M.C.Q (1 Marks)

MCQ 11 Mark

The radius of curvature of the curved surface of a plano$-$convex lens is $20 \ cm.$ If the refractive index of the material of the lens be $1.5,$ it will

A
act as a concave lens irrespective of side on which the object lies
B
act as a convex lens only for the objects that lie on its curved side
C
act as a concave lens for the objects that lie on its curved side
✓
act as a convex lens irrespective of the side on which the object lies

Answer

Correct option: D.

act as a convex lens irrespective of the side on which the object lies

The relation between focal length $f ,$ the refractive index of the given material $\mu, R _1$ and $R _2$ is known as lens maker's formula and it is $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$R_1=\infty, R_2=-R$
$f=\frac{R}{(\mu-1)}$

Here, $R = 20 \ cm, \mu= 1.5.$ On substituting the values, we get
$ f=\frac{R}{\mu-1}=\frac{20}{1.5-1}=40 \ cm $
As $f >0$ means converging nature.
Therefore, the lens act as a convex lens irrespective of the side on which the object lies.

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MCQ 21 Mark

Assume that each diode shown in the figure has a forward bias resistance of $50 \Omega$ and an infinite reverse bia resistance. The current through the $150 \Omega$ resistance is

✓
$0.04$
B
zero
C
$0.05$
D
$0.66$

Answer

Correct option: A.

$0.04$

Diode $D_1$ is forward biased and offers $50 \Omega$ resistance. Diode $D_2$ is reverse biased and offers infinite resistance. The equivalent circuit is

Current through the $150 \Omega$ resistance,
$I=\frac{10}{50+50+150}$
$=\frac{10}{250}=0.04 A$

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MCQ 31 Mark

Select the correct statements. Coulomb’s law correctly describes the electric force that:
i. binds the electrons of an atom to its nucleus.
ii. binds the protons and neutrons in the nucleus of an atom.
iii. binds atoms together to form molecules.

A
(i), (ii), and (iii)
B
(i) and (iii)
C
(ii) and (iii)
D
(i) and (ii)

Answer

(b) (i) and (iii)
Explanation: According to Coulomb’s law, electric force binds the electrons of an atom to its nucleus and atoms together to form molecules.

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MCQ 41 Mark

Consider a ray of light incident from air onto a slab of glass (refractive index $n$ ) of width d , at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

A
$\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}$
B
$\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+2 \pi$
C
$\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\frac{\pi}{2}$
D
$\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\pi$

Answer

(d) $\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\pi$
Explanation: The ray reflected by the top surface of the glass and the bottom surface is:-
$\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\pi$

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MCQ 51 Mark

A paramagnetic sample shows a net magnetisation of $8 Am ^{-1}$ when placed in an external magnetic field of $0.6 \ T$ at a temperature of $4 \ K .$ When the same sample is placed in an external magnetic field of $0.2 \ T$ at a temperature of $16 \ K ,$ the magnetisation will be

A
$6 Am ^{-1}$
✓
$\frac{2}{3} Am ^{-1}$
C
$2.4 Am ^{-1}$
D
$\frac{32}{3} Am ^{-1}$

Answer

Correct option: B.

$\frac{2}{3} Am ^{-1}$

On increasing the temperature magnetic susceptibility of paramagnetic material decreases or vice versa . According to Curie law, we can deduce a formula for the relation between magnetic field induction, temperature and magnetisation.
i.e., $I ($magnetization$) \propto \frac{B(\text { magnetic field induction })}{t(\text { temperature in kelvin })}$
$\Rightarrow \frac{I_2}{I_1}=\frac{B_2}{B_1} \times \frac{t_1}{t_2}$
Let us suppose, here $I _1=8 Am ^{-1}$
$ B _1=0.6 T, t _1=4 K$
$B _2=0.2 T, t _2=16 K$
$\Rightarrow \frac{0.2}{0.6} \times \frac{4}{16}=\frac{I_2}{8}$
$\Rightarrow I_2=8 \times \frac{1}{12}=\frac{2}{3} Am ^{-1}$

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MCQ 61 Mark

A conducting circular loop is placed in a uniform magnetic field, $B =0.025 T$ with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of $1 \ mm \ s ^{-1}$. The induced emf when the radius is $2 \ cm ,$ is

A
$2 \mu V$
✓
$\pi \mu V$
C
$2 \pi \mu V$
D
$\frac{\pi}{2} \mu V$

Answer

Correct option: B.

$\pi \mu V$

$\phi=B \pi r^2 \cos 0^{\circ}=B \pi r^2$
$|\varepsilon|=\frac{d \phi}{d t}=\frac{d}{d t}\left(\pi r^2 B\right)$
$=2 \pi r B \frac{d r}{d t}$
When $r =2 \ cm=2 \times 10^{-2} m \frac{d r}{d t}$
$=1 \ mm \ s^{-1}=10^{-3} ms^{-1}$
$|\varepsilon|=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 10^{-3}$
$=0.100 \times \pi \times 10^{-5}$
$=\pi \mu V$

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MCQ 71 Mark

The magnetic moment of a current (l) carrying circular coil of radius (r) and number of turns (n) varies as

A
$\frac{1}{r^2}$
B
r
C
$\frac{1}{r}$
D
$r^2$

Answer

(d) $r ^2$
Explanation: $M = IA = I \times \pi r ^2$ i.e., $M \propto r^2$

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MCQ 81 Mark

Submarine cables act as

A
spherical capacitor
B
cylindrical capacitor with inner cylinder earthed
C
cylindrical capacitor with inner cylinder earthed
D
cylindrical capacitor with outer cylinder earthed

Answer

(d) cylindrical capacitor with outer cylinder earthed
Explanation: A submarine cable consists of an inner conductor which carries power. This conductor is covered by an insulator, which acts as a dielectric. The dielectric material is covered by a metal coating called shield, which is connected to ground. The cable acts as a cylindrical capacitor, with the conductor acting as the inner cylinder, and the metal shield as the outer cylinder which is connected to earth.

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MCQ 91 Mark

The $\mu_0$ is also known as :