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^{07}C
- (iv) (b) 2
- (v) (b) ε_{r} F
- (vi) (d) 16r
- (vii) (d) gamma rays
(B) (i) Magnetic induction due to an infinitely long straight wire =µ_{0}I/2πa ...(i)
Magnetic induction due to the circular coil carrying current at its centre=µ_{0}nI/2a ...(ii)
Dividing equations (i) and (ii)
$$=\frac{\frac{\mu_0I/2\pi}{2\pi a}}{\frac{\mu_0nI}{2a}}=\frac{1}{n\pi}$$
(ii) Magnetic induction due to circular coil carrying current at its centre = µ_{0}nI/2a ...(i)
According to question,
Magnetic induction due to circular coil = Magnetic induction due to the arc
$$\frac{\mu_0nI}{2a}=\frac{\mu_0Il}{4\pi R^2}\\\qquad a=\frac{2\pi R^2n}{l}$$
(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.
- (i) Potential ∝ (1/r)
- (ii) Potential is not depend on r.
Answer 3.
(i) In self-pollinated species the seeds are not formed because of self-incompatibility where there is no growth of pollen tube on the stigma of a flower that prevents the fusion of gametes and thereby developing into an embryo. Hence, in such species seed could not be formed.
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^{–6} m/s. It is very small. | Thermal velocity is very large and is of the order of 10^{6} 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
$$\text{(ii) The unit of resistance R is Ohm =}\frac{volt}{ampere}=\frac{joule/coulomb}{ampere}\\=\frac{(newton × metre)/(ampere × sec.)}{ampere}\\=\frac{(kg ×metre ×sec^{-2} )×metre}{ampere^{2} × sec}\\=kg metre^{\normalsize –2} sec^{\normalsize –3} ampere^{\normalsize –2}\\\therefore\text{Dimensions of resistance are} [ML^2T^{\normalsize–3}A^{\normalsize–2}]\\\text{Electric conductance} (\sigma)=\frac{1}{R}\\\Rarr\text{Dimensions of}\space \sigma = \frac{1}{ML^2T^{\normalsize-3}A^{\normalsize-2}}\\\text{Dimensions of electric conductance are}[M^{\normalsize-1}L^{\normalsize-2}T^3A^{2}] $$
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.
$$\oint B.dl=\mu_0I$$
- 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.
- (iii) Voltage lags behind current by π/2 in a purely capacitive a.c. circuit.
Answer 6.
$$k=\frac{2RB_H}{\mu_0N}\\\text{N is doubled i.e., 2N}\\B_H \text{is reduced to half i.e.}, B_H/2\\\text{Then},\space k^{‘}=\frac{2RB_H}{2\mu_02N}\\\space k^{‘}=\frac{2RB_H}{4\mu_0N}=\frac{k}{4}$$
Answer 7.
- Conditions for sustained interference:
- (i) The two sources of light must be coherent to emit light of constant phase difference.
- (ii) The amplitude of electric field vector of interfering wave should be equal to have greater contrast between intensity of constructive and destructive interference. When monochromatic light is replaced by white light, then coloured fringe pattern is obtained on the screen.
Answer 8.
(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.
$$\text{Magnitude of electric dipole is}\\p = q × 2l\\= (10 × 10^{\normalsize–6}) × (10 × 10^{\normalsize–3}) = 10^{\normalsize–7} Cm\\\text{In End-on position,}\space E=\frac{1}{4\pi\epsilon_0}.\frac{2p}{r^3}=\frac{9×10^{9}×2×10^{\normalsize-7}}{(15×10^{-2})^3}\\=2.66 × 10^5 N/C,\text{and the direction is antiparallel to the dipole moment.}\\\text{In Broad side-on position,}\space\text{E}=\frac{1}{4\pi\epsilon_0}.\frac{p}{r^{3}}\\=9×10^{9}×\frac{10^{-7}}{(15 ×10^{-2})^{3}}\\= 2.66 × 10^5\text{N/C and the direction is antiparallel to the dipole moment.}$$
Answer 10.
$$\text{(i) (a) Resistance of bulb is given by R}=\frac{V^2}{P}\\\text{For same voltage}\space\frac{R_1}{R_2}=\frac{V^2/P_1}{V^2/P_2}=\frac{P_2}{P_1}=\frac{100}{60}=\frac{5}{3}\\\text{(b) When bulbs are connected in series, current (I) is same}\\\frac{P_1}{P_2}=\frac{I^2R_1}{I^2R_2}=\frac{R_1}{R_2}=\frac{5}{3}\\\text{(c) Ratio of potential difference across bulbs in series (I is same)}\\\frac{V_1}{V_2}=\frac{IR_1}{IR_2}=\frac{5}{3}$$
OR
$$\text{(ii) Given : J =}2.5 A/m^{2},E=15V/m\\\text{Resistivity}\rho=\frac{E}{J}=\frac{15}{2.5}=6\varOmega m$$
Answer 11.
$$\text{(i) Given magnetic flux,}\space\phi_1=47 × 10^{-3}\space Wb\\\phi_2=27 × 10^{-3}\space Wb\\\text{Change in flux} (d\phi) =\phi_2-\phi_1\\d\phi=(23 × 10^{\normalsize–3}) – (47 × 10^{\normalsize–3})\\d\phi = – 24 × 10^{-3}\\dt = 0.06 sec.\\\text{Emf induced in the coil (e)=}-N\frac{d\phi}{dt}\\e=-100×\frac{(-24×10^{-3})}{0.06}\\=100 × 400 × 10^{\normalsize–3} = 40 V\\\text{Current flowing through coil will be}\\I=\frac{V}{R}\\I=\frac{40}{16}=2.5A$$
$$\text{(ii) First, we need to check}\space\frac{v}{f}\text{ratio of transformer.}\\\frac{160}{40}=\frac{200}{50}=4\\\qquad\qquad\space\frac{v}{f}\space\text{ratio of transformer is constant.}\\\text{So, eddy current loss in transformer will be directly proportional to square of frequency}\\\therefore\space P_e∝f^2\\\frac{Pe_2}{Pe_1}=\bigg(\frac{f_2}{f_1}\bigg)^2\\\frac{Pe_2}{Pe_1}=\bigg(\frac{50}{40}\bigg)^2\\P_{e2}=\frac{25}{16}×240=375 W\\\text{New eddy current loss will be 375 W.}$$
Answer 12.
$$\text{We know}\space\omega=\frac{n_V-n_R}{n_Y-1}\\\text{Given}\space n_V = 1.62, nR = 1.52 \text{and} n_Y = 1.59\\\text{So,}\space\space\omega=\frac{1.62-1.52}{1.59-1}\\= 0.1695\\\text{Angular dispersion between red and violet rays is :}\\\theta=\delta_v-\delta_R\\\theta=\omega×\delta_Y\\\text{So,}\space\theta=0.169 × 40º = 6.78º$$
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.
OR
$$\frac{1}{v^{‘}}-\frac{1}{u}=\frac{1}{f_1}\space\space….(i)\\\text{I′ serves as a virtual object for the second lens L2 which forms a final image I at a distance v (say) from it.}\\\text{Then, we have}\\\frac{1}{v}-\frac{1}{v^{‘}}=\frac{1}{f_2}\space…(ii)\\\text{From equations (i) and (ii), we get}\\\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}\space\space…(iii)\\\text{If we replace these two lenses by a single lens which forms the image of an object placed distant u from it at a distance v, then the focal length F of this equivalent lens would be given by}\\\frac{1}{v}-\frac{1}{u}=\frac{1}{F}\space\space…(iv)\\\text{From equations (iii) and (iv), we get}\\\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\\\textbf{(b) One lens is convex and the other is concave :}\text{Suppose the focal length of the convex lens is} f_1 \text{and that of the concave lens is} f_2.\\ \text{If F be the focal length of the equivalent lens, then :}\\\frac{1}{F}=\frac{1}{f_1}+\frac{1}{-f_2}=\frac{1}{f_1}-\frac{1}{f_2}\space \text{or}\space\text{F}=\frac{f_1f_2}{f_2-f_1}\\\text{If}\space f_1\text{\textgreater}\space f_2,\text{then F is negative and the combination will behave like a concave lens.}\\\text{If}\space f_1\text{\textless}\space f_2,\text{then F is positive and the combination will behave like a convex lens.}\\\text{If}\space f_1=f_2,\text{then F is infinite and the combination will behave like a plane glass plate.}$$
Answer 14.
$$\text{Let} a_1 and a_2 \text{be the amplitudes and} I_1 \text{and} I_2 \text{the intensities of the light waves emitted from the sources.}\\\space\text{The intensity ratio is 100 : 1. Then, we have}\\\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{100}{1}\\\text{or}\space\space\frac{a_1}{a_2}=\frac{10}{1}\space ….(i)\\\text{In interference, the maximum and minimum resultant amplitudes are} (a_1 + a_2) \text{and} (a_1 – a_2) \text{respectively}.\\\therefore\space\frac{\text{Maximum intensity} I_{max}}{\text{Minimum intensity} I_{min}}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}\\\text{But from equation (i),\space\space\space} a_1 = 10a_2\\\therefore\space\frac{I_{max}}{I_{min}}=\frac{(10a_2+a_2)^2}{(10a_2-a_2)^2}=\frac{121}{81}$$
Answer 15.
$$\text{(i) Angular momentum} = nh/2 \pi\\\therefore n=\frac{3.17×10^{-34}×2×3.14}{(6.63×10^{-34})}\\n=3\\\text{ (ii) Work function} W_0 = hv_0\\W_0=5.1 eV = 5.1 × 1.6 × 10^{-19} J (\because 1 eV = 1.6 × 10^{\normalsize–19} J)\\\therefore \text{The threshold frequency,}v_0=\frac{W_0}{h}=\frac{(5.1 × 1.6 ×10^{-19})}{(6.6 ×10^{-34})}=12×10^{15}Hz$$
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,
$$\frac{mv^2}{r}=\frac{1}{4\pi\epsilon_0}\frac{Ze×e}{r^2}\\mv^2=\frac{Ze^2}{4\pi\epsilon_0r}\space\space…(i)\\\text{By first postulate:}\\mvr=\frac{nh}{2\pi}\space\space…(ii)\\\text{Where n is the quantum number.}\\\text{Squaring equation (ii) and dividing by equation (i), we get :}\\\frac{m^2v^2r^2}{mv^2}=\frac{\frac{n^2h^2}{4\pi^2}}{\frac{Ze^2}{4\pi\epsilon_0r}}\\\text{Then},r=\frac{n^2h^2\epsilon_0}{\pi Ze^2m}\\\text{For, hydrogen atom, Z = 1,}\\r=\frac{n^2h^2\epsilon_0}{\pi Ze^2m}\\\text{For, hydrogen atom, Z = 1,}\\r=\frac{n^2h^2\epsilon_0}{ne^2m}$$
Answer 17.
(i) Light emitting diode (LED)
(ii) 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.
(iii) The diffusion of electrons from n-region to the p-region of the diode causes depletion region in a p-n junction.
Section-D
Answer 18.
$$\text{(i) (a) For given coil,}\\\text{Area} = 100 m^2, \theta = 90° – 60° = 30°\\\text{B}=10^{-1}\text{T}\\\text{Flux,}\phi=?\\\text{When a coil having area A is placed in magnetic field B},[\text{where}\space\theta \text{is the angle between area vector and magentic field vector}]\\\phi=BAcos\space\theta\\=10^{-1}×100 \text\space{cos}\space30\degree\\=10\frac{\sqrt{3}}{2}=5\sqrt{3}wb\\\text{Now, magnetic field is reduced to zero in 10}^{-3}s.\\\text{i.e.}\Delta\phi=5\sqrt{3}-0=5\sqrt{3}wb\\\text{Induced emf =-}\frac{\Delta\phi}{\Delta t}=-\frac{5\sqrt{3}}{10^{-3}}=-5000\sqrt{3}\space\text{volt}\\ \text{(b) Magnetic dipole moment.}\\\text{(c) Current sensitivity of a galvanometer is defined as the defection produced in the galvanometer when a unit current flows through it.}$$
$$\text{If a current I produces of deflection}\space\phi\space\text{in the galvanometer, the current sensitivity is}\frac{\phi}{t}\\C_s=\frac{\phi}{i}=\frac{NAB}{C},\text{where C is the corsional constant of the curve.}$$
OR
$$\text{(ii)}\space\text{(a) The magnetic field at any point due to an element of a current carrying conductor is:}\\\text{(1) directly proportional to}\\\text{1. Strength of current, I}\\\text{2. Length of element}\space \vec{dl}\\\text{3. sine of angle between the elements in the direction of current and the line joining the element to the point P.}\\\text{i.e dB}∝I, dB∝dl, dB∝\text{sin}\theta\\\text{(2) inversely proportional to the square of the distance r of the point P from the centre of the element i.e.}\\dB∝\frac{1}{r^2}\\\text{The magnetic field at P is}\\dB∝\frac{Idlsin\theta}{r^2}\\dB=\frac{\mu_0}{4\pi}.\frac{Idlsin\theta}{r^2}\\\text{(b) To convert galvanometer to ammeter}$$
To convert galvanometer to voltmeter
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.
(a) There are two important radioactive decay laws :
- (1) In any radioactive transformation, either an a-particle or a b-particle is emitted but it is never, that particles are ejected simultaneously. (i.e., never both or more than one of each kind is emitted)
- (2) At any instant, the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant.
(b) Let at time t = 0, the number of radioactive atoms present be N0 and at time t, the number be N. Let dN atoms disintegrate in a time dt.
According to Rutherford and Soddy law, we have :
$$\frac{dN}{dt}∝-N\\\qquad \frac{dN}{dt}=-\lambda N or \frac{dN}{N}=-\lambda dt,\\\text{where}\lambda\text{is called as decay constant or disintegreation constant.}\\\text{On integreating we get}\\\text{log}_eN=-\lambda t+A\\\text{when}\space t=0, N=N_0 \therefore\space\text{log}_eN_0=A\\\text{or}\space \text{log}_eN=-\lambda+\text{log}_eN_0\\\text{log}_eN=-\lambda+\text{log}_eN_0\\\text{log}_eN-\text{log}_eN_0=-\lambda t\\\text{or}\space\text{log}_e\frac{N}{N_0}=-\lambda t\\N=N_0e^{-\lambda t}$$
(ii)
(a) Transitions labelled A shows absorption lines of Lyman series.
(b) Transitions labelled B show emission lines of Balmer series.
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.
- (iv) It is called arc.
- (v) No, the energy carried by the wave remains constant. As it depends on the amplitude of wire. And the frequency remains constant.
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