M.C.Q (1 Marks)

MCQ 11 Mark

A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

A
225
B
450
C
1800
D
1000

Answer

(c) 1800
Explanation: Power radiated, $P =\sigma A T^4$
$\begin{aligned} & \therefore P^{\prime}=\sigma\left(\frac{A}{4}\right)(2 T)^4\left[A \propto r ^2\right] \\ & =4 P =4 \times 450 W=1800 W\end{aligned}$

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MCQ 21 Mark

A thin uniform rod of length $2 l$ and mass M is acted upon a constant torque. The angular velocity changes from zero to $\omega$ in time $t$. The value of torque is:

A
$\frac{M l^2 \omega}{3 t}$
B
$\frac{2 M l^2 \omega}{3 t}$
C
$\frac{M l^2 \omega}{12 t}$
D
$\frac{M l^2 \omega}{t}$

Answer

(a) $\frac{M 2^2 \omega}{3 t}$
Explanation:
As Torque $(\tau)$ is equal to the product of Moment of Inertia (I) and Angular acceleration ( $\alpha$ )
$
\begin{aligned}
\tau & =I \alpha \\
\tau & =I \frac{\Delta \omega}{\Delta t} \\
\tau & =\left[\frac{M(2 l)^2}{12}\right]\left[\frac{\omega}{t}\right] \\
\tau & =\frac{M l^2 \omega}{3 t}
\end{aligned}
$

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MCQ 31 Mark

Satellites orbiting the earth have a finite life and sometimes debris of satellites fall to the earth. This is because,

A
of viscous forces causing the speed of the satellite and hence height to gradually decrease.
B
the solar cells and batteries in satellites run out.
C
of collisions with other satellites.
D
the laws of gravitation predict a trajectory spiralling inwards.

Answer

(d) the laws of gravitation predict a trajectory spiralling inwards.
Explanation: Due to the viscous atmosphere, friction force due to the atmosphere acts on the satellite which reduces its orbital speed and hence the energy of revolution around a planet. Due to the decrease in the energy of the satellite, its height gradually decreases.

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MCQ 41 Mark

An open tank filled with water (depsity $\rho$ ) has a narrow hole at a depth of h below the water surface. The velocity of water flowing out is

A
2 gh
B
$\sqrt{2 g h}$
C
$h \rho g$
D
gh

Answer

(b) $\sqrt{2 g h}$
Explanation: Velocity of efflux, $v =\sqrt{2 g h}$

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MCQ 51 Mark

The phase difference between two waves, represented by
$
\begin{aligned}
& y_1=10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right] m \\
& y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] m
\end{aligned}
$
where x is expressed in metres and $t$ is expressed in seconds, is approximately

A
1.07 rad
B
1.5 rad
C
0.5 rad
D
2.07 rad

Answer

1.07 rad
$
\begin{aligned}
& \text { Explanation: } y_1=10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right] m \\
& y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] m \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+\left(\frac{\pi}{2}\right)\right] \\
& \Delta \phi=\frac{\pi}{2}-0.05=\frac{3.14}{2}-0.5=1.57-0.5 \\
& =1.07 rad
\end{aligned}
$

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MCQ 61 Mark

A body $A$ is thrown up vertically from the ground with a velocity $v _0$ and another body $B$ is simultaneously dropped from a height H . They meet at a height $\frac{H}{2}$, if $v _0$ is equal to:

A
$\sqrt{2 g H}$
B
$\sqrt{\frac{2 g}{H}}$
C
$\sqrt{g H}$
D
$\frac{1}{2} \sqrt{g H}$

Answer

$\sqrt{g H}$
Explanation:
Let the two bodies A and B respectively meet at a time, at a height $\frac{H}{2}$ ground.

Using $S = ut +\frac{1}{2} a t^2$
For a body $A , u = V _0, a =- g , S =\frac{H}{2}$
$
\therefore \frac{H}{2}=v_0 t-\frac{1}{2} g t^2\ldots(i)
$
For a body B, u $=0, a =+\frac{g}{g}, S=\frac{H}{2}$
$
\therefore \frac{H}{2}=\frac{1}{2} gt^2\ldots(ii)
$
Equating equation (i) and (ii) we get
$V_0 t-\frac{1}{2} g t^2=\frac{1}{2} g t^2$
$
V_0 t=gt^2 \text { or } t=\frac{V_0}{g}
$
Substituting the value of $t$ is equation (i), we get
$
\begin{aligned}
& \frac{H}{2}=V_0 \times\left(\frac{V_0}{g}\right)-\frac{1}{2} g\left(\frac{V_0}{g}\right)^2=\frac{V_0^2}{V_0}-\frac{1}{2} \frac{V_0^2}{g} \\
& \frac{H}{2}=\frac{1}{2} \frac{V_0^2}{g} \text { or } v_0^2=gH \\
& V_0=\sqrt{g H}
\end{aligned}
$

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MCQ 71 Mark

A body is moving forwards and backward. Change in frequency observed of source is 2%. What is velocity of the body? (Speed of sound is 300 m/s)

A
3 m/s
B
2.5 m/s
C
2 m/s
D
6 m/s

Answer

$3 m / s$
Explanation: When the source is moving forward towards the observer, the apparent frequency is $f_1=\frac{v}{v-v_{ s }} \times f$
When source moves backwards $f_2=\frac{v}{v+v_s} \times f$
$
f_2-f_1=f v\left[\frac{1}{v+v_s}-\frac{1}{v-v_s}\right]=f v\left[\frac{-2 v_s}{v^2-v_s^2}\right]
$
As $v_{ s }<< v$, so
$
\begin{aligned}
& \frac{f_2-f_1}{f}=\left|\frac{2 v_s}{v}\right|=\frac{2}{100} \\
& v_s=\frac{v}{100}=\frac{300}{100}=3 m / s
\end{aligned}
$

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MCQ 81 Mark

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is $400 \ km$ above the earth's surface; at the high point, or apogee, it is $4000 \ km$ above the earth's surface. Using conservation of energy, find the speed at perigee and the speed at apogee. It is necessary to have the spacecraft escape from the earth completely.

✓
$9840\ m / \sec ($perigee$), 9760 m / \sec ($apogee$)$
B
$7840\ m / \sec ($ perigee), $5760\ m / \sec ($apogee$)$
C
$10840\ m / \sec ($perigee$), 8760\ m / \sec ($apogee$)$
D
$8840\ m / \sec ($perigee$), 6760\ m / \sec ($apogee$)$

Answer

Correct option: A.

$9840\ m / \sec ($perigee$), 9760 m / \sec ($apogee$)$

To escape Earth, we need total energy of zero.
$\text { ( } E_{\text {final }}=0 \text { because } U \longrightarrow 0 \text { as } R \longrightarrow \infty \text { and } K \longrightarrow 0 \text { as } v=0 \text { at } R \longrightarrow \infty \text { ) }$
So,
$K_{p}+U_{p}=0$
Looking for the new velocity at perigee;
$\frac{1}{2} m v_{p, \text { escapc }}^2=\frac{G M m}{R_p}$
$ v_{p, \text { escape }}=\sqrt{\frac{2 G M}{R_p}}$
$=\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.78 \times 10^6}}$
$ =1.084 \times 10^4 m / \sec$
$=10840 m / \sec$
The similar calculation at apogee gives
$v_{\text {d, scaps }}=\sqrt{\frac{2 G GM}{R_a}}$
$=\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{10.38 \times 10^6}}$
$=8.76 \times 10^3 m / \sec$
$=8760 m / \sec$

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MCQ 91 Mark

At critical temperature, the surface tension of a liquid

A
is zero
B
is infinity
C
cannot be determined
D
is same as that any other temperature

Answer

(a) is zero
Explanation: At the critical temperature, the surface tension of a liquid becomes zero.

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MCQ 101 Mark

If a gymnast sitting on a rotating stool with his arms outstretched, suddenly lowers his hands:

A
The angular velocity decreases
B
His moment of inertia decreases
C
The angular momentum increases
D
The angular velocity stays constant

Answer

(b) His moment of inertia decreases
Explanation: When gymnast lowers his hand the distance of the mass from rotational axis decrease. Hence his moment of inertia decreases and angular velocity increases to conserve angular momentum.

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MCQ 111 Mark

Two closed organ pipes, when sounded simultaneously gave 4 beats per sec. If longer pipe has a length of 1 m , then length of shorter pipe is $( v =300 m / s )$

A
80 cm
B
94.9 cm
C
90 cm
D
185.5 cm

Answer

(b) 94.9 cm
Explanation: $\nu=\frac{v}{4 L}=\frac{300}{4 \times 1}=75 Hz$
For shorter pipe,
$
\begin{aligned}
& \nu+4=\frac{300}{4 L^{\prime}} \\
& \text { or } 75+4=\frac{300}{4 L} \\
& L^{\prime}=\frac{300}{4 \times 79} m=94.9 cm
\end{aligned}
$

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MCQ 121 Mark

According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\frac{\Delta v}{\Delta x}$ is given by $F=-\eta A \frac{\Delta v}{\Delta x}$, where $\eta$ is constant called coefficient of viscosity. The dimensional formula of $\eta$ is

A
$\left[ ML ^{-2} T^{-2}\right]$
B
$\left[ M ^0 L^0 T^0\right]$
✓
$\left[ ML ^{-1} T^{-1}\right]$
D
$\left[ ML ^2 T^{-2}\right]$

Answer

Correct option: C.

$\left[ ML ^{-1} T^{-1}\right]$

$|\eta|=\left[\frac{F \Delta r}{A \Delta v}\right]$
$=\frac{\left[M L T^2 \mid L\right]}{\left[L^2\right]\left[L T^{-1}\right]}$
$=\left[M L^{-1} T^{-1}\right]$

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Physics M.C.Q (1 Marks)

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