M.C.Q (1 Marks)

MCQ 11 Mark

If boiling point of a liquid is $95^{\circ} F$, what will be the reading at Celsius scale?

A
$63^{\circ} C$
B
$65^{\circ} C$
C
$7^{\circ} C$
D
$35^{\circ} C$

Answer

(d) $35^{\circ} C$
$
\begin{aligned}
& \text { Explanation: } C=\frac{5}{9}(F-32) \\
& =\frac{5}{9}(95-32)=35^{\circ} C
\end{aligned}
$

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MCQ 21 Mark

A small object of uniform density rolls up a curved surface with an initial velocity v'. If reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

A
solid sphere
B
ring
C
disc
D
hollow sphere

Answer

(c) disc
Explanation: Loss in (translational K.E. + rotational K.E.) = Gain in P.E.
$
\begin{aligned}
& \frac{1}{2} m v^2+\frac{1}{2} T \omega^2=m g h_{\max } \\
& \frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2=m g \times \frac{3 v^2}{4 g}=\frac{3}{4} m v^2 \\
& \Rightarrow I=\frac{1}{2} m R^2
\end{aligned}
$

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MCQ 31 Mark

There are two bodies of masses 1 kg and 100 kg separated by a distance 1 m. At what distance from the smaller body, the intensity of gravitational field will be zero?

A
$\frac{10}{11} m$
B
$\frac{1}{9} m$
C
$\frac{1}{11} m$
D
$\frac{1}{10} m$

Answer

(c) $\frac{1}{11} m$
Explanation: At $\frac{1}{11} m$ from smaller body the intensity of gravitation field is zero.

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MCQ 41 Mark

In a streamline flow,
i. the speed of a particle always remains same
ii. the velocity of a particle always remains same
iii. the kinetic energies of all the particles arriving at a given point are the same iv. the potential energies of all the particles arriving at a given point are the same

A
Statement (iv) is correct.
B
Statement (ii) is correct.
C
Statement (i) is correct.
D
Statement (iii) is correct.

Answer

(b) Statement (ii) is correct.
Explanation: Both velocity and direction of flow remain same.

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MCQ 51 Mark

A string is tied on a sonometer, second end is hanging downward through a pulley with tension T. The velocity of the transverse wave produced is proportional to

A
$\frac{1}{T}$
B
T
C
$\sqrt{T}$
D
$\frac{1}{\sqrt{T}}$

Answer

(c) $\sqrt{T}$
Explanation: $v=\sqrt{\frac{T}{m}}$
$
\therefore v \propto \sqrt{T}
$

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MCQ 61 Mark

Given $a =2 t +5$. Calculate the velocity of the body after five sec if it starts from rest.

A
25 m/s
B
50 m/s
C
75 m/s
D
100 m/s

Answer

(b) $50 m / s$
$
\begin{aligned}
& \text { Explanation: } a=\frac{d v}{d t}=2 t+5 \\
& \int_0^v d v=\int_0^5(2 t+5) dt \\
& v=\left[t^2\right]_0^5+5[t]_0^5 \\
& =(25-0)+5(5-0) \\
& =50 m / s
\end{aligned}
$

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MCQ 71 Mark

Equation of progressive wave is given by $y=4 \sin \left[\pi\left(\frac{t}{5}-\frac{x}{9}\right)+\frac{\pi}{6}\right]$ where $y , x$ are in cm and t is in second. Then which of the following is correct?

A
$f =50 Hz$
B
$A =0.04 cm$
C
$\lambda=18 m$
D
$v =5 cm / sec$

Answer

(c) $\lambda=18 m$
Explanation: Given equation is $y =4 \sin \pi\left[\frac{t}{5}-\frac{x}{9}+\frac{\pi}{6}\right]$ $=4 \sin \left[\frac{\pi t}{5}-\frac{\pi x}{9}+\frac{\pi^2}{6}\right]$
Standard equation of wave is
$
y=r \sin \left[\frac{2 \pi}{T} t-\frac{2 \pi}{\lambda} x+\phi_0\right]
$
comparing these equations we get, $\frac{2 \pi}{\lambda}=\frac{-\pi}{9}$
or $\lambda=18 m$

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MCQ 81 Mark

The kinetic energy of the satellite in a circular orbit with speed $v$ is given as

✓
$KE =\frac{G m M_e}{2\left(R_e+h\right)}$
B
$KE =-\frac{1}{2} m v^2$
C
$KE =\frac{G m M_e}{\left(R_e+h\right)}$
D
$KE =\frac{-G m M_e}{2\left(R_e+h\right)}$

Answer

Correct option: A.

$KE =\frac{G m M_e}{2\left(R_e+h\right)}$

$\text { Explanation: } KE \text { of satellite }=\frac{1}{2} mv ^2$
$=\frac{1}{2} m\left(\sqrt{\frac{G M_e}{\left(R_e+h\right)}}\right)^2$
$=\frac{1}{2} \frac{G m M_e}{\left(R_e+h\right)}$

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MCQ 91 Mark

If the liquid neither rises nor falls in a capillary tube, then angle of contact is

A
$180^{\circ}$
B
$90^{\circ}$
C
$45^{\circ}$
D
$0^{\circ}$

Answer

(b) $90^{\circ}$
Explanation: $h =\frac{2 \sigma \cos \theta}{r \rho g}$
When $\theta=90^{\circ}, h =0$

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MCQ 101 Mark

A body of mass 2.0 kg and radius of gyration 0.5 m is rotating about an axis. If its angular speed is $2.0 rad / s$, the angular momentum of the body (in $kg - m ^2 / s$ ) is:

A
$2.0$
B
$0.5$
C
$1.5$
D
$1.0$

Answer

(d) 1.0
Explanation: Moment of inertia
$
\begin{aligned}
& I=mk^2 \\
& m=2 Kg \\
& k=0.5 m \\
& I=2 \times 0.5 \times 0.5=0.5 Kgm^2
\end{aligned}
$
angular momentum
$
\begin{aligned}
& L=I \omega \\
& \omega=2 rad / s \\
& L=0.5 \times 2=1.0 Kgm^2 / s
\end{aligned}
$

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MCQ 111 Mark

If the length of a closed organ pipe is 1 m and velocity of sound is $330 m / s$, then the frequency for the second note is

A
$3 \times \frac{330}{4} Hz$
B
$2 \times \frac{4}{330} Hz$
C
$4 \times \frac{330}{4} Hz$
D
$2 \times \frac{330}{4} Hz$

Answer

(a) $3 \times \frac{330}{4} Hz$
Explanation: Frequency of second note of a closed pipe, $\nu=\frac{3 v}{4 L}$ $=\frac{3 \times 330}{4 \times 1} Hz$

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MCQ 121 Mark

_________________ are units of physical quantity that can be expressed as a combination of fundamental physical quantities.

A
Complex units
B
Basic units
C
Derived units
D
Simple units

Answer

(c) Derived units
Explanation: Derived units are units which may be expressed in terms of base units by means of mathematical symbols of multiplication and division.

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Physics M.C.Q (1 Marks)

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