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Physics M.C.Q (1 Marks)

MODEL PAPER 2025 · Rajasthan Board (RBSE) STD 12 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.

Question 21 Mark
A long straight wire of circular cross section of radius'a carries a steady current $1.$ The current is uniformly distributed across its cross section. The ratio of magnitudes of the magnetic field at a point $a/2$ above the surface of wire to that of a point a $2$ below its surface is
Answer
$(c)$

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$\text { At } P_2, B_2=\frac{\mu_0 I}{2 \pi\left(\frac{3 a}{2}\right)}=\frac{\mu_0 I}{3 \pi a}$
$\text { At } P_1, B_1=\frac{\mu_0(I / 4)}{2 \pi(a / 2)}=\frac{\mu_0 I}{4 \pi a}$
$\therefore \frac{B_2}{B_1}=\frac{\left(\frac{\mu_0 I}{3 \pi a}\right)}{\left(\frac{\mu_0 I}{4 \pi a}\right)} $
$\Rightarrow \frac{B_2}{B_1}=\frac{4}{3}$
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Question 31 Mark
A conducting wire connects two charged conducting spheres such that they attain equilibrium with respect to each other. The distance of separation between the two spheres is very large as compared to either of their radii.
The ratio of the magnitudes of the electric fields at the surfaces of the two spheres is
Answer
(b) In the state of equilibrium,
The potential on the surface of bigger sphere the potential at the surface of the smaller sphere.
$\begin{aligned} & \frac{k q_1}{r_1}=\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{q_2}=\frac{r_1}{r_2} \\ \therefore & \frac{ E _1}{ E _2}=\frac{q_1}{q_2} \frac{r_2^2}{r_1^2}=\frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}\end{aligned}$
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Question 61 Mark
In a series $\text {LCR}$ circuit, the voltage across the resistance, capacitance and inductance is $10 V$ each. If the capacitance is short circuited the voltage across the inductance will be
Answer
$(c)$ When all the given components are connected
$I R=I X_C=I X_L=10 V$
$X_C=X_L=R$
$Z=\sqrt{R^2+\left(X_C-X_L\right)^2}$
$Z=\sqrt{R^2+(R-R)^2}$
$Z=R$
$V_S=I Z=I R=10 V$
So, the source voltage is also $10 V$
When the capacitor is short circuited then
$ Z =\sqrt{ R ^2+\left( X _{ L }\right)^2}$
$ =\sqrt{R^2+R^2}=R \sqrt{2}$
$V_{ L } = I ^{\prime} X _{ L }$
​​​​​​​$=\frac{10}{\sqrt{2} R } \times R =5 \sqrt{2} V$
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Question 71 Mark
The number of electrons made available for conduction by dopant atoms depends strongly upon
Answer
(a) doping level
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Question 81 Mark
If copper wire is stretched to make its radius decrease by 0.1%, then the percentage change in its resistance is approximately
Answer
(c) +0.4%
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Question 91 Mark
A capacitor consists of two parallel plates, with an area of cross-section of $0.001 m^2$, separated by a distance of 0.0001 m. If the voltage across the plates varies at the rate of $10^8 V / s$, then the value of displacement current through the capacitor is
Answer
(a)  $8.85 \times 10^{-3} A$
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Question 101 Mark
The distance of closest approach of an alpha particle is $d$ when it moves with a speed $V$ towards a nucleus.
Another alpha particle is projected with higher energy such that the new distance of the closest approach is $d/2$ . What is the speed of projection of the alpha particle in this case?
Answer
$(b)$ The distance of closest approach
$d=\frac{\text { const }}{V_1^2} ...(1)$
$\frac{d}{2}=\frac{\text { const }}{V_2^2} ...(2)$
From equations $(1)$ and $(2), $
$2=\frac{V_2^2}{V_1^2} $
$\Rightarrow V_2=\sqrt{2} V_1$
$\therefore V_2=\sqrt{2} V$
Given, $\left( V _1= V \right)$
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Question 121 Mark
A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index $1.5.$ The distance of virtual image from the surface of the sphere is
Answer

Image

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
$\frac{1}{v}-\frac{3}{2[-6]}=\frac{[1-3 / 2]}{-6}$
$\frac{1}{v}=\frac{-3}{12}+\frac{1}{12}=\frac{-2}{12}=\frac{-1}{6}$
$v=6 \ cm$
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