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Physics M.C.Q (1 Marks)

Model Paper 5 · 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 · self-marked practice — reveal the answer and mark yourself.

MCQ 11 Mark
Obtain the dimensional equation for universal gas constant.
  • A
    $\left[ M L ^2 T^{-2} mol^{-1} K^{-1}\right]$
  • B
    $\left[ M L ^3 T^{-1} mol^{-2} K^{-2}\right]$
  • C
    $\left[ M ^2 LT ^{-1} mol^{-1} K^{-1}\right]$
  • D
    $\left[ M ^3 LT ^{-2} mol^{-1} K^{-2}\right]$
Answer
(a) $\left[ M L ^2 T^{-2} mol^{-1} K^{-1}\right]$
Explanation: According to ideal gas equation for universal gas constant. i.e., $pV = nRT$, where n is the number of moles of gases.
$
\begin{aligned}
& R=\frac{(p)(V)}{(n)(T)}=\frac{\left[ML^{-1} T^{-2}\right]\left[L^3\right]}{[mol][K]} \\
& =\left[ML^2 T^{-2} mol^{-1} K^{-1}\right]
\end{aligned}
$
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MCQ 21 Mark
In figure, pressure inside a spherical drop is more than pressure outside. (S = surface tension and r = radius of bubble) 
Image
The extra surface energy if radius of bubble is in creased by $\Delta r$ is
  • A
    $2 \pi r \Delta r S$
  • B
    $4 \pi r \Delta r S$
  • C
    $10 \pi r \Delta r S$
  • D
    $8 \pi r \Delta r S$
Answer
(d) $8 \pi r \Delta r S$
Explanation: Suppose a spherical drop of radius $r$ is in equilibrium. If its radius increases by $\Delta r$. The extra surface energy is $\left|4 \pi(r+\Delta r)^2-4 \pi r^2\right| S=8 \pi r \Delta r S$
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MCQ 31 Mark
A chef, on finding his stove out of order, decides to boil the water for his wife's coffee by shaking it in a thermos flask. Suppose that he uses tap water at $15^{\circ} C$ and that the water falls 30 cm each shake, the chef makes 30 shakes each minute. Neglecting any loss of thermal energy by the flask, how long must he shake the flask until the water reaches $100^{\circ} C$ ?
  • A
    $4.00 \times 10^3 \min$
  • B
    $3.97 \times 10^3 min$
  • C
    $5.25 \times 10^3 min$
  • D
    $2.25 \times 10^3 min$
Answer
(a) $4.00 \times 10^3 min$
Explanation: Heat required to melt 50 g ice
$= ml =50 \times 80=4000 cal$
Heat given out by water in cooling from $80^{\circ} C$ to $0^{\circ} C$
$= mc \Delta T =50 \times 1 \times 80=4000 cal$
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MCQ 41 Mark
A mass is revolving in a circle, which is in the plane of the paper. The direction of angular acceleration if any, is:
  • A
    Upward from the plane of the paper
  • B
    Tangential
  • C
    At right angles to the plane of the paper
  • D
    Towards the radius
Answer
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MCQ 51 Mark
The ratio of escape velocity at earth $\left( v _{ e }\right)$ to the escape velocity at a planet $\left( u _{ p }\right)$ whose radius and mean density are twice as that of earth is:
  • A
    $1: 2 \sqrt{2}$
  • B
    $1: 4$
  • C
    $1: \sqrt{2}$
  • D
    $1: 2$
Answer
(a) $1: 2 \sqrt{2}$
Explanation: $v_e=\sqrt{2 g R}=R \sqrt{\frac{8}{3} \pi G p}$
$
\begin{aligned}
& \therefore \quad \frac{\nu_s}{v_p}=\frac{R_{\sqrt{\rho}}}{R_p \sqrt{P_p}} \\
& =\frac{R_{\sqrt{p}}}{2 R \times \sqrt{2 p}}=\frac{1}{2 \sqrt{2}} \\
& =1: 2 \sqrt{2}
\end{aligned}
$
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MCQ 81 Mark
Two waves each of amplitude a and have a phase difference $\frac{\pi}{2}$. The amplitude and frequency of a resultant wave due to their superposition will be
  • A
    2a, $\frac{f}{2}$
  • B
    $\sqrt{2} a , f$
  • C
    $\frac{a}{\sqrt{2}}, \frac{f}{2}$
  • D
    $\frac{a}{\sqrt{2}}, f$
Answer
(b) $\sqrt{2} a , f$
$
\begin{aligned}
& \text { Explanation: } y_1=a \sin 2 \pi f t \\
& y_2=a \sin \left(2 \pi f t+\frac{\pi}{2}\right) \\
& \therefore y=y_1+y_2=2 a \sin \left(2 \pi f t+\frac{\pi}{4}\right) \cos \frac{\pi}{4} \\
& =\frac{2 a}{\sqrt{2}} \sin \left(2 \pi f t+\frac{\pi}{4}\right) \\
& =\sqrt{2} a \sin \left(2 \pi f t+\frac{\pi}{4}\right)
\end{aligned}
$
Hence the resultant wave has amplitude $\sqrt{2} a$ and frequency f.
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MCQ 101 Mark
A cylindrical jar of cross-sectional area $0.01 m^2$ is filled with water to a height of 50 cm . It carries a tight-fitting piston of negligible mass. What is the pressure at the bottom of the jar when the mass of 1 kg is placed on the piston? Take $g =10 ms^{-2}$.
  • A
    6000 Pa
  • B
    4000 Pa
  • C
    2000 Pa
  • D
    5000 Pa
Answer
(a) 6000 Pa
Explanation: The pressure at the bottom of the jar is due to the weight of a column of water of height $h =50 cm=0.5 m$ and the weight of a load of $m=1 kg$.
The total force acting on the base $=h \rho g A+m g$
$
=0.5 \times 1000 \times 10 \times 0.01+1 \times 10=60 N
$
$\therefore$ Pressure $=\frac{\text { force }}{\text { area }}=\frac{60}{0.01}=6000 Nm ^{-2}$ or 6000 Pa
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MCQ 111 Mark
What will be the ratio of the distances moved by a freely falling body from rest in 4th and 5th seconds of journey?
  • A
    $1: 1$
  • B
    $16: 25$
  • C
    $7: 9$
  • D
    $4: 5$
Answer
$7: 9$
Explanation: Distance covered in $n ^{\text {th }}$ second is given by
$
s_{n}=u+\frac{a}{2}(2 n-1)
$
Given: $u =0, a = g$
$\begin{aligned} & \therefore s _4=\frac{g}{2}(2 \times 4-1)=\frac{7 g}{2} \\ & s_5=\frac{g}{2}(2 \times 5-1)=\frac{9 g}{2} \\ & \therefore \frac{s_4}{s_5}=\frac{7}{9}\end{aligned}$
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MCQ 121 Mark
A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is
  • A
    $\frac{M L^2 \omega}{2}$
  • B
    $M L \omega^2$
  • C
    $\frac{M L^2 \omega^2}{2}$
  • D
    $\frac{M L \omega^2}{2}$
Answer
(d) $\frac{M L \omega^2}{2}$
Explanation: The mass of the liquid acts at the centre of the tube.
Therefore, $r=\frac{L}{2}$
Force exerted by the liquid at the other end
$
\begin{aligned}
& =\text { Centrifugal force } \\
& =M r \omega^2=M\left(\frac{L}{2}\right) \omega^2=\frac{M L \omega^2}{2}
\end{aligned}
$
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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip