# Oswal 36 Sample Question Papers ISC Class 12 Physics Solutions

## Section-A

**Answer 1.**

(A) (i) (a) zero

(ii) (b) Alexander’s dark band is formed between primary and secondary rainbow because light is absorbed in this region.

(iii) (c) 0.3853 × 10^{6}C

(iv) (a) 1

(v) (b) ε_{r} F

$$\text{(vi)\space(b)\space}\frac{2q}{\epsilon_{0}}$$

**Explanation :**

$$\phi =\frac{q_{emc}}{\epsilon_{0}} =\frac{+3q -q}{\epsilon_{0}}\\=\frac{2q}{\epsilon_{0}}$$

(vii) (d) gamma rays

(B) (i) Magnetic induction due to an infinitely

$$\text{long straight wire} =\frac{\mu_{0}\text{I}}{2\pi a}\space\text{...(i)}$$

Magnetic induction due to the circular coil

$$\text{carrying current at its centre =}\frac{\mu_{0}\text{nI}}{\text{2a}}\\\text{...(ii)}$$

Dividing equations (i) and (ii),

$$=\frac{\frac{\mu_{0}\text{I}}{2\pi a}}{\frac{\mu_{0}\text{nI}}{2a}} =\frac{1}{n\pi}$$

(ii) Magnetic induction due to circular coil

$$\text{carrying current at its centre =}\frac{\mu_{0}n\text{I}}{2a}\\\space\text{...(i)}$$

Magnetic induction due to the arc =

$$\frac{\mu_{0}\text{Il}}{4\pi\text{R}^{2}}$$

According to question,

Magnetic induction due to circular coil =

Magnetic induction due to the arc

$$\frac{\mu_{0}nl}{2a} =\frac{\mu_{0\text{Il}}}{4\pi\text{R}^{2}}\\a =\frac{2\pi\text{R}^{2}n}{l}$$

(iii) In a n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers.

(iv) Reverse biased p-n junction has more resistance.

(v) Silver

(vi) If the source of light is linear, then the wavefront is cylindrical where square of the amplitude is inversely proportional to the distance.

If the source of light is a point source, then the wavefront is spherical where the amplitude is inversely proportional to the distance.

The plane wavefront is appeared at infinite distance from the source and the rays are always parallel straight lines.

(vii) Yes the rays which appear to diverge from the virtual image are real. Hence, they can be focussed on the screen of the camera to form a real image.

## Section-B

**Answer 2.**

S. No. | Drift velocity | Thermal velocity |

(a) | When a potential difference is applied across a metal, the free electrons drift with a small constant velocity in the direction opposite to the applied field is called drift velocity. | The conduction electrons move in all directions with all possible velocities called thermal velocities. |

(b) | Drift velocity is or the order or 10^{–4} m/s. It is very small. | Thermal velocity is very large and is of the order of 10^{5} m/s. |

(c) | When the electric field is zero, drift velocity is also zero. | Electrons move with thermal velocity even in the absence as well as in the presence of electric field. |

**OR**

(ii) The unit of resistance R is ohm =

$$\frac{\text{Volt}}{\text{ampere}} = \frac{\text{joule/coulomb}}{\text{ampere}}\\=\\\frac{\text{(newton × metre)}/(\text{ampere}×\text{sec})}{\text{ampere}}\\=\\\frac{(\text{kg}×\text{metre}×\text{sec}^{\normalsize-2})×\text{metre}}{\text{ampere}^{2}×\text{sec}}$$

= kg metre^{2} sec^{–3} ampere^{–2}

∴ Dimensions of resistance are [ML^{2}T^{–3}A^{–2}]

$$\text{Electric conductance (}\sigma) =\frac{1}{\text{R}}\\\Rarr\\\text{Dimensions of}\space\sigma =\frac{1}{\text{ML}^{\normalsize2}\text{T}^{-\normalsize3}\text{A}^{\normalsize-2}}$$

Dimensions of electric conductance are [M^{–1}L^{–2}T^{3}A^{2}]

**Answer 3.**

$$\text{(i) Potential due to point change}\\∝\frac{1}{r}$$

(ii) Potential is not depend on r (r < radius of sphere)

**Answer 4.**

It states that the line integral of the magnetic field along a closed path is proportional to the current

passing through that closed path.

$$\oint\space\text{B.dl =}\mu_0\text{I}$$

where μ_{0} is the permeability of free space and I is the current enclosed by the closed path.

**Answer 5.**

(i) Voltage and current are always in same phase in a purely resistive a.c. circuit. Angle between them is zero.

(ii) Voltage leads current by π/2 in a purely inductive a.c. circuit.

**Answer 6.**

$$k = \frac{\text{2RB}_{\text{H}}}{\mu_{0}\text{N}}$$

N is doubled i.e., 2N

B_{H} is reduced to half i.e., B_{H}/2

$$\text{Then,\space}k' =\frac{\text{2RB}_{\text{H}}}{2\mu_{0}\text{2N}}\\k' =\frac{\text{2RB}_{\text{H}}}{4\mu_{0}\text{N}} =\frac{k}{4}$$

**Answer 7.**

(a) Microwaves are reflected by metal surfaces.

(b) Microwaves have low frequency and high wavelength.

**Answer 8.**

(i)

**(ii) Formation of energy bands in solids:** An isolated atom has well defined energy levels. When large number of such atom get together to form a real soild, their individual energy levels overlap and get completely modified. Instead of discrete value of energy of electrons, the value lies in a certain range.

The collection of these closely packed energy levels are said to form an energy band. Two type of such bands formed in solid are called valence band and conduction band. The band formed by filled energy levels is known as valence band, whereas partially filled or unfilled band is known as conduction band.

The two bands are generally separated by a gap called energy gap or forbidden gap.

**Conductors :** In case of conductors, electrons fill the conduction band, partially the overlapping of both the bands i.e., valence and conduction band also take place. This shows that no forbidden gap is present.

**Insulators :** In case of insulator, forbidden gap is very large (> 6 eV). On the other hand, its valence band is fully filled with the electrons, whereas its conduction band is empty.

**Semiconductor :** In case of semiconductors the conduction band is empty and valence band is filled with electrons. Like insulators, forbidden energy gap is not so large in case of semiconductors. The gap is very small. The energy gap is nearly of 1 eV.

## Section-C

**Answer 9.**

Magnitude of electric dipole is

p = q × 2l

= (10 × 10^{–6}) × (10 × 10^{–3}) = 10^{–7} Cm

$$\text{In End-on position,}\space\\\text{E} =\frac{1}{4\pi\epsilon_{0}}.\frac{\text{2p}}{r^{3}}\\=\space\frac{9×10^{9}×2×10^{\normalsize-7}}{(15×10^{\normalsize-2})^{3}}$$

= 2.66 × 10^{5} N/C and the direction is antiparallel to the dipole moment.

**Answer 10.**

(i) (a) Resistance of bulb is given by

$$\text{R =}\frac{\text{V}^{2}}{\text{P}}\\\text{For same voltage}\\\frac{\text{R}_{1}}{\text{R}_{2}} =\frac{\frac{\text{V}^{2}}{\text{P}_{1}}}{\frac{\text{V}^{2}}{\text{P}_{2}}} =\frac{\text{P}_{2}}{\text{P}_{1}} =\frac{100}{60}=\frac{5}{3}$$

(b) When bulbs are connected in series, current (I) is same

$$\frac{\text{P}_{1}}{\text{P}_{2}} =\frac{\text{I}^{2}\text{R}_{1}}{\text{I}^{2}\text{R}_{2}} =\frac{\text{R}_{1}}{\text{R}_{2}} =\frac{5}{3}$$

(c) Ratio of potential difference across bulbs in series (I is same)

$$\frac{\text{V}_{1}}{\text{V}_{2}} =\frac{\text{IR}_{1}}{\text{IR}_{2}} =\frac{5}{3}$$

**OR**

(ii) Given : J = 2.5 A/m^{2}, E = 15 V/m

$$\text{Resistivity,\space ρ} =\frac{\text{E}}{\text{J}} \\=\frac{15}{2.5} = 6Ω\text{m}$$

**Answer 11.**

(i) Given magnetic flux, φ1 = 47 × 10^{–3} Wb

φ_{2} = 23 × 10^{–3} Wb

Change in flux (dφ) = φ_{2} – φ_{1}

∴ dφ = (23 × 10^{–3}) – (47 × 10^{–3})

dφ = – 24 × 10^{–3}

dt = 0.06 sec.

Emf induced in the coil (e) =

$$-\text{N}\frac{d\phi}{\text{dt}}$$

$$e =-100×\frac{(\normalsize-24)×10^{\normalsize-3}}{0.06}$$

= 100 × 400 × 10^{–3} = 40 V

Current flowing through coil will be

$$\text{I} =\frac{\text{V}}{\text{R}}\\\text{I =}\frac{40}{16} = 2.5\space\text{A}$$

(ii) First, we need to check

$$\frac{\text{V}}{\text{f}}\space\text{ratio of transformer.}\\\frac{160}{40} = \frac{200}{50} = 4\\\frac{\text{V}}{f}\space\text{ratio of transformer is constant.}$$

So, eddy current loss in transformer will be directly proportional to square of frequency

∴ P_{e} ∝ f^{2}

$$\frac{\text{P}e_{2}}{\text{P}e_{1}} =\bigg(\frac{f_{2}}{f_{1}}\bigg)^{2}\\\frac{\text{P}e_{2}}{\text{P}e_{1}} =\bigg(\frac{50}{40}\bigg)^{2}\\\text{P}e_{2} =\frac{25}{16}×240\\= 375\space\text{W}$$

New eddy current loss will be 375 W.

**Answer 12.**

We know

$$\omega =\frac{n_{\text{V}} - n_{\text{R}}}{n_{\text{Y}}-1}$$

Given n_{V} = 1.62, n_{R} = 1.52 and n_{Y} = 1.59

So,

$$\omega = \frac{1.62 -1.52}{1.59 -1} = 0.1695$$

Angular dispersion between red and violet rays is :

θ = δ_{V} – δ_{R}

θ = w × δ_{Y}

So, θ = 0.169 × 40° = 6.78°

**Answer 13.**

(i) When the sun shines upon falling raindrops, an observer with his back towards the sun sees concentric arcs of spectral colours hanging in the sky. These coloured arcs, which have their common centre on the line joining the sun and the observer, are called ‘rainbow’. Usually, two rainbows are seen, one above the other.

The lower one is called the ‘primary’ rainbow and the higher one is called the ‘secondary’ rainbow. The primary rainbow is brighter and narrower, having its inner edge violet and the outer edge red. The secondary rainbow, which is comparatively fainter, has reverse order of colours.

**Formation of Primary Rainbow :** Rainbows are formed by the dispersion of sunrays in raindrops. The primary rainbow is formed when sunrays, after suffering one internal reflection in the raindrops, emerge at minimum deviation and enters the observer’s eye. In figure, P_{1} and P_{2} are two raindrops, E is the observer’s eye and S is the sun. The sunrays fall on the drops parallel to SE. If the rays are deviated (and dispersed) by the drops so as to arrive at the observer, the observer would receive intense light in the those directions in which the rays suffer minimum deviation. It can be shown that he would receive red light in a direction making an angle of 43°, and intense violet light in a direction making an angle of 41° with the line SE produced. The drops

sending the intense red and violet light to the observer lie on concentric circles which generate cones of semivertical angles of 43° and 41° respectively with common vertex at E. Thus, the observer sees concentric coloured arcs of which the innermost is violet and the outermost is red. The intermediate colours lie in between. This is the primary rainbow.

**Formation of Secondary Rainbow :** The secondary fainter rainbow is formed by the sunrays undergoing two internal reflections in the raindrops and emerging at minimum deviation, as occurring in drops S_{1} and S_{2} in the figure. The semivertical angles for this bow are 51° for the red rays to 54° for the violet rays.

As such, the order of colours is reverse of that in the primary rainbow.

**OR**

(ii) (a) **Both the Lenses are Convex :** Suppose two thin convex lenses L_{1} and L_{2} of focal lengths

f_{1} and f_{2} are placed in contact in air having a common principal axis. A point object O is placed on the principal axis at a distance u from the first lens L_{1}. Its image would be formed by the lens L_{1} alone at I’. distant v’ (say) from L_{1}. Then, from the lens formula, we have,

$$\frac{1}{v'} - \frac{1}{u} =\frac{1}{f_{1}}\space\text{...(i)}$$

I′ serves as a virtual object for the second lens L_{2} which forms a final image I at a distance v (say) from it.

Then, we have

$$\frac{1}{v}-\frac{1}{v'} = \frac{1}{f_{2}}.\space\text{...(ii)}$$

From equations (i) and (ii), we get

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{1}} + \frac{1}{f_{2}}\space\text{...(iii)}$$

If we replace these two lenses by a single lens which forms the image of an object placed distant u from it at a distance v, then the focal length F of this equivalent lens would be given by

$$\frac{1}{v} - \frac{1}{u} =\frac{1}{f}\space\text{...(iv)}$$

From equations (iii) and (iv), we get

$$\frac{1}{\text{F}} = \frac{1}{\text{f}_{1}} + \frac{1}{\text{f}_{2}}.$$

**(b) One lens is convex and the other is concave :** Suppose the focal length of the convex lens is f_{1} and that of the concave lens is f_{2}. If F be the focal length of the equivalent lens, then :

$$\frac{1}{\text{F}} = \frac{1}{\text{f}_{1}} + \frac{1}{-\text{f}_{2}} = \frac{1}{\text{f}_{1}}-\frac{1}{\text{f}_{2}}\\\text{or\space F = }\frac{f_{1}f_{2}}{f_{2}-f_{1}}$$

If f_{1} > f_{2}, then F is negative and the combination will behave like a concave lens.

If f_{1} < f_{2}, then F is positive and the combination will behave like a convex lens.

If f_{1} = f_{2}, then F is infinite and the combination will behave like a plane glass plate.

**Answer 14.**

Let a_{1} and a_{2} be the amplitudes and I_{1} and I_{2} the intensities of the light waves emitted from the sources.

The intensity ratio is 100 : 1. Then, we have

$$\frac{\text{I}_{1}}{\text{I}_{2}} = \frac{\text{a}_{1}^{2}}{\text{a}_{2}^{2}} =\frac{100}{1}\\\text{or\space}\frac{a_{1}}{a_{2}} =\frac{10}{1}\space\text{...(i)}$$

In interference, the maximum and minimum resultant amplitudes are (a_{1} + a_{2}) and (a_{1} – a_{2}) respectively.

$$\therefore\space\frac{\text{Maximum intensity}\space\text{I} _\text{max}}{\text{Minimum intensity}\space\text{I}_{\text{min}}}=\\\frac{(a_{1} + a_{2})^{2}}{(a_{1} - a_{2})^{2}}$$

But from equation (i), a_{1} = 10a_{2}

$$\therefore\space\frac{\text{I}_{\text{max}}}{\text{I}_{\text{min}}} = \frac{(10a_{2} + a_{2})^{2}}{(10a_{2}-a_{2})^{2}} =\frac{121}{81}$$

**Answer 15.**

(i) Angular momentum = nh/2π

$$\therefore\space n=\frac{3.17× 10^{\normalsize-34}×2×3.14}{(6.63×10^{-34})}$$

n = 3

(ii) Work function W_{0} = hν_{0}

W_{0} = 5.1 eV = 5.1 × 1.6 × 10^{-19} J

(∵ 1 eV = 1.6 × 10^{–19} J)

∴ The threshold frequency, v_{0}

$$\frac{\text{W}_{0}}{h} =\frac{(5.1×1.6×10^{\normalsize-19})}{(6.6×10^{-34})}\\ = 1.2×10^{15}\space\text{Hz}$$

**Answer 16.**

Let e, m, v be the charge, mass and velocity of the electron and r be the radius of the orbit. Positive charge on the nucleus is Ze. In case of hydrogen atom, Z = 1. Centripetal force is provided by electrostatic force of attraction. Therefore,

$$\frac{mv^{2}}{r} = \frac{1}{4\pi\epsilon_{0}}\frac{Ze×e}{r^{2}}\\\text{mv}^{2} =\frac{Ze^{2}}{4\pi\epsilon_{0}r}\space\text{...(i)}$$

By first postulate :

$$\text{mvr} =\frac{nh}{2\pi}\space\text{...(ii)}$$

Where n is the quantum number.

Squaring equation (ii) and dividing by equation (i), we get

$$\frac{m^{2}v^{2}r^{2}}{mv^{2}} =\frac{\frac{n^{2}h^{2}}{4\pi^{2}}}{\frac{Ze^{2}}{4\pi\epsilon_{0}r}}\\\text{Then,\space r =}\frac{n^{2}h^{2}\epsilon_{0}}{\pi Ze^{2}m}$$

For, hydrogen atom, Z = 1,

$$r =\frac{n^{2}h^{2}\epsilon_{0}}{\pi e^{2}m}$$

**Answer 17.**

**(i) Correct:** Intrinsic semiconductors have no impurities in it. By doping it with phosphorus or arsenic, its conductivity can be increased.

**(ii) Correct:** The electrical conductivity of an extrinsic semiconductor increases with a rise in temperature.

**(iii) Correct:** At the absolute zero temperature, an intrinsic semiconductor behaves like an insulator. An intrinsic semiconductor at absolute zero temperature has electrons only in the valence bond.

## Section-D

**Answer 18.**

(i) (a) For given coil,

Area = 100 m^{2}, q = 90° – 60° = 30°

B = 10^{–1}T

Flux, Φ = ?

When a coil having area A is placed in magnetic field B [where θ is the angle between area vector and magnetic field vector]

Φ = BAcos θ

= 10^{–1} × 100 cos 30°

$$= 10\frac{\sqrt{3}}{2} =5\sqrt{3}\space\text{Wb} $$

Now, magnetic field is reduced to zero in 10^{–3}s.

$$\text{i.e.\space}\Delta\phi =5\sqrt{3} - 0 = 5\sqrt{3}\space\text{Wb}$$

$$\text{Induced emf =}-\frac{\Delta\phi}{\Delta t}=\\\frac{5\sqrt{3}}{10^{\normalsize-3}} =-5000\sqrt{3}\space\text{volt}$$

(b) Magnetic dipole moment.

(c) Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.

If a current i produces of deflection f in the galvanometer,

$$\text{the current sensitivity is}\space\frac{\phi}{i}$$

$$\text{C}_{s} =\frac{\phi}{t} = \frac{\text{NAB}}{\text{C}},$$

where C is the torsional constant of the curve.

**OR**

(ii) (a) The magnetic field at any point due to an element of a current carrying conductor is:

(1) directly proportional to

1. Strength of current, I

$$\text{2.\space Length of element}\space\vec{\text{dl}}$$

3. sine of angle between the elements in the direction of current and the line joining the element to the point P.

i.e. dB ∝ I, dB ∝ dl, dB ∝ sin θ

(2) inversely proportional to the square of the distance r of the point P from the centre of the element i.e.

$$\text{dB}∝\text{I}, \text{dB}∝\text{dl},\text{dB}∝\text{sin}\space\theta$$

(2) inversely proportional to the square of the distance r of the point P from the centre of the element i.e.

$$\text{dB}∝\frac{1}{r^{2}}$$

The magnetic field at P is

$$\text{dB}∝\frac{\text{Idl sin}\theta}{r^{2}}\\\text{dB = }\frac{\mu_{0}}{4\pi}.\frac{\text{Idl sin}\theta}{r^{2}}$$

(b) To convert galvanometer to ammeter

To convert galvanometer to voltmeter

(c) A combination of two isolated, equal and opposite magnetic poles separated by a small distance constitutes a magnetic dipole. A bar magnet suspended in a uniform magnetic field experiences a torque and so it sets with its axis parallel to the field. A current loop also experiences a torque in a magnetic field due to which it sets with its axis parallel to the field.

**Answer 19.**

(i) (a) According to Bohr’s theory, a hydrogen atom consists of a nucleus with a positive charge Ze, and a single electron of charge –e which revolve around it in a circular orbit or radius r. Here Z is the atomic number and for hydrogen Z = 1. The electrostatic force of attraction between the nucleus and the electron is

$$\text{F} = k\frac{(Ze)(e)}{r^{2}} =k\frac{Ze^{2}}{r^{2}}$$

To keep the electron in its orbit the centripetal force on the electron must be equal to the electrostatic attraction. Therefore,

$$\frac{mv^{2}}{r} =\frac{kZe^{2}}{r^{2}}\\\text{or}\qquad \text{mv}^{2} =\frac{kZe^{2}}{r}\space\text{...(i)}\\\text{or\space}r =\frac{kZe^{2}}{mv^{2}}\space\text{...(iii)}$$

where m is the mass of the electron and v, its speed in an orbit of radius r.

Bohr’s quantisation condition for angular momentum is

$$\text{L = mvr =}\frac{nh}{2\pi}\\\text{or\qquad}r =\frac{nh}{2\pi mv}\space\text{...(iii)}$$

From equations (ii) and (iii), we get

$$\frac{kZe^{2}}{mv^{2}} =\frac{nh}{2\pi mv}\\\text{or\space v =}\frac{2\pi kZe^{2}}{nh}\space\text{...(iv)}$$

Substituting this value of v is equation (iii), we get

$$r =\frac{nh}{2\pi m}.\frac{nh}{2\pi Ze^{2}}\\\text{or\space} r =\frac{n^{2}h^{2}}{4\pi^{2}mkZe^{2}}\\\Rarr\qquad\space r_{n} =\frac{n^{2}h^{2}}{4\pi^{2}mkZe^{2}}$$

(b)

(ii) (a) (1) The electrons are permitted to circulate only in those orbits in which the angular momentum of an

$$\text{electron is an integral multiple of}\space\bigg(\frac{h}{2\pi}\bigg);\\\text{h being Planck’s constant.}\\\text{L = mvr} =\frac{nh}{2\pi}, n=1,2,3$$

(2) An atom can emit or absorb radiation in the form of discrete energy photons only when an electron jumps from a higher to a lower orbit or from a lower to a higher orbit respectively. If E_{1} and E_{2} are the energies associated with these permitted orbits, then the frequency n of the emitted or absorbed radiation is given by,

hv = E_{2} – E_{1}

(b) (1) Mass defect of a nucleus: The difference between the rest mass of a stable nucleus and the sum of the rest masses of its constituent nucleons is called its mass defect.

$$\text{Consider the nucleus}\space^{\text{A}}_{\text{Z}}\text{X}.$$

It has Z protons and (A– Z) neutrons. Therefore its mass defect will be.

Δm = Zm_{p} + (A – Z)m_{n} – m

Where, m_{p}, m_{n} and m are the rest masses of a proton, neutron and

$$\text{the nucleus}\space^{\text{A}}_{\text{Z}}\text{X}\space\text{respectively.}$$

**(2) Binding energy of a nucleus :** The B.E. of a nucleus may be defined as the energy required to break up a nucleus into its constituent protons and neutrons and to separate them to such a large distance that they may not interact with each other.

If Δm is the mass defect of the nucleus then, binding energy is given by

ΔE_{b} = (Δm) × c^{2}

**Answer 20.**

(i) Total internal refection.

(ii) The thickness of optical fibers is of the order of micrometers.

(iii) For long distance transmissions light in optical fibres enters from one side and comes out from another side like water in a pipe that is why, its popular name is light pipe.

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