Selina ICSE Solutions for Class 9 Physics

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion

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Selina Concise Physics Class 9 ICSE Solutions Laws of Motion

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Selina ICSE Solutions for Class 9 Physics Chapter 3 Laws of Motion

Exercise 3(A)

Solution 1S.

a. The forces which act on bodies when they are in physical contact are called contact forces.
Example: frictional force and force exerted on two bodies during collision.

b. The forces experienced by bodies even without being physically touched are called non-contact forces.
Example: Gravitational force and Electrostatic force.

Solution 2S.

Contact force: (a) frictional force (b) normal reaction force (c) force of tension in a string
Non-contact force: (d) gravitational force (e) electric force (f) magnetic force

Solution 3S.

a. Force exerted on two bodies during collision.
b. Magnetic force between magnetic poles.

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 1
C. The forces acting on the block are its weight in the downward direction and the normal reaction force due to the table on the upward direction.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 2

Solution 5S.

The magnitude of non-contact force on two bodies depends on the distance of separation between them.
The force decreases as the distance of separation increases.
The force is inversely proportional to the square of the distance of separation.

Solution 6S.

The magnitude of gravitational force between two masses will become four times as gravitational force varies inversely as the square of distance of separation.

Solution 7S.

A force when applied on a non-rigid body changes the inter-spacing between its constituent particles and therefore causes a change in its dimensions and can also produce motion in it.
On the other hand, a force when applied on a rigid body, does not change the inter-spacing between its constituent particles and therefore it does not change the dimensions of the body but causes motion in it

Solution 8S.

i. A fielder on the ground stops a moving ball by applying a force with his hands.
ii. The pull exerted by horse makes a cart moves.
iii. In a cycle pump, when the piston is lowered, the air is compressed to occupy a less volume.
iv. On pressing a piece of rubber, its shape changes.

Solution 1M.

Frictional force is a contact force.

Solution 2M.

Force due to gravity is a non-contact force.

Exercise 3(B)

Solution 1S.

Force causes motion in a body.

Solution 2S.

Force is not needed to keep a moving body in motion.

Solution 3S.

The force of friction between the table and the ball opposes the motion of the ball.

Solution 4S.

In absence of any external force, its speed shall remain unchanged.

Solution 5S.

Galileo’s law of inertia states that a body continues to be in its state of rest or of uniform motion unless an external force is applied on it.

Solution 6S.

According to Newton’s first law of motion, if a body is in a state of rest, it will remain in the state of rest, and if the body is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

Solution 7S.

Solution 8S.

The property of an object by virtue of which it neither changes its state nor tends to change the state is called inertia.

Solution 9S.

Force is that external cause which can move a stationary object or which can stop a moving object.

Solution 10S.

Inertia of a body depends on its mass. Inertia is directly proportional to mass, i.e. greater the mass of a body, greater is its inertia.

Solution 11S.

Examples to show that greater the mass, greater is the inertia of the body are as shown below:

  1. If you want to start a car by pushing it, you find that it takes a very large force to overcome its inertia. On the other hand, only a small force is needed to start a child’s express wagon. The difference between the car and express wagon is the difference in mass. The car has a large mass, whereas the wagon has a small one.
  2. A cricket ball is more massive than a tennis ball. The cricket ball acquires a much smaller velocity than a tennis ball when the two balls are pushed with equal force for the same time.

Solution 12S.

It is difficult, i.e. a larger force is required to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass).

Solution 13S.

Two kinds of inertia are as listed below:

  1. Inertia of rest.
  2. Inertia of motion.

Solution 14S.

Examples of inertia of rest: A coin placed on top of a card remains in place when the card is slightly and quickly jerked horizontally.
Example of inertia of motion: A ball thrown vertically upwards by a person in a moving train comes back to his hand.

Solution 15S.

No, the body will not move because the net force acting on it is zero. Hence, it will remain stationary due to inertia of rest.

Solution 16S.

The motion remains unaffected because the net force acting on it is zero.

Solution 17S.

The net force on the airplane is zero or the upward force is equal to the downward force.

Solution 18S.

If a person jumps out of a moving train and tries to stop immediately, he falls due to inertia of motion. This is because his body tends to move forward with the velocity of the train while his feet are stationary.

Solution 19S.

The reason is that when the card is flicked, a momentary force acts on the card, so it moves away. However, the coin kept on it does not share the motion at once and it remains stationary at its place due to the inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.

Solution 20S.

The reason is that when the ball is thrown, the ball is in motion along with the person and train. Due to the inertia of motion, during the time the ball remains in air, the person and ball move ahead by the same distance. This makes the ball fall back into the thrower’s hand.

Solution 21S.

(a) When a train suddenly starts, the passengers tend to fall backwards. This is because the lower part of the body, which is in contact with the train, begins to move while the upper part of the body tends to maintain its position of rest. As a result, the upper part tends to fall backwards.

(b) The frame of the sliding door being in contact with the floor of the train also comes in motion on start of the train, but the sliding door remains in its position due to inertia. Thus, the frame moves ahead with the train, while the door slides opposite to the direction of motion of the train. Thus, the door may shut or open accordingly.

(c) When the branches of the tree are shaken, they come in motion, while the fruits due to inertia remain in a state of rest. Thus, the larger and weakly attached fruits get detached from the branches and fall down due to the pull of gravity.

(d) When people alight from a moving bus, they continue to run alongside the bus to avoid falling. If they were to stop at once, the feet would come to rest suddenly but the upper part of the body would still be in motion and they would tend to fall forward.

(e) The part of the carpet where the stick strikes comes in motion at once, while the dust particles settled on it remain in the state of rest due to inertia of rest. Thus, a part of the carpet moves ahead with the stick leaving behind the dust particles that fall down due to gravity.

(f) When running, the athlete brings his body in the state of motion. When the body is in motion, it becomes easier to take a long jump.

Solution 1M.

A truck

Solution 2M.

Less force is required for the tennis ball than for the cricket ball.

Solution 3M.

Change the state of motion or state of rest of the body.

Exercise 3(C)

Solution 1S.

Force needed to stop a moving body in a given time depends on its mass and velocity.

Solution 2S.

Linear momentum of a body is the product of its mass and velocity.
Its SI unit is kgms-1.

Solution 3S.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 3

Solution 4S.

Let a force ‘F’ be applied on a body of mass m for a time ‘t’ due to which its velocity changes from u to v. Then,
Initial momentum of body = mu
Final momentum of body = mv
Change in momentum of the body in ‘t’ seconds = mv – mu = m (v – u)
Rate of change of momentum = Change in momentum/time
= [m (v – u)]/t
However, acceleration a = Change in velocity/time = (v – u)/t
Therefore, rate of change of momentum = ma = mass × acceleration
This relation holds true when the mass of the body remains constant.

Solution 5S.

(i) Mass is the measure of inertia.
Let ‘m’ be the mass of the two bodies.
Inertia of body A:Inertia of body B :: m:m
Or, Inertia of body A:Inertia of body B :: 1:1

(ii) Momentum of body A = m (v)
Momentum of body B = m (2v) = 2mv
Momentum of body A:Momentum of body B :: mv:2mv
Or, Momentum of body A:Momentum of body B :: 1:2.

Solution 6S.

(i) Inertia of body A:Inertia of body B :: m:2m
Or, Inertia of body A:Inertia of body B :: 1:2.

(ii) Momentum of body A = m ( 2v) = 2mv
Momentum of body B = (2m) v = 2mv
Momentum of body A:Momentum of body B :: 2 mv:2mv
Or, Momentum of body A:Momentum of body B :: 1:1.

(iii) According to Newton’s 2nd law of motion, rate of change of momentum is directly proportional to the force applied on it. Therefore,
Force needed to stop A:Force needed to stop B :: 1:1.

Solution 7S.

According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it and the change of momentum takes place in the direction in which the force is applied.
It gives the quantitative value of force, i.e. it relates the force to the measurable quantities such as acceleration and mass.

Solution 8S.

Newton’s first law of motion gives the qualitative definition of force. It explains the force as the cause of acceleration only qualitatively but Newton’s second law of motion gives the quantitative value of force. It states force as the product of mass and acceleration. Thus, it relates force to the measurable quantities such as acceleration and mass.

Solution 9S.

Mathematical expression of Newton’s second law of motion is as shown below:
Force = Mass x Acceleration

Above relation holds for the following conditions:
(i) When the velocity of the body is much smaller than the velocity of light.
(ii) When the mass remains constant.

Solution 10S.

According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it, and the change of momentum takes place in the direction in which the force is applied.

The relation F=ma holds for the following conditions:
(i) When the velocity of the body is much smaller than the velocity of light.
(ii) When the mass remains constant.

Solution 11S.

From Newton’s second law of motion, F = ma.
If F = 0, then a = 0.

This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, then it will remain in the state of rest and if it is moving, then it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force, does not change its state of rest or motion. This is the statement of Newton’s first law of motion.

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 4

Solution 13S.

If a given force is applied on bodies of different masses, then the acceleration produced in them is inversely proportional to their masses.
A graph plotted for acceleration (a) against mass (m) is a hyperbola.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 5

Solution 14S.

The S.I. unit of force is newton.
One newton is the force which acts on a body of mass 1kg and produces an acceleration of 1 m/s2, i.e. 1 N = 1 kg × 1 m/s2.

Solution 15S.

The C.G.S. unit of force is dyne.
One dyne is the force which acts on a body of mass 1 gramme and produces an acceleration of 1 cms-2, i.e. 1 dyne = 1 g × 1 cms-2.

Solution 16S.

The S.I. unit of force is newton and the C.G.S. unit of force is dyne.
1 N = 105 dyne.

Solution 17S.

When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously, i.e. in a very short time, so the floor exerts a large force on the vessel and it breaks. However, if it falls on a carpet, then the time duration, in which the vessel comes to rest, increases, so the carpet exerts less force on the vessel and it does not break.

Solution 18S.

(a) We pull our hands back while catching a fast moving cricket ball, because by doing so, we increase the time of catch, i.e. increase the time to bring about a given change in momentum, and hence, the rate of change of momentum decreases. Thus, a small force is exerted on our hands by the ball.

(b) When an athlete lands from a height on a hard floor, his feet comes to rest instantaneously, so a very large force is exerted by the floor on his feet, but if he lands on sand, his feet push the sand for some distance; therefore, the time duration in which his feet comes to rest increases. As a result, the force exerted on his feet decreases and he is saved from getting hurt.

Solution 1M.

Mv

Solution 2M.

N s

Solution 3M.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 6

Solution 4M.

Mass of the body

Solution 1N.

Mass of the body, m = 5kg
Velocity, v = 2 m/s
Linear momentum = mv = (5)(2) kg m/s
= 10 kg m/s

Solution 2N.

Linear momentum = 0.5 kg m/s
Mass, m = 50 g = 0.05 kg
Velocity = Linear momentum/mass
= 0.5/0.05 m/s
= 10 m/s

Solution 3N.

Force, F = 15 N
Mass, m = 2kg
Acceleration, a = F/m [ From Newton’s second law]
Or, a = (15/2) ms-2
Or, a = 7.5 ms-2

Solution 4N.

Force, F = 10 N
Mass, m = 5kg
Acceleration, a = F/m [ From Newton’s second law]
Or, a = (10/5) ms-2
Or, a = 2 ms-2

Solution 5N.

Mass, m = 0.5 kg.
Acceleration, a = 5 ms-2
Force, F = ma   [ From Newton’s second law]
Or, F = (0.5) (5) N = 2.5 N.

Solution 6N.

Force, F = 10 N
Mass, m = 2 kg
Time, t = 3 s
Initial velocity, u = 0 m/s.

(i) Let v be the final velocity acquired.
From Newton’s second law,
F = ma.
Or, a = F/m = 10/2 = 5 ms-2.

From the 1st equation of motion,
a = (v – u)/t
Or, v = at + u.
Or, v = (5)(3) + 0 = 15 m/s.

(ii) Change in momentum = Final momentum – initial momentum
Δp = mv – mu.
Or, Δp = m (v – u).
Or, Δp = 2 ( 15 – 0) = 30 kg m/s.

Solution 7N.

Mass, m = 100 kg
Distance moved, s = 100 m
Initial velocity, u = 0

(i) Because the body moves through a distance of 100 m in 5 s,
Velocity of the body = Distance moved / time taken
Velocity = (100/5) = 20 m/s

(ii) From Newton’s third equation of motion,
v2  u2 = 2as.
Or, a = (v2  u2) /2s.
Or, a = [ (202  02)/ 2(100) ] ms-2.
Or, a = 2 ms-2.

(iii) Force, F = ma
Or, F = (100) (2) N.
Or, F = 200 N.

Solution 8N.

Slope of a velocity-time graph gives the value of acceleration.
Here, slope = 20/5 = 4 m/s2.
Or, acceleration, a = 4 m/s2.
Force = Mass × Acceleration.
Given mass, m = 100 g = 0.1 kg.
Force = (0.1) (4) = 0.4 N.

Solution 9N.

Mass, m = 2 kg
Initial velocity, u = 0
Final velocity, v = 2 m/s
Time, t = 0.1 s
Acceleration = Change in velocity/time
Or, a = (v – u) /t
Or, a = (2 – 0)/ 0.1 = 20 ms-2.
Force = Mass x Acceleration
Or, F = (2) (20) = 40 N.

Solution 10N.

Mass, m = 100g = 0.1 kg.
Initial velocity, u = 30 m/s.
Final velocity, v = 0.
Time, t = 0.03 s.
Acceleration = Change in velocity/time.
Or, a = (v – u)/t.
Or, a = (0 – 30)/ 0.03 = -1000 ms-2.
Here, negative sign indicates retardation.
Now, Force = Mass x Acceleration
Or, F = (0.1) (1000) = 100 N.

Solution 11N.

Mass, m = 480 kg.
Initial velocity, u = 54 km/hr = 15 m/s.
Final velocity, v = 0.
Time, t = 10 s.
Acceleration = Change in velocity/time.
Or, a = (v – u)/t.
Or, a = (0 – 15)/10 = -1.5 ms-2.
Here, negative sign indicates retardation.
Now, Force = Mass x Acceleration
Or, F = (480) (1.5) = 720 N.

Solution 12N.

Mass, m = 50 gm = 0.05 kg.
Initial velocity, u = 100 m/s.
Final velocity, v = 0.
Distance, s = 2cm = 0.02 m.

(i) Initial momentum = mu = (0.05) (100) = 5 kg m/s.
(ii) Final momentum = mv = (0.05) (0) = 0 kg m/s.

(iii) Acceleration, a = (v2 – u2)/2s.
Or, a = (02 – 1002)/ 2(0.02).
Or, a = – 2.5 x 105 ms-2.
Therefore, retardation is 2.5 x 105 ms-2.

(iv) Force, F = ma
Or, F = (0.05 kg) (2.5 x 105 ms-2)
Or, F = 12500 N

Solution 13N.

Let the force be F.
Force F causes an acceleration, a = 10 m/s2 in a body of mass, m = 500 g or 0.5 kg
Thus, F = ma
Or, F = (0.5) (10) = 5 N
Let a’ be the acceleration which force F (=5N) cause on a body of mass, m’ = 5 kg.
Then, a’ = F/m’.
Or, a’ = (5/5) ms-2.
Or, a’ = 1 ms-2.

Solution 14N.

Initial velocity, u = 30 m/s
Final velocity, v = 0
Time, t = 2s
Force, F = 1500 N

Here, a = (v – u)/t = (0 – 30)/ 2 = – 15 ms-2. Here, negative sign indicates retardation.
Now, F = ma.
Or, m = F/a = (1500/ 15) = 100 kg.

(a) Change in momentum = Final momentum – Initial momentum
Or, Δp = m (v – u)
Or, Δp = 100 (0 – 30)
Or, Δp = 3000 kg m/s

(b) Acceleration, a = (v – u)/t.
Or, a = (0 – 30)/ 2 = – 15 ms-2,
Here, negative sign indicates retardation.
Thus, retardation = 15 ms-2.

(c) From Newton’s second law of motion,
F = ma
Or, m = F/a = (1500/ 15) = 100 kg.

Exercise 3(D)

Solution 1S.

Newton’s third law explains how a force acts on an object.

Solution 2S.

According to Newton’s third law of motion, to every action there is always an equal and opposite reaction. The action and reaction act simultaneously on two different bodies.

Solution 3S.

Law of action and reaction: In an interaction of two bodies A and B, the magnitude of action, i.e. the force FAB applied by the body B on the body A, is equal in magnitude to the reaction, i.e., the force FBA applied by the body A on the body B, but they are in directions opposite to each other.

 

Examples:

  1. When a book is placed on a table, it does not move downwards. It implies that the resultant force on the book is zero, which is possible only if the table exerts an upward force of reaction on the book, equal to the weight of the book.
  2. While moving on the ground, we exert a force by our feet to push the ground backwards; the ground exerts a force of the same magnitude on our feet forward, which makes it easier for us to move.

Explanation: In the above stated example, there are two objects and two forces. In the first example, the weight of the book acts downwards (action) and the force of the table acts upwards (reaction).
In the second example, our feet exerts a force on the ground (action) and the ground exerts an equal and opposite force (reaction) on our feet.

Solution 4S.

(a) Action: Force exerted on the bullet.
Reaction: Recoil experienced by the gun.

(b) Action: The force exerted by the hammer on the nail.
Reaction: The force applied by the nail on the hammer.

(c) Action: Weight of the book acting downwards.
Reaction: Force acted by the table upwards.

(d) Action: Force exerted by the rocket on the gases backwards.
Reaction: Force exerted by outgoing gases on the rocket in forward direction.
(e) Action: Force exerted by the feet on the ground in backward direction.
Reaction: Force exerted by the ground on feet in forward direction.

(f) Action: Force exerted by a moving train on a stationary train.
Reaction: Force exerted by a stationary train on a moving train.

Solution 5S.

When a rocket moves in space, it pushes gases outside, i.e. the rocket applies force on the gases in the backward direction. As a reaction, the gases put equal amount of force on the rocket in the opposite direction and the rocket moves in the forward direction.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 7

Solution 6S.

When a man fires a bullet from a gun, a force F is exerted on the bullet (action), and the gun experiences an equal and opposite recoil (reaction) and hence gets recoiled.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 8

Solution 7S.

When a man exerts a force (action) on the boat by stepping into it, its force of reaction makes him step out of the boat, and the boat tends to leave the shore due to the force exerted by the man (i.e. action).
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 9

Solution 8S.

Couple two spring balances A and B as shown in the figure. When we pull the balance B, both the balances show the same reading indicating that both the action and reaction forces are equal and opposite. In this case, the pull of either of the two spring balances can be regarded as action and that of the other balance as the reaction.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 10

Solution 9S.

To move a boat, the boatman pushes (action) the water backwards with his oar. In this response, the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.

Solution 10S.

A person pushing a wall hard (action) by his palm, experiences a force (reaction) exerted by the wall on his palm in the opposite direction; thus, he is liable to fall backwards.

Solution 11S.

Yes, action and reaction act simultaneously.

Solution 12S.

Yes, action and reaction are equal in magnitude.

Solution 13S.

When a falling ball strikes the ground, it exerts a force on the ground. The ground exerts a force back at the ball in the opposite direction. This is the reason the ball rises upwards.

Solution 14S.

The given statement is wrong.
Reason: According to Newton’s third law of motion, the action and reaction act simultaneously on different bodies. Hence they do not cancel each other.

Solution 1M.

Explains the way the force acts on a body.

Solution 2M.

Different bodies in opposite directions

Solution 1N.

The wall exerts an equal force of 10 N on the boy in the opposite direction, i.e. west.

Solution 2N.

(a) A block exerts 15 N force (weight) on the string downwards.
(b) The string exerts an equal force of 15 N on the block in the opposite direction, i.e. upward direction (tension).
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 11

Exercise 3(E)

Solution 1S.

Newton’s law of gravitation: Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, and the direction of the force is along the line joining the masses.

Solution 2S.

Gravitational force is always attractive.

Solution 3S.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 12
Here G is a constant of proportionality called the universal gravitational constant.

Solution 4S.

The gravitational force of attraction between two masses is inversely proportional to the square of distance between them.

Solution 5S.

If the distance between the masses becomes half, the force reduces to one-fourth.

Solution 6S.

The gravitational constant is defined as the force of attraction between two bodies of unit mass separated by a unit distance.

Solution 7S.

The value of G in the S.I. system is 6.67 x 10-11Nm2kg-2.

Solution 8S.

The gravitational force of attraction is significant to explain the motion of heavenly bodies, e.g. motion of planets around the Sun, motion of the Moon around the Earth etc.

Solution 9S.

The force with which the Earth attracts a body towards its centre is called the force due to gravity.

Solution 10S.

The force due to gravity on a body of mass m kept on the surface of Earth (mass=M and radius=R) is equal to the force of attraction between the Earth and that body.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 13

Solution 11S.

The rate at which the velocity of a freely falling body increases is called acceleration due to gravity. Its S.I. unit is m/s2.

Solution 12S.

The average value of ‘g’ on the Earth’s surface is 9.8 m/s2.

Solution 13S.

Let g be the acceleration due to gravity on the Earth’s surface (mass = M and radius = R).
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 14

Solution 14S.

Acceleration due to gravity (g) is directly proportional to universal gravitational constant (G).

Solution 15S.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 15

Solution 16S.

If a body is thrown vertically up with an initial velocity u to a height h, then there will be retardation (a = – g).
At the highest point, the final velocity v = 0.
Thus, from the third equation,
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 16

Solution 17S.

Mass: The mass of a body is the quantity of matter it contains.
Weight: The weight of a body is the force with which the Earth attracts it.

Solution 18S.

Mass is a scalar quantity, but weight is a vector quantity. Mass is the measure of the quantity of matter contained in a body, but weight is the measure of force with which the Earth attracts the body. Mass of a body is always constant but weight varies from place to place.

Solution 19S.

The S.I. unit of mass is kg and that of weight is newton.

Solution 20S.

W = mg
At the centre of Earth, g = 0.
Therefore, W = 0.

Solution 21S.

Mass of a body is always constant.

Solution 22S.

1 kgf = 9.8 N.
One kilogramme force is the force due to gravity on a mass of 1 kilogramme.

Solution 1M.

Always attractive

Solution 2M.

6.7 x 10-11 N m2 kg-2

Solution 3M.

6.7 x 10-11 N

Solution 4M.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 19
METHOD: Let’t’ be the time in which the body reaches its maximum height.
Initial velocity = u.
Final velocity (at the highest point) = 0.
Acceleration due to gravity = g (negative sign indicates the body is moving against gravity).
Using the first equation of motion,
v = u + gt.
We get,
0 = u gt
Or t = u/g
Now total time for which the ball remains in air = Time of ascent + Time of descent
Because time of ascent = Time of descent,
Total time taken = u/g + u/g = 2u/g

Solution 5M.

19.6 m s-1

METHOD: Given, u = 0
g = 9.8 m/s2
Time t = 2s
Let ‘v’ be the velocity of object on reaching the ground.
Using the first equation of motion,
v = u + gt
We get,
v = 0 + (9.8) (2)
Or, v = 19.6 m/s.

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 17

Solution 2N.

Weight = mg
W = (5) (9.8) = 50 N.
Assumption: Value of acceleration due to gravity = 9.8 m/s2.

Solution 3N.

Mass = 10 kg
(i) Weight (in kgf) = 10 x 1 kgf = 10 kgf
[1 kgf = 9.8 N]
(ii) Weight (in newton) = 10 x 9.8 = 98 N.

Solution 4N.

Mass = 5 kg.
g = 9.8 m/s2.
Let F be the force of gravity,
F = mg.
F = (5) (9.8) = 49 N.
Force of gravity always acts downwards.

Solution 5N.

Weight, W = 2.0 N
g = 9.8 m/s2
Let ‘m’ be the mass of the body.
W = mg
Or, m = W/g = (2/9.8) kg = 0.2 kg.

Solution 6N.

Weight of the body on Earth = 98 N.
Acceleration due to gravity on Earth = 9.8 m/s2.
Let ‘m’ be the mass of the body on Earth.
m = W/g
m = (98/9.8) = 10 kg
Thus, the mass of the body is 10 kg, which always remains constant.
(a) Mass on moon = mass on Earth = 10 kg
(b) Let weight on moon is W’.
W’ = mass x acceleration due to gravity on the Moon.
[Given, acceleration due to gravity on the Moon = 1.6 m/s2]
W’ = 10 x 1.6 =16 N.

Solution 7N.

Man’s weight on Earth = 600 N
Man’s weight on the Moon = (1/6) man’s weight on Earth;
Because the acceleration due to gravity on the Moon is 1/6th that of Earth and w = mg.
Therefore, man’s weight on Moon = (600/6) = 100 N.

Solution 8N.

Mass, m = 10.5 kg
G = 10 m/s2
(a) Force, F = mg
F = (10.5) (10) = 105 N
(b) Weight, w = mg
w = (10.5) (10) = 105 N

Solution 9N.

Let ‘S’ be the height.
Time taken, t = 3s; g = 9.8 m/s2
Initial velocity, u = 0 (because the body starts from rest)

(a) Using the second equation of motion,
S = ut + (1/2) gt2
We get,
S = 0 + (1/2) (9.8) (3) (3)
S = 44.1 m

(b) Let ‘v’ be the velocity with which the ball strikes the ground.
Using the third equation of motion,
v2 – u2 = 2gs
or, v2 – 02 = 2(9.8) (44.1)
or, v2 = 864.36
or, v = 29.4 m/s

Solution 10N.

Mass, m = 5kg
Force, F = mg
F = (5) (9.8) = 49 N
Assumption: Value of acceleration due to gravity is 9.8 m/s2.

Solution 11N.

Given, maximum height reached, s = 20 m
Acceleration due to gravity, g = 10 m/s2

(a) Let ‘u’ be the initial velocity.
At the highest point, velocity = 0
Using the third equation of motion,
v2   u2 = 2gs
or, 0  u2 = 2 (10) (20) m/s
or, u2 =  (400) m/s [Negative sign indicates that the motion is against gravity]
or, u = 20 m/s

(b) Let v’ be the final velocity of the ball on reaching the ground.
Considering the motion from the highest point to ground,
Velocity at highest point = 0 = Initial velocity for downward journey of the ball.
Distance travelled, s = 20m
Using the third equation of motion,
v2  u2 = 2gs
or, v2  0 = 2 (10) (20) m/s
or, v2 = 400 m/s
or, v = 420 m/s

(c) Now total time for which the ball remains in air, t = 2u/g.
Or, t = 2 (20)/(10).
Or, t = 4s.

Solution 12N.

Initial velocity u = 0
Final velocity = 20 m/s
g = 10 m/s2
Let ‘h’ be the height of the tower.
Using the third equation of motion,
v2 – u2 = 2gs
or, (20)2 – 0 = 2 (10) h
or, h = 20 m

Solution 13N.

Total time of journey = 6 s
g = 10 m/s2

(i) Let ‘H’ be the greatest height.
Time of ascent, t = 6/2 = 3 s,
For ascent, initial velocity, u = 0
Using the second equation of motion,
H = ut + (1/2) gt2
H = 0 + (1/2) (10) (3) 2
H = 45 m

(ii) Let u’ be the initial velocity.
Final velocity, v = 0
Using the third equation of motion,
v2 – u2 = 2gH
or, v2 – 0 = 2(10) (45)
or, v2 = 900
or, v = 30 m/s

Solution 14N.

Initial velocity, u = 20 m/s
Time, t = 2s
g = 10 m/s2
Maximum height reached in 2s, H = (1/2) gt2
Or, H = (1/2) (10) (2) 2
Or, H = 20 m

Solution 15N.

(a) Height, s = 80m
g = 10 m/s2
Using the second equation of motion,
S = ut + (1/2) gt2
Or, 80 = 0+ (1/2) (10) (t) 2
Or, (t) 2 = 16
Or, t = 4s

(b) Let ‘v’ be the velocity on reaching the ground.
Using the third equation of motion,
v2 – u2 = 2gH
or, v2 – 0 = 2(10) (80)
or, v2 = 1600
or, v = 40 m/s

Solution 16N.

Given time t = 2.5, g = 9.8 m/s2
Height, H = (1/2) gt2
Or, H = (1/2) (9.8) (2.5)2
Or, H = 30.6 m

Solution 17N.

Initial velocity, u = 49 m/s
g = 9.8 m/s2

(i) Let H be the maximum height attained.
At the highest point, velocity = 0.
Using the third equation of motion,
v2 – u2 = 2gH
or, 0 – 492 = 2(-9.8) (H)
or, H = (492)/ 19.6
or, H = 122.5 m

(ii) Total time of flight is given by t = 2u/g
Or, t = 2(49)/ 9.8
Or, t = 10 s

Solution 18N.

Initial velocity u = 0
Time t = 4 s
g = 10 m/s2
Let ‘H’ be the height of the tower.
Using the second equation of motion,
H = ut + (1/2) gt2
Or, H = 0 + (1/2)(10)(4)2
Or, H = 80 m

Solution 19N.

(i) Time t =20 s
g = 10 m/s2
Let ‘D’ be the depth of the well.
Using the second equation of motion,
D = ut + (1/2) gt2
D = 0 + (1/2)(10)(20) 2
D = 2000 m

(ii) Speed of sound = 330 m/s
Depth of well = 2000 m
Time taken to hear the echo after the pebble reaches the water surface = Depth/speed
= (2000/330) s
= 6.1 s
Time taken for pebble to reach the water surface = 20 s.
Therefore, the total time taken to hear the echo after the pebble is dropped = 20 + 6.1 = 21.6 s.

Solution 20N.

Let x be the height of the tower.
Let h be the distance from the top of the tower to the highest point as shown in the diagram.
Selina Concise Physics Class 9 ICSE Solutions Laws of Motion 18
Initial velocity u = 19.6 m/s
g = 9.8 m/s2
At the highest point, velocity = 0
Using the third equation of motion,
v2 – u2 = 2gh
Or, – (19.6) 2 = 2 (-9.8) h
Or, h = 19.6 m
If the ball takes time t1 to go to the highest point from the top of building, then for the upward journey from the relation, v = u – gt,
0 = 19.6 – (9.8) (t1)
Or, t1 = 2s

(ii) Let us consider the motion for the part (x+h)
Time taken to travel from highest point to the ground = (5 – 2) = 3s
Using the equation s = ut + (1/2) gt2
We get,
(x + h) = 0 + (1/2) (9.8) (3) 2
Or, (x + 19.6) = 44.1 m
Or, x = 44.1 – 19.6 = 24.5 m
Thus, height of the tower = 24.5 m

(iii) Let v be the velocity of the ball on reaching the ground.
Using the relation, v = u + gt
We get:
v = 0 + (9.8) (3)
Or, v = 29.4 m/s

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Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation

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Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation

Exercise 1(A)

Solution 1S.

Measurement is the process of comparing a given physical quantity with a known standard quantity of the same nature.

Solution 2S.

Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.

Solution 3S.

The three requirements for selecting a unit of a physical quantity are

  1. It should be possible to define the unit without ambiguity.
  2. The unit should be reproducible.
  3. The value of units should not change with space and time.

Solution 4S.

Definitions of three fundamental quantities:

  1. S.I. unit of length (m): A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy made of 90% platinum and 10% iridium) rod kept at 0°C in the International Bureau of Weights and Measures at serves near Paris.
  2. S.I. unit of mass (kg): In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at International Bureau of Weights and Measures at serves near Paris.
  3. S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.

Solution 5S.

Three systems of unit and their fundamental units:

  1. C.G.S. system (or French system): In this system, the unit of length is centimeter (cm), unit of mass is gramme (g) and unit of time is second (s).
  2. F.P.S. system (or British system): In this system, the unit of length is foot (ft), unit of mass is pound (lb) and unit of time is second (s).
  3. M.K.S. system (or metric system): In this system, the unit of length is metre (m), unit of mass is kilogramme (kg) and unit of time is second (s).

Solution 6S.

A fundamental (or basic) unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit.

Solution 7S.

Fundamental quantities, units and symbols in S.I. system are

Quantity Unit Symbol
Length metre m
Mass kilogramme kg
Time second s
Temperature kelvin K
Luminous intensity candela cd
Electric current ampere A
Amount of substance mole mol
Angle radian rd
Solid angle steradian st-rd

Solution 8S.

The units of quantities other than those measured in fundamental units can be obtained in terms of the fundamental units, and thus the units so obtained are called derived units.

Example:
Speed = Distance/time
Hence, the unit of speed = fundamental unit of distance/fundamental unit of time
Or, the unit of speed = metre/second or ms-1.
As the unit of speed is derived from the fundamental units of distance and time, it is a derived unit.

Solution 9S.

A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy with 90% platinum and 10% iridium) rod kept at 0o C in the International Bureau of Weights and Measures at serves near Paris.

Solution 10S.

Astronomical unit (A.U.) and kilometer (km) are units of length which are bigger than a metre.
1 km = 1000 m
1 A.U. = 1.496 × 1011 m

Solution 11S.

Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.
1 cm = 10-2 m
1 mm = 10-3 m

Solution 12S.

1 nm = 10 Å

Solution 13S.

Three convenient units of length and their relation with the S.I. unit of length:

  1. 1 Angstrom (Å) = 10-10 m
  2. 1 kilometre (km) = 103 m
  3. 1 light year (ly) = 9.46 × 1015 m

Solution 14S.

S.I. unit of mass is ‘kilogramme’.
In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at serves near Paris.

Solution 15S.

(a) 1 light year = 9.46 × 1015 m
(b) 1 m = 1010 Å
(c) 1 m = 106 µ (micron)
(d) 1 micron = 104 Å
(e) 1 fermi = 10-15 m

Solution 16S.

The units ‘gramme’ (g) and ‘milligramme’ (mg) are two units of mass smaller than ‘kilogramme’.
1 g = 10-3 kg
1 mg = 10-6 kg

Solution 17S.

The units ‘quintal’ and ‘metric tonne’ are two units of mass bigger than ‘kilogramme’.
1 quintal = 100 kg
1 metric tonne = 1000 kg

Solution 18S.

(a) 1 g = 10-3 kg
(b) 1 mg = 10-6 kg
(c) 1 quintal = 100 kg
(d) 1 a.m.u (or u) = 1.66 x 10-27 kg

Solution 19S.

The S.I. unit of time is second (s).
A second is defined as 1/86400th part of a mean solar day, i.e.

Solution 20S.

The units ‘minute’ (min) and ‘year’ (yr) are two units of time bigger than second(s).
1 min = 60 s
1 yr = 3.1536 × 107 s

Solution 21S.

A leap year is the year in which the month of February has 29 days.

Solution 22S.

Yes, the given statement is true.

Solution 23S.

One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.

Solution 24S.

(a) 1 nanosecond = 10-9 s
(b) 1 µs = 10-6 s
(c) 1 mean solar day = 86400 s
(d) 1 year = 3.15 × 107 s

Solution 25S.

(a) Mass (b) Distance (or length) (c) Time (d) Length

Solution 26S.

(a) ms-1 (b) kg ms-2 (c) kg m2s-2 (d) kg m-1s-2

Solution 27S.

(a) kg ms-2 (b) kg m2s-3
(c) kg m2s-2 (d) kg m-1s-2

Solution 28S.

(a) Area (b) Force (c) Energy
(d) Pressure (f) Power

 

Solution 1M.

Second

Solution 2M.

Litre

Solution 3M.

Leap year

Solution 4M.

0.1 nm

Solution 5M.

Length

Solution 1N.

Wavelength of light of particular colour = 5800 Å

(a) (i) 1 Å = 10-1 nm
5800 Å = 5800 × 10-1 nm
= 580 nm

(ii) 1 Å = 10-10 m
5800 Å = 5800 × 10-10 m
= 5.8 × 10-7 m

(b) The order of its magnitude in metre is 10-6 m because the numerical value of 5.8 is more than 3.2.

Solution 2N.

Size of a bacteria = 1 µ
Since 1 µ = 10-6 m
Number of the particle = Total length/size of
one bacteria
= 1 m/10-6 m
= 106

Solution 3N.

Distance of galaxy = 5.6 × 1025 m
Speed of light = 3 × 108 m/s

(a) Time taken by light = Distance travelled/speed of light
= (5.6 × 1025 / 3 × 108) s
= 1.87 × 1017 s

(b) Order of magnitude = 100 × 1017 s = 1017 s
(This is because the numerical value of 1.87 is less than the numerical value 3.2)

Solution 4N.

Wavelength of light = 589 nm
= 589 × 10-9 m
= 5.89 × 10-7 m
Order of magnitude = 101 × 10-7 m
= 10-6 m
(This is because the numerical value of 5.89 is more than the numerical value 3.2)

Solution 5N.

Mass of an oxygen atom = 16.00 u
Now, 1 u = 1.66 × 10-27 kg
Hence, mass of oxygen in kg = 16 × 1.66 × 10-27 kg
= 26.56 × 10-27 kg
Because the numerical value of 26.56 is greater than the numerical value of 3.2, the order of magnitude of mass of oxygen in kg
= 101 × 10-27 kg
= 10-26 kg

Solution 6N.

Time taken by light to reach from the Sun to the Earth = 8 min = 480 s.
Speed of light = 3 × 108 m/s
Distance from the Sun to the Earth = Speed × time
= 3 × 108 × 480 m
= 1440 × 108 m
= 1440 × 108 × 10-3 km
= 1440 × 105 km
= 1.44 × 108 km
Because the numerical value of 1.44 is less than the numerical value of 3.2, the order of magnitude of distance from the Sun to the Earth in km = 100 × 108 km
= 108 km

Solution 7N.

The statement ‘the distance of a star from the Earth is 8.33 light minutes’ means that the light from the star takes 8.33 minutes to reach Earth.

Exercise 1(B)

Solution 1S.

The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.

Solution 2S.

Total length of the scale = 1 m = 100 cm
No. of divisions = 100
Length of each division = Total length/total no. of divisions
= 100 cm/100
= 1 cm
Thus, this scale can measure with an accuracy of 1 cm.
To increase the accuracy, the total number of divisions on the scale must be increased.

Solution 3S.

The least count of a metre rule is 1 cm.
The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter.

Solution 4S.

The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let n divisions on vernier callipers be of length equal to that of (n – 1) divisions on the main scale and the value of 1 main scale division be x. Then,
Value of n divisions on vernier = (n – 1) x
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 1

Solution 5S.

Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the least count of vernier callipers.

Solution 6S.

A vernier calipers is said to be free from zero error, if the zero mark of the vernier scale coincides with the zero mark of the main scale.

Solution 7S.

Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale, the vernier callipers is said to have zero error.

It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale.
The zero error is of two kinds

  1. Positive zero error
  2. Negative zero error

1. Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 2

To find this error, we note the division of the vernier scale, which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count of the vernier callipers, gives the zero error.

For example, for the scales shown, the least count is 0.01 cm and the 6th division of the vernier scale coincides with a main scale division.

Zero error = +6 × L.C. = +6 × 0.01 cm
= +0.06 cm

2. Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 3

To find this error, we note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.

For example, for the scales shown, the least count is 0.01 cm and the sixth division of the vernier scale coincides with a certain division of the main scale. The total number of divisions on vernier callipers is 10.

Zero error = – (10 – 6) × L.C.
= – 4 × 0.01 cm = – 0.04 cm

Correction:
To get correct measurement with vernier callipers having a zero error, the zero error with its proper sign is always subtracted from the observed reading.
Correct reading = Observed reading – zero error (with sign)

Solution 8S.

Selina Concise Physics Class 9 ICSE Soluti

Solution 9S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 24

Main parts and their functions:

  • Main scale: It is used to measure length correct up to 1 mm.
  • Vernier scale: It helps to measure length correct up to 0.1 mm.
  • Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
  • Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.
  • Strip: It helps to measure the depth of a beaker or a bottle.

Solution 10S.

Three uses of vernier callipers are

  1. Measuring the internal diameter of a tube or a cylinder.
  2. Measuring the length of an object.
  3. Measuring the depth of a beaker or a bottle.

Solution 11S.

Two scales of vernier calipers are

  1. Main scale
  2. Vernier scale

The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.
Value of 1 division on the main scale= 1 mm
Total no. of divisions on the vernier scale = 10
Thus, L.C. = 1 mm /10 = 0.1 mm = 0.01 cm.
Hence, a vernier callipers can measure length correct up to 0.01 cm.

Solution 12S.

Measuring the length of a small rod using vernier calipers:
The rod whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 6
In this position, the zero mark of the vernier scale is ahead of 1.2 cm mark on main scale. Thus the actual length of the rod is 1.2 cm plus the length ab (i.e., the length between the 1.2 cm mark on the main scale and 0 mark on vernier scale).
To measure the length ab, we note the pth division of the vernier scale, which coincides with any division of main scale.
Now, ab + length of p divisions on vernier scale = length of p divisions on main scale
Alternatively, ab = length of p divisions on the main scale – length of p divisions on the vernier scale.
= p (length of 1 division on main scale – length of 1 division on vernier scale)
= p × L.C.
Therefore, total reading = main scale reading + vernier scale reading
= 1.2 cm + (p × L.C.)

Solution 13S.

(a) Outside jaws
(b) Inside jaws
(c) Strip
(d) Outer jaws

Solution 14S.

(i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation.
(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.
Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.
If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.
Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm

Solution 15S.

The least count of a screw gauge can be increased by decreasing the pitch and increasing the total number of divisions on the circular scale.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 7

Main parts and their functions:

  1. Ratchet:It advances the screw by turning it until the object is gently held between the stud and spindle of screw.
  2. Sleeve:It marks the main scale and base line.
  3. Thimble:It marks the circular scale.
  4. Main scale:It helps to read the length correct up to 1 mm.
  5. Circular scale:It helps to read length correct up to 0.01 mm.

Solution 17S.

A screw gauge is used for measuring diameter of circular objects mostly wires with an accuracy of 0.001 cm.

Solution 18S.

Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw.

Solution 19S.

Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of main scale. It is either above or below the base line of the main scale, in which case the screw gauge is said to have a zero error. It can be both positive and negative.
It is accounted by subtracting the zero error (with sign) from the observed reading in order to get the correct reading.
Correct reading = Observed reading – zero error (with sign)

Solution 20S.

Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 8

Solution 21S.

Backlash error: If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.

Reason: It happens due to wear and tear of the screw threads.
To avoid the backlash error, while taking the measurements the screw should be rotated in one direction only. If the direction of rotation of the screw needs to be changed, then it should be stopped for a while and then rotated in the reverse direction.

Solution 22S.

Measurement of diameter of wire with a screw gauge:

The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated until the wire is firmly held between the anvil and spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in the figure below.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 9
The pitch of the screw = 1 mm
L.C. of screw gauge = 0.01 mm
Main scale reading = 2.5 mm
46th division of circular scale coincides with the base line.
Therefore, circular scale reading = 46 × 0.01 = 0.46 mm
Total reading = Main scale reading + circular scale reading
= (2.5 + 0.46) mm
= 2.96 mm

Solution 23S.

(a) Screw gauge
(b) Screw gauge
(c) Vernier calipers
(d) Screw gauge

Solution 24S.

Screw gauge measures a length to a high accuracy.

Solution 25S.

(a) Vernier callipers (b) Metre scale (c) Screw gauge.

Solution 1M.

The least count of a vernier calipers is 0.01 cm

Solution 2M.

0.002 cm

Solution 3M.

A screw gauge

Solution 1N.

Range of the stop watch = 5s
Total number of divisions = 10
L.C. = 5/10 = 0.5 s

Solution 2N.

Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm

Solution 3N.

There are 20 divisions in 1 cm on the main scale.
Therefore, the value of 1 m.s.d. (x) = 1/20 cm = 0.05 cm
No. of divisions on the vernier scale (n) = 25
Hence, the L.C. of the microscope = x/n = (0.05 / 25) cm
= 0.002 cm

Solution 4N.

Thickness of the pencil (observed reading) = 1.4 mm
Zero error = + 0.02 cm = + 0.2 mm
Correct reading = observed reading – zero error (with sign)
= 1.4 mm – 0.2 mm
= 1.2 mm

Solution 5N.

(i) Value of 1 m.s.d. =1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm

(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.
3rd vernier division coincides with a main scale division.
Total no. of vernier divisions = 10
Zero error = +3 × L.C.
= +3 × 0.01 cm
= +0.03 cm

Solution 6N.

(i) Value of 1 m.s.d = 1 mm = 0.1 cm
20 vernier divisions = 19 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1mm/20
= (0.1/20) cm
= 0.005 cm

(ii) Main scale reading = 35 mm = 3.5 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.005 cm = 0.02 cm
Total reading = Main scale reading + vernier scale reading
= (3.5 + 0.02) cm
= 3.52 cm
Radius of the cylinder = Diameter (Total reading) / 2
= (3.52/2) cm
= 1.76 cm

Solution 7N.

(a) L.C. of vernier callipers = 0.01 cm
Main scale reading = 1.8 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.01 cm = 0.04 cm
Total reading = Main scale reading + vernier scale reading
= (1.8 + 0.04) cm
= 1.84 cm

(b) Observed reading = 1.84 cm
Zero error = -0.02 cm
Correct reading = Observed reading – Zero error (with sign)
= [1.84 – (-0.02)] cm
= 1.86 cm

Solution 8N.

L.C. of vernier callipers = 0.01 cm
In the shown scale,
Main scale reading = 3.3 mm
6th vernier division coincides with an m.s.d.
Therefore, vernier scale reading = 6 × 0.01 cm = 0.06 cm
Total reading = m.s.r. + v.s.r.
= 3.3 + 0.06
= 3.36 cm

Solution 9N.

Pitch of a screw gauge = 0.5 mm
No. of divisions on the circular scale = 100
L.C. = (0.5/100) mm
= 0.005 mm or 0.0005 cm

Solution 10N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm

(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm

Solution 11N.

Pitch of the screw gauge = 1mm
No. of divisions on the circular scale = 100

(i) L.C. = Pitch/No. of divisions on the circular head
= (1/100) mm
= 0.01 mm or 0.001 cm

(ii) Main scale reading = 2mm = 0.2 cm
No. of division of circular head in line with the base line (p) = 45
Circular scale reading = (p) × L.C.
= 45 x 0.001 cm
= 0.045 cm
Total reading = M.s.r. + circular scale reading
= (0.2 + 0.045) cm
= 0.245 cm

Solution 12N.

(i) L.C. of screw gauge = 0.01 mm or 0.001 cm
Main scale reading = 1 mm or 0.1 cm
No. of division of circular head in line with the base line (p) = 27
Circular scale reading = (p) × L.C.
= 27 × 0.001 cm
= 0.027 cm
Diameter (Total reading) = M.s.r. + circular scale reading
= (0.1 + 0.027) cm
= 0.127 cm

(ii) Zero error = 0.005 cm
Correct reading = Observed reading – zero error (with sign)
= [0.127 – (+0.005)] cm
= 0.122 cm

Solution 13N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm

(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm

(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm

Solution 14N.

No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.

(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm

(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm

Solution 15N.

Pitch of the screw gauge = 0.5 mm
L.C. of the screw gauge = 0.001 mm
No. of divisions on circular scale = Pitch / L.C.
= 0.5 / 0.001
= 500

Exercise 1(C)

Solution 1S.

A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.

No, the pendulum used in pendulum clock is not a simple pendulum because the simple pendulum is an ideal case. We cannot have a heavy mass having the size of a point and string having no mass.

Solution 2S.

(i) Oscillation: One complete to and fro motion of the pendulum is called one oscillation.
(ii) Amplitude: The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation. It is measured in metres (m).
(iii) Frequency: It is the number of oscillations made in one second. Its unit is hertz (Hz).
(iv)Time period: This is the time taken to complete one oscillation. It is measured in second (s).

Solution 3S.

Simple Pendulum:
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 10

Solution 4S.

Two factors on which the time period of a simple pendulum depends are

  1. Length of pendulum (l)
  2. Acceleration due to gravity (g)
    Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 11

Solution 5S.

Two factors on which the time period of a simple pendulum does not depend are

  1. Material of the bob
  2. Amplitude

Solution 6S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 12
Therefore, the time period is doubled.

(b) If the acceleration due to gravity is reduced to one-fourth,
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 13
Therefore, the time period is doubled.

Solution 7S.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 14

Solution 8S.

Measurement of time period of a simple pendulum:

  1. To measure the time period of a simple pendulum, the bob is slightly displaced from its mean position and is then released. This gives a to and fro motion about the mean position to the pendulum.
  2. The time ‘t’ for 20 complete oscillations is measured with the help of a stop watch.
  3. Time period ‘T’ can be found by dividing ‘t’ by 20.

To find the time period, the time for the number of oscillations more than 1 is noted because the least count of stop watch is either 1 s or 0.5 s. It cannot record the time period in fractions such as 1.2 or 1.4 and so on.

Solution 9S.

The time period of a simple pendulum is directly proportional to the square root of its effective length.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 15
From this graph, the value of acceleration due to gravity (g) can be calculated as follows.
The slope of the straight line can be found by taking two points P and Q on the straight line and drawing normals from these points on the X- and Y-axis, respectively. Then, the value of T2 is to be noted at a and b, the value of l at c and d. Then,
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 16
where g is the acceleration due to gravity at that place. Thus, g can be determined at a place from these measurements using the following relation:
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 17

Solution 10S.

The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.

Solution 11S.

Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportional to the square root of the length of the pendulum. Therefore, the pendulum B will have a greater time period and thus making lesser oscillations.

Solution 12S.

(a) The time period of oscillations is directly proportional to the square root of the length of the pendulum.
(b) The time period of oscillations of simple pendulum does not depend on the mass of the bob.
(c) The time period of oscillations of simple pendulum does not depend on the amplitude of oscillations.
(d) The time period of oscillations of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

Solution 13S.

A pendulum with the time period of oscillation equal to two seconds is known as a seconds pendulum.

Solution 14S.

The frequency of oscillation of a seconds’ pendulum is 0.5 s-1. It does not depend on the amplitude of oscillation.

Solution 1M.

Half

Solution 2M.

2 s

Solution 3M.

Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 18

Solution 1N.

(a) Frequency = Oscillations per second
= (40/60) s-1
 = 0.67 s-1

(b) Time period = 1/frequency
= (1/0.67) s
= 1.5 s

Solution 2N.

Time period = 2 s
Frequency = 1/time period
= (½)s-1
 = 0.5 s-1
Such a pendulum is called the seconds’ pendulum.

Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

Let T1 and T2 be the time periods of the two pendulums of lengths l1 and l2, respectively.
Then, we know that the time period is directly proportional to the square root of the length of the pendulum.
Selina Concise Physics Class 9 ICSE Solutions Measurements and Experimentation image - 23

Solution 8N.

Time period = Time taken to complete 1 oscillation
= (4 × 0.2) s
= 0.8 s

Solution 9N.

Time period of a seconds’ pendulum = 2 s
Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.

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Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

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Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 8 Propagation of Sound Waves. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 8 Propagation of Sound Waves

Exercise 8(A)

Solution 1S.

Sound is caused due to vibrations of a body.

Solution 2S.

Sound is a form of energy that produces the sensation of hearing in our ears. Sound is produced by a vibrating body.

Solution 3S.

Vibrating

Solution 4S.

Experiment: A tuning fork is taken and its one arm is struck on a rubber pad and it is brought near a tennis ball suspended by a thread as shown in figure.
Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 1
It is noticed that as the arm of the vibrating fork is brought close to the ball, it jumps back and forth and sound of the vibrating tuning fork is heard. When its arm stop vibrating, the ball becomes stationary and no sound is heard.

Solution 5S.

Experiment to demonstrate that a material medium is necessary for the propagation of sound:
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An electric bell is suspended inside an airtight glass bell jar. The bell jar is connected to a vacuum pump as shown in figure. As the circuit of electric bell is completed by pressing the key, the hammer of the electric bell begins to strike the gong repeatedly due to which sound is heard.

Keeping the key pressed, air is gradually withdrawn from jar by starting the vacuum pump. It is noticed that the loudness of sound goes on decreasing as the air is taken out from the bell jar and finally no sound is heard when all the air from the jar has been drawn out. The hammer of the electric bell is still seen striking the gong repeatedly which means that sound is still produced but it is not heard.

When the jar is filled with air, the vibrations produced by the gong are carried by the air to the walls of jar which in turn set the air outside the jar in vibration and sound is heard by us but in absence of air, sound produced by bell could not travel to the wall of the jar and thus no sound is heard. It proves that material medium is necessary for the propagation of sound waves.

Solution 6S.

We cannot hear each other on moon’s surface because there is no air on moon and for sound to be heard, a material medium is necessary.

Solution 7S.

Requisites of the medium for propagation of sound:

  1. The medium must be elastic.
  2. The medium must have inertia.
  3. The medium should be frictionless.

Solution 8S.

Take a vertical metal strip with its lower end fixed and upper end being free to vibrate as shown in fig (a).

As the strip is moved to right from a to b as shown in Fig (b), the air in that layer is compressed (compression is formed at C). The particles of this layer compress the layer next to it, which then compresses the next layer and so on. Thus, the disturbance moves forward in form of compression without the particles themselves being displaced from their mean positions.

As the metal strip returns from b to a as shown in Fig (c) after pushing the particles in front, the compression C moves forward and particles of air near the strip return to their normal positions.

When the strip moves from a to c as shown in Fig (d), it pushes back the layer of air near it towards left and thus produces a low pressure space on its right side i.e. layers of air get rarefied. This region is called rarefaction (rarefaction is formed at R).

When the strip returns from C to its mean position A in Fig (e), the rarefaction R travels forward and air near the strip return to their normal positions.

Thus, one complete to and fro motion of the strip forms one compression and one rarefaction, which together form one wave. This wave through which sound travels in air is called longitudinal wave.
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Solution 9S.

the disturbance

Solution 10S.

Sound travels in a medium in form of longitudinal and transverse waves.

Solution 11S.

A type of wave motion in which the particle displacement is parallel to the direction of wave propagation is called a longitudinal wave. It can be produced in solids, liquids as well as gases.

Solution 12S.

A type of wave motion in which the particle displacement is perpendicular to the direction of wave propagation is called a transverse wave. It can be produced in solids and on the surface of liquids.

Solution 13S.

A longitudinal wave propagates by means of compressions and rarefactions.

When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (C), as shown in Fig. 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 (R), as shown in Fig.
Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 4
Compressions are the regions of high density where the particles of the medium come very close to each other and rarefactions are the regions of low density where the particles of the medium move away from each other.

Solution 14S.

A crest is a point on the transverse wave where the displacement of the medium is at a maximum.
A point on the transverse wave is a trough if the displacement of the medium at that point is at a minimum.
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Solution 15S.

Experiment to show that in a wave motion, only energy is transferred, but particles of the medium do not move:

If we drop a piece of stone in the still water of pond, we hear a sound of stone striking the water surface. Actually a disturbance is produced in water at the point where the stone strikes it. This disturbance spreads in all directions radially outwards in form of circular waves on the surface of water.

If we place a piece of cork on water surface at some distance away from the point where the stone strikes it, we notice that cork does not move ahead, but it vibrates up and down, while the wave moves ahead. The reason is that particles of water (or medium) start vibrating up and down at the point where the stone strikes. These particles then transfer their energy to the neighboring particles and they themselves come back to their mean positions. Thus only energy is transferred but the particles of the medium do not move.

Solution 16S.

The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave.
Its SI unit is metre.

Solution 17S.

The number of vibrations made by the particle of the medium in one second is called the frequency of the wave. It can also be defined as the number of waves passing through a point in one second.
Its SI unit is hertz (Hz).

Solution 18S.

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Solution 19S.

The distance travelled by a wave in one second is called its wave velocity.
Its SI unit is metre per second (ms-1).

Solution 20S.

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Solution 21S.

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 8

Solution 22S.

Let the velocity of a wave be V, time period T, frequency ν and wavelength λ. By the definition of wavelength,
Wavelength = Distance travelled by the wave in one time period i.e., in T second
Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 9

Solution 23S.

The speed of sound in a medium depends upon its elasticity and density.

Solution 24S.

Vg < Vl < Vs

Solution 25S.

(i) Speed of light in air = 3 x 108 m s-1 (ii) Speed of sound in air = 330 m s-1.

Solution 26S.

1 : 4 : 15

Solution 27S.

(i) No, sound cannot travel in vacuum as it requires a material medium for its propagation.
(ii) Speed of sound is maximum in solids, less in liquids and least in gases.

Solution 28S.

This happens because the light travels much faster than sound.

Solution 29S.

Sound travels in iron faster than in air so first the sound travelled in iron rail is heard and then the sound travelled in air is heard.

Solution 30S.

(i) The diver would hear the sound first.
(ii) This is because sound travels faster in water than in air.
(iii) It would take 0.25t to reach the diver because sound travels almost four times faster in water.

Solution 31S.

(i) Frequency of sound has no effect on the speed of sound.
(ii) Speed of sound increases with the increase in the temperature of sound.
(iii) Pressure of sound has no effect on the speed of sound.
(iv) Speed of sound increases with the increase in presence of moisture in air.

Solution 32S.

(i) Speed of sound does not change with a change in amplitude.
(ii) Speed of sound does not change with a change in wavelength.

Solution 33S.

Speed of sound is more in humid air because in presence of moisture, the density of air decreases and sound travels with greater speed.
Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 10

Solution 34S.

The speed of sound increases by 0.61 m s-1 for each 1°C rise in temperature.

Solution 35S.

The simple experiment that a person can do to calculate the speed of sound in air is that a person stands at a known distance (d meter) from the cliff and fires a pistol and simultaneously start the stopwatch. He stops the stopwatch as soon as he hears an echo. The distance travelled by the sound during the time (t) seconds is 2d. So, speed of sound = distance travelled / time taken = 2d/t

The approximation made is that speed of sound remains same for the time when the experiment is taking place.

Solution 36S.

(a) Vacuum, medium (b) do not move, moves (c) rarefaction (d) trough.

Solution 1M.

Sound needs medium, but light does not need medium for its propagation.

Solution 2M.

Longitudinal wave

Solution 3M.

330 m s-1

Solution 4M.

3 x 108 m s-1

Solution 1N.

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Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 12

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 13

Solution 4N.

Wave velocity = 0.3 m/s
Frequency = 20 Hz
Separation between two consecutive compressions is the wavelength of a wave.
We know that,
Wave velocity = Frequency x Wavelength
Or, wavelength = Wave velocity / frequency
Or, λ = 0.3 / 20 = 1.5 x 10-2 m

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Propagation of Sound Waves image - 14

Solution 6N.

Distance between the two observers = 1650 m
Speed of sound = 330 m/s
Time in which B hears the sound = Distance / speed = 1650/330 = 5s
Thus, B will hear the sound 5s after the gun is shot.

Solution 7N.

Speed of sound in air (V) = 330 m/s
Time in which thunder is heard after lighting is seen (t) = 5s
Thus, distance between flash and observer = V x t = (330 x 5) = 1650 m

Solution 8N.

Speed of sound in air (V) = 340 m/s
Time in which sound of fire is heard after flash is seen (t) = 2.5s
Thus, distance between flash and observer = V x t = (340 x 2.5) = 850 m

Solution 9N.

Time taken by the observer to hear the sound of the first tank A= 3.5s
Time taken by the observer to hear the sound of the second tank B = 2s
Time taken by the tank B to hear the sound of tank A= (3.5 – 2)s = 1.5s
Distance between the two tanks = 510m
Speed = 510/1.5=340m/s

Solution 10N.

(a) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in iron (V) = 5280 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, t = (3300 / 5280) s = 0.625 s

(b) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in air (V) = 330 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, t = (3300/330) s = 10 s

Solution 11N.

(i) Distance travelled (D) = 1700
Speed of sound in air (V) = 340 m/s
Time taken (t) = D/V = (1700 / 340) s = 5 s

(ii) Distance travelled (D) = 1700
Speed of sound in water (V’) = 1360 m/s
Time taken (t) = D/V = (1700 / 1360) s = 1.25 s

Exercise 8(B)

Solution 1S.

The range of frequency within which the sound can be heard by a human being is called the audible range of frequency.

Solution 2S.

The audible range of frequency for humans is 20 Hz to 20 kHz.

Solution 3S.

Human ears are most sensitive for the range 2000 Hz to 3000 Hz.

Solution 4S.

Ultrasonic has higher frequency.

Solution 5S.

(a) 20 Hz, 20 kHz (b) above 20 kHz (c) below 20 Hz (d) ultrasonic (e) infrasonic.

Solution 6S.

(a) Infrasonic (b) Audible (c) Audible (d) Ultrasonic.

Solution 7S.

No, we cannot hear the sound produced due to vibrations of a seconds pendulum because the frequency of sound produced due to vibrations of seconds pendulum is 0.5 Hz which is infrasonic.

Solution 8S.

Sounds of frequency above 20 kHz are called ultrasound.

Solution 9S.

The approximate speed of ultrasound in air is 330 m/s.

Solution 10S.

Two properties of ultrasound which make it useful to us are:

  1. High energy contents
  2. High directivity

Solution 11S.

Bats locate the obstacles and prey in their path by producing and hearing the ultrasound. They emit an ultrasound which returns after striking an obstacle in their way. By hearing the reflected sound and from the time interval (when they produce ultrasound and they receive them back), they can judge the direction and the distance of the obstacle in their way.

Solution 12S.

Two applications of ultrasound:

  1. Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
  2. Ultrasound is used in surgery to remove cataract and in kidneys to break the small stones into fine grains.

Solution 1M.

1000 Hz

Solution 2M.

High power and good directivity

Solution 3M.

Ultrasound

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Selina Concise Physics Class 9 ICSE Solutions Reflection of Light

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Selina Concise Physics Class 9 ICSE Solutions Reflection of Light

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 7 Reflection of Light. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 7 Reflection of Light

Exercise 7(A)

Solution 1S.

The return of light into the same medium after striking a surface is called reflection.

Solution 2S.

Black silvered surface reflects most of the light incident on it.

Solution 3S.

(a) Plane mirror: Plane mirror is a highly polished and smooth reflecting surface made from a clear plane glass sheet, usually thin and silvered with suitable reflecting abrasive (for example, mercury) on one side. Once this pasting is done, then the glass becomes opaque but due to the reflecting property of the abrasive, the plane glass sheet becomes a plane glass reflector or a plane glass mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 1
(b) Incident ray: The light ray striking a reflecting surface is called the incident ray.

(c) Reflected ray: The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.

(d) Angle of incidence: The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter i.

(e) Angle of reflection: The angle which the reflected ray makes with the normal at the point of incidence is called the angle of reflection. It is denoted by the letter r.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 2

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 3
Regular reflection occurs when a beam of light falls on a smooth and polished surface and irregular reflection occurs when a beam of light falls on a rough surface. Since the surface is uneven, from different points light rays get reflected in different directions and give rise to irregular reflection.

Solution 5S.

Reflection of light from a plane mirror is regular reflection and reflection of light from plane sheet of paper is irregular reflection of light.

Solution 6S.

Laws of reflection:

  1. The angle of incidence is equal to the angle of reflection.
  2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.

Solution 7S.

Laws of reflection:

  1. The angle of incidence is equal to the angle of reflection.
  2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.

Experiment to verify the laws of reflection:

Fix a white sheet of paper on a drawing board and draw a line MM1 as shown in figure. On this line, take a point O nearly at the middle of it and draw a line OA such that ∠MOA is less than 90o. Then draw a normal ON on line MM1 at the point O, and place a small plane mirror vertical by means of a stand with its silvered surface along MM1.

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Next fix two pins P and Q at some distance (≈5 cm) apart vertically on line OA, on the board. Keeping eye on the other side of normal (but on the same side of mirror), see clearly images P’ and Q’ of the pins P and Q. Next fix a pin R such that it is in line with the images of pins P and Q as observed in the mirror. Next, fix one more pin S such that the pin S is in line with the pin R as well as images P’ and Q’ of pins P and Q.

Draw small circles on paper around the positions of pins as shown in figure. Remove the pins and draw a line OB joining the pin points S and R, which meets the surface of mirror at O. The angles AON and BON are measured and recorded.

The experiment is then repeated for the angle of incidence ∠AON equal to 40o, 50o, 60o.
From results, it is observed that angle of incidence is equal to the angle of reflection. This verifies the first law of reflection.

The experiment has been performed on a flat drawing board, with mirror normal to the plane of board on which white sheet of paper is being fixed. Since the lower tips of all the pins also lie on the same plane (i.e., the plane of paper), it proves the second law of reflection.

Solution 8S.

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Solution 9S.

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Solution 10S.

(a) Angle of incidence = 90o – 30o = 60o
(b) Angle between the incident ray and reflected ray = Angle of incidence + Angle of reflection
Angle of reflection = Angle of incidence = 60o
Therefore, Angle between the incident ray and reflected ray = 60o + 60o = 120o

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 7

Solution 11S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 8

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 9

Solution 13S.

(a) Three characteristics of image formed by plane mirror:

  1. Image formed in erect (upright)
  2. Image formed is virtual
  3. Image formed is of the same size as the object

(b) The image is situated at the same perpendicular distance behind the mirror as the object in front of it.

Solution 14S.

Real Image Virtual image
1. A real image is formed due to actual intersection of the reflected rays. 1. A virtual image is formed when the reflected rays meet if they are produced backwards.
2. A real image can be obtained on a screen. 2. A virtual image cannot be obtained on a screen.
3. A real image is inverted with respect to the object. 3. A virtual image is erect with respect to the object.

Solution 15S.

The interchange of the left and right sides in the image of an object in a plane mirror is called lateral inversion.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 10
Figure above shows the image formation of a letter P in a plane mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 43
Solution 16S.

The letters on the front of a ambulance are written laterally inverted, so that the driver of the vehicle moving ahead of the ambulance reads these words laterally inverted as AMBULANCE, in his rear view mirror, and gices side to pass the ambulance first.

Solution 17S.

Due to lateral inversion, , it becomes difficult to read the image of the text of a page formed due to reflection by a plane mirror.

Solution 1M.

i = r

Solution 2M.

Erect and of same size

Solution 3M.

virtual with lateral inversion

Solution 1N.

Angle of incidence (i) + Angle of reflection(r) = 90o 
But, as per the laws of reflection, i = r
Therefore, 2 i = 90o
Or, i = r = 45o

Solution 2N.

Distance between man and his image = 6m
Distance between man and mirror + distance between mirror and image = 6m
But, Distance between man and mirror (object distance) = distance between mirror and image (image distance)
Therefore, distance of man from mirror = 6/2 = 3m

Solution 3N.

(a) Image of the insect is formed 1m behind the mirror.
(b) Distance between the insect and his image = 1 + 1 = 2 m

Solution 4N.

Initially, distance of the object from the mirror = 60 cm.
Therefore, image is formed at a distance 60 cm from the mirror, behind it.
Thus, initial distance between the object and image = 60 + 60 = 120 cm
If the mirror is moved 25 cm away from the object,
The new distance of the object from the mirror = 60 + 25 = 85 cm
The new image is now at a distance 85 cm from the mirror behind it.
Thus, new distance of the image from the object = 85 + 85 = 170 cm
Taking the position of the object as reference point, the distance between the two positions of the image = new distance of image from the object – initial distance of the image from the object
= (170 – 120) cm = 50 cm
Thus, the image shifts 50 cm away.

Solution 5N.

Distance between man and chart = 3m
Distance between man and mirror = 2m
Therefore, distance between chart and mirror = 5 m
Now, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore, the chart as seen by patient is (5m + 2m =) 7m away.

Exercise 7(B)

Solution 1S.

If two mirrors make an angle θ with each other and object is placed in between the two mirrors, the number of images formed is n or (n – 1) depending upon n = 360o / θo is odd or even.

(a) If n = 360o / θo is odd,
(i) The number of images formed is n, when the object is placed asymmetrically between the mirrors.
(ii) The number of images formed is n-1, when the object is placed symmetrically between the mirrors.

(b) If n = 360o / θo is even, the number of images is always n-1.

Solution 2S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 44

Solution 3S.

For two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 11

Solution 4S.

For two mirrors kept parallel to each other, an infinite number of images are formed for an object kept in between them.Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 12

Solution 5S.

Two uses of plane mirror:

  1. In barber’s shop for seeing the hairs at the back of head, two mirrors facing each other are fixed on opposite walls at the front and back of the viewer.
  2. In solar heating devices such as a solar cooker, solar water heater, etc., a plane mirror is used to reflect the incident light rays from sun on the substance to be heated.

Solution 1M.

5

Solution 2M.

In a barber’s shop, two plane mirrors are placed parallel to each other.

Solution 1N.

(a) Angle between the mirrors, θ = 90o
Now, n = 360o / θo = 360o / 90o = 4, which is even.
Hence number of images formed will be (n-1); i.e., 4-1 = 3 images

(b) Angle between the mirrors, θ = 60o
Now, n = 360o / θo = 360o / 60o = 6, which is even.
Hence number of images formed will be (n-1); i.e., 6-1 = 5 images

Solution 2N.

Angle between the mirrors, θ = 50o
Now, n = 360o / θo = 360o / 50o = 7.2  7, which is odd.
(i) When placed asymmetrically, number of images formed will be n, i.e. 7.
(ii) When placed symmetrically, number of images formed will be (n-1); i.e. 7-1 = 6 images

Exercise 7(C)

Solution 1S.

A reflecting surface which is a part of a sphere is called a spherical mirror.

Solution 2S.

Two kinds of spherical mirrors are concave and convex.

Distinction between concave and convex mirror: A concave mirror’s bulging surface is silvered and reflection takes place from the hollow surface but a convex mirror’s inner surface is silvered and reflection takes place from the bulging surface.

Solution 3S.

Pole: The geometric centre of the spherical surface of mirror is called the pole of mirror.
Principal axis: It is the straight line joining the pole of the mirror to its centre of curvature.
Centre of curvature: The centre of curvature of a mirror is the centre of the sphere of which the mirror is a part.

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 13

Solution 5S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 14

Solution 6S.

Focus of a concave mirror: The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to principal axis, pass after reflection from the mirror.
Focal length of a concave mirror: The distance of the focus from the pole of the concave mirror is called its focal length.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 15

Solution 7S.

Focus of a convex mirror: The focus of a convex mirror is a point on the principal axis from which, the light rays incident parallel to principal axis, appear to come, after reflection from the mirror.
Focal length of a convex mirror: The distance of the focus from the pole of the convex mirror is called its focal length.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 16

Solution 8S.

Incident ray is directed towards the centre of curvature because the ray is normal to the spherical mirror, so ∠i = ∠r = O.

Solution 9S.

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Solution 10S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 18

Solution 11S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 19

Solution 12S.

Two convenient rays that are chosen to construct the image by a spherical mirror for a given object:

  1. A ray passing through the centre of curvature: A ray of light passing through the centre of curvature of a concave mirror or a ray directed in the direction of centre of curvature of a convex mirror is reflected back along the same path after reflection.
    Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 20
  2. A ray parallel to the principal axis: A ray of light parallel to the principal axis, after reflection pass through the principal focus in case of a concave mirror or appears to diverge from it in case of convex mirror.
    Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 21

Solution 13S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 22

Solution 14S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 23

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 24
The image formed is virtual, erect and magnified.

Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 25
The image formed is real, inverted and magnified.

Solution 18S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 26
The image formed is virtual, erect and diminished.

Solution 19S.

Convex mirror always produces erect and virtual images. The images formed are diminished, i.e. the size of the image is shorter than the size of the object.

Solution 20S.

(a) If the object is placed between the pole and focus of a concave mirror, the image formed is magnified and erect.
(b) The image is virtual.

Solution 21S.

(a) If the object is placed at the centre of curvature of a concave mirror, the image formed is of same size.
(b) The image formed is real and inverted.

Solution 22S.

(a) An image which can be obtained on a screen is called a real image.
(b) A concave mirror can be used to obtain a real image of an object.
(c) No, it does not form real image for all locations of the object.

Solution 23S.

When an object is moved from infinity towards the pole of mirror, the image formed moves away from the mirror. The image formed is real and inverted.

Solution 24S.

In a convex mirror, the image formed is always virtual, upright and diminished. It is always situated between its pole and focus, irrespective of the distance of object in front of the mirror.

Solution 25S.

(a) Concave, (b) Concave, (c) Convex and (d) Concave

Solution 26S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 27

Solution 27S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 28

Solution 28S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 29

Solution 29S.

The image formed by a convex mirror is always between pole and focus. Hence, the maximum distance that can be obtained in convex mirror is the focal length. For this case the object has to be at infinity.

Solution 30S.

The maximum distance that can be obtained in concave mirror is infinity. For this case the object has to be at focus.

Solution 31S.

To distinguish between a plane mirror, concave mirror and convex mirror, the given mirror is held near the face and image is seen. There can be following three cases:

Case (i): If the image is upright, of same size and it does not change in size by moving the mirror towards or away from the face, the mirror is plane.
(ii) If the image is upright and magnified, and increases in size on moving the mirror away, the mirror is concave.
(iii) If the image is upright and diminished and decreases in size on moving the mirror away, the mirror is convex.

Solution 32S.

Two uses of concave mirror:

  1. It is used as a shaving mirror.
  2. It is used as reflector in torch, head light of automobiles etc.

Solution 33S.

(a) Concave mirror
(b) Concave mirror

Solution 34S.

(a) The person’s face is between the pole and focus of the mirror.
(b) The image formed is erect, virtual and magnified.

Solution 35S.

A convex mirror is preferred as a rear view mirror because it has a wider field of view as compared to a plane mirror of same size.

Solution 36S.

A convex mirror diverges the incident beam and always forms a virtual, small and erect image between its pole and focus. Thus, a driver can see all the traffic approaching from behind. This fact enables the driver to use it as a rear view in vehicles to see all the traffic approaching from behind.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 30

Solution 1M.

Retraces its path

Solution 2M.

Erect and diminished

Solution 3M.

Concave mirror

Solution 1N.

Focal length = ½ (Radius of curvature)
Or, f = 40/2 = 20 cm

Solution 2N.

Radius of curvature = 2 x focal length
Or, R = 2f = 2 x 10 = 20 cm

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 31
The image is 30 cm in front of the mirror, 3 cm high, real, inverted and magnified.

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 32
The image is 6 cm behind the mirror.
Yes the image is magnified.

Solution 5N.

The size of the image is equal to the size of the object if the object is placed at the centre of curvature of a concave mirror.
Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 33
Hence, the object should be placed at 50 cm.

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 45
The position of the object is 12 cm in front of the mirror.
Its size is 1 cm.

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 35

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 36
A ray passing parallel to the principal axis passes through the focal point after reflection. Hence, the focal length is 12 cm.

Solution 9N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 37

Solution 10N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 38

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 39

Solution 12N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 40

Solution 13N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 41

Solution 14N.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 42

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Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure

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Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure

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Selina ICSE Solutions for Class 9 Physics Chapter 4 Pressure in Fluids and Atmospheric Pressure

Exercise 4(A)

Solution 1S.

Thrust is the force acting normally on a surface.
Its S.I. unit is ‘newton’.

Solution 2S.

Pressure is the thrust per unit area of the surface.
Its S.I. unit is ‘newton per metre2‘ or ‘pascal’.

Solution 3S.

(a) Pressure is measured in ‘bar’.
(b) 1 bar = 105 pascal.

Solution 4S.

One pascal is the pressure exerted on a surface of area 1 m2 by a force of 1N acting normally on it.

Solution 5S.

Thrust is a vector quantity.

Solution 6S.

Pressure is a scalar quantity.

Solution 7S.

Thrust is the force applied on a surface in a perpendicular direction and it is a vector quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.

Solution 8S.

Pressure exerted by thrust is inversely proportional to area of surface on which it acts. Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it.
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that sand, our body does not sink into the sand. In both the cases, the thrust exerted on the sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust acts on a large area and when we stand, the same thrust acts on a small area.

Solution 9S.

The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and it can be driven into with less effort.

Solution 10S.

(a) It is easier to cut with a sharp knife because even a small thrust causes great pressure at the edges and cutting can be done with less effort.
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure exerted by the rails on the ground becomes less.

Solution 11S.

A substance which can flow is called a fluid.

Solution 12S.

Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the fluid is called fluid pressure.

Solution 13S.

A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a fluid exerts pressure at all points in all directions.

Solution 14S.

Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a series of holes in the wall of the vessel anywhere below the free surface of the liquid. The water spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the bottle.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 3
Liquid exerts pressure at all points in all directions

Solution 15S.

Pressure at a point in a liquid depends upon the following three factors:

  1. Depth of the point below the free surface.
  2. Density of liquid.
  3. Acceleration due to gravity.

Solution 16S.

P = Po + hρg
Here, P = Pressure exerted at a point in the liquid
Po = Atmospheric pressure
h = Depth of the point below the free surface
ρ = Density of the liquid
g = Acceleration due to gravity

Solution 17S.

Consider a vessel containing a liquid of density ρ. Let the liquid be stationary. In order to calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a depth h below the free surface XY of the liquid. The pressure on the surface PQ will be due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base and top face RS lying on the frees surface XY of the liquid.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 2
Total thrust exerted on the surface PQ
= Weight of the liquid column PQRS
= Volume of liquid column PQRS x density x g
= (Area of base PQ x height) x density x g
= (A x h) x ρ x g

This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown below.
P = Thrust on surface / Area of surface
P = Ah ρg / A = hρg
Thus, Pressure = depth x density of liquid x acceleration due to gravity

Solution 18S.

Due to dissolved salts, density of sea water is more than the density of river water, so pressure at a certain depth in sea water is more than that at the same depth in river water.

Solution 19S.

(a) P2 = P1 + h ρ g,
(b) P2 > P1

Solution 20S.

The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the atmospheric pressure plus the pressure due to water column. As the gas bubble rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., the bubble grows in size.

Solution 21S.

The pressure exerted by a liquid increases with its depth. Thus as depth increases, more and more pressure is exerted by water on wall of the dam. A thicker wall is required to withstand greater pressure, therefore, the thickness of the wall of dam increases towards the bottom.

Solution 22S.

The sea divers need special protective suit to wear because in deep sea, the total pressure exerted on the diver’s body is much more than his blood pressure. To withstand it, he needs to wear a special protective suit.

Solution 23S.

Laws of liquid pressure:

  1. Pressure at a point inside the liquid increases with the depth from its free surface.
  2. In a stationary liquid, pressure is same at all points on a horizontal plane.
  3. Pressure is same in all directions about a point in the liquid.
  4. Pressure at same depth is different in different liquids. It increases with the increase in the density of liquid.
  5. A liquid seeks its own level.

Solution 24S.

The liquid from hole B reaches a greater distance on the horizontal surface than that from hole A.
This explains that liquid pressure at a point increases with the depth of point from the free surface.

Solution 25S.

(i) As the diver moves to a greater depth, pressure exerted by sea water on him also increases.
(ii) When the diver moves horizontally, his depth from the free surface remains constant and hence the pressure on him remains unchanged.

Solution 26S.

Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Solution 27S.

Two applications of Pascal’s law:

  1. Hydraulic press
  2. Hydraulic jack

Solution 28S.

The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Hydraulic press and hydraulic brakes work on this principle.

Solution 29S.

Hydraulic press works on principle of hydraulic machine.
It states that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Use: It is used for squeezing oil out of linseed and cotton seeds.

Solution 30S.

(i) X : Press Plunger; Y: Pump Plunger

(ii) When the lever is moved down, valve B closes and valve A opens, so the water from cylinder P is forced into the cylinder Q.

(iii) Valve B closes due to an increase in pressure in cylinder P. This pressure is transmitted to the connecting pipe and when the pressure in connecting pipe becomes greater than the pressure in the cylinder Q, valve A opens up.

(iv) When the release valve is opened, the ram (or press) plunger Q gets lowered and water of the cylinder Q runs out in the reservoir.

Solution 31S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 1
Working: When handle H of the lever is pressed down by applying an effort, the valve V opens because of increase in pressure in cylinder P. The liquid runs out from the cylinder P to the cylinder Q. As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of the lever is no longer pressed. The valve V gets closed (since the pressure on the either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to cylinder P.

Solution 32S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 4

Working: To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q. As a result, the pressure is transmitted equally and undiminished through the liquid to the pistons B1and B2 of the wheel cylinder. Therefore, the pistons B1 and B2 get pushed outwards and brake shoes get pressed against the rim of the wheel due to which the motion of the vehicle retards. Due to transmission of pressure through the liquid, equal pressure is exerted on all the wheels of the vehicle connected to the pipe line R.

On releasing the pressure on the pedal, the liquid runs back from the wheel cylinder Q to the master cylinder P and the spring pulls the break shoes to their original position and forces the pistons B1 and B2 to return back into the wheel cylinder Q. Thus, the brakes get released.

Solution 33S.

(a) h ρ g (b) same (c) the same (d) directly proportional (e) directly proportional.

Solution 1M.

Pa

Solution 2M.

h ρ g

Solution 3M.

P1 < P2

Solution 4M.

P2 – P1 = h ρ g

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 5

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 6

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 7

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 8

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 9

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 10

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 11

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 12

Solution 9N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 13

Solution 10N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 14

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 15

Solution 12N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 16

Solution 13N.

Data is incomplete

Solution 14N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 17

Exercise 4(B)

Solution 1S.

The thrust exerted per unit area of the earth surface due to column of air, is called the atmospheric pressure on the earth surface.

Solution 2S.

1.013 x 10 5 pascal

Solution 3S.

Atmospheric pressure is measured in ‘torr’.
1 torr = 1 mm of Hg.

Solution 4S.

At normal temperature and pressure, the barometric height is 0.76 m of Hg at sea level which is taken as one atmosphere.
1 atmosphere = 0.76 m of Hg = 1.013 x 105 pascal

Solution 5S.

We do not feel uneasy under enormous pressure of the atmosphere above as well as around us because of the pressure of our blood, known as blood pressure, is slightly more than the atmospheric pressure. Thus, our blood pressure balances the atmospheric pressure.

Solution 6S.

Experiment to demonstrate that air exerts pressure:

Take a thin can fitted with an airtight stopper. The stopper is removed and a small quantity of water is boiled in the can. Gradually the steam occupies the entire space of can by expelling the air from it [Fig (a)]. Then stopper is then tightly replaced and simultaneously the flame beneath the can is removed. Cold water is then poured over the can.

It is observed that the can collapses inwards as shown in fig (b).

The reason is that the pressure due to steam inside the can is same as the air pressure outside the can [Fig (a)]. However, on pouring cold water over the can, fitted with a stopper [fig (b)], the steam inside the can condenses producing water and water vapour at very low pressure. Thus, the air pressure outside the can becomes more than the vapour pressure inside the closed can.

Consequently, the excess atmospheric pressure outside the can causes it to collapse inwards.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 18

Solution 7S.

(i) When air is removed from the balloon, the pressure inside the balloon (which was due to air in it) is much less than the atmospheric pressure outside and hence the balloon collapses.

(ii) Water is held inside the dropper against the atmospheric pressure because the pressure due to height column of liquid inside the dropper is less than the atmospheric pressure. By pressing the dropper we increase the pressure inside the dropper and when it becomes greater than the atmospheric pressure the liquid comes out of the dropper.

(iii) There is no air inside a completely filled and sealed can. When a single hole is made to drain out the oil from the can, some of the oil will come out and due to that the volume of air above the oil will increase and hence the pressure of air will decrease. But if two holes are made on the top cover of the can, air outside the can will enter it through one hole and exert atmospheric pressure on the oil from inside along with the pressure due to oil column, and it will come out of the can from the other hole.

Solution 8S.

When syringe is kept with its opening just inside a liquid and its plunger is pulled up in the barrel, the pressure of air inside the barrel below the plunger becomes much less than the atmospheric pressure acting on the liquid. As a result, the atmospheric pressure forces the liquid to rise up in the syringe.

Solution 9S.

In a water pump, on pulling the piston up, the pressure of air inside the siphon decreases and the atmospheric pressure on the water outside increases. As a result, the atmospheric pressure pushes the water up in pump.

Solution 10S.

(a) Pressure increases inside the bell jar.
(b) Pressure decreases inside the balloon.

Solution 11S.

A barometer is used to measure atmospheric pressure.

Solution 12S.

A barometer is an instrument which is used to measure the atmospheric pressure.

Construction of a simple barometer:

A simple mercury barometer can be made with a clear, dry, thick-walled glass tube about 1 metre ling. The glass tube is sealed at one end and is filled with mercury completely. While filling the tube with mercury care has to be taken so that there are no air bubbles present in the mercury column. Close the open end with thumb and turn the tube upside down carefully over a trough containing mercury. Dip the open end under the mercury level in the trough and remove the thumb.

The mercury level in the tube falls until it is about 76 cm (h =760 mm) vertically above the mercury level. It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. The empty space above the mercury column is called the ‘Torricellian vacuum’.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 19

Solution 13S.

In given figure, at all points such as C on the surface of mercury in trough, only the atmospheric pressure acts. When the mercury level in the tube becomes stationary, the pressure inside tube at the point A, which is at the level of point C, must be same as that at the point C. The pressure at point A is due to the weight (or thrust) of the mercury column AB above it. Thus, the vertical height of the mercury column from the mercury surface in trough to the level in tube is a measure of the atmospheric pressure.
The vertical of the mercury column in it (i.e., AB = h) is called the barometric height.
Had the pressure at points A and C be not equal, the level of mercury in the tube would not have been stationary.
Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 20

Solution 14S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 21

Solution 15S.

It is the atmospheric pressure acting on the surface of the mercury in the trough that supports the vertical mercury column. Hence, barometric height is used as unit to express the atmospheric pressure.

Solution 16S.

The atmospheric pressure at a place is 76 cm of Hg means at normal temperature and pressure, the height of the mercury column supported by the atmospheric pressure is 76 cm.
76 cm of Hg = 1.013 x 105 pascal

Solution 17S.

The space above mercury is a vacuum. This empty space is called ‘Torricellian vacuum’.

This can be shown by tilting the tube till the mercury fills the tube completely. Again when the mercury column becomes stationary, the empty space is created above the mercury column. If somehow air enters into the empty space or a drop of water gets into the tube, it will immediately vaporize and the air will exert pressure on mercury column due to which the barometric height will decrease.

Solution 18S.

(a) The barometric height remains unaffected.
(b) The barometric height remains unaffected.
(c) The barometric height decreases.

Solution 19S.

Two uses of barometer:

  1. To measure the atmospheric pressure.
  2. For weather forecasting

Solution 20S.

Two advantages of using mercury as barometric liquid:

  1. The density of mercury is greater than that of all the liquids, so only 0.76m height of mercury column is needed to balance the normal atmospheric pressure.
  2. The mercury neither wets nor sticks to the glass tube therefore it gives the correct reading.

Solution 21S.

Water is not a suitable barometric liquid because:

  1. The vapour pressure of water is high, so its vapours in the vacuum space will make the reading inaccurate.
  2. Water sticks with the glass tube and wets it, so the reading becomes inaccurate.

Solution 22S.

In a simple barometer, there is no protection for the glass tube but in Fortin’s barometer, this defect has been removed by enclosing the glass tube in a brass case.
In a simple barometer, a scale cannot be fixed with the tube (or it cannot be marked on the tube) to measure the atmospheric pressure but Fortin’s barometer is provided with a vernier calipers to measure the accurate reading.

Solution 23S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 22
To measure the atmospheric pressure, first the leather cup is raised up or lowered down with the help of the screw S so that the ivory pointer I just touches the mercury level in the glass vessel. The position of the mercury level in the barometer tube is noted with the help of main scale and the vernier scale. The sum of the vernier scale reading to the main scale reading gives the barometric height.

Solution 24S.

A barometer calibrated to read directly the atmospheric pressure is called an aneroid barometer. It has no liquid, it is light and portable.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 23

Construction: Figure above shows the main parts of an aneroid barometer. It consists of a metallic box B which is partially evacuated. The top D of the box is springy and corrugated in form of a diaphragm as shown. At the middle of diaphragm, there is a thin rod L toothed at its upper end. The teeth of rod fit well into the teeth of a wheel S attached with a pointer P which can slide over a circular scale. The circular scale is initially calibrated with a standard barometer so as to read the atmospheric pressure directly in terms of the barometric height.

Working: When atmospheric pressure increases, it presses the diaphragm D and the rod L gets depressed. The wheel S rotates clockwise and pointer P moves to the right on the circular scale. When atmospheric pressure decreases, the diaphragm D bulges out due to which the rod L moves up and the wheel S rotates anti-clockwise. Consequently, the pointer moves to the left.

Solution 25S.

Aneroid barometer has no liquid and it is portable. It is calibrated to read directly the atmospheric pressure.

Solution 26S.

(i) In a mine, reading of a barometer increases.
(ii) On hills, reading of barometer decreases.

Solution 27S.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 24

Solution 28S.

Factors that affect the atmospheric pressure are:

  1. Height of air column
  2. Density of air

Solution 29S.

A fountain pen filled with ink contains some air at a pressure equal to atmospheric pressure on earth’s surface. When pen is taken to an altitude, atmospheric pressure is low so the excess pressure inside the rubber tube forces the ink to leak out.

Solution 30S.

On mountains, the atmospheric pressure is quite low. As such, nose bleeding may occur due to excess pressure of blood over the atmospheric pressure.

Solution 31S.

An altimeter is a device used in aircraft to measure its altitude.

Principle: Atmospheric pressure decreases with the increase in height above the sea level; therefore, a barometer measuring the atmospheric pressure can be used to determine the altitude of a place above the sea level.

The scale of altimeter is graduated with height increasing towards left because the atmospheric pressure decreases with increase of height above the sea level.

Solution 32S.

(a) It indicates that the moisture is increasing i.e., there is a possibility of rain.
(b) It indicates the coming of a storm or cyclone.
(c) It indicates that the moisture is decreasing i.e., it indicates dry weather.

Solution 1M.

1 torr = 1 mm of Hg

Solution 2M.

76 cm of Hg

Solution 3M.

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 25

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 26

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 27
Assumption: Atmospheric pressure falls linearly with ascent.

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 28

Solution 5N.

Selina Concise Physics Class 9 ICSE Solutions Pressure in Fluids and Atmospheric Pressure - 29

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Selina Concise Physics Class 9 ICSE Solutions Heat and Energy

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Selina Concise Physics Class 9 ICSE Solutions Heat and Energy

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 6 Heat and Energy. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 6 Heat and Energy

Exercise 6(A)

Solution 1S.

Heat is the energy of random motion of molecules constituting the body.
Its S.I. unit is ‘joule’.

Solution 2S.

Heat will flow from a hot body (body at a higher temperature) to a cold body (body at a lower temperature).

Solution 3S.

S.I. unit of heat is ‘joule’.
1 joule = 0.24 cal

Solution 4S.

Temperature is the parameter which tells the thermal state of a body (i.e. the degree of hotness or coldness).
The S.I. unit of temperature is ‘kelvin’.

Solution 5S.

On touching a piece of ice, heat flows from our hand (hot body) to the ice (cold body), and hence, it appears cold.

Solution 6S.

Heat is a form of energy obtained due to the random motion of molecules in a substance but temperature is a quantity which decided the direction of flow of heat when two bodies at different temperature are placed in contact. Two quantities having the same amount of heat may differ in temperature.

Solution 7S.

The expansion of a substance on heating is called thermal expansion.

Solution 8S.

Brass and iron expand on heating.

Solution 9S.

Water contracts on heating from 0°C to 4°C.

Silver iodide contracts on heating from 80°C to 141°C.

Solution 10S.

The expansion of water when it is cooled from 4°C to 0°C is known as the anomalous expansion of water.

Solution 11S.

Density of water is maximum at 4°C. Its value is 1000 kgm-3.

Solution 12S.

When a given mass of water is heated from 0°C to 4°C, it contracts, i.e. its volume decreases.
On heating from 4°C to 10°C, it expands, i.e. its volume increases.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 1

Solution 13S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 2

Solution 14S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 3

Solution 15S.

Hope’s experiment to demonstrate that water has maximum density at 4°C:

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 4
Hope’s apparatus consists of a tall metallic cylinder provided with two side openings P and Q, P near the top and Q near the bottom, fitted with thermometers T1 and T2 in them. The central part of the cylinder is surrounded with a cylindrical trough containing a freezing mixture of ice and salt. The cylinder is fitted with pure water at room temperature.

Observations: (i) Initially, both thermometers T1 and T2 are at the same temperature.

(ii) First, the temperature recorded by the lower thermometer T2 starts decreasing and finally it becomes steady at 4°C, while the temperature recorded in the upper thermometer T1 remains almost unchanged during this time.

(iii) Then, the temperature recorded by the lower thermometer T2 remains constant at 4°C and upper thermometer T1 records a continuous fall in temperature up to 0°C and then it becomes steady.

Thus, finally, the temperature recorded by the upper thermometer is 0°C and that by lower thermometer is 4°C.

As the freezing mixture cools water in the central portion of the cylinder, water contracts and its density increases, consequently it sinks to the bottom, thereby causing the reading of the lower thermometer T2 to fall rapidly. The reading of the upper thermometer T1 does not change as the temperature of water in the upper part does not change. This continues till the entire water below the central portion reaches 4°C. On cooling further below 4°C, due to anomalous expansion, water of the central portion expands, so its density decreases and hence it rises up. As a result, reading of the upper thermometer T1 falls rapidly to 0°C and water freezes to form ice at 0°C near the top. This proves that water has maximum density at 4°C.

This anomalous expansion of water helps in preserving the aquatic life during the very cold weather. In winters, when the temperature falls, the top layer of water in a pond contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the pond reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layers are frozen, the water near the bottom is at 4°C and the fishes can survive in it easily.

Solution 16S.

(i) Water just in contact with ice is at 0°C.
(ii) Water at the bottom of the pond is at 4°C.

Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 5

Solution 18S.

(a) On winter nights, as the atmospheric pressure starts falling below 4°C, water in the pipe lines expand and exert a great pressure on the pipes, causing them to burst.

(b) In winters, when temperature falls, the surface of water in the tank contracts, becomes denser and sinks to the bottom. A circulation is thus set up until the entire water in the tank reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, water in a tank starts freezing from the top and not from the bottom.

(c) The anomalous expansion of water helps preserve aquatic life during very cold weather. When temperature falls, the top layer of water in a pond contracts becomes denser and sinks to the bottom. A circulation is thus set up until water in the pond reaches its maximum density at 4°C. If the temperature falls further, then the top layer expands and remains on the top till it freezes. Thus, even though the upper layer are frozen, the water near the bottom is at 4°C and the fishes can survive in it easily.

(d) On heating water above 4°C, the density of water decreases. As a result, the upthrust acting due to water on hollow glass sphere also decreases, which causes it to sink.

(e) Inside the freezer, when the temperature of water falls below 4°C, the water in the bottle starts expanding. If the bottle is completely filled and tightly closed, there is no space for water to expand, and hence, the bottle may burst.

Solution 1M.

Calorie is the unit heat.

Solution 2M.

1 calorie = 4.186 J
Therefore,
1 J = 1/4.186 = 0.24 cal

Solution 3M.

The SI unit of temperature is kelvin (K).

Solution 4M.

Water shows anomalous behavior between 0°C and 4°C. Hence, when it is cooled it expands.

Solution 5M.

Water shows anomalous behavior between 0°C and 4°C. It has lowest volume at 4°C. Hence, its density will be maximum at 4°C.

Exercise 6(B)

Solution 1S.

A unit composed of biotic components (i.e. producers, consumers and decomposers) and abiotic components (i.e. light, heat, rain, and humidity, inorganic and organic substances) is called an ecosystem.

Solution 2S.

The source of energy for all ecosystems is the Sun.

Solution 3S.

Green plants absorb most of the energy falling on them and by the process of photosynthesis they produce food for the consumers. Plants, being primary producers are of great importance in the ecosystem. They also maintain the balance of oxygen and carbon dioxide on earth.

Solution 4S.

Producers like plants and some bacteria are capable of producing its own food using the energy of sun but consumers are not capable of producing their own food. They depend on producers for food.

Solution 5S.

The role of a decomposer is to break down dead organisms and then feed on them. The nutrients created by the dead organisms are returned to the soil to be later used by the producers. Once these deceased organisms are returned to the soil, they are used as food by bacteria and fungi by transforming the complex organic materials into simpler nutrients. The simpler products can then be used by producers to restart the cycle. These decomposers play an important role in every ecosystem.

Solution 6S.

A food chain shows the feeding relationship between different living things in a particular environment or habitat. Often, a plant will begin a food chain because it can make its own food using energy from the Sun. In addition, a food chain represents a series of events in which food and energy are transferred from one organism in an ecosystem to another. Food chains show how energy is passed from the sun to producers, from producers to consumers, and from consumers to decomposers.

Solution 7S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 6

Solution 8S.

Ecosystems maintain themselves by cycling energy and nutrients obtained from external sources. At the first trophic level, primary producers (plants, algae, and some bacteria) use solar energy to produce organic plant material through photosynthesis. Herbivores-animals that feed solely on plants-make up the second trophic level. Predators that eat herbivores comprise the third trophic level; if larger predators are present, they represent still higher trophic levels. Decomposers, which include bacteria, fungi etc. break down wastes and dead organisms and return nutrients to the soil.
Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 7

On average about 10 percent of net energy production at one trophic level is passed on to the next level. Processes that reduce the energy transferred between trophic levels include respiration, growth and reproduction, defecation, and non-predatory death.

The low rate of energy transfer between trophic levels makes decomposers generally more important than producers in terms of energy flow. Decomposers process large amounts of organic material and return nutrients to the ecosystem in inorganic forms, which are then taken up again by primary producers.

Solution 9S.

The laws of thermodynamics govern the energy flow in the ecosystem.
According to the first law of thermodynamic, the energy can be transformed from one form to the other form, but it can neither be created nor destroyed.
According to the second law of thermodynamics, when energy is put to work, a part of it is always converted in un-useful form such as heat mainly due to friction and radiation.

Solution 10S.

The energy flow in ecosystem is linear i.e., it moves in a fixed direction. The solar energy is absorbed by plants and a part of it is converted into food. These plants (or primary producers) are then eaten by the primary consumers, which are consumed by secondary consumers and the secondary by tertiary consumers. This cycle is unidirectional. The dead and decomposed are fed by decomposers, which return the nutrients to the soil. At the end, the energy reaches the degraded state. It does not return to the sun to make the process cyclic, thus energy flow is linear.

Solution 11S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 8

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Heat and Energy image - 9

Solution 1M.

Photosynthesis

Solution 2M.

Sun

Solution 3M.

Producers

Solution 4M.

Consumer

Exercise 6(C)

Solution 1S.

  1. A source of energy should be safe and convenient to use.
  2. A source of energy should be economical and easy to store and transport.

Solution 2S.

The two groups in which various sources of energy are classified are renewable or non-conventional sources of energy and non-renewable or conventional sources of energy.
These sources are classified on the basis of their availability and utility.

Solution 3S.

Renewable: The natural sources providing us energy continuously are called renewable sources of energy.
Non-renewable: The sources of energy which have accumulated in nature over a very long period of time and cannot be quickly replaced when exhausted are called non-renewable sources of energy.

Difference:

Renewable sources Non-renewable sources
They can be utilised continuously. They cannot be utilised once exhausted.
Examples: Sun, Wind Example: Coal, Petroleum

Solution 4S.

Renewable: Wood, Water and Wind
Non-renewable: Coal, Diesel and Oil

Solution 5S.

Wood is obtained from trees. Hence, trees need to be cut down for wood to be used as a fuel.
Also, burning wood releases a lot of smoke which pollutes the atmosphere.

Solution 6S.

Renewable:

  1. Sun
  2. Wind
  3. Flowing water
  4. Tides
  5. Nuclear fuel

Non-renewable:

  1. Coal
  2. Petroleum
  3. Natural gas

Solution 7S.

  1. Tidal energy: The energy possessed by rising and falling water in tides is known as tidal energy.Dams are constructed across a narrow opening to the sea to harness tidal energy and produce electricity. However, it is not a major source of energy as the rise and fall of seawater during tides is not enough to generate electricity on a large scale.
  2. Ocean energy: Water in the oceans possesses energy in two forms:
    1. Ocean thermal energy- The energy available due to the difference in temperature of water at the surface and at deeper levels of ocean is called the ocean thermal energy. This energy is harnessed for producing electricity by a device called ocean thermal energy conversion power plant (OCTEC power plant).
    2. Oceanic waves energy- The kinetic energy possessed by fast moving oceanic (or sea) waves is called oceanic waves energy. Though models have been made to generate electricity from oceanic waves, but so far it has not been put to practical use.
  3. Geo thermal energy: The heat energy possessed by the rocks inside the Earth is called geothermal energy.The hot rocks present at the hot spots deep inside the Earth, heat the underground water and turn it into steam. This steam is compressed at high pressure between the rocks. Holes are drilled deep into the Earth up to the hot spots to extract the steam through pipes, which is utilized to rotate the turbines connected to the armature of an electric generator to produce electricity.

Solution 8S.

Sun is the main source of energy on Earth.

Solution 9S.

The energy obtained from Sun is called solar energy.

A solar power plant is a device in which heat energy of sun is used to generate electricity. It consists of a large number of concave reflectors, at the focus of which there are black painted water pipes. The reflectors concentrate the heat energy of the sun rays on the pipes due to which water inside the pipes starts boiling and produces steam. The steam thus produced is used to rotate a steam turbine which drives a generator producing electricity.

Solution 10S.

A solar cell is an electrical device that converts light energy directly into electricity with the help of photovoltaic effect. Solar cells are usually made from semiconductors like silicon and gallium with some impurity added to it. When sunlight is made incident on a solar cell, a potential difference is produced between its surface, due to which a current flows in the circuit connected between the opposite faces of the semiconductor.

Two uses of solar cells are as listed below:

  1. They do not require maintenance and last over a long period of time at zero running cost.
  2. They are very useful for remote, inaccessible and isolated places where electric power lines cannot be laid.Solar cell produces d.c. (direct current).

One disadvantage of solar cell is listed below:

  1. The initial cost of a solar panel is sufficiently high.

Solution 11S.

Advantages of using solar panels:

  1. They do not cause any pollution in the environment.
  2. Running cost of solar panel is almost zero.
  3. They last over a long period of time.
  4. They do not require any maintenance.
  5. They are suitable for remote and inaccessible places where electricity power lines cannot be laid.

Disadvantages of using solar panels:

  1. The initial cost of a solar panel is sufficiently high.
  2. The efficiency of conversion of solar energy to electricity is low.
  3. A solar panel produces d.c. electricity which cannot be directly used for many household purposes.

Solution 12S.

The kinetic energy of the moving large masses of air is called the wind energy. Wind energy is used in a wind generator to produce electricity by making use of wind mill to drive a wind generator.

At present in India, more than 1025 MW electric power is generated using wind energy.

Solution 13S.

Advantages of using wind energy:

  1. It does not cause any kind of pollution.
  2. It is an everlasting source.

Limitations of using wind energy:

  1. The establishment of a wind farm is expensive.
  2. A large area of land is needed for the establishment of a wind farm.

Solution 14S.

The kinetic energy possessed by flowing water is called the water or hydro energy.
Principle of a hydroelectric power plant is that the water flowing in high altitude rivers is collected in a high dam (or reservoir). The water from the dam is then allowed to fall on a water turbine which is located near the bottom of the dam. The shaft of the turbine is connected to the armature of an electric generator or dynamo.

At present only 23% of the total electricity is generated by the hydro energy.

Solution 15S.

Advantages of producing the hydro electricity:

  1. It does not produce any environmental pollution.
  2. It is a renewable source of energy.

Disadvantages of producing hydroelectricity:

  1. Due to the construction of dams over the rivers, plants and animals of that place get destroyed or killed.
  2. The ecological balance in the downstream areas of rivers gets disturbed.

Solution 16S.

When a heavy nucleus is bombarded with slow neutrons, it splits into two nearly equal light nuclei with a release of tremendous amount of energy. In this process of nuclear fission, the total sum of masses of products is less than the total sum of masses of reactants. This lost mass gets converted into energy. The energy so released is called nuclear energy.

Principle: The heat energy released due to the controlled chain reaction of nuclear fission of uranium-235 in a nuclear reactor is absorbed by the coolant which then passes through the coils of a heat exchanger containing water. The water in heat exchanger gets heated and converts into steam. The steam is used to rotate the turbine which in turn rotates the armature of a generator in a magnetic field and thus produces electricity.

Solution 17S.

At present only about 3% of the total electrical power generated in India is obtained from the nuclear power plants.
Tarapur in Maharahtra and Narora in Uttar Pradesh are the places where electricity is produced using nuclear energy.

Solution 18S.

Advantages of using nuclear energy:

  1. A very small amount of nuclear fuel can produce a tremendous amount of energy.
  2. Once the nuclear fuel is loaded into nuclear power plant, it continues to release energy for several years.

Disadvantages of using nuclear energy:

  1. It is not a clean source of energy because very harmful nuclear radiations are produced in the process.
  2. The waste causes environmental pollution.

Solution 19S.

i. Light energy into electrical energy
ii. Mechanical energy into electrical energy.
iii. Mechanical energy into electrical energy.
iv. Nuclear energy (or heat energy) into electrical energy.

Solution 20S.

Four ways for the judicious use of energy are:

  1. The fossil fuels such as coal, petroleum, natural gas should be used only for the limited purposes when there is no other alternative source of energy available.
  2. The wastage of energy should be avoided.
  3. Efforts must be made to make use of energy for community or group purposes.
  4. The cutting of trees must be banned and more and more new trees must be roped to grow.

Solution 21S.

The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:

  1. When we cook food over fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
  2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.

Solution 22S.

The conversion of part of energy into a non-useful form of energy is called degradation of energy.

Solution 1M.

The ultimate source of energy is the Sun.

Solution 2M.

Renewable source of energy is the Sun.

Exercise 6(D)

Solution 1S.

Greenhouse effect is the process of warming of planet’s surface and its lower atmosphere by absorbtion of infrared radiations of longer wavelength emitted out from the surface of planet.

Solution 2S.

Carbon-di-oxide, water vapour and methane are greenhouse gases.

Solution 3S.

Visible light rays and short infrared radiation pass through the atmosphere of earth.

Solution 4S.

Infrared radiations of long wavelength are absorbed by the green house gases.

Solution 5S.

The concentration of carbon-di-oxide content’s of earth’s atmosphere has increased due to industrial growth, combustion of fossil fuels and clearing of forests.

Solution 6S.

In absence of green house gases, the average temperature on earth would be -18°C.

Solution 7S.

The greenhouses gases have an average warming effect on Earth’s surface of about 15.5°C (or 60°F ).

Solution 8S.

Global warming means the increase in average effective temperature near the earth’s surface due to an increase in the amount of green house gases in its atmosphere.

Solution 9S.

With activities industrialization, deforestation, excess burning of fossil fuel, the concentration of green house gases has increased on earth’s atmosphere. This increase in the amount of greenhouse gases present in atmosphere has caused the rise in atmospheric temperature.

Solution 10S.

The increase in green house gases due to activities like industrialization, deforestation, natural gas exploration, burning of biomass, natural gas exploration, more use of gadgets like refrigerators has caused the increase of green house effect.

Solution 11S.

At the poles, due to increase in temperature, the snow and ice will melt which will cause flood in coastal countries. The icebergs of dark land and oceans will melt, so the dark land and oceans will become uncovered and will absorb more heat radiations coming from sun, increasing the green house effect further.

Solution 12S.

Due to global warming, the snow and ice around the poles will melt and cause flood in coastal countries.

Solution 13S.

Due to melting of polar ice and glaciers, there will be rise in sea level on coastal wet lands. It would raise worldwise sea level, thereby, many big cities in the coastal areas will be covered by sea water.

Solution 14S.

Global warming will cause drastic changes in the patterns of wind, rainfall etc. Thus it will result in low agricultural yield.

Solution 15S.

  1. Use of renewable sources of energy to generate electricity in place of generating electricity from the fossil fuels based power plants.
  2. Controlling population through family planning, welfare reforms and the empowerment of women.

Solution 16S.

The tax calculated on the basis of carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.
This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.

Solution 1M.

Carbon dioxide

Solution 2M.

Increase in temperature

Solution 3M.

Without green house effect, the average temperature of Earth’s surface would have been -18 °C.

Solution 4M.

The global warming has resulted in the increase in sea levels.

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Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension

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Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 2 Motion in One Dimension. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 2 Motion in One Dimension

Exercise 2(A)

Solution 1S.

Scalar Vector
They are expressed only by their magnitudes. They are expressed by magnitude as well as direction.
They can be added, subtracted, multiplied or divided by simple arithmetic methods. They can be added, subtracted or multiplied following a different algebra.
They are symbolically written by English letter. They are symbolically written by their English letter with an arrow on top of the letter.
Example: mass, speed Example: force, velocity

Solution 2S.

a) Pressure is a scalar quantity.
b) Momentum is a vector quantity.
c) Weight is a vector quantity.
d) Force is a vector quantity.
e) Energy is a scalar quantity.
f) Speed is a scalar quantity.

Solution 3S.

A body is said to be at rest if it does not change its position with respect to its immediate surroundings.

Solution 4S.

A body is said to be in motion if it changes its position with respect to its immediate surroundings.

Solution 5S.

When a body moves along a straight line path, its motion is said to be one-dimensional motion.

Solution 6S.

The shortest distance from the initial to the final position of the body is called the magnitude of displacement. It is in the direction from the initial position to the final position.
Its SI unit is metre (m).

Solution 7S.

Distance is a scalar quantity, while displacement is a vector quantity. The magnitude of displacement is either equal to or less than the distance. The distance is the length of path travelled by the body so it is always positive, but the displacement is the shortest length in direction from initial to the final position so it can be positive or negative depending on its direction. The displacement can be zero even if the distance is not zero.

Solution 8S.

Yes, displacement can be zero even if the distance is not zero.

For example, when a body is thrown vertically upwards from a point A on the ground, after sometime it comes back to the same point A. Then, the displacement is zero, but the distance travelled by the body is not zero (it is 2h; h is the maximum height attained by the body).

Solution 9S.

The magnitude of displacement is equal to distance if the motion of the body is one-dimensional.

Solution 10S.

The velocity of a body is the distance travelled per second by the body in a specified direction.
Its SI unit is metre/second (m/s).

Solution 11S.

The speed of a body is the rate of change of distance with time.
Its SI unit is metre/second (m/s).

Solution 12S.

Speed is a scalar quantity, while velocity is a vector quantity. The speed is always positive-it is the magnitude of velocity, but the velocity is given a positive or negative sign depending upon its direction of motion. The average velocity can be zero but the average speed is never zero.

Solution 13S.

Velocity gives the direction of motion of the body.

Solution 14S.

Instantaneous velocity is equal to average velocity if the body is in uniform motion.

Solution 15S.

If a body travels equal distances in equal intervals of time along a particular direction, then the body is said to be moving with a uniform velocity. However, if a body travels unequal distances in a particular direction in equal intervals of time or it moves equal distances in equal intervals of time but its direction of motion does not remain same, then the velocity of the body is said to be variable (or non-uniform).

Solution 16S.

Average speed is the ratio of the total distance travelled by the body to the total time of journey, it is never zero. If the velocity of a body moving in a particular direction changes with time, then the ratio of displacement to the time taken in entire journey is called its average velocity. Average velocity of a body can be zero even if its average speed is not zero.

Solution 17S.

The motion of a body in a circular path with uniform speed has a variable velocity because in the circular path, the direction of motion of the body continuously changes with time.
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 1

Solution 18S.

If a body starts its motion from a point and comes back to the same point after a certain time, then the displacement is zero, average velocity is also zero, but the total distance travelled is not zero, and therefore, the average speed in not zero.

Solution 19S.

Acceleration is the rate of change of velocity with time.
Its SI unit is metre/second2 (m/s2).

Solution 20S.

Acceleration is the increase in velocity per second, while retardation is the decrease in velocity per second. Thus, retardation is negative acceleration. In general, acceleration is taken positive, while retardation is taken negative.

Solution 21S.

The acceleration is said to be uniform when equal changes in velocity take place in equal intervals of time, but if the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable.

Solution 22S.

Retardation is the decrease in velocity per second.
Its SI unit is metre/second2 (m/s2).

Solution 23S.

Velocity determines the direction of motion.

Solution 24S.

(a) Example of uniform velocity: A body, once started, moves on a frictionless surface with uniform velocity.
(b) Example of variable velocity: A ball dropped from some height is an example of variable velocity.
(c) Example of variable acceleration: The motion of a vehicle on a crowded road is with variable acceleration.
(d) Example of uniform retardation: If a car moving with a velocity ‘v’ is brought to rest by applying brakes, then such a motion is an example of uniform retardation.

Solution 25S.

Initially as the drops are equidistant, we can say that the car is moving with a constant speed but later as the distance between the drops starts decreasing, we can say that the car slows down.

Solution 26S.

When a body falls freely under gravity, the acceleration produced in the body due to the Earth’s gravitational acceleration is called the acceleration due to gravity (g). The average value of g is 9.8 m/s2.

Solution 27S.

No. The value of ‘g’ varies from place to place. It is maximum at poles and minimum at the Equator on the surface of the Earth.

Solution 28S.

In vacuum, both will reach the ground simultaneously because acceleration due to gravity is same (=g) on both objects.

Solution 1M.

Velocity is a vector quantity. The others are all scalar quantities.

Solution 2M.

m s-1

Solution 3M.

m s-2

Solution 4M.

The displacement is zero.

Solution 5M.

5 m s-1

Solution 1N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 2

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 3

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 4

Solution 4N.

18 km h-1 < 10 m s-1 < 1 km min-1

Solution 5N.

Total time taken = 3 hours
Speed of the train = 65 km/hr
Distance travelled = speed x time
= 65 x 3 = 195 km

Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 5

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 6
(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.

Solution 8N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 7

Solution 9N.

Here, final velocity = 10 m/s
Initial velocity = 0 m/s
Time taken = 2s
Acceleration = (Final Velocity – Initial Velocity)/time
= (10/2) m/s2
 = 5 m/s2

Solution 10N.

Here, final velocity = 180 m/s
Initial velocity = 0 m/s
Time taken = 0.05 h or 180 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (180-0)/180 m/s2
 = 1 m/s2

Solution 11N.

Here, final velocity = 20 m/s
Initial velocity = 50 m/s
Time taken = 3 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (20 – 50)/3 m/s2
 = -10 m/s2
Negative sign here indicates that the velocity decreases with time, so retardation is 10 m/s2.

Solution 12N.

Here, final velocity = 18 km/h or 5 m/s
Initial velocity = 0 km/h
Time taken = 2 s
Acceleration = (Final Velocity – Initial Velocity)/time
= (5 – 0) / 2 m/s2
 = 2.5 m/s2

Solution 13N.

Acceleration = Increase in velocity/time taken
Therefore, increase in velocity = Acceleration × time taken
= (5 × 2) m/s
= 10 m/s

Solution 14N.

Initial velocity of the car, u = 20 m/s
Retardation = 2 m/s2
Given time, t = 5 s
Let ‘v’ be the final velocity.
We know that, Acceleration = Rate of change of velocity /time
= (Final velocity – Initial velocity)/time
Or, -2 = (v – 20) / 5
Or, -10 = v – 20
Or, v = -20 + 10 m/s
Or, v = -10 m/s
Negative sign indicates that the velocity is decreasing.

Solution 15N.

Initial velocity of the bicycle, u = 5 m/s
Acceleration = 2 m/s2
Given time, t = 5 s
Let ‘v’ be the final velocity.
We know that, acceleration = Rate of change of velocity/time
= (Final velocity – Initial velocity)/time
Or 2 = (v – 5)/5
Or, 10 = (v – 5)
Or, v = -5 – 10
Or, v = -15 m/s
Negative sign indicates that the velocity is decreasing.

Solution 16N.

Initial velocity of the bicycle, u = 18 km/hr
Time taken, t = 5 s
Final velocity, v = 0 m/s (As the car comes to rest)
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 8
(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes.
Then, Acceleration = (V – 5)/ 2
Or, -1 = (V – 5)/2
Or, V = -2 + 5
Or, V = 3 m/s2

Exercise 2(B)

Solution 1S.

For the motion with uniform velocity, distance is directly proportional to time.

Solution 2S.

From displacement-time graph, the nature of motion (or state of rest) can be understood. The slope of this graph gives the value of velocity of the body at any instant of time, using which the velocity-time graph can also be drawn.

Solution 3S.

(a) Slope of a displacement-time graph represents velocity.
(b) The displacement-time graph can never be parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time, which is not possible.

Solution 4S.

(a) There is no motion, the body is at rest.
(b) It depicts that the body is moving away from the starting point with uniform velocity.
(c) It depicts that the body is moving towards the starting point with uniform velocity.
(d) It depicts that the body is moving with variable velocity.

Solution 5S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 9

Solution 6S.

(i) The slope of the velocity-time graph gives the value of acceleration.
(ii) The total distance travelled by a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (without any sign).
(iii) The displacement of a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (with proper signs).

Solution 7S.

Vehicle A is moving with a faster speed because the slope of line A is more than the slope of line B.

Solution 8S.

(a) Fig. 4.33 (a) represents uniformly accelerated motion. For example, the motion of a freely falling object.
(b) Fig. 4.33 (b) represents motion with variable retardation. For example, a car approaching its destination.

Solution 9S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 10
In this graph, initial velocity = u
Velocity at time t = v
Let acceleration be ‘a’
Time = t
Then, distance travelled by the body in t s = area between the v-t graph and X-axis
Or distance travelled by the body in t s = area of the trapezium OABD
= (1/2) × (sum of parallel sides) × (perpendicular distance between them)
= (1/2) × (u + v) × (t)
= (u + v)t /2

Solution 10S.

The slope of the velocity-time graph represents acceleration.

Solution 11S.

Car B has greater acceleration because the slope of line B is more than the slope of line A.

Solution 12S.

For body A: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body A is zero.
For body B: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body B is also zero.
For body C: The slope of the graph is decreasing with time. Hence, the acceleration is negative.
For body D: The slope of the graph is increasing with time. Hence, the acceleration is positive.

Solution 13S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 11
Velocity-time for a body moving with uniform velocity and uniform acceleration.

Solution 14S.

Retardation is calculated by finding the negative slope.

Solution 15S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 12
The area enclosed between the straight line and time axis for each interval of time gives the value of change in speed in that interval of time.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 13

Solution 17S.

For motion under uniform acceleration, such as the motion of a freely falling body, distance is directly proportion to the square of the time.

Solution 18S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 14

The value of acceleration due to gravity (g) can be obtained by doubling the slope of the S – t2 graph for a freely falling body.

Solution 1M.

B
Acceleration is uniform.

Solution 2M.

The solution is C.
The speed-time graph is a straight line parallel to the time axis.

Solution 3M.

D
A straight line inclined to the time axis.

Solution 1N.

Velocity of body at t = 1s is 2 m/s
Velocity of body at t = 2s is 4 m/s
Velocity of body at t = 3s is 6 m/s
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 15

Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 16
Displacement-time graph
From the part AB of the graph,
Average velocity = (Displacement at B – Displacement at A)/Time taken
= (30 – 20) m/( 6 – 4) s
= (10/2) m/s
= 5 m/s
(b) (i) From the graph, the displacement of car at 2.5 s is 12.5 m.
(ii) From the graph, the displacement of car at 4.5 s is 22.5 m.

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 17

Solution 4N.

(a) (i) Velocity from 0 to 5 s = Displacement /time
= (3/5) m/s
= 0.6 m/s

(ii) Velocity from 5 s to 7 s = Displacement /time
= (0/2) m/s
= 0 m/s.

(iii) Velocity from 7 s to 9 s = Displacement /time
= (7 – 3)/(9 – 7) m/s
= (4/2) m/s
= 2 m/s

(b) From, 5 s to 9 s, displacement = 7m – 3m = 4m.
Time elapsed between 5 s to 9 s = 4 s
Average velocity = Displacement/time
= (4/4) m/s
= 1 m/s

Solution 5N.

(i) Displacement in first 4s = 10 m
Therefore, the average velocity = Displacement/time
= (10/4) m/s
= 2.5 m/s

(ii) Initial position = 0 m
Final position at the end of 10 s = -10m
Displacement = Final position – Initial position
= (-10) m – 0
= -10 m

(iii) At 7 s and 13 s, the cyclist reaches his starting point.

Solution 6N.

(i) Initially, the car B was 40 km ahead of car A.
(ii) Straight line depicts that cars A and B are moving with uniform velocities.
For car A
Displacement at t = 1 h is 40 m
Velocity = Displacement /time
= (40/1) km/h
= 40 km/h

For car B
Displacement at t = 4 h is (120 – 40) km, i.e. 80 km
Velocity = Displacement /time
= (80/4) km/h
= 20 km/h

(iii) Car A catches car B in 2 hours.
(iv) After starting, car A will catch car B at 80 km.

Solution 7N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 18

Solution 8N.

Velocity of the body at t = 1 s is 1 m/s.
Displacement of the body at t = 1 s is velocity × time = (1) × (1) m or 1 m.

Velocity of the body at t = 2s is 2 m/s.
Displacement of the body at t = 1 s is velocity × time = (2) × (2) m or 4 m.

Velocity of the body at t = 3 s is 3 m/s.
Displacement of the body at t = 3 s is velocity × time = (3) × (3) m or 9 m

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 19

Solution 9N.

(i) Distance travelled in any part of the graph can be determined by finding the area enclosed by the graph in that part with the time axis.

(ii) Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t – t) × vo
 = vot

Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × vo
 = (1/2) vo t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.

(iii) (a) BC shows motion with uniform velocity.
(b) AB shows motion with uniform acceleration.
(c) CD shows motion with uniform retardation.

(iv) (a) The magnitude of acceleration is lower as the slope of line AB is less than that of line CD.
(b) Slope of line AB = vo/t
Slope of line CD = vo/0.5t
Slope of line AB/Slope of line CD = (vo /t)/(vo /0.5t)
Slope of line AB:Slope of line CD :: 1:2.

Solution 10N.

(i) Acceleration in the part AB = Slope of AB
= tan (∠BAD)
= (30/4) ms-2
 = 7.5 ms-2

Acceleration in the part BC = 0 ms-2
Acceleration in the part CD = slope of CD = -tan (∠CDA)
= -(30/2) ms-2
 = -15 ms-2

(ii) Displacement of part AB = Area of ΔAB4 = (1/2) (4) (30)
= 60 m
Displacement of part BC = Area of rectangle 4BC8
= (30) × (4) = 120 m
Displacement of part CD = Area of ΔC8D = (1/2) (2) (30)
= 30 m

(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD
= 60 + 120 + 30 = 210 m

Solution 11N.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 20
Distance travelled in first 6 s = velocity × time
= 10 m/s × 6
= 60 m/s

Distance travelled in next 6 s = velocity × time
= 10 m/s × 6
= 60 m/s

Total distance travelled in 12 s = (60 + 60) m = 120 m
Total displacement = 0, as the ball returns its starting point.

Solution 12N.

(i) From 0 to 4 seconds, the motion is uniformly accelerated and from 4 to 6 seconds, the motion is uniformly retarded.

(ii) Displacement of the particle at 6 s = (1/2) (6) (2) = 6 m

(iii) The particle does not change its direction of motion.

(iv) Distance travelled by the particle from 0 to 4s (D1) = (1/2) (4) (2) = 4 m
Distance travelled by the particle from 4 to 6s (D2) = (1/2) (2) (2) = 2 m
D1:D2:: 4:2
D1:D2:: 2:1

(v) Acceleration from 0 to 4 s = (2/4) ms-2 or 0.5 ms-2
Retardation from 4 s to 6 s = (2/2) ms-2 or 1 ms-2.

Exercise 2(C)

Solution 1S.

Three equations of a uniformly accelerated motion are
v = u + at
s = ut + (1/2)at2
v2 = u2 + 2as

Solution 2S.

Derivation of equations of motion

First equation of motion:
Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity, and at t = t, let v be its final velocity.
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 21
Acceleration = Change in velocity/Time
a = (v – u)/t
at = v – u
v = u+ at … First equation of motion.

Second equation of motion:
Average velocity = Total distance traveled/Total time taken Average velocity = s/t …(1)
Average velocity can be written as (u+v)/2 Average velocity = (u+v)/2 …(2)
From equations (1) and (2) s/t = (u+v)/2 …(3)
The first equation of motion is v = u + at.
Substituting the value of v in equation (3), we get
s/t = (u + u + at)/2 s = (2u + at) t/2 = 2ut + at2/2 = 2ut/2 + at2/2
s = ut + (1/2) at2 …Second equation of motion.

Third equation of motion:
The first equation of motion is v = u + at. v – u = at … (1)
Average velocity = s/t …(2)
Average velocity =(u+v)/2 …(3)
From equation (2) and equation (3) we get,
(u + v)/2 = s/t …(4)
Multiplying eq (1) and eq (4) we get,
(v – u)(v + u) = at × (2s/t) (v – u)(v + u) = 2as
[We make the use of the identity a2 – b2 = (a + b) (a – b)]
v2 – u2 = 2as …Third equation of motion.

Solution 3S.

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 22

Solution 1M.

v = u + at

Solution 2M.

5 km

Solution 1N.

Initial velocity u = 0
Acceleration a = 2 m/s2
Time t = 2 s
Let ‘S’ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at2
S = 0 + (1/2) (2) (2) 2
S = 4 m

Solution 2N.

Initial velocity u = 10 m/s
Acceleration a = 5 m/s2
Time t = 5s
Let ‘S’ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at2
S = (10)(5) + (1/2) (5) (5) 2
S = 50 + 62.5
S = 112.5 m

Solution 3N.

Acceleration = Change in velocity/time taken
In the first two seconds,
Acceleration = [(33.6 – 30)/2] km/h2
 = 1.8 km/h2
 = 0.5 m/s2 …(i)

In the next two seconds,
Acceleration = [(37.2 – 33.6)/2] km/h2
 = 1.8 km/h2
 = 0.5 m/s2…(ii)
From (i) and (ii), we can say that the acceleration is uniform.

Solution 4N.

Initial velocity u = 0 m/s
Acceleration a = 2 m/s2
Time t = 5 s

(i) Let ‘v’ be the final velocity.
Then, (v – u)/5 = 2
v = 10 m/s

(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(10) 2 – (0) 2 = 2(2) (s)
Thus, s = (100/4) m = 25 m

Solution 5N.

Initial velocity u = 20 m/s
Final velocity v = 0
Distance travelled s = 10 cm = 0.1 m
Let acceleration be ‘a’.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (20) 2 = 2(a) (0.1)
a = -(400/0.2) m/s2
a = -2000 m/s2
Thus, retardation = 2000 m/s2

Solution 6N.

Initial velocity u = 20 m/s
Final velocity v = 0
Time taken t = 5 s
Let acceleration be ‘a’.
Using the first equation of motion,
v = u + at
0 = 20 + 5a
a = -4 m/s2
Thus, retardation = 4 m/s2

Solution 7N.

Let ‘s’ be the distance between stations A and B.
(i) Average speed = Total distance/total time taken
Here, total distance = s + s = 2s
Total time taken = Time taken to travel from A to B + Time taken to travel from B to A.
= [(s/ 60) + (s/ 80)] s
= [ 140 s / 4800] s

Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 23

(ii) Average velocity = Displacement/total time taken
Because the train starts and ends at the same station, the displacement is zero. Thus the average velocity is zero.

Solution 8N.

Initial velocity u = 90 km/h = 25 m/s
Final velocity v = 0 m/s
Acceleration a = -0.5 m/s2

(i) Let ‘V’ be the velocity after time t = 10 s
Using the first equation of motion,
v = u + at
We get,
V = 25 + (-0.5) (10) m/s
V = 25 – 5 = 20 m/s

(ii) Let t’ be the time taken by the train to come to rest.
Using the first equation of motion,
v = u + at
We get,
t’ = [(0 – 25)/ (-0.5)] s
t’ = 50 s

Solution 9N.

Distance travelled s = 100 m
Average velocity V = 20 m/s
Final velocity v = 25 m/s

(i) Let u be the initial velocity.
Average velocity = (Initial velocity + Final velocity)/2
V = (u + v)/2
20 = (u + 25)/2
u = 40 – 25 = 15 m/s

(ii) Let ‘a’ be the acceleration of the car.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(25) 2 – (15) 2 = 2 (a) (100)
625 – 225 = 200 a
a = (400/200) m/s2 = 2 m/s2

Solution 10N.

Final velocity v = 0
Acceleration = -25 cm/s2 or -0.25 m/s2
Time taken t = 20 s

(i) Let ‘u’ be the initial velocity.
Using the first equation of motion,
v = u + at
We get,
u = v – at
u = 0 – (-0.25)(20) = 5 m/s

(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (5)2 = 2 (-0.25) (s)
s = (25/0.5) = 50 m.

Solution 11N.

Initial velocity u = 0 m/s
Distance travelled s = 270 m
Time taken to travel s distance = 3 s
Let ‘a’ be the uniform acceleration.
Using the second equation of motion,
S = ut + (1/2) at2
We get,
270 = 0 + (1/2) a (3)2
270 = 9a/2
a = 60 m/s2

Let v be the velocity of the body 10 s after the start.
Using the first equation of motion, we get
v = u + at
v = 0 + (60)(10) = 600 m/s

Solution 12N.

Let the constant acceleration with which the body moves be ‘a’.
Given, the body travels distance S1 = 3 m in time t1 = 1 s.
Same body travels distance S2 = 8 m in time t2 = 2 s.

(i) Let ‘u’ be the initial velocity.
Using the second equation of motion,
S = ut + (1/2) at2
Substituting the value for S1 and S2, we get
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 24

(ii) Putting u = 2 m/s in the equation
Selina Concise Physics Class 9 ICSE Solutions Motion in One Dimension - 25

Solution 13N.

Initial velocity u = 25 m/s
Final velocity v = 0

(i) Before the brakes are applied, let S be the distance travelled.
Distance = Speed × time
S = (25) × (5) m
S = 125 m

(ii) Acceleration = (Final velocity – Initial velocity)/Time taken
= [(0 – 25)/15] ms-2
= (-5/2) ms-2
 = -2.5 ms-2
Therefore, retardation = 2.5 ms-2

(iii) After applying brakes, the time taken to come to stop = 10 s
Let S’ be the distance travelled after applying the brakes.
Initial velocity u = 25 m/s
Final velocity v = 0
Using the third equation of motion,
v2 – u2 = 2as
We get,
(0) 2 – (25)2 = 2 (-2.5) (S’)
625 = 5(S’)
S’ = 125 m

Solution 14N.

Given, the initial velocity u = 75 km/s
Final velocity v = 120 km/s
Time taken = 6 s

(i) Acceleration = (Final velocity – Initial velocity)/time taken
= [(120 – 75)/6] kms-2
= (45/6) kms-2
 = 7.5 kms-2

(ii) Distance travelled by the aircraft in the first 10 s = Distance travelled in the first 6 s + Distance travelled in the next 4 s.
Distance travelled in the first 6s (S1) = ut + (1/2) at2
(S1) = ut + (1/2) at2
(S1) = (75)(6) + (1/2) (7.5)(6)2
(S1) = 450 + 135
(S1) = 585 km

Distance travelled in the next 4 s (S2) = Speed × time
Speed at the end of 6 s is 120 km/s.
(S2) = (120) (4)
(S2) = 480 km
Thus, the distance travelled by the aircraft in the first 10 s = (S1) + (S2) = 585 + 480 = 1065 km.

Solution 15N.

(i) For the first 10 s, initial velocity u = 0
Acceleration a = 2 m/s2
Time taken t = 10 s
Let ‘v’ be the maximum velocity reached.
Using the first equation of motion
v = u + at
We get
V = (0) + (2) (10) = 20 m/s

(ii) For the last 50 s: Final velocity = 0 m/s, initial velocity = 20 m/s.
Acceleration = (Final velocity – Initial velocity)/time
= (0 – 20)/50 = -0.4 m/s2
Retardation = 0.4 m/s2

(iii) Total distance travelled = Distance travelled in the first 10 s + Distance travelled in 200 s + Distance travelled in last 50 s
Distance travelled in first 10s (s1) = ut + (1/2) at2
S1= (0) + (1/2) (2) (10)2
S1= 100 m
Distance travelled in 200s (s2) = speed × time
S2 = (20) (200) = 4000 m

Distance travelled in last 50s (s3) = ut + (1/2) at2
Here, u = 20 m/s, t = 50 s and a = -0.4 m/s2
S3= (20)(50) + (1/2) (-0.4) (50)2
S3= 1000 – 500
S3= 500 m
Therefore, total distance travelled = S1 + S2 + S3 = 100 + 4000 + 500 = 4600 m

(iv) Average velocity = Total distance travelled/total time taken
= (4600/260) m/s
= 17.69 m/s

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Selina ICSE Solutions for Class 9 Physics Chapter 10 Magnetism

Exercise 10(A)

Solution 1S.

Lodestone is an ore of iron oxide (Fe3O4). This ore attracts small pieces of iron and it sets itself along a definite direction when it is suspended freely. It is a natural magnet which was used for the navigation by the mariners.

Solution 2S.

The pieces of lodestone found in nature are called the natural magnets. Limitations of a natural magnet are as listed below:

  1. They are irregular and odd shaped.
  2. They are not magnetically very strong.

Solution 3S.

An artificial magnet is a magnetized piece of iron (or other magnetic material). Artificial magnets are required because natural magnets have odd and irregular shape and they are not magnetically very strong. Artificial magnets can be given desired shape and made very strong.

Solution 4S.

Iron rod is magnetized when placed near a bar magnet by magnetic induction, while copper rod is not magnetized.

Solution 5S.

A magnet when suspended freely will rest only in north-south direction, but the soft iron bar will rest in any direction.

Solution 6S.

(a) Poles, (b) Attract, repel,
(c) At the middle, (d) North – South

Solution 7S.

If a small magnet is suspended by a silk thread such that it can swing freely then it rests itself in the geographic north-south direction.
Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 1

Solution 8S.

The magnetism acquired by a magnetic material when it is kept near (or in contact with) a magnet, is called induced magnetism.

Solution 9S.

The process in which a piece of magnetic material acquires the magnetic properties temporarily in presence of another magnet near it is called the magnetic induction.

When a piece of iron is placed near or in contact with a magnet, the piece of iron becomes a magnet i.e., it acquires the property of attracting iron filings when they are brought near its ends. Thus, a piece of iron behaves as a magnet as long as it is kept near (or in contact with) a magnet.

Solution 10S.

When iron nails are brought near one end of a magnet, the nearer end of piece acquires an opposite polarity by magnetic induction. Since unlike poles attract each other, therefore, iron nails are attracted towards the end of the magnet. Thus, the iron nail first becomes a magnet by induction and then it is attracted.

Solution 11S.

(a) When two pins are hung by their heads from the same pole of a magnet, they acquire same polarity. Because like poles repel each other, their pointed ends move apart.

(b) Several soft iron pins can cling one below the other from the pole of a magnet because the magnet induces magnetism in an iron nail which gets attracted by the magnet and clings to it. This magnetized nail magnetizes the other nail near it by magnetic induction and attracts it. This process continues until force of attraction on first nail is sufficient to balance the total weight of all nails in chain.

(c) When a piece of soft iron is placed a little distance away from the needle, the needle induces magnetism to the piece of soft iron. Thus, soft iron piece starts behaving like a magnet and it attracts the magnetic needle towards it.

Solution 12S.

The iron bar acquires magnetism due to magnetic induction.
Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 2
If the magnet is removed, the iron bar loses its magnetism.

Solution 13S.

Induced magnetism is temporary as it lasts as long as the magnet causing induction remains in it vicinity.

Solution 14S.

When a piece of magnetic material is brought near a magnet, it first becomes a magnet by induction and then it is attraction. Thus, we say that induction precedes attraction.

Solution 15S.

A magnetic field line is a continuous curve in a magnetic field such that tangent at any point of it gives the direction of the magnetic field at that point.

Solution 16S.

Properties of magnetic field lines:

  1. They are closed and continuous curves.
  2. They are directed from the North Pole towards the South Pole outside the magnet.
  3. The tangent at any point on a field line gives the direction of magnetic field at that point.
  4. Two magnetic lines never intersect each other.

Solution 17S.

The iron filings take up a definite pattern (curved lines). This happens because each piece of iron filing becomes a magnet to the magnetic induction of the magnet. It thus experiences a force in the direction of magnetic field of the bar magnet at that point and aligns itself along curved lines.

Solution 18S.

Method of plotting the magnetic field lines using a compass needle:

Fix a sheet of paper on a drawing board by means of board pins. Place a small compass needle at position 1 as shown in fig (a) and looking from the top of the needle, mark two pencil dots exactly at two ends of the needle. Then move the compass needle to position 2 in such a way that one end of needle coincides with the second pencil dot. Repeat the process of moving the compass needle to positions 3, 4,… to obtain several dots. On joining the different dots, you will get a straight line. Thus one line of magnetic field of earth is traced.
Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 3
This process is repeated starting from a different point and tracing out another line of magnetic field. In this manner, several lines of magnetic field can be drawn. Each line should be labeled with an arrow from the south pole of the needle towards the north pole to indicate the direction of the magnetic field. Fig (b) shows several magnetic lines so obtained.

Solution 19S.

No two magnetic field lines can intersect each other. If they do, there would be two directions of the field at that point which is not possible.

Solution 20S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 4

Solution 21S.

Two evidences of existence of earth’s magnetic field:

  1. A freely suspended magnetic needle always rests in geographic north-south direction.
  2. Neutral points are obtained on plotting the field lines of a magnet.

Solution 22S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 5

Solution 23S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 6

Solution 24S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 7
(c) Magnitude of magnetic field at neutral points is zero. It is so because at these points, the magnetic field of the magnet is equal in magnitude to the earth’s horizontal magnetic field, but it is in opposite direction. Hence, they cancel each other.

Solution 25S.

It can be concluded that magnetic field at that point is zero. This is because the earth’s magnetic field at that point is neutralized by the magnetic field of some other magnetized material.

Solution 26S.

Neutral points are the points where the magnetic field of the magnet is equal in magnitude to the earth’s horizontal magnetic field, but it is in opposite direction. Thus the net magnetic field at the neutral points is zero.

Since the net magnetic field is zero at neutral points, the compass needle remains unaffected (i.e. it comes to rest pointing in any direction) at these points and hence, they can be detected.

Solution 27S.

(i) Neutral points will be in east-west direction.
(ii) Neutral points will be north-south direction.

Solution 28S.

(a) Uniform, (b) Zero and (c) On either side of the magnet in east and west.

Solution 1M.

Repel each other

Solution 2M.

Parallel equidistant straight lines

Exercise 10(B)

Solution 1S.

An electromagnet is a temporary strong magnet made from a piece of soft iron when current flows in the coil wound around it. It is an artificial magnet.

Solution 2S.

The material used for preparing an electromagnet is soft iron.

Solution 3S.

An electromagnet is made by winding an insulated copper wire around a soft iron core either in the shape of a solenoid or U-shape.
The strength of magnetic field of an electromagnet depends on:

  1. The number of turns of wire wound around the coil, and
  2. The amount of current flowing through the wire.

Solution 4S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 8

Solution 5S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 9
By increasing the number of turns of winding in the solenoid, the strength of the electromagnet can be increased.

Solution 6S.

The device formed is an electromagnet.
Use: For separating the magnetic substances such as iron from other debris.

Solution 7S.

Selina Concise Physics Class 9 ICSE Solutions Magnetism image - 10

Solution 8S.

The strength of an electromagnet can be increased by following ways:

  1. Increasing the number of turns of winding in the solenoid.
  2. Increasing the current through the solenoid.

Solution 9S.

The electromagnet is used in an electric relay.

Solution 10S.

  1. An electromagnet can produce a strong magnetic field.
  2. The strength of the magnetic field of an electromagnet can easily be changed by changing the current in its solenoid.

Solution 11S.

Electromagnet Permanent magnet
It is made up of soft iron It is made up of steel
The magnetic field strength can be changed The magnetic field strength cannot be changed
Electromagnets of very strong field can be made. Permanent magnets are not so strong.

Solution 12S.

The soft iron bar acquires the magnetic properties only when an electric current flows through the solenoid and loses the magnetic properties as the current is switched off. Hence, soft iron is used as the core of the electromagnet in an electric bell.

Solution 13S.

If an a.c. source is used in place of a battery, the core of the electromagnet will get magnetized, but the polarity at its ends will change. Since attraction of armature does not depend on the polarity of the electromagnet, the bell will still ring on pressing the switch.

Solution 14S.

The material used for making the armature of an electric bell is soft iron which can induce magnetism rapidly.

Solution 1M.

Electromagnets are made up of soft iron.

Solution 2M.

The strength of an electromagnet can be increased by

  1. increasing the number of turns of coil, and
  2. increasing the current through the coil
  3. Hence, the correct answer is option c.

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Selina Concise Physics Class 9 ICSE Solutions Current Electricity

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Selina Concise Physics Class 9 ICSE Solutions Current Electricity

ICSE SolutionsSelina ICSE Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 9 Current Electricity. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 9 Current Electricity

Exercise 9(A)

Solution 1S.

Source of D.C.: Cell
Source of A.C.: Mains

Solution 2S.

Direct current (D.C.) is a current of constant magnitude flowing in one direction but alternating current (A.C.) is a current which reverses its magnitude and direction with time.

Solution 3S.

An electric cell is a device which converts chemical energy into electrical energy. When connected in a circuit, it acts as a source of D.C. current.

Solution 4S.

Chemical energy changes into electrical energy.

Solution 5S.

Constituents of cell: Two electrodes and an electrolyte in a vessel.

Solution 6S.

Two kinds of cells:

  1. Primary cell: e.g. Leclanche cell
  2. Secondary cell: Lead (or acid) accumulator

Solution 7S.

Primary cells are cells which provide current as a result of irreversible chemical current.
Examples: Simple Voltaic cell and Leclanche cell.

Solution 8S.

Secondary cells are cells which provide current as a result of reversible chemical reactions. It converts electrical energy into chemical energy when current is passed in it (i.e. during charging), while it converts chemical energy into electrical energy when current is drawn from it (i.e., during discharging).
Example: Lead (or acid) accumulator.

Solution 9S.

Primary Cell Secondary cell

1. Chemical reaction is irreversible.

2. Only chemical energy is converted into electrical energy when current is drawn from it.

3. It cannot be recharged and its internal resistance is high.

1. Chemical reaction is reversible.

2. It converts electrical energy into chemical energy when current is passed in it (i.e., during charging), while converts chemical energy into electrical energy when current is drawn from it (i.e., during discharging).

3. It can be recharged and its internal resistance is low.

Solution 10S.

Current is the rate of flow of charge across a cross-section. It is a scalar quantity.
Its S.I. unit is ampere (coulomb per second).
If 1 ampere current flows through a conductor, it means that 6.25 x 1018 electrons pass in 1 second across that cross-section of conductor.

Solution 11S.

Charge on an electron is -1.6 x 10-19 coulomb.

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 8

Solution 13S.

A rheostat is used to control current in an electric circuit.

Solution 14S.

A: Ammeter – It measures the current flowing through the circuit.
B: Cell – It acts as a source of direct current for the circuit.
C: Key – It is used to put the current on and off in the circuit.
D: Load – It is an appliance connected in a circuit. It may just be a resistance (e.g., bulb) or a combination of different electrical components.
E: Voltmeter – It is used to measure the potential difference between two points of a circuit.
F: Rheostat – It is used to control the current in the circuit.
Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 1

Solution 15S.

A key or switch is used to put on or off, the current in the circuit.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 2
It is used to measure the potential difference between two points of a circuit.

Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 3

Solution 18S.

The substances which allow electric current to flow through them easily are called conductors. Examples: Impure water and metals.
The substances which do not allow the electric current to flow through them are called insulators. Examples: Rubber and wood.

Solution 19S.

Copper wire, acidulated water and human body.

Solution 20S.

Conductors have a large number of free electrons and they offer a very small resistance in the path of current but insulators have no free electrons and they offer a very high resistance in the path of current.

Solution 21S.

A circuit is said to be closed when every part of it is made of a conductor and on plugging in the key or on being complete, current flows through the circuit.
A circuit is said to be open when no current flows through it. It can happen when the key is not plugged in or when any one of its components is not made of a conductor or when the circuit is broken.
Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 4

Solution 22S.

For an electric circuit to be closed, every part of it must be made of conductors.

Solution 1M.

provide current in a circuit

Solution 2M.

ampere

Solution 3M.

silk

Solution 1N.

Current (I) = Charge (q)/time (t)
Or, I = 0.5/ 5 = 0.1 A

Solution 2N.

Charge (q) = Current (I) x time (t)
Or, q = 1.5 x 2 = 3 C

Solution 3N.

Current (I) = Charge (q)/time (t)
Or, I = 24/ 0.8 = 30 A

Exercise 9(B)

Solution 1S.

When both the conductors are joined by a metal wire:

  1. Electrons will flow from A to B.
  2. Current will flow from B to A.

Solution 2S.

Current always flows from high potential to low potential.

Solution 3S.

Electric potential difference between two conductors is equal to the work done in transferring a unit positive charge from one conductor to other conductor.

Solution 4S.

Electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is
Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 5

Solution 5S.

S.I. unit of potential difference is volt (joule per coulomb).
Potential difference between two points is said to be 1 volt if work done in transferring 1 coulomb of charge from one point to the other point is 1 joule.

Solution 6S.

Potential difference between two points is 1 volt; it means 1 joule of work is done in transferring 1 coulomb of charge from one point to the other point.

Solution 7S.

The obstruction offered to the flow of current by the filament or wire is called its electrical resistance.

Solution 8S.

A metal wire has free electrons which move in random directions. When the ends of the wire are connected to a cell, the electrons start moving from the negative terminal of the cell to its positive terminal through the metal wire. During their movement, they collide with the free electrons and fixed ions of the wire. This causes them to lose their speed and change their direction. As a result, the electrons slow down and slowly drift towards the positive terminal. Thus, the wire offers resistance to the flow of current (or electrons) through it.

Solution 9S.

The S.I. unit of resistance is ‘ohm’ (volt per ampere).
The resistance of a conductor is said to be 1 ohm if a current of 1 ampere flows through it when the potential difference across it is 1 volt.

Solution 10S.

Ohm’s law states that the electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same.

Solution 11S.

Potential difference = Current x Resistance
i.e., V = I R

Solution 12S.

The resistance of a wire is 1 ohm; it means a current of 1 ampere will flow through the wire when the potential difference across it is 1 volt.

Solution 13S.

V = IR or, I = V/R….(i)
If R is doubled,
Then, I’ = V/2R = I/2……………(ii)
From (i) and (ii), it is clear that current will be halved.

Solution 14S.

Resistance of a wire depends upon:

  1. Length of wire: Resistance is directly proportional to the length of a wire.
  2. Area of cross-section of wire: Resistance is inversely proportional to the area of cross-section the wire.
  3. The temperature of wire: Resistance of a wire is directly proportional to the temperature of the wire.

Solution 15S.

(a) Resistance of a wire is directly proportional to the length of a wire; so if the length is doubled, resistance is also doubled.
(b) Resistance of a wire is inversely proportional to the area of cross-section the wire. Thus, if radius is doubled, area increases four times and hence the resistance becomes one-fourth.

Solution 16S.

The temperature of the filament increases when it glows. So, when the temperature of the wire (bulb filament) increases, ions in it vibrate violently. As a result, the number of collisions increases and hence the resistance increases.

Solution 17S.

(i) Potential difference (ii) charge (iii) resistance (iv) current.

Solution 1M.

In direction from high potential to low potential.

Solution 2M.

volt

Solution 3M.

Decreases

Solution 1N.

Potential difference (V) = work done (W) / charge (q)
Or, V = 9/1.5 = 6 volt.

Solution 2N.

Given, potential difference (V) = 12 V
Resistance, R = 24 Ω
Therefore, current (I) = V / R
Or, I = 12/24 = 0.5 A

Solution 3N.

Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 6

Solution 4N.

Selina Concise Physics Class 9 ICSE Solutions Current Electricity image - 7

Exercise 9(C)

Solution 1S.

Efficient use of energy means to reduce cost and amount of energy to be used to provide us the various products and services.

Solution 2S.

Two ways to save energy:

  1. Instead of fossil fuels, other renewable sources of energy such as the biogas prepared from animal dung should be used.
  2. The use of hydroelectric energy, wind energy etc. should be given priority.

Solution 3S.

By properly insulating a home, it is possible to maintain a comfortable temperature inside. It will reduce the cost of heating devices in winter and cooling devices in summer.

Solution 4S.

LED or light emitting diodes are most efficient for lighting purposes.

Solution 5S.

Modern appliances like refrigerators make use of significantly less energy than older appliances as they have star rating according to their efficient use of electricity. Higher the star rating, higher is the efficiency.

Solution 6S.

Three ways to use energy efficiently:

  1. The use of compact fluorescent lights (CFL) saves 67% energy and may last 6 to 10 times longer than the incandescent lamps.
  2. The use of advanced boilers and furnaces in industry can save sufficient amount of energy in attaining high temperatures while burning less fuel. Such technologies are more efficient and less polluting.
  3. The fuel efficiency in the vehicles can be increased by reducing the weight of the vehicle, using the advanced tyres and computer controlled engines.

Solution 7S.

The following social initiatives need to be taken:

  1. Public awareness can be improved through mass-media and children’s participation in campaigns and eco-club activities.
  2. Community involvement need to be done to reduce the misuse of electricity.
  3. NGO’s can be used to create social awareness of the sensitive use of resources.

Solution 1M.

The most non-polluting and efficient lighting device is the LED.

Solution 2M.

IEA is the short form of international energy agency.

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Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes’ Principle and Floatation

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Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes’ Principle and Floatation

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation. You can download the Selina Concise Physics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 9 Physics Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation

Exercise 5(A)

Solution 1S.

When a body is partially or wholly immersed in a liquid, an upward force acts on it. This upward force is known as an upthrust.

Upthrust can be demonstrated by the following experiment:

Take an empty can and close its mouth with an airtight stopper. Put it in a tub filled with water. It floats with a large part of it above the surface of water and only a small part of it below the surface of water. Push the can into the water. You can feel an upward force and you find it difficult to push the can further into water. It is noticed that as the can is pushed more and more into the water, more and more force is needed to push the can further into water, until it is completely immersed. When the can is fully inside the water, a definite force is still needed to keep it at rest in that position. Again, if the can is released in this position, it is noticed that the can bounces back to the surface and starts floating again.

Solution 2S.

Buoyant force on a body due to a liquid acts upwards at the centre of buoyancy.

Solution 3S.

The property of a liquid to exert an upward force on a body immersed in it is called buoyancy.

Solution 4S.

The upward force exerted on a body by the fluid in which it is submerged is called the upthrust. Its S.I. unit is ‘newton’.

Solution 5S.

A liquid contained in a vessel exerts pressure at all points and in all directions. The pressure at a point in a liquid is the same in all directions – upwards, downwards and sideways. It increases with the depth inside the liquid.
Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 1
When a body is immersed in a liquid, the thrusts acting on the side walls of the body are neutralized as they are equal in magnitude and opposite in direction. However, the magnitudes of pressure on the upper and lower faces are not equal. The difference in pressure on the upper and lower faces cause a net upward force (= pressure x area) or upthrust on the body.
It acts at the centre of buoyancy.

Solution 6S.

Upthrust due to water on block when fully submerged is more than its weight. Density of water is more than the density of cork; hence, upthrust due to water on the block of cork when fully submerged in water is more than its weight.

Solution 7S.

A piece of wood if left under water comes to the surface of water because the upthrust on body due to its submerged part is equal to its own weight.

Solution 8S.

Experiment to show that a body immersed in a liquid appears lighter:
Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 2
Take a solid body and suspend it by a thin thread from the hook of a spring balance as shown in the above figure (a). Note its weight. Above figure (a) shows the weight as 0.67 N.
Then, take a can filled with water. Immerse the solid gently into the water while hanging from the hook of the spring balance as shown in figure (b). Note its weight. Above figure (b) shows the weight as 0.40 N.
The reading in this case (b) shall be less than the reading in the case (a), which proves that a body immersed in a liquid appears to be lighter.

Solution 9S.

The readings in the spring balance decreases.
As the cylinder is immersed in the jar of water, an upward force acts on it, which is in opposition to the weight component of the cylinder. Hence the cylinder appears to be lighter.

Solution 10S.

A body shall weigh more in vacuum because in vacuum, i.e. in absence of air, no upthrust will act on the body.

Solution 11S.

Upthrust on a body depends on the following factors:

  1. Volume of the body submerged in the liquid or fluid.
  2. Density of liquid or fluid in which the body is submerged.

Solution 12S.

Larger the volume of body submerged in liquid, greater is the upthrust acting on it.

Solution 13S.

A stone falls faster.
Because the volume of stone is less than the volume of bunch of feathers of the same mass, the upthrust due to air on stone is less than that on the bunch of feathers, and hence, the stone falls faster.
However, in vacuum, both shall fall together because there will be no upthrust.

Solution 14S.

F2 > F1; Sea water is denser than river water; therefore, the upthrust due to sea water will be greater than that due to river water at the same level. This shall make the body to appear lighter in the sea water.

Solution 15S.

Observation: Volume of a block of wood immersed in glycerine is smaller as compared to the volume of block immersed in water.
Explanation: Density of glycerine is more than that of water. Hence, glycerine exerts more upthrust on the block of wood than water, causing it to float in glycerine with a smaller volume.

Solution 16S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 48

Solution 17S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 3

Solution 18S.

(a) Both have equal volumes.
(b) Bounce back to the surface.
(c) More than

Solution 19S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 4

Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density ρ as shown in the figure above. Let the upper surface PQ of the body is at a depth h1 while its lower surface RS is at depth h2 below the free surface of liquid.

At depth h1, the pressure on the upper surface PQ,
P1 = h1 ρg.

Therefore, the downward thrust on the upper surface PQ,
F1 = Pressure x Area = h1 ρgA ……………….(i)

At depth h2, pressure on the lower surface RS,
P2 = h2 ρg

Therefore, the upward thrust on the lower surface RS,
F2 = Pressure x Area = h2 ρgA …………………(ii)

The horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.

From the above equations (i) and (ii), it is clear that F2 > F1 because h2 > h1 and therefore, body will experience a net upward force.

Resultant upward thrust or buoyant force on the body,

FB = F2 – F1
 = h2 ρgA – h1 ρgA
= A (h2 – h1) ρg

However, A (h2 – h1) = V, the volume of the body is submerged in a liquid.
Therefore, upthrust FB = V ρg.

Now, V g = Volume of solid immersed x Density of liquid x Acceleration due to gravity
= Volume of liquid displaced x Density of liquid x Acceleration due to gravity
= Mass of liquid displaced x Acceleration due to gravity
= Weight of the liquid displaced by the submerged part of the body

Thus, Upthrust FB = weight of the liquid displaced by the submerged part of the body…..(iii)

Now, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.

Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 5

When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.

From the diagram, it is clear that
Loss in weight (Weight in air – Weight in water) = Volume of water displaced.
Or, Loss in weight = Volume of water displaced x 1 gcm-3 [Because the density of water = 1 gcm-3]
Or, Loss in weight = Weight of water displaced ……………(iv)

From equations (iii) and (iv),
Loss in weight = Upthrust or buoyant force

Solution 20S.

Since the spheres have the same radius, both will have an equal volume inside water, and hence, the upthrust acted by water on both the spheres will be the same.
Hence, the required ratio of upthrust acting on two spheres is 1:1.

Solution 21S.

Sphere of iron will sink.

Density of iron is more than the density of water, so the weight of iron sphere will be more than the upthrust due to water in it; thus, it causes the iron sphere to sink.

Density of wood is less than the density of water, so the weight of sphere of wood shall be less than the upthrust due to water in it. So, the sphere of wood will float with a volume submerged inside water which is balanced by the upthrust due to water.

Solution 22S.

The bodies of average density greater than that of the liquid sink in it. While the bodies of average density equal to or smaller than that of liquid float on it.

Solution 23S.

(i) The body will float if ρ ≤ ρL
(ii) The body will sink if ρ > ρL

Solution 24S.

It is easier to lift a heavy stone under water than in air because in water, it experiences an upward buoyant force which balances the actual weight of the stone acting downwards. Thus, due to upthrust there is an apparent loss in the weight of the heavy stone, which makes it lighter in water, and hence easy to lift.

Solution 25S.

Archimedes’ principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of liquid displaced by it.

Solution 26S.

Let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight (Fig a).
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid gets collected in the measuring cylinder.
Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 6

When water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
From diagram, it is clear that
Loss in weight (Weight in air – weight in water) = 300 gf – 200 gf = 100 gf
Volume of water displaced = Volume of solid = 100 cm3
Because density of water = 1 gcm-3 
Weight of water displaced = 100 gf = Upthrust or loss in weight
This verifies Archimedes’ principle.

Solution 1M.

Turpentine

Solution 2M.

N

Solution 3M.

ρ > ρL

 

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 12

Solution 7N.

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Solution 8N.

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Solution 9N.

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Solution 10N.

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Exercise 5(B)

Solution 1S.

The density of a substance is its mass per unit volume.

Solution 2S.

(i) The C.G.S. unit of density is gcm-3.
(ii) The S.I. unit of density is kgm-3.

Solution 3S.

1 gcm-3 = 1000 kgm-3

Solution 4S.

It means the mass of 1 m-3 of iron is 7800 kg.

Solution 5S.

Density of water at 4°C in S.I. units is 1000 kgm-3.

Solution 6S.

(i) Mass of a metallic body remains unchanged with increase in temperature.
(ii) Volume of metallic body increases with an increase in temperature.
(iii) Density (= Mass/volume) of a metallic body decreases with an increase in temperature.

Solution 7S.

On heating from 0°C, the density of water increases up to 4°C and then decreases beyond 4°C.

Solution 8S.

(i) Volume, (ii) kg m-3, (iii) 1000 and (iv) 1000

Solution 9S.

The relative density of a substance is the ratio of density of that substance to the density of water at 4°C.

Solution 10S.

Relative density is the ratio of two similar quantities; thus, it has no unit.

Solution 11S.

Density of a substance is the ratio of its mass to its volume but R.D. of a substance is the ratio of density of that substance to the density of water at 4°C.

Solution 12S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 17

Steps:

  1. With the help of a physical balance, find the weight, W1 of the given solid.
  2. Immerse the solid completely in a beaker filled with water such that it does not touch the walls and bottom of beaker, and find the weight W2 of solid in water.

Observations:

Loss in weight of solid when immersed in water = (W1 – W2) gf
R.D. = Weight of solid in air/Loss of weight of solid in water
R.D. = W1/(W1 – W2).
If the solid is soluble in water, then instead of water, take a liquid in which the solid is insoluble and it sinks in the liquid.
Then, R.D. = (Weight of solid in air/Loss of weight of solid in liquid) x R.D. of the liquid

Solution 13S.

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Solution 14S.

Experimental determination of R.D. of a solid lighter than water (such as cork):

  1. Take a sinker (i.e. a piece of metal or stone).
  2. Place a beaker nearly two-third filled with water on a wooden bridge kept over the left pan of a physical balance.
  3. Suspend the sinker with a thread from the hook of the left pan of balance so that it is completely immersed in water (as shown in the figure below). Find the weight W1 of the sinker in water.
  4. Tie the given solid (say, a cork) in the middle of a thread, and then measure the weight W2 of a solid in the air along with the sinker in water.
  5. Tie the cork with the sinker and immerse both of them completely in water of beaker and measure the weight W3 of the solid and sinker both in water.
    Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 19

Observations:

Weight of the sinker in water = W1gf
Weight of the sinker in water and cork in air = W2gf
Weight of sinker and cork together in water = W3gf

Calculations:

Weight of cork in air = (W2 – W1) gf
Weight of cork in water = (W3 – W1) gf
Loss in weight of the cork in water = Weight of cork in air  Weight of cork in water.
= [(W2 – W1) – (W3 – W1)] gf
= (W2 – W3) gf
R.D. of cork = Weight of cork in air/Loss of weight of cork in water
Or, R.D. of cork = (W2 – W1)/(W2 – W3).

Solution 15S.

The weight of the sinker and cork combined, in water will be less than the weight of the sinker alone in water because the upthrust due to water on cork (when completely immersed) is more than the weight of cork itself.

 

Solution 1M.

Water

Solution 2M.

No unit.

Solution 3M.

1 g cm-3 

Solution 1N.

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Solution 2N.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 21

Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

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Solution 8N.

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Solution 9N.

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Solution 10N.

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Solution 11N.

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Solution 12N.

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Solution 13N.

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Solution 14N.

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Solution 15N.

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Solution 16N.

a. The mass of stone is 15.1 g. Hence, its weight in air will be Wa = 15.1 gf

b. When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 – 9.7 = 5.4 gf, Since the density of water is 1 g cm-3, the volume of stone is 5.4 cm3.

 

c. Weight of stone in liquid is Wl = 10.9 gf
Weight of stone in water is Ww = 9.7 gf
Therefore, the relative density of stone is

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Exercise 5(C)

Solution 1S.

According to the principle of floatation, the weight of a floating body is equal to the weight of the liquid displaced by its submerged part.

Solution 2S.

(i) Two forces acting on the body are as listed below:
(a) Weight of the body (downwards)
(b) Upthrust of the liquid (upwards)
Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 36
(ii) If the weight of the body is greater than the upthrust acting on it, the body will sink
If the weight of the body is equal to or less than the upthrust acting on it, the body will float.

(iii) (a) The net force acting on the body when it sinks is body’s own weight.
(b) The net force acting on the body when it floats is the upthrust due to the liquid.

Solution 3S.

The reading on the spring balance will be zero because wood floats on water and while floating the apparent weight = 0.

Solution 4S.

(a) The ball will float because the density of ball (i.e. iron) is less than the density of mercury.
(b) While floating, the apparent weight = 0.

Solution 5S.

The body will float if its density is less than or equal to the density of the liquid ρS ≤ ρL.
The body will sink if its density is greater than the density of the liquid ρS > ρL.

Solution 6S.

Density of iron is less than the density of mercury; hence, an iron nail floats in mercury and density of iron is more than the density of water; hence, an iron nail sinks in water.

Solution 7S.

(i) Weight of the floating body is equal to the upthrust.
(ii) While floating, the apparent weight is zero.

Solution 8S.

When the body is partially immersed, its centre of buoyancy will be below the centre of gravity of the block.
When the body is completely immersed, its centre of buoyancy will coincide the centre of gravity.

Solution 9S.

The upthrust on the body by each liquid is the same and equal to the weight of the body.
However, upthrust = Volume submerged × ρL × g,
For the liquid C, since the volume submerged is least so the density ρ3 must be maximum.

Solution 10S.

Selina Concise Physics Class 9 ICSE Solutions Upthrust in Fluids, Archimedes' Principle and Floatation image - 37
The forces acting are as listed below:

  1. Weight of the body acting downwards.
  2. Upthrust due to water acting upwards.

Weight of water displaced by the floating body = Weight of the body

Solution 11S.

Centre of buoyancy: It is the point through which the resultant of the buoyancy forces on a submerged body act; it coincides with the centre of gravity of the displaced liquid, if the body is completely immersed.
For a floating body with its part submerged in the liquid, the centre of buoyancy is at the centre of gravity of the submerged part of the body and it lies vertically below the centre of gravity of the entire body.

Solution 12S.

Observation : The balloon will sink.
Explanation : As air is pumped out from jar, the density of air in jar decreases, so the upthrust on balloon decreases. As weight of balloon exceeds the upthrust on it, it sinks.

Solution 13S.

(a) It will float with some part outside water.
Reason : On adding some salt to water, the density of water increases, so upthrust on a block of wood increases, and hence, the block rises up till the weight of salty water displaced by the submerged part of block becomes equal to the weight of the block.

(b) The block will sink.
Reason: On heating, the density of water decreases, so upthrust on the block decreases and the weight of block exceeds upthrust due to which it sinks.

Solution 14S.

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Solution 15S.

Density of brine is more than the density of water. Hence, the upthrust exerted by brine is more than the upthrust exerted by water on ice. Therefore, floating ice is less submerged in brine.

Solution 16S.

(i) 1:1; The weight of the water displaced by the man in sea and river will be same and will be equal to his own weight.
(ii) He finds it easier to swim in the sea because the density of sea water is more than the density of river water. So his weight is balanced in sea water with a part of his body submerged in the water.

Solution 17S.

An iron nail sinks in water because density of iron is more than the density of water, so the weight of the nail is more than the upthrust of water on it.

On the other hand, ships are also made of iron, but they do not sink. This is because the ship is hollow and the empty space in it contains air, which makes its average density less than that of water. Therefore, even with a small portion of ship submerged in water, the weight of water displaced by the submerged part of ship becomes equal to the total weight of ship and it floats.

Solution 18S.

Due to the hollow and empty space in the ship, the average density of a ship is less than the density of water.

Solution 19S.

When a floating piece of ice melts into water, it contracts by the volume equal to the volume of ice pieces above the water surface while floating on it. Hence, the level of water does not change when ice floating on it melts.

Solution 20S.

Forces acting on the body are listed below:

  1. Weight of the body vertically downwards.
  2. Upthrust of water on body vertically upwards.
  3. Tension in thread vertically downwards.

Solution 21S.

A ship submerges more as it sails from sea water to river water.
Density of river water is less than the density of sea water. Hence, according to the law of floatation, to balance the weight of the ship, a greater volume of water is required to be displaced in river water of lower density.

Solution 22S.

(a) Icebergs are dangerous for ships as they may collide with them. Icebergs being lighter than water, float on water with a major part of their surfaces laying under the water surface and only a small part lies outside water. Thus, it becomes difficult for the driver of the ship to estimate the size of the iceberg.

(b) Density of a strong salt solution is more than the density of fresh water. Hence, the salt solution exerts a greater upthrust on the egg which balances the weight of the egg, so the egg floats in a strong salt solution but sinks in fresh water.

(c) Density of hydrogen is much less than the density of carbon dioxide. When a balloon is filled with hydrogen, the weight of the air displaced by an inflated balloon (i.e. upthrust) becomes more than the weight of a gas filled balloon, and hence, it rises. In case of a balloon filled with carbon dioxide, weight of the balloon becomes more than the upthrust of the air, and hence, it sinks to the floor.

(d) As a ship in harbor is unloaded, its weight decreases. As a result, it displaces less water, and the ship’s hull rises in water till the weight of the water displaced balances the weight of the unloaded ship.

(e) The reason is that the density of air decreases with altitude. Therefore, as the balloon gradually goes up, the weight of the displaced air (i.e. uphrust) decreases. It keeps on rising as long as the upthrust exceeds its weight. When upthrust becomes equal to its weight, it stops rising.

(f) Density of river water is less than the density of sea water. Hence, according to the law of floatation, to balance the weight of the ship, a great volume of water is required to be displaced in river water having a comparitively lower density.

Solution 1M.

W= FB

Solution 2M.

Zero

Solution 3M.

ρ1 > ρ2

Solution 1N.

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Solution 2N.

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Solution 3N.

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Solution 4N.

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Solution 5N.

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Solution 6N.

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Solution 7N.

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Solution 8N.

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Solution 9N.

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