M.C.Q (1 Marks)

MCQ 11 Mark

Two identical metallic balls, whose temperatures are $200^{\circ} C$ and $400^{\circ} C$ respectively, are placed in an enclosure at $27^{\circ} C$. The ratio of heat-loss of the balls will be

A
$1: 4$
B
$\frac{(473)^4-(300)^4}{(673)^4-(300)^4}$
C
$1: 2$
D
$\frac{(200)^4-(27)^4}{(400)^4-(27)^4}$

Answer

(b) $\frac{(473)^4-(300)^4}{(673)^4-(300)^4}$
Explanation: From Stefan Boltzmann's law,
$
\begin{aligned}
& \frac{E_1}{E_2}=\frac{T_1^4-T_0^4}{T_2^4-T_0^4} \\
& =\frac{(273+200)^4-(273+27)^4}{(273+400)^4-(273+27)^4} \\
& =\frac{(473)^4-(300)^4}{(673)^4-(300)^4}
\end{aligned}
$

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MCQ 21 Mark

The angular momentum of a moving body remains constant, if

A
net pressure is applied
B
net external torque is applied
C
net external force is applied
D
net external torque is not applied

Answer

(d) net external torque is not applied
Explanation: Angular momentum is conserved only in the absence of external torque.

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MCQ 31 Mark

An astronaut is standing on an asteroid when he accidentally drops a wrench. He observes that the gravitational acceleration on the asteroid is $2.4 m / s ^2$. If he had thrown the wrench at an upward angle instead, he would have found the gravitational acceleration on the asteroid to be:

A
less than $2.4 m / s ^2$
B
downward at $2.4 m / s ^2$
C
greater than $2.4 m / s ^2$
D
toward him at $2.4 m / s ^2$

Answer

(b) downward at $2.4 m / s ^2$
Explanation: The acceleration will be downward at $2.4 m / s ^2$.

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MCQ 41 Mark

One large soap bubble of diameter D breaks into 27 bubbles having surface tension T. The change in surface energy is

A
$2 \pi TD ^2$
B
$8 \pi TD ^2$
C
$4 \pi TD ^2$
D
$\pi TD ^2$

Answer

(c) $4 \pi TD ^2$
Explanation: Volume of 27 small bubbles = Volume of larger bubble
$
\begin{aligned}
& 27 \times \frac{4}{3} \times \pi r^3=\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \\
& \therefore r=\frac{D}{6}
\end{aligned}
$
Increase in surface area,
$
=2\left[27 \times 4 \pi\left(\frac{D}{6}\right)^2-4 \pi\left(\frac{D}{2}\right)^2\right]=4 \pi D^2
$
Increase in surface energy, $=$ Increase in surface area $\times$ surface tension $=4 \pi D ^2 T$

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MCQ 51 Mark

A standing wave is represented by:
$
y=a \sin (100 t) \cos (0.01 x)
$
where y and a are in millimetre, t in second and x in metre. Velocity of wave is:

A
not derived
B
$1 m / s$
C
$10^4 m / s$
D
$10^{-4} m / s$

Answer

$\begin{aligned} & \text { } (c) 10^4 m / s \\ & \text { Explanation: } y = A \sin (100 t ) \cos (0.01 x ) \\ & y=A \sin \left(\frac{2 \pi}{T} t\right) \cos \left(\frac{2 \pi}{\lambda} x\right) \\ & \therefore \frac{2 \pi}{T}=100 \text { or } T =\frac{\pi}{50} \\ & \frac{2 \pi}{\lambda}=0.01 \text { or } \lambda=200 \pi \\ & v=\frac{\lambda}{T}=\frac{200 \pi}{\frac{\pi}{50}}=10^4 m / s \end{aligned}$

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MCQ 61 Mark

A police jeep is chasing with velocity $45 km / h$. A thief in another jeep is moving with $155 km / h$. Police fires a bullet with a muzzle velocity of $180 m / s$. The bullet strikes the jeep of the thief with a velocity

A
$27 ms^{-1}$
B
$150 ms^{-1}$
C
$450 ms^{-1}$
D
$250 ms^{-1}$

Answer

(b) $150 ms^{-1}$
Explanation: Relative speed of bullet with respect to thief's jeep,
$
\begin{aligned}
& =\left(v_{b}+v_{p}\right)-v_{t} \\
& =180 ms^{-1}+45 ms^{-1}-155 ms^{-1} \\
& =180 ms^{-1}+45 \times \frac{5}{18} ms^{-1}-155 \times \frac{5}{18} ms^{-1} \\
& =180 ms^{-1}-305 ms^{-1} \\
& =150 ms^{-1}
\end{aligned}
$

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MCQ 71 Mark

Sound waves of wavelength $\lambda$ travelling in a medium with a speed of $v m / s$ enter into another medium where its speed is $2 v m / s$. The wavelength of sound waves in the second medium is

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

Answer

(b) $2 \lambda$
Explanation: When wave passes from one medium to another its frequency (v) does not change, but its velocity and wavelength changes.
$
\begin{aligned}
& v=v \lambda \text { or } v=\frac{v}{\lambda} \\
& \frac{v}{\lambda}=\frac{2 v}{\lambda_2} \Rightarrow \lambda_2=2 \lambda
\end{aligned}
$

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MCQ 81 Mark

A man waves his arms, while walking. This is:

A
to balance the effect of earth's gravity
B
to keep constant velocity
C
to increase the velocity
D
to ease the tension

Answer

(a) to balance the effect of earth's gravity
Explanation: The waving of arms helps in keeping the centre of gravity at a suitable position. This helps the man to walk comfortably.

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MCQ 91 Mark

A liquid drop of radius R is broken up into N small droplets. The work done is proportional to( take $N^{\frac{1}{3}}>>1$ ).

A
$N^{\frac{1}{3}}$
B
N
C
$N^{\frac{-2}{3}}$
D
$N^{-0.5}$

Answer

(a) $N^{\frac{1}{3}}$
Explanation: When a droplet of radius R is broken into N small droplets total volume will remain constant. Let the radius of small droplets be r . Then
$
\begin{aligned}
& \frac{4}{3} \pi R^3=N \frac{4}{3} \pi r^3 \\
& r=\frac{R}{N^{\frac{1}{3}}}
\end{aligned}
$
work done will be equal to the change in surface energy, thus
$
\begin{aligned}
& W=S_f-S_i=N 4 \pi r^2 T-4 \pi R^2 T \\
& W=N 4 \pi\left(\frac{R}{N^{\frac{1}{3}}}\right)^2 T-4 \pi R^2 T \\
& W=4 \pi R^2 T\left(N^{\frac{1}{3}}-1\right)
\end{aligned}
$
if $N^{\frac{1}{3}}$ is very large thus it becomes
$
W=4 \pi R^2 T N^{\frac{1}{3}}
$
thus
$
W \propto N^{\frac{1}{3}}
$

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MCQ 101 Mark

A ladder is leaned against a smooth wall and it is allowed to slip on a frictionless floor. Which figure represents trace of its centre of mass?

Answer

(a)

Explanation: The centre of mass remains at the centre of the ladder. Hence Fig. (a) represents the correct trace of CM.

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MCQ 111 Mark

The displacement of the wave given by equation $y ( x , t )= a \sin ( kx -\omega t +\phi)$, where $\phi=0$ at point x and $t =0$ is same as that at point

A
Both $x+\frac{2 n \pi}{k}$ and $k x+2 n \pi$
B
$x+\frac{2 n \pi}{k}$
C
$x+2 n \pi$
D
$k x+2 n \pi$

Answer

(b) $x+\frac{2 n \pi}{k}$
Explanation: $y ( x , 0)= a \sin kx = a \sin (k x+2 \pi n)$
$
=a \sin k\left(x+\frac{2 n \pi}{k}\right)
$
$\Rightarrow$ The displacement at points x and $\left(x+\frac{2 n \pi}{k}\right)$ are the same where, $n =1,2,3, \ldots$

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MCQ 121 Mark

Number of degrees present in one radian is

A
$57.3^{\circ}$
B
$56.3^{\circ}$
C
$58^{\circ}$
D
$56^{\circ}$

Answer

(a) $57.3^{\circ}$
Explanation: We know that,
$
\begin{aligned}
& \pi \text { radian }=180^{\circ} \\
& 1 \text { radian }=\frac{180}{\pi}=\frac{180}{22} \times 7=57.3^{\circ}
\end{aligned}
$

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Physics M.C.Q (1 Marks)

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