# Oswal Practice Papers CBSE Class 12 Physics Solutions (Practice Paper - 7)

## Section-A

**1.** (b) −Q/4

** Explanation : **

For equilibrium, net force on Q = 0.

Let, the distance between line joining two equal charges Q is x unit.

$$\frac{KQQ}{(2x)^2}+\frac{kqQ}{x^2}=0\\q=–Q/4$$

**2. **(c) remains constant from centre to surface

** Explanation : **

Electric potential inside a conductor is constant and it is equal to that on the surface of the conductor.

**3.** (c) – i

** Explanation : **

According the Kirchhoff’s law, in any analytic circuit, if the direction of current assumed to be opposite, then the value of current will be – i.

**4.** (c)

$$\frac{mv}{Bq}$$

** Explanation : **

$$Force,F=qvB=\frac{mv^2}{R}\\R=\frac{mv}{Bq}$$

**5.** (d) 7.85 × 10^{−12} N

** Explanation : **

We know that,

F = qvB

$$= 1.6 × 10^{–19} ×(\sqrt{\frac{2E}{m}})2.5\\=4 × 10^{–19}\sqrt{\frac{2 ×2 ×1 .6×10^{-19}×10^{6}}{1.66×10^{-27}}}\\= 7.85 × 10^{–12} N$$

**6.** (a) NIA

** Explanation : **

Each turn of the solenoid behaves as a small dipole having dipole moment IA. Hence, the magnetic dipole moment of a sole noid having N turns = NIA

**7. **(b)

** Explanation : **

We know that,

$$XC =\frac{1}{\omega C}=\frac{1}{2\pi v C}\\XC ∝\frac{1}{v}.$$

**8.** (c) A charge moving in a circular orbit

** Explanation : **

A charge moving in a circular orbit can act as a source of electromagnetic waves as the e.m. waves are produced by accelerated charged particles only. A charge moving in a circular orbit has an accelerated motion.

**9.** (c) 5 H

** Explanation : **

$$e=−M\frac{dI}{dt}\\1500=-M(\frac{0-3}{0.01})\\M=\frac{1500×0.01}{3}\\=5 H$$

**10.** (d) there will be no effect

** Explanation : **

Young’s double slit arrangement is contained in a transparent chamber. If the chamber is evacuated, then the position of maxima and minima will change but there will be no effect on fringe width.

**11.** (b) 8 × 10^{14} Hz

** Explanation : **

$$f=\frac{3.3×1.6×10^{-19}J}{6.626×10^{-34}J-s}=8×10^{14}Hz$$

**12.** (a) E_{k} = –\frac{E_p}{2}

** Explanation : **

$$\text{Potential energy of an orbital electron, Ep =}\frac{-Ze^2}{r}\\\text{Kinetic energy of an electron,}E_k=\frac{-Ze^2}{2r}\\So,E_k=−\frac{E_p}{2}$$

**13.**(c) A is true but R is false.

** Explanation : **

Photoelectric effect supports quantum nature of light because when metal surface is faintly illuminated, photoelectrons leave surface immediately.

**14.** (c) A is true but R is false.

** Explanation : **

When the speed of light is independent of direction, the secondary waves are spherical.

**15.** (c) A is true but R is false.

** Explanation : **

If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons in conduction band then it makes the material an insulator. Resistance of insulators is very high.

**16.** (c) Both assertion and reason are true, but reason is not be correct explanation of assertion.

** Explanation : **

According to electromagnetic induction, whenever the magnetic flux changes, an emf will be induced in the coil.

## Section-B

**17.** (i) Infra-red waves are called heat waves because they raise the temperature of the object on which they fall. They also affect the photographic plate and are readily absorbed by most materials.

(ii) Electromagnetic waves transport momentum. This means that when an e.m. wave travels through space with energy U and speed c, then it transports linear momentum p=U/C If a surface absorbs the waves completely, then momentum ‘p’ is delivered to the surface. If the surface reflects the wave, then momentum delivered by both incident and reflected wave adds on to give ‘2p’ momentum.

**18.** Magnetic field produced by the two coils at their common centre having currents I1 and I2, radius a1 and a2, number of turns N1 and N2 respectively are given by :

$$B_1=\frac{\mu_0N_1I_1}{2a_1}\\and\space B_2=\frac{\mu_0N_2I_2}{2a_2}\\\text{The resultant field at the common centre is :}\\B=\sqrt{B^2_1+B^2_2}\\=\sqrt{(\frac{\mu_0N_1I_1}{2a_1})+\frac{\mu_0N_2I_2}{2a_2}}\\Here, N_1 = N_2 = 1, I_1 = I_2 = I, a_1 = a_2 = R\\B =\sqrt{(\frac{\mu_0×1×I}{2R})+(\frac{\mu_0×1×I}{2R})}\\=\frac{\mu_0I}{2R}\sqrt{1+1}=\sqrt{2}(\frac{\mu_0I}{2R})\\\text{Direction of B,}\\tan\theta=\frac{B_1}{B_2}=1\\\theta = 45°\\\text{Hence, the net magnetic field is directed at an angle of 45° with either of the fields.}$$

**19.** In an equiconcave lens, radius of curvature of both surfaces are equal

$$\frac{1}{f}=(\mu-1)(-\frac{1}{R}-\frac{1}{R})\\Since, P = –5 D\frac{1}{f (In meter)}=P = – 5D\\⇒f =\frac{-100 cm}{5}=-20cm\\and\space\mu_e=1.5\space and\space\mu_m = 1.4\\\frac{1}{-20}=(\frac{1.5}{1.4}-1)(\frac{-2}{R})\\⇒\frac{1}{-20}=\frac{0.1}{1.4}×\frac{-2}{R}\\⇒ R=\frac{0.1×(-2)×(-20)}{1.4}\\=\frac{20}{7}= 2.86 cm$$

**20.** As c = vλ,

$$\frac{\Delta v}{v}=-\frac{\Delta\lambda}{\lambda}\text{(for small changes in v and λ)}\\For Δλ = 589.6 – 589.0 = + 0.6 nm\\\frac{\Delta v}{v}=-\frac{\Delta\lambda}{\lambda}=-\frac{v_{radial}}{c}\\or\\v_{radial} =+c(\frac{0.6}{589.0})=\frac{3×10^8×0.6}{589.0}\\= + 3.06 × 10^5 ms{–1}\\= 306 km/s\\\text{Therefore, the galaxy is moving away from us.}$$

**21.** When the p-n junction diode, is forward biased, the diode’s depletion layer or potential barrier shrinks. This can be considered as a practical short circuit.

Hence, resistance is negligible and current flow in forward bias is maximum. Similarly, when it is reverse bias, depletion layer width (or potential barrier) increases. Hence, resistance increases and very small amount of current flow in reverse bias.

**OR**

Resistance of a material can be found out by the slope of the V-I curve. Part BC of the curve shows the negative resistance as with the increase in voltage, current decreases.

## Section-C

**22. **Let a series LCR circuit is connected to an a.c. source V (Fig). We take the voltage of the source to be V = V_{0} sin wt

The a.c. current in each element is the same at any time, having the same amplitude and phase. It is given by,

$$I = I_0 sin (\omega t+\varPhi)\\Let V_L, V_R, V_C \text{and V represent the voltage across the inductor, resistor, capacitor and the source respectively.}\\Let V_L > V_C\\V^{2}_{0}=V^{2}_{R}+ (VL – VC)^2\\V^{2}_{0}=(I_0R)^2 + (I_0XL – I_0X_C)^2\\V^{2}_{0}=I^2_0 [R^2 + (XL – XC)^2]\\and\space I_0 =\frac{V_0}{\sqrt{R^2+(X_L-X_C)^2}}\\⇒I_0=\frac{V_0}{Z}\\Where\space Z =\sqrt{R^2+(X_l-X_c)^2}\\\text{It is called the impedance in an a.c. circuit.From the figure,}\\tan\theta=\frac{V_L-V_C}{V_R}\\=\frac{I_0X_L-I_0X_C}{I_0R}\\⇒\theta=tan^{-1}\frac{X_L-X_C}{R}\\\text{Condition : The current will be in phase with the voltage at resonance condition.}\\\text{At resonance condition,}\\X_L = X_C\\\omega L=\frac{1}{\omega C}\\w =\frac{1}{\sqrt{LC}}\\2\pi f=\frac{1}{\sqrt{LC}}\\f=\frac{1}{2\pi\sqrt{LC}}$$

**OR**

$$As\space cos\theta =\frac{R}{Z}\\\text{In LR circuit, Power factor }P_1\\P_1=cos\varPhi\\P_1=\frac{R}{\sqrt{R^2+X^2_L}}=\frac{R}{\sqrt{2R^2}}[˙.˙ XL = R]\\P_1=\frac{1}{\sqrt{2}}\\\text{In LCR circuit when,}\\XL = XC\\P_2=\frac{R}{\sqrt{R^2+(X_L-X_C)^2}}\\=\frac{R}{R}=1[˙.˙ XL = XC]\\\frac{P_1}{R}=\frac{1}{\sqrt{2}}$$

**23.** When a charged particle having charge q moves inside a magnetic field$$\vec{B}$$ having velocity v, it experiences a force, which is given by :

$$\vec{F}= q (\vec{v}×\vec{B})\\\text{Here,}\vec{v}\space\text{is perpendicular to}\vec{B}\vec{F}\\\text{is the force on the charged particle which behaves as the centripetal force and make it move in a circular path.}\\\text{Let m be the mass of the charged particle and r be the radius of the circular path.}\\q(\vec{v}×\vec{B})=\frac{mv^2}{r}\\\text{v and B are at right angl}\\qvB =\frac{mv^2}{r}\\r =\frac{mv}{Bq}$$

Time period of circular motion of the charged particle can be calculated by,

$$T=\frac{2\pi r}{v}\\⇒T=\frac{2\pi}{v}\frac{mv}{Bq}\\T=\frac{2\pi m}{Bq}\\\text{\ Angular frequency is}\\w =\frac{2\pi}{T}\\w=\frac{Bq}{m}\\\text{Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.}$$

**24.** Magnetic field due to loop P at centre

$$B_1=\frac{\mu_0}{2R}.I_1\\B_1=\frac{\mu_0}{2}×\frac{3}{5×10^{-2}}\\\text{Similarly, Magnetic field due to loop Q at centre}\\B_2 =\frac{\mu_0}{2}×\frac{4}{5×10^{-2}}\\\text{Resultant magnetic field is,}\\B =\sqrt{B^2_1+B^2_2}\\B=\sqrt{(\frac{\mu_0}{2}×\frac{3}{5×10^{-2}})^2+(\frac{\mu_0}{2}×\frac{4}{5×10^{-2}})^2}\\B =\frac{\mu_0}{2×5×10^{-2}}×5\\B =\frac{\mu_0}{2}×100=50\mu_0\\= 50 × 4\pi × 10^{– 7}\\= 62.8 × 10^{– 6} T\\tan\theta =\frac{B_1}{B_2}=\frac{3}{4}\\\theta=tant^{-1}(\frac{3}{4})\\\text{Thus, direction of magnetic field makes an angle q with the vertical.}$$

**25.** Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for 1, 2 and 3, 4 are the same. For the given frequency of the incident radiation, the stopping potential is independent of its intensity. So, the pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation.

**OR**

Einstein photoelectric equation :

$$K_{max}=\frac{1}{2}mv^2=hv-\phi_0$$

Where φ0 is work function. If the energy of the photon absorbed by the electron is less than the work function φ0 of the metal, then the electron will not be emitted. Therefore, for the given metal, the threshold frequency of light be n0, then an amount of energy hn0 of the photon of light will be spent in ejecting the electron out of the metal, that is, it will be equal to the work function W. Thus,

W = hv_{0}

Thus EK = hv – hv_{0}$$K_{max}=\frac{1}{2}mv^2=h(v-v_0)$$

This equation is called ‘Einstein Photoelectric equation’.

Salient features :

(i) The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation.

(ii) There exists a minimum cut-off frequency n0, for which the stopping potential is zero.

(iii) Photoelectric emission is instantaneous.

**26.** Let

$$E_n = 13.6\begin{bmatrix}\frac{1}{1^2}-\frac{1}{n^2}\end{bmatrix}eV\\\text{Work function = 2 eV}\\\text{∴Maximum kinetic energy = 13.6}\begin{bmatrix}\frac{1}{4}-\frac{1}{n^2}\end{bmatrix}eV-2eV\\\frac{13.6}{4}-\frac{13.6}{n^2}– 2 = 0.55\\⇒\frac{13.6}{4}– 2 = 0.55=\frac{13.6}{n^2}\\⇒ n^2 =\frac{13.6}{0.85}=16\\n = 4$$

**27.** Distance of closest approach is defined as the minimum distance of the charged particle from the nucleus at which initial kinetic energy of the particle is equal to the potential energy due to the charged nucleus.

At this distance,

Kinetic Energy (K) = Potential Energy

$$k=\frac{1}{4\pi\epsilon_0}\frac{Ze(2e)}{r}\\r=\frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{K}…(i)\\\text{If K is doubled,}\\r′ ==\frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{K}…(ii)\\\text{Dividing (ii) by (i), we get}\\\frac{r'}{r}=\frac{1}{2}\\⇒ r′ =\frac{r}{2}$$

**28.** Consider a thin infinite uniformly charged plane sheet having the surface charge density of σ. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction.

Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A.

The cylinder is made from three surfaces A, S_{2}, and A’ and the electric flux linked with S_{2} is 0. So, the total electric flux linked through the Gaussian surface is

φ_{E} = Electric flux through A + Electric flux through S_{2} + Electric flux through A’

φ_{E} = EA cos 0° + 0 + EA cos 0°.

φ = 2EA …(i)

According to Gauss theorem,

$$\phi=\frac{q}{\epsilon_0}\\\phi=\frac{\sigma A}{\epsilon_0}(·.· q = σA) …(ii)\\\text{From equations (i) and (ii),}\\2EA =\frac{\sigma A}{\epsilon_0}\\E=\frac{\sigma }{2\epsilon_0}$$

The direction of field for positive charge density is in outward direction and perpendicular to the plane infinite sheet whereas for the negative charge density the direction becomes inward and perpendicular to the sheet.

**OR**

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (q/m) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

## Section-D

**29.****(i)** (b) Its critical angle with reference to air is too small**(ii)** (a) 2.42**(iii)** (b) High refractive index**(iv)** (d) Increases

**OR**

(a) Less than the first

**30.****(i)** (c) nh >> ne**(ii)** (d) insulator**(iii)** (c) insulators

**OR**

(c) zero**(iv)** (d) (a) and (b) both

## Section-E

**31.** (a) The system of two protons and one electron is represented in the given figure.

The potential energy of the system is,

$$V=\frac{q_1q_2}{4\pi\epsilon_0d_1}+\frac{q_2q_3}{4\pi\epsilon_0d_3}+\frac{q_3q_1}{4\pi\epsilon_0d_2}\\\text{Substituting}\frac{1}{4\pi\epsilon_0}= 9 × 10^9 Nm^2C^{–2}\\\text{we obtain}\\V =\frac{9×10^9×10^{-19}×10^{-19}}{10^{-10}}[-(1.6)^2+\frac{(1.6)^2}{1.5}-(1.6)^2]\\= – 30.7 × 10^{–19} J\\= – 19.2 eV\\\text{Therefore, the potential energy of the system is – 19.2 eV.}$$

(b) Two charges placed at points A and B are shown in the given figure. O is the mid-point of the line joining the two charges.

$$\text{Magnitude of charge located at A,}\\q_1 = 1.5 \mu C\\\text{Magnitude of charge located at B,}\\q_2 = 1.5 \mu C\\\text{Distance between the two charges,}\\d = 30 cm= 0.3 m\\(i) Let V_1 and E_1\text{ are the electric potential and electric field respectively at O.}\\V_1 =\text{ Potential due to charge at A + Potential due to charge at B}\\V_1=\frac{q_1}{4\pi\epsilon_0(\frac{d}{2})}+\frac{q_2}{4\pi\epsilon_0(\frac{d}{2})}\\=\frac{1}{4\pi\epsilon_0(\frac{d}{2})}[q_1+q_2]\\Where, \epsilon_0 = \text{Permittivity of free space}\\\frac{1}{4\pi\epsilon_0}= 9 × 10^9 NC^2m^{– 2}\\V_1=\frac{9×10^9×10^{-6}}{(\frac{0.30}{2})}[2.5 + 1.5]\\=2.4×10^5V\\E_1=\text{Electric field due to}\space q_2–\text{Electric field due to}\space q_1\\\frac{q_2}{4\pi\epsilon_0(\frac{d}{2})^2}-\frac{q_1}{4\pi\epsilon_0(\frac{d}{2})^2}\\=\frac{9×10^9×10^{-6}}{(\frac{0.30}{2})^2}[2.5 – 1.5]\\=4×10^5Vm– 1$$

Therefore, the potential at mid-point is 2.4 × 10^{5} V and the electric field at mid-point is 4 × 10^{5} Vm^{– 1}.

The field is directed from the larger charge to the smaller charge.

(ii) Consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure:

$$Let, V_2 and E_2\text{are the electric potential and electric field respectively at Z.}\\\text{It can be observed from one figure that distance,}\\BZ=AZ\\=\sqrt{(0.10)^2+(0.15)^2}\\=0.18m\\V_2 =\text{Electric potential due to A+Electric potential due to B}\\=\frac{q_1}{4\pi\epsilon_0(AZ)}+\frac{q_2}{4\pi\epsilon_0(BZ)}\\=\frac{9×10^9×10^{-6}}{(0.18)^2}\\=2×10^5V\\\text{Electric field due to }q_1 at Z,\\E_A=\frac{q_1}{4\pi\epsilon_0(AZ)^2}\\=\frac{9×10^9×1.5×10^{-6}}{(0.18)^2}\\=0.416 × 10^6 V m^{–1}\\\text{Electric field due to }q_2 at Z,\\E_B=\frac{q_2}{4\pi\epsilon_0(BZ)^2}\\=\frac{9×10^9×2.5×10^{-6}}{(0.18)^2}\\== 0.69 × 10^6 V m^{–1}\\\text{The resultant field intensity at Z,}\\E =\sqrt{E^2_A+{E^2_B}+2E_AE_Bcos2\theta}\\cos\theta=\frac{0.10}{0.18}=\frac{5}{9}= 56.25°\\2\theta = 112.5°\\cos 2\theta = – 0.38\\E =\sqrt{(0.416×10^6)^2+(0.69×10^6)^2+2×0.416×0.069×10^{12}×(-0.38)}\\= 6.6 × 10^5 Vm^{– 1}\\\text{Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is}2.0 × 10^5 V and electric field is 6.6 × 10^5 V m^{– 1}.$$

**OR**

(a) (i) Electric force at A due to charge 2q,

$$F_C=\frac{1}{4\pi\epsilon_0}\frac{q×2q}{l^2}along\vec{CA}\\\text{Electric force at A due to charge (– 4q),}\\F_B=\frac{1}{4\pi\epsilon_0}\frac{q×(-4q)}{l^2}along\vec{AB}\\\text{Resultant force,}F=\sqrt{F^2_B+F^2_C+2F_BF_C cos120}\\=\frac{1}{4\pi\epsilon_0}\frac{q^2}{l^2}\sqrt{(-4)^2+(2)^2+2(-4)(2)(-\frac{1}{2})}\\=\frac{1}{4\pi\epsilon_0}\frac{2\sqrt{7}q^2}{l^2}$$

(ii) Work done to separate the charge to infinity :

Initial potential energy,

$$U_i=\frac{1}{4\pi\epsilon_0}[\frac{(-4q)q}{l}+\frac{(-4q)2q}{l}+\frac{q(2q)}{l}]\\=\frac{1}{4\pi\epsilon_0}\frac{(q)^2}{l}[-4-8+2]\\=\frac{1}{4\pi\epsilon_0}\frac{(-10q^2)}{l}\\\text{Final potential energy,}\\U_f= 0\\\text{Thus, work done,}\\U_f – U_i = 0 -(\frac{-10q^2}{{4\pi\epsilon_0l}})$$

(b) The forces acting on charge q at A, due to charges q at B and – q at C are F_{12} along BA and F_{13} along AC respectively, as shown in figure from the parallelogram law, the total force F1 on the charge q at A is given

$$by F_1=F\space\hat{r_1} where\space\hat{r_1}\text{is a unit vector along BC.}\\\text{The force of attraction or repulsion for each pair of charges has the same magnitude}\\F=\frac{q^2}{4\pi\epsilon_0l^2}\\\text{The total force}F_2 \text{on charge q at B is thus}\\F_2=F\hat{r_2},where\space\hat{r_2}\text{is a unit vector along AC.}\\\text{Similarly, the total force on charge – q at C is }F_3=\sqrt{3}F\hat{n}\space where\space \hat{n}\text{is the unit vector along the direction bisecting the ∠BCA.}\\\text{It is interesting to see that the sum of the forces on the three charges is zero, i.e.,}\\F_1+F_2+F_3=0$$

(a) PQ = incident ray

QR = refracted ray

RS = emergent ray

∠A = angle of prism

∠ i = angle of incidence

∠ r1 = angle of refraction

∠ e = angle of emergence

∠δ = angle of deviation

Consider, the above diagram and μ as the refractive index of the prism.

In case of minimum deviation, ∠ r 1 = ∠ r 2 = ∠ r

and A = ∠ r 1 + ∠ r 2

So, A = ∠ r + ∠ r = 2 ∠ r

$$or ∠r=\frac{A}{2}\\\text{Now again,}\\A+\delta=i+e=2i\\or\space i=\frac{A+\delta}{2}\\\text{Now from Snell’s law}\\\mu=\frac{sini}{sinr}=\frac{sin(\frac{A+\delta}{2})}{sin(\frac{A}{2})}\\\text{is the required expression.}\\\text{(b) (i) For critical angle,}\\sin C =\frac{1}{\mu}=\frac{1}{1.5}=\frac{2}{3}\\C=sin^{-1}(\frac{2}{3})=41°8’\\\text{So, if incident angle is greater than 41°8’,}\\\text{Total internal reflection will occur.}$$

$$\text{(ii) If μ changes to 1.4}\\Then, sin C’ =\frac{1}{\mu}=\frac{1}{1.4}=\frac{5}{7}\\C = sin^{-1}(\frac{5}{7})= 45°9’\\\text{Since, incident angle is less than the critical angle. Then total internal reflection will not occur.}$$

**OR**

(a)

$$\text{From figure, d = dm, i = e which implies}r_1=r_2\\2r = A, or\space r =\frac{A}{2}\\Using\delta = i + e – A\\\delta_m = 2i – A\\i =\frac{A+\delta_m}{2}\\µ =\frac{sin i}{sin r}=\frac{sin\frac{a+\delta_m}{2}}{sinA/2}$$

(b) (i) Dispersion of Light : Dispersion is often observed as light passes through a triangular prism. Upon passing through the prism, the white light is separated into its component colours, red, orange, yellow, green, blue, indigo and violet. The separation of visible light into its different colours is known

as dispersion. Dispersion occurs because for different colour of light, a transparent medium will have different refractive indices (m) as different colours have different speed of light in transparent medium.**(ii)** For total internal reflection :

$$i≥Q_c\\sini≥sin\theta_c\\sin 45°≥\frac{1}{\mu}\\\frac{1}{\sqrt{2}}≥\frac{1}{\mu}\\µ≥\sqrt{2}\\µ_{min}=\sqrt{2}$$

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