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Physics M.C.Q (1 Marks)

Model Paper 7 · 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
In rotation of a rigid body about a fixed axis is that in which:
  • A
    every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.
  • B
    every particle of the body moves in an ellipse, which lies in a plane perpendicular to the axis and has its focii on the axis.
  • C
    particles close to the axis have larger velocities.
  • D
    every particle of the body moves at the same speed.
Answer
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MCQ 21 Mark
Two sound sources each emitting waves of wavelength $\lambda$ are fixed a given distance apart and an observer moves from one source to another with velocity $u$. Then number of beats heard by him per second is
  • A
    $\frac{2 u}{\lambda}$
  • B
    $\frac{u}{2 \lambda}$
  • C
    $\frac{u}{\lambda}$
  • D
    $\sqrt{u \lambda}$
Answer
(a) $\frac{2 u}{\lambda}$
Explanation:
Image
Number of extra waves received from $S_2$ per second $=+\frac{u}{\lambda}$
Number of lesser waves received from $S _1$ per second $=-\frac{u}{\lambda}$
$\therefore$ Beat frequency $=\frac{u}{\lambda}-\left(-\frac{u}{\lambda}\right)=\frac{2 u}{\lambda}$
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MCQ 31 Mark
Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. The speed v of each particle is:
  • A
    $\sqrt{\frac{4 G m}{R}}$
  • B
    $\frac{1}{2} \sqrt{\frac{G m}{R}}$
  • C
    $\sqrt{\frac{G m}{R}}$
  • D
    $\frac{1}{2 R} \sqrt{\frac{1}{G m}}$
Answer
(b) $\frac{1}{2} \sqrt{\frac{G m}{R}}$
Explanation: The two masses, separated by a distance 2 R , revolve about the common centre of mass O .
Image

Centripetal force $=$ Mutual gravitational attraction
$
\begin{aligned}
& \frac{m v^2}{R}=\frac{G m \times m}{(2 R)^2} \\
& \text { or } v^2=\frac{GM}{4 R} \\
& \therefore v=\frac{1}{2} \sqrt{\frac{G m}{R}}
\end{aligned}
$
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MCQ 41 Mark
The time period T of a small drop of liquid (due to surface tension) depends on density $\rho$, radius r and surface tension S. The relation is:
  • A
    $T \propto \frac{S}{\rho r}$
  • B
    $T \propto \rho r S$
  • C
    $T \propto \frac{\rho r}{S}$
  • D
    $T \propto\left(\frac{\rho r^3}{S}\right)^{1 / 2}$
Answer
(d) $T \propto\left(\frac{\rho r^3}{S}\right)^{1 / 2}$
Explanation: Let $T =K \rho^a r^b S^c$
$
\therefore[T]=\left[ML^{-3}\right]^{a}[L]^{b}\left[MT^{-2}\right]^{c}
$
or
$
\begin{aligned}
& M^0 L^0 T^1=M^{a+c} L^{-3 a+b^{-}} T^{-2 c} \\
& \therefore a+c=0,-3 a+b=0,-2 c=1
\end{aligned}
$
On solving, $a =\frac{1}{2}, b=\frac{3}{2}, c =-\frac{1}{2}$
Hence, $T \propto\left(\frac{\rho r^3}{S}\right)^{1 / 2}$
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MCQ 51 Mark
The reading of a spring balance corresponds to 100 N while situated at the north pole and a body is kept on it.
The weight recorded on the same scale if it is shifted to the equator (take, $g=10 m / s ^2$ and radius of the earth, R=$6.4 \times 10^3 m$) is :
  • A
    106 N
  • B
    97.66 N
  • C
    110 N
  • D
    99.66 N
Answer
(d) 99.66 N
$
\begin{aligned}
& \text { Explanation: } g_{p}-g_{e}=R \omega^2=3.37 \times 10^{-2} ms^{-2} \\
& g_{e}=g_{p}-3.37 \times 10^{-2} ms^{-2} \\
& =(10-0.0337) ms^{-2} \\
& W_{e}=mg_{e}=\frac{100}{10} \times 9.9663 N=99.66 N
\end{aligned}
$
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MCQ 61 Mark
If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
  • A
    $\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{-1 / 2}$
  • B
    $\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{1 / 4}$
  • C
    $\left(\frac{4 \pi R^2 Q}{\sigma}\right)^{1 / 4}$
  • D
    $\frac{Q}{4 \pi R^2 \sigma}$
Answer
(b) $\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{1 / 4}$
Explanation: From Stefan' law, the rate of energy radiated by the star acting as a black body is
$
\begin{aligned}
& Q=\sigma A T^4=\sigma \times 4 \pi R^2 \times T^4 \\
& \therefore T=\left[\frac{Q}{4 \pi R^2 \sigma}\right]^{\frac{1}{4}}
\end{aligned}
$
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MCQ 71 Mark
When the adhesive force in the case of liquid and glass is greater than the cohesive forces between the liquid molecules, the shape of the meniscus of liquid in a capillary tube is?
  • A
    Plane
  • B
    Circular
  • C
    Convex
  • D
    Concave
Answer
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MCQ 111 Mark
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Qare at distances of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is:
  • A
    $4: 9$
  • B
    $2: 3$
  • C
    $3: 2$
  • D
    $9: 4$
Answer
(d) $9: 4$
Explanation: Intensity $=\frac{\text { energy }}{\text { time } \times \text { area }}=\frac{\text { power }}{\text { area }}$
From a point source, energy spreads over the surface of a sphere of radius $r$.
$\therefore$ Intensity, $I =\frac{P}{A}=\frac{P}{4 \pi r^2}$
or $I \propto \frac{1}{r^2}$
$
\frac{I_1}{I_2}=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}
$
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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip