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Physics M.C.Q (1 Marks)

Model Paper 1 · 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 21 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:
  • A

    Image
  • B

    Image
  • C

    Image
  • D

    Image
Answer
(a)
Image
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.
Image
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MCQ 41 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
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MCQ 81 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 91 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 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
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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M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip