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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.
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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.

Electromagnet Permanent magnet
It is made up of soft iron It is made up of steel.
The magnetic field strength can be changed. The magnetic field strength cannot be changed.
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.
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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.

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(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. Generator D.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.
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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,
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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.
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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.
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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.
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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 transformer Step 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.
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Solution 3.
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Solution 4.
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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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ICSE SolutionsSelina ICSE Solutions

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.
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Solution 6.
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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.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 7
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.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 8
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.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 13

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 name Colour (Old convention) Colour (New convention)
1 Neutral wire Black Light blue
2 Earth wire Green Green or yellow
3 Live wire Red Brown

(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 Current Electricity

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Selina Concise Physics Class 10 ICSE Solutions Current Electricity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 8 Current Electricity. 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 8 Current Electricity

Exercise 8(A)

Solution 1.

Current is defined as the rate of flow of charge.
I=Q/t
Its S.I. unit is Ampere.

Solution 2.

Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.

Solution 3.

The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other.
It’s S.I. unit is Volt.

Solution 4.

One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.

Solution 7.

In a metal, the charges responsible for the flow of current are the free electrons. The direction of flow of current is conventionally taken opposite to the direction of motion of electrons.

Solution 8.

It states that electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same. This is called Ohm’s law.
V = IR
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 1

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 2

Solution 12.

Ohmic Resistor: An ohmic resistor is a resistor that obeys Ohm’s law. For example: all metallic conductors (such as silver, aluminium, copper, iron etc.)
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 3

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 4

Solution 14.

  1. Ohmic resistor obeys ohm’s law i.e., V/I is constant for all values of V or I; whereas Non-ohmic resistor does not obey ohm’s law i.e., V/I is not same for all values of V or I.
  2. In Ohmic resistor, V-I graph is linear in nature whereas in non-ohmic resistor, V-I graph is non-linear in nature.

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

In the above graph, T1 > T2. The straight line A is steeper than the line B, which leads us to conclude that the resistance of conductor is more at high temperature T1 than at low temperature T2. Thus, we can say that resistance of a conductor increases with the increase in temperature.

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 6

Solution 18.

Resistance of a wire is directly proportional to the length of the wire.
R ∝ l
The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

Solution 19.

With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature.
The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).

Solution 20.

Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.

Solution 21.

  1. Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.
    R ∝ l
  2. Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.
    R ∝ 1/A
  3. Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.
  4. Resistance depends on the nature of conductor because different substances have different concentration of free electrons.Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.

Solution 22.

The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section.
Its S.I. unit is ohm metre.

Solution 23.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 7

Solution 24.

Metal < Semiconductor < Insulator

Solution 26.

Manganin

Solution 28.

‘Copper or Aluminium’ is used as a material for making connection wires because the resistivity of these materials is very small, and thus, wires made of these materials possess negligible resistance.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 8

Solution 30.

Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.

Solution 31.

Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.

Solution 32.

A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity.

A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.

Solution 33.

A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.

Solution 34.

Superconductor

Solution 1 (MCQ).

Nichrome is an ohmic resistance.
Hint: Substances that obey Ohm’s law are called Ohmic resistors.

Solution 2 (MCQ).

For carbon, resistance decreases with increase in temperature.
Hint: For semiconductors such as carbon and silicon, the resistance and resistivity decreases with the increase in temperature.

Numericals

Solution 1.
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Solution 9.
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Solution 11.
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Solution 12.
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Exercise 8(B)

Solution 1.

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).
Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.
Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.

Solution 2.

e.m.f. of cell Terminal voltage of cell
1. It is measured by the amount of  work done in moving a unit positive charge in the complete circuit inside and outside the cell. 1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell.
2. It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell 2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage.
3. It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use. 3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

 

Solution 3.

Internal resistance of a cell depends upon the following factors:

  1. The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.
  2. The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

Solution 4.

Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 13

Solution 5.

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.
(b) e.m.f. is equal to the terminal voltage when no current is drawn.

Solution 6.

When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Solution 7.
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Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 15

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 16

Solution 10.

(a) series
(b) parallel
(c) parallel
(d) series

Solution 11.

For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.

Solution 1 (MCQ).

In series combination of resistances, current is same in each resistance.
Hint: In a series combination, the current has a single path for its flow. Hence, the same current passes through each resistor.

Solution 2 (MCQ).

In parallel combination of resistances, P.D. is same across each resistance.
Hint: In parallel combination, the ends of each resistor are connected to the ends of the same source of potential. Thus, the potential difference across each resistance is same and is equal to the potential difference across the terminals of the source (or battery).

Solution 3 (MCQ).
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 17

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 18

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 19

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 20

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 21

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 22

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

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 24

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 25

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 26

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 27

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 28

Solution 12.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 29

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 30

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 31

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 32

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 33

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

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 35

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 36

Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 37

Solution 21.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 38

Solution 23.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 39

Solution 24.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 40

Solution 25.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 41

Solution 26.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 42

Solution 27.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 43

Solution 28.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 44

Solution 29.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 45

Solution 30.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 46

Exercise 8(C)

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 47

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 48

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 49

Solution 4.

The S.I. unit of electrical energy is joule.
1Wh = 3600 J

Solution 5.

The power of an appliance is 100 W. It means that 100 J of electrical energy is consumed by the appliance in 1 second.

Solution 6.

The S.I. unit of electrical power is Watt.

Solution 7.

(i) The household unit of electricity is kilowatt-hour (kWh).
One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
(ii) The voltage of the electricity that is generally supplied to a house is 220 Volt.

Solution 8.

(i) Electrical power is measured in kW and
(ii) Electrical energy is measured in kWh.

Solution 9.

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.
Its value in SI unit is 1kWh = 3.6 x 106J

Solution 10.

Kilowatt is the unit of electrical power whereas kilowatt-hour is the unit of electrical energy.

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 50
Solution 12.

An electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V) which is known as its power rating. For example: If an electric bulb is rated as 50W-220V, it means that when the bulb is lighted on a 220 V supply, it consumes 50 W electrical power.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 51

Solution 13.

It means that if the bulb is lighted on a 250 V supply, it consumes 100 W electrical power (which means 100J of electrical energy is converted in the filament of bulb into the light and heat energy in 1 second).

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 52

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 53

Solution 16.

When current is passed in a wire, the heat produced in it depends on the three factors:

  1. on the amount of current passing through the wire,
  2. on the resistance of wire and
  3. on the time for which current is passed in the wire.
  • Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire,  i.e., H ∝ I2
  • Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R
  • Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire, i.e., H ∝ t

 

Solution 1 (MCQ).
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 77

Solution 2 (MCQ).
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 55

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 56

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 57

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 58

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 59

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 60

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 61

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 62

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 63

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 64

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 65
When one lamp is connected across the mains, it draws 0.25 A current, while if two lamps are connected in series across the mains, current through each bulb becomes
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 66
(i.e., current is halved), hence heating (= I2Rt) in each bulb becomes one-fourth, so each bulb appears less bright.

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 67

Solution 12.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 68

Solution 13.
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Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 70

Solution 15.
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Solution 16.
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Solution 17.
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Solution 18.
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Solution 19.
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Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Current Electricity img 76

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Selina Concise Physics Class 10 ICSE Solutions Sound

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Selina Concise Physics Class 10 ICSE Solutions Sound

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 7 Sound. 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 7 Sound

Exercise 7(A)

Solution 2.

(a) Amplitude: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its S.I. unit is metre (m).
(b) Frequency: The number of vibrations made by a particle of the medium in one second is called the frequency of the waves.
It is also defined as the number of waves passing through a point in one second. Its S.I. unit is hertz (Hz).
(c) Wavelength: The distance travelled by the wave in one time period of vibration of particle of medium is called its wavelength. Its S.I. unit is metre (m).
(d) Wave velocity: The distance travelled by a wave in one second is called its wave velocity. Its S.I. unit is metre per second (ms-1).

Solution 3.

(i) Wavelength (or speed) of the wave changes, when it passes from one medium to another medium.
(ii) Frequency of a wave does not change when it passes from one medium to another medium.

Solution 4.

Two factors on which the speed of a wave travelling in a medium depends are:

  1. Density: The speed of sound is inversely proportional to the square root of density of the gas.
  2. Temperature: The speed of sound increases with the increase in temperature.

Solution 5.

The light waves can travel in vacuum while sound waves need a material medium for propagation.
The light waves are electromagnetic waves while sound waves are the mechanical waves.

Solution 6.

If a person stands at some distance from a wall or a hillside and produces a sharp sound, he hears two distinct sounds: one is original sound heard almost instantaneously and the other one is heard after reflection from the wall or hillside, which is called echo.

The condition for the echo: An echo is heard only if the distance between the person producing the sound and the rigid obstacle is long enough to allow the reflected sound to reach the person at least 0.1 second after the original sound is heard.

Solution 7.

t = 2d/V = 2 x 12/340 = 24/340 < 0.1 seconds so the man will not be able to hear the echo. This is because the sensation of sound persists in our ears for about 0.1 second after the exciting stimulus ceases to act.

Solution 8.

The applications of echo:

  1. Dolphins detect their enemy and obstacles by emitting the ultrasonic waves and hearing their echo.
  2. In medical science, the echo method of ultrasonic waves is used for imaging the human organs such as the liver, gall bladder, uterus, womb etc. This is called ultrasonography.

Solution 9.

Sound is produced from a place at a known distance say, d at least 50 m from the reflecting surface. The time interval t in which the echo reaches the place from where the sound was produced, is noted by a stop watch having the least count 0.01 s. then the speed of sound is calculated by using the following relation
Selina Concise Physics Class 10 ICSE Solutions Sound img 1

Solution 10.

Bats, dolphin and fisherman detect their enemies or obstacles or position of fish by emitting/sending the ultrasonic waves and hearing/detecting the echo.

Solution 11.

Bats can produce and detect the sound of very high frequency up to about 1000kHz. The sounds produced by flying bats get reflected back from any obstacle in front of it. By hearing the echoes, bats come to know even in the dark where the obstacles are. So they can fly safely without colliding with the obstacles.

Solution 12.

The process of detecting obstacles with the help of echo is called sound ranging. It’s used by the animals like bats, dolphin to detect their enemies.

Solution 13.

The ultrasonic waves are used for the sound ranging. Ultrasonic waves have a frequency more than 20,000 Hz but the range of audibility of human ear is 20Hz to 20,000 Hz

Solution 14.

Sonar is sound navigation and ranging. Ultrasonic waves are sent in all directions from the ship and they are received on their return after reflection from the obstacles. They use the method of echo.

Solution 15.

In medical science, echo method of ultrasonic waves is used for the imaging of human organs such as liver, gall bladder, uterus, womb; which is called ultrasonography.

Solution 1 (MCQ).

17 m.

Explanation: An echo is heard distinctly if it reaches the ear at least 0.1 s after the original sound.
If d is the distance between the observer and the obstacle and V is the speed of sound, then the total distance travelled by the sound to reach the obstacle and then to come back is 2d and the time taken is,
t = Total distance travelled/Speed of sound = 2d/V
or, d = V t/2
Putting t = 0.1 s and V = 340 m/s in air at ordinary temperature, we get:
d = (340 x 0.1)/2 = 17 m
Thus, to hear an echo distinctly, the minimum distance between the source and the reflector in air is 17 m.

Solution 2 (MCQ).

Ultrasonic waves

Numericals

Solution 1.

(i) Frequency or the number of waves produced per second
= Velocity/Wavelength
= 24 / 20 x 10-2
= 120
(ii) Time = 1/ frequency = 1/ 120= 8.3 x 10-3 seconds

Solution 2.

Velocity = 2D/Time
350 = 2 x D/ 0.1
D = 350 x 0.1 / 2 = 17.5 m

Solution 3.

Velocity = 2D/Time
1400 = 2 x D/ 0.1
D = 1400 x 0.1/ 2 = 70 m

Solution 4.

(a) Velocity = 2D/Time
Time = 2 x 25 / 350 = 0.143 seconds
(b) Yes, because the reflected sound reaches the man 0.1 second after the original sound is heard and the original sound persists only for 0.1 second.

Solution 5.

Velocity = 2D/Time
3 x 108 = 2 x 45 x 1000/ Time
Time = 90000/ 3 x 108 = 3 x 10-4 second

Solution 6.

Velocity = 2 x D/Time
Time after which an echo is heard = 2 D/Velocity = 2 x 48 / 320 = 0.3 seconds

Solution 7.

2 D = velocity x time
D = (velocity x time) / 2 = 1450 x 4 / 2 = 2900 m = 2.9 km

Solution 8.

5 vibrations by pendulum in 1 sec
So 8 vibrations in 8/5 seconds = 1.6 sec
Velocity = 2 x D/ time
340 = 2 x D/ 1.6
D = 340 x 1.6 / 2 = 272 m

Solution 9.

The distance of first cliff from the person, 2 x D1 = velocity x time
D1 = 320 x 4 / 2 = 640 m
Distance of the second cliff from the person, D2 = 320 x 6 / 2 = 960 m
Distance between cliffs = D1 + D2 = 640 + 960 = 1600 m

Solution 10.

Distance of hill from the man
D1 = velocity x time/ 2 =v x 5 / 2—————– (eqn 1)
Now, D1 – 310 = v x 3 / 2————————— (eqn 2)
By subtracting eqn 2 form eqn 1 ,we get310 = v x (5/2-3/2)
So, v = 310m/s

Solution 11.

Depth of the sea = velocity x time/2 = 1400 x 1.5 / 2 = 1050 m

Exercise 7(B)

Solution 1.

The vibrations of a body in the absence of any external force on it are called the free vibrations. Eg.: When we strike the keys of a piano, various strings are set into vibration at their natural frequencies.

Solution 2.

When each body capable of vibrating is set to vibrate freely and it vibrates with a frequency f. It is the natural frequency of vibration of the body.
The natural frequency of vibration of a body depends on the shape and size of the body.

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

Solution 4.

The free vibrations of a body occur only in vacuum because the presence of medium offer some resistance due to which the amplitude of the vibration does not remain constant, but it continuously decreases.

Solution 5.

The frequency of sound emitted due to vibration in an air column depends on the length of the air column.

Solution 6.

The frequency of the note produced in the air column can be increased by decreasing the length of the air column.

Solution 7.

The frequency of vibration of the stretched string can be increased by increasing the tension in the string, by decreasing the length of the string.

Solution 8.

A stringed instrument is provided with the provision for adjusting the tension of the string. By varying the tension, we can get the desired frequency.

Solution 9.

a) (i) Diagram is showing the principal note.
b) (iii)Diagram has frequency four times that of the first.
d) Ratio is 1:2

Solution 10.

Strings of different thickness are provided on a stringed instrument to produce different frequency sound waves because the natural frequency of vibration of a stretched string is inversely proportional to the radius (thickness) of the string.

Solution 11.

The frequency of vibrations of the blade can be lowered by increasing the length of the blade or by sticking a small weight on the blade at its free end.

Solution 12.

The presence of the medium offers some resistance to motion, so the vibrating body continuously loses energy due to which the amplitude of the vibration continuously decreases.

Solution 13.

The periodic vibrations of a body of decreasing amplitude in the presence of resistive force are called the damped vibrations.
The amplitude of the free vibrations remains constant and vibrations continue forever. But, the amplitude of damped vibrations decreases with time and ultimately the vibrations ceases.
For eg, When a slim branch of a tree is pulled and then released, it makes damped vibrations.
A tuning fork vibrating in air excute damped vibrations.

Solution 14.

  1. Damped vibrations
  2. Example: When a slim branch of a tree is pulled and then released, it makes damped vibrations.
  3. The amplitude of vibrations gradually decreases due to the frictional (or resistive) force which the surrounding medium exerts on the body vibrating in it. As a result, the vibrating body continuously loses energy in doing work against the force of friction causing a decrease in its amplitude.
  4. After sometime, the vibrating body loses all of its energy and stops vibrating.

Solution 15.

The tuning fork vibrates with the damped oscillations.

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Sound img 3

Solution 17.

The vibrations of a body which take place under the influence of an external periodic force acting on it, are called the forced vibrations. For example: when guitar is played, the artist forces the strings of the guitar to execute forced vibrations.

Solution 18.

The vibrations of a body in the absence of any resistive force are called the free vibrations. The vibrations of a body in the presence of an external force are called forced vibrations.

In free vibrations, the frequency of vibration depends on the shape and size of the body. In forced vibrations, the frequency is equal to the frequency of the force applied.

Solution 19.

Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
Mount two identical tuning forks A and B of same frequency upon two separate sound boxes such that their open ends face each other as shown.
Selina Concise Physics Class 10 ICSE Solutions Sound img 4
If the prong A is struck on a rubber pad, it starts vibrating. On putting A on its sound box, tuning fork B also starts vibrating and a loud sound is heard. The vibrations produced in B are due to resonance.

Solution 20.

Condition for resonance:
Resonance occurs when the frequency of the applied force is exactly equal to the natural frequency of the vibrating body.

Solution 21.

forced,equal to the

Solution 22.

Solution 23.

At resonance, the body vibrates with large amplitude thus conveying more energy to the ears so a loud sound is heard.

Solution 24.

a) The vibrating tuning fork A produces the forced vibrations in the air column of its sound box. These vibrations are of large amplitude because of the large surface area of air in the sound box. They are communicated to the sound box of the fork B. The air column of B starts vibrating with the frequency of the fork A. Since the frequency of these vibrations is same as the natural frequency of the fork B, the fork B picks up these vibrations and starts vibrating due to resonance.
b) On putting the tuning fork A to vibrate, the other tuning fork B will also start vibrating. The vibrations produced in the second tuning fork B are due to resonance.

Solution 25.

(a) Set the pendulum A into vibration by displacing it to one side, normal to its length. It is observed that pendulum D also starts vibrating initially with a small amplitude and ultimately it acquires the same amplitude as the pendulum A initially had. When the amplitude of the pendulum D becomes maximum, the amplitude of the pendulum A becomes minimum since the total energy is constant. After some time the amplitude of the pendulum D will decreases and amplitude of A increases. The exchange of energy takes place only between the pendulums A and D because their natural frequencies are same. The pendulums B and C also vibrate, but with very small amplitudes.
(b) The vibrations produced in pendulum A are communicated as forced vibrations to the other pendulums B, C and D through XY. The pendulums B and C remain in the state of forced vibrations, while the pendulum D comes in the state of resonance.

Solution 26.

The phenomenon responsible for producing a loud audible sound is named resonance. The vibrating tuning fork causes the forced vibrations in the air column. For a certain length of air column, a loud sound is heard. This happens when the frequency of the air column becomes equal to the frequency of the tuning fork.

Solution 27.

(a) No loud sound is heard with the tubes A and C, but a loud sound is heard with the tube B.
(b) Resonance occurs with the air column in tube B whereas no resonance occurs in the air column of tubes A and C. The frequency of vibrations of air column in tube B is same as the frequency of vibrations of air column in tube D because the length of the air column in tube D is 20-18 = 2cm and that in tube B is 20-14 = 6 cm (3 times). On the other hand, the frequency of vibrations of air column in tubes A and C is not equal to the frequency vibrations of air column in tube B.
(c) When the frequency of vibrations of air column is equal to the frequency of the vibrating tuning fork, resonance occurs.

Solution 28.

When a troop crosses a suspension bridge, the soldiers are asked to break steps. The reason is that when soldiers march in steps, all the separate periodic forces exerted by them are in same phase and therefore forced vibrations of a particular frequency are produced in the bridge. Now, if the natural frequency of the bridge happens to be equal to the frequency of the steps, the bridge will vibrate with large amplitude due to resonance and suspension bridge could crumble

Solution 29.

The sound box is constructed such that the column of the air inside it, has a natural frequency which is the same as that of the strings stretched on it, so that when the strings are made to vibrate, the air column inside the box is set into forced vibrations. Since the sound box has a large area, it sets a large volume of air into vibration, the frequency of which is same as that of the string. So, due to resonance a loud sound is produced.

Solution 30.

When we tune a radio receiver, we merely adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive. When the two frequencies match, due to resonance the energy of the signal of that particular frequency is received from the incoming waves. The signal received is then amplified in the receiver set.
The phenomenon involved is resonance. It is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body rapidly begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

Solution 1 (MCQ).

It executes free vibrations.
Hint: The periodic vibrations of a body of constant amplitude in the absence of any external force on it are called free vibrations.

Solution 2 (MCQ).

Forced vibrations
Hint: The vibrations of a body which take place under the influence of external periodic force acting on it are called the forced vibrations.

Solution 3 (MCQ).

A tuning fork of frequency 256 Hz will resonate with another tuning fork of frequency 256 Hz.
Hint: Resonance occurs when the frequency of an externally applied periodic force on the body is equal to its natural frequency.

Exercise 7(C)

Solution 1.

The following three characteristics of sound are:

  1. Loudness
  2. Pitch or shrillness
  3. Quality or timber.

Solution 2.

(a) Amplitude – The louder sound corresponds to the wave of large amplitude.
(b) Loudness is directly proportional to the square of amplitude.

Solution 3.

Loudness will be four times because loudness is directly proportional to the square of amplitude.

Solution 4.

(a) Ratio of loudness will be 1:9
(b) The ratio of frequency will be 1:1

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

Solution 6.

The unit of loudness is phon.

Solution 7.

Because the board provides comparatively a large area and forces a large volume of air to vibrate and thereby increases the sound energy reaching our ears.

Solution 8.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point. Its unit is microwatt per metre2.

Solution 9.

Relationship between loudness L and intensity I is given as:
L = K log I, where K is a constant of proportionality.

Solution 10.

The intensity at any point of the medium is the amount of sound energy passing per second normally through unit area at that point.
The loudness of a sound depends on the energy conveyed by the sound wave near the eardrum of the listener. Loudness, being a sensation, also depends on the sensitivity of the ears of the listener. Thus the loudness of sound of a given intensity may differ from listener to listener. Further, two sounds of the same intensity but of different frequencies may differ in loudness even to the same listener because of the sensitivity of ears is different for different frequencies. So, loudness is a subjective quantity while intensity being a measurable quantity is an objective quantity for the sound wave.

Solution 11.

The loudness of the sound heard depends on:

  1. Loudness is proportional to the square of the amplitude.
  2. Loudness is inversely proportional to the square of distance.
  3. Loudness depends on the surface area of the vibrating body.

Solution 12.

Decibel is the unit used to measure the sound level

Solution 13.

Upto 120 dB

Solution 14.

The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB from the various sources such as loudspeaker, moving vehicles etc. is called noise pollution.

Solution 15.

Pitch of sound is determined by its wavelength or the frequency. Two notes of the same amplitude and sounded on the same instrument will differ in pitch when their vibrations are of different wavelengths or frequencies.

Solution 16.

Pitch

Solution 17.

Pitch is the characteristic of sound which enables us to distinguish different frequencies sound. Pitch is the characteristic of sound by which an acute note can be distinguished from a grave or flat note.

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Sound img 6

Solution 19.

As the water level in a bottle kept under a water tap rises, the length of air column decreases, so the frequency of sound produced increases i.e., sound becomes shriller and shriller. Thus by hearing sound from a distance, one can get the idea of water level in the bottle.

Solution 20.

Trumpet. Because its frequency is highest.

Solution 21.

(a) increases
(b) one-fourth

Solution 23.

Quality or timber of sound.

Solution 24.

The two sounds of same loudness and same pitch produced by different instruments differ due to their different waveforms.
The waveforms depend on the number of the subsidiary notes and their relative amplitude along with the principal note.
Diagram below shows the wave patterns of two sounds of same loudness and same pitch but emitted by two different instruments. They produce different sensation to ears because they differ in waveforms: one is a sine wave, while the other is a triangular wave.
Selina Concise Physics Class 10 ICSE Solutions Sound img 7

Solution 25.

Since the guitars are identical, they will have a similar waveform and so the similar quality.

Solution 26.

Different instruments emit different subsidiary notes. A note played on one instrument has a large number of subsidiary notes while the same note when played on other instrument contains only few subsidiary notes. So they have different waveforms.

Solution 27.

It is because the vibrations produced by the vocal chord of each person have a characteristic waveform which is different for different persons.

Solution 28.

  1. Frequency
  2. Amplitude
  3. Waveform

Solution 29.

  1. Loudness
  2. Quality or timbre
  3. Pitch

Solution 30.
Selina Concise Physics Class 10 ICSE Solutions Sound img 8

Solution 31.

  1. IV
  2. I
  3. II

Solution 33.

(i) b, since amplitude is largest
(ii) a, since frequency is lowest

Solution 34.

Musical note is pleasant, smooth and agreeable to the ear while noise is harsh, discordant and displeasing to the ear.
In musical note, waveform is regular while in noise waveform is irregular.

Solution 1 (MCQ).

By reducing the amplitude of the sound wave, its loudness decreases.
Hint: Loudness of sound is proportional to the square of the amplitude.

Solution 2 (MCQ).

Waveforms
Explanation: The waveform of a sound depends on the number of the subsidiary notes and their relative amplitude along with the principal note. The resultant vibration obtained by the superposition of all these vibrations gives the waveform of sound.

Solution 3 (MCQ).

B is shrill, A is grave
Explanation: Shrillness or pitch of a sound is directly proportional to the frequency of the sound wave. Greater the frequency, shriller will be the note.

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Selina Concise Physics Class 10 ICSE Solutions Spectrum

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Selina ICSE Solutions for Class 10 Physics Chapter 6 Spectrum

Exercise 6(A)

Solution 1.

The deviation produced by the prism depends on the following four factors:

  1. The angle of incidence – As the angle of incidence increases, first the angle of deviation decreases and reaches to a minimum value for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
  2. The material of prism (i.e., on refractive index) – For a given angle of incidence, the prism with a higher refractive index produces a greater deviation than the prism which has a lower refractive index.
  3. Angle of prism- Angle of deviation increases with the increase in the angle of prism.
  4. The colour or wavelength of light used- Angle of deviation increases with the decrease in wavelength of light.

Solution 2.

The deviation caused by a prism increases with the decrease in the wavelength of light incident on it.

Solution 3.

Speed of light increases with increase in the wavelength.

Solution 4.

Red colour travels fastest and Blue colour travels slowest in glass.

Solution 5.

Colour of light is related to its wavelength.

Solution 6.

(i) 4000 Å to 8000 Å
(ii) 400 nm to 800 nm

Solution 7.

(i) For blue light, approximate wavelength = 4800 Å
(ii) For red light, approximate wavelength = 8000 Å

Solution 8.

Seven prominent colours of the white light spectrum in order of their increasing frequencies:
Red, Orange, Yellow, Green, Blue, Indigo, Violet

Solution 10.

Green, Yellow orange and red have wavelength longer than blue light.

Solution 11.

A glass prism deviates the violet light most and the red light least.

Solution 12.

(a) In vacuum, both have the same speeds.
(b) In glass, red light has a greater speed.

Solution 13.

The phenomenon of splitting of white light by a prism into its constituent colours is known as dispersion of light.

Solution 14.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism. Thus the cause of dispersion is the change in speed of light with wavelength or frequency.

Solution 15.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism.
Selina Concise Physics Class 10 ICSE Solutions Spectrum img 1
On the second surface, only refraction takes place and different colours are deviated through different angles. As a result, the colours get further separated on refraction at the second surface (violet being deviated the most and red the least).

Solution 16.

The colour band obtained on a screen on passing white light through a prism is called the spectrum.

Solution 17.

(a) Violet, Indigo, Blue, Green, Yellow, Orange, Red.
(b) No, different colours have different widths in the spectrum.
(c) (i) Violet colour is deviated the most. (ii) Red colour is deviated the least.

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Spectrum img 2

Solution 19.

  • Constituent colours of white light are seen on the screen after dispersion through the prism.
    Selina Concise Physics Class 10 ICSE Solutions Spectrum img 3
  • When a slit is introduced in between the prism and screen to pass only the light of green colour, only green light is observed on the screen.
  • From the observation, we conclude that prism itself produces no colour.

Solution 20.

  • If a monochromatic beam of light undergoes minimum deviation through an equi-angular prism, then the beam passes parallel to the base of prism.
  • White light splits into its constituent colours i.e., spectrum is formed.
  • We conclude that white light is polychromatic.

Solution 1 (MCQ).

Both deviation and dispersion.
Hint: When a white light ray falls on the first surface of a prism, light rays of different colours due to their different speeds in glass get refracted (or deviated) through different angles. Thus, the dispersion of white light into its constituent colours takes place at the first surface of prism.

Solution 2 (MCQ).

The colour of the extreme end opposite to the base of the prism is red.
Hint: The angle of deviation decreases with the increase in wavelength of light for a given angle of incidence. Since the red light has greatest wavelength, it gets deviated the least and is seen on the extreme end opposite to the base of prism.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Spectrum img 4
Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Spectrum img 5

Exercise 6(B)

Solution 1.

(a) Five radiations in the order of their increasing frequencies are:
Infrared waves, Visible light, Ultraviolet, X-rays and Gamma rays.
(b) Gamma rays have the highest penetrating power.

Solution 2.

(a) Gamma rays, X-rays, infrared rays, micro waves, radio waves.
(b) Microwave is used for satellite communication.

Solution 3.

(a) Gamma ray.
(b) Gamma rays have strong penetrating power.

Solution 5.

(a) X-rays are used in the study of crystals.
(b) It is also used to detect fracture in bones.

Solution 8.

4000 Å to 8000 Å

Solution 9.

(i) Infrared
(ii) Ultraviolet

Solution 10.

The part of spectrum beyond the red and the violet ends is called the invisible spectrum as our eyes do not respond to the spectrum beyond the red and the violet extremes.

Solution 11.

(a) infrared radiation
(b) ultra violet radiation

Solution 12.

(i) Ultraviolet rays-wavelength range 100Å to 4000Å

(ii) Visible light-wavelength range 4000Å to 8000Å

(iii) Infrared radiations-wavelength range 8000Å to 107Å

Solution 13.

(i) Infrared radiations are longer than 8 x 10-7m.
(ii) ultraviolet radiations are shorter than 4 x 10-7 m.

Solution 14.

Solution 15.

(i) Microwaves are used for satellite communication.
(ii) Ultraviolet radiations are used for detecting the purity of gems, eggs, ghee etc.
(iii) Infrared radiations are used in remote control of television and other gadgets.
(iv) Gamma rays are used in medical science to kill cancer cells.

Solution 16.

Solution 17.

(a) A- Gamma rays, B-infrared radiations
(b) Ratio of speeds of these waves in vacuum is 1:1 as all electromagnetic waves travel with the speed of light in vacuum.

Solution 18.

All heated bodies such as a heated iron ball, flame, fire etc., are the sources of infrared radiations.
The electric arc and sparks give ultraviolet radiations.

Solution 19.

Infrared radiations are the electromagnetic waves of wavelength in the range of 8000Å to 107Å.

Detection: If a thermometer with a blackened bulb is moved from the violet end towards the red end, it is observed that there is a slow rise in temperature, but when it is moved beyond the red region, a rapid rise in temperature is noticed. It means that the portion of spectrum beyond the red end has certain radiations which produce a strong heating effect, but they are not visible. These radiations are called the infrared radiations.

Use: The infrared radiations are used for therapeutic purposes by doctors.

Solution 20.

The electromagnetic radiations of wavelength from 100Å to 4000Å are called the ultraviolet radiations.

Detection: If the different radiations from the red part of the spectrum to the violet end and beyond it, are made incident on the silver-chloride solution, it is observed that from the red to the violet end, the solution remains unaffected. However just beyond the violet end, it first turns violet and finally it becomes dark brown. Thus there exist certain radiations beyond the violet end of the spectrum, which are chemically more active than visible light, called ultraviolet radiations.

Use: Ultraviolet radiations are used for sterilizing purposes.

Solution 21.

  • Ultraviolet radiations travel in a straight line with a speed of 3 x 108 m in air (or vacuum).
  • They obey the laws of reflection and refraction.
  • They affect the photographic plate.

Solution 22.

  • Ultraviolet radiations produce fluorescence on striking a zinc sulphide screen.
  • They cause health hazards like cancer on the body.

Solution 23.

  • Infrared radiations travel in straight line as light does, with a speed equal to 3 x 108m/s in vacuum.
  • They obey the laws of reflection and refraction.
  • They do not cause fluorescence on zinc sulphide screen.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

  1. Infrared radiations are used in photography in fog because they are not much scattered by the atmosphere, so they can penetrate appreciably through it.
  2. Infrared radiations are used as signals during the war as they are not visible and they are not absorbed much in the medium.
  3. Infrared lamps are used in dark rooms for developing photographs since they do not affect the photographic film chemically, but they provide some visibility.
  4. Infrared spectrum can be obtained only with the help of a rock-salt prism since the rock-salt prism does not absorb infrared radiations whereas a glass prism absorbs them.
  5. A quartz prism is used to obtain the spectrum of the ultraviolet radiations as they are not absorbed by quartz, whereas ordinary glass absorbs the ultraviolet light.
  6. Ultraviolet bulbs have a quartz envelope instead of glass as they are not absorbed by quartz, whereas ordinary glad absorbs the ultraviolet light.

Solution 1 (MCQ).

Gamma rays

Solution 2 (MCQ).

Carbon arc-lamp

Solution 3 (MCQ).

Infrared radiation
Hint: Infrared radiations produce strong heating effect.

Numericals

Solution 1.

(a) Frequency =500MHz =500 x 106Hz
Wavelength= 60 cm=0.6 m
Velocity of wave= frequency x wavelength
= 500x 106 x 0.6=3 x 108m/s
(b) Electromagnetic wave is travelling through air.

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Spectrum img 6

Exercise 6(C)

Solution 1.

When white light from sun enters the earth’s atmosphere, the light gets scattered i.e., the light spreads in all directions by the dust particles, free water molecules and the molecules of the gases present in the atmosphere. This phenomenon is called scattering of light.

Solution 2.

The intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light. This relation holds when the size of air molecules is much smaller than the wavelength of the light incident.

Solution 3.

Violet colour is scattered the most and red the least as the intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light.

Solution 6.

Since the wavelength of red light is the longest in the visible light, the light of red colour is scattered the least by the air molecules of the atmosphere and therefore the light of red colour can penetrate to a longer distance. Thus red light can be seen from the farthest distance as compared to other colours of same intensity. Hence it is used for danger signal so that the signal may be visible from the far distance.

Solution 7.

On the moon, since there is no atmosphere, therefore there is no scattering of sun light incident on the moon surface. Hence to an observer on the surface of moon (space), no light reaches the eye of the observer except the light directly from the sun. Thus the sky will have no colour and will appear black to an observer on the moon surface.

Solution 8.

Scattering property of light is responsible for the blue colour of the sky as the blue colour is scattered the most due to its short wavelength.

Solution 9.

As the light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path. The blue light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus the light reaching our eye directly from sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light. Therefore, the sky in direction other than in the direction of sun is seen blue.

Solution 10.

At the time of sunrise and sunset, the light from sun has to travel the longest distance of atmosphere to reach the observer. The light travelling from the sun loses blue light of short wavelength due to scattering, while the red light of long wavelength is scattered a little, so is not lost much. Thus blue light is almost absent in sunlight reaching the observer, while it is rich in red colour.

Solution 11.

At noon, the sun is above our head, so we get light rays directly from the sun without much scattering of any particular colour. Further, light has to travel less depth of atmosphere; hence the sky is seen white.

Solution 12.

The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of sizes bigger than the wavelength of visible light. Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light reaches our eye, the clouds are seen white.

Solution 1 (MCQ).

Blue colour
Hint: When light of certain frequency falls on that atom or molecule, this atom or molecule responds to the light, whenever the size of the atom or molecule comparable to the wavelength of light. The sizes of nitrogen and oxygen molecules in atmosphere are comparable to the wavelength of blue light. These molecules act as scattering centers for scattering of blue light. This is also the reason that we see the sky as blue.

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Selina ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens

Exercise 5(A)

Solution 1.

A lens is a transparent refracting medium bounded by two curved surfaces which are generally spherical.

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 1

Solution 3.

Convex lens:

  1. It converge the incident rays towards the principal axis.
  2. It has a real focus.

Concave lens:

  1. It diverges the incident rays away from the principal axis.
  2. It has a virtual focus.

Solution 4.

Equiconvex lens is converging.

Solution 5.

Concave lens will show the divergent action on a light beam.

Solution 6.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 2
As shown in the figure the convex lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 7.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 3
As shown in the figure the concave lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 8.

If a parallel beam of light is incident on a convex lens then the upper part of the lens bends the incident ray downwards. The lower part bens the ray upwards while the central part passes the ray undeviated.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 4
But in case of a concave lens the upper part of the lens bends the incident ray upwards and lower part bends the ray downwards while the central part passes the ray undeviated.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 5

Solution 9.

It is the line joining the centers of curvature of the two surfaces of the lens.

Solution 10.

It is point on the principal axis of the lens such that a ray of light passing through this point emerges parallel to its direction of incidence.
It is marked by letter O in the figure. The optical centre is thus the centre of the lens.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 6

Solution 11.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 7

Solution 12.

A lens is called an equiconvex or equiconcave when radii of curvature of the two surfaces of lens are equal.

Solution 13.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a convex lens, the first focal point is a point F1 on the principal axis of the lens such that the rays of light starting from it or passing through it, after refraction through lens, become parallel to the principal axis of the lens.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 8
The second focal point for a convex lens is a point F2 on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 9

Solution 14.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a concave lens, the first focal point is a point F1 on the principal axis of the lens such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 10
The second focal point for a concave lens is a point F2 on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 11

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 75

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 12

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 13

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 14

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 15

Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 17

Solution 21.

The distance from the optical centre O of the lens to its second focal point is called the focal length of the lens.

Solution 22.

A plane passing through the focal point and normal to the principal axis of the lens is called the first focal plane.

Solution 23.

(i) If a lens has both its focal length equal medium is same on either side of lens.
(ii)If a ray passes undeviated through the lens it is incident at the optical centre of the lens.

Solution 24.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 18

Solution 25.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 19

Solution 26.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 20

Solution 27.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 21

Solution 28.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 22

Solution 29.

(a) If half part of a convex lens is covered, the focal length does not change, but the intensity of image decreases.
(b) A convex lens is placed in water. Its focal length will increase.
(c) The focal length of a thin convex lens is more than that of a thick convex lens.

Solution 1 (MCQ).

First focus

Solution 2 (MCQ).

Its second focus

Exercise 5(B)

Solution 1.

  1. A ray of light incident at the optical centre O of the lens passes undeviated through the lens.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 23
  2. A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F2 (in a convex lens) or appears to come from the second focus F2 (in a concave lens).
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 24
  3. A ray of light passing through the first focus F1 (in a convex lens) or directed towards the first focus F1 (in a concave lens), emerges parallel to the principal axis after refraction.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 25

Solution 2.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 26

Solution 3.

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

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 27

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 28

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 29

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 30
(ii) The position of the images will be more than twice the focal length of lens.
(iii) The image will be magnified, real and inverted.
(iv) As the object move towards F1 the image will shift away from F2 and it is magnified. At F1 the image will form at infinity and it is highly magnified. Between F1 and optical centre, the image will form on the same side of object and will be magnified.

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 31

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 32

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 33

Solution 13.

Let the candle is placed beyond 2F1 and its diminished image which is real and inverted is formed between F2 and 2F2.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 34
Here the candle is AB and its real and inverted image is formed between F2 and 2F2.

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 35

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 36

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 37

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 38
The object is placed between focal point F1 and convex lens and its image is formed at the same side of the lens which is enlarged.
So this lens can be used as a magnifying lens.

Solution 18.

The sun is at infinity so convex lens forms its image at second focal point which is real and very much diminished in size.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 39
While using the convex lens as burning glass, the rays of light from the sun (at infinity) are brought to focus on a piece of paper kept at the second focal plane of the lens. Due to sufficient heat of the sun rays, the paper burns. Hence this lens is termed as ‘burning glass’.

Solution 19.

(a) This is convex lens.
(b) The nature of the image is real.

Solution 20.

(a) Convex lens.
(b) Virtual.

Solution 21.

(a) Concave lens
(b) Image is diminished

Solution 23.

Image formed by a concave lens is virtual and diminished.

Solution 24.

The virtual image formed by a convex lens will be magnified and upright.

Solution 25.

(a) at focus,
(b) at 2F,
(c) between F and 2F,
(d) between optical centre and focus.

Solution 26.

Type of lens Position of object Nature of image Size of image
Convex Between optic centre and focus Virtual and upright Magnified
Convex At focus Real and inverted Very much magnified
Concave At infinity Virtual and upright Highly diminished
Concave At any distance Virtual and upright Diminished

Solution 27.

  1. When the object is situated at infinity, the position of image is at F2, it is very much diminished in size and it is real and inverted.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 40
  2. When the object (AB) is situated beyond 2F1, the position of image (A’B’) is between F2 and 2F2, it is diminished in size and real and inverted.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 41
  3. When the object (AB) is situated at 2F1, the position of image (A’B’) is at 2F2, it is of same size as the object and real and inverted.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 42
  4. When the object (AB) is situated between 2F1and F1, the position of image (A’B’) is beyond 2F2, it is magnified in size and real and inverted.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 43
  5. When the object (AB) is situated at F1, the position of image is at infinity; it is very much magnified in size and real and inverted.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 44
  6. When the object (AB) is situated between lens and F1, the position of image (CD) is on the same side, behind the object; it is magnified in size and virtual and upright.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 45

Solution 28.

  1. When object (AB) is situated at infinity then parallel rays from object appears to fall on concave lens. Due to which image forms at focus. This image is highly diminished in size and virtual and upright.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 46
  2. When object (AB) is situated at any point between infinity and optical centre of the lens then image forms between focus and optical centre. This image is diminished in size and virtual and upright.
    Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 47

Solution 29.

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.
(b) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is equal to that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.

Solution 30.

(a) False
(b) False
(c) False
(d) True
(e) False

Solution 1 (MCQ).

The focal length of the convex lens is 10 cm.
Hint: As the object distance = image distance, the object must be kept at 2f.
Therefore, 2f = 20 cm or f = 10 cm.

Solution 2 (MCQ).

Virtual and enlarged.
Explanation: When the object is kept between optical centre and focus of a convex lens, the image is formed on the same side, behind the object. The image thus formed is virtual, enlarged and erect.

Solution 3 (MCQ).

Virtual, upright and diminished
Hint: Concave lens forms virtual, upright and diminished image for all positions of the object.

Exercise 5(C)

Solution 1.

  1. The axis along which the distances are measured is called as the principal axis. These distances are measured from the optical centre of the lens.
  2. All the distances which are measured along the direction of the incident ray of the light are taken positive, while the distances opposite to the direction of the incident ray are taken as negative.
  3. All the lengths that are measured above the principal axis are taken positive, while the length below the principal axis is considered negative.
  4. The focal length of the convex lens is taken positive and that of concave lens is negative.

Solution 1 (MCQ).

Magnification is -0.5. The negative sign of magnification indicates that the image is real while 0.5 indicates that the image is diminished. A convex lens only forms a real and diminished image of an object. Hence, the correct answer is option (d).

Solution 2.

(i) The positive focal length of a lens indicates that it is a convex lens.
(ii) The negative focal length of a lens indicates that it is a concave lens.

Solution 3.

Lens formula:
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 48

  • The distance of the object from the optical centre is called the object distance (u).
  • The distance of the image from the optical centre is called the image distance (v).
  • The distance of the principal focus from the optical centre is called the focal length (f).

Solution 3 (MCQ).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 49

Solution 4.

The term magnification means a comparison between the size of the image formed by a lens with respect to the size of the object.
For a lens: Magnification ‘m’ is the ratio of the height of the image to the height of the object.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 50

Solution 4 MCQ.

Power of a lens is +1.0 D. The positive sign indicates that the focal length of the lens is positive which indicates the lens is a convex lens.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 51

Solution 5.

(i) Positive sign of magnification indicates that the image is virtual while negative sign indicates that the image is real.
(ii) Positive sign of magnification indicates that the image is erect while negative sign indicates that the image is inverted.

Solution 6.

The power of a lens is a measure of deviation produced by it in the path of rays refracted through it.
Its unit is Dioptre (D).

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 52

Solution 8.

If focal length of a lens doubled then its power gets halved.

Solution 9.

The sign of power depends on the direction in which a light ray is deviated by the lens. The power could be positive or negative. If a lens deviates a ray towards its centre (converges), the power is positive and if it deviates the ray away from its centre (diverges), the power is negative.

Solution 10.

It is a concave.

Solution 9 (MCQ).

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 53

Solution 10 (MCQ).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 54

Solution 1 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 55

Solution 2 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 56

Solution 3 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 57

Solution 4 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 58

Solution 5 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 59

Solution 6 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 60

Solution 7 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 61

Solution 8 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 62

Solution 9 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 63

Solution 10 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 64

Solution 11 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 65

Solution 11 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 64

Solution 13 (Num).
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 67

Exercise 5(D)

Solution 1.

Magnifying glass is a convex lens of short focal length. It is mounted in a lens holder for practical use.
It is used to see and read the small letters and figures. It is used by watch makers to see the small parts and screws of the watch.

Solution 2.

Let the object (AB) is situated between focal length and optical centre of a convex lens then its image (A’B’) will form on the same side of lens.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 68
The image formed will be virtual, magnified and erect.

Solution 3.

The object is placed between the lens and principal focus.
The image is obtained between the lens and principal focus.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 69

Solution 4.

The magnifying power of the microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object (assumed to be placed at the least distance of distinct vision D = 25 cm) at the eye, i.e.,
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 76
where F is the focal length of the lens.
The magnifying power of a microscope can be increased by using the lens of short focal length. But it cannot be increased indefinitely.

Solution 5.

The two applications of a convex lens are:-

  1. It is used as an objective lens in a telescope, camera, slide projector, etc.
  2. With its short focal length it is also used as a magnifying glass.

The two applications of a concave lens are:-

  1. A person suffering from short sightedness or myopia wears spectacles having concave lens.
  2. A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

Solution 6.

The approximate focal length of a convex lens can be determined by using the principle that a beam of parallel rays incident from a distant object is converged in the focal plane of the lens.
In an open space, against a white wall, a metre scale is placed horizontally with its 0 cm end touching the wall.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 71
By moving the convex lens to and fro along the scale, focus a distant object on wall. The image which forms on the wall is very near to the focus of the lens and the distance of the lens from the image is read directly by the metre scale. This gives the approximate focal length of the lens.

Solution 7.

Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 72

Solution 8.

To determine focal length by using plane mirror we need a vertical stand, a plane mirror, a lens and a pin.
Place the lens L on a plane mirror MM’ horizontally. Arrange a pin P on the clamp of a vertical stand such that the tip of pin is vertically above the centre O of the lens.
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 73
Adjust the height of the pin until it has no parallax (i.e., when the pin and its image shift together) with its inverted image as seen from vertically above the pin.
Now measure the distance x of the pin from the lens and the distance y of the pin from the mirror, using a metre scale and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, i.e.,
Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens img 74

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Selina Concise Biology Class 10 ICSE Solutions Pollution- A Rising Environmental Problems

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Selina Concise Biology Class 10 ICSE Solutions Pollution- A Rising Environmental Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 15 Pollution A Rising Environmental Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology 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 Biology Chapter 15 Pollution – A Rising Environmental Problems

Exercise 1

Solution A.1.
(d) The dust raised during road-cleaning

Solution A.2.
(b) Disposing of corpses in rivers

Solution A.3.
(b) Ozone

Solution B.1.
(i) SO2
(ii) Bromochlorodifluoromethane and chlorofluoromethane
(iii) Mercury

Solution B.2.

Column I

 Column II (Answers)
(i) Chlorofluocarbons (CFCs)

(ii) Flyash

(iii) Cow dung

(iv) CO2 and methane

(v) Sulphur dioxide

(vi) Iodine – 131

(f) Ozone depletion

(e) Industrial Waste

(b) Biodegradable

(a) Global Warming

(d) Acid Rain

(c) Nuclear Radiation Pollutant

Solution B.3.
(i) vehicular air
(ii) X-ray
(iii) hot
(iv) domestic activities

Solution C.1.

(i) Rivers contaminated with sewage:

  • A number of waterborne diseases are produced by the pathogens present in polluted water, affecting humans as well as animals.
  • The flora and fauna of rivers, sea and oceans is adversely affected.

(ii) Too much gaseous exhausts containing CO2 and SO2:

  • The high concentration of CO2 in atmosphere has been the main component of the green house effect that has caused global warming i.e. the rise of atmospheric temperature in recent years. Global warming causes melting of snow caps rise in sea levels.
  • SO2 is poisonous and irritates the respiratory system of animals and humans. A continuous exposure to SO2 has been reported to damage the lungs and increase the rate of mortality.
  • SO2 is also responsible for acid rain

(iii) Pesticides such as DDT used in agriculture:

  • Pesticides kill soil microbes which are responsible to recycle the nutrients in the soil.
  • Pesticides can enter the food chain and affect the health of humans as well as animals. It can cause damage to the lungs and central nervous system, failures of reproductive organs and dysfunctions of the immune system, endocrine system, and exocrine system, as well as potential cancer risks and birth defects.

(iv) Prolonged noise such as the one produced by crackers throughout night:

  • Prolonged exposure to the high decibel noise damages ear drums and can bring permanent hearing impairment.
  • Noise pollution can lead to high blood pressure (hypertension), constant headache, lack of concentration.

Solution C.2.
Three major constituents of sewage:

  1. Kitchen wastes
  2. Sanitary waste
  3. Waste from agricultural lands

Solution C.3.

  • The common sources of oil spills are: The overturned oil tankers, offshore oil mining and Oil Refineries.
  • The sea birds and sea animals sometimes get thick, greasy coating on their bodies due to oil spills.
  • Sea birds may ingest their oil coated. This may irritate their digestive system, may damage liver and kidney.
  • Oil spills lead to the death of sea birds as well as sea animals.

Solution C.4.
Measures to minimise noise pollution:

  1. Use of loud speakers should be banned.
  2. Airports should be located away from the residential area.

Solution D.1.

  1. Industrial Waste:
    Large number of industries produces waste water which contains various types of chemical pollutants. Such wastes are commonly discharged into the rivers. These chemicals cause irritation to the body systems of fish.
  2. Thermal Pollution:
    Many industries such as thermal power plants, oil refineries, nuclear plants use water for cooling their machinery. This hot waste water may be 8-10oC warmer than the intake water. This hot water is released into the nearby streams, rivers or the sea and causes warming. The sudden fluctuation in the temperature of water kills the fishes and harms the plant life growing in it.

Solution D.2.
(i) Noise Pollution
(ii) Industrial machines, workshops, trains, loud conversation, loudspeakers, etc.
(iii) Effects of noise pollution:

  1. It lowers efficiency of work.
  2. It disturbs sleep and leads to nervous irritability.

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Selina Concise Biology Class 10 ICSE Solutions Health Organisations

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Selina Concise Biology Class 10 ICSE Solutions Health Organisations

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 14 Health Organisations. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology 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 Biology Chapter 14 Health Organisations

Exercise 1

Solution A.1.
(a) May 8

Solution A.2.
(c) Geneva

Solution B.1.

  1. A
  2. B
  3. A
  4. B
  5. A
  6. A
Column I
(Activity)		 Column II
(Organisation)
(i)To extend relief to victims of earthquake. Red Cross
(ii)To lay pharmaceutical standards for Important drugs. WHO
(iii)Arranging ambulance in emergencies. Red Cross
(iv)To suggest quarantine measures. WHO
(v)Training of midwives. Red Cross
(vi)Procuring and supplying blood for transfusion. Red Cross

Solution B.2.
Geneva

Solution B.3.
(a) WHO: World Health Organisation
(b) UNO: United Nations Organisation

Solution C.1.
(i) Sanitation – Removal and proper disposal of garbage, sewage and other wastes, elimination of breeding places of flies, mosquitoes, etc.
(ii) Supply of safe drinking water.
(iii) Keeping statistical records – Apart from the registration of births and deaths, to maintain the information about the health and diseases of the people in their area need regularly.

Solution C.2.

  1. Food and water borne diseases:
    The contaminated food and water cause several diseases. Water borne diseases occur due to contaminated water from hand pumps or mixing of untreated sewage with river water.
  2. Insect and air-borne diseases:
    Lack of cleanliness leads to breeding of houseflies, mosquitoes which are the carries of certain diseases.
  3. Lack of medical facilities:
    Lack of medical facilities especially in rural areas, leads to unavoidable deaths and damage to health. Lack of knowledge and superstitions beliefs also delay timely treatment which may result in serious consequences.

Solution C.3.
Functions of WHO:

  • To promote and support projects for research on diseases.
  • To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.

Solution C.4.
Functions of Red Cross:

  • To extend relief and help to the victims of any calamity – flood, fire, famine, earthquakes, etc.
  • To procure and supply blood for the needy victims of war and other calamities.
  • To extend all possible first-aid in any accident.
  • To arrange for ambulance services in all emergencies.

Solution C.5.
Functions of World Health Organisation (WHO):

  • To promote and support projects for research on diseases.
  • To collect and supply information about the occurrence of diseases of epidemic nature such as cholera, plague, yellow fever, etc.
  • To lay pharmaceuticals standards for important drugs, to ensure purity and size of the dose.
  • To organize campaigns for the control of epidemic (widespread) and endemic (local) diseases.

Solution C.6.

  1. To extend relief and help to victims of any calamity
  2. To procure and supply blood for the needy victims of war or calamity
  3. To extend all possible first aid in any accident
  4. To educate people in accident prevention
  5. To arrange for ambulance services in emergencies
  6. To look after maternal and child welfare centres
  7. To train midwives
    (any four)

Solution D.1.

  • The World Health Organization (WHO) is a specialized agency of the United Nations (UN) that is concerned with international public health.
  • It was established on 7 April 1948, with headquarters in Geneva, Switzerland and is a member of the United Nations Development Group.
  • There were several reasons for the formation of WHO:
  • Member countries of the UNO focused on the need for creating an international body to look after the health problem of people of the world.
  • This was particularly felt in the field of research on the causes and cures of the diseases.
  • The combined efforts in this direction were to give better and faster results.
  • The poor and developing countries were to benefit quickly.

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Selina Concise Biology Class 10 ICSE Solutions Aids To Health

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Selina Concise Biology Class 10 ICSE Solutions Aids To Health

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 13 Aids To Health. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology 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 Biology Chapter 13 Aids To Health

Exercise 1

Solution A.1.
(c) An antibiotic

Solution A.2.
(c) Tetanus

Solution A.3.
(c) April 7

Solution B.1.
(a) Salvarson
(b) Penicillin
(c) Passive Acquired Immunity
(d) Antiseptics – Lysol, Iodine, Boric acid and Carbolic acid;
Disinfectants – Cresol and Phenol;
Antibiotics – Ampicillin and Penicillin
(e) Oral polio vaccine (OPV)

Solution B.2.
(i) Acquired Immuno Deficiency Syndrome
(ii) Bacillus Calmette Guerin
(iii) Diphtheria, Pertussis and Tetanus
(iv) World Health Organisation

Solution B.3.
Antibodies which are immunoglobulins are produced in the blood to fight and destroy harmful microbes.

Solution C.1.
(a) False
(b) True
(c) False
(d) False
(e) False
(f) False

Solution C.2.
(a) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas an antibiotic is a chemical substance produced by a micro-organism, which can kill or inhibit the growth of some other disease producing microorganisms.

(b) Antiseptic is a mild chemical substance which, when applied on the body, kills germs whereas a disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply.

(c) Disinfectant is a strong chemical, which is applied on spots and places on the body where germs thrive and multiply whereas deodorants are neither antiseptics nor disinfectants; they aerosols used to mask a bad smell.

(d) Vaccination is the introduction of any kind of dead or weakened germs into the body of a living being to develop immunity (resistance) against the respective disease or diseases whereas sterilization is a process of eliminating or killing all the microbes present on a surface, contained in a fluid, in medication, or in a compound such as biological culture media.

(e) Active immunity is the immunity developed by an individual due to a previous infection or antigen which enters his body naturally whereas passive immunity is the immunity provided to an individual from an outside source in the form of ”readymade” antibodies.

(f) Innate immunity is the immunity by the virtue of genetic constitutional makeup i.e. it is inherited from parents. It is present in the body without any external stimulation or a previous infection whereas acquired immunity is the resistance to a disease which an individual acquires during his lifetime. It may be the result of either a previous infection or readymade antibodies supplied from outside.

Solution C.3.

  1. TAB vaccine for typhoid
  2. BCG vaccine for measles
  3. DTP vaccine for diphtheria, tetanus and whooping cough

Solution C.4.
(a) Lysol, Benzoic acid, DDT, mercurochrome
Antiseptics. DDT is a wrong example for this category as it belongs to disinfectant which is not good for human skin.

(b) Formalin, iodine, Lysol, phenol.
Disinfectants. Iodine is a wrong example as it is an antiseptic.

(c) BCG, DTP, ATP.
Vaccines. ATP is a wrong example as it is used as an energy carrier in the cells of all known organisms.

(d) Tears, skin, nasal secretion, HCl (in stomach).
Germ Killing Secretions. Skin is a wrong example as it is a protective mechanical barrier. It prevents the entry of microorganisms at first place.

Solution C.5.

Vaccine

 Disease (s) The Nature of Vaccine
TAB Typhoid

Killed germs

Salk’s Vaccine

 poliomyelitis Killed germs
BCG tuberculosis

Living weakened germs

Vaccines for Measles

 Measles Living weakened germs
Cowpox Virus small pox

Living fully poisonous germs

Toxoids

 Diphtheria Extracts of toxins
Tetanus

Secreted by bacteria

Solution C.6.

  1. Innate Immunity
  2. Acquired Immunity
  3. Specific Immunity
  4. Active Acquired Immunity
  5. Passive Acquired Immunity
  6. Natural Acquired Active Immunity
  7. Artificial Acquired Active Immunity
  8. Natural Acquired Passive Immunity
  9. Artificial Acquired Passive Immunity

Solution C.7.
(a) antibiotics have a wide use in medicine to fight infections.
(b) Certain antibiotics are used as food preservatives, especially for fresh meat and fish.
(c) Some antibiotics are used in treating animal feed to prevent internal infections.
(d) Some antibiotics are used for controlling plant pathogens.

Solution C.8.

Merits of the Local Defence Systems:

  1. Local defence systems start working instantaneously.
  2. These systems are not dependent on previous exposure to infections.
  3. They are effective against a wide range of potentially infectious agents.

Solution C.9.

  • Diphtheria is a serious bacterial infectious disease. It leads to cold, coughing, sneezing and in severe cases if undiagnosed it might result in heart failure or paralysis.
  • Treatment includes a combination of medications and supportive care. The most important step is prompt administration of diphtheria toxoid which is made harmless is given intravenously. The harmless toxoid once administered in patient’s body triggers the production of antibodies against the pathogens causing diphtheria.

Solution C.10.

(a) Bleeding from a cut in the skin:

  • In case of bleeding, raise the affected part to minimize the blood flow.
  • Wash the cut surface with clean water.
  • Press the area with a piece of clean cotton and apply some antiseptic.

(b) A fractured Arm:

  • Lay the victim comfortably, loosen or remove the clothes from the affected part.
  • Do not move the part fractured.
  • If the affected limb is an arm, then tie a sling around the neck to rest the arm in it.

(c) Stoppage of breathing due to electrical shock:

  • Lay the victim flat on his back and put a pillow or folded towel under his shoulders in a way that his chest is raised and the head thrown back.
  • Hold and draw his arms upwards and backwards. This will cause his chest to expand and draw the air.
  • Next, fold the victim’s arms and press them against the ribs. The air will now be expelled.
  • Repeat the two steps at the rate of about 15 times per minute. Continue till the victim starts breathing without any extra help or till the doctor arrives.

Solution D.1.
Vaccination is the practice of artificially introducing the germs or the germ substance into the body for developing resistance to particular diseases. Scientifically, this practice is called prophylaxis and the material introduced into the body is called the vaccine. The vaccine or germ substance is introduced into the body usually by injection and sometimes orally (e.g. polio drops). Inside the body, the vaccine stimulates lymphocytes to produce antibodies against the germs for that particular disease. Antibodies are the integral part of our immunity. Their function is to destroy unwanted particles entered in the body. Vaccines give our immunity a signal to produce specific antibodies. Hence, the principle of vaccination is to produce immunity against a disease.

Solution D.2.
Whenever a germ or infection invades the body. A signal is sent to the immune system to produce specific antibodies. In order to cope up with the number of germs being multiplied inside the body, white blood cells start multiplying rapidly. This enables them to produce more number of antibodies and invade infection in time. Therefore, ”Abnormally, large numbers of WBCs in the blood are usually an indication of some infection in our body”.

Solution D.3.
(a) Antiseptics:  Antiseptics are mild chemical substances applied to the body, which prevent the growth of some bacteria and destroy others.
Example: Lysol and Iodine

(b) Disinfectants:  Disinfectants are chemicals which will kill all micro-organisms they come in contact with. Disinfectants are usually too strong to be used on body.
Example: Cresol and Phenol

(c) Vaccines: Vaccines are the materials used to administer in the body to provide passive immunity. The materials are generally germs or the substances secreted by the germs.
Example: OPV (Oral polio vaccine) and DTP (Diphtheria, Tetanus and Pertussis)

Solution D.4.
First aid is the immediate care given to a victim of an accident, sudden illness or other medical emergency before the arrival of an ambulance, doctor or other qualified help.

(a) Little toe in the foot is pierced by a thorn and is bleeding:

  • In the case of bleeding, raise the affected part to minimise gravitational outflow of blood.
  • Wash the cut surface with clean water, press the area with a piece of clean cotton wool, and if possible, apply some mild antiseptic.

(b) An elderly woman walking on the footpath during a hot mid-day has fallen unconscious:

  • Immediately lay the woman comfortably on a side of the road.
  • Loosen the clothes.
  • Let fresh air be around the woman.
  • Give some fluids to drink to the woman.

(c) A young boy has burnt his finger tip while firing crackers:

  • Immediately wash his burnt finger with sufficiently cold water for a few minutes.
  • Do not rub the burnt region.
  • Apply creams/ointments specially recommended, in case they are readily available.

(d) Your gardener has been bitten by a snake while digging soil in the flower bed:

  • Immediately squeeze out some blood from the wound.
  • Tie a tourniquet above the site to prevent spreading of venom into the body.

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Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems

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Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 12 Population The Increasing Numbers and Rising Problems. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology 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 Biology Chapter 12 Population – The Increasing Numbers and Rising Problems

Exercise 1

Solution A.1.
(d) Use of antibiotics and prophylactic vaccinations

Solution A.2.
(a) Per 1000 people per year

Solution B.1.
Demography

Solution B.2.
(i) Tubectomy (For Female)
(ii) Vasectomy (For Male)

Solution C.1.
(a) True
(b) False

Solution C.2.
It is approximately 7 billion.

Solution C.3.

(a) Birth Rate: It is the number of live births per 1000 people of population per year.

(b) Death Rate: It is the number of deaths per 1000 of population per year.

(c) Rate of growth of population: It is the difference between the birth rate and the death rate. As long as the birth rate exceeds the death rate, the population grows. If the birth rate is lower than the death rate, the population declines.

(d) Population density: It is the number of individuals per square kilometre (Km2) at any given time.

(e) Exhaustible resource: Exhaustible resources are the once which are perished by the prolonged us and they can never be renewed or replaced.

Solution C.4.
(a) False.
Dog was the first domesticated animal.

(b) True.
Rapidly growing industries made human life more and more comfortable, with greater opportunities of jobs and with more production of food. All this favoured population rise.

(c) False.
Present human population growth is following geometrical progression.

(d) False.
Birth Rate (natality) is the number of live births per 1000 people of population per year.

(e) False.
Vasectomy is the surgical method of contraception in human males while tubectomy is the surgical method used in females.

Solution C.5.
The rate of growth of population of the world is 1.092% (this rate results in about 145 net additions to the worldwide population every minute or 2.4 every second {2011 estimates}) and for India is 1.344%.

Solution C.6.
By law the minimum age is 21 years for boys and 18 years for girls.

Solution C.7.
Two advantages of small family are:

  1. Parents can give more attention to their children.
  2. Small family helps every country in controlling the growth of population.

Solution C.8.

  1. Illiteracy:
    Most of the rural population which forms the bulk of our society is still illiterate, ignorant and superstitious.
    They also do not know the functioning of the human reproductive system.
  2. Traditional Beliefs:
    Among the people from lower strata of the society, children are regarded as a gift of God and a sign of prosperity.
    Therefore, they make no effort to avoid pregnancy.

Solution C.9.
Population growth is not the only threat humanity is facing, but it will be a major contributor to the crises that await us and the planet in the coming century. Overpopulating the planet puts us all at risk of extreme environmental and social consequences that we are beginning to witness today. The extreme growth in human population is mortally taxing the Earth and its resources. Each individual person has a unique impact on the planet’s environment. Some people may be relatively less damaging than others, but no living individual is without an ecological footprint. In other words, each person needs basic resources and almost all people aspire to utilize significantly more resources than are required by their basic needs. As a result, the Earth is attempting to impose its own checks on human population. We can witness these “checks” in the form of widespread disease and the emergence of new disease strains, food and water shortages, poor harvests and violent and destructive weather caused by climate change. While it should be obvious that the Earth is a finite sphere and cannot endure infinite growth by any single species, we should also remember that Earth’s current web-of-life is the result of billions of years of complex evolution. It is irreplaceable. When we look forward to the next 40 years, the most significant population increases will take place in the areas of our world where natural resources and the infrastructure of modernity are already the scarcest. 95% of the human population growth is occurring in countries already struggling with poverty, illiteracy and civil unrest. It will further stress, the already strained ecological systems and worsen poverty in much of the developing world, thus aggravating threats to international security.

Thus, the statement made by an author ‘some great author has said that a population explosion is far more dangerous than an atomic explosion’ is true.

Solution C.10.
Poverty and population have been closely linked ever since the world faced changes due to the major revolutions. Poverty has its own effects on the population and vice versa. Poverty prevails because of illiteracy and traditional beliefs in the economically weaker strata. Since illiteracy and traditional beliefs prevail the people from this stratum, they regard children as gift of God and a sign of prosperity. They consider children to be helping hands in increasing the family income, hence they keep producing more children forgetting that their current situation would do no good for the children and they would add more to this already overburdened poverty strata. Hence, the population keeps on rising and so does poverty. As the population increases the quality of life goes down.

Solution C.11.

  1. Tool making revolution.
  2. Agricultural revolution.
  3. Scientific industrial revolution.

Solution C.12.
According to census, the Indian population in 1981 was 685 million and it was 846 million in 1991.

Solution C.13.
Yes, there could be a corresponding operation made in women. The name of the surgical procedure in females is ‘tubectomy’. In tubectomy, the abdomen is opened and the fallopian tubes (oviducts) are cut or ligated i.e. tied with nylon thread to close the passage of the egg.
Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems image - 1

Solution C.14.
Family welfare centres are those places where any help or advice about family planning is available free of cost. These places could be any hospitals, dispensaries, etc. The inverted red triangle is the symbol of family welfare in India.
Selina Concise Biology Class 10 ICSE Solutions Population- The Increasing Numbers and Rising Problems image - 2

Solution C.15.
Below are some of the advantages of having a small family:

  1. Financial condition of family is deeply related to the size of the family. A living cost of a large family is surely much higher than a small family. A large family has more expenses on cloth, toys, education and food whereas expenses in small family are very low.
  2. Parents can easily fulfill the needs of one or two children. They can provide them best education and look after them very well whereas when there are many children to look after parents just cannot fulfill even the basic needs of the children properly. Therefore, as a result, children suffer, the parents suffer and the nation suffers.
  3. A child in a small family receives more support from their parents than in a large family. In large family, parents have many children to look after, so they cannot give their best support to everyone whereas in small family parents can give more support to children as they have only one or two children to look after.
  4. Family size also affects the health, especially that of the mother and the child. Frequent pregnancies can cause illness to both the mother and the children. It can disrupt the health of the women. It puts mother and baby’s health at risk. So having a small family definitely leads to healthy and happy family.

Solution D.1.

  1. Food: The first and most important need of the humans (or any living organism) is food. But with the production of food rising by arithmetic progression and population growing by geometric progression i.e. the number at each step is being multiplied. At the same time growing population is increasing the use of more and more agricultural land to build houses. Thus it is evident that food would be running short for the unchecked rising population.
  2. Water: Availability of clean and germ-free water for drinking purposes would be more and more scarce with increase in population; the reason would be mainly, the pollution of rivers, ponds, lakes etc.
  3. Land: Man is bringing more and more land under cultivation and also using up land for building more residential colonies, factories and industries. Usable land would thus become less and less available.

Solution D.2.

  1. The orthodox view, to have at least one son especially in Indian society, should be modified with education. People should be educated that their greed for a son can lead to numerous children in the household which would worsen both their family’s health and wealth. They should focus on proper upbringing of the child, be it a son or a daughter.
  2. Married couples should be educated to delay the birth of their first child, to space the second with a sufficient interval for proper upbringing and to stop the third. They should also be educated to adopt family planning methods by which they can prevent pregnancy after two children. These include devices for both men and women, for example: Condoms, intrauterine devices (IUD) and oral pills.

Solution D.3.
For developing countries like India, population explosion is a curse and is damaging the development of the country and its society. The developing countries are already facing a lack in their resources, and with the rapidly increasing population, the resources available per person are reduced further, leading to increased poverty, malnutrition, and other large population-related problems. The literal meaning of population is “the whole number of people or inhabitants in a country or region”, and the literal meaning of population explosion is “a pyramiding of numbers of a biological population”. As the number of people in a pyramid increases, so do the problems related to the increased population. Some of the reasons for this population explosion are poverty, better medical facilities, and immigration from the neighboring countries. The population in India continues to increase at an alarming rate. The effects of this population increase are evident in the increasing poverty, unemployment, air and water pollution, and shortage of food, health resources and educational resources.

Solution E.1.
(a) B; 1981; 1991.
(b) B; 1981; 1991.
(c) B; 1971; 1981.

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Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

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Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 11 The Reproductive System. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology 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 Biology Chapter 11 The Reproductive System

Exercise 1

Solution A.1.
(d) Epididymis → vas deferens → urethra

Solution A.2.
(d) 28 days

Solution A.3.
(d) About seven days

Solution B.1.
(a) Scrotum
(b) Seminiferous Tubules
(c) GraafianFollicle
(d) Seminal vesicle
(e) Epididymis

Solution B.2.
(a) Testosterone
(b) Ureter
(c) Ovum
(d) After birth

Solution B.3.
(a) Testes → Sperms → Sperm duct → Semen → Penis
(b) Menarche Puberty → Reproductive age → Menstruals → Menopause
(c) Graafian follicle → Ostium → Fallopian tube → Uterus

Solution B.4.
Seminiferous tubule → Epididymis → Vas deferens → Penis

Solution C.1.
Semen is the mixture of sperms and secretions from seminal vesicles, prostate gland and Cowper’s (bulbo-urethral gland).

Solution C.2.
(a) Inguinal canal: It is the canal which allows the descent of testes along with their ducts, blood vessels and nerves into the abdomen.
(b) Prostate gland: It is a bilobed structure which surrounds the urethra and pours an alkaline secretion into the semen.
(c) Testis: Testis is a male reproductive organ. There a pair of testes present in a scrotal sac descended outside the body cavity. Testes produce sperms which are the male gametes.
(d) Ovary: Ovary is a female reproductive organ.It produces ova i.e. female gametes.
(e) Oviduct: A pair of oviduct is present on either side of the uterus. Oviduct carries the released ovum from the ovary to the uterus.

Solution C.3.
Secondary sexual characters in males:

  1. Beard and moustache
  2. Stronger muscular built
  3. Deeper voice

Secondary sexual characters in females:

  1. Breasts in females
  2. Large hips
  3. High pitched voice

Solution C.4.
The accessory reproductive organs include all those structures which help in the transfer and meeting of two kinds of sex cells leading to fertilization and growth and development of egg up to the birth of the baby.
For example: uterus in females, penis in males.

Solution C.5.

Primary Reproductive Organs

 Accessory Reproductive Organs
The primary reproductive organs produce sex cells.

The accessory reproductive organs help in the transfer and meeting of two kinds of sex cells leading to fertilization.

The primary reproductive organs do not help in the development of baby.

 The accessory organs help in the growth and development of egg up to the birth of baby.
Example: Testes in males and ovaries in females.

Example: penis in males, Uterus, vagina in female.

Solution C.6.
Hymen is a thin membrane which partially covers the opening of the vagina in young females.

Solution C.7.
(a) Hernia: It is an abnormal condition which is caused when the intestine due to the pressure in abdomen bulges into the scrotum through the inguinal canal.
(b) Ovulation: It is the release of the mature ovum by the rupture of the Graafian follicle.
(c) Puberty: It is the period during which immature reproductive system in boys and girls matures and becomes capable of reproduction.

Solution C.8.
Changes in human male:

  1. Development of Beard and moustache
  2. Voice becomes deeper

Changes in human female:

  1. Development of Breasts in females
  2. Development of high pitched voice

Solution C.9.
(a) Menarche is the onset of menstruation in young females at about 13 years of age whereas menopause is the permanent stoppage of menstruation at about 45 years of age.

(b) Cowper’s gland opens into urethra in human males and its secretion serves as a lubricant whereas the prostate gland surrounds the urethra in males and its alkaline secretion neutralizes acid in female’s vagina.

(c) Hymen is a thin membrane that partially covers the opening of vagina in young females whereas clitoris is a small erectile structure located in the uppermost angle of vulva in front of the urethral opening.

(d) Uterus is a hollow, pear shaped muscular organ located in the pelvic cavity. It is the site of implantation for the embryo after fertilisation whereas the vagina is the muscular tube extending from the cervix to the outside. At the time of sexual intercourse, the vagina receives the male penis and provides entry for the sperms.

(e) Efferent ducts join to form the epididymis whereas the epididymis is continued by the side of the testes to give rise to the sperm duct or vas deferens.

Solution D.1.

  • Testes are responsible for the production of male gametes i.e. sperms. The normal body temperature does not allow the maturation of the sperms. Being suspended outside the body cavity, the temperature in the scrotal sac is 2 to 3oC which is the suitable temperature for the maturation of the sperms.
  • When it is too hot, the skin of the scrotum loosens so that the testes hang down away from the body. When it is too cold, the skin contracts in a folded manner and draws the testes closer to the body for warmth.
  • In an abnormal condition, in the embryonic stage, the testes do not descend into the scrotum. It can lead to sterility or incapability to produce sperms.

Solution D.2.
Testosterone is the male reproductive hormone produced by the interstitial cells or the Leydig cells. These cells are located in the testes. They serve as a packing tissue between the coils of the seminiferous tubules. Therefore, it can be said that the testes produce the male hormone testosterone.

Solution D.3.
otal reproductive period = 45 – 13 = 32 years
Total eggs produced = 32 x 12 = 384 eggs approximately

Solution E.1.
(a) Excretory system and Female Reproductive system

(b)

  1. Kidney
  2. Ureter
  3. Fallopian Tube
  4. Infundibulum
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Cervix
  9. Vagina
  10. Vulva

(c)

  • Function of Fallopian Tube (part 3): The fallopian tubes carry the ovum released from the ovary to the uterus.
  • Function of Infundibulum (part 4): Infundibulum is the funnel shaped distal end of the ovary which picks up the released ovum and pushes it further on its passage into the fallopian tube.
  • Function of Ovary (part 5): Ovary produces female gametes i.e. ova.
  • Function of Uterus (part 6): Uterus allows the growth and development of the embryo.

Solution E.2.
(a)

  1. Fallopian Tube
  2. Infundibulum
  3. Ureter
  4. Vagina
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Urethra

(b) Oestrogen secreted by the corpus luteum secrets oestrogen. Oestrogen stimulates the thickening of the endometrial wall of the uterus. The uterine wall becomes thickened and is supplied with a lot of blood to receive the fertilized egg.
(c) If fertilization fails to take place, the endometrial lining of the uterus starts shedding on the 28th day of the menstrual cycle. Finally it is discharged out along with the unfertilised ovum as the menstrual flow.

Solution E.3.
a.

  1. Seminal vesicles
  2. Prostate gland
  3. Bulbo-urethral gland
  4. Epididymis
  5. Testis
  6. Scrotum
  7. Urinary bladder
  8. Vas deferens
  9. Erectile tissue
  10. Penis
  11. Urethra

b. Functions of

  1. Seminal vesicles
    They produce the fluid which serves as the transporting medium for sperms.
  2. Prostate gland
    It produces an alkaline secretion which mixes with the semen and helps neutralise the vaginal acids.
  3. Bulbo-urethral gland
    It produces a secretion which serves as a lubricant for the semen to pass through the urethra.
  4. Testis
    It produces the male gamete sperm and the male sex hormone testosterone.
  5. Vas deferens
    They carry the sperms from the epididymis to the urethra.
  6. Urethra
    It serves as an outlet for delivering the sperms into the vagina.

Exercise 2

Solution A.1.
(c) fallopian tube

Solution A.2.
(a) energy

Solution A.3.
(c) 280 days

Solution B.1.
(a) Amniotic fluid
(b) Uterus
(c) Amniotic membrane
(d) Inguinal canal

Solution B.2.
(a) Sperm
(b) Follicle

Solution B.3.
(a) Ovulation → fertilization → implantation → gestation → child birth
(b) Sperm → sperm duct → urethra → coitus → vagina → ovum

Solution B.4.
(a) Menarche
(b) Ovulation
(c) Menstruation
(d) Fertilization
(e) Implantation

Solution B.5.

Column I

 Column II
(a) Acrosome

(v) spermatozoa

(b) Gestation

 (vii) Time taken by a fertilized egg till the delivery of baby
(c) Menopause

(vi) complete stoppage of menstrual cycle

(d) Foetus

 (i) An embryo which looks like human baby
(e) Oogenesis

(iii) ovum producing cells

(f) Ovulation

(ii) Luteinizing hormone

 

Solution C.1.
(a)

  1. False
  2. False
  3. False
  4. False

(b)

  1. Fertilization occurs in the fallopian tube.
  2. Vagina is also known as the birth canal.
  3. Nutrition and oxygen diffuse from the mother’s blood into the foetus’s blood through placenta.
  4. Gestation period in humans is about 280 days.

Solution C.2.

Structure

 Function
1. Corpus luteum

1. secretes progesterone & other hormones to prepare the uterine wall for the receival of the embryo.

2. Testes

 2. produces male gametes in mass
3. Placental disc

3. supplies oxygen and nutrients to embryo

4. Oxytocin

 4. increases the force in uterine contractions during child birth
5. Umbilical cord

5. connects placenta with foetus

6. Fallopian tube

6. The site of fertilization for the sperm and ovum

Solution C.3.

(a) Foetus:

  1. It is contained in the uterus.
  2. In foetus, limbs have appeared and resembles the humans unlike the embryo which is a growing or dividing zygote.

(b) Hyaluronidase:

  1. Enzyme
  2. It is an enzyme secreted by the sperm that allows the sperm to penetrate the egg.

(c) Morula:
It is the stage in the development of human embryo which consists of a spherical mass of cells. Blastocyst

(d) Amniotic fluid:

  1. Between amnion and embryo
  2. It protects the embryo from physical damage, keeps the pressure all around embryo and prevents sticking of foetus to amnion.

(e) Gestation:
Gestation is the full term of the development of an embryo in the uterus. 280 days in humans.

(f) Placenta:

  1. Placenta is formed by two sets of minute finger like processes called the villi. One set of villi is from the uterine wall and the other set is from the allantois.
  2. Oxygen and amino acids.
  3. Progesterone and oestrogen.

(g) Implantation:

  1. Blastocyst
  2. It occurs in about 5-7 days after ovulation.

Solution D.1.

(a) Sperm is the male gamete produced by the testes. Semen on the other hand is the mixture of sperms and alkaline secretions from the seminal vesicle, prostate gland and Cowper’s gland.

(b) Implantation is the fixing of embryo in the wall of uterus. The state that implantation produces is known as pregnancy.

(c) Follicle is the cellular sac containing a maturing egg. Corpus luteum on the other hand is the remnant of the follicle the release of ovum during ovulation.

(d) Amnion is a sac which develops around the embryo whereas allantois is an extension from the embryo which forms villi of placenta.

(e) Sterility is the incapability to produce sperms whereas impotency is the inability to copulate.

(f) Prostate gland pours alkaline secretions into the semen to neutralize the acid in female’s vagina whereas the secretion of Cowper’s gland serves as a lubricant.

(g) Identical twins are produced from one ovum i.e. one developing zygote splits and grows into two foetuses whereas fraternal twins are produced when two ova get fertilized at a time.

Solution D.2.

  1. After fertilization zygote is formed inside the fallopian tube.
  2. The zygote then divides repeatedly to form a spherical mass of cells known as ‘Morula’.
  3. The morula then develops into a hollow sphere of cells with a surrounding cellular layer and an inner cell mass projecting from it centrally. This stage is known as the ‘blastocyst’. It implants itself into the uterine wall.
    Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -1
  4. From the blastocyst arises an embryo which is around 3 weeks old. It is a tiny organism that hardly resembles human being.
  5. By the end of 5 weeks, the embryo is with a develoed heart and blood vessels.
  6. By the end of 8 weeks, limbs are developed. This stage is known as ‘foetus’.
  7. At the end of nearly 40 weeks i.e. end of gestation period, the infant is born.

Solution D.3.
(a) Amnion:

  1. Amnion contains the amniotic fluid which surrounds the embryo.
  2. This fluid protects the embryo from physical damage.
  3. It maintains even pressure all around the embryo.
  4. It also prevents sticking of foetus to amnion.

(b) Placenta:

  1. The placenta allows the diffusion of oxygen and nutrients such as glucose, vitamins and amino acids from mother to foetus.
  2. Similarly, it also allows the diffusion of carbon dioxide, urea and waste products from foetus to mother.
  3. Placenta also acts as an endocrine tissue. It secretes oestrogen and progesterone.

Solution E.1.
a. A – ovum
B – sperm

b. Sperms are produced in the testis.
The ovum is produced in the ovary.

c. The reproductive cells unite in the fallopian tubes of the female reproductive system.

d. Ovary – Oestrogen and progesterone
Testis – Testosterone

e. Accessory glands:

  • Seminal vesicle – Seminal fluid
  • Prostate gland – Alkaline secretion
  • Bulbo-urethral gland – Lubricant

Solution E.2.
(a)

  1. umbilical cord,
  2. placenta,
  3. amnion,
  4. mouth of uterus,
  5. muscular wall of uterus

(b) Gestation
(c) 280 days
(d) Placenta provides the foetus with oxygen and nutrients. In addition, the placenta also removes carbon dioxide and waste products of the foetus.
(e) Progesterone

Solution E.3.

Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -2

Solution E.4.
(a) A – Muscular wall of uterus,
B – Oviduct,
C – Ovary,
D – Cervix
(b) If part B will get blocked, ovum released from the ovary will not get fertilized by the sperm and hence pregnancy will be prevented.

Solution E.5.

  1. Prostate gland
  2. Bulbo-urethral gland
  3. Urethra
  4. Vas deferens
  5. Testis

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Selina Concise Biology Class 10 ICSE Solutions Endocrine Glands

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Selina Concise Biology Class 10 ICSE Solutions Endocrine Glands

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Selina ICSE Solutions for Class 10 Biology Chapter 10 Endocrine Glands

Exercise 1

Solution A.1.
(c) pancreas

Solution A.2.
(b) thyroid

Solution A.3.
(c) alcohol

Solution B.1.
(a) Insulin, glucagon, somatostatin
(b) Adrenaline
(c) Hypoglycemia
(d) Insulin
(e) Adrenaline
(f) Anti-diuretic hormone (Vasopressin)
(g) Adrenaline

Solution B.2.
If there was hyposecretion of the thyroid gland in a child; the child will suffer from cretinism. The symptoms of cretinism are dwarfism, mental retardation, etc.

Solution B.3.
(a) Prostate
(b) Scurvy
(c) Cretinism
(d) Cortisone

Solution B.4.
(a) Larynx
Reason- Larynx is the sound box while the rest three i.e. glucagon; testosterone and prolactin are hormones.

(b) Penicillin
Reason – Penicillin is an antibiotic while adrenaline; insulin; thyroxine are hormones.

(c) Adrenaline
Reason – Adrenaline is a hormone while the stomach, ileum and liver are the organs of the digestive system.

(d) Insulin
Reason- Insulin is secreted by the pancreas while TSH, GH, ADH are the hormones secreted by the pituitary gland.

(e) Iodine
Reason- Iodine is required for the synthesis of thyroxine hormone. While cretinism, goitre, myxoedema are the deficiencies occur due to the deficiency of thyroxine.

Solution B.5.

Column I

 Column II
1. Beta cells of islets of Langerhans

(g) Insulin

2. Thyroid

 (c) Exophthalmic goitre
3. Cretinism

(h) Under secretion of thyroxine in a child

4. Addison’s disease

 (b) Glucocorticoids
5. Hypothyroidism

(e) Thyroxine

6. Myxoedema

 (a) condition due to under     secretion of thyroxine in adults
7. Adrenaline

(d) Increases heart beat

8. Cortisone

(f) Adrenal cortex

Solution B.6.

A (Condition)

 B (Cause)
(a) Dwarfism and mental retardation

v. Hypothyroidism

(b) Diabetes mellitus

 i. Excess of glucose in blood
(c) Shortage of glucose in blood

iii. Insulin shock

(d) Gigantism

 ii. Over secretion of growth hormone
(e) Enlargement of breasts in adult males

vi. Over secretion of cortical hormones

(f) Exophthalmic goitre

iv. Over secretion of thyroxine

Solution C.1.
(a) True
Reason- Adrenaline is described as emergency hormone because during any emergency situation more adrenaline is secreted which makes the heart beat faster, increases the breathing, releases more glucose into the blood stream to fulfill the energy requirement.

(b) False
Reason- The two different kinds of diabetes are diabetes insipidus caused due to insufficient secretion of vasopressin and the other is ‘diabetes mellitus’ caused due to hyposecretion of insulin but they cannot be described as mild and severe.

(c) True
Reason-Iodine is an active ingredient in the production of the thyroxine hormone.

(d) True
Reason- Pituitary gland controls the functioning of all the other endocrine glands.

(e) True
Reason- Hormones are poured directly into blood the blood stream and control physiological processes by chemical means. Their action depends on the feedback mechanism.

(f) True
Reason- Gigantism and dwarfism are controlled by the growth hormone from the pituitary gland. Growth hormone is much more active in children for their normal body growth along with which necessary substance required for the synthesis of growth hormone need to be consumed.

Solution C.2.
Endocrine glands are ductless glands, means they pour their secretion i.e. hormones directly into the blood stream while the other glands are exocrine glands which have ducts. Through ducts they pour their secretions (not hormones) into the blood stream.

Solution C.3.
Hormones unlike enzymes are secreted by the endocrine glands only. Also the hormones unlike the enzymes are poured directly into the blood. Hormones can be peptides, steroids, amine but all enzymes are proteins.

Solution C.4.
Chemically hormones are peptides, amines or steroids. They are involved in regulating the metabolism of the body. They can bring about specific chemical changes during metabolic process. Therefore hormones can be termed as ‘chemical messengers’.

Solution C.5.
Iodine is an active ingredient in the production of the thyroxine hormone secreted by the thyroid gland. Thyroxine hormone is a very essential hormone for our body. In case of its abnormal secretions a person may suffer certain sever disorders. Therefore, it is an important nutrient for our body.

Solution C.6.
Adrenaline is the hormone which prepares the body to meet any emergency situation. Adrenaline makes the heart beat faster. At the same time, it stimulates the constriction of the arterioles of the digestive system reducing the blood supply of the digestive system which makes the mouth dry.

Solution C.7.
If one adrenal gland is removed, the other one gets enlarged. This is to meet the requirement of hormones produced by the body.

Solution C.8.

  1. Diabetes mellitus:
    Cause – under secretion of Insulin hormone
    Symptoms – excretion of great deal of urine with sugar, Person feels thirsty and loss of weight. In severe cases, the person may lose the eye sight.
  2. Diabetes insipidus:
    Cause – Under secretion of Anti-diuretic hormone
    Symptoms – frequent urination resulting in loss of water from body and the person feels thirsty.

Solution C.9.
The Himalayan soil is deficient in iodine. Thus, the food grown in such soil also becomes iodine deficient. Due to this reason, when Himalayan people consume iodine deficient food, they do not get the proper intake of iodine. Therefore, people living in the low Himalayan hilly regions often suffer from goitre.

Solution C.10.

S.No.

 Source

Gland cells

 Hormone produced Chief function Effect of over secretion

Effect of under secretion

1.

 Thyroid thyroxine Regulates basal metabolism Exophthalmic goiter Simple goiter, cretinism in children and myxoedema in adults
2. Beta cells of Islets of Langerhans Insulin Promotes glucose utilization by the body cells Hypoglycemia

Diabetes mellitus

3.

 Anterior pituitary Growth hormone Promotes growth of the whole body Gigantism Dwarfism
4. Posterior pituitary Vasopressin Increases reabsorption of water from kidney tubule More concentrated and less amount of urine

Diabetes insipidus

Solution C.11.

Gland

 Hormone secreted

Effect on body

Thyroid

 Thyroxine Regulates basal metabolism
Pancreas (“beta” cells) Insulin

Controls blood sugar level

Adrenal gland

 Adrenaline Increases heart beat
Anterior pituitary Thyroid stimulating hormone

Stimulates thyroxine secretion

Solution C.12.

Gland

 Secretions

Effect on body

Ovary

 oestrogen development of secondary sexual characteristics
Alpha cells of islets of Langerhans Glucagon

Raises blood sugar level

Thyroid

 Hypersecretion of thyroxine Protruding eyes
Anterior pituitary Hypersecretion of Growth hormone

Gigantism

Solution D.1.

Hormonal Response

 Nervous Response
Hormonal response is slow.

Nervous response is immediate.

Hormones are chemical messengers transmitted through blood stream.

 Nerve impulses are transmitted in the form of electro-chemical responses through nerve fibres.
This response brings about a specific chemical changes. Therefore it regulates the metabolism.

This response does not bring any chemical change during metabolism.

Solution D.2.

Action of Hormones

 Action of Nerves
The effect of hormones is wide spread in the body. They can show their effect on more than one target site at a time.

The nerve response affects only particular glands.

The effect of hormones can be short-lived or long lasting.

 The effect of nervous response is always short-lived.
Cannot be modified by the previous learning experiences.

Can be modified by the previous learning experiences.

Solution E.1.
a. Glucagon: Alpha cells of the islets of Langerhans
Insulin: Beta cells of the islets of Langerhans
b. Insulin: It maintains the levels of glucose (sugar) in the blood.
Glucagon: It raises the blood glucose levels by stimulating the breakdown of glycogen to glucose in the liver.
c. An endocrine gland is one which does not pour its secretions into a duct, while an exocrine gland is a gland which pours its secretions into a duct. Because the pancreas produces hormones such as insulin, glucagon and somatostatin directly into the blood and not into a duct, it functions as an endocrine gland. Because it secretes the pancreatic juices for digestion via a duct, it functions as an exocrine gland. Hence, the pancreas is an exo-endocrine gland.
d. Insulin is not administered orally because the digestive juices degrade insulin, and thus the insulin is ineffective in the body.
e. Islets of Langerhans
f. The pancreas is located in the abdomen behind the stomach.

Solution E.2.
(a) This portion is located in the neck region above the sternum.
(b) 1- Larynx, 2 – Thyroid gland, 3 – Trachea
(c) Larynx is the voice box containing vocal cords. It helps in producing sound.
Thyroid gland produces thyroxine and calcitonin which are essential hormones.
Trachea is the wind pipe that helps in passing air to and from the respiratory system while breathing.
(d) Structure 2 is the thyroid gland. It is an endocrine gland, so it is ductless and pours its secretions directly into the blood. Hence, there is no duct.

Solution E.3.
(a) 1- Pituitary gland, 2 – thyroid gland, 3 – pancreas, 4 – adrenal glands
(b) All the glands shown in the above diagram are endocrine glands. They secrete essential hormones and pour their secretions directly into the blood.
(c) Iodine is essential for the normal working of thyroxine.

Solution E.4.
(a) A
(b) Hormone secreted by the endocrine gland is shown in the image A to be moving only in one direction i.e. towards the target organ. But actually the hormones poured into the blood stream may have one or more target sites at a time. The arrows shown are carried to all parts by the blood and their effect is produced only in one or more specific parts.

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