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Physics M.C.Q (1 Marks)

Model Paper 9 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 12 questions.

Questions

M.C.Q (1 Marks)

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12 questions · self-marked practice — reveal the answer and mark yourself.

MCQ 11 Mark
The displacement of the wave given by equation $y ( x , t )= a \sin ( kx -\omega t +\phi)$, where $\phi=0$ at point x and $t =0$ is same as that at point
  • A
    Both $x+\frac{2 n \pi}{k}$ and $k x+2 n \pi$
  • B
    $x+\frac{2 n \pi}{k}$
  • C
    $x+2 n \pi$
  • D
    $k x+2 n \pi$
Answer
(b) $x+\frac{2 n \pi}{k}$
Explanation: $y ( x , 0)= a \sin kx = a \sin (k x+2 \pi n)$
$
=a \sin k\left(x+\frac{2 n \pi}{k}\right)
$
$\Rightarrow$ The displacement at points x and $\left(x+\frac{2 n \pi}{k}\right)$ are the same where, $n =1,2,3, \ldots$
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MCQ 21 Mark
Number of degrees present in one radian is
  • A
    $57.3^{\circ}$
  • B
    $56.3^{\circ}$
  • C
    $58^{\circ}$
  • D
    $56^{\circ}$
Answer
(a) $57.3^{\circ}$
Explanation: We know that,
$
\begin{aligned}
& \pi \text { radian }=180^{\circ} \\
& 1 \text { radian }=\frac{180}{\pi}=\frac{180}{22} \times 7=57.3^{\circ}
\end{aligned}
$
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MCQ 31 Mark
One large soap bubble of diameter D breaks into 27 bubbles having surface tension T. The change in surface energy is
  • A
    $2 \pi TD ^2$
  • B
    $8 \pi TD ^2$
  • C
    $4 \pi TD ^2$
  • D
    $\pi TD ^2$
Answer
(c) $4 \pi TD ^2$
Explanation: Volume of 27 small bubbles = Volume of larger bubble
$
\begin{aligned}
& 27 \times \frac{4}{3} \times \pi r^3=\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \\
& \therefore r=\frac{D}{6}
\end{aligned}
$
Increase in surface area,
$
=2\left[27 \times 4 \pi\left(\frac{D}{6}\right)^2-4 \pi\left(\frac{D}{2}\right)^2\right]=4 \pi D^2
$
Increase in surface energy, $=$ Increase in surface area $\times$ surface tension $=4 \pi D ^2 T$
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MCQ 41 Mark
Two identical metallic balls, whose temperatures are $200^{\circ} C$ and $400^{\circ} C$ respectively, are placed in an enclosure at $27^{\circ} C$. The ratio of heat-loss of the balls will be
  • A
    $1: 4$
  • B
    $\frac{(473)^4-(300)^4}{(673)^4-(300)^4}$
  • C
    $1: 2$
  • D
    $\frac{(200)^4-(27)^4}{(400)^4-(27)^4}$
Answer
(b) $\frac{(473)^4-(300)^4}{(673)^4-(300)^4}$
Explanation: From Stefan Boltzmann's law,
$
\begin{aligned}
& \frac{E_1}{E_2}=\frac{T_1^4-T_0^4}{T_2^4-T_0^4} \\
& =\frac{(273+200)^4-(273+27)^4}{(273+400)^4-(273+27)^4} \\
& =\frac{(473)^4-(300)^4}{(673)^4-(300)^4}
\end{aligned}
$
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MCQ 61 Mark
Sound waves of wavelength $\lambda$ travelling in a medium with a speed of $v m / s$ enter into another medium where its speed is $2 v m / s$. The wavelength of sound waves in the second medium is
  • A
    $4 \lambda$
  • B
    $2 \lambda$
  • C
    $\frac{\lambda}{2}$
  • D
    $\lambda$
Answer
(b) $2 \lambda$
Explanation: When wave passes from one medium to another its frequency (v) does not change, but its velocity and wavelength changes.
$
\begin{aligned}
& v=v \lambda \text { or } v=\frac{v}{\lambda} \\
& \frac{v}{\lambda}=\frac{2 v}{\lambda_2} \Rightarrow \lambda_2=2 \lambda
\end{aligned}
$
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MCQ 71 Mark
A police jeep is chasing with velocity $45 km / h$. A thief in another jeep is moving with $155 km / h$. Police fires a bullet with a muzzle velocity of $180 m / s$. The bullet strikes the jeep of the thief with a velocity
  • A
    $27 ms^{-1}$
  • B
    $150 ms^{-1}$
  • C
    $450 ms^{-1}$
  • D
    $250 ms^{-1}$
Answer
(b) $150 ms^{-1}$
Explanation: Relative speed of bullet with respect to thief's jeep,
$
\begin{aligned}
& =\left(v_{b}+v_{p}\right)-v_{t} \\
& =180 ms^{-1}+45 ms^{-1}-155 ms^{-1} \\
& =180 ms^{-1}+45 \times \frac{5}{18} ms^{-1}-155 \times \frac{5}{18} ms^{-1} \\
& =180 ms^{-1}-305 ms^{-1} \\
& =150 ms^{-1}
\end{aligned}
$
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MCQ 81 Mark
A standing wave is represented by:
$
y=a \sin (100 t) \cos (0.01 x)
$
where y and a are in millimetre, t in second and x in metre. Velocity of wave is:
  • A
    not derived
  • B
    $1 m / s$
  • C
    $10^4 m / s$
  • D
    $10^{-4} m / s$
Answer

$\begin{aligned} & \text { } (c) 10^4 m / s \\ & \text { Explanation: } y = A \sin (100 t ) \cos (0.01 x ) \\ & y=A \sin \left(\frac{2 \pi}{T} t\right) \cos \left(\frac{2 \pi}{\lambda} x\right) \\ & \therefore \frac{2 \pi}{T}=100 \text { or } T =\frac{\pi}{50} \\ & \frac{2 \pi}{\lambda}=0.01 \text { or } \lambda=200 \pi \\ & v=\frac{\lambda}{T}=\frac{200 \pi}{\frac{\pi}{50}}=10^4 m / s \end{aligned}$
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MCQ 101 Mark
A liquid drop of radius R is broken up into N small droplets. The work done is proportional to( take $N^{\frac{1}{3}}>>1$ ).
  • A
    $N^{\frac{1}{3}}$
  • B
    N
  • C
    $N^{\frac{-2}{3}}$
  • D
    $N^{-0.5}$
Answer
(a) $N^{\frac{1}{3}}$
Explanation: When a droplet of radius R is broken into N small droplets total volume will remain constant. Let the radius of small droplets be r . Then
$
\begin{aligned}
& \frac{4}{3} \pi R^3=N \frac{4}{3} \pi r^3 \\
& r=\frac{R}{N^{\frac{1}{3}}}
\end{aligned}
$
work done will be equal to the change in surface energy, thus
$
\begin{aligned}
& W=S_f-S_i=N 4 \pi r^2 T-4 \pi R^2 T \\
& W=N 4 \pi\left(\frac{R}{N^{\frac{1}{3}}}\right)^2 T-4 \pi R^2 T \\
& W=4 \pi R^2 T\left(N^{\frac{1}{3}}-1\right)
\end{aligned}
$
if $N^{\frac{1}{3}}$ is very large thus it becomes
$
W=4 \pi R^2 T N^{\frac{1}{3}}
$
thus
$
W \propto N^{\frac{1}{3}}
$
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MCQ 121 Mark
An astronaut is standing on an asteroid when he accidentally drops a wrench. He observes that the gravitational acceleration on the asteroid is $2.4 m / s ^2$. If he had thrown the wrench at an upward angle instead, he would have found the gravitational acceleration on the asteroid to be:
  • A
    less than $2.4 m / s ^2$
  • B
    downward at $2.4 m / s ^2$
  • C
    greater than $2.4 m / s ^2$
  • D
    toward him at $2.4 m / s ^2$
Answer
(b) downward at $2.4 m / s ^2$
Explanation: The acceleration will be downward at $2.4 m / s ^2$.
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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip