M.C.Q [1M]

MCQ 11 Mark

An electron is revolving around a proton in a circular orbit of diameter 0.1 nm . It produces a magnetic field of $14 Wb / m ^2$ at the proton. What is angular speed of the electron?

A
$4.4 \times 10^{16} rads ^{-1}$
B
$8.8 \times 10^{16} rads ^{-1}$
C
$6.4 \times 10^{16} rads ^{-1}$
D
$1.4 \times 10^{16} rads ^{-1}$

Answer

(a): $4.4 \times 10^{16} rads ^{-1}$
Explanation: The revolving electron is similar to a loop carrying current. Field at the center of the loop of radius r is $B=\frac{\mu_0 I}{2 r}$.
The current due to the revolving electron $I=\frac{B(2 r)}{10}=\frac{14 \times 0.1 \times 10^{-9}}{4 \times 10^{-7}}=\frac{7 \times 10^{-3}}{2 \pi}$
The current can also be written as, $I=\frac{e}{T}$
where, T is the time taken to complete one revolution.
Since $T=\frac{2 \pi}{\omega}$ where $\omega$ is the angular speed of the electron,
$\begin{array}{l}I=\frac{\in}{T}=\frac{\in \omega}{2 \pi} \\ \frac{\in \omega}{2 \pi}=\frac{7 \times 10^{-3}}{2 \pi} \\ \omega=\frac{7 \times 10^{-3}}{\in}=\frac{7 \times 10^{-3}}{1.5 \times 10^{-19}} \\ =4.38 \times 10^{16} \approx 4.4 \times 10^{16} rad / s \end{array}$

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MCQ 21 Mark

In the diagram, a prism of angle $30^{\circ}$ is used. A ray PQ is incident as shown. An emergent ray RS emerges perpendicular to the second face. The angle of deviation is:

A
$60^{\circ}$
B
$0^{\circ}$
C
$30^{\circ}$
D
$45^{\circ}$

Answer

(c): $30^{\circ}$
Explanation:

In $\triangle AQR$
$\begin{array}{l}\angle A +\angle Q +\angle R =180^{\circ} \\
30^{\circ}+90- r +90=180^{\circ}\end{array}$
$\begin{array}{l} r =30^{\circ} \\
\delta=90^{\circ}-30^{\circ}- r \\
\delta=90^{\circ}-30^{\circ}-30^{\circ} \\
\delta=30^{\circ}\end{array}$
so angle of deviation is 30 degree.

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MCQ 31 Mark

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r . The Coulomb force $\vec{F}$ between the two is:

A
$\kappa \frac{e^2}{r^3} \vec{r}$
B
$-\kappa \frac{e^2}{r^3} \vec{r}$
C
$-\kappa \frac{e^3}{s^3} \hat{r}$
D
$\kappa \frac{e^2}{n^3} \hat{r}$

Answer

(b): $-\kappa \frac{e^2}{r^3} \vec{r}$
Explanation: Charge on an electron = -e
Charge on nucleus of hydrogen = +e
$\therefore \vec{F}=\kappa \frac{(-e) \times e}{r^2} \hat{r}=-\frac{k e^2}{r^3} \vec{r}$
Here $\hat{r}=\frac{\vec{r}}{r}$ is unit vector along the line joining electron to the nucleus. The negative sign shows that the force is of attraction.

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MCQ 41 Mark

How does the magnetic susceptibility $\chi$ of a paramagnetic material change with absolute temperature T?

A
$\chi \propto e^T$
B
$\chi \propto e^T$
C
$\chi=$ constant
D
$\chi \propto T^{-1}$

Answer

(d): $\chi \propto T^{-1}$
Explanation: $\chi \propto T^{-1}$

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MCQ 51 Mark

A biconvex lens of focal length f is cut into two identical plano convex lenses. The focal length of each part will be

A
2f
B
$\frac{f}{2}$
C
4f
D
f

Answer

(a): 2f
Explanation: The focal length of each part will be 2f

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MCQ 61 Mark

The wavefront of a distant source of unknown shape is approximately:

A
plane
B
elliptical
C
cylindrical
D
spherical

Answer

(a): plane
Explanation: When the point source or linear source of light is a very large distance, a small portion of the spherical or cylindrical wavefront appears to be plane. Such a wavefront is plane wavefront.

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MCQ 71 Mark

An electric bulb marked 40 W - 200 V is used in a circuit of supply voltage 100 V. Now its power is:

A
10W
B
40W
C
20W
D
100W

Answer

(a): 10 W
Explanation: $P =\frac{V^2}{R}$ i.e., $P \propto V^2$
$\begin{array}{l}\text { or } \frac{P_1}{P_2}=\frac{V_1^2}{V_2^2} \text { or } \frac{40}{P_2}=\frac{(200)^2}{(100)^2} \\ \text { or } P _2=10 W\end{array}$

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MCQ 81 Mark

Which of the following has its permeability less than that of free space?

A
Copper
B
Nickel
C
Copper chloride
D
Aluminium

Answer

(a): Copper
Explanation: as Copper is diamagnetic substance.

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MCQ 91 Mark

The dimension of $\frac{1}{2} \varepsilon_0 E^2$ where $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field, is

A
$ML ^{-1} T^{-2}$
B
$MLT ^{-1}$
C
$ML ^2 T^{-2}$
D
$ML ^2 T^{-1}$

Answer

(a): $ML ^{-1} T^{-2}$
Explanation: $\left[\frac{1}{2} \varepsilon_0 E^2\right]=$ energy density
$=\frac{ ML ^2 T^{-2}}{L^3}=\left[ ML ^{-1} T ^{-2}\right]$

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MCQ 101 Mark

For MRI, a patient is slowly pushed in a time of 10 s within the coils of the magnet where magnetic field is B = 2.0 T. If the patient's trunk is 0.8 m in circumference, the induced emf around the patient's trunk is

A
$10.18 \times 10^{-3} V$
B
$10.18 \times 10^{-2} V$
C
$1.51 \times 10^{-2} V$
D
$9.66 \times 10^2 V$

Answer

(a): $10.18 \times 10^{-3} V$
Explanation: Change in magnetic field in $10 s=2.0 T$
As $\varepsilon=\frac{-d \phi}{d t}=-A \frac{d B}{d t} \quad(\because \phi=B A)$
Circumference of patient's trunk,
$\begin{array}{l}2 \pi r=0.8 m \text { (given) } \\ \therefore r=\frac{0.8}{2 \pi} m=\frac{0.4}{\pi} m\end{array}$
Area of cross-section,
$\begin{array}{l}A=\pi r^2=\pi\left(\frac{0.4}{\pi}\right)^2=\frac{0.16}{\pi} m^2 \\ \therefore|\varepsilon|=\frac{0.16}{\pi} \times \frac{2}{10} V \\ \approx 10.18 \times 10^3 V\end{array}$

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MCQ 111 Mark

The electrical conductivity of semiconductor increases when electromagnetic radiation of wavelength shorter than 2800 nm is incident on it. The band gap in (eV) for the semiconductor is:

A
0.5 eV
B
0.7 eV
C
2.5 eV
D
1.2 eV

Answer

(a): 0.5 eV
Explanation: $E_g=\frac{h c}{\lambda}=\frac{1240 eVnm }{2800 nm}=0.5 eV$

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MCQ 121 Mark

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

A
zero
B
0.5A
C
0.75A
D
0.25A

Answer

(b): 0.5 A
Explanation: Diode $D _1$ conducts as it is forward biased.
Diode $D _2$ does not conduct as it is reverse biased.
$\therefore I=\frac{5 V}{100}=0.5 A$

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Physics M.C.Q [1M]

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