Selina ICSE Solutions for Class 9 Physics

Selina Concise Physics Class 9 ICSE Solutions Current Electricity

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Selina Concise Physics Class 9 ICSE Solutions Current Electricity

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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 CellSecondary 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 Magnetism

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Selina Concise Physics Class 9 ICSE Solutions Magnetism

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 10 Magnetism. 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 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.

ElectromagnetPermanent magnet
It is made up of soft ironIt is made up of steel
The magnetic field strength can be changedThe 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 Laws of Motion

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Selina Concise Physics Class 9 ICSE Solutions Laws of Motion

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 9 Physics Chapter 3 Laws of Motion. 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 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 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 Reflection of Light

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Selina Concise Physics Class 9 ICSE Solutions Reflection of Light

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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.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 4

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.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 5

Solution 9S.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 6

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 ImageVirtual 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.

Selina Concise Physics Class 9 ICSE Solutions Reflection of Light image - 17

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.
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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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