# Oswal Practice Papers CBSE Class 12 Physics Solutions (Practice Paper - 9)

## Section-A

1. (c)
$$\frac{q}{4\epsilon_0}$$

Explanation :

In the figure, when a charge 2q is placed at corner A of the cube, it is being shared equally by eight cubes. So, total flux through the faces of the given cube =
$$\frac{q}{4\epsilon_0}$$

2.(a) zero

Explanation :

As distance of point M and N from C is constant so They are at same potential.
$$V_M=V_N=\frac{1}{4\pi\epsilon_0}.\frac{q}{r}\\V_M – V_N = 0\\\text{The work done in this process} W=(V_M–V_N)q=0$$

3. (a) be double the initial velocity

Explanation :

As we know that
I = nevd
I ∝ vd
From Ohm’s law,
V ∝ I ∝ vd
If the potential difference is doubled, drift velocity of electrons will also double.

4. (b)
$$\frac{nr^2}{(x^2+r^2)^{3/2}}$$

Explanation :

Magnetic field on the axis of circular coil carrying current,
$$B=\frac{\mu_o}{4\pi}\frac{2\pi n Ir^2}{(x^2+r^2)^{3/2}}\\⇒B ∝\frac{nr^2}{(x^2+r^2)^{3/2}}$$

5. (d) 5πμT

Explanation :

$$B=10^{–7}×\frac{\pi×I}{r}\\=10^{–7}×\frac{\pi×10}{20×10^{-2}}\\⇒B=5\pi\mu T$$

6. (d) 50 × 10–3 N

Explanation :

Force between magnetic poles in air is given by,
$$F=\frac{\mu_0}{4\pi}×\frac{m_1m_2}{r^2}\\\text{Given that,}m_1=50Am\\m_2=100Am\\r=10 cm=0.1m\\\mu_0=\text{ permeability of air}\\=4\pi×10^{–7} Hm^{–1}\\F=\frac{4\pi×10^{-7}}{4\pi}.\frac{50×100}{0.1×0.1}\\=50×10^{–3}N$$

7. (b) 1.256 mH

Explanation :

Self-inductance of solenoid coil having N number of turns, length l and area of cross-section
A is given by,
$$L=μ_0\frac{N^2}{l}A\\Here,N=1000,\\A=10cm^2\\orA=10×10^{–4} m^2\\and\space l=1m\\\text{Substituting these values in the above equation, we get}\\L=4π×10^{–7}×\frac{(1000)^2}{1}×10×10^{-4}\\⇒ L=1.256 mH$$

8. (c) L is large and R is small

Explanation :

Since, selectivity depends on the quality of resonance. The quality factor is given by Q = ω0L/R. High value of quality factor make sure that the resonance curve is sharp, sharper the resonance curve, more selective is the LCR circuit.

9. (d) Frequency remain unchanged

Explanation :

The frequency of an electromagnetic wave is independent of the medium in which the wave travels.

10. (d) red

Explanation :

Since v ∝ l, the light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerge first.

11. (b) only as half wave rectifier

Explanation :

A diode is a p-n junction which acts as open switch in reverse biasing and closed switch in forward biasing. A diode can be used for rectification i.e. to convert A.C. voltage into D.C. voltage. There are two type of rectifying circuit-full wave rectifier and half wave rectifier. But a single diode works as a
half wave rectifier.

12. (d) all the above conditions

Explanation :

1913, Niels Bohr's proposed a model for the hydrogen atom. The Bohr's model was based on the following assumptions:
(1) The electron in a hydrogen atom travels around the nucleus in a circular orbit.
(2) The nucleus is of infinite mass and is in rest.
(3) Electrons in a quantized orbit will not radiate energy.
(4) Mass of electron remains constant.

13. (d) A is false, and R is also false.

Explanation :

Critical angle is minimum for violet colour, critical angle =
$$\theta_c = sin^{–1}(\frac{1}{\mu})$$

14. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

$$\text{As de-Broglie wavelength,}\space\lambda=\frac{h}{mv}\\\text{so, for constant velocity,}\space\lambda∝\frac{1}{m}\\\text{Lesser the mass greater will be its de-Broglie wavelength.}$$

15. (b) Both A and R are true, but R is not the correct explanation of A.

Explanation :

As we know that,
V = 5 – 2 = 3 V
$$I=\frac{V}{R}=\frac{3}{300}=\frac{1}{100}A\\= 10 mA$$

16. (a) Assertion is true, but reason is false.

Explanation :

The manner in which the two coils are oriented, determines the coefficient of coupling between them.
$$M=K.\sqrt{L_1L_2}$$

When the two coils are wound on each other, the coefficient of coupling is maximum and hence mutual inductance between the coil is maximum.

## Section-B

17. (i) Microwaves
(ii) X-rays
(iii) Gamma rays

18. Figure 1 shows magnetic field lines when (i) diamagnetic materials are placed in external magnetic field.

Figure 2 shows magnetic field lines, when (ii) paramagnetic materials are placed in external magnetic field.

Magnetic dipole moment of materials are responsible for this diamagnetic material has zero magnetic moment. Paramagnetic material has non-zero magnetic moment.

19. The two differences between the interference patterns obtained in Young’s double slit experiment and the diffraction pattern due to a single slit are as follows:
(i) The fringes in the interference pattern obtained from diffraction are of varying width, while in case of interference, all are of the same width.
(ii) The bright fringes in the interference pattern obtained from diffraction have a central maximum followed by fringes of decreasing intensity, whereas in case of interference, all the bright fringes are of equal width.

20. Intensity pattern for single slit diffraction :

Intensity pattern for double slit interference :

21. Forward biasing of a diode : In this arrangement, the positive terminal of battery is connected to p-end and negative terminal to n-end of the crystal, so that an external electric field E is established directed from p to n-end to oppose the internal field Ei. Thus, the junction is said to be conducting.

Reverse biasing of a diode : In this arrangement the positive terminal of battery is connected to n-end and negative to p-end of the crystal, so that the external field is established to support the internal field Ei.

OR

The accumulation of negative charges in the p-region and positive charges in the n-region set up a potential difference across the junction. This acts as a barrier and is called potential barrier.

## Section-C

22. Principle of ac generator : The ac generator is based on the principle of electromagnetic induction. When closed coil is rotated in a uniform field with its axis perpendicular to field, then magnetic flux changes and emf is induced.
Working : When the armature coil rotates, the magnetic flux linked with it changes and produces induced current. If initially, coil PQRS is in vertical position and rotated clockwise, then PQ moves down and SR moves up. By Fleming’s right hand rule, induced current flows from Q to P and S to R which is the first half rotation of coil. Brush B1 is positive terminal and B2 is negative. In second half rotation, PQ moves up and SR moves down. So induced current reverses and the alternating current is produced in this manner by the generator.

Expression for emf induced : If number of turns in coil = N, cross-section = A, angular speed of rotation = ω,
$$\text{magnetic field =}\vec{B}\space\text{then to find emf induced,}\\\text{Flux through the coil when its normal makes angle q with the field,}\\\phi=BAcos\theta\\\text{When coil rotates with angular velocity w and turns through q in time t then}\theta=\omega t\\\phi=BAcos\omega t\\\text{When coil rotates, f changes to set an induced emf.}\\\epsilon=\frac{-d\phi}{dt}\\=\frac{-d\phi}{dt}(BAcos\omega t)\\=BA\omega sin\omega t\\\text{For N turns, total induced emf,}\\\epsilon=NBA\omega sin\omega t$$

OR

$$We have,\\e_s=-N_s\frac{d\phi}{dt}…(i)\\e_p=-N_p\frac{d\phi}{dt}…(ii)$$

Dividing (i) by (ii), we get
$$We have,\\\frac{e_s}{e_p}=\frac{N_s}{N_p}…(iii)\\\text{According to the law of conservation of energy,}\\e_sI_s = e_pI_p\\\frac{e_s}{e_p}=\frac{I_p}{I_s}…(iv)\\\text{From equations (iii) and (iv),}\\\frac{e_s}{e_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$$

23. (i) The magnetic field due to a current carrying solenoid :

where, n = number of turns per unit length, i = current through the solenoid.
Now, the magnetic field due to solenoid S1 will be in the upward direction and the magnetic field due to S2 will be in the downward direction (by right-hand screw rule).
Bnet = B1 – B2
Bnet = µ0n1I – µ0n2I
= µ0I (n1 – n2)
(ii) The magnetic field is zero outside a solenoid.

24. (i) A charge particle having charge q is moving with velocity ‘v’ in a magnetic field of field strength ‘B’ then the force acting on it is given by the formula
$$F=q(\vec{v}×\vec{B })andF=qvBsin θ\\\text{(where θ is the angle between velocity vector of magnetic field).}\\\text{Direction of force is given by the cross product of velocity and magnetic field.}$$

(ii) α-particle will trace circular path in anticlockwise direction as it’s deviation will be in the direction
of
$$(\vec{v}×\vec{B })$$

Neutron will pass without any deviation as magnetic field does not exert force on neutral particle.

Electron will trace circular path in clockwise direction as its deviation will be in the direction opposite to
(\vec{v}×\vec{B })

with a smaller radius due to large charge/mass ratio as
$$r=\frac{mv}{qB}$$

25. (i) Reverse biased.
(ii) Full wave Rectifier :
Diode conducts only when the junction is forward biased. Hence, during first half cycle of input A.C., D1 will conduct while D2 will not and current in RL will flow from A to B. During second half cycle of input A.C., diode D2 will conduct while D1 will not conduct and current will again flow from A to B in
RL. Hence, complete cycle will become unidirectional.

26. 2 deuterium nuclei fuse to produce 3.27 MeV
= 3.27 × 1.6 × 10–13 J
= 5.232 × 10–13 J
Energy produced per nuclei
$$=\frac{5.23×10^{-13}}{2}\\= 2.61 × 10^{–13} J\\\text{2 kg deuterium atoms will have}\\=\frac{6.023×10^{23}}{2}×2000\\=6.023×10^{26}\text{deuterium atom}\\=2.61×10^{–13}×6.023×10^{26}J\\=15.77× 10^{13}J\\\text{Power=}\frac{E}{t}\\800=\frac{15.72×10^{13}}{t}\\⇒t=\frac{15.72×10^{13}}{800}\\=1.965×10^{11}\text{seconds}\\t=\frac{1.965×10^{11}}{365×24×60×60}\\=6.23×10^{3}years.\\\text{Hence, electric lamp will glow for} 6.23×10^{3} years$$

27. When the plane is parallel to the Y-Z plane
$$\phi=\vec{E}.\vec{A}\\\vec{E}=2×10^3\hat{i}\\A = (20 cm)^2\hat{i}=0.04m^2\hat{i}\\\phi=(2×10^3\hat{i}.(0.04i))\\⇒ \phi= 80 Wb\space or\space Nm^2C^{– 1}.\\\text{When the plane makes a 30° angle with the X-axis, the area vector makes a 60° angle with the X-axis.}\\\phi=\vec{E}.\vec{A}\\⇒\phi=EAcos\theta\\⇒\phi=2×10^3×0.04×cos60°\\⇒\phi=2×10^3×0.04×\frac{1}{2}\\⇒\phi=40 Wb\space or\space Nm^2C^{–1.}$$

28. (i) We know that
$$eV_0=\frac{hc}{\lambda}-\phi\\or\space V_0=\frac{hc}{e}(\frac{1}{\lambda})-\frac{\phi}{e}…(i)\\\text{where, e=Charge on electron}\\\text{h=Planck’s constant}\\\phi \text{=Work function of metal surface}\\\text{Equation (i) is a straight line as shown in the graph.}\\\text{Here, slope of line =}tan\theta=\frac{hc}{e}\\or h=\frac{e\space tan\theta}{c}…(ii)$$

Planck’s constant can easily be determined by substituting the values of the slope of graph, speed of light and electronic charge in equation (ii).
(ii) Stopping potential only depends on frequency of incident light. If the distance is increased, intensity will decrease but no change takes place in stopping potential.

OR

From the observations made (surface A, B and C) on the basis of photoelectric effect:
(i) For surface A, threshold frequency is more than frequency of incident light, so no emission occurs.
(ii) For surface B, threshold frequency is equal to frequency of incident light, so emission just occurs.
(iii) For surface C, the threshold frequency is less than frequency of incident light. So, photo-emission occurs and photo-electron have some K. E.
Threshold frequency is nA > nB > nC.

## Section-D

29.
(i) (c) Diffraction
(ii) (c) Very large
(iii) (c) Decreases
(iv) (d) Both (a) and (b)

OR

(c) d < l

30.
(i) (b) Both I and III
(ii) (b) 3 and 1 keV
(iii) (d) Bohr's model

OR

(c) absorption
(iv) (a) less than 1

## Section-E

31. (a) (i) Electric flux : Electric flux through an area is defined as the product of electric field strength E and area dS perpendicular to the field. It represents the field lines crossing the area. It is a scalar  quantity.
Imagine a cube of edge d, enclosing the charge. The square surface is one of the six faces of this cube. According to Gauss’ theorem in electrostatics.

$$\text{Total electric flux through the cube}=\frac{q}{\epsilon_0}\\\text{\ Electric flux through the square surface}=\frac{q}{6\epsilon_0}$$

(ii) On moving the charge to distance d from the centre of square and making side of square 2d, does not change the flux at all because flux is independent of side of square or distance of charge in this case.
(b) Consider a long thin wire XY of uniform linear charge density λ [Fig. (a)].

Take a point P at perpendicular distance l from mid-point O of wire as shown in [Fig. (b)].

Let E be the electric field at point P due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
q = λ dx
Electric field due to the piece,
$$dE=\frac{1}{4\pi\epsilon_0}.\frac{\lambda dx}{(PZ)^2}\\However, PZ=\sqrt{l^2+x^2}\text{(by pythagoras theorem)}\\\ dE=\frac{1}{4\pi\epsilon_0}.\frac{\lambda dx}{(l^2+x^2)}$$
The electric field is resolved into two rectangular components. dE cos q is the perpendicular component and dE sin q is the parallel component. When the whole wire is considered, the component dE sin q is cancelled. Only the perpendicular component dE cos q affects point P.
Hence, effective electric field at point P due to the element dx is dE1
$$d_E=\frac{1}{4\pi\epsilon_0}.\frac{\lambda dxcos\theta}{(l^2+x^2)}…(i)\\In Δ PZO, tan\theta =\frac{x}{l}\\⇒ x = l\space tan\theta …(ii)\\\frac{dx}{d\theta}= l sec^2\theta\text{[from differentiating (ii)]}\\x^2 + l^2 = l^2 tan^2\theta + l^2\\= l^2 sec^2\theta…(iii)\\\text{Now,}\space dE_1=\frac{1}{4\pi\epsilon_0}.\frac{\lambda l sec^2\theta d\theta.cos\theta}{l^2sec^2\theta}=\frac{1}{4\pi\epsilon_0}.\frac{\lambda cos\theta d\theta}{l}…(iv)\\\text{The wire is so long that q tends from}-\frac{\pi}{2}to\frac{\pi}{2}\\\text{By integrating equation (iv), we obtain the value of field }E_1 as,\\\int^{\pi/2}_{-\pi/2}dE_1=\int^{\pi/2}_{-\pi/2}\frac{1}{4\pi\epsilon_0}\frac{\lambda}{l}cos\theta d\theta\\⇒ E_1=\frac{1}{4\pi\epsilon_0}\frac{\lambda}{l}[sin\theta]^{\pi/2}_{-\pi/2}\\⇒ E_1=\frac{1}{4\pi\epsilon_0}\frac{\lambda}{l}×2\\⇒ E_1 =\frac{\lambda}{2\pi\epsilon_0l}$$

OR

(a) Suppose we have a ring of radius a that carries a uniformly distributed positive charge q.

As the total charge q is uniformly distributed, the charge dq on the element dl is
$$d_q=\frac{q}{2\pi a}.dl$$

The magnitude of the electric field produced by the element dl at the axial point P is
$$dE=k.\frac{dq}{r^2}=\frac{kq}{2\pi a}.\frac{dl}{r^2}$$

The electric field dE has two components:
(i) The axial components dE cos θ and
(ii) The perpendicular component dE sin θ.
Since, the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field. E at point P is given by,
$$E=\int^{2\pi a}_{0}dEcos\theta\space\text{[ Only the axial components contribute towards E]}\\E=\int^{2\pi a}_{0}\frac{kq}{2\pi a}.\frac{dl}{r^2}.\frac{x}{r}[cos\theta=\frac{x}{r}]\\=\frac{kqx}{2\pi a}.\frac{1}{r^3}\int^{2\pi a}_{0}dl\\=\frac{kqx}{2\pi a}.\frac{1}{r^3}l^{2\pi a}_{0}\\=\frac{kqx}{2\pi a}.\frac{1}{(x^2+a^2)^{3/2}}l^{2\pi a}_{0}\\=\frac{1}{4\pi\epsilon_0}.\frac{qx}{(x^2+a^2)^{3/2}}\\If x >> a, then x^2 + a^2 ≈ x^2\\E =\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2)^{3/2}}\\E =\frac{1}{4\pi\epsilon_0}\frac{q}{(x^2)}$$

This expression is similar to electric field due to a point charge

(b) The surface charge density for a spherical conductor of radius R1 is given by :
$$\sigma_1=\frac{q_1}{4\pi R^2_1}\\\text{Similarly, for spherical conductor }R_2, \text{the surface charge density is given by :}\\\sigma_2=\frac{q_2}{4\pi R^2_2}\\\frac{\sigma_1}{\sigma_2}=\frac{q_1R^2_1}{q_1R^2_2}…(i)\\\text{As the spheres are connected so the charges will flow between the spherical conductors till their potential become equal.}\\i.e.,\frac{kq_1}{R_1}=\frac{kq_2}{R_2}\\⇒\frac{q_1}{q_2}=\frac{R_1}{R_2}…(ii)\\\text{Using (ii) in (i), We have}\\\frac{\sigma_1}{\sigma_2}=\frac{R_1}{R_2}.\frac{R^2_2}{R^2_1}=\frac{R_2}{R_1}\\\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}$$

32. According to the question,
(a) Work done :
$$W=\frac{kQ_1Q_3}{r/2}-\frac{kQ_3Q_2}{r/2}\\W=\frac{2kQ_1Q_3}{r}-\frac{2kQ_3Q_2}{r}\\=\frac{2kQ_3(Q_1-Q_2)}{r}\\=\frac{2Q_3(Q_1-Q_2)}{4\pi\epsilon_0r}$$

(b) For work done to be zero : Work done will be zero for the point where the electric potential is zero
consider a point P to be lying in between the two charges, where electric potential is zero. Let the distance
of the point P from the charge Q1 be x.
$$V=\frac{kQ_1}{x}-\frac{kQ_2}{(r-x)}=0\\⇒\frac{Q_1}{x}-\frac{Q_2}{(r-x)}=0\\⇒ rQ_1–xQ_1–xQ_2=0\\⇒x=\frac{rQ_1}{(Q_1+Q_2)}$$

OR

(a) The electric field at any point A having
$$\vec{r}$$

as its position-vector due to a point charge q placed at the
origin of the coordinate system is given by,
$$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}=\frac{kq}{r^2}\hat{r}[...k=\frac{1}{4\pi\epsilon_0}=9×10^9Nm^2/C^2]\\\text{The electric potential at point A as shown in the figure is given by,}\\V (A)=–\int^A_∞\vec{E}.\vec{dr}\\\text{= Amount of work done to bring unit positive charge from ∞ to A.}\\V (A) = –\int^A_∞|\frac{kq}{r^2}\hat{r}|.(d.r\hat{r})\\= –kq\int^A_∞\frac{dr}{r^2}=-kq[-\frac{1}{r}]^r_∞\\=\frac{kq}{r}\\V(A)=V(r)=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$$

The electric potential is positive if q is positive and negative if q is negative. Hence, for negative charge negative sign should be used for q in the above equation.

(b) Electrostatic potential due to electric dipole: Let the two charges – q and + q be placed at A and B respectively with a distance 2a between them.
Let P be a unit at a distance r from the centre of this dipole and making an angle q with its direction such that AP = r1 and BP = r2

The electric potential at P is given by
$$V (r)=\frac{1}{4\pi\epsilon_0}\frac{q}{r_1}-\frac{1}{4\pi\epsilon_0}\frac{q}{r_2}…(i)\\\text{Now, by geometry,}\\r^2_1=r^2+a^2–2ar cos\theta\\r^2_2=r^2+a^2+2ar cos\theta\\\text{We take r much greater than a (r > > a) and retain terms only upto the first order in}\frac{a}{r}\\r^2_1=r^2(1-\frac{2a cos\theta}{r}+\frac{a^2}{r^2})\\=r^2(1-\frac{2a cos\theta}{r})\\\text{Similarly,}r^2_2=r^2\\\text{Using the Binomial theorem and retaining terms upto the first order in}\frac{a}{r}\text{we obtain,}\\\frac{1}{r_1}\cong\frac{1}{r}(1-\frac{2a cos\theta}{r})^{\frac{-1}{2}}\\\cong\frac{1}{r}(1+\frac{a cos\theta}{r})…(ii)\\\frac{1}{r_2}\cong\frac{1}{r}(1+\frac{2a cos\theta}{r})^{\frac{-1}{2}}\\\cong\frac{1}{r}(1-\frac{a cos\theta}{r})…(iii)\\\text{Putting equations (ii) and (iii) in equation (i), we get}\\\ V(r)=\frac{q}{4\pi\epsilon_0}(\frac{2acos\theta}{r^2})=\frac{1}{4\pi\epsilon_0}(\frac{pcos\theta}{r^2})(˙.˙ p=2aq)\\\text{As p cos}\theta=\vec{p}\hat{r}\\V(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{\vec{p}\hat{r}}{r^2}\space[for|\vec{r}|>>2a]$$

33. (a)
f1 = fconvex = 30 cm
f2 = fconcave = –20 cm
$$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}\\\frac{1}{f}=-\frac{1}{60}\space[... concave (diverging) lens]$$

(b) Let C is the critical angle,
$$sin\space C=\frac{1}{\mu}\\r_2=\text{C (for total internal reflection)}\\r_1+r_2=A\\r_1+C=A\\r_1=A–C\\\mu=\frac{sini}{sinr_1}\\⇒ sini=\mu sinr_1\\sini=\mu sin(A–C)\\i=sin^{-1}[\mu sin(A–C)]\\i=sin^{-1}[\frac{sin(A-C)}{sin C}]$$

OR

(a) For first lens,
$$\frac{1}{f_1}=\frac{1}{v'}-\frac{1}{u}…(i)$$

$$\frac{1}{f_1}=\frac{1}{v'}-\frac{1}{u}…(i)\\\text{For second lens,}\\\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v'}…(ii)\\\text{Adding (i) and (ii),}\\\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}…(iii)\\\text{If f is the combined focal length then}\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\text{From equation (iii),}\\\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\P =\frac{1}{f_1}+\frac{1}{f_2}[...\frac{1}{f}=p]\\P =\frac{f_1+f_2}{f_1f_2}$$

(b) (i)

(ii) An astronomical telescope should have an objective of large aperture and longer focal length while
an eyepiece of small aperture and small focal length.
Therefore, we will use L1 as an objective and L3 as an eyepiece.

#### CBSE Practice Paper Physics Class 12

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