Selina Class 10 Physics Solutions

Selina Concise Physics Class 10 ICSE Solutions Spectrum

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Selina Concise Physics Class 10 ICSE Solutions Spectrum

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 6 Spectrum. 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 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 Concise Physics Class 10 ICSE Solutions Radioactivity

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Selina Concise Physics Class 10 ICSE Solutions Radioactivity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 12 Radioactivity. 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 12 Radioactivity

Exercise 12(A)

Solution 1.

Three constituent of an atom are:

  1. Electrons: mass is 9.1 x 10-31 kg, charge is -1.6 x 10-19C
  2. Neutron: mass is 1.6749 x 10-27 kg, charge is zero.
  3. Protons: mass is 1.6726 x 10-27 kg, charge is +1.6 x 10-19 C

Solution 2.

Atomic number – the number of protons in the nucleus is called atomic number.
Mass number – the total number of nucleons in the nucleus is called mass number.

Solution 3.

  • The nucleus at the centre of atom, whose size is of the order of 10-15 m to 10-14 m.
  • The size of a nucleus is 10-5 to 10-4 times the size of an atom. It consists of protons and neutrons.
  • If Z is the atomic number and A is the mass number of an atom, then the atom contains Z number of electrons; Z number of protons and A – Z number of neutrons.
  • The atom is specified by the symbol ZXA where X is the chemical symbol for the element.

Solution 4.

Atomic number Z = 11
Mass number A = 23
Number of neutrons A – Z = 12

Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 1

Solution 5.

Isotopes: the atoms of the same element which have the same atomic number Z but differ in their mass number A are called isotopes.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 2

Solution 6.

Isobars: the atoms of different elements which have the same mass number A, but differ in their atomic number Z are called isobars.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 3

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 4

Solution 8.

Radioactivity: Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of α or β and γ radiations from the nuclei of atoms during their decay.
Example: uranium, radium.

Solution 9.

There will be no change in the nature of radioactivity. This is because radioactivity is a nuclear phenomenon.

Solution 10.

(a) Three types of radiations: Alpha, beta and gamma.
(b) Alpha and beta radiations
(c) Gamma radiations
(d) Gamma radiations
(e) Alpha radiations
(f) Beta radiations

Solution 11.

(a) Gamma radiations have zero mass.
(b) Gamma radiations have the lowest ionizing power.
(c) Alpha particles have lowest penetrating power.
(d) Alpha particle has positive charge equal to 3.2 x 10-19C and rest mass equal to 4 times the mass of proton i.e. 6.68 x 10-27 kg.
(e) The gas is Helium.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 5
(f) These radiations come from nucleus of the atom.

Solution 12.

  • Radiations labeled A, B and C are α, β and γ respectively.
  • Radiation labeled A is gamma radiation because they have no charge and hence under action of magnetic field they go undeflected.
  • Radiation B is alpha radiation because its mass is large and it would be deflected less in comparison to beta radiation. The direction of deflection is given by Fleming’s left hand rule. Also directions of deflection of alpha and beta radiations are opposite as they have opposite charge.

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 6

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 7
(b) The radioactive substances are kept in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions because they may cause biological damage.

Solution 15.

This is because alpha and beta particles are charged particles, but gamma rays are neutral particles.

Solution 16.

No, it is not possible to deflect gamma radiation in a way similar to alpha and beta particles, using the electric or magnetic field because they are neutral and hence do not deflected under the action of electric or magnetic field.

Solution 17.

Property

α-particleβ-particle

γ-particle

Nature

Stream of positively charged particles, i.e. helium nuclei.Stream of negatively charged particles, i.e. energetic electrons.Highly energetic electromagnetic radiation.
ChargePositive charge (Two times that of a proton) = + 3.2 x 10-19 C (or +2e)Negative charge = – 1.6 x 10-19 C (or -e)

No charge

Mass

Four times the mass of proton i.e., 6.68 x 10-27 kgEqual to the mass of electron, i.e. 9.1 x 10-31 kgNo mass (Rest mass is zero)
Effect of electric fieldLess deflectedMore deflected than alpha particles but in direction opposite to those of α particles

Unaffected

Solution 18.

Ionizing power of alpha radiation is maximum i.e., 10000 times of gamma radiation while beta particles have lesser ionizing power i.e., 100 times of gamma radiation and gamma radiation have least ionizing power.
Penetration power is least for alpha particle and maximum for gamma radiation.

Solution 19.

  • Speed of α radiation is nearly 107 m/s.
  • Speed of β radiation is about 90% of the speed of light or 2.7 x 108 m/s.
  • Speed of γ radiation is 3 x 108 m/s in vacuum.

Solution 20.

  • Alpha radiations are composed two protons and two neutrons.
  • Beta particles are fast moving electrons.
  • Gamma radiations are photons or electromagnetic waves like X rays.
  • Alpha radiations have the least penetrating power.

Solution 21.

  • Gamma radiation are produced when a nucleus is in a state of excitation (i.e., it has an excess of energy). This extra energy is released in the form of gamma radiation.
  • Gamma radiations like light are not deflected by the electric and magnetic field.
  • Gamma radiations have the same speed as that of light.

Solution 22.

It will become singly ionized helium He+.

Solution 23.

Any physical changes (such as change in pressure and temperature) or chemical changes (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) do not alter the rate of decay of the radioactive substance. This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. The radioactivity should therefore be the property of the nucleus. Thus radioactivity is a nuclear phenomenon.

Solution 24.

On emitting a β particle, the number of nucleons in the nucleus (i.e. protons and neutrons) remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.

If a radioactive nucleus P with mass number A and atomic number Z emits a beta particle to form a daughter nucleus Q with mass number A and atomic number Z+1, then the change can be represented as follows:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 8
(a) Atomic number ‘Z’ is not conserved. It is increased by 1.
(b) Mass number A is conserved.

Solution 25.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 9

Solution 26.

(a) Atomic number decreases by 2.
(b) Atomic number increases by 1.
(c) Atomic number does not change.

Solution 27.

(a) After emitting an alpha particle the daughter element occupies two places to the left of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of α-decay, then the α-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 10
Thus, the resulting nucleus has an atomic number equal to (Z-2). Hence, it shifts two places to the left of the parent element in the periodic table.

(b) After emitting a -particle, the daughter element occupies one place to the right of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of β-decay, then the β-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 11
Thus, the resulting nucleus has an atomic number equal to (Z+1). Hence, it shifts one place to the right of the parent element in the periodic table.

(c) After emitting -radiation, the element occupies the same position in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of γ-decay, then the γ-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 12
Thus, the resulting nucleus has atomic number equal to Z. Hence, it occupies the same position as the parent element in the periodic table.

Solution 28.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 13

Solution 29.

(a) The composition of B – 82 protons and 126 neutrons.
(b) The composition of C – 83 protons and 125 neutrons.
(c) The mass number of nucleus A = no. of protons +no. of neurons = 84+128=212.
(d) Their will be no change in the composition of nucleus C.

Solution 30.

(a) The alpha particle was emitted.
(b) This is because the atomic number has decreased by 2 and mass number has decreased by 4.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 14

Solution 31.

(a) This is allowed.
(b) This is not allowed because mass number is not conserved.

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

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

Solution 34.

The atomic number of P decreases by 2 and mass no. decreases by 4 due to the emission of one alpha particle and then increases by 1 due to the emission of each beta particle, so the atomic number of Q formed after the emission of one alpha and two beta particles is same as that of P. Hence P and Q are the isotopes.

Solution 36.

(a) The mass number (A) of an element is not changed when it emits beta and gamma radiations.
(b) The atomic number of a radioactive element is not changed when it emits gamma radiations.
(c) During the emission of a beta particle, the mass number remains same.

Solution 37.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 17

Solution 38.

Radio isotopes: The isotopes of some elements with atomic number Z
Example: carbon (Z=6, A=14).
Radio isotopes are used in medical and scientific and industrial fields. Radio isotopes such as 92U232 are used as fuel for atomic energy reactors.

Solution 39.

Because they cannot penetrate the human skin.

Solution 40.

Gamma radiations have very high penetration power and can easily pass through the human body. Therefore they are used as radioactive tracers in medical science.

Solution 41.

When the number of neutrons exceeds much than the number of protons in a nuclei, it become unstable or radioactive.

Solution 42.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 18

Solution 43.

Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. Radiations from cobalt -60 are used to treat cancer by killing the cells in the malignant tumor of the patient.
The salt of weak radioactive isotopes such as radio-sodium chloride, radio-iron and radio-iodine are used for diagnosis. Such radio isotopes are called the tracers.

Solution 44.

a  <  β  <  γ

An α-particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

The penetrating power of β-particles is more than that of the α-particles. They can pass through nearly 5 m in air, through thin card sheet, and even through thin aluminium foil, but a 5 mm thick aluminium sheet can stop them.

Whereas, the penetrating power of γ-rays is high. It is about 104 times that of α-particles and 102 times that of β-particles. They can pass through 500 m in air or through 30 cm thick sheet of iron. Thick sheet of lead is required to stop them.

Solution 45.

Two main sources of nuclear radiations are:

  1. Radioactive fallout from nuclear plants and other sources.
  2. Disposal of nuclear waste.

These radiations are harmful because when these radiations falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 46.

The following safety measures must be taken in a nuclear power plant:

  1. The nuclear reactor must be shielded with lead and steel walls so as to stop radiations from escaping out to the environment during its normal operation.
  2. The nuclear reactor must be housed in an airtightbuilding of strong concrete structure which can withstand earthquakes, fires and explosion.
  3. There must be back up cooling system for the reactor core, so that in case of failure of one system, the other cooling system could take its place and the core is saved from overheating and melting.

Solution 47.

The radioactive material after its use is known as nuclear waste.
It must be buried in the specially constructed deep underground stores made quite far from the populated area.

Solution 48.

Three safety precautions that we would take while handling the radioactive substances are:

  1. Put on special lead lined aprons and lead gloves.
  2. Handle the radioactive materials with long lead tongs.
  3. Keep the radioactive substances in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions.

Solution 49.

Radioactive substance should not be touched by hands because these radiations are harmful; when radiation falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 50.

Background radiation: These are the radioactive radiations to which we all are exposed even in the absence of an actual visible radioactive source.
There are two sources of background radiation:

  1. Internal source: potassium, carbon and radium are present inside our body.
  2. External sources: cosmic rays, naturally occurring radioactive elements such as radon-222 and solar radiation.

It is not possible for us to keep ourselves away from the background radiations.

Solution 1 (MCQ).

α or β
Hint: In a single radioactive decay, α and β particles are never emitted simultaneously. There will be either an α-emission or a α β-emission, which may be accompanied by γ emission.

Solution 2 (MCQ).

The nucleus of the atom.
Hint: Radioactivity is a nuclear phenomenon. Hence, electrons come out from the nucleus. Electron is created as a result of decay of one neutron into a proton inside the nucleus and it is not possible for the electron to stay inside the nucleus; thus, it is spontaneously emitted.

Solution 3 (MCQ).

(a) α – particles

An α – particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

Solution 4 (MCQ).

(b) β-particles

β-particles are negatively charged, so they get deflected by the electric and magnetic fields. The deflection of β-particle is more than that of a-particle since a β-particle is lighter than the α-particle. Whereas, gamma radiations are not deflected by the electric and magnetic fields since they are not charged particles.

Exercise 12(B)

Solution 1.
Energy released by combining of nuclei of an atom or by decay of an unstable radioactive nucleus during a nuclear reaction i.e., during fusion or fission is known as nuclear energy.

Solution 1 (MCQ).
(d) neutron
A neutron is used in nuclear fission for bombardment.

Solution 1 (Num).
1 a.m.u. = 1.66 × 10-27 kg
→ 0.2 a.m.u. = 0.2 × 1.66 × 10-27 kg
Δm = 0.332 Δ 10-27 kg
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 19

Solution 2.
Einstein’s mass-energy equivalence relation : E = Δmc2
Where E is the energy released due to the loss in the mass Δm and c is the speed of light.

Solution 2 (MCQ).
(d) 107 K
To make the fusion possible, a high temperature of approximately 107 K and high pressure is required.

Solution 2 (Num).
Given that Δm = 0.0265 a.m.u.
1 a.m.u. liberates 931.5 MeV of energy. Thus, energy liberated equivalent to 0.0265 a.m.u. is
= 0.0265 a.m.u. × 931.5 MeV
= 24.7 meV

Solution 3.
(a) The mass of atomic particles is expressed in terms of atomic mass unit (a.m.u.). 1 a.m.u. of mass is equivalent to 931 MeV of energy.
(b) Mass of proton = 1.00727 a.m.u.
Mass of neutron = 1.00865 a.m.u.
Mass of electron = 0.00055 a.m.u.

Solution 4.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 20

Solution 5.
(a) \(_{ 92 }^{ 235 }{ U }\) and \(_{ 92 }^{ 235 }{ U }\)
(b) Experimentally it is found that isotope of \(_{ 92 }^{ 235 }{ U }\) is more easily fissionable because the fission of
is \(_{ 92 }^{ 235 }{ U }\) possible by sloe neutron unlike \(_{ 92 }^{ 238 }{ U }\) where fission is possible only by the fast neutrons.
(c) Slow and fast both.

Solution 6.
Nearly 190 MeV of energy is released due to fission of one nucleus of \(_{ 92 }^{ 235 }{ U }\) The cause of emission of this energy is the loss in mass i.e., the sum of masses of product nuclei is less than the sum of mass of the parent nucleus and neutron.

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 21

Solution 8.
A chain reaction is a series of nuclear fissions whereby the neutrons produced in each fission cause additional fissions, releasing enormous amount of energy.
It is controlled by absorbing some of the neutrons emitted in the fission process by means of moderators like graphite, heavy water, etc. then the energy obtained in fission can be utilized for the constructive purposes

Solution 9.
(i) It is used in a nuclear bomb.
(ii) It is used in a nuclear reactor where the rate of release of energy is slow and controlled which is used to generate electric power.

Solution 10.

Radioactive decayNuclear Fission
It is a self process.It does not occur by itself. Neutrons are bombarded on a heavy nucleus.
The nucleus emits either the a or b particles

with the emission of energy in form of g rays which is not very large.

A tremendous amount of energy is released when a heavy nucleus is bombarded with neutrons and the nucleus splits in two nearly equal fragments.
The rate of radioactive decay cannot be controlled.The rate of nuclear fission can be controlled.

Solution 11.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.
When uranium with Z = 92 is bombarded with neutron, it splits into two fragments namely barium (Z = 56) and krypton (Z = 36) and a large amount of energy is released which appears due to decrease in the mass.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 22
Nuclear fusion is also known as thermo-nuclear reaction. This is because nuclear fusion takes place at very high temperature.

Solution 12.
When two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion between them becomes too strong that they do not fuse. Thus, nuclear fusion is not possible at ordinary temperature and ordinary pressure.
Hence to make the fusion possible, a high temperature of approximately 107 K and high pressure is required. At such a high temperature, due to thermal agitations both nuclei acquire sufficient kinetic energy so as to overcome the force of repulsion between them when they approach each other, and so they get fused.

Solution 13.

Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 23
(b) In all three deuterium nuclei fuse to form a helium nucleus with a release of 21·6 MeV energy.
(c) When two deuterium nuclei (\(_{ 1 }^{ 2 }{ H }\)) fuse, nucleus of helium isotope (\(_{ 3 }^{ 2 }{ He }\) ) is formed and 3·3 MeVenergy is released. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus (\(_{ 4 }^{ 2 }{ He }\)
) and 18·3 MeV of energy is released in this process.

Solution 14.
(a)
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 24

Solution 15.
(a) Nuclear fusion
(b) Nuclear fission

Solution 16.
Both fission and fusion create release of neutrons and large amount of energy.
Nuclear fission: A heavy nucleus splits in two nearly equal light fragments when bombarded with neutrons. It is possible at very ordinary temperature and pressureNuclear fusion: Two light nuclei combine to form a heavy nucleus at very high temperature and high pressure. Possible only at a very high temperature (≈107 K) and a very high pressure.

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 25

Solution 18.
The source of energy in the Sun and stars is the nucleus fusion of light nuclei such as hydrogen present in them in their inner part. This takes place at a very high temperature and high pressure due to which helium nucleus is formed with the release of high amount of energy.

Solution 19.
(a) Nuclear fission
(b) Nuclear fusion

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Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens

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Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens. 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 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 imageVirtual 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 lensPosition of objectNature of imageSize of image
ConvexBetween optic centre and focusVirtual and uprightMagnified
ConvexAt focusReal and invertedVery much magnified
ConcaveAt infinityVirtual and uprightHighly diminished
ConcaveAt any distanceVirtual and uprightDiminished

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 Physics Class 10 ICSE Solutions Sound

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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 Current Electricity

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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.
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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.
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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 cellTerminal 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 cell2. 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.
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Solution 9.
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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.
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Solution 2.
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Solution 3.
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Solution 4.
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Solution 5.
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Solution 6.
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Solution 7.
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Solution 8.
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Solution 9.
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Solution 10.
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Solution 11.
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Solution 12.
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Solution 13.
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Solution 14.
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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.
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Solution 21.
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Solution 23.
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Solution 24.
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Solution 25.
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Solution 26.
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Solution 27.
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Solution 28.
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Solution 29.
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Solution 30.
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Exercise 8(C)

Solution 1.
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Solution 2.
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Solution 3.
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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.
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Solution 15.
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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.
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Solution 2.
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Solution 3.
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Solution 4.
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Solution 5.
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Solution 6.
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Solution 7.
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Solution 8.
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Solution 9.
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Solution 10.
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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
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(i.e., current is halved), hence heating (= I2Rt) in each bulb becomes one-fourth, so each bulb appears less bright.

Solution 11.
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Solution 12.
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Solution 13.
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Solution 14.
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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.
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