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Physics M.C.Q (1 Marks)

Model Paper 2 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 12 questions.

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M.C.Q (1 Marks)

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12 questions · timed · auto-graded

MCQ 11 Mark
Average distance of the earth frorn the sun is $L _1$. If one year of the earth $= D$ days, one year of another planet whose average distance from the sun is $L _2$ will be:
  • A
    $D\left(\frac{L_2}{L_1}\right)^{\frac{1}{2}}$ days
  • B
    $D\left(\frac{L_2}{L_1}\right)^{\frac{2}{3}}$
  • C
    $D\left(\frac{L_2}{L_1}\right)^{\frac{3}{2}}$ days
  • D
    $D\left(\frac{L_2}{L_1}\right)$
Answer
(c) $D\left(\frac{L_2}{L_1}\right)^{\frac{3}{2}}$ days
Explanation: According to Kepler's law of periods,
$T ^2 \propto r ^3 \Rightarrow T \propto r ^{\frac{3}{2}}$
$
\frac{D}{D}=\left(\frac{L_2}{4}\right)^{\frac{3}{2}}
$
$\Rightarrow D^{\prime}=D\left(\frac{L_2}{L_1}\right)^{\frac{3}{2}}$ days
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MCQ 21 Mark
A solid sphere of volume V and density $\rho$ floats at the interface of two immiscible liquids of densities $\rho_1$ and $\rho_2$ respectively. If $\rho_1<\rho<\rho_2$, then the ratio of the volume of the parts of the sphere in upper and lower liquids is
  • A
    $\frac{\rho+\rho_2}{\rho+\rho_1}$
  • B
    $\frac{\rho-\rho_1}{\rho_2-\rho}$
  • C
    $\frac{\rho_2-\rho}{\rho-\rho_1}$
  • D
    $\frac{\rho+\rho_1}{\rho+\rho_2}$
Answer
(c) $\frac{\rho_2-\rho}{\rho-\rho_1}$
Explanation: Let $V _1$ and $V _2$ be the volumes of the parts of the sphere immersed in liquids of densities $\rho_1$ and $\rho_2$ respect
According to the law of floatation,
Weight of sphere $=$ Weight of liquid 1 displaced + Weight of liquid 2 displaced
$
\begin{aligned}
& V \rho g=V_1 \rho_1 g+V_2 \rho_2 g \\
& \Rightarrow\left(V_1+V_2\right) \rho g=V_1 \rho_1 g+V_2 \rho_2 g \\
& \Rightarrow V_1\left(\rho-\rho_1\right)=V_2\left(\rho_2-\rho\right) \\
& \therefore \frac{V_1}{V_2}=\frac{\rho_2-\rho}{\rho-\rho_1}
\end{aligned}
$
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MCQ 31 Mark
When 1 m, 1 kg and 1 min are taken as the fundamental units, the magnitude of the force is 36 units. What will be the value of this force in CGS system?
  • A
    $10^4$ dyne
  • B
    $10^8$ dyne
  • C
    $10^3$ dyne
  • D
    $10^5$ dyne
Answer
(c) $10^3$ dyne
Explanation: As, dimensional formula of force $=\left[ MLT ^{-2}\right]$
$
\begin{aligned}
& n_1=36, M_1=1 kg, L_1=1 m, T_1=1 min=60 s \\
& n_2=?, M_2=1 g, L_2=1 cm, T_2=1 s
\end{aligned}
$
So, conversion of 36 units into CGS system
$
\begin{aligned}
& \text { i.e., } n_2=n_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c \\
& n_2=n_1\left[\frac{1 kg}{1 g}\right]^1\left[\frac{1 m}{1 cm}\right]^1\left[\frac{1 \min }{1 s}\right]^{-2} \\
& =36\left[\frac{1000 g}{1 g}\right]\left[\frac{100 cm}{1 cm}\right]^1\left[\frac{60 s}{1 s}\right]^{-2}=10^3 \text { dyne }
\end{aligned}
$
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MCQ 61 Mark
A rod has length 3 m and its mass acting per unit length is directly proportional to distance x from one of its end, then its centre of gravity from that end will be at
  • A
    1.5 m
  • B
    2.5 m
  • C
    2 m
  • D
    3.0 m
Answer
(c) 2 m
Explanation:
Suppose the rod is placed along .x-axis. Consider a small element of thickness dx at distance x from its left end.
As the mass acting per unit length is directly proportional to distance x from one end, mass of the small element is $dm = kx dx$
Image
The position of CM of the rod will be
$
\begin{aligned}
& x_{C M}=\frac{\int_0^3 x d m}{\int_0^3 d m} \\
& =\frac{\int_0^3 k x^2 d x}{\int_0^3 k x d x} \\
& =\frac{\left[\frac{x^3}{3}\right]_0^3}{\left[\frac{x^2}{2}\right]_0^3}=\frac{27}{3} \times \frac{2}{9}=2 m
\end{aligned}
$
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MCQ 71 Mark
The wavelength of ultrasonic waves in air is of the order of:
  • A
    $0.00000001 cm$
  • B
    $1 cm$
  • C
    $0.1 cm$
  • D
    $0.0001 cm$
Answer
1 cm
Explanation: The frequency of ultrasonic waves is above 20000 Hz and the speed of waves in air is $330 m / s$.
$
\begin{aligned}
& \lambda=\frac{v}{f} \\
& \lambda=\frac{330}{20000} \\
& \lambda=1 cm
\end{aligned}
$
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MCQ 81 Mark
Two sources of sound placed close to each other, are emitting progressive waves given by $y _1=4 \sin 600 \pi t$ and $y _2=5 \sin 608 \pi t$. An observer located near these two sources of sound will hear:
i. 8 beats per second with intensity ratio $81: 1$ between waxing and waning
ii. 4 beats per second with intensity ratio $81:1$ between waxing and waning
iii. 4 beats per second with intensity ratio $25: 16$ between waxing and waning
iv. 8 beats per second with intensity ratio $25: 16$ between waxing and waning
  • A
    i and ii
  • B
    only iv
  • C
    only ii
  • D
    iii and iv
Answer
(c) only ii
$\begin{aligned} & \text { Explanation: } \omega_1=600 \pi \text { or } n _1=\frac{600 \pi}{2 \pi}=300 s^{-1} \\ & \omega_2=608 \pi \text { or } n _2=\frac{600 \pi}{2 \pi}=304 s^{-1} \\ & \therefore \text { Number of beats }= n _2- n _1=304-300=4 s^{-1} \\ & \text { Intensity ratio }=\frac{I_{\max }}{I_{\min }}=\left(\frac{a_2+a_1}{a_2-a_1}\right)^2=\left(\frac{5+4}{5-4}\right)^2=\frac{81}{1}\end{aligned}$
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MCQ 101 Mark
A truck has a velocity of $2 m / s$ at time $t=0$. It accelerates at $2 m / s ^2$ on seeing police. What is its velocity in $m / s$ at a time of 2 sec ?
  • A
    6
  • B
    3
  • C
    4
  • D
    7
Answer
6
Explanation: Initial velocity is given by, $u =2 m / s$
Final velocity is given by,$= v m / s$
Time duration is $=$ final time - initial time $=2-0=2 s$
Acceleration, $a=2 m / s ^2$
We know, $v = u +$ at
$
\begin{aligned}
& \Rightarrow v=2+2 \times 2 \\
& \Rightarrow v=6 m / s
\end{aligned}
$
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MCQ 111 Mark
A tuning fork produces 4 beats $/ sec$. with 50 cm and 40 cm of a stretched wire, of a sonometer. The frequency of fork is:
  • A
    90 Hz
  • B
    36 Hz
  • C
    110 Hz
  • D
    50 Hz
Answer
(b) 36 Hz
Explanation: the beat frequency is given by
$
\begin{aligned}
& f_{\text {beat }}=f_1-f_2 \\
& 4=f_1-f_2 \rightarrow(1)
\end{aligned}
$
also frequency
$
\begin{aligned}
& f \propto \frac{1}{L} \\
& \frac{f_1}{f_2}=\frac{L_2}{L_1} \\
& \frac{f_1}{f_2}=\frac{40}{50} \rightarrow(2)
\end{aligned}
$
on solving equation 1 and 2
$
\begin{aligned}
& f_1=16 Hz \\
& f_2=-20 Hz \\
& \left|f_1-f_2\right|=36 Hz
\end{aligned}
$
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MCQ 121 Mark
The radius of gyration of a disc of mass 100 g and radius 5 cm about an axis passing through its centre of gravity and perpendicular to the plane is
  • A
    3.54 cm
  • B
    2.51 cm
  • C
    6.54 cm
  • D
    1.52 cm
Answer
3.54 cm
Explanation: M.I. of a disc about the central perpendicular axis, $M k^2=\frac{1}{2} M R^2$
$
\begin{aligned}
& \therefore \quad k=\frac{R}{\sqrt{2}}=0.707 \quad R=0.707 \times 5 cm \\
& =3.54 cm .
\end{aligned}
$
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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip