M.C.Q (1 Marks)

MCQ 11 Mark

Green light of wavelength 5460 $\stackrel{\circ}{A}$ is incident on an air-glass interface. If the refractive index of glass is 1⋅ 5, the wavelength of light in glass would be (Given that the velocity of light in air, $c =3 \times 10^8 m s ^{-1}$ )

A
6,731 $\stackrel{\circ}{A}$
B
3,640 $\stackrel{\circ}{A}$
C
5,446 $\stackrel{\circ}{A}$
D
4,861 $\stackrel{\circ}{A}$

Answer

(b) : 3,640 $\stackrel{\circ}{A}$
Explanation: Now, $\lambda^{\prime}=\frac{\lambda}{\mu}=\frac{5460}{1.5}=$3,640 $\stackrel{\circ}{A}$

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MCQ 21 Mark

In the circuit given in the figure, an a.c. source of 200 V is connected through a diode D to a capacitor. The potential difference across the capacitor will be

A
283 V
B
100 V
C
310 V
D
200 V

Answer

(a) : 283 V
Explanation: A diode conducts only during the positive half cycle of a.c. Accordingly, the capacitor charges and discharges. During charging, the p.d. across capacitor
$=200 \times \sqrt{2}=283 V$

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MCQ 31 Mark

If an electron is accelerated by $8.8 \times 10^{14} m / s ^2$ , then electric field required for acceleration is $($given specific charge of the electron $=1.76 \times 10^{11} Ckg ^{-1} )Z$

A
$52 V cm ^{-1}$
✓
$50 V cm ^{-1}$
C
$54 V cm ^{-1}$
D
$56 V cm ^{-1}$

Answer

Correct option: B.

$50 V cm ^{-1}$

$a=8.8 \times 10^{14} m / s ^2$
$\frac{e}{m}=1.76 \times 10^{11} C^{-1} \ kg^{-1}$
$a=\frac{F}{m}=\frac{e E}{m}=\left(\frac{e}{m}\right) E$
$8.8 \times 10^{14}=1.76 \times 10^{11} \times E$
$E=\frac{8.8 \times 10^{14}}{1.76 \times 10^{11}}$
$=5000 Vm ^{-1}$
$=50 Vcm ^{-1}$

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MCQ 41 Mark

The shape of the wavefront of the portion of the wavefront of light from a distant star intercepted by the earth is

A
plane
B
spherical
C
conical
D
hyperboloid

Answer

(a) : plane
Explanation: Stars are very far away from earth. Near the star the shape is spherical but by the time its light reaches earth, the portion of the wavefront is plane due to increase in radius.

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MCQ 51 Mark

An aeroplane having a wingspan of $35m$ files due north with the speed of $90 m/s,$ given $B=4 \times 10^{-5} T$ The potential difference between the tips of the wings will be

✓
$0.126 V$
B
$1.26 V$
C
$0.013 V$
D
$12.6 V$

Answer

Correct option: A.

$0.126 V$

$0.126 V$
Explanation:
$\varepsilon=B l v$
$=4 \times 10^{-5} \times 35 \times 90$
$=126 \times 10^{-3} V=0.126 V$

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MCQ 61 Mark

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, change of flux linkage with the other coil is

A
45 Wb
B
35 Wb
C
30 Wb
D
40 Wb

Answer

(c) : 30 Wb
Explanation: $\Delta \phi=M \Delta i=1.5 \times 20=30 Wb$

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MCQ 71 Mark

An electron is travelling along the X-direction. It encounters a magnetic field in the Y-direction. Its subsequent motion will be:

A
a circle in the YZ-plane
B
straight line along the X-direction
C
a circle in the XZ-plane
D
a circle in the XY-plane

Answer

(c) :a circle in the XZ-plane
Explanation: $\vec{F}=q(\vec{v} \times \vec{B})=-e(v \hat{i} \times B \hat{j})$
$=-e v B i \times \hat{j}=-e v B \hat{k}$

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MCQ 81 Mark

A parallel plate capacitor of plate area A has a charge Q. The force on each plate of the capacitor is

A
$\frac{2 q^2}{\epsilon_o A}$
B
zero
C
$\frac{q^2}{\epsilon_o A}$
D
$\frac{q^2}{2 \epsilon_o A}$

Answer

(d) : $\frac{q^2}{2 \epsilon_\sigma A}$
Explanation: Force between two plates of the capacitor
$F = u A$, where $u=\frac{1}{2} \varepsilon_0 E^2$ is the energy density and $A =$ Area of each plate Also the electric field, $E=\frac{\sigma}{\varepsilon_0}$ and the charge density, $\sigma=\frac{q}{A}$
Therefore, $F=\frac{1}{2} \varepsilon_0 E^2 A=\frac{1}{2} \varepsilon_0\left(\frac{\sigma}{\varepsilon_0}\right)^2 A=\frac{1}{2} \frac{\sigma^2 A}{\varepsilon_0}=\frac{1}{2}\left(\frac{q}{A}\right)^2 \frac{A}{\varepsilon_0}=\frac{q^2}{2 A \varepsilon_0}$

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MCQ 91 Mark

A magnet of magnetic moment M is suspended in a uniform magnetic field B. The maximum value of torque acting on the magnet is

A
zero
B
MB
C
2MB
D
$\frac{1}{2} M B$

Answer

(b) : MB
Explanation: $\tau=M B \sin \theta$
$\tau_{\max }=M B \sin 90^{\circ}=M B$

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MCQ 101 Mark

The focal length of a concave mirror is $f$. An object is placed at a distance $x$ from the focus. The magnification is

A
$\frac{f}{(f+x)}$
B
$\frac{(f+x)}{f}$
✓
$\frac{f}{x}$
D
$\frac{x}{f}$

Answer

Correct option: C.

$\frac{f}{x}$

Explanation:
$u = f + x$
Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Or $,\frac{1}{v}-\frac{1}{(f+x)}=-\frac{1}{f}$
$\therefore v=-\frac{f(f+x)}{x}$
So, the magnification $=| m |=\frac{v}{u}=\frac{f}{x}$

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MCQ 111 Mark

Current density of a conductor is

A
Is always zero
B
the net charge flowing through the area
C
measure of the flow of electric charge in amperes per unit area of cross-section
D
the net charge flowing through the area per unit time

Answer

(c) : measure of the flow of electric charge in amperes per unit area of cross-section
Explanation: Current density $J =\frac{I}{A}$
In electromagnetism, current density is the electric current per unit area of cross section. Or It is the measure of the flow of electric charge in amperes per unit area of cross-section .It is a vector and has a direction along the area vector.

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MCQ 121 Mark

The cations and anions are arranged in alternate form in

A
ionic crystal
B
semiconductor crystal
C
covalent crystal
D
metallic crystal

Answer

(a) : ionic crystal
Explanation: In an ionic crystal, cations and anions are arranged in alternate form.

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Physics M.C.Q (1 Marks)

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