/m³). It is a scalar quantity. The density of a substance under specified conditions always remains the same.
4. Hence, the density of a substance is one of its characteristic properties. It can help us to determine its purity. It is different for different substances. The lightness and the heaviness of different substances can be described by using the word density.
5. Objects having a density less than that of a liquid, float on the liquid. Objects having greater density than that of liquid, sink in the liquid. It decreases with an increase in temperature.

Example 12. A sealed can of mass 700 g has a volume of 500 cm³. Will this can sink in water? (The density of water is 1 g cm^-3)

Answer:

Given, the mass of the can, m=700 g

Volume of can, V=500 \(\mathrm{~cm}^3\)

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

=\(\frac{700}{500}=1.4 \mathrm{~g} \mathrm{~cm}^{-3}\)

Since the density of the can is greater than the density of the water, the can will sink into the water.

Archimedes’ Principle

The statement ‘When an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust, which is equal to the weight of the liquid displaced by the object’, i.e.

Buoyant force or upthrust acting on an object = Weight of liquid displaced by the object

Even gases like air, exert an upward force or buoyant force on the objects placed in them. It is buoyant force or upthrust due to displaced air which makes a balloon rise in air.

Applications of Archimedes’ Principle

1. Archimedes’ principle is used in:
2. Designing ships and submarines.
3. lactometer (a device used to determine the purity of milk). hydrometer (a device used for determining the density of liquid).

Example 13 If an iron object is immersed in water, it displaces 8 kg of water. How much is the buoyant force acting on the iron object in Newton? (Given, g = 10 ms^-2)

Answer:

According to Archimedes’ principle, “the buoyant force acting on this iron object will be equal to the weight of water displaced by this iron object.”

We know that, weight, w = mx g. Here, the mass of water, m= 8kg

Acceleration due to gravity, g =10 ms^-2

On putting values, w = 8 x 10 = 80 N

Since the weight of water displaced by the iron object is 80 N, therefore the buoyant force acting on the iron object (due to water) will also be 80 N.

Activity Zone

Activity 1

Objective: To describe a circular path in the motion of a stone.

Materials Required:

Thread and stone.

Procedure

1. A stone describing a circular path with a velocity of constant magnitude
2. Take a piece of thread.
3. Tie a small stone at one end and hold the other end of the thread. Whirl it as shown in the figure.
4.  Observe the motion of the stone.
5. Release the thread.
6. Again note the direction of motion of the stone.

Observation

1. The stone moves in a circular path with a certain speed and changes direction at every point.
2. The change in direction involves a change in velocity or acceleration.
3. The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre.
4. This force is called centripetal force. When the thread is released, the stone makes a tangent to the circle and falls down.

Question 1. Name the motion of the stone.
Answer:

The motion of the stone is along a circular path.

Question 2. What happens to the stone if centripetal force vanishes?
Answer:

Due to the absence of centripetal force, the stone flies off along a tangent of a circle.

Question 3. Name the straight line that meets the circle at one point.
Answer:

The tangent is a straight line that meets the circle at one point.

Activity 2

Objective: To study free-falling or rising bodies.

Material Required: Stone

Procedure

1. Take a stone.

2. Throw it upwards.

3. It reaches a certain height and then it starts falling down.

Observations

1. The stone is attracted by the earth hence it falls down due to the earth’s gravitational force.
2. The stone gets accelerated, when it goes upwards, g is negative when it falls towards the earth, g is positive.

Question 1. Which force accelerates a body in free fall?
Answer:

The force of gravity on earth accelerates a body in free fall.

Question 2. What is the direction of motion of an object if g is taken negative?
Answer:

The direction of motion of an object is upwards if g is taken negative.

Question 3. What is the direction of acceleration due to gravity?
Answer:

The direction of acceleration due to gravity is always downward.

Activity 3

Objective: To study any object that may be hollow or solid, big or small should fall at the same rate.

Materials Required

1. Glass jar
2. One sheet of paper

Procedure

1. Take a sheet of paper and a stone.
2. Drop them simultaneously from the first floor of a building.
3. Observe whether both of them reach the ground simultaneously.
4. Also, repeat in a glass jar and then observe what happens.

Observation

1. The paper reaches the ground a little later than the stone. This is due to air resistance. The air offers resistance due to friction to the motion of the falling objects.
2. The resistance offered by air to the paper is more than the resistance offered to the stone.
3. In a glass jar, the air is sucked out, the paper and the stone falls at the same rate. When the earth attracts the paper sheet and crumbled paper, they experience the same acceleration during free fall.
4. This acceleration experienced by an object is independent of its mass. Thus, any object whether it be hollow or solid, big or small should fall at the same rate.

Question 1. Which one reaches the ground first?
Answer:

The stone reaches the ground first.

Question 2. The paper reaches the ground later, why?
Answer:

The paper reaches the ground later due to air resistance.

Question 3. What happens when air is sucked in along a glass jar?
Answer:

When air is sucked into a long glass jar both stone and paper reach at the bottom at the same time.

Activity 4

Objective: To understand buoyant force.

Materials Required: Plastic bottle, stopper, bucket and water.

Procedure

1. Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You will see that the bottle will float.
2. Push the bottle into the water. You will feel an upward push. Try to push it further down.
3. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bride is pushed deeper till it is completely immersed.
4. Now, release the bottle. It bounces back to the surface.

Observation

1. The force due to the gravitational attraction of the Earth acts on this bottle in a downward direction. But the water exerts an upward force on the bottle. This upthrust is known as buoyant force.
2. When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore, it rises up when released.
3. To keep the bottle completely immersed, the upward force on the bottle due to water must be balanced. This can be achieved by an externally applied force acting downwards.
4. This force must at least be equal to the difference between the upward force and the weight of the bottle.

Conclusion

It is the buoyant force which keeps the object float.

Question 1. When a body is fully immersed in a liquid, what happens to its weight?
Answer:

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

Question 2. What are the factors affecting buoyant force?
Answer:

The buoyant force depends on

1.  the density of the liquid
2. the volume of the solid immersed in liquid.

Question 3. What happens, if the density of an object is more than the density of the liquid in which it is immersed?
Answer:

The object will sink into the liquid.

Question 4. Who exerts buoyant force?
Answer:

The buoyant force is exerted by the liquid molecules on the body when it is dipped in that liquid.

Question 5. By which force does a body start floating?
Answer:

Buoyant force helps the body to float in a liquid in which it is immersed.

Activity 5

Objective: To know why objects float or sink when placed on the surface of water.

Materials Required: Two beakers filled with water, an iron nail and a piece of cork.

Procedure

1. Take a beaker filled with water.
2. Take an iron nail.
3. Place it on the surface of the water.
4. Take another beaker filled with water.
5. Take a piece of cork and an iron nail of equal mass.
6. Place them on the surface of the water.

Observation

1. The nail sinks.
2. The cork floats (as shown alongside).

Reasons

1. The force due to the gravitational attraction of the Earth on the iron nail, pulls it downwards. There is an upthrust of water on the nail, which pushes it upwards.
2. However, the downward force acting on the nail is greater than the upthrust of water on the nail. So, it sinks. This happens because of the difference in their densities.
3. The density of cork is less than the density of water. This means that the upthrust of water on the cork is greater than the weight of the cork. So, it floats.
4. The density of an iron nail is more than the density of water. This means that the upthrust of water on the iron nail is less than the weight of the nail. So, it sinks.

Conclusion

The object of density less than that of a liquid floats on the surface of the liquid. Whereas, the object of density greater than that of a liquid sinks in the liquid.

Question 1. What happens to buoyant force, when we increase the density of the fluid?
Answer:

The buoyant force increases with an increase in the density of the fluid.

Question 2. What is the condition for floating of a body?

Answer:

If the density of a body is less than that of the liquid in which it is immersed, then the body floats in liquid.

Question 3. Give the condition for the sinking of a body.

Answer:

If the density of the body is greater than that of the liquid, it sinks into the liquid.

Question 4. If an object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of its volume
Answer:

1. Outside the surface of the liquid of densities \(d_1, d_2\) and \(d_3\). Then, arrange the densities in increasing order.
2. Upthrust due to liquid on an object is directly proportional to the density of the liquid. Therefore, densities in increasing order are of \(d_1<d_2<d_3\)

Question 5. A nail sinks while a cork floats. Why?
Answer:

The density of the nail is greater than that of liquid, so it sinks. But the density of cork is less than that of water, so it floats.

Gravitation Question And Answer

Question 1. State the universal law of gravitation.
Answer:

The universal law of gravitation states that the force of attraction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law is applicable to any two objects anywhere in the universe.

Question 2. Write the formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:

The formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth is given by  F=G \(\frac{M m}{R^2}\)

where, M = mass of the earth

m = mass of an object

G = gravitational constant

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

and R = distance between the centres of the earth and an object.

Question 3. What do you mean by free fall?
Answer:

The falling of a body from a height towards the earth under the gravitational force of the earth is called free fall. Hence, the motion of a particle falling down or going up under the action of gravity means the body is in free fall.

Question 4. What do you mean by acceleration due to gravity?
Answer:

The acceleration of a body during free fall towards a celestial body is called acceleration due to gravity. Its value is 9.8 m/s².

For objects on or near the surface of the earth,

m g=G \(\frac{M \times m}{R^2}\)

where, g = acceleration due to gravity

M = mass of the earth

m = mass of an object

and R = radius of the earth

Hence, g=\(\frac{G M}{R^2}\)

Question 5. What are the differences between the mass of an object and its weight?
Answer:

Differences between the mass of an object and its weight are as follows:

Question 6. Why is the weight of an object on the moon 1/6 its weight on the earth?
Answer:

Weight of an object, w = mg [where, m = mass of an object and g — acceleration due to gravity]

The mass of an object m remains constant at all places. Acceleration due to gravity changes from place to place. So, we can say that the weight of an object depends on the acceleration due to gravity.

On the moon, the acceleration due to gravity is l/6th that of the earth, this is the reason why the weight of an object on the moon is 1 /6th its weight on the earth.

Question 7. Why is it difficult to hold a school bag having a strap made of thin and strong string?
Answer:

It is difficult to hold a school bag having a strap made of a thin and strong string because the area under the strap is small. Hence, a large pressure is exerted by the strap on the fingers or shoulders. Due to the large pressure, the strap tends to cut the skin and hence pain is caused.

Question 8. What do you mean by buoyancy?
Answer:

The upward force exerted by a liquid on any object immersed in it is called buoyancy or upthrust.

Question 9. Why does an object float or sink when placed on the surface of water?
Answer:

When an object is placed on the surface of water, two forces act on the object.

1. The weight of the object acts vertically downwards.
2. The upthrust of the water acts vertically upwards.

The object will float on the surface of the water if the upthrust is greater than the weight of the object. The object will sink if the weight of the object is more than the upthrust of the water.

Question 10. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:

Mass is more than 42 kg. As the buoyant force due to air is acting on us, the reading of the weighing machine will be less than the actual mass of a person.

Question 11. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer:

The cotton bag is heavier as the buoyant force acting on the cotton bag is more as its surface area is more than that of the iron piece.

Exercises

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:

The force of gravitation between two objects is given by

F=G \(\frac{m_1 m_2}{r^2}\)

If the distance is reduced to half, i.e. \(r^{\prime}=r / 2\). Then, a new force of gravitation,

⇒ \(F^{\prime}=\frac{G m_1 m_2}{r^{\prime 2}}=\frac{G m_1 m_2}{(r / 2)^2}=4 \times \frac{G m_1 m_2}{r^2}=4 F\)

i.e. The force of gravitation becomes 4 times the original value.

Question 2. Gravitational force acts on all objects in proportion- to their masses. Why, then a heavy object does not fall faster than a light object?
Answer:

Acceleration due to gravity (g) is independent of the mass of the falling object and is equal for all objects at a point. So, a heavy object falls with the same acceleration as a light object.

Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Take, the mass of the earth is 6 \(\times 10^{24}\) kg and the radius of the earth is 6.4 \(\times 10^6 \mathrm{~m}\).)
Answer:

The gravitational force between the earth and an object is given by F=\(\frac{G M m}{R^2}\).

where, G = gravitational constant

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

M = mass of the earth =6 \(\times 10^{24} \mathrm{~kg}\)

R = radius of the earth =\(6.4 \times 10^6 \mathrm{~m}\)

and m = mass of an object =1 kg

F =\(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2}\)

=\(9.77 \approx 9.8 \mathrm{~N}\)

Thus, the magnitude of the gravitational force between the earth and a 1 kg object is 9.8 N.

Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater than smaller than or equal to the force with which the moon attracts the earth? Why?
Answer:

When two objects attract each other, then the gravitational force of attraction applied by the first object on the second object is the same as the force applied by the second object on the first object. So, both the earth and moon attract each other by the same gravitational force of attraction.

Question 5. If the moon attracts the earth, then why does the earth not move towards the moon?
Answer:

The earth does not move towards the moon in spite of the attraction by the moon because the mass of the earth is much greater than the mass of the moon and for a given force, acceleration is inversely proportional to the mass of the object.

Question 6. What happens to the force between two objects, if

1. the mass of one object is doubled?
2. the distance between the objects is doubled and tripled?
3. the masses of both objects are doubled?

Answer:

The force of attraction between two objects is given by

F=\(\frac{G m_1 m_2}{r^2}\)

where, \(m_1\) and \(m_2\) = masses of the objects

r= distance between the objects

and

G= gravitational constant

(1) If the mass of one object is doubled, then the new force,

=2 \(\times \frac{G m_1 m_2}{r^2}\)=2 F  i.e. Force becomes double.

(2) If the distance between the objects is doubled, then the new force,

⇒ \(F^{\prime}=\frac{G m_1 m_2}{(2 r)^2}=\frac{G m_1 m_2}{4 r^2}=\frac{1}{4} \cdot \frac{G m_1 m_2}{r^2}\)

=\(\frac{F}{4}\)

i.e. Force becomes one-fourth. If the distance between the objects is tripled, then new force,

⇒ \(F^{\prime} =\frac{G m_1 m_2}{(3 r)^2}=\frac{G m_1 m_2}{9 r^2}\)

= \(\frac{1}{9}\left(\frac{G m_1 m_2}{r^2}\right)=\frac{F}{9}\)

i.e. Force becomes one-ninth.

(3) If the masses of both objects are doubled, then the new force,

⇒ \(F^{\prime} =\frac{G\left(2 m_1\right)\left(2 m_2\right)}{r^2}\)

=4 \(\times \frac{G m_1 m_2}{r^2}=4 F\)

i.e. Force becomes four times.

Question 7. Calculate the force of gravitation between the earth and the sun given that the mass of the earth = 6 \(\times 10^{24}\) kg and of the sun = 2 \(\times 10^{30}\) kg. The average distance between the two is 1.5 \(\times 10^{11}m\).
Answer:

The force of attraction between the Earth and the sun is given by

F=\(\frac{G M_s M_e}{r^2}\)

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

Mass of the sun, \(M_s=2 \times 10^{30} \mathrm{~kg}\)

Mass of the earth, \(M_e=6 \times 10^{24} \mathrm{~kg}\)

Average distance between the Earth and the sun

F =\(\frac{6.67 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2}=1.5 \times 10^{11} \mathrm{~m}\)

=3.6 \(\times 10^{22} \mathrm{~N}\)

Thus, the force between the earth and the sun is 3.6 \(\times 10^{22} \mathrm{~N}\).

Question 8. What is the importance of the universal law of gravitation?
Answer:

1. The universal law of gravitation successfully explained several phenomena given below:
2.  The force that binds us to the earth.
3. The motion of the moon around the earth.
4. The motion of planets around the sun.
5. The tides are due to the moon and the sun.
6. The flow of water in rivers is also due to the gravitational force of the earth on water.

Question 9. What is the acceleration of free fall?
Answer:

The acceleration of free fall is the acceleration produced in the motion of an object when it falls freely towards the Earth. It is also called acceleration due to gravity. Its value on the earth’s surface is 9.8 m/s².

Question 10. What do we call gravitational force between the earth and an object?
Answer:

The gravitational force between the earth and an object is called the force of gravity or gravity.

Question 11. Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why?

[Hint The value of g is greater at the poles than at the equator.]

Answer:

No, his friend will not agree with the weight of gold. w = mg

⇒ \(g \propto \frac{1}{R^2}\)

The value of g is greater at the poles than at the equator. Therefore, gold at the equator weighs less than that at the poles. Thus, Amit’s friend will not agree with the weight of the gold bought.

Question 12. Why does a sheet of paper fall slower than one that is crumpled into a ball?
Answer:

The sheet of paper falls slower than one that is crumpled into a ball because in the first case, the area of the sheet is more, so it experiences a large opposing force due to air. In contrast, the sheet crumpled into a ball experienced less opposing force due to the small area. This opposing force arises due to air resistance or air friction.

Question 13. The gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in Newton of a 10 kg object on the moon and on the earth?
Answer:

Given, the mass of the object, m=10 kg

Weight on the earth, w=m g=10 \(\times 9.8=98 \mathrm{~N}\)

Weight on the moon =\(\frac{1}{6}\) of the weight on the earth

=\(\frac{1}{6} \times\) 98=16.33 N

Question 14. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

1. the maximum height to which it rises.
2. the total time it takes to return to the surface of the earth.

Answer:

Given, initial velocity, u = 49 m/s

(1) At the maximum height velocity becomes zero.

Final velocity, v = 0

⇒ \(v^2 =u^2-2 g h\)

0 =\((49)^2-2 \times 9.8 \times h\)

h =\(\frac{(49)^2}{2 \times 9.8}=122.5\) m

Maximum height attained =122.5 m

(2) Time taken by the ball to reach the maximum height.

From the first equation of motion, v=u-g t or

0=49-9.8 \(\times\) t

t=\(\frac{49}{9.8}\)=5 s

For the motion against gravity, the time of descent is the same as the time of ascent. So, the time taken by the ball to fall from maximum height is 5 s

Total time taken by the ball to return to the surface of the earth =5+5=10 s.

Question 16. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g=10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:

Given, initial velocity, u = 40 m/s

Final velocity becomes zero, i.e. v = 0 [at maximum height]

From the third equation of upward motion,

⇒ \(v^2 =u^2-2 g h\)

⇒ \((0)^2 =(40)^2-2 \times 10 \times h\)

0 =1600-20 h

h =\(\frac{1600}{20}=80 \mathrm{~m}\)

The maximum height reached by the stone = 80 m.

After reaching the maximum height, the stone will fall towards the earth and will reach the earth’s surface covering the same distance.

So, the distance covered by the stone = 80 + 80 = 160 m. Displacement of the stone = 0.

Because the stone starts from the earth’s surface and finally reaches the earth’s surface again, i.e. the initial and final positions of the stone are the same.

Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:

Let after time t both stones meet and s be the distance travelled by the stone dropped from the top of the tower at which the stones will meet.

Distance travelled by the stone dropped = s

Distance travelled by the stone projected upwards = (100-5) m

For the stone dropped from the tower,

s=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2}(10) t^2\)

[ u=0 because the stone is dropped, i.e. it starts from rest] s=5 \(t^2\)

For the stone projected upwards, \(s^{\prime}=u t-\frac{1}{2} g t^2\)

[due to upward motion, negative sign is taken] (100-s) =25 \(t-\frac{1}{2} \times 10 t^2\)

100-s = \(25 t-5 t^2\)

From Eq. (1), we get

⇒ \(100-5 t^2=25 t-5 t^2\)

25 t=100

t=4 s

So, the stones will meet after 4 s.

s=5 \(t^2=5 \times(4)^2=80 \mathrm{~m}\)

So, the stones will be at a distance of 80 m from the top of the tower or 20 m (100 m – 80 m) from the base of the tower.

Question 18. A ball thrown up vertically returns to the thrower after 6 seconds, find

1. the velocity with which it was thrown up,
2. the maximum height it reaches and
3. its position after 4 s.

Answer:

Total time taken =6 s

Time taken to reach the maximum height = \(\frac{6}{2}\) = 3 s [ time of ascent = time of descent]

(1) From the first equation of motion, v = u = gt [negative sign is taken due to upward motion]

0 =U -9.8 x 3 [v at maximum height, v = 0]

⇒ u- 29.4 m/s

(2) From the third equation of motion,

⇒ \(v^2=u^2-2\) g h [negative sign is taken due to upward motion]

0 =\((29.4)^2-2 \times 9.8 \times h\)

h =\(\frac{(29.4)^2}{2 \times 9.8}=44.1 \mathrm{~m}\)

The maximum height attained by the ball is 44.1 m.

(3) In the initial 3 s, the ball will rise, and then in the next 3 s it falls toward the earth.

The position after 4 s

= Distance covered in Is in the downward motion

From the second equation of motion,

h=u t+\(\frac{1}{2} g t^2=0+\frac{1}{2} \times 9.8 \times(1)^2=4.9 \mathrm{~m}\)

i.e. The ball will be at 4.9 m below the top of the tower or the height of the ball from the ground will be at (44.1-4.9) =39.2

Question 19. In what direction, does the buoyant force on an object immersed in a liquid act?
Answer:

The buoyant force on an object immersed in a liquid always acts in the vertically upward direction.

Question 20. Why does a block of plastic released under water, come up to the surface of water?
Answer:

The upthrust or buoyant force acting on the block of plastic by the water is greater than the weight of the plastic block. So, the plastic block comes up to the surface of the water.

Question 21. The volume of 50 g of a substance is 20 \(\mathrm{~cm}^3\). If the density of water is 1 g \(\mathrm{cm}^{-3}\), will the substance float or sink?
Answer:

Given, the mass of the substance, m = 50 g

Volume of substance, V=20 \(\mathrm{~cm}^3\)

Density of substance,

⇒ \(\rho=\frac{\text { Mass }}{\text { Volume }}=\frac{50}{20}=2.5 \mathrm{~g} \mathrm{~cm}^{-3}\)

i.e. The density of the substance is greater than the density of water, so it will sink in water.

Question 22. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water, if the density of water is 1 g \(\mathrm{~cm}^{-3}\)? What will be the mass of the water displaced by this packet?
Answer:

Given, the mass of the packet =500 g

Volume of packet =350 \(\mathrm{~cm}^3\)

Density of packet, \(\rho=\frac{\text { Mass }}{\text { Volume }}\)

=\(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^3}\)

=1.43 \(\mathrm{~g} \mathrm{~cm}^{-3}\)

i.e. The density of the packet is greater than the density of the water, so it will sink in the water.

The mass of water displaced by the packet

= Volume of packet x Density of water =350 x 1 = 350 g

Summary

• Gravitation is defined as the non-contact force of attraction between any two bodies in the universe.
• According to the universal law of gravitation, the attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
• Mathematically, F = \(\frac{G M m}{d^2}\) where G is called the universal gravitational constant and its value is 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
• Free fall Whenever objects falls towards the earth under the earth’s gravitational force alone, then such a motion is called free fall.
• The acceleration with which an object falls towards the earth due to the earth’s gravitational pull is called acceleration due to gravity. It is denoted by g.
• At the surface of earth, g = \(\frac{G M}{R^2}\)
• The SI unit of g is \(\mathrm{ms}^{-2}\) and its value is 9.8 \(\mathrm{~m} / \mathrm{s}^2\).

Equations of motion for freely falling bodies

v =u+g t

h =\(u t+\frac{1}{2} g t^2\)

⇒ \(v^2 =u^2+2 g h\)

• where h is the height from which the object falls, t is the time of fall, u is the initial velocity and v is the final velocity when the body accelerates at g.
• The total amount of matter contained in an object is called its mass.
• The SI unit of mass is kilogram (kg). Mass is a scalar quantity.
• The weight of an object is the force with which it is attracted towards the earth i.e. weight of an object, w = mg The SI unit of weight is Newton (N) Weight is a vector quantity.
• The weight of an object on the moon is 1 / 6 th of its weight on the earth.
• Thrust is the force, acting on an object perpendicular to its surface.
• The SI unit of thrust is Newton (N). Thrust is a vector quantity.
• Pressure is the force acting perpendicularly on a unit area of an object.
• Pressure can be calculated as, (p)=\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{\text { Thrust }}{\text { Area }}\)
• The SI unit of pressure is \(\mathrm{Nm}^{-2}\) or pascal(Pa).
• Pressure is a scalar quantity.
• All the liquids and gases are called fluids.
• Buoyant force is an upward force, which acts on an object when it is immersed in a liquid. It is also called upthrust. Factors Affecting the Buoyant Force Density of the Fluid.
• The volume of the object immersed in the liquid.
• If the buoyant force exerted by the liquid is less than the weight of the object, the object will sink in the liquid.
• If the buoyant force is equal to the weight of the object, the object will float in the liquid.
• If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.
• The density of a substance is defined as mass per unit volume. Density = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\)
• The SI unit of density is kilogram per metre cube (kg/m3) and it is a scalar quantity.
• According to Archimedes’ principle, when an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust which is equal to the weight of liquid displaced by the object.

Class 9 Science Notes For Chapter 10  Work, Energy, And Power

Work:  Work is said to be done if by applying a force on an object, it is displaced from its position in the direction of force.

Scientific Conception of Work

From the point of view of science, the following two conditions need to be satisfied for work to be done.

1. A force should act on an object.
2. The object must be displaced.

If any one of the above conditions does not exist, work is not done.

For Example. A girl pulls a trolley and the trolley moves through a distance. In this way, she has exerted a force on the trolley and it is displaced. Hence, work is done.

Work Done by a Constant Force

‘Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force.’

Read and Learn  More Class 9 Science Notes

Let us assume if a constant force F acts on an object at point A, due to which the object gets displaced through a distance s in the direction of the force and reaches point B, then the work done (W) by force {F) on that object will be equal to the product of the force and displacement.

Work done = Force x Displacement in the direction of force or W = F x s

SI Unit of Work

1. If F = 1 N and s = l m. then the work clone hy the force will he 1 N-m.
2. The SI unit of work is newton-metre (N-m) which is also called joule (J)
3. Thus 1 J is the amount of work done on an object when a three of 1 N displaces it by 1 m along the line of action of the force.
4. 1 joule = 1 newton x 1 metre ⇒ 1J = lN-m
5. Work is a scalar quantity, it has only magnitude and no direction.

Example 1. A force of 10 N is acting on an object. The object is displaced through 5 m in the direction of force. What is the work done in this case?
Answer:

Given, force, F = 10N, displacement, s=5m

Work done, W = f x J = 10 Nx 5m = 50 N-m or 50J

Positive, Negative, and Zero Work

When the force F and displacement s are in the same direction (the angle between the direction of force and displacement is 0°), work done will be positive, i.e. work is done by the force.

For Example. A boy pulls an object towards himself.

W = + F xs

When the force F and displacement s are in opposite directions (the angle between the direction of force and displacement is 180°), work done will be negative, i.e. work is done against the force.

For Example. The frictional force acts in the direction opposite to the direction of displacement, so work done by friction will be negative. W=-Fxs

When the force and displacement are in the perpendicular direction (the angle between the direction of force and displacement is 90°), the work done is zero.

For Example. A coolie carrying a load on his head. In this case, the gravitational force is acting vertically downward (weight of load) and displacement is along the horizontal direction, i.e. force and displacement are perpendicular to each other. So, in this case, the work done by the gravitational force is zero. W= 0

Example 2. A crane lifts” a crate upwards through a height of 20 m. The lilting force provided by the crane is 5 kN. How much work Is done by the force?

Answer: 

Given, force, F = 5 kN = 5000 N Displacement, s = 20 m

Work done, W =?

We know that work done, W = F s

Here, force and displacement are in the same direction.

So, W = Fs ⇒ W =5000Nx 20m = 100000 J

So, the work done by the force is 100000 J or 100 kj

Energy

1. It is the ability to do work It is always essential for performing any mechanical work. An object having the capability to do work is said to possess energy.
2. The object that does the work, loses energy, and the object on which work is done, gains energy.
3. The energy of an object is measured in terms of its capacity to do work.
4. The SI unit of energy is the same as that of work, i.e. joule (J). 1 joule of energy is required to do 1 J of work. A larger unit of energy is kJ.
5. 1 kilo joule (kJ) =10³J
6. Work done against a force is therefore stored as energy.
7. For Example, When a fast-moving cricket ball hits a stationary wicket, the wicket is thrown away.
8. When a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood.
9. Sun is the biggest natural source of energy for us. We can also get energy from the nuclei of atoms, the interior of the Earth, and the tides in the ocean.

Forms of Energy

Energy exists in various forms mechanical energy (the sum of potential energy and kinetic energy), heat energy, chemical energy, electrical energy, and light energy.

Kinetic Energy

1. The energy which is possessed by an object due to its motion is called kinetic energy.
2. Its SI unit is joule (J). The kinetic energy of a body moving with a certain velocity” is equal to the work done on it to make it acquire that velocity. The kinetic energy of an object increases with its speed.
3. Due to kinetic energy”, a bullet fired from a gun can pierce a target.
4. A moving hammer drives a nail into the wood. Due to its motion, it has kinetic energy or ability to do work.
5. The kinetic energy possessed by an object of mass m, moving with a uniform velocity v is given by,
6. \(\mathrm{KE}\) or \((E_K)=\frac{1}{2} m v^2\)

Calculation of Kinetic Energy

The kinetic energy of an object is measured by the amount of work, it can do before coming to rest. Consider an object of mass m moving with a uniform velocity u.

A force F is applied to it which displaces it through a distance s and it attains a velocity v.

Then, work is done to increase its velocity from u to v. W =Fs

According to the equation of motion,

⇒ \(v^2-u^2 =2 a y\)

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

where a is uniform acceleration, u is initial velocity and v is final velocity,

Also from, F=m a

Substituting the values of F and t in Eq. (1), we have

W = ma \(\cdot \frac{v^2-u^2}{2 a}\) of W=\(\frac{1}{2} m\left(v^2-u^2\right)\)

This is known as the work-energy theorem (i.e. total work is equal to the change in kinetic energy).

If initial velocity,

Then, u = 0

W =\(\frac{1}{2} m v^2\)

This work is equal to the kinetic energy of the object.

⇒ \(\mathrm{KE}\)(or \(E_K)=\frac{1}{2} m v^2\)

Some Important Results can be Derived from the Formula KE=\(\frac{1}{2} m v^2\)

These are given below :

1.  If the mass of an object is doubled, its kinetic energy also gets doubled.
2. If the mass of an object is halved, its kinetic energy also gets halved.
3.  If the speed of an object is doubled, its kinetic energy becomes four times.
4. If the speed of an object is halved, its kinetic energy becomes one-fourth.
5. Heavy objects moving with high speed have more kinetic energy than small objects moving with less speed.

Example 3. A bullet of mass 8 g is fired with a velocity of 80 \(\mathrm{~ms}^{-1}\). Calculate its kinetic energy.
Answer:

Given, mass, m=8 \(\mathrm{~g}=\frac{8}{1000} \mathrm{~kg}, velocity, v=80 \mathrm{~ms}^{-1}\)

KE of the bullet =\(\frac{1}{2} m v^2=\frac{1}{2} \times \frac{8}{1000} \times(80)^2 \)

= \(\frac{1}{2} \times \frac{8}{1000} \times 80 \times 80=25.6 \mathrm{~J}\)

Example 4. If a body of mass 5 kg is moving along a straight line with a velocity of 10 ms^-1 and acceleration of 20 ms^-2. Find its kinetic energy (KE) after 10 s.
Answer:

Given, the mass of the body, m = 5 kg

Initial velocity, u = 10 ms\(\mathrm{~ms}^{-1}\)

Acceleration, a = 20 \(\mathrm{~ms}^{-2}\)

Time, r = 10s Velocity, v = ?; KE = ?

First, we use the equation of meeting, N=N+ at to hinsl p, Then, we see \(KI-\frac{1}{2} m v^7\) to find kinetic energy.

r=\(\Delta+\Delta f\)

r=\((v+a t) m^{-1}\)

As we know, kinetic erg, KE =\(\frac{1}{2} m w^2\)

So. \(\mathrm{KE}=\frac{1}{2} m \times(w+a t)^2\)

KI=\(\frac{1}{2} \times 5 \times(10+20 \times 10)^2\)

∴ \(\mathrm{KE}=\frac{1}{2} \times 5 \times 210 \times 210=110250 \mathrm{j}\)

Example 5. What is the work to be done to increase the velocity of a van from 10 m/s to 20 m/s, if the mass of the is 2000 kg?
Answer:

Given,m=2000 kg, \(v_1=10 ms^{-1}, p_2=20 \mathrm{~ms}^{-1}\)

The initial Kinetic energy of the van

⇒ \(\mathrm{KE}_1.=\frac{1}{2} m \nu_1^2\) \([\mathrm{KE}=\frac{1}{2} m \nu^2]\)

= \(\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(10 m s^{-1}\right)^2=100000 \mathrm{~J}=100 \mathrm{~kJ}\)

The final kinetic energy of the van

⇒ \(\mathrm{KE}_2=\frac{1}{2} m{ }_2^2=\frac{1}{2} \times 2000 \mathrm{~kg} \times\left(20 \mathrm{~ms}^{-1}\right)^2\)

= 400000J = 400 kJ

The work done = Change in kinetic energy = 400 kJ -100 kJ = 300 kJ

So, the kinetic energy of the van increases by 300 kJ when it speeds up from 10 \(m s^{-1}\) to 20 \(m s^{-1}\).

Potential Energy

1. The energy possessed by a body due to its change in position or shape is called potential energy. Its SI unit is joule (J).
2. We can say that the potential energy possessed by a body is the energy present in it by its position or configuration,
3. For Example. a stretched rubber band, spring, string on the bow, etc. Now, we can say that a body possesses energy even when it is not in motion.

Examples of potential energy are

Water stored in a dam has potential energy due to its position at the height.

A stone lying on the roof of the building has potential energy due to its height.

A wound spring of a watch has potential energy due to the change in its shape.

Potential Energy of an Object at a Height

1. When an object is raised through a certain height above the ground, its energy increases. This is because work is done on it against gravity while it is being raised.
2. The energy present in such an object is the gravitational potential energy. The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.

Expression for Potential Energy

• Consider an object of mass m, lying at point A on the Earth’s surface. Here, its potential energy is zero and its weight mg acts vertically downwards.
• To lift the object to another position B at a height of h, we have to apply a minimum force that is equal to mg in the upward direction. So, work is done on the body against the force of gravity. Therefore,
• Work done = Force x Displacement or W = F x s
• As, F = mg [weight of the body]
• Here, s = h
• Therefore, W = mg x h = mg i.e. PE = mg
• This work is equal to the gain in energy of the body. This is the potential energy (PE) of the body.
• The potential energy of an object at a height depends on the ground level or the zero level you choose.
• An object in a given position can have a certain potential energy concerning one level and a different value of potential energy concerning another level.
• The work done by gravity depends on the difference in vertical heights of the initial and final positions of the objects and not on the path along which the object is moved.
• It is clear from the given
• In both the above situations, the work done on the object is much.

Example 6. Suppose you have a body of mass 1 kg in your hand. Tb what height will you raise it, so that it may acquire a gravitational potential energy of 1 J? (Take, g = 10 ms^-2)
Answer:

Given, PE =1J, mass, m = 1 kg,

Acceleration due to gravity, g = 10 ms^-2, h =?

We know that, PE = mg or 1 = 1 x 10 x h

Height, b = \(\frac{1}{1 \times 10}\)=0.1 m = 10 cm

Example 7. A boy weighing 40 kg climbs up a vertical height of 200 m. Calculate the amount of work done by him. How much potential energy does he gain? (Take, g =9.8 ms^-2)
Answer:

Given that, mass, m = 40 kg

Acceleration due to gravity, g = 9.8 ms^-2, height, h = 200 m

Work done by the body = mgt = 40 x 9.8 x 200 = 78400J = 7.84 x 10⁴ J

Gain in PE = Work done = 7.84 x 10⁴J

Example 8. Suppose two bodies A and B having equal masses are kept at heights of h and 3 h, respectively. Find the ratio of their potential energies.
Answer:

Let the mass of each body be m.

PE of body A= mg PE of body B=m g \(\times\) 3 h

Ratio of their potential energies =\(\frac{m g b}{m g \times 3 b}=\frac{1}{3}\)=1: 3

Law of Conservation of Energy

The law of conservation of energy states that energy can neither be created nor be destroyed, it can only be transformed from one form to another. The total energy before and after transformation always remains constant.

Conservation of Energy During the Free Fall of a Body

Consider an object of mass m, lying at position B. It is made to fall freely from a height (h) above the ground

At point B: At the start, the potential energy is mgh and kinetic energy is zero (as its velocity is zero),

i.e. PE = mg

KE = 0

Total energy, TE = PE + KE = mgh

At point A: As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, its kinetic energy would be \(\frac{1}{2} m v^2\).

PE = mg(h – x)

From Newton’s third equation of motion, v

⇒ \(v^2=u^2+2 g x\)

⇒ \(v^2\)=2 g x[u=0]

⇒ \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g\) x=m g x [\(v^2=2 g x]\)

Total energy, TE = mg

At point C: As the fall of the object continues, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h =0 and v will be the highest.

PE =0

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m\left(2 g^h\right)\)=m g h [ \(v^2=2 g^h\)]

Total energy, TE = mg

Thus, the sum of the potential energy and kinetic energy of the object would be the same at all points,

PE + KE = Constant Or mgh+\(\frac{1}{2} m v^2\)= constant

This verifies the law of conservation of energy.

Example 9. An object of mass 10 kg is dropped from a height of 5 m. Fill in the blanks by computing the potential energy and kinetic energy in each case.(Take, g = 10 ms^-2]
Answer:

Given, mass, m =10 kg Height, h = 5 m

Acceleration due to gravity, g = 10 ms^-2

At height b = 5 m,

KE = 0, as v = 0

PE = mgt =10x 10x 5 = 500J

Total energy (KE + PE) at height 5 m = 500 J

At height h = 4m,

KE =\(\frac{1}{2} m v^2=\frac{1}{2} m \times 2 g y [ v^2=2 g]\)

Distance covered, s = 5-4 = 1 m

KE =\(\frac{1}{2} \times 10 \times 2 \times 10 \times 1=100 \mathrm{~J}\)

PE = mgt =10 \(\times 10 \times 4=400 \mathrm{~J}\)

Total energy \((\mathrm{KE}+\mathrm{PE})\) at height 4 \(\mathrm{~m}=(100+400)=500 \mathrm{~J}\)

At height b=3 m,

s =5-3=2 m

KE =\(\frac{1}{2} m \nu^2=\frac{1}{2} m \times 2 g r=m g s=10 \times 10 \times 2=200 \mathrm{~J}\)

Therefore, PE =\(\mathrm{mgh}=10 \times 10 \times 3=300 \mathrm{~J}\)

Total enetgy (KE + PE) at height 3 m =(200+300)=500 J

At height b=2 m,

s =5-2=3 m

KE=m g y=10 \(\times 10 \times 3=300 \mathrm{~J}\)

PE = mgh =10 \(\times 10 \times 2=200 \mathrm{~J}\)

Tatal encrgy (KE + PE) at height 2 m = 300+200=500 J

At height k=1 m,

s = 5-1 = 4 m

KE = \(\frac{1}{2} m v^2=\frac{1}{2} \times m \times 2 \mathrm{~g}\)

= mgr =10 \(\times 10 \times\) 4=400 J

PE = mgt =10 \(\times 10 \times 1=100 \mathrm{~J}\)

Total energy( KE + PE ) at height 1 m =(400+100)=500 J

At just above the ground, h=0,

s=5-0=5 m

KE=\(\frac{1}{2} m v^2=m_S=10 \times 10 \times 5=500 J\)

PE = m \(\mathrm{mg}^{\prime}=10 \times 10 \times\) 0=0

Total energy (KE + PE) at just above the ground

=(500+0)=500 J

Thus, total mechanical energy remains constant at each height, which proves that energy is always conserved.

Example 10. A man is moving with a high velocity of 30 ms^-1. Determine the total mechanical energy of the man weighing 60 kg, if he is on a height of 50 m at this speed. [Take, g =10 ms^-2)

Answer:

Given, the mass of man, m=60 kg

Velocity, v=30 \(ms^{-1}\), height, b=50 m

The total energy (TE) of the man at a height of 50 m is given by

TE = PE + KE

where, PE = potential energy ( = mg )

and KE = kinetic energy \(\left(=\frac{1}{2} m y^2\right)\)

TE =m g h+\(\frac{1}{2} m v^2\) [from Eq. (1)]

=60 \(\times 10 \times 50+\frac{1}{2} \times 60 \times(30)^2\)

=30000+27000

=57000 J

Transformation of Energy (Are Various Energy Forms Interconvertible?)

1. One form of energy can be converted into another form of energy and this phenomenon is called transformation of energy.
2. When an object is dropped from some height, its potential energy continuously converts into kinetic energy.
3. When an object is thrown upwards, its kinetic energy continuously converts into potential energy.

For Example:

1. Green plants prepare their food (stored in the form of chemical energy) by using solar energy through the process of photosynthesis.
2. When we throw a ball, the muscular energy which is stored in our body gets converted into the kinetic energy of the ball.
3. 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.
4. In a stretched bow, potential energy is stored. As it is released, the potential energy of the stretched bow gets converted into the kinetic energy of the arrow which moves in the forward direction with large velocity.

Rate of Doing Work: Power

1. The rate of doing work or the rate at which energy is transferred used or transformed to other forms is called power.
2. If work W is done in time t, then
3. Power, P=\(\frac{\text { Work }}{\text { Time }}\)
4. P=\(\frac{\mathbb{W}}{t}\)
5. The SI unit of power is watt in honor ofJarr.c: Wait having the symbol W. We express a larger rate of energy transfer in kilowatt (kW).
6. 1 W=1 \(\mathrm{Js}^{-1}\) or 1 kW=1000 W=1000 \(\mathrm{Js}^{-1}\)
7. 1 MW = \(10^6\) W, 1 (horsepower) HP =746 W

Average Power

1. Average power is defined as the ratio of total work done to the total time taken. An agent may perform work at different rates at different intervals of time.
2. In such a situation, the average power is considered by dividing the total energy consumed by the total time taken.
3. Average power =\(\frac{\text { Tocal energy consumed }}{\text { Total time raken }}\)

Example 11. A boy does 400 J of work in 20 s and then he does 100 J or work in 2s. Find the ratio of the power delivered by the boy in two cases.
Answer:

Case 1 Work done by the boy. \(W_1\)=400 J

Time taken, \(z_1\)=20 s

power, \(P_1\)=z

Case 2 Work done by the bor, \(W_2\)=100 J

Time taken, \(s_2\)=2 s,

power, \(R_2\)= ?

Power, P=\(\frac{\text { Work dont }(W)}{\text { Time taken }(t)}\)

⇒ \(R_1=\frac{W}{t_1}=\frac{400}{20}=20 W\)

and \(R_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

Time raken,\(s_2=2 s\),

Power, P=\(\frac{\text { Work done }(W)}{\text { Time taken }(t)}\)

and \(P_2=\frac{W_2}{t_2}=\frac{100}{2}=50 W\)

So, \(\frac{B_1}{P_2}=\frac{20}{50}=2: 5\)

Example 12. A boy of mass of 55 kg runs up a staircase of 50 steps in 10 s. If the height of each step is 10 cm, find his power. Take g =10 m/s².
Answer:

Weight of the boy = mg

= 55×10 = 550 N

Height of the stains, h=\(\frac{50 \times 10}{100}\)=5 m

Time taken to dim =10 s

Power, p =\(\frac{\text { Work done }}{\text { Time taken }}\)

=\(\frac{\text { mgt }}{t}=\frac{550 \times 5}{10}=275 W\)

Question 13. The heart does 1.2 J of work in each _g Hp heartbeat. How many times per minute does it beat, if its power is 2 W?
Answer:

Here, work done in each heartbeat =1.2 J

t=1 min =60 s, power, P = 2W = 2 \(Js^{-1}\)

Total work done =P x r = 2 x 60=120 J

Number of times heart beats per minute

= \(\frac{\text { Total work done }}{\text { Work done in each beart beat }}\)

= \(\frac{120}{1.2}=100 \text { times }\)

Example 14. A horse exerts a pull on a cart of 500 N so that the horse cart system moves with a uniform velocity of 36 km^-1. What is the power developed by the horse in watts as well as in horsepower?
Answer:

Given, F=500 N

v=\(36 \mathrm{kmh}^{-1}=\frac{36 \times 1000}{3600} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}\)

As, P=\(\frac{W}{t}=\frac{F s}{t}\)

\(p^2 =\frac{W}{t}=\frac{F_s}{t}\) [ W=\(F_s\) and v = t/t]

=F v=500 x 10=5000 W

In horsepower.

P=\(\frac{5000}{746}\)=6.70 HP

Activity Zone

Activity 1

Objective: To understand why we use the term work differently in science.

Procedure

We come across several activities which are normally considered as work in our day-to-day life.

For each of these activities, the teacher should ask the following questions

1.  What is the work being done?
2. What is happening to the object?
3. Who (what) is doing the work?

Conclusion

1.  When an object moves a distance by applying force on it, work is said to be done.
2. The object moves under the influence of the applied force.
3. The agency that exerts the force is doing work.

Question 1. Work done by the force depends on.
Answer:

Work done by the force depends on the path or displacement of the body.

Question 2. Net work during the motion of the Earth around the Sun Is zero. Why?
Answer:

Net work during the motion of the Earth around the Sun is zero because the angle between force and displacement is 90°.

Question 3. What Is the SI unit of work?
Answer:

The SI unit of work is N-m or joule (J).

Question 4. Give mathematical expression for work done.
Answer:

The mathematical expression for work done (W) is as follows: W = Fs cos θ

where F is force, s is displacement, and θ is the angle between F and s.

Question 5. During a circular motion, what will be the work done?
Answer:

During a circular motion, force is perpendicular to the displacement, so the net work done is zero.

Activity 2

Objective: To understand how work is done.

Procedure:

1. Think of some situations from your daily life involving work.
2.  List them.
3.  Discuss with your friends whether work is being done in each situation or not.
4. Try to reason out your response by taking an example of an engine pulling a train.

Conclusion

The engine exerts a force on the train. Work is being done on the train due to which the train begins to move.

Question 1. 5 N of force Is applied to an object, but the object does not move. Then, how much work Is being done?
Answer:

If there is no displacement due to the application of force, then the net work done will be zero.

Question 2. What are the two conditions that need to be satisfied during work?
Answer:

The two conditions that need to be satisfied during work are

1. A force should act on an object.
2. There must be displacement due to applied force.

Question 3. When an engine pulls a train, who exerts force on It?
Answer:

When an engine pulls a train, the engine exerts a force on it

Question 4. 2 N of force displaces a body by 2 m, the work done will be.
Answer:

Work done by the given force = Force x Displacement =2 x2 = 4 J

Question 5. N-m Is the SI unit of which physical quantity?
Answer:

Ans N-m is the SI unit of work done.

Activity 3

Objective: To understand how work is not done.

Procedure:

1. Think of some situations when the object is not displaced despite a force acting on it Also, think of situations when an object gets displaced in the absence of a force acting on it
2. List all the situations that you can think of for each.
3. Discuss with your friends whether work is done in these situations. Suppose, we work hard to push a huge rock.
4. The rock does not move despite all the effort and we get completely exhausted. However, we have not done any work on the rock as there is no displacement of the rock.
5. Suppose a car is moving at a constant speed. The car moves without any external force acting on it However, there is no work done as there is no external force acting on the car.

Conclusion

1. When a force is applied to an object and it is not displaced from its position, the work done will be zero.
2. When an object gets displaced in the absence of a force acting on it, the work done will be zero.

Question 1. When a bullock pulls a cart, who does work?
Answer:

When a bullock pulls a cart, work is being done by the bullock.

Question 2. We pick a suitcase from the Earth to a height of h. Net work done by us will be (where, mass of suitcase =m, acceleration due to gravity =g).
Answer:

Net work done during the pickup of a suitcase to a height h is W = Fxs = mg

Question 3. Net work done during motion in a circular path is
Answer:

Net work done during motion in a circular path is zero as the angle between force and displacement is 90°.

Question 4. If a body does not move by the application of force, the net work done will be
Answer:

If a body does not move by the application of force, then the work done will be zero because of zero displacement.

Question 5. A waiter is carrying a tray to the table, the net work done by the waiter will be
Answer:

A waiter is carrying a tray to the table, the net work done by him will be zero as the weight of the tray is acting downwards and he is moving horizontally. The angle between force and displacement is 90°.

Hence, work done = 0.

Activity 4

Objective: To understand that the work done by a force can be either positive or negative.

Procedure

1. Lift an object. Work is done by the force exerted by you on the object. The object moves upwards.
2. The force exerted by you is in the direction of displacement. However/there is the force of gravity acting on the object

Conclusion

1. The work done by the force applied by you on the object is positive because it acts in the direction of displacement of the object
2. As the object moves up the force of gravity acts in the downward direction, i.e. in the opposite direction of the displacement of the object.
3. Hence, the work done by the force of gravity on the object is negative.

Question 1. What is the direction of force and displacement, if net work done is zero?
Answer:

If the net work done is zero, it means the direction of force is perpendicular to the direction of displacement.

Question 2. Give the work done in the given diagram.

Answer:

Work done when force and displacement are in opposite direction will be negative, i.e. W = -Fs

Question 3. Give the work done in case of a given diagram

Answer:

Work done in the given diagram is (W) = Fs, as the angle between F and s is 0°.

Question 4. What is the angle between force and displacement for maximum work?
Answer:

Work done will be maximum, if

cos θ = 1 or 0 = 0°, i.e. W = Fs cosθ°=1

Question 5. What is the work done by frictional force when the body is dragged along a rough surface?
Answer:

Work done is negative because the displacement of the body and frictional force is in the opposite direction

Activity 5

Objective: To understand that a falling body possesses kinetic energy due to its motion.

Procedure

1. Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be a better way than any other material.
2. Drop the ball on the sand bed from a height of about 25 cm. The ball creates a depression.
3. Repeat this activity from heights of 50 cm, lm, and 1.5 m.
4. Ensure that all the depressions are distinctly visible.
5. Mark the depressions to indicate the height from which the ball was dropped. Compare their depths.

Observation

1. The depression is deepest when the ball is dropped from 1.5 m height.
2. The depression is shallowest when the ball is dropped from 25 cm height, as it is the minimum height from which, the ball is dropped.

Conclusion

The larger the height from which the ball is dropped, the larger the kinetic energy gained by the ball on reaching the ground and more will be its capability to do work.

Question 1. A stone Is dropped from some height. What kind of energy is present in stone on reaching the ground?
Answer:

On reaching the ground, the stone has kinetic energy.

Question 2. A moving object has velocity due to which it has which type of energy?
Answer:

A moving object has velocity due to which it has kinetic energy.

Question 3. If two balls A and 8 are dropped from the same height. Which one has maximum kinetic energy?
Answer:

If two balls are dropped from the same height, then the heavier ball has greater kinetic energy w.r.t. the lighter one.

Question 4. If two bodies A and 8 of the same mass fall from height hJ and h, respectively, on the sand, where h₂ >hv Which body has more energy?
Answer:

Body B falling from height h₂ has more energy.

Question 5. A ball Is allowed to fall freely from a tower. Which energy Is gained at every point during the fall?
Answer:

Kinetic energy.

Activity 6

Objective: To understand the conversion of energy from one form to another.

Procedure

1. Many of the human activities and the gadgets we use involve the conversion of energy from one form to another.
2. Make a list of such activities and gadgets.
3. Identify in each activity/gadget the kind of energy conversion that takes place.

Observation and Conclusion

1. A list of such activities and gadgets is as follows:
2. Electric cell → Chemical energy changes into electrical energy.
3. Electric motor  → Electrical energy changes into mechanical energy.
4. Electric heater → Electrical energy changes into heat energy.
5. Dynamo → Mechanical energy changes into electrical energy.
6. Headphone → Electrical energy changes into sound energy.
7. Hydroelectric power station → Mechanical energy changes into electrical energy.
8. Microphone → Sound energy changes into electrical energy.
9. Steam engine → Heat energy changes into mechanical energy.
10. Photoelectric cell → Light energy changes into electrical energy.

Question 1. In a photocell, which type of energy conversion takes place?
Answer:

In photocell, light energy is converted into electrical energy.

Question 2. In solar cells, light energy Is converted to what?
Answer:

In solar cells, Sun energy is converted into electrical energy.

Question 3. Give the sequence of energy changes during the production of electricity In the dam.
Answer:

In the production of electricity in the dam, firstly stored potential energy of water is converted into mechanical energy to move the turbine and this mechanical energy gets converted into electricity or electrical energy by a dynamo.

Question 4. Give the energy change In the electric motor.
Answer:

In an electric motor, electrical energy is converted into mechanical energy (to move the body).

Question 5. In the headphones, which type of energy gets converted into sound energy?
Answer:

In the headphones, electrical energy gets converted into sound energy.

Activity 7

Objective: To understand that the sum of kinetic energy and potential energy of an object is its total mechanical energy.

Procedure

An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.

For simplifying the calculations, take the value of g as 10 ms^-2.

Calculation:

At height h=4 m ,

⇒ \(E_P\)=m g h=20 x 10 x 4=800 J

V = 0

⇒ \(E_K=\frac{1}{2} m v^2\)=0

Total energy =\(E_p+E_K\)=800+0=800 J

At height h=3 m,

⇒ \(E_P=m g t=20 \times 10 \times 3\)=600 J

Distance covered, s=4-3=1 m

From \(v^2-w^2\)=2 g s,

⇒ \(v^2 =u^2+2 g s \Rightarrow v^2\)=0+2 g s=2 g s

⇒ \(E_K =\frac{1}{2} m v^2=\frac{1}{2} m \times\) 2 g s=m g s

=20 \(\times 10 \times 1=200 \mathrm{~J}\)

Total energy =\(E_P+E_K\)=600+200=800 J

At height h=2 m,

⇒ \(E_P =m g h=20 \times 10 \times 2\)=400 J

s =4-2=2 m

⇒ \(E_E =m g r=20 \times 10 \times\) 2=400 J

Total energy =\(E_P+E_K\)=400+400=800 J

At height h =1 m,

⇒ \(E_P =m g h=20 \times 10 \times 1\)=200 J

s =4-1=3 m

⇒ \(E_K =m g s=20 \times 10 \times\) 3=600 J

Total energy =\(E_{\Gamma}+E_K\)=200+600=800 J

At just above the ground h=0,

s =4-0=4 m

⇒ \(E_r\)=m g h=0

⇒ \(E_K\) =m g s=20 x 10 x 4=800 J

Total energy =\(E_R+E_K\)

=0+800=800 J

Therefore, the table will be as follows:

Conclusion:

The total mechanical energy remains constant at each height, hence energy is conserved.

Question 1. When a body falls from a certain height, which energy transformation takes place?
Answer:

When a body falls suddenly, then its potential energy gradually gets converted into kinetic energy. On just reaching the ground, the whole potential energy of the body gets converted into kinetic energy.

Question 2. A car accelerates up a hill. What happens to its kinetic energy and its potential energy?
Answer:

The speed of the car is increasing so its kinetic energy increases. Going up the hill implies a gain in the vertical height so its potential energy increases.

Question 3. Give the variation of KE with height (h) graphically.
Answer:

Question 4. Give the variation of PE with height (h) graphically.
Answer:

Question 5. A ball Is thrown upwards from point A. It reaches up to the highest point 8 and then returns what Is Its PE at 8?
Answer:

From the law of conservation of energy, PE at point B = KE at point A.

Activity 8

Objective: To understand the agents that transfer energy to do work at different rates.

Procedure

1. Consider two children, say A and B. Let us say, they have the same weight. Both start climbing up a rope separately and both reach a height of 8 m.
2. For this A takes 15 s while B takes 20 s to accomplish the task.
3. Work done by each child is mgh, where mg is the weight of each child.

Observation

The same amount of work is being done by both but A takes less time (15 s) while B takes more time (20 s) to do the same work.

Conclusion

1. So, A has done more work in 1 s.
2. Hence, A has more power than B.

Question 1. Give the relation between work and power.
Answer:

The relation between work and power is as follows:

Power, P=\(\frac{\text { Work }}{\text { Time }}=\frac{W}{t}
Time t\)

Question 2. A body takes 2 s to do 10 J of work. What is its power?
Answer:

Power =\(\frac{\text { Work done }}{\text { Time taken }}=\frac{10}{2}=5 \mathrm{~W}\)

Question 3., Boy A does 400 J of work In 10 min her boy 8 does 500 J of work in 20 min. Who expends more power A or 8?
Answer:

Power expended by A=\(\frac{W}{t}=\frac{400}{10 \times 60}=0.67 W[ 1 \mathrm{~min}=60 \mathrm{~s}]\)

and power expended by B=\(\frac{W}{t}=\frac{500}{20 \times 60}\)=0.42 W

A expends more power.

Question 4. If a force F acts on the body and moves it with constant velocity v. Then, the power of the body will be
Answer: 

The power of a body moving with velocity v by a force F is given by

Power, P = Force x Velocity ⇒ P = Fv

Question 5. The unit kWh Is used for which physical quantity?
Answer:

kWh is the unit of energy. It is used for commercial purposes.

Work, Energy, And Power Question And Answers

Question 1. A force of 7 N acts on an object. The displacement is say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:

As work done, W = Fs[Work done is positive because force and displacement are in the same direction] W=7×8=56J

Question 2. When do we say that work is done?

Answer:

If a force acting on a body causes some displacement, then we can say that work is being done by the force on the body that is displaced.

Question 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:

Expression for the work done is given by

Work done, W = + Fs [F and s are in the same direction] where, F = force, s = displacement,

So, W = Fs

Question 4. Define 1 J of work.

Answer:

1 J of work is defined as the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5. A pair of bullocks exerts a force of 140 N on a plow. The field being ploughed is 15 m long. How much work is done in plowing the length of the field?

Answer:

Given, force, F =140 N, displacement, s =15 m

Work done, W = + F . s [Fand s are in the same direction]

So, W =F-s = 140×15 =2100J

Question 6. What is the kinetic energy of an object? kinetic energy at the surface of the earth 

Answer:

The kinetic energy of an object is defined as the energy due to its motion.

Question 7. Write an expression for the kinetic energy of an object.

Answer:

The expression for kinetic energy for an object is given by KE=\(\frac{1}{2} m v^2\)

where, KE = kinetic energy, w = mass of the body, v – velocity of the body.

Question 8. The kinetic energy of an object of mass m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is increased three times?

Answer:

As for the question,

Kinetic energy \(\left(\mathrm{KE}_i\right)\) initially is given by \(\frac{1}{2} m v_i^2\) where, m= mass of the body, v_i= initial velocity

⇒ \(\mathrm{KE}_l \left.=\frac{1}{2} m v_i^2\left[\text { given, } \mathrm{KE}_i=25\right], w_i=5 \mathrm{~ms}^{-1}\right]\)

25 =\(\frac{1}{2} m\left(5^2\right)\)

m =\(\frac{25 \times 2}{5 \times 5}\)

m =2 kg

Now, as from the question, final velocity \(\left(v_f\right)\) becomes 3 times its initial velocity, i.e.

⇒ \(v_f=3 v_i \Rightarrow v_f=3 \times 5=15 \mathrm{~ms}^{-1}\)

Now, kinetic energy, KE =\(\frac{1}{2} m v_f^2=\frac{1}{2} \times 2 \times 15 \times 15\)

=225

Question 9. What is power?

Answer:

Power is defined as the rate of doing work. If the work done by an object in time t is W Then, power, P = \(\frac{W}{t}\)

Its unit is Js^-1 or watt.

Question 10. Define 1 watt of power.

Answer:

Power, P=\(\frac{W}{t}\), If W =1 J, t=1 s

Then, P=\(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}=1 \mathrm{~W}\)

The power of an object is said to be 1 watt if it does 1 J of work in 1 s.

Question 11. A lamp consumes 1000 J of electrical energy in 10 seconds. What is the power?

Answer:

Given energy = 1000

i.e. wark done, W=1000

Time, r=10 s

Power of lamp, P=\(\frac{\mathrm{W}}{t}=\frac{1000}{10}\)=100 W

Question 12. Define average power.
Answer:

Average power is defined as the ratio of total work done to the total time taken.

Exercises

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work.’

1. Seema is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the Sun.
7. A sailboat is moving due to wind energy.

Answer:

1. Work is being done by Seema because she displaces the water by applying force.
2. No work is being done by the gravitational force because the direction of force, i.e. load is vertically downward and displacement is horizontally. If displacement and force are perpendicular, then no work is done.
3. Work is done because the mill is lifting the water, i.e. it is changing the position of water.
4. No work is done because there is no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work is done because there is no force and no displacement.
7. Work is done because of the force acting on the boat, it starts moving.

Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer:

As we know work done is the product of force and displacement and here in this case, displacement in the direction of gravitational force (change in height) is zero, so work done by the force of gravity on the object is zero.

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer:

A battery contains chemicals and supplies electrical energy. So, a battery converts chemical energy into electrical energy.

In an electrical bulb, the electrical energy is first converted into heat energy. This heat energy causes the filament of the bulb to become white-hot and produce light energy.

Thus, the energy changes are Chemical energy →  Electrical energy → Heat energy → Light energy

Question 4. Certain force acting on a mass 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work done by the force.

Answer:

Given, mass, m = 20 kg Initial velocity, u – 5 ms 1 Final velocity, v = 2 ms^-1

Work done by the force = Change in kinetic energy = Final kinetic energy – Initial kinetic energy

= \(\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

= \(\frac{1}{2} m\left(y^2-v^2\right)=\frac{1}{2} \times 20\left[(2)^2-(5)^2\right]\)

= 10 (4-25)=10 x (- 2,1) = – 210 J

Question 5. A mass of 10 kg is at point A on the table. It is moved to a point. If the line joining A and 6 is horizontal, what is the work done on the object by the gravitational force? Explain your Solwer.

Answer:

Here, both the initial and final positions are on the same horizontal line. So, there is no difference in height,

i.e. h =0. where, h = difference in the heights of initial and final positions of the object.

We know that work done by gravitational force, W = mg
Work done, W = mg x 0 = 0

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:

Indeed, the potential energy of a freely falling object decreases progressively. But as the object falls Ground^ down, its speed increases, i.e. the kinetic energy of the object increases progressively (kinetic energy will increase with the increase in speed).

Now, we can say that the law of conservation of energy is not violated, because the decrease in potential energy results in the increase of kinetic energy.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?

Answer:

In the case of riding a bicycle, the muscular energy is converted into the kinetic energy of the bicycle.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer:

When we push a huge rock, then the rock also exerts a large force on us (according to Newton’s third law of motion). The muscular energy spent by us in the process is used to oppose the huge force acting on us due to the rock.

Question 9. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy, if the object is allowed to fall? Find its kinetic energy when it is halfway down.

Answer:

Given, mass, m = 40 kg, height, h = 5 m As potential energy is given by PE = mgh

So, PE= 40 x 10 x 5 =2000J [v =10 ms^-2]

When it is allowed to fall, its PE gets converted into kinetic energy KE. So, when it reaches to half-way, half of its PE gets converted to KE.

So, mg \(\frac{b}{2}=\frac{1}{2} m v^2\) [where, v= velocity at the bottom]

So,KE =m g \(\frac{b}{2}\)

=40 \(\times 10 \times \frac{5}{2}\)=1000 J

Question 10. What is the work done by the force of gravity on a satellite moving around the Earth? Justify your answer.

Answer:

Work done by the force of gravity on a satellite moving around the Earth is zero. Because of the angle between force (centripetal) and displacement in the case of circular motion.

So, work done, W = 0

Question 11. Can there be displacement of an object in the absence of any force acting on it? Think, and discuss this question with your friends and teacher.

Answer:

Yes, if an object moves with a constant velocity, i.e. there is no acceleration, then no force acts on it. As the object is moving, so it gets displaced from one position to another position.

Question 12. A person holds a bundle of hay over his head for 30 min and gets tired. Has he done some work or not? Justify your answer.

Answer:

On holding a bundle of hay over the head, the work done by the person is zero because there is no displacement.

Question 13. Illustrate the law of s conservation of energy by discussing the energy changes which occur when we draw pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:

1. Let a simple pendulum be suspended from a rigid support S and OS be the equilibrium position of the pendulum. Let the pendulum be displaced to a position P, where it is at rest.
2. At position P, the pendulum has potential energy {mgh). When the pendulum is released from position P, it begins to move towards position O.
3. The speed of the pendulum increases and its height decreases which means the potential energy is converted into kinetic energy.
4. At position O, the whole of the potential energy of the pendulum is converted into its kinetic energy.
5. Then, the pendulum swings to the other side due to the inertia of motion. As the pendulum begins to move towards position Q, the speed of the pendulum decreases and height increases which means kinetic energy is converted into potential energy.
6. At point Q, the whole of the kinetic energy is converted into potential energy. Thus, we find that the potential energy is converted into kinetic energy and vice-versa during the motion of the pendulum. But the total energy remains constant.
7. When the pendulum oscillates in air, the air friction opposes its motion. So, some part of the kinetic energy of the pendulum is used to overcome this friction.
8. With time, the energy of the pendudecreasessing and finally becomes zero.
9. The energy of the pendulum is transferred to the atmosphere. So, energy is being transferred, i.e. is converted from one form to another. So, no violation of the law of conservation of energy takes place.

Question 14. An object of mass m is moving with a constant velocity How much work should be done on the object to bring the object to rest?

Answer:

Concept Change in kinetic energy (KE) =Work done

Given, mass = m, initial velocity, u = v

Final velocity, v = 0

So, W=\(\frac{1}{2} m v^2-\frac{1}{2} m s^2\)

=\(\frac{1}{2} m(0)^2-\frac{1}{2} m v^2\)

W=-\(\frac{1}{2} m v^2\)

Hence, the work that should be done to bring the object to rest is \(\frac{1}{2} m v^2\).

Question 15. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km-1.

Answer:

Concept Change in kinetic energy is equal to the work done W.

Given, initial velocity, u=60 \(\mathrm{kmh}^{-1}\)

=60 \(\times \frac{5}{18}=\frac{50}{3} \mathrm{~ms}^{-1} {\left[1 \mathrm{kmh}^{-1}=5 / 18 \mathrm{~ms}^{-1}\right] }\)

Final velocity v=0

So, the magnitude of change in kinetic energy =W

=\(\frac{1}{2} m v^2-\frac{1}{2} m v^2\)

W =\(\frac{1}{2} m\left(v^2-s^2\right)\)

=\(\frac{1}{2} \times 1500 \times\left(\frac{-50 \times 50}{9}\right)\)

=-\(\frac{1}{2} \times \frac{1500 \times 50 \times 50}{9}\)

W\(\left.=-\frac{625000}{3}=-208333.3\right)\)

Hence, the work required to be done to stop a car is 208333.3 J.

Question 16. In each of the following, a force F is acting on an object of mass m. The direction of displacement is from West to East shown by the longer arrow. Observe the figure and state whether the work done by the force is negative, positive, or zero.

Answer:

In given. (1), the angle between F and s is 90°, so the work done is zero.

In given. (2), the angle between F and s is 0°, so the work done is positive.

In given. (3), the angle between F and s is 180°, so the work done is negative.

Question 17. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer:

Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are actonng onacting the resultant of all the forces acting on an object is zero.

Question 18. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:

When a freely falling object strikes the Earth, some sound and heat are produced. So, the kinetic energy of the object converts into sound energy and heat energy.

Summary

Work is said to bo done in a physical activity involving n form and movement in the direction of force and the process an equal amount of energy is used up.

Two conditions need to be satisfied for work to be done.

1. A force should act on the object.
2. The object must be displaced.

Work done by a force on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of force, i.e. U’ = F xs.

• The SI unit of work is joule (J).
• 1 J is the amount of work done on an object when a force of 1 X displaces it by 1 m along the line of action of force.
• Energy is the ability to do work.
• The SI unit of energy is the same as that of work, i.e. joule.
• A larger unit of energy is kilojoule (kJ).
• Relation between Joule and Kilojoule
• 1 kilojoule = 1000 joule.
• The SI unit of work and energy is named after British Physicist, James Prescott Joule.
• Then energy that is possessed by an object due to its motion is called kinetic energy.
• The SI unit of kinetic energy is joule (J).
• The kinetic energy possessed by an object of mass m, moving
  noth a uniform velocity v is KE = \(\frac{1}{2} m v^2\)
• The work-energy theorem is given asW = \(\frac{1}{2} m\left(v^2-u^2\right)\). If initial velocity u = 0, then W = \(\frac{1}{2} m v^2\)
• The energy possessed by a body due to its position or change in its configuration (i.e. shape or position) is known as potential energy.
• The SI unit of potential energy is the joule (J).
• Tho potential energy of an object can be given ns PE = mg.
• According to the law of conservation of energy, energy can neither be created nor be destroyed but it can be transformed from one form to another.
• During the free fall of a body, its energy is always conserved. The sum of the potential energy and kinetic energy of an object is the same at all points.
• One form of energy can be converted into other forms of energy, this phenomenon is known as transformation of energy.

Some Energy Transformations

1. Electric Motor Electrical energy into mechanical energy.
2. Electric Generator Mechanical energy into electrical energy.
3. Steam Engine Heat energy into kinetic energy.
4. Electric Bulb Electrical energy into light energy.
5. Dry Cell Chemical energy into electrical energy.
6. Solar Cell Light energy into electrical energy.
7. The rate of doing work or the rate at which the energy is
   transformed is known as power (P).
8. Power =\(\frac{\text { Work done }}{\text { Time }}\)
9. P =\(\frac{W}{t}\).
10. The SI unit of power is watt (W).
11. The relation between watt and horsepower can be given as 1 (horsepower) HP = 746 W.
12. Average power is defined as the ratio of total work done to total time taken.

Class 9 Science Notes For Chapter 11 Sound

We hear sound from various sources, For Example. from humans, birds, machines, vehicles, TV, radio, etc. Sound is a form of energy which produces a sensation of hearing in our ears.

Production of Sound

• A sound is produced by vibrating objects. Vibration means a kind of rapid-to-and-fro motion of an object. The sound of the human voice is produced due to vibrations in the vocal cords.
• We can produce sound by striking the tuning fork, by plucking, scratching, rubbing, blowing or shaking different objects. They all produce sound due to vibrations.

Propagation of Sound

1. When an object vibrates, it sets the particles of the medium (solid, liquid or gas) around it in vibrations. The particles do not travel from the vibrating object to the ear. A particle of the medium in contact with the vibrating object is first displaced from its equilibrium position.
2. It then exerts a force on the adjacent particle. As a result of which, the adjacent particle gets displaced from its position of rest.
3. After displacing the adjacent particle, the first particle comes back to its original position. This process continues in the medium till the sound reaches our ear.
4. The source of sound creates a disturbance in the medium which travels through the medium. The particles of the medium do not move forward but the disturbance is carried forward.
5. This is the propagation of sound in a medium, hence sound can be visualised as a wave. Sound waves require a medium to travel, so they are called mechanical waves.

Read and Learn  More Class 9 Science Notes

Formation of Compression and Rarefaction in Air

1. When a vibrating object moves forward in the air, it pushes and compresses the air in front of it, creating a compression which starts to move away from the vibrating object. When the vibrating object moves backwards, it creates rarefaction.
2. Compression is the part of a longitudinal wave in which the particles of the medium are closer to one another than they normally are and it is the region of high pressure. It is denoted by C in the figure given here.
3. Rarefaction is the part of a longitudinal wave in which the panicles of the medium arc are farther apart than they normally arc and it is the region of low pressure. It is denoted by R in the figure given here.
4. A vibrating object creates a series of compressions (C) and rarefactions (R) in the medium
5. As the object moves back and forth rapidly, a series of compressions and rarefactions are created in the air.
6. Thus, the propagation of sound can be visualised as the propagation of density variations or pressure variations in the medium as pressure is related to the number of particles of a medium in a given volume. More density of the particles in the medium gives more pressure and vice-versa.

Types of Waves

There are mainly two types of waves.

Longitudinal Waves

1. In longitudinal waves, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance.
2. The particles do not move from one place to another but they simply oscillate back and forth about their positions of rest.
3. This is exactly how a sound wave propagates, hence sound waves are longitudinal waves. Longitudinal waves can be produced in all three media such as solids, liquids and gases.
4. The waves which travel along a spring when it is pushed and pulled at one end, are the longitudinal waves.
5. When coils are closer together than normal, compressions (C) are observed in spring. When coils arc farther apart than normal, rarefactions (R) are observed.
6. A long flexible spring which can be compressed or extended easily is called a slinky.

Transverse Waves

• In transverse waves, the individual particles of the medium move about their mean position in a direction perpendicular to the direction of wave propagation, For Example. Light is a transverse wave (but it is not a mechanical wave i.e. it does not require a medium for its propagation).
• Transverse waves can be produced only in solids and liquids but not in gases.
• The waves produced by moving one end of a long spring or rope, up and down rapidly, whose other end is fixed, are transverse waves.

Graphical Representation of a Sound Wave

• When a sound wave passes through the air, the density of air changes continuously.
• A sound Wave in hr has been represented using a density-distance graph as shown below.
• A sound wave in the air by density-distance graph

Terms to Describe Sound Waves

Sound waves can be described by its

1. Wavelength
2. Frequency
3. Time period
4. Amplitude
5. Speed

Wavelength

1. The distance between the two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength. Wavelength is the minimum distance in which a sound wave repeats itself.
2. In other words, it is the combined length of a compression and an adjacent rarefaction. It is represented by a Greek letter lambda X. Its SI unit is metre (m).

Frequency

1. The number of complete waves (or oscillations) produced in one second is called the frequency of the wave. It is the number of vibrations that occur per second.
2. Or the number of compressions or rarefactions that cross a point per unit of time.
3. The frequency of a wave is fixed and does not change even when it passes through different substances. It is denoted by v (Greek letter, nu). Its SI unit is hertz (symbol, Hz) named in honour of Heinrich Rudolf Hertz who discovered the photoelectric effect.

Time Period

• The time taken by two consecutive companion* or rarefactions to cross a fixed point is called the period of the wave.
• In other words, the time required to produce one complete wave (or oscillations) is called the period of the wave. It is denoted by the symbol T. Its SI unit is second (s).
• The time period of a wave is the reciprocal of its frequency, Time period(T)=\(\frac{1}{\text { Frequency }(v)}\)

Amplitude

1. The maximum displacement of the particles of the medium from their original mean positions on passing a wave through the medium is called the amplitude of the wave.
2. It is used to describe the size of the wave. It is usually denoted by the letter A. Its SI unit is metre (m). The amplitude of a wave is the same as the amplitude of the vibrating body producing the wave.

Speed

• The distance travelled by a wave in one second is called the speed of the wave or the velocity of the wave. Under the same physical conditions, the speed of sound remains the same for all frequencies.
• It is represented by the letter v. Its SI unit is metres per second (m/s or ms^-1).
• Relationship between speed, frequency and wavelength of a wave: Speed =\(\frac{\text { Distance travelled }}{\text { Time taken }}\)
• Suppose the distance travelled by a wave is X (wavelength), in time T, then the speed is given byv=XJT We know that, frequency, v = 1 IT Therefore, y=A.xvor v=vX

Example 1. Sound waves travel in the air with a speed of about 330 ms^-1. Calculate the wavelength of sound whose frequency is 550 Hz. Calculate the period of one oscillation.

Answer:

Given, the frequency of sound, v = 550 Hz

Speed of sound wave, v=330 ms^-1

The wavelength of the sound wave, X =?

Time period, T=?

We know that, r-v \(\lambda\) and \(\tau-1 / v\)

As. r-v \(\lambda \Rightarrow \lambda=\frac{r}{v}-\frac{330}{550}=\frac{3}{5}=0.6 \mathrm{~m}\)

Asain, T=\(\frac{1}{v}=\frac{1}{550}=0.001 \mathrm{~s}\)

Characteristics of Sound

A sound has three characteristics. These are loudness, pitch and quality (or timbre).

Loudness

• It is the measure of the sound energy reaching the ear per second. The greater the sound energy reaching our ear per second, the louder the sound will appear to be.
• If the sound waves have a small amplitude, then the sound will be faint or soft but, if waves have a large amplitude then the sound will be loud. The figure given above shows the wave shapes of a loud and a soft sound of the same frequency.
• Since the amplitude of a sound wave is equal to the amplitude of vibrations of the source producing the sound waves, hence the loudness of sound depends on the amplitude of vibrations of the source producing the sound waves. Loud sound can travel a larger distance as it is associated with higher energy.
• A sound wave spreads out from its source, as it moves away from the source, its amplitude as well as its loudness decreases. The loudness of sound is measured in decibels dB). It depends on the sensitivity or the response of our ears.

Intensity

• The amount of sound energy passing each second through the unit area is known as the intensity of sound. Loudness and intensity are not the same terms. Loudness is a measure of the response of the car to the sound.
• Even when two sounds are of equal intensity, we may hear one as louder than the other, simply because our car detects it in a better way.
• The SI unit of intensity is watt per square metre (W/m²).

(3) Quality or Timbre

The quality or timbre of sound enables us to distinguish one sound from another having the same pitch and loudness. A sound of a single frequency is called a tone.

The sound produced due to a mixture of several frequencies is called a note and is pleasant to listen to. Noise is unpleasant to the ear. Music is pleasant to hear and is of rich quality.

Speed of Sound and Light

• The speed of sound in air is about 344 ms^-1 at 22°C and 331 ms^-1 at 0°C and the speed of light in air is 300000000 ms 1 or 3 x 108 ms^-1. Thus, the speed of light is very high as compared to the speed of sound.
• This is the reason why in the rainy season, the flash of lightning is seen first and the sound of thunder is heard a little later, though both are produced at the same time in clouds.

Speed of Sound in Different Media

The medium through which sound propagates can be a solid substance, a liquid or a gas. The speed of sound depends on the properties of the medium through which it travels and the temperature of the medium.

The speed of sound decreases when we go from a solid to a gaseous state. If the temperature of the medium increases then the speed of sound also increases.

Reflection Of Sound Wave

The bouncing back of sound when it strikes a hard surface is known as the reflection of sound. It can be reflected from any surface whether it is smooth or rough.

Sound is reflected in the same way as light and follows the same laws of reflection, which are as follows:

1. The incident sound wave (AO), the reflected sound wave (OB) and the normal (ON) at the point of incidence, all lie in the same plane.
2. The angle of incidence (∠AON) of sound is equal to the angle of reflection (∠NOB) of sound.

Echo

When a person shouts in a big empty hall, we first hear his original sound, after that, we hear the reflected sound of that shout. So, the repetition of sound caused by the reflection of sound waves is called an echo.

The sensation of sound persists in our brain for about 0.1 seconds. Thus, to hear a distinct echo, the time interval between the original sound and the reflected one must be at least 0.1 s.

The distance travelled by the sound in 0.1 s

= speed x time

= 344 x 0.1 = 34.4 m

So, the echo will be heard, if the minimum distance between the source of sound and the obstacle is =\(\frac{34.4}{2} \mathrm{~m}\)=17.2 m

This distance will change with the change in temperature. Echoes may be heard more than once due to successive multiple reflections.

The rolling of thunder is due to successive reflections of sound from a number of reflecting surfaces, such as clouds and the land.

Reverberation

1. The persistence of a sound in a big hall due to repeated reflections from the walls, ceiling and floor of the wall is known as reverberation.
2. This occurs when the original sound and reflected sound overlap. For reverberation to occur, reflection occurs at less than 17 metres distance.
3. A short reverberation is desirable in a concert hall, where music is being played, as it boosts the sound level. However excessive reverberation is highly undesirable because sound becomes blurred, distorted and confusing due to overlapping of different sounds.
4. To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibre board, rough plaster or draperies.

Uses of Multiple Reflections of Sound

The reflection of sound is used in the working of devices such as megaphones, horns, stethoscopes and soundboards. These devices involve multiple reflections of sound waves.

(1) Megaphone and Horn

A megaphone is a large cone-shaped device used to amplify and direct the voice of a person who speaks into it. When a person speaks into the narrow end of the megaphone tube, the sound waves produced are prevented from spreading by successive reflections from the wider end of the megaphone tube, hence sound of his voice can be heard over a longer distance.

(2) Stethoscope

It is a medical instrument used by doctors to listen to the sounds produced within the heart and the lungs in the human body. The sound of heartbeats (or lungs) reaches the doctor’s ears through the multiple reflections of sound waves through the stethoscope tube.

Sound Board

• A concave board (curved board) is placed behind the stage in big halls; so that sound after reflecting from the sound hoard, spreads evenly across the width of the hall.
• Generally, the ceilings of concert halls, conference halls and cinema halls are curved, so that sound after reflection reaches all corners of the hall.

Range of Hearing

The average frequency range over which the human ear is sensitive is called the audible range. The audible range of sound for human beings is from 20 Hz to 20,000 Hz (20 kilohertz). Children under the age of 5 and some animals, such as dogs can hear up to 25000 Hz. As people grow older, their ears become less sensitive to higher and lower frequencies.

Infrasonic Sound

• The sound of frequencies lower than 20 Hz is known as infrasonic sounds or infrasound, which cannot be heard by human beings.
• Earthquakes and some animals like whales, elephants and rhinoceroses produce the infrasonic sounds of frequency 5 Hz.
• It is observed that some animals get disturbed and start running here and there just before the earthquakes occur. This is because earthquakes produce low-frequency infrasound before the main shock waves begin which possibly alert the animals and they get disturbed.

Ultrasonic Sound

The sounds of frequencies higher than 20000 Hz are called ultrasonic sounds or ultrasounds which cannot be heard by human beings. Dogs can hear ultrasonic sounds of frequencies up to 50,000 Hz. This is why dogs are used for detective work by the police. Bats, dolphins, and porpoises can produce ultrasonic sounds.

Hearing Aid

This is a device used by people who are hard of hearing. It is an electronic, battery-operated device. It receives sound through a microphone which converts the sound waves to electrical signals. These electrical signals are amplified by an amplifier.

The amplified electrical signals are given to the speaker of the hearing aid. The speaker converts the amplified electrical signals to sound and then sends it to the ear for clear hearing.

Ultrasound and Its Applications

Ultrasounds are high-frequency waves. They travel in a straight line without bending around the corners. They can penetrate into matter to a large extent. Due to these properties, ultrasound is used in industry and in hospitals for medical purposes. Some of the important applications of ultrasound are given below.

In Cleaning Minute Parts of Machines

1. Ultrasound is used to clean parts located in hard-to-reach- places, such as spiral tubes, odd-shaped machines electronic components, etc.
2. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution.
3. Due to their high frequency, the ultrasound waves stir up the solution, hence the particles of dust, grease and dirt vibrate too much, become loose, get detached from the object and fall into the solution. The objects, thus get thoroughly cleaned.

In Internal Investigation of the Human Body

• Ultrasound is used to investigate the internal organs of the human body such as the liver, gall bladder, pancreas, kidneys, uterus heart, etc.
• Ultrasound waves can penetrate the human body and different types of tissues get reflected in different ways from a region where there is a change of tissue density.
• In this way, ultrasound helps us to see inside the human body and to give pictures of the inner organs by converting them into electrical signals.
• These pictures or images are then displayed on a monitor or printed on a film. This technique is called ultrasonography.
• Ultrasonography is used for the examination of the foetus during pregnancy to detect any growth abnormalities, which helps in taking the necessary action to rectify the abnormalities.
• The ultrasonic scanner is an instrument that helps the doctor to detect abnormalities, such as stones in the gall bladder and kidney or tumours in different organs and many other ailments.
• Ultrasound is also used for diagnosing heart diseases by scanning the heart from inside. This technique is echocardiography. Ultrasound may be employed to break small stones formed in the kidneys into fine grains which later get flushed out with urine. This way, the patient gets relief from pain.

In Industries

1. Ultrasound is used in industry for detecting flaws (cracks, etc.) in metal blocks without damaging them.
2. Metal blocks are used in the construction of big structures like bridges, machines scientific equipment, etc.
3. If there are some cracks and flaws in the metal blocks, which are invisible from the outside reduces the strength of the structure. These can be detected by using ultrasound.
4. This is based on the fact that an internal crack (or hole) does not allow ultrasound to pass through it.
5. It reflects the ultrasound. Ultrasound waves are allowed to pass through one face of the metal block (to be tested) and detectors are placed on the opposite face of the metal block to detect the transmitted ultrasound waves.
6. If there is even a small defect, the ultrasound waves get reflected back indicating the presence of the flaw or defect, as shown
7. Ultrasound is reflected from a part of the block, which shows that this metal block has a flaw or defect (like a crack] inside it.
8. Ordinary sound waves cannot be used for detecting the flaws in metal blocks because they will bend around the corners of the defective location and therefore enter the detector.

Use of Ultrasonic Waves by Bats

1. Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves.
2. The method used by some animals like bats, tortoises and dolphins to locate objects by hearing the echoes of their ultrasonic squeaks is known as echolocation.
3. Bats emit high-frequency or high-pitched ultrasonic squeaks while flying and listen to the echoes produced by the reflection of their squeaks from the obstacles or prey in their path.
4. From the time taken by the echo to be heard, bats can determine the distance of the obstacle or prey and can avoid the obstacle by changing the direction or catching the prey.
5. However, certain moths can hear the high-frequency ultrasonic squeaks of a bat and can know where the bat is flying nearby and are able to escape from being captured.

Sound Questions and Answers

Question 1. How does the sound produced by a vibrating object in a medium reach your ear?

Answer:

• Sound is produced by vibrating objects. When an object vibrates, it sets the particles of the medium around it in vibration.
• Their vibrations are passed or transmitted to neighbouring particles in all directions.
• When vibrations are transmitted by medium particles to our ears, we get the sensation of hearing.

Question 2. Explain, how sound is produced by your school bell.

Answer:

• The bell produces the sound when the gong of the bell is struck by a hammer. When the gong is struck by the hammer, it starts vibrating.
• Since the vibrating objects produce sound, the bell produces sound.

Question 3. Why are sound waves called mechanical waves?

Answer:

Sound waves are called mechanical waves because they are produced by the motion of particles of a medium and require a material medium for their propagation.

Question 4. Suppose you and your friend are on the Moon. Will you be able to hear any sound produced by your friend?

Answer:

• No, I will not be able to hear any sound produced by my friend because the sound waves require some material medium like air to travel.
• There is no atmosphere or air on the Moon, so the sound produced by my friend will not reach me and I will not be able to hear.

Question 5. Which wave property determines (1) loudness (2) pitch?

Answer:

• The loudness of a sound wave is determined by its amplitude.
• The pitch of the sound wave is determined by its frequency.

Question 6. Guess which sound has a higher pitch, guitar or car horn?

Answer:

The pitch of a guitar sound is higher because the frequency of the sound produced by a guitar is higher than that of a car horn.

Question 7. How are the wavelength and frequency of a sound wave related to speed?

Answer:

The relation between wavelength (λ), frequency (v) and speed of wave {v) is v=vλ

Question 8. Calculate the wavelength of a sound wave whose frequency is 200Hz and speed is 440 \(\mathrm{~ms}^{-1}\) in a given medium.

Answer:

Given, frequency, v=200 Hz, velocity, v=440 \(\mathrm{~ms}^{-1}\)

According to the relation, v=\(\mathrm{v}\rangle \Rightarrow \lambda=\frac{v}{\mathrm{v}}\)

= \(\frac{440}{200}=2.2 \mathrm{~m}\)

Question 9. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound, what is the time interval between successive compressions from the source?

Answer:

The time interval between two successive compressions or rarefactions is equal to the time period of the wave.

Required time interval = Time period

=\(\frac{1}{\text { Frequency }}=\frac{1}{500}=0.002 \mathrm{~s}\)

=2 \(\times 10^{-3} \mathrm{~s}=2 \mathrm{~ms}\)

Question 10. Distinguish between loudness and intensity of sound.

Answer:

Question 11. An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 \(\mathrm{~ms}^{-1}\)?

Answer:

Given, speed of sound, v=342 \(\mathrm{~ms}^{-1}\)

Time taken, t = 3s

So, distance travelled by sound =v x t

= 342 x 3

=1026 m

Hence, the distance between the reflecting surface and the source should be \(\frac{1026}{2}=513 \mathrm{~m}\)

Question 12. Why are the ceilings of concert halls curved?

Answer:

The Curved ceiling of the conference hall focuses on reflecting sound from the walls to the audience so that every corner of the hall gets sound equivalently.

Question 14. What is the audible range of the average human ear?

Answer:

The audible frequency range for the average human car is 20 Hz to 20000 Hz’.

Question 15. What is the range of frequencies associated with (1)Infrasound and (2) ultrasound?

Answer:

1. Sound waves having frequencies less than 20 Hz and greater than zero are called infrasound.
2. Sound waves having frequencies more than 20000 Hz are called ultrasound.

Exercises

Question 1. What is sound and how is it produced?
Answer:

Sound is a form of energy, which produces the sensation of hearing in our ears. It is produced when an object is set to vibrate or we can say that vibrating objects produce sound.

Question 2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:

1. When a vibrating object moves forward, it pushes the air in front of it creating a region of high pressure. This region is called compression.
2. This compression starts to move away from the vibrating object. When the vibrating object moves backwards, it creates a region of low pressure called rarefaction.
3. As the object moves back and forth rapidly, a series of compressions and rarefactions is created. These make the sound wave that propagates through the medium.

Question 3. Why is a sound wave called a longitudinal wave?
Answer:

The sound wave is called a longitudinal wave because, on the propagation of a sound wave in a medium, the particles of the medium vibrate to and fro about their equilibrium positions and parallel to the direction of propagation of the wave.

Question 4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:

Timbre, a quality of sound is the characteristic by which we can identify the person by his voice.

Question 5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:

Thunder is heard a few seconds after the flash is seen because the speed of light in the atmosphere (or air) is 3 \(\times 10^8 \mathrm{~ms}^{-1}\) which is very high as compared to the speed of sound which is only 330 \(\mathrm{~ms}^{-1}\). So, the sound of thunder reaches us later than the flash.

Question 6. A person has hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves Tair corresponding to these two frequencies? (Take, the speed of sound in air as \(\mathrm{~ms}^{-1}\)
Answer:

The relation between speed (v), wavelength \((\lambda)\) and frequency (v) of a wave is \(\nu=\mathrm{v} \lambda\)

⇒ \(\lambda=\frac{y}{v}\)

(1) Here, v=344 \(\mathrm{~ms}^{-1}, \mathrm{v}=20 \mathrm{~Hz}\)

Corresponding wavelength,

⇒ \(\lambda=\frac{p}{v}=\frac{344}{20}=17.2 \mathrm{~m}\)

⇒ (2) Here, v=344 \(\mathrm{~ms}^{-1}, \mathrm{v}=20 \mathrm{kHz}=20 \times 10^3 \mathrm{~Hz}\)

Corresponding wavelength, \(\lambda=\frac{v}{v}=\frac{344}{20 \times 10^3}\)

=1.72 \(\times 10^{-2} \mathrm{~m}\)

Question 7. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:

Let l be the length of the rod.

Time taken by the sound to travel through the aluminium rod is given by = \(t_1=\frac{\text { distance }}{\text { speed }}=\frac{1}{v_{\mathrm{N}}}\)

Similarly, the time taken by the sound to travel through the air is given by r, \(t_2=\frac{\text { distance }}{\text { speed }}=\frac{1}{v_{\text {air }}}\)

Required ratio, \(t_1: t_2=\frac{v_{\text {air }}}{v_{\mathrm{Al}}}\)

Question 8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:

Given, frequency, v = 100 Hz

From the definition of frequency, we can say that

Number of oscillations in l s = 100

Number of oscillations in l min or 60 s

= 100×60 = 6000

Thus, the tyre source of sound vibrates 6000 times in a minute.

Question 9. Does sound follow the same laws of reflection as light does? Explain.
Answer:

Yes, sound wave follows the same laws as in the case of laws of reflection of light.

The laws of reflection of sound are as follows:

(1) The incident sound wave (AO), the reflected sound wave (OB) and the normal (ON) at the point of incidence, all lie in the same plane.

{2) The angle of incidence (∠AON) of sound is equal to the angle of reflection (∠NOB) of sound.

Question 10. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remain the same. Do you hear an echo sound on a hotter day?
Answer:

The time taken by echo to be heard, t = \(\frac{2 d}{v}\)

where d = distance between the reflecting surface and source of sound and v = speed of sound in air.

• As we know, the speed of sound increases with an increase in temperature, so on a hotter day, the speed of sound will be higher, so the time after which the echo is heard will decrease.
• If the time taken by the reflected sound is less than 0.1 s after the production of the original sound, then the echo is not heard.

Question 11. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? [Given, g = 10 ms^-2 and speed of sound = 340ms^-1]
Answer:

Time after which the splash is heard = Time taken by the stone to reach the pond + Time taken by splash sound to reach the top of the tower.

(1) For the time taken by the stone to reach the pond.

Here, u=0 [stone is dropped from rest] From the equation of motion, h=\(u t+\frac{1}{2} g t^2\)

500=0 \(\times t+\frac{1}{2} \times 10(t)^2\)

500=\(5 t^2 \Rightarrow t^2=100 \Rightarrow t=\sqrt{100}=10 \mathrm{~s}\)

(2) Time taken by splash sound to reach the top of the tower \(t^{\prime}=\frac{\text { Distance }}{\text { Speed }}=\frac{500}{340}=1.47 \mathrm{~s}\)

Time after which a splash is heard

=10+1.47=11.47 s

Question 12. A sound wave travels at a speed of 339 \(\mathrm{~ms}^{-1}\). If its wavelength is 1.5 cm, then what is the frequency of the wave? Will it be audible?
Answer:

Given, speed, v=339 \(\mathrm{~ms}^{-1}\)

Wavelength, \(\lambda=1.5 \mathrm{~cm}=1.5 \times 10^{-2} \mathrm{~m}\)

Frequency, v=\(\frac{v}{\lambda}=\frac{339}{1.5 \times 10^{-2}}=22600 \mathrm{~Hz}\)

This frequency is greater than 20000 Hz, so it will not be audible. The audible range for the human ear is 20 Hz to 20000 Hz.

Question 13. What is reverberation? How can it be reduced?
Answer:

• The persistence of a sound in a big hall due to repeated reflections from the walls, ceiling and floor of the wall is known as reverberation.
• It can be reduced by covering the roofs and walls of the hall with sound-absorbing materials.

Question 14. What is the loudness of sound? What factors does it depend on?
Answer:

Loudness of a sound is a subjective quantity, it is the measure of the sound energy reaching the ear per second. Loudness depends on

1.  the amplitude of the vibrating body and
2. the sensitivity of the human ear.

Question 15. Explain how bats use ultrasound to catch prey.
Answer:

1. Bats can produce ultrasonic waves by flapping their wings, they can also detect these waves. The ultrasonic waves produced by a bat spread out.
2. These waves after reflecting from a prey like an insect reach the bat. So, the bat can locate its prey.

Summary

• Sound is produced by vibrating objects.
• Vibration means a kind of rapid-to-and-fro motion of an object.
• Compression is the part of a longitudinal wave in which the particles of a medium are closer to one another than they normally are and it is the region of high pressure.
• Rarefaction is the part of a longitudinal wave in which the particles of the medium are farther apart than they normally are and it is the region of low pressure.
• Waves are of two types (1) longitudinal waves and (2) transverse waves.
• In longitudinal waves, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance.
• In transverse waves, the individual particles of the medium move about their mean position in a direction perpendicular to the direction of wave propagation.
• The distance between the two consecutive compressions (C) or two consecutive rarefactions (fl) is called the wavelength. Its SI unit is metre (m).
• The number of vibrations that occur per second is known as frequency. Its SI unit is Hertz (Hz).
• The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave.
• The time period of a wave is reciprocal of its frequency, i.e. T=\(\frac{1}{\mathrm{v}}\) .
• The maximum displacement of the particles of the medium from their original mean positions on passing a wave through the medium is called the amplitude of the wave. Its SI unit is metre (m).
• The relationship between speed (v), frequency (v)and wavelength of a wave is given by v = vλ
• The measure of sound energy reaching the ear per second is known as loudness. It is measured in decibels (dB).
• The amount of sound energy passing each second through the unit area is known as the intensity of sound. Its SI unit is watt per square metre (W/m²).
• Pitch is that characteristic of sound by which we can distinguish between the different sounds of the same loudness.
• The quality or timbre of sound enables us to distinguish one sound from another having the same pitch and loudness.
• The speed of sound is greatest in solids, then in liquids and least in gases.
• When the speed of any object exceeds the speed of sound, it is said to have supersonic speed.

The bouncing back of sound when it strikes a hard surface is known as the reflection of sound. c Laws of Reflection of Sound

1. The incident sound wave, reflected sound wave and the normal at the point of incidence, all lie in the same plane.
2. The angle of incidence of sound is equal to the angle of reflection of sound.

The repetition of sound caused by the reflection of sound waves is called an Echo.

• The persistence of sound in a big hall due to repeated reflections from the walls, ceiling and floor is known as reverberation.
• The audible range of sound for the human ear is from 20 Hz to 20000 Hz.
• The sound of frequencies lower than 20 Hz are known as infrasonic sounds, which cannot be heard by humans.
• The sounds of frequencies higher than 20000 Hz are called ultrasonic sounds, which cannot be heard by human beings.
• Ultrasound is used to clean parts located in hard-to-reach places such as spiral tubes, and odd-shaped machines.
• Ultrasound is also used for diagnosing heart diseases by scanning the heart from the inside. This technique is called echocardiography.
• The method used by some animals like bats, tortoises and dolphins to locate objects by hearing the echoes of their ultrasonic squeaks is known as echolocation.

Improvement In Food Resources

Food is an essential organic substance required for the growth and proper functioning of all living organisms. It provides nutrients like carbohydrates, proteins, fats, vitamins, and minerals for the growth, development, and maintenance of large populations present on the Earth. To meet ever ever-increasing demand for food, improvement in food resources is required.

Need for Improvement In Food Resources in India

1. India is a populous country. Its population is over one billion people and it is still growing. To feed this growing population, we will soon require more than a quarter of a billion tonnes of grain every year.
2. This requirement can be fulfilled by farming on more land, but India is already intensively cultivated. So, we do not have a scope of increasing the area of land under cultivation. It is, therefore, necessary to increase our production efficiency for both crops and livestock.
3. To meet these requirements, the Green Revolution has contributed to increasing food grain production. The white revolution has led to better availability of milk. Scientific research has also contributed to these revolutions.
4. These revolutions mean that our natural resources are being used more intensively. Due to this, there are more chances of causing damage to our natural resources, and environment and disturbing its balance completely.
5. We should increase our food production without degrading our environment and disturbing its balance. This can be done by incorporating sustainable practices in agriculture and animal husbandry.
6. High yields from farms can be obtained easily by undertaking scientific management practices. It includes mixed farming, intercropping, and integrated farming practices (For Example. a combination of agriculture practices with livestock/ poultry/ fisheries/ bee-keeping).

Types of Crops

1. Crops are cultivated By- humans hemps hu their own benefit. The important types of crops ate:
2. Cereal crops These plants are cultivated to provide daily energy requirements, For Example. wheat, rice, maize, millet, and Sorghum (provide carbohydrates for energy requirement).
3. Pulses These plants are cultivated to fulfill protein requirements, For Example. gram (chana), pea (matar), black gram (urad), green gram (moong), pigeon pea (arhar), lentil (masoor), etc.
4. Oilseed crops These plants provide necessary fats and oils, For Example, soybean, groundnut, sesame, castor, mustard, linseed, and sunflower.
5. Vegetables, spices, and fruits These plants fulfill the requirement of a variety of vitamins and minerals with small quantities of proteins, carbohydrates, and fats, For Example. cabbage, onion, pepper, etc.
6. Fodder crops These plants are raised as food for the livestock, For Example. berseem, oats sudan grass, etc.
7. Each crop requires different climatic conditions, temperature and photoperiods for their growth and completion of life cycle. Growth of plants and flowering depend on the duration of sunlight. Plants also need sunlight to perform photosynthesis (photoperiods).

Classification of Crops

1. Crops are classified on the basis of seasons as follows
2. Kharif crops These crops are grown in hot and rainy season from the month of June to October. For e.g. paddy, soybean, pigeon pea, maize, cotton, green gram, groundnut, black gram, etc.
3. Rabi crops These crops are grown in dry and winter season from the month of November to April. For Example. wheat, gram, pea, mustard, linseed, barley, etc.

Improvement In Crop Yield

The practices involved in farming to increase crop production can be divided into three stages:

The first is the choice of seeds for planting. The second is the nurturing of the crop plants. The third is the protection of the growing and harvested crops from loss.

Accordingly, the major groups of activities for crop yields i an be classified as:

1. Crop variety improvement
2. Crop production management
3. Crop protection management.

Crop Variety Improvement

• The main aim of this practice is to find a variety of crop, which can withstand different situations like high soil salinity, diverse climatic conditions and water availability (drought and flood).
• In order to accept the new varieties of crops, it is necessary that the variety should produce high yields under different conditions found in different areas. For this, farmers should be provided with good quality seeds of a particular variety. The seeds should be of the same variety and germinate under the same conditions.
• Varieties or strains of crops can be selected by breeding for various traits such as disease resistance, response to fertilizers, product quality, and high yields. A new variety developed with all such features is highly acceptable.

Ways for Improvement in Crop Variety

There are two ways to incorporate desirable characteristics into crop varieties. These are:

(1) Hybridisation It is the crossing between genetically dissimilar plants to produce a new type (hybrid) or High Yielding Variety (HYV).
It is further of following types:

•  AIntervarietal The cross is made between two plants belonging to different varieties of crops. It is the most common method used in plant breeding.
• Interspecific The cross is made between two plants belonging to different species of same genus.
•  Intergeneric The cross is made between plants belonging to different genera.

(2) Genetically modified crops It involves the manipulation of crop plants for increasing their yield, improving quality, sustainability, etc. Genetic manipulation provides desired characteristics in the crop.

Factors of Crop Variety Improvement

Some of the factors for which crop variety improvement is done are as follows

• Higher yield Variety improvement is done to increase the productivity of the crop per acre.
• Improved quality The definition of quality is different for different crops. For e.g. baking quality is important in wheat, protein quality in pulses, oil quality in oilseeds and preserving quality in fruits and vegetables.
• Biotic and abiotic resistance Biotic stresses (diseases, insects and nematodes) and abiotic stresses (drought, salinity, water logging, heat, cold and frost) affect crop production to a great extent. Varieties resistant to such conditions are always preferred as they help to improve crop production.
• Change in maturity duration Short duration or a period between sowing and harvesting makes a crop more economical. It allows the farmers to grow multiple rounds of crops in a year. It also reduces the cost of crop production. Uniform maturity makes the harvesting process easy. It also reduces losses during harvesting.
• Wider adaptability Developing varieties that can grow and adapt to different conditions help in stabilizing crop production. Thus, a single variety can be grown in different regions with different climatic conditions.
• Desirable agronomic characteristics These characteristics depict good growth and higher productivity in plants. Plants showing such characteristics are preferred more than others, e.g. tallness and profuse branching are preferred characters for fodder crops. Dwarfness is desired in cereals

Crop Production Management

1. It involves the control of various aspects of crop production for the best yield. It requires skilful dealing with almost all aspects of crop production.
2. It is the money or financial condition which allows farmers to take up different forming practices and agricultural technologies. There is a correlation between higher inputs and yields.
3. The purchasing capacity of former for inputs decides cropping system and production practices. Thus production practices can be grouped at three levels, ,.e. no cost, low cost and high cost production practices.

Crop production management includes the management of

1. Nutrient Management

• Like animals, plants also require nutrients for their growth and development. Nutrients are the inorganic elements, which are supplied to plants by air, water, and soil. There are sixteen essential nutrients for plants.
• Essential plant nutrients are divided into two categories which are as follows:

(1) Macronutrients These include six nutrients. They are utilized by plants in large quantities, hence, known as macronutrients. They are nitrogen, phosphorus, potassium, calcium, magnesium and sulfur.

(2) Micronutrients These include seven nutrients. They are required by plants in smaller quantities. They are iron, manganese, boron, zinc, copper, molybdenum, and chlorine.

• Deficiency of any of these nutrients affects physiological processes in plants including reproduction, growth, and susceptibility to disease. Nutrients can be supplied to the soil m the form of manure and fertilizers. It helps to increase the yield of crops.
• Manures are natural fertilizers. These are the organic sustances formed by the decomposition of animal-excited plant wastes. It supplies small quantities of nutrients to
• Based on the type of biological marching used, manure can be classified as:

(1) Compost and vermicompost The process in which Farm waste materials like livestock excreta (cow dung, etc), vegetable waste, animal refuse, domestic waste, sewage waste, straw, eradicated weeds, etc., are decomposed in pits is known as composting.

• The compost is rich in organic matter and nutrients.
• The preparation of compost by using earthworms to hasten the process of decomposition of plant and animal refuse is called vermicomposting.

(2) Green manure Some plants like sunhemp or guar are grown and mulched by ploughing into the soil before sowing of the crop seeds. These green plants turn into green manure. It helps in increasing nitrogen and phosphorus content in the soil. It also helps to improve hydration, aeration and crumb structure of the soil.

The advantages of manure are as follows:

• Manure enriches soil with nutrients and organic matter (called humus).
• Manure increases soil fertility and decreases the harmful effects of pesticides and insecticides on soil.
• It helps in improving soil structure by increasing the water-holding capacity in sandy soil. In clayey soil, a large quantity of organic matter helps in drainage and avoiding water logging.
• By the use of biological waste material (manure), we can protect the environment from excessive use of fertilisers.
• Manure helps in the recycling of farm waste.

Fertilizers: They are commercially produced plant nutrients. They supply Nitrogen, Phosphorus, and Potassium (NPK) to the soil.

The advantages of fertilizers are as follows:

• They are easily available, easy to use, and store.
• They help in the higher yields of high-cost farming.
• They are used to ensure good vegetative growth (leaves, branches, and flowers) and give rise to healthy plants.

The disadvantages of fertilizers are as follows:

1. They need to be applied carefully in terms of proper dose, time, and looking after the pre and post-application precautions for their complete utilization.
2. For example, excessive use of fertilizers can cause water pollution as they get washed away when they arc not absorbed fully by the plants due to excessive irrigation.
3. Continuous use of fertilizers can destroy soil fertility because the organic matter in the soil does not get replenished. Hence, microorganisms in the soil are harmed by the use of fertilizers.
4. They provide short-term benefits. Thus, for maintaining soil fertility, short-term benefits of using fertilisers and long-term benefits of using manure must be considered in order to aim optimum yields in crop production.

2. Irrigation

• The process of supplying water to crop plants in fields by means of canals, reservoirs, wells, tubewells, etc., is called irrigation. Agricultural practices in India are rain-dependent.
• The success of a crop mainly depends on timely monsoons and sufficient rainfall during its growing season. Ensuring that water will be supplied to the crops at the right stages and in the required amounts, the expected yields of any crop can be increased.
• Farmers depend on various natural resources like ponds, wells, canals, etc., for the irrigation of their farmlands.
• Some commonly used irrigation systems depending on the type of water resources available for agricultural purposes are as follows:
• Wells: They are constructed wherever groundwater is present for irrigation. They are of two types:
• Dug wells Water is collected from water-bearing strata.
• Tube wells Water can be drawn from deeper strata using pumps.

Canals: They are an elaborate and the extensively used method of receiving water from reservoirs like dams or rivers. The main canal is further divided into other branches that have distributaries to irrigate fields.

River Lift Systems: This method is used in areas, where canal flow is insufficient or irregular due to insufficient reservoir release. Here, water is directly drawn from the rivers for supplementing irrigation in areas close to rivers.

Tanks: These are small storage reservoirs. They catch and store the run-off of smaller catchment areas.

Modern Techniques

• These are fresh initiatives for increasing water availability for agriculture by augmenting groundwater. They include
• Rainwater harvesting Rainwater is collected into the ground by digging tunnels, etc. This water percolates into the soil, thus maintaining the water table.
• Watershed development Small check-dams are built to increase groundwater levels. The purpose of check-dams is to stop the rainwater from flowing away and also to reduce soil erosion.

Cropping Patterns: It involves raising crops so as to obtain maximum benefit from the same piece of land. It reduces the risk of crop failure, disease, etc. For this purpose, crops can be grown in different ways. Some of them are:

Mixed Cropping: In this practice, two or more crops are grown simultaneously on the same piece of land. For Example Wheat + Gram, Wheat + Mustard, Groundnut -(-Sunflower, etc. Some advantages of mixed cropping are:

• Improves soil fertility.
• The risk of total crop failure due to uncertain monsoon is reduced.
• Gives some insurance against the failure of one of the crops.

Intercropping

• In this, two or more crops are grown simultaneously on the same field in a definite pattern. A few rows of one crop alternate with a few rows of another crop. The crops are selected be on the basis of their nutrient requirements.
• Two crops must have different nutrient requirements from each other. For example. Soybean + Maize, Finger millet (bajra) 4-Cowpea (lobia), etc.

Some advantages of intercropping are:

1. It ensures maximum utilization of supplied nutrients and better returns.
2. It prevents the spread of pests and diseases to all the plants of one crop in a field.
3. Both crops give better returns in it.

Crop Rotation

1. In this type of practice, different crops are grown on a piece of land in a pre-planned succession. The crop combination depends upon the duration of the crops.
2. One crop is grown on a field and after its harvest, a second crop is grown on the same field. This can also follow a third crop. The crop to be chosen after one harvest depends upon the availability of moisture and irrigation facilities.

Crop Protection Management

• In fields, crops have to be protected from weeds, insects, pests, and diseases. Crop protection management includes methods to reduce such kinds of infestation.
• If not controlled in time, they can cause heavy damage to the crops in a way that most of the crops are lost.
  Various threats to crops include:

1. Weeds

These are the unwanted plants in the cultivated field. They compete with the crops for food, space, and light.

Weeds take up nutrients and reduce the growth of the crop. Therefore, they should be removed during the early stages of crop growth in order to obtain a good harvest.

Examples of weeds are:

• Xanthium (gokhroo), Parthenium (gajar ghas), Cyperinus rotundus (motha), Amaranthus, Chenopodium, wild oat, etc. The following are the methods to control weeds:
• Mechanical methods Uprooting, weeding with harrow or hand, ploughing, burning, and flooding.
• preventive methods of Proper seedbed preparation, timely sowing of crops, intercropping, and crop rotation.

2. Insect Pests

• They affect the health of the crop and reduce its yield. Inseam pests attack the plants in the following ways:
• They cut the root, stem, and leaf, For Example. locusts.
• They suck the cell sap from various parts of the plant, For Example. aphids.
• They bore into stems and fruits, For Example. shoot borer larvae. Pests can be controlled in many ways such as:

Use of resistant varieties.

• Summer plowing In this method, fields are
• plowed deep during summers to destroy weeds and pests.

3. Crop Diseases

• Diseases in plants are caused by pathogens such as bacteria, fungi, and viruses. These pathogens are present in and transmitted through the soil, water, and air.
• Crop diseases can be controlled by the use of pesticides like herbicides, insecticides, and fungicides. They are sprayed on crop plants in limited amounts. Excessive use of these chemicals is harmful to many species of plants and animals. It also causes environmental pollution.

Storage of Grains

• During the storage of grains, high losses can occur in agricultural produce.
• Factors responsible for such losses can be categorized as:
• Biotic factors These include rodents, fungi, insects, mites, and bacteria.
• Abiotic factors These include inappropriate moisture and temperature conditions in the place of storage.
• The effects of these factors on grains are as follows:
• Degradation in quality.
• Poor germinative capacity.
• Discoloration of the produce.
• Loss in weight.
• All these lead to poor marketability and heavy economic losses. Some of the preventive and control measures during storage are:
• The proper storage of grains can be done by proper treatment and systematic management of warehouses.
• Strict cleaning of the produce before storage.
• Proper drying of the produce in sunlight and then in shade.
• Fumigation should be done to kill pests. In fumigation, the insect pests are exposed to fumes of chemicals.

Animal Husbandry

Animal husbandry is the scientific management of livestock. It can be defined as the science of rearing, feeding, breeding, disease control, and utilization of animals. Animal-based farming includes cattle, goat, sheep, poultry, and fish farming.

Need of Animal Husbandry

It is required to meet the increasing demands of animal-based goods like milk, meat, eggs, leather, etc., according to the size of the population and the living standards of the people.

It sets guidelines for proper management and a systematic approach to animal rearing.

Cattle Farming: In India, cattle husbandry is done for two purposes; milk and draught labor for agricultural work (such as tilling, irrigation, and caning).

Cattle in India belong to two different species:

• Bos indices (cows)
•  Bos bubalis (buffaloes).

On the basis of the work done by cattle, they can be divided into two categories:

• Milch animals These are milk-producing females or dairy animals.
• Draught animals These are used to do labor work on farms.

Breeds of Cattle

1. Indigenous or local breeds are selected because of their high resistance to disease, For Example. Redsindhi and Sahiwal.
2. Exotic or foreign breeds They are selected because of their long lactation period, For Example. Jersey and Brown Swiss.

These two breeds can be cross-bred to get both the desirable qualities in animals.

Lactation Period: It is the period of milk production after the birth of a calf. Milk production largely depends on the duration of the lactation period. We can increase the milk production by increasing the lactation period.

Farm Management for Cattle

• Efficient farm management is essential for humane farming, better health of animals, and production of clean milk. Various measures for farm management are as follows:
• Proper cleaning and shelter facilities are required for cattle.
• Regular brushing of animals should be done to remove dirt and loose hair.
• The cattle should be sheltered in well-ventilated roofed sheds in order to protect them from rain, heat, and cold.
• The floor of the catde shed should be sloping so as to keep it dry and facilitate cleaning.

Food Requirements of Cattles

Food is required for dairy catdes for following two purposes:

• For maintenance, Food is required to support the animal to live a healthy life.
• For producing milk The type of food is required during the lactadon period.

Different types of animal feeds are:

• Roughage This is largely fibrous and contain low nutrients. For example. green fodder, silage, hay and legumes. ‘
•  Concentrates These are low in fibre. They contain relatively high levels of proteins and other nutrients. For example. cereals like gram and bajra.

Apart from the above mentioned products, some feed additives containing micronutrients promote the health and milk output of dairy animals. It should also be noted that catde should be given balan^d rations with all the nutrients in proportionate amounts.

Diseases in Cattles

Like other animals, catdes also suffer from a number of diseases. These besides causing death, also reduce milk production.

The parasites of catde can be of following types:

• External parasites They live on the skin and cause skin diseases, For Example. lice, mites, etc.
•  Internal parasites They include worms that affect stomach and intestine and flukes that damage the liver.

Cattles also get infectious iliseascs from various bacteria and v,ms. As preventive measure, vaccinations are given to farm animals against many viral and bacterial diseases.

Poultry Farming

It involves rearing of domestic fowl for the production of eggs and chicken meat. Therefore, improved poultry breeds are developed and Firmed to produce layers for eggs and broilers for meat.

For the improvement of poultry breeds, cross-breeding is done successfully between Indian or indigenous (For Example. Aseel) and foreign or exotic (For Example. Leghorn) breeds.

These cross-breeding programs focus to develop desirable traits like:

• Quality and quantity (number) of chicles.
•  Dwarf broiler parent for commercial chick production.
• Summer adaptation capacity/tolerance to high temperature.
•  Low maintenance requirements.
• Reduction in the size of egg-laying bird with the ability to utilise more fibrous and cheaper diets. This diet is formulated using agricultural byproducts.

Egg and Broiler Production

• Broiler chickens are fed with vitamin-rich supplementary feed for good growth rate and better feed efficiency.
• Care is taken to avoid mortality and to maintain feathering and carcass quality. They are produced as broilers and sent to market for meat purposes.
• Broilers and egg layers have different housing, nutritional and environmental requirements.
• The diet of broilers is rich in protein with adequate fat. In the poultiy feed, the level ofvitamin-A and K is kept high.

Maintenance of the Shelter

The following practices are required for the maintenance of shelter:

• Proper cleaning and sanitation of the shelter.
• Maintenance of temperature and hygiene in the shelter.
• Proper ventilation.
• Prevention and control of diseases and pests.
• Poultry Diseases and Their Prevention
• Poultry fowl suffer from various diseases caused by virus, bacteria, fungi and parasites.
• They also suffer from nutritional deficiency diseases.
• These diseases can be prevented by:
• Providing nutritional diet to poultry birds.
• Proper cleaning and sanitation of shelter.
• Vaccination of poultry birds can prevent the occurrence of infectious diseases. Loss of poultry during an outbreak of disease can also be reduced by proper vaccination.
• Spraying of disinfectants at regular intervals in the shelter.

Marine fishes of high economic value that are formed in seawater are:

1. Finned fishes Mullets, bhetki and pearl spots.
2. Shell fishes Prawns, mussels, oysters and seaweeds. Oysters are also cultivated for the pearls they produce.

Meld of fishes can be increased by locating large schools of fish in the open sea with the use of satellites and echosounders.

Inland Fisheries

• It includes fishery in freshwater and brackish water. Freshwater resources include canals, ponds, reservoirs and rivers. Brackish water resources are those where seawater and freshwater mix together, For Example. estuaries and lagoons. These are also important fish reservoirs.
• The yield of capture fishing is not high in such inland water bodies. Thus, most fish production from these resources is done through aquaculture.
• Sometimes fish culture is done in combination with rice crops. In this, paddy crop gets ample of water and fishes get food.

Composite Fish Culture (Polyculture)

Fish production by cultivating a single species (monoculture) gives a low yield and demands higher cost. In composite fish culture, a combination of 5 or 6 fish species is cultivated in a single pond having different food habits. Due to this, they do not compete for food with each other. Thus, it helps in more intensive fish farming.

Advantages of Composite Fish Culture

1. Both local and imported fish species are used.
2. Due to different food habits all the food in pond is consumed by the fishes.
3. The fish yield from pond is high as their is no competition for food. For example. Catla is surface feeder, Rohu feeds in the middle-zone of the pond, Mrigal and common carps are bottom feeders, grass carp feeds on weeds in the pond.

Disadvantages of Composite Fish Culture

Many of the fishes breed only during monsoon. Thus, one of the major problem of fish farming is the lack of availability of good quality seed. To overcome this problem, fishes are breed in ponds using hormonal stimulation. It ensures the supply of pure fish seed in desired quantities.

Bee-Keeping

Honey is being widely used for various purposes. Thus, its production has become an agricultural enterprise these days. It is scientifically known as apiculture.

It is the method of rearing, care and management of honeybees for obtaining bee products like honey, bee wax (used in medicinal preprations) etc. For commercial honey production, apiaries or bee farms are established.

Advantages of Bee-Keeping

1. Requires low investment.
2. Provides varied products like honey (for eating or making other products), wax (used in medicinal and cosmetic preparations), bee venom, etc.
3. Acts as an additional source of income for farmers. (hi) Helps in increasing crop yield by better pollination.
4. Out of the above mentioned species, A. mellifera has been brought in the country in order to increase the yield of honey. This is the main variety used for the commercial honey production.

Advantages of Italian Bees

1. They have high honey collection capacity.
2. They sting somewhat less.
3. They can stay in a given beehive for long periods and breed well.

Honey

1. It is the major product that is obtained from apiculture.
2. Value of honey It depends on pasturage or flowers available to bees for nectar and pollen collection.
3. Taste of honey It depends on adequate quantity of pasturage and kind of flowers available.

Improvement In Food Sources Question And Answers

Question 1. What do we get from cereals, pulses, fruits and vegetables?
Answer:

Cereals (wheat, rice, maize, etc) are the sources of carbohydrates, which provide energy. Pulses (pea, gram and soybean, etc.) are the source of proteins. Vegetables and fruits provide us vitamins and minerals in addition to small amount of carbohydrates, proteins and fats.

Question 2. How do biotic and abiotic factors affect crop production?
Answer:

Factors which affect crop production are as follows:

1. Biotic factors that cause loss of grains are rodents, pests, insects, etc.
2.  Abiotic factors include drought, salinity, water logging, heat, cold and frost.

Both biotic and abiotic factors cause stresses on crop and affect crop production in the following ways:

1. Poor seed germination
2. Infestation of insects
3. Low yield
4. Discolouration of leaves

Question 3. What are the desirable agronomic characteristics for crop improvements?
Answer:

Desirable agronomic characteristics in crop plants help to give higher productivity. For example:

1. Tallness and profused branching are desirable characters for fodder crops.
2. Dwarfness is desired in cereals, so that less nutrients are consumed by these crops.

Question 4. What are macronutrients and why are they called macronutrients?
Answer:

Macronutrients are essential nutrients required for growth, functioning and survival of plants. They are so called because they are required in large amounts by plants. They are six in number, i.e.

• Nitrogen
• Phosphorus
• Potassium
• Calcium
• Magnesium
• sulphur

Question 5. How do plants get nutrients?
Answer:

The nutrients to the plants are supplied by soil, which are absorbed by roots of plants. Some nutrients are provided by air and water too.

Question 6. Compare the use of manure and fertilisers in maintaining soil fertility.
Answer:

Effects of the use of manure in maintaining soil fertility are as follows:

1. Manures provide a lot of organic matter (humus) to the soil. Humus helps to restore water retention capacity of sandy soil and drainage in clayey soil.
2.  These are the sources of soil organisms like soil friendly bacteria.

Effects of fertilisers on soil quality are:

1. Use of excess fertilisers leads to dryness of soil and the rate of soil erosion increases.
2. Due to continuous use of fertilisers, the organic matter decreases. It reduces porosity of the soil and the plant roots do not get sufficient oxygen.

Question 7. Which of the following conditions will give the most benefits? Why?

1. Farmers use high quality seeds, do not adopt irrigation or use fertilisers.
2. Farmers use ordinary seeds, adopt irrigation and use fertilisers.
3. Farmers use quality seeds, adopt irrigation, use fertilisers and use crop protection measures.

Answer:

Condition will give most benefits because

good quality seeds will give good yield.

irrigation methods will overcome drought and flood situation.

fertilisers fulfil the nutrient requirement of the soil, providing high yield.

crop protection method protects the plants from weeds, pests and pathogens.

Question 8. Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:

Preventive measures and biological control methods should be preferred for protecting crops because :

1. they are simple.
2. they are economical as they involve less financial investment.
3. they minimise pollution and are ecologically safe.
4. they minimise the adverse effects on soil fertility.
5. they are harmless to other living organisms.
6. they are target specific.

Question 9. What factors may be responsible for losses of grain during storage?
Answer:

The major factors responsible for losses of grain during storage are:

1. Biotic factors They include attack from insects, rodents, fungi, mites and bacteria.
2. Abiotic factors They include inappropriate moisture and temperature in the place or storage.

Question 10. Which method is commonly used for improving cattle breeds and why?
Answer:

Cross breeding is a method commonly used for improving cattle breeds. It is the process in which indigenous varieties of cattle are crossed with exotic breeds to get a cross-breed which is of desired qualities.

Question 11. Discuss the implications of the following statement: “It is interesting to note that poultry is India’s most efficient converter of low fibre foodstuff (which is unfit for human consumption) into highly nutritious animal protein food”.
Answer:

• Poultry birds are efficient converters of agricultural byproducts and fibrous wastes into high quality meat. As, the waste which is unfit for human consumption can be formulated into cheaper diets for poultry Birds.
• Also, they help in providing eggs, feather and nutrient rich manure. So, the mentioned statement is implicit for poultry birds.

Question 12. What management practices are common in dairy and poultry farming?
Answer:

The common management practices include:

1. Keeping the shelter well-designed, ventilated and hygienic.
2. The animals and birds are given healthy feed with balanced nutrition.
3. Both animals and birds must be protected from various diseases. Regular check-up should be done.

Question 13. What are the differences between broilers and layers and in their management?
Answer:

• A broiler is a poultry bird specially kept for obtaining meat. Layer is a poultry bird that gives eggs. There is a difference in their housing, nutrition and environmental requirements.
• The daily food requirement of broilers is somewhat different from those of layers. Broilers require protein rich food with adequate fat and high amount of vitamin-A and K. Layers require feed with vitamins, minerals and micronutrients and enough space and proper lighting.

Question 14. How are fish obtained?
Answer:

Fishes are obtained either by capturing them from their natural resources or by culturing them by fish farming.

Question 15. What are the advantages of composite fish culture?
Answer:

Advantages of composite fish culture are as follows:

1. Fishes selected for this culture differ in their feeding habits and thus, avoid competition for food between them.
2. All these species together use all the food in the pond without competing with each other.
3. This increases the fish yield from the pond.

Question 16. What are the desirable characters of bee varieties suitable for honey production?
Answer:

The desirable characters of bee for honey production are as follows:

1. The bee should have good honey collection capacity.
2. They should be stingless and breed very well.
3. They should be able to stay in a beehive for long periods.

Question 17. What is pasturage and how is it related to honey production?
Answer:

Pasturage includes the plants and trees found around an apiary. From them, nectar and pollen are collected by bees to form honey.

Pasturage plays an important role in determining the quantity and quality of honey.

1. The quality of honey depends upon the pasturage.
2.  Kinds of flowers determine the taste of honey.

Exercises

Question 1. Explain any one method of crop production, which ensures high yield.
Answer:

Intercropping is a method of crop production, which ensures high yield. During this, two or more crops having different nutrient requirements are grown simultaneously on the same field in a definite pattern.

Question 2. Why are manures and fertilisers used in fields?
Answer:

Manures and fertilisers are used to improve soil fertility and increase crop productivity. They replenish deficient nutrients in the soil.

Question 3. What are the advantages of intercropping and crop rotation?
Answer:

Advantages of intercropping are:

1. Crops selected in this method differ in their nutrient requirements. This ensures maximum utilisation of the supplied nutrients.
2.  It prevents pests and diseases from spreading to all the plants belonging to one crop in a field. Advantages of crop rotation are:
3. It makes the soil fertile and increases the yield from a single field.
4. It reduces the use of fertilisers. For example. use of nitrogenous fertilisers is not required as leguminous plants that are grown in crop rotation help in biological nitrogen-fixation.
5. It helps in the replenishment of soil fertility

Question 4. What is genetic manipulation? How is it useful in agricultural practices?
Answer:

• Genetic manipulation is the incorporation of desirable characters into an organism by hybridisation, mutation, DNA recombination, etc.
• By genetic manipulation, improved varieties of seeds can be obtained having desirable characters like high yield, disease resistance and better adaptability.

Question 5. How do good animal husbandry practices benefit farmers?
Answer:

Animal husbandry involves the scientific management of the farm animals. Its benefits to farmers are:

1. Improvement of the breeds having good desirable characters.
2. Better yield in quantity and quality.
3.  Reduction of input cost.

Question 6. What are the benefits of cattle farming?
Answer:

The main benefits of cattle farming are as follows:

1. We get milk from cattle. Various milk products can be manufactured with this milk.
2.  Cattle can be employed for labour work in agricultural fields for tilling, irrigation and carting.

Question 7. For increasing production, what is common in poultry, fisheries and bee-keeping?
Answer:

Cross-breeding is the most important practice to increase the production and that too of desired characteristics.

Question 8. How do you differentiate between capture fishing, mariculture and aquaculture?
Answer:

Differences between capture fishing, mariculture and aquaculture are as follows:

Summary

• Improvement in food resources is essential to obtain higher yield to fulfil the need of food for continuously increasing population.
• The green revolution has contributed in increasing the food grain production, while the white revolution has led to better availability of milk.
• High yields from farm can be obtained by undertaking scientific management practices like mixed cropping, intercropping and integrated farming practices.
• Improvement in crop yield are practices involved in farming to increase crop production. In includes crop variety improvement, crop production management and crop protection management.
• Crop variety improvement is a practice to find an improved variety of crop, which can withstand different situations like soil quality, different weather conditions, water availability (drought and flood) and can ultimately give good yield.
• Hybridisation and genetic manipulations are the two ways to incorporate desirable characteristics into crop varieties. Factors for which crop variety improvement is done are higher yield, improved quality, resistance against biotic and abiotic factors, change in maturity duration, wider adaptability and desirable agronomic characteristics.
• Crop production management involves the control of various aspects of crop production for the best yield. It includes nutrient management, irrigation and cropping patterns
• Nutrient management includes adopting various methods to increase the nutrient level in the soil. This is done by adding manures and fertilisers in the field.
• Irrigation is the process of supplying water to crop plants in fields by means of canals, reservoirs, wells and tube wells. Cropping patterns involve raising crops so as to obtain maximum benefit from the same piece of land, reducing the risk of crop failure, disease, etc. It can be done by mixed cropping, intercropping and crop rotation.
• Crop protection management involves the protection of crops from weeds, insects, pests and disease causing organisms. It includes methods to reduce such kinds of infestation. If not controlled in time, they can cause heavy damage to crops.
• Grains are affected by biotic and abiotic factors. Thus, proper measures should be adopted for their storage, For Example. fumigation. Animal husbandry is the scientific management of livestock.
• It is animal based farming of cattle, goat, sheep, poultry and fish.
• 1 Cattle farming is done for milk and drought labour by cattles. Poultry farming is the method of rearing fowls for the production of meat and egg. It aims to improve poultry breeds.
• Fish production refers to capturing and culturing of fishes as a suppliment of animal protein for humans. It is a cheep source of animal protein for our food.
• Bee keeping is scientifically known as apiculture. It is rearing, care and management of honeybees for obtaining honey, wax, etc. For commercial honey production, apiaries or bee farms are established.