M.C.Q (1 Marks)

MCQ 11 Mark

Ratio of the amount of heat radiation, transmitted through the body to the amount of heat radiation incident on it, is known as

A
absortance
B
inductance
C
transmittance
D
conductance

Answer

(c) transmittance
Explanation: The ratio of the amount of heat transmitted through an object to the amount of heat incident on it is called transmittance.

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MCQ 21 Mark

A ring of radius r and mass m rotates about its central axis. The kinetic energy is :

A
$\frac{m r \omega^2}{2}$
B
$\frac{\operatorname{mr}^2 \omega^2}{2}$
C
$mr \omega^2$
D
$mr ^2 \omega^2$

Answer

(b) $\frac{\operatorname{mr}^2 \omega^2}{2}$
Explanation: The kinetic energy of the body in rotational motion is $K E=\frac{1}{2} I \omega^2=\frac{1}{2} m r^2 \omega^2$ as a moment of inertia of ring about its central axis is $I = mr ^2$.

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MCQ 31 Mark

A satellite is revolving around the earth with a kinetic energy E. The minimum addition of kinetic energy needed to make it escape from its orbit is :

A
$\sqrt{E}$
B
$\frac{E}{2}$
C
E
D
2E

Answer

(c) E
Explanation: $v _0=\sqrt{\frac{G M}{r}}$ and $v _{ e }=\sqrt{\frac{2 G M}{r}}$
Initial K.E., $E =\frac{1}{2} m v_0^2=\frac{1}{2} \frac{ CMm }{r}$
Total K.E. neede for escaping,
$
E=\frac{1}{2} m v_e^2=\frac{1}{2} m \times \frac{2 G M}{r}=\frac{G M m}{r}=2 E
$
Additional K.E. needed for escaping
$
=E^{\prime}-E=2 E-E=E
$

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MCQ 41 Mark

The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon

A
acceleration due to Force
B
acceleration due to gravity
C
height of liquid
D
size of orifice

Answer

(d) size of orifice
Explanation: Velocity of efflux, $v =\sqrt{2 g h}$
Clearly, it does not depend on the size of the orifice.

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MCQ 51 Mark

A wave of frequency 100 Hz travels along a string towards its fixed end. When this wave travels back, after reflection, a node is formed at a distance of 10 cm from the fixed end. The speed of the wave (incident and reflected) is

A
$40 m / s$
B
$10 m / s$
C
$5 m / s$
D
$20 m / s$

Answer

(d) $20 m / s$
Explanation: Distance between two nodes, i.e.,
$\left(\frac{\lambda}{2}\right)=10$ or $\lambda=20 cm$
Now, $v = n \lambda=100 \times 20 cm / sec =20 m / sec$

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MCQ 61 Mark

A body moving with uniform acceleration has a velocity of $12.0 cm / s$ in the positive x -direction when its x coordinate is 3.00 cm. If its $x$ coordinate 2.00 seconds later is -5.00 cm , what is the magnitude of its acceleration in $cm / s ^2$ ?

A
$-14.0$
B
$-12.0$
C
$-16.0$
D
$-18.0$

Answer

(c) $-16.0$
Explanation: Distance covered s = Final position - initial position $=-5-3=-8 cm$
Initial velocity $u =12.0 cm / s$
Time taken $t =2.0 s$
We know
$
\begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& \Rightarrow-8=2 \times 12.0+\frac{1}{2} a \times 4
\end{aligned}
$
$
\Rightarrow-8=24+2 a
$
$\Rightarrow a =\frac{-8-24}{2}=-16.0 cm / s ^2$ hence this is required result

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MCQ 71 Mark

In a resonance tube, the first resonance with a tuning fork occurs at $16 \ cm$ and the second at $49 \ cm$ . If the velocity of sound is $330\ m/s,$ the frequency of the tuning fork is

A
$330$
B
$300$
✓
$500$
D
$165$

Answer

Correct option: C.

$500$

$v =2 \nu\left( l _2- l _1\right)$
$330=2 \nu(0.49-0.16)$
$\nu=\frac{330}{2 \times 0.33}=500\ Hz$

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MCQ 81 Mark

A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to:

A
$\frac{ GM _P}{D_P^2}$
B
$\frac{4 GM _P m}{D_P^2}$
C
$\frac{4 GM _P}{D_P^2}$
D
$\frac{ GM _P m}{D_P^2}$

Answer

(c) $\frac{4 GM _P}{D_P^2}$
Explanation: Gravitational attraction on the particle,
$
F=G \frac{M_p m}{\left(\frac{D_p}{2}\right)^2}
$
Acceleration due to gravity,
$
g=\frac{F}{m}=\frac{4 G M_P}{D_P^2}
$

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MCQ 91 Mark

Soldering of two metals possible because of the property of:

A
osmosis
B
cohesion
C
surface tension
D
viscosity

Answer

(b) cohesion
Explanation: Cohesion is the attractive force among the same kind of materials. Thus soldering of two metals is possible due to cohesion.

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MCQ 101 Mark

If $\overrightarrow{ a }= a _{ x } \hat{ i }+ a _{ y } \hat{ j }+ a _{ z } \hat{ k }$ and $\overrightarrow{ b }= b _{ x } \hat{ i }+ b _{ y } \hat{ j }+ b _{ z } \hat{ k }$ then the cross product $\overrightarrow{ a } \times \overrightarrow{ b }$ is given by:

Answer

(a)

Explanation: $\overrightarrow{ a } \times \overrightarrow{ b }=\left(a_y b_z-a_z b_y\right) \hat{ i }-\left(a_x b_z-a_z b_x\right) \hat{ j }+\left(a_x b_y-a_y b_x\right) \hat{ k }$
As the cross product is the determinant of a $3 \times 3$ matrix.

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MCQ 111 Mark

Two tuning forks of frequency 250 Hz and 256 Hz produce beats. If a maximum is produced just now, after how much time the minimum is produced at the same place:

A
$\frac{1}{6} s$
B
$\frac{1}{12} s$
C
$\frac{1}{24} s$
D
0.25 s

Answer

(d) 0.25 s
Explanation: The time interval between two consecutive maxima and minima is $
\triangle t=\frac{1}{n_1-n_2}
$
For consecutive maxima and minima, the time interval will be $\frac{1}{4} s$ or 0.25 s
Thus time interval between maxima and minima at the same place will be $\frac{1}{4} s$ or 0.25 s

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MCQ 121 Mark

In SI system the fundamental units are

A
meter, kilogram, second, ampere, Kelvin, mole and watt
B
meter, Newton, second, ampere, Kelvin, mole and candela
C
meter, kilogram, second, coulomb, Kelvin, mole, candela and horse power
D
meter, kilogram, second, ampere, Kelvin, mole and candela

Answer

(d) meter, kilogram, second, ampere, Kelvin, mole and candela
Explanation: The SI base units and their physical quantities are the metre for measurement of length, the kilogram for mass, the second for time, the ampere for electric current, the kelvin for temperature, the candela for luminous intensity, and the mole for amount of substance.

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Physics M.C.Q (1 Marks)

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