Worked out from the 105 one-mark questions of the 2020–2026 public exams. The table shows how many board questions came from each chapter; below it are the questions the board asked more than once, with the answer.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to :
- A. \(\sqrt{2}\,D\)
- B. \(\frac{D}{2}\)
- C. \(\frac{D}{\sqrt{2}}\)
- D. 2DCorrect
Explanation. Fringe width \(\beta=\frac{\lambda D}{d}\). If d becomes 2d, D must become 2D to keep \(\beta\) unchanged.
Emission of electrons by the absorption of heat energy is called ________ emission.
- A. field
- B. thermionicCorrect
- C. photoelectric
- D. secondary
Explanation. When a metal is heated, electrons gain enough thermal energy to escape; this is thermionic emission.
The transverse nature of light is shown in :
- A. scattering
- B. interference
- C. polarisationCorrect
- D. diffraction
Explanation. Only transverse waves can be polarised; interference and diffraction occur for longitudinal waves too. So polarisation shows that light is transverse.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is :
- A. \(\frac{Q}{\sqrt{2}}\)Correct
- B. \(\frac{Q}{2}\)
- C. Q
- D. \(\frac{Q}{\sqrt{3}}\)
Explanation. Total energy \(\frac{Q^2}{2C}\). When the electric energy is half of it, \(\frac{q^2}{2C} = \frac{1}{2}\cdot\frac{Q^2}{2C}\), so \(q = \frac{Q}{\sqrt{2}}\).
Two identical conducting balls having positive charges \(q_1\) and \(q_2\) are separated by a centre to centre distance r. If they are made to touch each other and then separated to the same distance, the force between them will be :
- A. zero
- B. less than before
- C. more than beforeCorrect
- D. same as before
Explanation. After touching, each ball carries \(\frac{q_1+q_2}{2}\). Since \(\left(\frac{q_1+q_2}{2}\right)^2 \ge q_1q_2\), the force (proportional to the product of charges) becomes more than before (equal only if \(q_1=q_2\)).
The speed of light in an isotropic medium depends on :
- A. the nature of propagation
- B. its intensity
- C. the motion of the source with respect to medium
- D. its wavelengthCorrect
Explanation. In a medium the refractive index varies with wavelength (dispersion), so the speed \(v=c/n\) depends on the wavelength of light.
The principle based on which a solar cell operates is :
- A. Photovoltaic actionCorrect
- B. Diffusion
- C. Carrier flow
- D. Recombination
Explanation. A solar cell converts light energy directly into electrical energy by the photovoltaic effect.
In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6 A, then that in the secondary coil is :
- A. 12 A
- B. 2 ACorrect
- C. 1 A
- D. 18 A
Explanation. \(I_s = I_p\frac{N_p}{N_s} = 6\times\frac{410}{1230} = 2\) A.
For light incident from air on a slab of refractive index 2, the maximum possible angle of refraction is :
- A. 60°
- B. 30°Correct
- C. 90°
- D. 45°
Explanation. The angle of refraction is largest for grazing incidence (i = 90°): \(\sin r = \frac{\sin 90^\circ}{2} = \frac{1}{2}\), so r = 30°.
If the amplitude of the magnetic field is \(3\times10^{-6}\) T, then the amplitude of the electric field for a electromagnetic wave is :
- A. 600 \(Vm^{-1}\)
- B. 100 \(Vm^{-1}\)
- C. 900 \(Vm^{-1}\)Correct
- D. 300 \(Vm^{-1}\)
Explanation. \(E_0 = cB_0 = 3\times10^{8}\times3\times10^{-6} = 900\ Vm^{-1}\).
The value of forbidden energy gap for Si at room temperature is :
- A. 1.1 V
- B. 0.7 eV
- C. 1.1 eVCorrect
- D. 0.7 V
Explanation. The forbidden energy gap is an energy, measured in eV; for silicon at room temperature it is about 1.1 eV (0.7 V is its barrier potential).
The Zener diode is primarily used as :
- A. Oscillator
- B. Rectifier
- C. Voltage regulatorCorrect
- D. Amplifier
Explanation. In reverse breakdown the voltage across a Zener diode stays nearly constant, so it is mainly used as a voltage regulator.
A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field \(\vec{B}\).
- A. \(\sqrt{\frac{2q^3B^2V}{m}}\)Correct
- B. \(\sqrt{\frac{2q^3BV}{m}}\)
- C. \(\sqrt{\frac{2q^3BV}{m^3}}\)
- D. \(\sqrt{\frac{q^3B^2V}{2m}}\)
Explanation. \(\frac{1}{2}mv^2=qV\Rightarrow v=\sqrt{\frac{2qV}{m}}\). Force \(F=qvB=qB\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2q^3B^2V}{m}}\).
