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The questions your board exam loves to ask
800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
PhysicsClass 123 marksmedium
Ray Optics and Optical Instruments
Write the expression for the magnifying power of an astronomical telescope in normal adjustment and state the required focal lengths.
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An astronomical telescope has an objective lens of large focal length fo and an eyepiece of small focal length fe. In normal adjustment (final image at infinity), the magnifying power is M = fo/fe, and the length of the telescope is fo + fe. For high magnification, the objective should have a large focal length (and large aperture) and the eyepiece a small focal length. The final image is inverted.
M = fo/fe
Marking-scheme points
- ✓Magnifying power (normal adjustment): M = fo/fe
- ✓Length of telescope = fo + fe
- ✓Objective: large fo; eyepiece: small fe; image inverted
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Wave Optics
Write the expression for the fringe width in Young's double slit experiment and explain the terms.
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In Young's double slit experiment, two coherent sources produce alternate bright and dark fringes on a screen. The fringe width (the distance between two consecutive bright or dark fringes) is beta = lambda D/d, where lambda is the wavelength of light used, D is the distance between the slits and the screen, and d is the separation between the two slits. All fringes are of equal width, and the width increases with wavelength and D but decreases as d increases.
beta = lambda D/d
Marking-scheme points
- ✓Fringe width beta = lambda D/d
- ✓lambda = wavelength, D = slit-to-screen distance, d = slit separation
- ✓Fringes are equally spaced; beta increases with lambda and D
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Wave Optics
In Young's double slit experiment, light of wavelength 600 nm is used with a slit separation of 1 mm and a screen 1 m away. Calculate the fringe width.
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Given lambda = 600 nm = 600 x 10^-9 m, d = 1 mm = 1 x 10^-3 m, D = 1 m. Fringe width beta = lambda D/d = (600 x 10^-9 x 1)/(1 x 10^-3) = 600 x 10^-6 = 6 x 10^-4 m = 0.6 mm.
beta = lambda D/d
Marking-scheme points
- ✓Convert units: lambda = 6e-7 m, d = 1e-3 m
- ✓beta = lambda D/d = (6e-7 x 1)/1e-3
- ✓beta = 6 x 10^-4 m = 0.6 mm
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Wave Optics
What is diffraction of light? Write the condition for minima in single slit diffraction.
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Diffraction is the bending of light around the edges of an obstacle or aperture and its spreading into the geometrical shadow region. In diffraction at a single slit of width a, a central bright maximum is flanked by alternate dark and bright fringes. The condition for the dark fringes (minima) is a sin theta = n lambda, where n = 1, 2, 3, ... and theta is the angle of diffraction. The central maximum is the brightest and widest.
a sin theta = n lambda
Marking-scheme points
- ✓Diffraction: bending/spreading of light around obstacles/apertures
- ✓Single slit minima: a sin theta = n lambda (n = 1, 2, 3...)
- ✓Central maximum is the brightest and widest
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Dual Nature of Radiation and Matter
Write Einstein's photoelectric equation and explain each term.
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Einstein's photoelectric equation is h f = W0 + KE(max), where h f is the energy of the incident photon (h is Planck's constant and f the frequency of light), W0 is the work function (the minimum energy needed to eject an electron from the metal surface), and KE(max) is the maximum kinetic energy of the emitted photoelectron. It expresses conservation of energy: the photon's energy is used partly to free the electron and the rest appears as its kinetic energy. Thus KE(max) = h f - W0.
h f = W0 + KE(max)
Marking-scheme points
- ✓h f = W0 + KE(max)
- ✓h f = photon energy; W0 = work function
- ✓KE(max) = h f - W0 (energy conservation)
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Dual Nature of Radiation and Matter
Light of wavelength 400 nm is incident on a metal of work function 2 eV. Calculate the maximum kinetic energy of the emitted photoelectrons. (Use hc = 1240 eV nm)
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The energy of the incident photon E = hc/lambda = 1240/400 = 3.1 eV. Using Einstein's equation, the maximum kinetic energy KE(max) = E - W0 = 3.1 - 2.0 = 1.1 eV. Since the photon energy (3.1 eV) is greater than the work function (2 eV), emission occurs and the photoelectrons have a maximum kinetic energy of 1.1 eV.
KE(max) = hc/lambda - W0
Marking-scheme points
- ✓Photon energy E = hc/lambda = 1240/400 = 3.1 eV
- ✓KE(max) = E - W0 = 3.1 - 2.0
- ✓KE(max) = 1.1 eV
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Atoms
State the postulates of Bohr's model of the hydrogen atom.
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Bohr's postulates are: (1) The electron revolves around the nucleus only in certain fixed circular orbits called stationary states, in which it does not radiate energy. (2) Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/(2 pi), that is, m v r = n h/(2 pi) (quantisation of angular momentum). (3) Energy is emitted or absorbed only when the electron jumps from one orbit to another, the energy of the emitted or absorbed photon being h f = E2 - E1.
m v r = n h/(2 pi)
Marking-scheme points
- ✓Electrons revolve in fixed stationary orbits without radiating
- ✓Angular momentum quantised: m v r = n h/(2 pi)
- ✓Energy change on jump: h f = E2 - E1
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Atoms
Calculate the energy of the photon emitted when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Use En = -13.6/n^2 eV)
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Energy of the electron in n = 3: E3 = -13.6/9 = -1.51 eV. Energy in n = 2: E2 = -13.6/4 = -3.4 eV. The energy of the emitted photon = E3 - E2 = -1.51 - (-3.4) = 1.89 eV. This corresponds to the H-alpha line of the Balmer series (visible red light).
E(photon) = E(higher) - E(lower)
Marking-scheme points
- ✓E3 = -13.6/9 = -1.51 eV; E2 = -13.6/4 = -3.4 eV
- ✓Photon energy = E3 - E2 = 1.89 eV
- ✓This is the H-alpha (Balmer series) line
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Nuclei
The mass defect of a helium nucleus is 0.0304 u. Calculate its binding energy. (1 u = 931 MeV)
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The binding energy is the energy equivalent of the mass defect. BE = (mass defect in u) x 931 MeV = 0.0304 x 931 = 28.3 MeV. Thus the binding energy of the helium nucleus is about 28.3 MeV, and the binding energy per nucleon = 28.3/4 = 7.1 MeV.
BE (MeV) = (delta m in u) x 931
Marking-scheme points
- ✓BE = (mass defect) x 931 MeV
- ✓= 0.0304 x 931
- ✓BE = 28.3 MeV (about 7.1 MeV per nucleon)
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Nuclei
State the radioactive decay law. What fraction of a radioactive sample remains after 3 half-lives?
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The radioactive decay law states that the rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present at that instant: N = N0 e^(-lambda t), where lambda is the decay constant. The half-life T is related to it by T = 0.693/lambda. After each half-life, half of the sample remains, so after 3 half-lives the fraction remaining = (1/2)^3 = 1/8 of the original sample.
N = N0 (1/2)^(t/T)
Marking-scheme points
- ✓Decay law: N = N0 e^(-lambda t); half-life T = 0.693/lambda
- ✓Fraction after n half-lives = (1/2)^n
- ✓After 3 half-lives: (1/2)^3 = 1/8 remains
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Semiconductor Electronics
Explain how a p-n junction diode works as a half-wave rectifier.
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A rectifier converts alternating current (AC) into direct current (DC). In a half-wave rectifier, a single diode is connected in series with the AC source and a load resistor. During the positive half-cycle of the AC input, the diode is forward biased and conducts, so current flows through the load. During the negative half-cycle, the diode is reverse biased and does not conduct, so no current flows. As a result, output is obtained only during one half of each cycle, giving a pulsating DC. Its efficiency is low because half the input is wasted.
Marking-scheme points
- ✓Rectifier converts AC into DC; uses one diode
- ✓Positive half-cycle: diode forward biased -> conducts
- ✓Negative half-cycle: diode reverse biased -> no output (pulsating DC)
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Semiconductor Electronics
Explain the working of a full-wave rectifier using two diodes.
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A full-wave rectifier converts both halves of the AC input into DC. It uses two diodes with a centre-tapped transformer and a load resistor. During the positive half-cycle, one diode is forward biased and conducts while the other is reverse biased; during the negative half-cycle, the second diode conducts while the first does not. In both half-cycles, the current through the load flows in the same direction, so output is obtained during the whole cycle. This gives a smoother, more efficient pulsating DC than a half-wave rectifier.
Marking-scheme points
- ✓Uses two diodes and a centre-tapped transformer
- ✓Each diode conducts during one half-cycle
- ✓Current through load is in the same direction for both halves (full-wave DC)
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Semiconductor Electronics
Write the truth tables of the OR, AND and NOT logic gates for inputs A and B.
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A logic gate is a digital circuit that gives an output based on its inputs following a logical rule (using binary 0 and 1). OR gate (output Y = A + B): Y = 1 if any input is 1; for inputs (0,0),(0,1),(1,0),(1,1) the outputs are 0,1,1,1. AND gate (Y = A.B): Y = 1 only if both inputs are 1; outputs are 0,0,0,1. NOT gate (Y = not A): it has a single input and inverts it, so input 0 gives output 1 and input 1 gives output 0.
OR: Y = A + B; AND: Y = A.B; NOT: Y = not A
Marking-scheme points
- ✓OR (Y = A + B): output 1 if any input is 1 -> 0,1,1,1
- ✓AND (Y = A.B): output 1 only if both inputs 1 -> 0,0,0,1
- ✓NOT: single input, inverts it (0 -> 1, 1 -> 0)
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