MCQ 11 Mark
If copper wire is stretched to make its radius decrease by 0.1%, then the percentage change in its resistance is approximately
- A-0.4%
- B+0.8%
- C+0.4%
- D+0.2%
Answer
View full question & answer→(c) +0.4%
Questions
M.C.Q (1 Marks)
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12 questions · timed · auto-graded
MCQ 21 Mark
The number of electrons made available for conduction by dopant atoms depends strongly upon
- Adoping level
- Bincrease in ambient temperature
- Cenergy gap
- Doptions (a) and (b) both
Answer
View full question & answer→(a) doping level
MCQ 31 Mark
Colours observed on a CD (Compact Disk) is due to
- AReflection
- BDiffraction
- CDispersion
- DAbsorption
Answer
View full question & answer→(b) Diffraction
MCQ 41 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
- A$2 \ cm$
- B$4 \ cm$
- ✓$6 \ cm$
- D$12 \ cm$
Answer
$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
$\frac{1}{v}-\frac{3}{2[-6]}=\frac{[1-3 / 2]}{-6}$
$\frac{1}{v}=\frac{-3}{12}+\frac{1}{12}=\frac{-2}{12}=\frac{-1}{6}$
$v=6 \ cm$
View full question & answer→Correct option: C.
$6 \ cm$
$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$
$\frac{1}{v}-\frac{3}{2[-6]}=\frac{[1-3 / 2]}{-6}$
$\frac{1}{v}=\frac{-3}{12}+\frac{1}{12}=\frac{-2}{12}=\frac{-1}{6}$
$v=6 \ cm$
MCQ 51 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?
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?
- A$V / 2$
- ✓$\sqrt{2} V$
- C$2 V$
- D$4 V$
Answer
View full question & answer→Correct option: B.
$\sqrt{2} V$
$d=\frac{\text { const }}{V_1^2} ...(1)$
$\frac{d}{2}=\frac{\text { const }}{V_2^2} ...(2)$
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, $\left( V _1= V \right)$
$\frac{d}{2}=\frac{\text { const }}{V_2^2} ...(2)$
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, $\left( V _1= V \right)$
MCQ 61 Mark
Correct match of column I with column II is
| C-1 (waves) | C-II (Production) |
| (1) Infra-red | P. Rapid vibration of electrons in aerials |
| (2) Radio | Q. Electrons in atoms emit light when they move from higher to lower energy level. |
| (3) Light | R. Klystron valve |
| (4) Microwave | S. Vibration of atoms and molecules |
- A1-P, 2-R, 3-S, 4-Q
- B
- C1-Q, 2-P, 3-S, 4-R
- D1-S. 2-R, 3-P, 4-Q
Answer
View full question & answer→(b) 1-S, 2-P, 3-O, 4-R
MCQ 71 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
- A$10 V$
- B$10 \sqrt{2} V$
- ✓$10 / \sqrt{2} V$
- D$20 V$
Answer
View full question & answer→Correct option: C.
$10 / \sqrt{2} V$
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}$
$V_{ L } = I ^{\prime} X _{ L }$
$=\frac{10}{\sqrt{2} R } \times R$
$=5 \sqrt{2} V$
$I R=I X_C=I X_L=10 V$
$X_C=X_L=R$
$Z=\sqrt{R^2+\left(X_C-X_L\right)^2}$
$Z=\sqrt{R^2+(R-R)^2}$
$Z=R$
$V_S=I Z=I R=10 V$
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}$
$V_{ L } = I ^{\prime} X _{ L }$
$=\frac{10}{\sqrt{2} R } \times R$
$=5 \sqrt{2} V$
MCQ 81 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
- A$8.85 \times 10^{-3} A$
- B$8.85 \times 10^{-4} A$
- C$7.85 \times 10^{-3} A$
- D$9.85 \times 10^{-3} A$
Answer
View full question & answer→MCQ 91 Mark
The diffraction effect can be observed in
- Asound waves only
- Blight waves only
- Cultrasonic waves only
- Dsound waves as well as light waves
Answer
View full question & answer→(d) Sound waves as well as light waves
MCQ 101 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
- A$4:1$
- B$1:1$
- ✓$4:3$
- D$3:4$
Answer
At $P_2, B_2=\frac{\mu_0 I}{2 \pi\left(\frac{3 a}{2}\right)}=\frac{\mu_0 I}{3 \pi a}$
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)} $
$\Rightarrow \frac{B_2}{B_1}=\frac{4}{3}$
View full question & answer→Correct option: C.
$4:3$
At $P_2, B_2=\frac{\mu_0 I}{2 \pi\left(\frac{3 a}{2}\right)}=\frac{\mu_0 I}{3 \pi a}$
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)} $
$\Rightarrow \frac{B_2}{B_1}=\frac{4}{3}$
MCQ 111 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
The ratio of the magnitudes of the electric fields at the surfaces of the two spheres is
- A$\frac{r_1}{r_2}$
- B$\frac{r_2}{r_1}$
- C$\frac{r_2^2}{r_1^2}$
- D$\frac{r_1^2}{r_2^2}$
Answer
View full question & answer→(b) In the state of equilibrium,
The potential on the surface of bigger sphere the potential at the surface of the smaller sphere.
$\begin{aligned} & \frac{k q_1}{r_1}=\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{q_2}=\frac{r_1}{r_2} \\ \therefore & \frac{ E _1}{ E _2}=\frac{q_1}{q_2} \frac{r_2^2}{r_1^2}=\frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}\end{aligned}$
The potential on the surface of bigger sphere the potential at the surface of the smaller sphere.
$\begin{aligned} & \frac{k q_1}{r_1}=\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{q_2}=\frac{r_1}{r_2} \\ \therefore & \frac{ E _1}{ E _2}=\frac{q_1}{q_2} \frac{r_2^2}{r_1^2}=\frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}\end{aligned}$
MCQ 121 Mark
A uniform electric field pointing in positive X-direction exists in a region. Let A be the origin, B be the point on the X-axis at x = +1 cm and C be the point on the Y-axis at y 1 cm. Then the potential at points A, B and C satisfy
- A$V _{ A }< V _{ B }$
- B$V _{ A }> V _{ B }$
- C$V _{ A }< V _{ C }$
- D$V_A>V_C$
