M.C.Q (1 Marks)

MCQ 11 Mark

The value of coefficient of volume expansion of glycerin is $5 \times 10^{-4} K^{-1}$. The fractional change in the density of glycerin for a rise of $40^{\circ} C$ in its temperature, is

A
$0.020$
B
$0.010$
C
$0.025$
D
$0.015$

Answer

(c) 0.020
Explanation: Fractional change in density is
$
\frac{\rho_0-\rho_t}{\rho_0}=\gamma \Delta T=5 \times 10^{-4} \times 40=0.020
$

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MCQ 21 Mark

A ladder of length L leans against a wall at an angle of $\theta$ from the horizontal, as shown in the figure.

What torque is applied about the ladder's centre of mass by the normal force $F _{ N }$ exerted by the ground on the ladder?

A
$F _{ N }\left(\frac{L}{2}\right) \cos \theta$
B
$F_N\left(\frac{L}{2}\right)$
C
$F _{ N } L \sin \theta$
D
$F _{ N } L \cos \theta$

Answer

(a) $F _{ N }\left(\frac{L}{2}\right) \cos \theta$
Explanation:
$\tau= F _{ N } \times$ perpendicular distance from CM
$
\begin{aligned}
& =F_{N} \times \frac{L}{2} \sin \left(90^{\circ}-\theta\right) \\
& =F_{N}\left(\frac{L}{2}\right) \cos \theta
\end{aligned}
$

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MCQ 31 Mark

A body is projected from earth's surface to become its satellite. Its time period of revolution will not depend upon:

A
mass of earth
B
its own mass
C
gravitational constant
D
radius of earth

Answer

(b) its own mass
Explanation: $T =2 \pi \sqrt{\frac{(R+h)^3}{G M}}$, Clearly, T does not depend on the mass m of the satellite.

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MCQ 41 Mark

A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

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

Answer

(d) $R^2$
Explanation: $v =\frac{2}{9} \frac{R^2}{\eta}(\rho-\sigma) g$
$
\therefore v \propto R^2
$

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MCQ 51 Mark

If wave $y=A \cos (\omega t+k x)$ is moving along $x$-axis, the shape of pulse at $t=0$ and $t=2 s$ :

A
may not be different
B
may not be same
C
are same
D
are different

Answer

(c) are same
Explanation: At $t =0$ and $t =2 s$, the shapes of $y - x$ graphs are same.

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MCQ 61 Mark

A truck on a straight road starts from rest, accelerating at $2.00 m / s ^2$ until it reaches a speed of $20.0 m / s$. Then the truck travels for 20.0 s at a constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s . What is the average velocity in $m / s$ of the truck for the motion described?

A
$15.7$
B
$16.2$
C
$154$
D
$17.5$

Answer

(a) 15.7
Explanation: As start from rest,
So Initial velocity $u =0 m / s$
Final velocity $v =20 m / s$
Acceleration $a =2 m / s ^2$
Let Time during this period $=t_1$
Also let distance covered $= s _1$
We know,
$
v-u=at
$
So, $20-0=2 t_1$
$
t_1=\frac{20}{2}=10 s
$
Also,
$
\begin{aligned}
& v^2-u^2=2 as_1 \\
& \Rightarrow 400-0=2 \times 2 \times s_1 \\
& s_1=\frac{400}{4}=100 m
\end{aligned}
$
Now travel with constant speed of $20 m / s$ for time $t _2=20 s$
Distance covered
$
s_2=20 \times 20=400 m
$
Time taken to stop
$
t_3=5 s
$
Before stopping it covers distance $=s_3$
$
s_3=\frac{1}{2}(20-0) 5=50 m
$
Total distance covered $=100+400+50=550 m$
Total time of motion $t =10+20+5=35 s$
Average velocity $v_{\text {avg }}=\frac{\text { total distance }}{\text { total time }}$
$
=\frac{550}{35}=15.7 m / s
$

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MCQ 71 Mark

Two vibrating tuning forks produce waves given by $y _1=4 \sin 500 \pi t$ and $y _2=2 \sin 506 \pi t$. A number are beats produced per minute is

A
$360$
B
$180$
C
$60$
D
$3$

Answer

(b) 180
Explanation: $y _1=4 \sin 500 \pi t$ and $y _2=2 \sin 506 \pi t$
$
\begin{aligned}
& \omega_1=2 \pi v_1=500 \pi \\
& \therefore \nu_1=250 Hz \\
& \omega_2=2 \pi v_2=506 \pi \\
& \therefore \nu_2=253 Hz
\end{aligned}
$

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MCQ 81 Mark

The total energy of a satellite is E. What is its P.E.?

A
$-E$
B
E
C
$-2 E$
D
2 E

Answer

(d) 2 E
Explanation: Potential energy of satellite $U =-\frac{G M m}{r}$ and total energy of satellite $E =-\frac{G M m}{2 r}$ Therefore, $U =2 E$

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MCQ 91 Mark

The raindrops are in spherical shape due to

A
thrust on drop
B
viscosity
C
surface tension
D
residual pressure

Answer

(c) surface tension
Explanation: Rain drops are spherical due to surface tension.

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MCQ 101 Mark

The vector product of two vectors a and b is a vector c such that c is perpendicular to the plane containing a and b and the direction is given by:

A
left hand rule
B
left-handed screw rule
C
index finger rule
D
right-handed screw rule

Answer

(d) right-handed screw rule
Explanation: If vectors A and B lie in the plane of this page, the vector C will be perpendicular to this plane.
The sense (upward or downward) of the direction of the vector product is given by the direction of the advance of the tip of a right-handed screw when rotated from A to B through angle $\theta$ between them, the screw being placed with its axis perpendicular to the plane containing the two vectors.

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MCQ 111 Mark

What is the minimum length of a tube, open at both ends, that resonates with a tuning fork of frequency 350 Hz ? [velocity of sound in air $=350 m / s$ ]

A
100 cm
B
50 cm
C
25 cm
D
75 cm

Answer

(b) 50 cm
Explanation: The fundamental frequency of open pipe $=\frac{v}{2 L}$
$
\begin{aligned}
& 350=\frac{350}{2 L} \\
& L=\frac{1}{2} m=50 cm
\end{aligned}
$

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MCQ 121 Mark

The equation $\left(P+\frac{a}{V^2}\right)(V-b)=$ Constant. The units of a are

A
dyne $cm ^{-2}$
B
dyne $\times cm ^5$
C
dyne $\times cm ^4$
D
dyne $cm ^{-3}$

Answer

$\begin{aligned} & \text (c) { dyne } \times cm ^4 \\ & \text { Explanation: Unit of } a =\text { Unit of } P \times \text { Unit of } V ^2 \\ & =\text { dyne } cm ^{-2} \times\left( cm ^3\right)^2 \\ & =\text { dyne } cm ^4\end{aligned}$

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Physics M.C.Q (1 Marks)

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