List of top Questions asked in CBSE CLASS XII

“You cannot see a person standing on the other side of a boundary wall but can hear him.” Explain with reason.
“There is a limit to the amount of charge that can be stored on a given capacitor.” Explain.
Define the term, ‘distance of closest approach’. A proton of 3.95 MeV energy approaches a target nucleus \( Z = 79 \) in head-on position. Calculate its distance of closest approach.
A current of 5 A is passing along +X direction through a wire lying along X-axis. Find the magnetic field \( \vec{B} \) at a point \( \vec{r} = (3 \hat{i} + 4 \hat{j}) \, \text{m} \) due to 1 cm element of the wire, centered at the origin.
Find the effective resistance of the network of resistors between points A and F as shown in the figure.
The ends of six wires, each of resistance R
Four point charges \( Q \) each, are held at the four corners of a square of side \( l \). The amount of work done in bringing a charge \( Q \) from infinity to the centre of the square will be:
A capacitor of capacitance \( C \) has reactance \( X \) in an AC circuit. If the capacitance and the frequency of the applied voltage are doubled, the new reactance will become:
A beam of light of wavelength 720 nm in air enters water (refractive index \( n = \frac{4}{3} \)). Its wavelength in water will be:
Why is Ampere’s law used for the derivation in magnetic field due to a long straight current-carrying wire of infinite length and not Biot-Savart’s law? Explain.
Use Ampere’s law to derive the expression for the magnetic field due to a long straight current-carrying wire of infinite length.
How is Coulomb repulsion between protons in a nucleus overcome? Explain.
Why is the mass of a nucleus always less than the sum of the masses of its constituents, i.e. free neutrons and free protons?
A circular coil of wire having 200 turns, each radius 4.0 cm is placed in a horizontal plane. It carries a current of 0.40 A in the clockwise direction. Find the magnitude and direction of the magnetic field at the center of the coil.
A glass slab (\( \mu = 1.5 \)) of thickness 6 cm is placed over a paper. The shift in the letters printed on the paper will be:
A transformer is a device used for converting:
The effective capacitance of the network between points A and B shown in the figure is:
effective capacitance of the network between
A thin plano-concave lens with its curved face of radius of curvature R is made of glass of refractive index \( n_1 \). It is placed coaxially in contact with a thin equiconvex lens of same radius of curvature of refractive index \( n_2 \). Obtain the power of the combination lens.
Write any two advantages of a compound microscope over a simple microscope. Draw a ray diagram for the image formation at the near point by a compound microscope and explain it.
In a refracting telescope, the focal length of the objective is 50 times the focal length of the eyepiece. When the final image is formed at infinity, the length of the tube is 102 cm. Find the focal lengths of the two lenses.
Draw a ray diagram of a reflecting telescope (Cassegrain) and explain the formation of the image. State two important advantages that a reflecting telescope has over a refracting telescope.
The current through an inductor is uniformly increased from zero to 2 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at t = 10 s.
Define the term mutual inductance. Deduce the expression for the mutual inductance of two long coaxial solenoids of the same length having different radii and different number of turns.
The currents passing through two inductors of self-inductances 10 mH and 20 mH increase with time at the same rate. Draw graphs showing the variation of:
(I) The magnitude of emf induced with the rate of change of current in each inductor:
(II) The energy stored in each inductor with the current flowing through it:
Define self-inductance of a coil. Derive the expression for the energy required to build up a current \(I\) in a coil of self-inductance \(L\).
Current I (= 1 A) is passing through a copper rod (n = \(8.5 \times 10^{28} \, \text{m}^{-3}\)) of varying cross-sections as shown in the figure. The areas of cross-section at points A and B along its length are \(1.0 \times 10^{-7} \, \text{m}^2\) and \(2.0 \times 10^{-7} \, \text{m}^2\) respectively. Calculate:
Current I (= 1 A) is passing through a copper rod
(I) the ratio of electric fields at points A and B.
(II) the drift velocity of free electrons at point B.