Selina Class 10 Physics Solutions

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits

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Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 9 Electrical Power and Household Circuits. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 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 10 Physics Chapter 9 Electrical Power and Household Circuits

Solution 1.

The electric power is generated at 11 KV, 50Hz at the power generating station.

Solution 2.

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 1
At a power generating station, the electric power is generated at 11 kV. From here, the alternating voltage is transmitted to the grid sub-station and stepped up to 132 kV using a step-up transformer. It is then transmitted to the main sub-station where the voltage is stepped down to 33 kV using a step-down transformer and is then transmitted to the intermediate sub-station. At the intermediate sub-station, the voltage is stepped down to 11 kV using a step-down transformer and is transmitted to the city sub-station, where the voltage is further stepped down to 220 V and is supplied to our houses.

Solution 3.

Electric power from the generating station is transmitted at 11 kV because voltage higher than this causes insulation difficulties, while the voltage lower than this involves high current and loss of energy in form of heat (I2Rt).

Solution 4.

At 220 V of voltage and 50 Hz of frequency, the a.c. is supplied to our houses.

Solution 5.

(a) Step-up transformer
(b) Step-down transformer

Solution 6.

(a) The three connecting wires used in a household circuit are:

  1. Live (or phase) wire (L),
  2. Neutral wire (N), and
  3. Earth wire (E).

(b) Among them neutral and earth wires are at the same potential.
(c) The switch is connected in the live wire.

Solution 7.

Before the electric line is connected to the meter in a house, a fuse of rating (≈ 50 A) is connected in the live wire at the pole or just before the meter. This fuse is called the pole fuse.
Its current rating is ≈ 50 A.

Solution 8.

  1. After the company fuse, the cable is connected to a kWh meter and from this meter; connections are made to the distribution board through a main fuse and a main switch.
  2. Main fuse is connected in the live wire and in case of high current it gets burnt and cut the connections to save appliances.
  3. Main switch is connected in the live and neutral wires. It is used to cut the connections of the live as well as the neutral wires simultaneously from the main supply.

Solution 9.

The electric meter in a house measures the electrical energy consumed in kWh.
Its value in S.I. unit is 1kWh = 3.6 x 106J.

Solution 10.

The main fuse in a house circuit is connected on the distribution board, in live wire before the main switch.

Solution 12.

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 2
Advantages of ring system over tree system

  1. In a ring system the wiring is cheaper than tree system.
  2. In ring system the sockets and plugs of same size can be used while in a tree system sockets and plugs are of different size.
  3. In ring system, each appliance has a separate fuse due to which if there is a fault and the fuse of one appliance burns it does not affect other appliances; while in a tree system when fuse in one distribution line blows, it disconnects all the appliances connected to that distribution circuit.

Solution 13.
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Solution 14.

All the electrical appliances in a building should be connected in parallel at the mains, each with a separate switch and a separate fuse connected in the live wire so that the switching on or off in a room has no effect on other lamps in the same building.

Solution 16.

In set A, the bulbs are connected in series. Thus, when the fuse of one bulb blows off, the circuit gets broken and current does not flow through the other bulbs also.
In set B, the bulbs are connected in parallel. Thus, each bulb gets connected to its voltage rating (= 220 V) and even when the fuse of one bulb blows off, others remain unaffected and continue to glow.

Solution 1 (MCQ).

The main fuse is connected in live wire.
Hint: The main fuse is connected in live wire so that if the current exceeds its rating, the fuse melts and breaks the circuit; thus, preventing the excessive current from flowing into the circuit.

Solution 2 (MCQ).

Electrical appliances in a house are connected in parallel.
Hint: On connecting the electrical appliances in parallel, each appliance works independently without being affected whether the other appliance is switched on or off.

Solution 3 (MCQ).

Energy
Hint: The electric meter in a house records the amount of electrical energy consumed in a house.

Exercise 9(B)

Solution 1.

An electric fuse is a safety device, which is used to limit the current in an electric circuit. The use of fuse safeguards the circuit and appliances connected in that circuit from being damaged.
An alloy of lead and tin is used as a material of fuse because it has low melting point and high resistivity.

Solution 2.

‘Fuse’ is used to protect electric circuits from overloading and short circuiting. It works on heating effect of current.

Solution 3.

(a) A fuse is a short piece of wire of material of high resistance and low melting point.
(b) A fuse wire is made of an alloy of lead and tin. If the current in a circuit rises too high, the fuse wiremelts
(c) A fuse is connected in series with the live wire.
(d) Higher the current rating, Thicker is the fuse wire.

Solution 4.

The fuse wire is fitted in a porcelain casing because porcelain is an insulator of electricity.

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 4

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 5

Solution 7.

The fuse wire is always connected in the live wire of the circuit because if the fuse is put in the neutral wire, then due to excessive flow of current when the fuse burns, current stops flowing in the circuit, but the appliance remains connected to the high potential point of the supply through the live wire. Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Solution 8.

The 20 A fuse wire will be thicker so that its resistance be low.

Solution 9.

It means that the line to which this fuse is connected has a current carrying capacity of 5 A.

Solution 10.

The safe limit of current which can flow through the electrical appliance is I = P/V = 5000/200 = 25 A; which is greater than 8 A. So, such fuse cannot be used.

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 16

Solution 12.

A switch is an on-off device for current in a circuit (or in an appliance). The switch should always be connected in the live wire so that the appliance could be connected to the high potential point through the live wire. In this position the circuit is complete as the neutral wire provides the return path for the current. When the appliance does not work i.e., in off position of the switch, the circuit is incomplete and no current reaches the appliance.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 6
On the other hand, if switch is connected in the neutral wire, then in ‘off’ position, no current passes through the bulb. But the appliance remains connected to the high potential terminal through the live wire.
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Thus, if the switch is connected in the neutral wire, it can be quite deceptive and even dangerous for the user.
Precaution while handling a switch: A switch should not be touched with wet hands.

Solution 13.

A switch should not be touched with wet hands. If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand and the person may get a fatal shock.

Solution 14.

Let a switch S1 be fitted at the bottom and a switch S2 at the top of the staircase. Fig. (a) shows the off position of the bulb.
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The bulb can now be switched on independently by either the switch S1 or the switch S2. If the switch S1 is operated, the connection ‘ab’ is changed to ‘bc’, which completes the circuit and the bulb lights up [Fig. (b)].
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 9

Similarly, on operating the switch S2, the connection ‘bc’ changes to ‘ba’, which again completes the circuit [Fig. (c)].
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 10
Similarly if the bulb is in on position as shown in Fig. (b) or (c), one can switch off the bulb either from the switch S1 or the switch S2.

Solution 15.

All electrical appliances are provided with a cable having a plug at one end to connect the appliance to the electric supply.
In this three way pin plug, the top pin is for earthing (E), the live pin (L) in on the left and the neutral pin (N) is on the right.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 11

Solution 16.

The three pins in the plug are labelled as
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 12
Here E signifies the earth pin,
L is for live wire, and
N is for neutral wire.

  1. The earth pin is made long so that the earth connection is made first. This ensures the safety of the user because if the appliance is defective, the fuse will blow off. The earth pin is thicker so that even by mistake it cannot be inserted into the hole for the live or neutral connection of the socket.
  2. The pins are splitted at the end to provide spring action so that they fit in the socket holes tightly.

Solution 17.
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Solution 18.

(a) 1 – Earth, 2 – Neutral, 3 – Live
(b) Terminal 1 is connected to the outer metallic case of the appliance.
(c) The fuse is connected to live wire joined to 3 so that in case of excessive flow of current fuse melts first and breaks down the circuit to protect appliances.

Solution 19.

Local earthing is made near kWh meter. In this process a 2 – 3 metre deep hole is dug in the ground. A copper rod placed inside a hollow insulating pipe, is put in the hole. A thick copper plate of dimensionsis 50 cm x 50 cm welded to the lower end of the copper rod and it is buried in the ground. The plate is surrounded by a mixture of charcoal and salt to make a good earth connection.
To keep the ground damp, water is poured through the pipe from time to time. This forms a conducting layer between the plate and the ground. The upper end of the copper rod is joined to the earth connection at the kWh meter.

Solution 20.

If the live wire of a faulty appliance comes in to direct contact with the metallic case due to some reason then the appliance acquires the high potential of live wire. This may results in shock if any person touches the body of appliance. But if the appliance is earthed then as soon as the live wire comes in to contact with the metallic case, high current flows through the case to the earth. The fuse connected to the appliance will also blows off, so the appliance get disconnected.

Solution 21.

(a) The fuse must be connected in the live wire only. If the fuse is in the neutral wire, then although the fuse burns due to the flow of heavy current, but the appliance remains at the supply voltage so that on touching the appliance current flows through the appliance to the person touching it.
(b) Metallic case of the appliance should be earthed.

Solution 22.

The paint provides an insulating layer on the metal body of the appliance. To make earth connection therefore, the paint must be removed from the body part where connection is to be made.

Solution 23.

  1. According to new international convention
  2. Live wire is brown in colour.
  3. Neutral is light blue and
  4. Earth wire is yellow or green in colour.

Solution 25.

(a) The three wires are: Live wire, Earth wire and Neutral wire.
(b) The heating element of geyser should be connected to live wire and neutral wire.
(c) The metal case should be connected to earth wire.
(d) The switch and fuse should be connected to live wire.

Solution 26.

One may get an electric shock from an electrical gadget in the following two cases:

  1. If the fuse is put in the neutral wire instead of live wire and due to fault, if an excessive current flows in the circuit, the fuse burns, current stops flowing in the circuit but the appliance remains connected to the high potential point of the supply through the live wire. In this situation, if a person touches the faulty appliance, he may get an electric shock as the person will come in contact with the live wire through the appliance.
    Preventive measure: The fuse must always be connected in the live wire.
  2. When the live wire of a faulty appliance comes in direct contact with its metallic case due to break of insulation after constant use (or otherwise), the appliance acquires the high potential of the live wire. A person touching it will get a shock because current flows through his body to earth.
    Preventive measure: Proper ‘earthing’ of the electric appliance should be done.

Solution 27.

Power circuit carries high power and costly devices. If there is some unwanted power signal (noise) in the wire it can damage the device. To reduce this effect earth is necessary.
Lighting circuit carries low power (current).So, we ignore the earth terminal.

Solution 28.

A high tension wire has a low resistance and large surface area.

Solution 29.

To carry larger current, the resistance of the wire should be low, so its area of cross section should be large. Therefore 15 A current rated wire will be thicker.

Solution 30.

(a) Switches 2 and 3.
(b) The lamps are connected in series.

Solution 31.
(a)
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 14
(b)

Wire no.Wire nameColour (Old convention)Colour (New convention)
1Neutral wireBlackLight blue
2Earth wireGreenGreen or yellow
3Live wireRedBrown

(c) The bulbs are joined in parallel.

Solution 1 (MCQ).

5 A
Hint: The electric wiring for light and fan circuit uses a thin fuse of low current rating (= 5 A) because the line wire has a current carrying capacity of 5 A.

Solution 2 (MCQ).

A switch must be connected in live wire.
Explanation: A switch must be connected in live wire, so that when it is in ‘off’ position, the circuit is incomplete and no current reaches the appliance through the live wire.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 15

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Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism

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Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 10 Electro-magnetism. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 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 10 Physics Chapter 10 Electro-magnetism

Exercise 10(A)

Solution 1.

Experiment:
In Fig , AB is a wire lying in the north- south direction and connected to a battery through a rheostat and a tapping key. A compass needle is placed just below the wire. It is observed that

  1. When the key is open i.e., no current passes through the wire, the needle shows no deflection and it points in the N-S direction (i.e. along the earth’s magnetic field). In this position, the needle is parallel to the wire as shown in Fig. (a).
    Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 1
  2. When the key is pressed, a current passes in the wire in the direction from A to B (i.e. From south to north) and the north pole(N) of the needle deflects towards the west [Fig. (b)].
    Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 2
  3. When the direction of current in the wire is reversed by reversing the connections at the terminals of the battery, North Pole (N) of the needle deflects towards the east [Fig. (c)].
    Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 3
  4. If the compass needle is placed just above the wire, the North Pole (N) deflects towards east when the direction of current in wire is from A to B [Fig. (d)], but the needle deflects towards west as in fig (e), if the direction of current in wire is from B to A.
    Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 4

The above observations of the experiment suggest that a current carrying wire produces a magnetic field around it.

Solution 2.
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Solution 3.

(a) On decreasing the current the magnetic field lines become rarer.
(b) The direction of magnetic field lines will get reversed.

Solution 4.

Right hand thumb rule determines the direction of magnetic field around a current carrying wire.
It states that if we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 5.

(a) The direction of magnetic field at a point just underneath is towards east.
(b) Right hand thumb rule.

Solution 6.

A current carrying conductor produces a magnetic field around it and the magnetic needle in this magnetic field experience a torque due to which it deflects to align itself in the direction of magnetic field.

Solution 7.
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Solution 8.

Face of the coil exhibit North polarity.

Solution 9.

(i) Along the axis of coil inwards.
(ii) Along the axis of coil outwards.

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 7

Solution 11.

Right hand thumb rule: If we hold the current carrying conductor in right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic fields lines.

Solution 12.

(a) A – North pole, B – South pole.
(b) The magnet will be repelled because the end of the solenoid near the north pole of magnet becomes the north pole as current at this face is anticlockwise and the two like poles repel.

Solution 13.

(a) A – North pole, B – South pole.
(b) The north pole of compass needle will deflect towards west.
Reason: The end A of the coil behaves like north pole which repels north pole of compass needle towards west.

Solution 14.

The magnetic field due to a solenoid can be made stronger by using:

  1. By increasing the number of turns of winding in the solenoid.
  2. By increasing the current through the solenoid.

Solution 15.

A current carrying freely suspended solenoid at rest behaves like a bar magnet. It is because a current carrying solenoid behaves like a bar magnet.

Solution 16.

The needle of the compass will rest in in the direction of magnetic field due to solenoid at that point.

Solution 17.

Magnetic fielddue to a solenoid carrying current increases if a soft iron bar is introduced inside the solenoid.

Solution 18.

(A) When current flows in a wire, it creates magnetic field around it.
(B) On reserving the direction of current in a wire, the magnetic field produced by it gets reversed.
(C) A current carrying solenoid behaves like a bar magnet
(D) A current carrying solenoid when freely suspended, it always rest in north-south direction.

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 8

Solution 20.

(a) X-north pole, Y –south pole.
(b) By reducing resistance of circuit by mean of rheostat to increase current.

Solution 21.

(a) Solenoid is a cylindrical coil of diameter less than its length.
(b) The device so obtained is electromagnet.
(c) It is used in electric relay.

Solution 22.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 9

Solution 23.

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.

The strength of magnetic field of an electromagnet depends on:

  1. Number of turns: The strength of magnetic increases on increasing the number of turns of winding in the solenoid.
  2. Current: The strength of magnetic field increases on increasing the current through the solenoid.

Solution 24.

At A-south pole and at B-north pole.

Solution 25.

The strength of an electromagnet can be increased by following ways:

  1. By increasing the number of turns of winding in the solenoid.
  2. By increasing the current through the solenoid.

Solution 27.

  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 28.

ElectromagnetPermanent magnet
It is made up of soft ironIt is made up of steel.
The magnetic field strength can be changed.The magnetic field strength cannot be changed.
The electromagnets of very strong field can be made.The permanent magnets are not so strong.

Solution 29.

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. That’s why soft iron is used as the core of the electromagnet in an electric bell.

Solution 30.

If an a.c. source is used in place of battery, the core of electromagnet will get magnetized, but the polarity at its ends will change. Since attraction of armature does not depend on the polarity of electromagnet, so the bell will still ring on pressing the switch.

Solution 31.
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Solution 32.

The material used for making the armature of an electric bell is soft iron which can induce magnetism rapidly.

Solution 1 (MCQ).

The presence of magnetic field at a point can be detected by a compass needle.
Note: In the presence of a magnetic field, the needle of compass rests only in the direction of magnetic field and in the absence of any magnetic field, the needle of compass can rest in any direction. In the earth’s magnetic field alone, the needle rests along north-south direction.

Solution 2 (MCQ).

By reversing the direction of current in a wire, the magnetic field produced by it gets reversed in direction.
Hint: On reversing the direction of current in a wire, the polarity of the faces of the wire also reverses. Thus, the direction of magnetic field produced by it also gets reversed.

Exercise 10(B)

Solution 1.

The magnitude of force on a current carrying conductor placed in a magnetic field depends on:

  1. On strength of magnetic field B.
  2. On current I in the conductor.
  3. On length of conductor.

Magnitude of force on a current carrying conductor placed in a magnetic field depends directly on these three factors.

Solution 2.

(a) When current in the conductor is in the direction of magnetic field force will be zero.
(b) When current in the conductor is normal to the magnetic field.

Solution 3.

Direction of force is also reversed.

Solution 4.

Fleming’s left hand rule: Stretch the forefinger, middle finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the middle finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor.

Solution 6.

Unit is: Newton/ampere x meter (or NA-1m-1).

Solution 7.

(a) The coil will experience a torque due to which it will rotate.
(b) The coil will come to rest when their plane become normal to the magnetic field.
(c) (i) When plane of a oil is parallel to the magnetic field,
(ii) When plane of coil is normal to the magnetic field.
(d) The instrument which makes use of the principle stated above is d.c. motor.

Solution 8.

(a) The coil begins to rotate in anticlockwise direction.
(b) This is because, after half rotation, the arms AB and CD get interchanged, so the direction of torque on coil reverses. To keep the coil rotating in same direction, commutator is needed to change the direction of current in the coil after each half rotation of coil.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 11

Solution 9.

Electric motor: An electric motor is a device which converts the electrical energy into the mechanical energy.
Principle: An electric motor (dc motor) works on the principle that when an electric current is passed through a conductor placed normally in a magnetic field, a force acts on the conductor as a result of which the conductor begins to move and mechanical energy is obtained.

Solution 10.
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Solution 11.

Electrical energy converts into mechanical energy.

Solution 12.

The speed of rotation of an electric motor can be increased by:

  1. Increasing the strength of current.
  2. Increasing the number of turns in the coil.

Solution 13.

Electric motor is used in electrical gadgets like fan, washing machine, juicer, mixer, grinder etc.

Solution 1 (MCQ).

In an electric motor, the energy transformation is from electrical to mechanical.
Note: An electric motor is a device which converts electrical energy into mechanical energy.

Exercise 10(C)

Solution 1.

(a) Electromagnetic induction: whenever there is change in number of magnetic field lines associated with conductor, an electromotive force is developed between the ends of the conductor which lasts as long as the change is taking place.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 13

  1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig. (a)]
  2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in [Fig (b)]
  3. As the motion of magnet stops, the pointer of the galvanometer comes to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
  4. If the magnet is moved away from the solenoid, the current again flows in the solenoid, but now in a direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left[ Fig. (d)].
  5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
  6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

Solution 2.

Faraday’s formulated two laws of electromagnetic induction:

  1. Whenever there is a change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts so long as there is a change in the magnetic flux linked with the coil.
  2. The magnitude of the e.m.f. induced is directly proportional to the rate of change of the magnetic flux linked with the coil. If the rate of change of magnetic flux remains uniform, a steady e.m.f. is induced.

Solution 3.

Magnitude of induced e.m.f depend upon:

  1. The change in the magnetic flux.
  2. The time in which the magnetic flux changes.

Solution 4.

(a) Mechanical energy changes to the electrical energy.
(b) Phenomenon is called electromagnetic induction.

Solution 5.

Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 14
(a) When there is a relative motion between the coil and the magnet, the magnetic flux linked with the coil changes. If the north pole of the magnet is moved towards the coil, the magnetic flux through the coil increases as shown in above figure. Due to change in the magnetic flux linked with the coil, an e.m.f. is induced in the coil. This e.m.f. causes a current to flow in the coil if the circuit of the coil is closed.
(b) The source of energy associated with the current obtained in part (a) is mechanical energy.

Solution 6.

Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 15

  1. When the magnet is stationary there is no deflection in galvanometer. The pointer read zero. [Fig (a)]
  2. When the magnetwith north pole facing the solenoid is moved towards the solenoid, the galvanometer shows a deflection towards the right showing that a current flows in the solenoid in the direction as shown in Fig (b)
  3. As the motion of magnet stops, the pointer of the galvanometer to the zero position [Fig (c)]. This shows that the current in the solenoid flows as long as the magnet is moving.
  4. If the magnet is moved away from the solenoid , the current again flows in the solenoid , but now in the direction opposite to that shown in [Fig. (b)] and therefore the pointer of the galvanometer deflects towards left [Fig. (d)].
  5. If the magnet is moved away rapidly i.e. with more velocity, the extent of deflection in the galvanometer increases although the direction of deflection remains the same.It shows that more current flows now.
  6. If the polarity of the magnet is reversed and then the magnet is brought towards the solenoid, the current in solenoid flows in the direction opposite to that shown in Fig (b) and so the pointer of galvanometer deflect towards left [Fig. (e)].

(b) Magnitude of induced e.m.f depend upon:

  1. The change in the magnetic flux.
  2. The time in which the magnetic flux changes.

(c) The direction of induced e.m.f depends on whether there is an increase or decrease in the magnetic flux.

Solution 7.

The current induced in a closed circuit only if there is change in number of magnetic field lines linked with the circuit.

Solution 8.

  1. Yes.
  2. Yes.
  3. Yes.
  4. No.

Solution 9.

Fleming’s right hand rule determines the direction of current induced in the conductor.

Solution 10.

Fleming’s right hand rule: Stretch the forefinger, middle finger and the thumb of your right hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the thumb will indicates the direction of motion of conductor, then the middle finger indicates the direction of induced current.

Solution 11.

Lenz’s law: It states that the direction of induced e.m.f. (or induced current) is such that it always tends to oppose the cause which produces it.

Solution 12.

When a coil has a large number of turns, then magnitude of induced e.m.f. in the coil become more and then by Lenz’s law it will oppose more.

Solution 13.

So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Solution 14.

Lenz’s law implies the law of conservation of energy. It shows that the mechanical energy spent in doing work, against the opposing force experienced by the moving magnet, is transformed into the electrical energy due to which current flows in the solenoid.

Solution 15.

The pointer of galvanometer deflects. The deflection last so long as the coil moves.

(a) Deflection becomes twice.
(b) Deflection becomes thrice

Solution 16.

  1. The pointer of galvanometer deflects towards left. The deflection lasts so long as the coil moves.
  2. (a) Deflection becomes twice (b) Deflection becomes thrice.

Solution 17.

(a)

  1. When switch is closed suddenly, the galvanometer needle deflects for a moment.
  2. If switch is kept closed then galvanometer needle returns to zero.
  3. If switch is opened again then galvanometer needle deflects again but in opposite to the direction of deflection in case (a).

(b) This can be explained by Faraday’s law which states that whenever there is change in the magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts as long as there is a change in the magnetic flux linked with the coil.

Solution 18.

An A.C. generator works on the principle of ‘electromagnetic induction’.
Statement: Whenever a coil is rotated in a magnetic field, the magnetic flux linked with the coil changes, and therefore, an EMF is induced between the ends of the coil. Thus, a generator acts like a source of current if an external circuit containing load is connected between the ends of its coil.

Solution 19.

The number of rotations of the coil in one second or the speed of rotation of the coil.

Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 16

Solution 21.

(a)In an a.c generator, if the speed at which the coil rotates is doubled, the frequency is also doubled.
(b) Maximum output voltage is also doubled.

Solution 22.

Two ways in an a.c generator to produce a higher e.m.f. are:

  1. By increasing the speed of rotation of the coil.
  2. By increasing the number of turns of coil.

Solution 23.

Mechanical energy changes into the electrical energy.

Solution 24.

Two dissimilarities between D.C. motor and A.C. generator:

A.C. GeneratorD.C. Motor
1. A generator is a device which converts mechanical energy into electrical energy.1. A D.C. motor is a device which converts electrical energy into mechanical energy.
2. A generator works on the principle of electromagnetic induction.2. A D.C. works on the principle of force acting on a current carrying conductor placed in a magnetic field.

Similarity: Both in A.C generator and D.C motor, a coil rotates in a magnetic field between the pole pieces of a powerful electromagnet.

Solution 25.

The voltage of a.c. can be stepped up by the use of step-up transformer at the power generating station before transmitting it over long distances. It reduces the loss of electrical energy as heat in the transmission line wires. On the other hand, if d.c. is generated at the power generating station, its voltage cannot be increased for transmission, and so due to passage of high current in the transmission line wires, there will be a huge loss of electrical energy as heat in the line wires.

Solution 26.

The purpose of the transformer is to step up or step down the a.c. voltage.
No, a transformer cannot be used with a direct current source.

Solution 28.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 25

Solution 29.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 17

Solution 30.

The device is step up transformer.
It works on the principle of electromagnetic induction.

Solution 31.

Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 18
Step-up transformer: The step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency.

Working: When the terminals of primary coil are connected to the source of alternating e.m.f., a varying current flows through it which also produces a varying magnetic field in the core of the transformer. Thus, the magnetic field lines linked with the secondary coil vary and induce an e.m.f. in the secondary coil. The induced e.m.f. varies in the same manner as the applied e.m.f. in the primary coil varies, and thus, has the same frequency as that of the applied e.m.f.
The magnitude of e.m.f. induced in the secondary coil depends on the ‘turns ratio’ and the magnitude of the applied e.m.f.
For a transformer,
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 26
Two characteristics of the primary coil as compared to its secondary coil:

  1. The number of turns in the primary coil is less than the number of turns in the secondary coil.
  2. A thicker wire is used in the primary coil as compared to that in the secondary coil.

Solution 32.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 19

Working: In a step down transformer, the number of turns in secondary coil are less than the number of turns in the primary coil i.e., turns ratio NS/NP<1.
As Es/Ep = NS/NP.
So Es/Eps is less than Ep.

Two uses of step down transformer are:

  1. With electric bells
  2. At the power sub-stations to step-down the voltage before its distribution to the customers.

Solution 34.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 20
A is the primary coil. B is the secondary coil.
We have drawn laminated core in the diagram.
The material of this part is soft iron.
This transformer a step-down transformer because the number of turns in primary coil is much greater than that in the secondary coil.

Solution 35.

(a) Soft iron core is used. The core is made up from the thin laminated sheets of soft iron of T and U shape, placed alternately one above the other and insulated from each other by paint or varnish coating over them.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 21

Solution 36.

The secondary windings of a transformer in which the voltage is stepped down are usually made up of thicker than the primary because more current flows in the secondary coil. The use of thicker wire reduces its resistance and therefore the loss of energy as heat in the coil.

Solution 37.

To reduce the energy losses due to eddy currents.

Solution 38.

  1. In a step-up transformer, the number of turns in the primary is less than the number of turns in the secondary.
  2. The transformer is used in alternating current circuits.
  3. In a transformer, the frequency of A.C. voltage remain same.

Solution 39.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 27

Solution 40.

The energy loss in a transformer is called ‘copper loss’.

Copper losses: Primary and secondary coils of a transformer are generally made of copper wire. These copper wires have resistance. When current flows through these wires, a part of the energy is lost in the form of heat. This energy lost through the windings of the transformer is known as copper loss.

This loss can be minimized by using thick wires for the windings. Use of thick wire reduces its resistance and therefore reduces the loss of energy as heat in the coil.

Solution 41.

Step up transformerStep down transformer
It increases the a.c. voltage and decrease the current.It decreases the a.c. voltage and increase the current.
The wire of primary coil is thicker than that in the secondary coil.The wire in the secondary coil is thicker than that in the primary coil.

Solution 42.

Soft iron is used in all.

Solution 1 (MCQ).

Fleming’s right hand rule
Statement: According to Fleming’s right hand rule, if we stretch the thumb, middle finger and forefinger of our right hand mutually perpendicular to each other such that the forefinger indicates the direction of magnetic field and thumb indicates the direction of motion of conductor, then the middle finger will indicate the direction of induced current.

Solution 2 (MCQ).

NS > NP
Hint: Since a step-up transformer is used to change a low voltage alternating e.m.f. to a high voltage alternating e.m.f. of same frequency, the number of turns in the secondary coil is more than the number of turns in the primary coil, i.e. NS > NP.

Numericals

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 22

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 23

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Electro Magnetism img 24

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Selina Concise Physics Class 10 ICSE Solutions Calorimetry

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Selina Concise Physics Class 10 ICSE Solutions Calorimetry

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Selina ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry

Exercise 11(A)

Solution 1.

The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.

Solution 2.

S.I. unit of heat is joule (symbol J).

Solution 3.

One calorie of heat is the heat energy required to raise the temperature of 1 g of water from 14.5oC to 15.5 oC.
1 calorie = 4.186 J

Solution 4.

One kilo-calorie of heat is the heat energy required to raise the temperature of 1 kg of water from 14.5oC to 15.5oC.

Solution 5.

The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.
S.I. unit kelvin (K).

Solution 6.

HeatTemperature
The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.
S.I. unit joule (J).S.I. unit kelvin (K).
It is measured by the principle of calorimetry.It is measured by a thermometer.

Solution 7.

The measurement of the quantity of heat is called calorimetry.

Solution 8.

The heat capacity of a body is the amount of heat energy required to raise its temperature by 1oC or 1K.
S.I. unit is joule per kelvin (JK-1).

Solution 9.

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1oC (or 1K).
S.I. unit is joule per kilogram per kelvin (Jkg-1K-1).

Solution 10.

Heat capacity = Mass x specific heat capacity

Solution 11.

  • Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1oC whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1oC.
  • Heat capacity of a substance depends upon the material and mass of the body. Specific heat capacity of a substance does not depend on the mass of the body.
  • S.I. unit of heat capacity is JK-1 and S.I. unit of specific heat capacity is Jkg-1K-1.

Solution 12.

Water has the highest specific heat capacity.

Solution 13.

Specific heat capacity of water=4200 J kg-1 K-1.

Solution 14.

  1. The heat capacity of a body is 50JK-1 means to increase the temperature of this body by 1K we have to supply 50 joules of energy.
  2. The specific heat capacity of copper is 0.4Jg-1K-1 means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.

Solution 16.

The quantity of heat energy absorbed by a body depends on three factors :

  1. Mass of the body – The amount of heat energy required is directly proportional to the mass of the substance.
  2. Nature of material of the body – The amount of heat energy required depends on the nature on the substance and it is expressed in terms of its specific heat capacity c.
  3. Rise in temperature of the body – The amount of heat energy required is directly proportional to the rise in temperature.

Solution 17.

The expression for the heat energy Q
Q = mcΔt (in joule)

Solution 18.

Heat capacity of liquid A is less than that of B.
As the substance with low heat capacity shows greater rise in temperature.

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 1

Solution 20.

The principle of method of mixture:
Heat energy lost by the hot body = Heat energy gained by the cold body.
This principle is based on law of conservation of energy.

Solution 21.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 2

Solution 22.

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0oC, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0oC.

Solution 23.

The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

Solution 24.

The reason is that water does not cool quickly due to its large specific heat capacity, so a hot water bottle provides heat energy for fomentation for a long time.

Solution 25.

By allowing water to flow in pipes around the heated parts of a machine, heat energy from such part is removed. Water in pipes extracts more heat from surrounding without much rise in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.

Solution 26.

  1. Radiator in car.
  2. To avoid freezing of wine and juice bottles.

Solution 28.

A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained or lost by a body when it is mixed with other body.
It is made up of thin copper sheet because:

  1. Copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents.
  2. Copper has low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

Solution 29.

By making the base of a cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps the food warm for a long time, after cooking.

Solution 1 (MCQ).

JK-1

Solution 2 (MCQ).

J kg-1K-1

Solution 3 (MCQ).

4200 J kg-1K-1

Numericals

Solution 1.

The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales.
Therefore, the corresponding rise in temperature on the Kelvin scale will be 15K.

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 3

Solution 3.

  1. We know that heat energy needed to raise the temperature by 15o is = heat capacity x change in temperature.
    Heat energy required= 966 J K-1 x 15 K = 14490 J.
  2. We know that specific heat capacity is = heat capacity/ mass of substance
    So specific heat capacity is = 966 / 2=483 J kg-1 K-1.

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 4

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 5

Solution 6.
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Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 7

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 8

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 9

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 10

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 11

Solution 12.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 12

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 13

Exercise 11(B)

Solution 1.

(a) The process of change from one state to another at a constant temperature is called the change of phase of substance.
(b) There is no change in temperature during the change of phase.
(c) Yes, the substance absorbs or liberates heat during the change of phase.

Solution 2.

Melting: The change from solid to liquid phase on heating at a constant temperature is called melting.
Melting point: The constant temperature at which a solid changes to liquid is called the melting point.

Solution 3.

  1. Average kinetic energy of molecules changes.
  2. Average potential energy of molecules changes.

Solution 4.

(a) Average kinetic energy does not change.
(b) Average potential energy increases.
Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.

However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.
This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.

Solution 5.

The melting point of ice decreases by the presence of impurity in it.
Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.

Solution 6.

The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072oC for every one atmosphere rise in pressure.

Solution 7.

(a) AB part shows rise in temperature of solid from 0 to T1oC, BC part shows melting at temperature T1oC, CD part shows rise in temperature of liquid from T1oC to T3oC , DE part shows the boiling at temperature T3oC.
(b) T1oC.
(c) T3oC.

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 14

Solution 9.

Boiling: The change from liquid to gaseous phase on heating at a constant temperature is called boiling.
Boiling point: The particular temperature at which vaporization occurs is called the boiling point of liquid.

Solution 10.

Volume of water wills increases when it boils at 100oC.

Solution 11.

The boiling point of water increases on adding salt.

Solution 12.

The boiling point of a liquid increases on increasing the pressure.

Solution 14.

In a pressure cooker, the water boils at about 120oC to 125oC.

Solution 15.

This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.

Solution 16.

(a) When ice melts, its volume decreases.
(b) Decrease in pressure over ice increases its meltingpoint.
(c) Increase in pressure increases the boiling point of water.
(d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.
(e) The boiling point of water is defined as the constant temperature at which water changes to steam.
(f) water can be made to boil at 115°C by increasing pressure over its surface.

Solution 17.

Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.

Solution 18.

The quantity of heat required to convert unit mass of ice into liquid water at 00C (melting point) is called the specific latent heat of fusion of ice.
Its S.I. unit is Jkg-1.

Solution 19.

Specific latent heat of ice: 336000 J kg-1.

Solution 20.

It means 1 g of ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.

Solution 21.

Latent heat is absorbed by ice.

Solution 22.

1 g of water at 0oC has more heat than 1 g of ice at 0oC. This is because ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.

Solution 23.

(a) 1 g ice at 0oC requires more heat because ice would require additional heat energy equal to latent heat of melting.
(b) 1 g ice at 0oC first absorbs 336 J heat to convert into 1 g water at 0oC.

Solution 24.

This is because 1 g of ice at 0oC takes 336 J of heat energy from the mouth to melt at 0oC. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1g of water at 0oC. Therefore cooling produced by 1 g of ice at 0oC is more than for 1g of water at 0oC.

Solution 25.

This is because 1 g of ice at 0oC takes 336 J of heat energy from the bottle to melt into water at 0oC. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1 g iced water at 0oC. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.

Solution 26.

The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.

Solution 27.

The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.

Solution 29.

  1. The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.
  2. Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.

Solution 1 (MCQ).

J kg-1

Solution 2 (MCQ).

80cal g-1

Numericals

Solution 1.

Mass of ice=10g = 0.01kg
Amount of heat energy absorbed, Q=5460J
Specific latent heat of fusion of ice=?
Specific heat capacity of water = 4200Jkg-1K-1
Amount of heat energy required by 10g (0.01kg) of water at 0oC to raise its temperature by 50oC= 0.01X4200X50=2100J.
Let Specific latent heat of fusion of ice=L Jg-1.
Then,
Q = mL + mcΔT
5460 J = 10 x L + 2100J
L = 336Jg-1.

Solution 2.

Mass of water m = 5.0 g
specific heat capacity of water c = 4.2 J g-1 K-1
specific latent heat of fusion of iceL =336 J g-1
Amount of heat energy released when 5.0 g of water at 20oC changes into water at 0oC = 5 x 4.2 x 20 = 420 J.
Amount of heat energy released when 5.0g of water at 0oC changes into ice at 0oC=5x336J=1680J.
Total amount of heat released =1680 J + 420 J = 2100 J.

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 15

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 16

Solution 5.

Mass of ice m1 =17 g
Mass of water m2 =40 g.
Change in temperature =34-0=34K
Specific heat capacity of water is 4.2Jg-1K-1.
Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water
Q = 40 x 34 x 4.2 = 5712 J.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 17

Solution 6.

Let whole of the ice melts and let the final temperature of the mixture be ToC.
Amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
Amount of heat energy gained by 10g of ice at 0oC to convert into water at 0oC=10×336=3360 J
Amount of heat energy gained by 10g of water (obtained from ice) at 0oC to raise its temperature to ToC = 10×4.2x(T-0)=42T
Amount of heat energy released by 10g of water at 10oC to lower its temperature to ToC = 10×4.2x(10-T)=420-42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420-42T
T = -37.5oC
This cannot be true because water cannot exist at this temperature.
So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0oC.
So, amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
Amount of heat energy gained by m gm of ice at 0oC to convert into water at 0oC=mx336=336m J
Amount of heat energy released by 10g of water at 10oC to lower its temperature to 0oC = 10×4.2x(10-0)=420
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 gm

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 18

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 19

Solution 9.

Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be 0 oC.
Amount of heat energy gained by 2 kg of ice at 0oC to convert into water at 0oC=2 x 336000 = 672000 J
Let amount of water poured = m kg.
Initial temperature of water = 100oC.
Final temperature of water = 0oC.
Amount of heat energy lost by m kg of water at 100oC to reach temperature 0oC =m x 4200 x 100 = 420000m J
We know that heat energy gained = heat energy lost.
672000J = mX420000J
m = 672000/420000=1.6kg

Solution 10.

Amount of heat energy gained by 100 g of ice at -10o C to raise its temperature to 0o C = 100 × 2.1 × 10 = 2100 J
Amount of heat energy gained by 100 g of ice at 0o C to convert into water at 0o C = 100 × 336 = 33600 J
Amount of heat energy gained when temperature of 100 g of water at 0o C rises to 100o C = 100 × 4.2 × 100 = 42000 J
Total amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 104 J

Solution 11.

Amount of heat energy gained by 1kg of ice at -10oC to raise its temperature to 0oC= 1 x 2100 x 10 = 21000 J
Amount of heat energy gained by 1kg of ice at 0oC to convert into water at 0oC = L
Amount of heat energy gained when temperature of 1kg of water at 0oC rises to 100oC= 1 x 4200 x 100 = 420000 J
Total amount of heat energy gained = 21000 + 420000 + L = 441000 + L.
Given that total amount of heat gained is = 777000J.
So,
441000 + L = 777000.
L = 777000 – 441000.
L = 336000JKg-1

Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 20

Solution 12.

Mass of ice, mice = 200 g

Time for ice to melt, t1 = 1 min = 60 s

Mass of water, mw = 200 g

Temperature change of water, ΔT = 20 °C

Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.

Exercise 11(C)

Solution 1.

The earth receives heat radiations from sun which reach us after passing through its atmosphere. The earth’s atmosphere is transparent for the visible and thermal radiations of short wavelengths coming from the sun. The earth’s surface and objects on it thus become warm in the day time. After the sunset, the earth’s surface and the objects on it radiate the infrared radiations of long wavelengths. A part of these radiations are reflected back by the clouds and a part of it is absorbed by the green house gases like carbon dioxide, methane, water vapours and chlorofluorocarbons. Thus the clouds and green house gases prevents a large fraction of radiations given out by the earth’s surface, from escaping into the space. This phenomenon is called greenhouse effect.

Solution 2.

Carbon dioxide, methane, water vapours and chlorofluorocarbons.

Solution 3.

(a) Short wavelength radiations
(b) Long wavelength radiations

Solution 4.

Coal, petroleum, natural gas.

Solution 5.

From sun, we receive 1366W m-2 energy at the top of our earth’s atmosphere, out of which only 235Wm-2 energy reaches near the earth’s surface. The earth and ocean surface absorbs 168 W m-2 energy and only 67 W m-2 energy remains in the lower atmosphere. With this much energy received on earth surface, its actual surface temperature would have been around -180C which is quite uncomfortable for human living. Fortunately the greenhouse gases present in the earth’s atmosphere contribute in trapping the heat energy within the atmosphere and they produce an average warming effect of about 330C to keep the effective temperature around 150C. So, greenhouse effect helps in keeping the temperature of earth’s surface suitable for living of human beings.

Solution 6.

Three reasons for increase of greenhouse gases:

  1. The burning of fuels, deforestation and industrial production
  2. Increase of population.
  3. Imbalance of carbon dioxide cycle

Solution 7.

The effect of enhancement of greenhouse effect are:

  1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
  2. It has affected the blooming season of the different plants.
  3. The climate changes have shown the immediate effect on simple organism and plants.
  4. It has affected the world’s ecology.
  5. It has increased the heat stroke deaths.

Solution 8.

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

Solution 9.

The effect of enhancement of greenhouse effect are:

  1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.
  2. It has affected the blooming season of the different plants.
  3. The climate changes have shown the immediate effect on simple organism and plants.
  4. It has affected the world’s ecology.
  5. It has increased the heat stroke deaths.

Solution 10.

Due to rise in sea level the building and roads in the coastal areas will get flooded and they could suffer damage from hurricanes and tropical storms.

Solution 11.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 12.

Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.

Solution 13.

At the present rate of increase of green house effect, it is expected that nearly 30% of the plant species will extinct by the year 2050 and up to 70% by the end of the year 2100. In the near future, warming of nearly 3oC will result in poor yield in farms in low latitude regions. This will increase the rise of malnutrition.

Solution 14.

At the present rate of increase in green house effect, is estimated that nearly 30% of the plant and animal species will extinct by the year 2050 and upto 70 % by the end of year 2100. This will disrupt ecosystem. The animals from the equatorial region will shift to higher latitude in search of cold regions. The absorption of carbon dioxide by ocean will cause acidification due to which marine species will migrate.

Solution 16.

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

Solution 1 (MCQ).

-18oC

Solution 2 (MCQ).

The increase in sea levels.
Explanation: Due to global warming, the average temperature of the Earth has increased and has lead to the melting of ice around both the poles. This melting of ice has lead to an increase in the level of water in sea.

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 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 10 Physics Chapter 2 Work, Energy and Power

Exercise 1(A)

Solution 1.

Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.

Solution 2.

(i) When force is in direction of displacement, then work done , W = F x S
(ii) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ

Solution 3.

(a) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ
(b) (i) For zero work done, the angle between force and displacement should be 90o as cos 90o = 0
W = FS cos90o= FS x 0 = 0
(ii) For maximum work done, the angle between force and displacement should be 0o as cos0o = 1
Hence, W=FScos 0o=FS

Solution 4.

Two conditions when the work done is zero are:

  1. When there is no displacement (S = 0) and,
  2. When the displacement is normal to the direction of the force (θ = 90o).

Solution 5.

(i) If the displacement of the body is in the direction of force, then work done is positive.
Hence, W= F x S
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.

(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative.
Hence, W =- F x S
For example: When a body moves on a surface, the frictional forces between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

Solution 6.

Work is done against the force.

Solution 7.

When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.

Solution 8.

Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.

Solution 9.

Work is done only in case of a boy climbing up a stair case.

Solution 10.

When a coolie carrying some load on his head moves, no work is done by him against the force of gravity because the displacement of load being horizontal, is normal to the direction of force of gravity.

Solution 11.

Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.

Solution 12.

When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.

Solution 13.

S.I unit of work is Joule.
C.G.S unit of work is erg.
Relation between joule and erg :
1joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm = 102 cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 14.

S. I unit of work is Joule.
1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.
Previous

Solution 15.

Relation between joule and erg :
1 joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm= 102 cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 16.

Let a body of mass m fall down through a vertical height h either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is F = mg acting vertically downwards and the displacement in the direction of force (i.e., vertical) is S=h. Therefore the work done by the force of gravity is
W = FS = mgh

Solution 17.

Let a boy of mass m climb up through a vertical height h either through staircase of using a lift. The force of gravity on the boy is F=mg acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is S=-h. Therefore the work done by the force of gravity on the boy is
W= FS =-mgh
or,the work W=mgh is done by the boy against the force of gravity.

Solution 18.

The energy of a body is its capacity to do work. Its S.I unit is Joule (J).

Solution 19.

eV measures the energy of atomic particles.
1eV= 1.6 x 10-19J

Solution 20.

1 J = 0.24 calorie

Solution 21.

Calorie measures heat energy.
1calorie = 4.18 J

Solution 22.

1kWh is the energy spent (or work done) by a source of power 1kW in 1 h.
1kWh = 3.6 x 106J

Solution 23.

The rate of doing work is called power. The S.I. unit of power is watt (W).

Solution 24.

Power spent by a source depends on two factors:
(i) The amount of work done by the source, and
(ii) The time taken by the source to do the said work.
Example: If a coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is the same, but the power spent by the coolie A is twice the power spent by the coolie B because the coolie A does work at a faster rate.

Solution 25.

WorkPower
1. Work done by a force is equal to the product of force and the displacement in the direction of force.1. Power of a source is the rate of doing work by it.
2. Work done does not depend on time.2. Power spent depends on the time in which work is done.
3. S.I unit of work is joule (J).3. S.I unit of power is watt (W).

Solution 26.

EnergyPower
1. Energy of a body is its capacity to do work.1. Power of a source is the energy spent by it in 1s.
2. Energy spent does not depend on time.2. Power spent depends on the time in which energy is spent.
3. S.I unit of energy is joule (J).3. S.I unit of power is watt (W).

Solution 27.

S.I unit of power is watt (W).
If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.

Solution 28.

Horse power is another unit of power, largely used in mechanical engineering. It is related to the S.I unit watt as :
1 H.P =746 W

Solution 29.

Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power x time = work.

Solution 30.

a. Energy is measured in kWh
b. Power is measure in kW
c. Energy is measured in Wh
d. Energy is meaused in eV
Concept insight: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt).
The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).

Solution 1 (MCQ).

746 W

Solution 2 (MCQ).

The unit kWh is the unit of energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 1

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 2

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 3

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 4

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 5

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 6

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 7

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 8

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 9

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 10

Solution 12.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 11

Solution 13.

(i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 12
(ii) Power developed by the person A and B is calculated as follows:
A takes 20 s to climb the stairs while B takes 15 s, to do the same. Hence B does work at a much faster rate than A; more power is spent by B.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 13

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 14

Exercise 2(B)

Solution 1.

Two forms of mechanical energy are:

  1. Kinetic energy
  2. Potential energy

Solution 2.

Elastic potential energy is possessed by wound up watch spring.

Solution 3.

(a) Kinetic energy (K)
(b) Potential energy (U)
(c) Kinetic energy (K)
(d) Potential energy (U)
(e) Kinetic energy (K)
(f) Potential energy (U)

Solution 4.

Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy.
Different forms of P.E. are as listed below:

  1. Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy.
    Example: A stone at a height has gravitational potential energy due to its raised height.
  2. Elastic potential energy: The potential energy possessed by a body in the deformed state due to change in its configuration is called its elastic potential energy.
    Example: A compressed spring has elastic potential energy due to its compressed state.

Solution 5.

Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.

Solution 6.

Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth.
For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity.
Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
= mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 7.

The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 8.

A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

Solution 9.

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 15

Solution 10.

According to the work-energy theorem, the work done by a force on a moving body is equal to the increase in its kinetic energy.

Solution 11.

Body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v and produces an acceleration a in moving a distance S.Then,
Work done by the force= force x displacement
W = F x S———(i)
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 16

Solution 12.

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 17
Both the masses have same momentum p. The kinetic energy, K is inversely proportional to mass of the body.
Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.

Solution 13.

Kinetic energy is related to momentum and mass as
p = √2mK
As the kinetic energy of both bodies are same, momentum is directly proportional to square root of mass.
Now, mass of body B is greater than that of body A.
Hence, body B will have more momentum than body A.

Solution 15.
The three forms of kinetic energy are:

  1. Translational kinetic energy- example: a freely falling body
  2. Rotational kinetic energy-example: A spinning top.
  3. Vibrational kinetic energy-example: atoms in a solid vibrating about their mean position.

Solution 16.

Potential energy (U)

Kinetic energy (K)
    1. The energy possessed by a body by virtue of its specific position or changed configuration is called potential energy.

    1.The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

     2.Two forms of potential energy are gravitational potential energy and elastic potential energy.

2. Forms of kinetic energy are translational, rotational and vibrational kinetic energy.
     3.Example: A wound up watch spring has potential energy.

3. For example: a moving car has kinetic energy.

Solution 17.

(a) Motion.
(b) Position.

Solution 18.

When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move.

Solution 19.

The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 18

Solution 20.

When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.

Solution 21.

Yes, when force is normal to displacement, no transfer of energy takes place.

Solution 22.

Kinetic energy.

Solution 23.

The six different forms of energy are:

  1. Solar energy
  2. Heat energy
  3. Light energy
  4. Chemical or fuel energy
  5. Hydro energy
  6. Nuclear energy

Solution 24.

(a) Potential energy of wound up spring converts into kinetic energy.
(b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy)
(c) Kinetic energy to potential energy
(d) Light energy changes into chemical energy
(e) Electrical energy changes into chemical energy
(f) Chemical energy changes into heat energy
(g) Chemical energy changes into heat and light energy
(h) Chemical energy changes into heat, light and sound energy

Solution 25.

(a) Electrical energy into sound energy
(b) Heat energy into mechanical energy
(c) Sound energy into electrical energy
(d) Electrical energy to mechanical energy
(e) Electrical energy into light energy
(f) Chemical energy to heat energy
(g) Light energy into electrical energy
(h) Chemical energy into heat energy
(i) Chemical energy into electrical energy
(j) Chemical energy to mechanical energy
(k) Electrical energy into heat energy
(l) Light energy into electrical energy
(m) Electrical energy into magnetic energy.

Solution 27.

No. This is because, whenever there is conversion of energy from one form to another apart of the energy is dissipated in the form of heat which is lost to surroundings.

Solution 1 (MCQ).

Potential energy
Hint: P.E. is the energy possessed by a body by virtue of its position.

Solution 2 (MCQ).

Chemical to electrical
Hint: When current is drawn from an electric cell, the chemical energy stored in it changes into electrical energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 19

Solution 2.

Mass , m=1kg
Height, h=5m
Gravitational potential energy = mgh
=1 x 10 x 5=50J

Solution 3.

Gravitational potential energy=14700 J
Force of gravity = mg= 150 x 9.8N/kg= 1470N
Gravitational potential energy= mgh
14700 =1470 x h
h=10m

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 20

Solution 5.

Mass =0.5 kg
Energy= 1 J
Gravitational potential energy= mgh
1=0.5 x10 x h
1=5h
Height, h= 0.2 m

Solution 6.

Force of gravity on boy=mg= 25 x 10 =250N
Increase in gravitational potential energy= Mg (h2-h1)
= 250 x (9-3)
=250 x6=1500 J

Solution 7.

Mass of water, m= 50kg
Height, h=15m
Gravitational potential energy= mgh
=50 x10 x 15
=7500 J

Solution 8.

Mass of man=50kg
Height of ladder, h2=10m

(i) Work done by man =mgh2
=50 x 9.8 x10= 4900J

(ii) increase in his potential energy:
Height, h2= 10m
Reference point is ground, h1=0m
Gravitational potential energy= Mg (h2-h1)
= 50 x9.8x (10-0)
= 50 x 9.8 x10= 4900J

Solution 9.

F=150N

(a) Work done by the force in moving the block 5m along the slope =Force x displacement in the direction of force
=150 x 5=750 J.

(b) The potential energy gained by the block U =mgh where h =3m
=200 x 3=600 J
he potential energy gained by the block

(c) The difference i.e., 150 J energy is used in doing work against friction between the block and the slope, which will appear as heat energy.

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 21

Solution 11.

If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity).

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 22

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 23

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 24

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 25

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 26

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 27

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 28

Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 29

Solution 21.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 30

Solution 22.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 31

Solution 23.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 32

Exercise 2(C)

Solution 1.

According to the law of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another.

Solution 2.

According to the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant i.e., K + U = constant when there are no frictional forces.

Mechanical energy is conserved only when there are no frictional forces for a given system (i.e. between body and air). Thus, conservation of mechanical energy is strictly valid only in vacuum, where friction due to air is absent.

Solution 3.

Motion of a simple pendulum and motion of a freely falling body.

Solution 4.

Kinetic energy of the body changes to potential energy when it is thrown vertically upwards and its velocity becomes zero.

Solution 5.

(a) Potential energy
(b) Potential energy and kinetic energy
(c) Kinetic energy

Solution 6.

Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance x) and at C (on the ground).
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 33

(i) At the position A (at height h above the ground):
Initial velocity of body= 0 (since body is at rest at A)
Hence, kinetic energy K =0
Potential energy U = mgh
Hence total energy = K + U= 0 + mgh =mgh—–(i)

(ii) At the position B (when it has fallen a distance x):
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 34
Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Solution 7.

When the bob swings from A to B, the kinetic energy decreases and the potential energy becomes maximum at B where it is momentarily at rest.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 35
From B to A, the potential energy again changes into the kinetic energy and the process gets repeated again and again.
Thus while swinging, the bob has only the potential energy at the extreme position B or C and only the kinetic energy at the resting position A. At an intermediate position (between A and B or between A and C), the bob has both the kinetic energy and potential energy, and the sum of both the energies (i.e., the total mechanical energy) remains constant throughout the swing.

Solution 8.

(a) At position A, pendulum has maximum kinetic energy and its potential energy is zero at its resting position. Hence, K=mgh and U= 0.
(b) At B, kinetic energy decreases and potential energy increases. Hence, K= 0 and U=mgh
(c) At C also, kinetic energy K= 0 and potential energy U=mgh.

Solution 9.

a) Extreme position: Potential energy
b) Mean position: Kinetic energy
c) Between mean and extreme: Both kinetic energy and potential energy

Solution 10.

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

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

Solution 1 (MCQ).

Potential energy of the ball at the highest point is mgh.
Hint: At the highest point, the ball momentarily comes to rest and thus its kinetic energy becomes zero.

Solution 2 (MCQ).

The sum of its kinetic and potential energy remains constant throughout the motion.
Hint: In accordance with law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 36

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 37

Solution 3.

(a)Potential energy of the ball =mgh
=2 x 10 x 5=100J
(b)Kinetic energy of the ball just before hitting the ground = Initial potential energy= mgh=2x10x5=100J
(c)Mechanical energy converts into heat and sound energy.

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 38

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 39

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 40

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Selina Concise Physics Class 10 ICSE Solutions Machines

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Selina Concise Physics Class 10 ICSE Solutions Machines

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 3 Machines. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Physics Chapter 3 Machines

Exercise 3(A)

Solution 1.

(a) A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.

(b) An ideal machine is a machine whose parts are weightless and frictionless so that which there is no dissipation of energy in any manner. Its efficiency is 100%, i.e. the work output is equal to work input.

Solution 2.

Machines are useful to us in the following ways:

  1. In lifting a heavy load by applying a less effort.
  2. In changing the point of application of effort to a convenient point.
  3. In changing the direction of effort to a convenient direction.
  4. For obtaining a gain in speed.

Solution 3.

(a) To multiply force: a jack is used to lift a car.
(b) To change the point of application of force: the wheel of a cycle is rotated with the help of a chain by applying the force on the pedal.
(c) To change the direction of force: a single fixed pulley is used to lift a bucket full of water from the well by applying the effort in the downward direction instead of applying it upwards when the bucket is lifted up without the use of pulley.
(d) To obtain gain in speed: when a pair of scissors is used to cut the cloth, its blades move longer on cloth while its handles move a little.

Solution 4.

The purpose of jack is to make the effort less than the load so that it works as a force multiplier.

Solution 6.

The ratio of the load to the effort is called mechanical advantage of the machine. It has no unit.

Solution 7.

The ratio of the velocity of effort to the velocity of the load is called the velocity ratio of machine. It has no unit.

Solution 8.

For an ideal machine mechanical advantage is numerically equal to the velocity ratio.

Solution 9.

It is the ratio of the useful work done by the machine to the work put into the machine by the effort.
In actual machine there is always some loss of energy due to friction and weight of moving parts, thus the output energy is always less than the input energy.

Solution 10.

(a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage of such machines is greater than 1.
(b) A machine acts a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage of such machines is less than 1.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Machines img 1

Solution 12.

Let a machine overcome a load L by the application of an effort E. In time t, let the displacement of effort be dE and the displacement of load be dL.
Work input = Effort X displacement of effort = E X dE
Work output = Load X displacement of load = L X dL
Selina Concise Physics Class 10 ICSE Solutions Machines img 2

Solution 13.

Selina Concise Physics Class 10 ICSE Solutions Machines img 3
The efficiency of such a machine is always less than 1, i.e. h<1. This is because there is always some loss in energy in form of friction etc.

Solution 14.

This is because the output work is always less than the input work, so the efficiency is always less than 1 because of energy loss due to friction.
Selina Concise Physics Class 10 ICSE Solutions Machines img 47

Solution 15.

A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.
Principle: A lever works on the principle of moments. For an ideal lever, it is assumed that the lever is weightless and frictionless. In the equilibrium position of the lever, by the principle of moments,
Moment of load about the fulcrum = Moment of the effort about the fulcrum.

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Machines img 5

Solution 17.

The three classes of levers are:

  1. Class I levers: In these types of levers, the fulcrum F is in between the effort E and the load L. Example: a seesaw, a pair of scissors, crowbar.
  2. Class II levers: In these types of levers, the load L is in between the effort E and the fulcrum F. The effort arm is thus always longer than the load arm. Example: a nut cracker, a bottle opener.
  3. Class III levers: In these types of levers, the effort E is in between the fulcrum F and the load L and the effort arm is always smaller than the load arm. Example: sugar tongs, forearm used for lifting a load.

Solution 18.

(a) More than one: shears used for cutting the thin metal sheets.
(b) Less than one: a pair of scissors whose blades are longer than its handles.

Solution 19.

When the mechanical advantage is less than 1, the levers are used to obtain gain in speed. This implies that the displacement of load is more as compared to the displacement of effort.

Solution 20.

A pair of scissors and a pair of pliers both belong to class I lever.
A pair of scissors has mechanical advantage less than 1.

Solution 21.

A pair of scissors used to cut a piece of cloth has blades longer than the handles so that the blades move longer on the cloth than the movement at the handles.
While shears used for cutting metals have short blades and long handles because as it enables us to overcome large resistive force by a small effort.

Solution 22.

(a) The weight W of the scale is greater than E.
It is because arm on the side of effort E is 30 cm and on the side of weight of scale is 10 cm. So, to balance the scale, weight W of scale should be more than effort E.
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Solution 23.

Class II lever always have a mechanical advantage more than one.
Example: a nut cracker.
To increase its mechanical advantage we can increase the length of effort arm.

Solution 24.
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Solution 25.

In these types of levers, the load L is in between the effort E and the fulcrum F. So, the effort arm is thus always longer than the load arm. Therefore M.A > 1.

Solution 26.
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Solution 27.
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Solution 29.
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Solution 30.

In these types of levers, the effort is in between the fulcrum F and the load L and so the effort arm is always smaller than the load arm. Therefore M.A. < 1.

Solution 31.

With levers of class III, we do not get gain in force, but we get gain in speed, that is a longer displacement of load is obtained by a smaller displacement of effort.

Solution 32.
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Solution 33.

(a) A bottle opener is a lever of the second order, as the load is in the middle, fulcrum at one end and effort at the other.
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(b) Sugar tongs is a lever of the third order as the effort is in the middle, load at one end and fulcrum at the other end.
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Solution 34.
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Solution 35.

a. Class II
b. Class I
c. Class II
d. Class III

Solution 36.

(a) Class III.
Here, the fulcrum is the elbow of the human arm. Biceps exert the effort in the middle and load on the palm is at the other end.
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(b) Class II.
Here, the fulcrum is at toes at one end, the load (i.e. weight of the body) is in the middle and effort by muscles is at the other end.
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Solution 37.
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Solution 38.

Class I lever in the action of nodding of the head: In this action, the spine acts as the fulcrum, load is at its front part, while effort is at its rear part.

Class II lever in raising the weight of the body on toes: The fulcrum is at toes at one end, the load is in the middle and effort by muscles is at the other end.

Class III lever in raising a load by forearm: The elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load on the palm is at the other end.

Solution 1 (MCQ).

M.A. x E = L

Solution 2 (MCQ).

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Solution 3 (MCQ).

It can have a mechanical advantage greater than the velocity ratio.
Reason: If the mechanical advantage of a machine is greater than its velocity ratio, then it would mean that the efficiency of a machine is more than 100%, which is practically not possible.

Solution 4 (MCQ).

Effort is between fulcrum and load
Hint: Levers, for which the mechanical advantage is less than 1, always have the effort arm shorter than the load arm.

Solution 5 (MCQ).

M.A > 1
Hint: In class II levers, the load is in between the effort and fulcrum. Thus, the effort arm is always longer than the load arm and less effort is needed to overcome a large load. Hence, M.A > 1.

Numericals

Solution 1.
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Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Machines img 19

Solution 3.
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Solution 4.
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Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Machines img 22

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Machines img 23

Solution 7.

Total length of rod=4 m = 400 cm

(a) 18kgf load is placed at 60 cm from the support.
W kgf weight is placed at 250 cm from the support.
By the principle of moments
18 x 60 = W x 250
W = 4.32 kgf

(b) Given W=5 kgf
18kgf load is placed at 60 cm from the support.
Let 5 kgf of weight is placed at d cm from the support.
By the principle of moments
18 x 60 = 5 x d
d = 216 cm from the support on the longer arm

(c) It belongs to class I lever.

Solution 9.
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Solution 10.
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Solution 11.

(a) The principle of moments: Moment of the load about the fulcrum=moment of the effort about the fulcrum
FB x Load = FA x Effort
(b) Sugar tongs the example of this class of lever.
(c) Given: FA=10 cm, AB = 500 cm, BF =500+10=510 cm.
The mechanical advantage
Selina Concise Physics Class 10 ICSE Solutions Machines img 26

Solution 12.
Selina Concise Physics Class 10 ICSE Solutions Machines img 27

Exercise 3(B)

Solution 1.

Fixed pulley: A pulley which has its axis of rotation fixed in position, is called a fixed pulley.
Single fixed pulley is used in lifting a small load like water bucket from the well.

Solution 2.

The ideal mechanical advantage of a single fixed pulley is 1.
It cannot be used as force multiplier.

Solution 3.

There is no gain in mechanical advantage in the case of a single fixed pulley. A single fixed pulley is used only to change the direction of the force applied that is with its use, the effort can be applied in a more convenient direction. To raise a load directly upwards is difficult.

Solution 4.

The velocity ratio of a single fixed pulley is 1.

Solution 5.

The load rises upwards with the same distance x.

Solution 6.

Single movable pulley: A pulley, whose axis of rotation is not fixed in position, is called a single movable pulley.
Mechanical advantage in the ideal case is 2.

Solution 8.

The efficiency of a single movable pulley system is not 100% this is because

  1. The friction of the pulley bearing is not zero ,
  2. The weight of the pulley and string is not zero.

Solution 9.

The force should be in upward direction.
The direction of force applied can be changed without altering its mechanical advantage by using a single movable pulley along with a single fixed pulley to change the direction of applied force.
Diagram:
Selina Concise Physics Class 10 ICSE Solutions Machines img 28

Solution 10.

The velocity ratio of a single movable pulley is always 2.

Solution 11.

The load is raised to a height of x/2.

Solution 12.

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Ideal mechanical advantage of this system is 2. This can be achieved by assuming that string and the pulley are massless and there is no friction in the pulley bearings or at the axle or between the string and surface of the rim of the pulley.

Solution 13.

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(b) The fixed pulley B is used to change the direction of effort to be applied from upward to downward.
(c) The effort E balances the tension T at the free end, so E=T
(d) The velocity ratio of this arrangement is 2.
(e) The mechanical advantage is 2 for this system (if efficiency is 100%).

Solution 14.

Single fixed pulleySingle movable pulley
1. It is fixed to a rigid support.1. It is not fixed to a rigid support.
2. Its mechanical advantage is one.2. Its mechanical advantage istwo.
3. Its velocity ratio is one.3. Its velocity ratio is two.
     4. The weight of pulley itself does not affect its mechanical advantage.4. The weight of pulley itself reduces its mechanical advantage.
     5. It is used to change the direction of effort5. It is used as force multiplier.

Solution 15.
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Solution 16.

Selina Concise Physics Class 10 ICSE Solutions Machines img 32
Mechanical advantage = MA = L/E = 23
As one end of each string passing over a movable pulley is fixed, so the free end of string moves twice the distance moved by the movable pulley.
If load L moves up by a distance x, dL = x, effort moves by a distance 23x, dE = 23xSelina Concise Physics Class 10 ICSE Solutions Machines img 33

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Machines img 34

Solution 18.

  1. In a single fixed pulley, some effort is wasted in overcoming friction between the strings and the grooves of the pulley; so the effort needed is greater than the load and hence the mechanical advantage is less than the velocity ratio.
  2. This is because of some effort is wasted in overcoming the friction between the strings and the grooves of the pulley.
  3. This is because mechanical advantage is equal to the total number of pulleys in both the blocks.
  4. The efficiency depends upon the mass of lower block; therefore efficiency is reduced due to the weight of the lower block of pulleys.

Solution 19.

(a) Multiply force: a movable pulley.
(b) Multiply speed: gear system or class III lever.
(c) Change the direction of force applied: single fixed pulley.

Solution 20.

  1. The velocity ratio of a single fixed pulley is always more than 1.(false)
  2. The velocity ratio of a single movable pulley is always 2.(true)
  3. The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2n.(true)
  4. The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. (true)

Solution 1 (MCQ).

It helps in applying effort in a convenient direction.
Explanation: A single fixed pulley though does not reduce the effort but helps in changing the direction of effort applied. As it is far easier to apply effort in downward direction, the single fixed pulley is widely used.

Solution 2 (MCQ).

The mechanical advantage of an ideal single movable pulley is 2.
Derivation: Consider the diagram given below:
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Here the load L is balance by the tension in two segments of the string and the effort E balances the tension T at the free end, so
L = T + T = 2T and E = T
Assumption: Weight of the pulley is negligible.
We know that,
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Solution 3 (MCQ).

Force multiplier
Explanation: The mechanical advantage of movable pulley is greater than 1. Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e. the single movable pulley acts as a force multiplier.

Numericals

Solution 1.
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Solution 2.
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Solution 3.
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Solution 4.
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Solution 5.
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Solution 6.
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Solution 7.
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(d) (i) There is no friction in the pulley bearings, (ii) weight of lower pulleys is negligible and (iii) the effort is applied downwards.

Solution 8.
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Solution 9.
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Solution 10.
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