MCQ 11 Mark
The minimum distance between an object and its real image formed by a convex lens of focal length f is:
- A4f
- Bf
- C2f
- D3f
Answer
View full question & answer→(a) 4f
Explanation: 4f
Explanation: 4f
Questions
M.C.Q (1 Marks)
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12 questions · timed · auto-graded
MCQ 21 Mark
In the energy band diagram of a material as given below, the open circles and filled circles denote holes and electrons respectively. The material is a/an
- Ainsulator
- Bmetal
- Cp-type semiconductor
- Dn-type semiconductor
Answer
View full question & answer→(c) p-type semiconductor
Explanation: One can see in the figure that number of holes are greater than number of electrons. Hence it is p−type semi conductor.
Explanation: One can see in the figure that number of holes are greater than number of electrons. Hence it is p−type semi conductor.
MCQ 31 Mark
When $10^{19}$ electrons are removed from a neutral metal plate, the electric charge on it is
- A-1.6 C
- B$10^{+19} C$
- C+1.6 C
- D$10^{-19} C$
Answer
View full question & answer→(c) +1.6 C
Explanation: $q = ne =10^{19} \times 1.6 \times 10^{-19} C =+1.6 C$.
Explanation: $q = ne =10^{19} \times 1.6 \times 10^{-19} C =+1.6 C$.
MCQ 41 Mark
Phase difference between any two points of a wavefront is
- A$\pi$
- B$0$
- C$\frac{\pi}{4}$
- D$\frac{\pi}{2}$
Answer
View full question & answer→(b) $0$
Explanation: Wavefront is the locus of all points those are in same phase.
Explanation: Wavefront is the locus of all points those are in same phase.
MCQ 51 Mark
A bar magnet having a magnetic moment of $2 \times 10^4 JT ^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B =6 \times 10^{-4} T$ exists in the space. The work done in taking the magnet slowly from a direction
- A0.6 J
- B12 J
- C2 J
- D6 J
Answer
View full question & answer→(d) 6 J
Explanation: $W =m B\left(\cos \theta_1-\cos \theta_2\right)$
$=2 \times 10^4 \times 6 \times 10^{-4}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)=6 J$
Explanation: $W =m B\left(\cos \theta_1-\cos \theta_2\right)$
$=2 \times 10^4 \times 6 \times 10^{-4}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)=6 J$
MCQ 61 Mark
There are two coils A and B as shown in the figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that
- Athere is a constant current in the counterclockwise direction in A.
- Bthere is a constant current in the clockwise direction in A.
- Cthere is a varying current in A.
- Dthere is no current in A.
Answer
View full question & answer→(a) there is a constant current in the counterclockwise direction in A.
Explanation: there is a constant current in the counterclockwise direction in A.
Explanation: there is a constant current in the counterclockwise direction in A.
MCQ 71 Mark
An electron with velocity $\vec{v}=\left(v_x \hat{i}+v_y \hat{j}\right)$ moves through a magnetic field $\vec{B}=\left(B_x \hat{i}-B_y \hat{j}\right)$. The force $\overrightarrow{ F }$ on the electron is: ( e is the magnitude of its charge)
- A$e\left(v_x B_y-v_y B_x\right) \hat{k}$
- B$-e\left(v_x B_y-v_y B_x\right) \hat{k}$
- C$-e\left(v_x B_y+v_y B_x\right) \hat{k}$
- D$e\left(v_x B_y+v_y B_x\right) \hat{k}$
Answer
View full question & answer→(d) $e\left(v_x B_y+v_y B_x\right) \hat{k}$
Explanation: $e\left(v_x B_y+v_y B_x\right) \hat{k}$
The force is given as $- e ( v \times B )$.
Explanation: $e\left(v_x B_y+v_y B_x\right) \hat{k}$
The force is given as $- e ( v \times B )$.
MCQ 81 Mark
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be:
- A920 keV
- B$\frac{1}{1840} keV$
- C1 keV
- D1840 keV
Answer
View full question & answer→(c) 1 keV
Explanation: K.E. gained $= qV = e \times 1 kV =1 keV$
Explanation: K.E. gained $= qV = e \times 1 kV =1 keV$
MCQ 91 Mark
A bar$-$magnet of the pole-strength $2 Amp - m$ is kept in a magnetic field of induction $4 \times 10^{-5} Wb / m ^2$ such that the axis of the magnet makes an angle $30^{\circ}$ with the direction of the field. If the couple acting on the magnet is found to be $80 \times 10^{-7} Nm$, then the distance between the poles of the magnet is:
- ✓$20m$
- B$4m$
- C$2m$
- D$8m$
Answer
View full question & answer→Correct option: A.
$20m$
$\tau=q_m \times 2 l \times B \sin \theta$
$\therefore 2 l=\frac{\tau}{q_m \times B \sin \theta}$
$=\frac{80 \times 10^{-7}}{2 \times 4 \times 10^{-5} \times \sin 30^{\circ}}=0.20 m=20 \ cm$
$\therefore 2 l=\frac{\tau}{q_m \times B \sin \theta}$
$=\frac{80 \times 10^{-7}}{2 \times 4 \times 10^{-5} \times \sin 30^{\circ}}=0.20 m=20 \ cm$
MCQ 101 Mark
For a glass prism, the angle of minimum deviation will be smallest for the light of
- Ablue colour
- Byellow colour
- Cgreen colour
- Dred colour.
Answer
View full question & answer→(d) red colour.
Explanation: For a glass prism, the angle of minimum deviation will be smallest for the light of red color. As wavelength of red color is maximum among all, hence, $\mu \propto \frac{1}{\lambda}$, hence $\mu$ is smaller. As $\mu$ decreases, angle of deviation decreases.
Explanation: For a glass prism, the angle of minimum deviation will be smallest for the light of red color. As wavelength of red color is maximum among all, hence, $\mu \propto \frac{1}{\lambda}$, hence $\mu$ is smaller. As $\mu$ decreases, angle of deviation decreases.
MCQ 111 Mark
$m ^2 V^{-1} s^{-1}$ is the SI unit of which of the following?
- APotential gradient
- BMobility
- CDrift velocity
- DResistivity
Answer
View full question & answer→(b) Mobility
Explanation: The charge carrier in most metals is the negatively charged electron. The mobility of the charge carrier is defined as the drift velocity of the charge carrier per unit electric field. It is denoted by $\mu$ and $\mu=v_d / E$ is given as. The Sl unit of $\mu$ is $m 2 V-1 s-1$.
Explanation: The charge carrier in most metals is the negatively charged electron. The mobility of the charge carrier is defined as the drift velocity of the charge carrier per unit electric field. It is denoted by $\mu$ and $\mu=v_d / E$ is given as. The Sl unit of $\mu$ is $m 2 V-1 s-1$.
MCQ 121 Mark
In the circuit shown, the current through the ideal diode is:
- A100mA
- B20mA
- C25mA
- D75mA
Answer
View full question & answer→(c) 25 mA
Explanation: The given diode is ideal and forward-biased so there is short circuit. No current flow through the $20 \Omega$ and whole current flow through the short circuit. The current flow through the circuit is, $I =\frac{2 V}{80 \Omega}=25 mA$
Explanation: The given diode is ideal and forward-biased so there is short circuit. No current flow through the $20 \Omega$ and whole current flow through the short circuit. The current flow through the circuit is, $I =\frac{2 V}{80 \Omega}=25 mA$