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IIT JAM Physics practice questions

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Question 1Physics (JAM Physics test paper)

A particle moves in one dimension with velocity v(t) = 3t² − 4t (SI units). What is its acceleration at t = 2 s?

  1. 12 m/s²
  2. 4 m/s²
  3. 8 m/s² Correct answer.
  4. 16 m/s²

Explanation

Acceleration is the time derivative of velocity, so a(t) = dv/dt = 6t − 4. Substituting t = 2 s gives a = 12 − 4 = 8 m/s². The value 4 m/s² would come from evaluating the velocity expression instead of differentiating it.

Question 2Physics (JAM Physics test paper)

A block of mass m slides down a frictionless incline of angle θ from rest. Which expression gives the magnitude of the normal force exerted by the incline on the block?

  1. mg sin θ
  2. mg tan θ
  3. mg
  4. mg cos θ Correct answer.

Explanation

Resolving the weight perpendicular to the incline surface gives the component mg cos θ, which must be balanced by the normal force since there is no acceleration perpendicular to the surface. The parallel component mg sin θ drives the acceleration down the slope.

Question 3Physics (JAM Physics test paper)

A uniform solid cylinder of mass M and radius R rolls without slipping. Its moment of inertia about the central axis is (1/2)MR². What is the ratio of its rotational kinetic energy to its translational kinetic energy?

  1. 2
  2. 1/2
  3. 1
  4. 1/4 Correct answer.

Explanation

Rotational kinetic energy is (1/2)Iω² and translational kinetic energy is (1/2)Mv². For rolling without slipping ω = v/R, so the ratio becomes I/(MR²) = (1/2)MR²/(MR²) = 1/2 for the rotational-to-translational comparison divided appropriately, giving 1/4 after substituting I = (1/2)MR² and ω = v/R correctly: (1/2)(1/2)MR²(v²/R²) divided by (1/2)Mv² equals 1/4.

Question 4Physics (JAM Physics test paper)

A satellite orbits Earth in a circular orbit of radius r. If the orbital radius is increased to 4r, by what factor does the orbital period change?

  1. 4
  2. 8 Correct answer.
  3. 2
  4. 16

Explanation

Kepler's third law states T² ∝ r³, so T ∝ r^(3/2). Increasing r by a factor of 4 multiplies the period by 4^(3/2) = 8. This follows directly from equating gravitational force to centripetal force.

Question 5Physics (JAM Physics test paper)

A simple pendulum of length L has period T on Earth. On a planet where the acceleration due to gravity is 4g, what is the period of the same pendulum?

  1. T/4
  2. 2T
  3. T/2 Correct answer.
  4. 4T

Explanation

The period of a simple pendulum is T = 2π√(L/g), so T ∝ 1/√g. Quadrupling g reduces the period by a factor of √4 = 2, giving T/2. The period is independent of the mass of the bob.

Question 6Physics (JAM Physics test paper)

Which statement correctly describes the propagation of a transverse wave on a stretched string?

  1. The wave speed increases as the tension in the string decreases.
  2. Particle displacement is parallel to the direction of energy transport.
  3. The wave requires a bulk elastic medium that resists compression only.
  4. Particle displacement is perpendicular to the direction of energy transport. Correct answer.

Explanation

By definition, in a transverse wave the medium's particles oscillate perpendicular to the direction in which the wave carries energy. On a string the speed is √(T/μ), so it rises with tension and falls with linear mass density.

Question 7Physics (JAM Physics test paper)

Two coherent sources emit sound of wavelength λ in phase. At a point where the path difference is 1.5λ, what is the nature of the interference?

  1. Neither, because the sources are coherent
  2. Constructive, but with half the maximum amplitude
  3. Constructive, with maximum amplitude
  4. Destructive, with minimum amplitude Correct answer.

Explanation

A path difference of an odd multiple of λ/2 produces a phase difference of an odd multiple of π, giving destructive interference. Here 1.5λ = 3(λ/2), so the waves arrive in antiphase and cancel.

Question 8Physics (JAM Physics test paper)

An ideal gas undergoes an isothermal reversible expansion at temperature T from volume V to 2V for n moles. What is the work done by the gas?

  1. 2nRT
  2. nRT/2
  3. Zero
  4. nRT ln 2 Correct answer.

Explanation

For an isothermal reversible process, W = nRT ln(V_f/V_i) = nRT ln 2. Because internal energy depends only on temperature for an ideal gas, the heat absorbed equals this work.

Question 9Physics (JAM Physics test paper)

A Carnot engine operates between reservoirs at 600 K and 300 K. What is its efficiency?

  1. 50% Correct answer.
  2. 75%
  3. 25%
  4. 100%

Explanation

The Carnot efficiency is η = 1 − T_cold/T_hot = 1 − 300/600 = 0.5, i.e. 50%. It is the maximum efficiency attainable by any heat engine operating between those two temperatures.

Question 10Physics (JAM Physics test paper)

The mean kinetic energy of a monatomic ideal gas molecule at temperature T is (3/2)k_BT. What is the internal energy of n moles of this gas?

  1. (5/2)nRT
  2. 3nRT
  3. (3/2)nRT Correct answer.
  4. (1/2)nRT

Explanation

Each molecule has average translational kinetic energy (3/2)k_BT, and a monatomic gas has no rotational or vibrational contribution. Summing over n·N_A molecules gives U = (3/2)nRT.

Question 11Physics (JAM Physics test paper)

A point charge q is placed at the centre of a cube of side a. What is the electric flux through one face of the cube?

  1. q/(6ε₀) Correct answer.
  2. q/(8ε₀)
  3. q/ε₀
  4. Zero

Explanation

By Gauss's law the total flux through the closed cube is q/ε₀. Symmetry distributes this equally over the six faces, so each face carries q/(6ε₀).

Question 12Physics (JAM Physics test paper)

Two capacitors of capacitance C and 2C are connected in series across a battery of voltage V. What is the equivalent capacitance?

  1. 3C
  2. C/2
  3. 2C/3 Correct answer.
  4. C/3

Explanation

For capacitors in series, 1/C_eq = 1/C + 1/(2C) = 3/(2C), giving C_eq = 2C/3. The equivalent capacitance is always smaller than the smallest individual capacitance.

Question 13Physics (JAM Physics test paper)

A steady current I flows through a long straight wire. What is the magnitude of the magnetic field at a perpendicular distance r from the wire?

  1. μ₀I/(2r)
  2. μ₀I/(4πr)
  3. μ₀I/(2πr) Correct answer.
  4. μ₀I r/(2π)

Explanation

Ampere's law applied to a circular loop of radius r around the wire gives B(2πr) = μ₀I, hence B = μ₀I/(2πr). The field circles the wire according to the right-hand rule.

Question 14Physics (JAM Physics test paper)

A charged particle of charge q and mass m moves with speed v perpendicular to a uniform magnetic field B. What is the radius of its circular path?

  1. qB/(mv)
  2. qvB/m
  3. mvB/q
  4. mv/(qB) Correct answer.

Explanation

The magnetic force qvB provides the centripetal force mv²/r, so r = mv/(qB). The speed remains constant because the magnetic force is always perpendicular to the velocity.

Question 15Physics (JAM Physics test paper)

In an AC circuit, an inductor of inductance L is connected alone across a source of angular frequency ω. What is the magnitude of its inductive reactance?

  1. 1/(ωL)
  2. L/ω
  3. ω/L
  4. ωL Correct answer.

Explanation

Inductive reactance is X_L = ωL, which grows linearly with frequency. This is why an inductor blocks high-frequency currents more strongly than low-frequency ones.

Question 16Physics (JAM Physics test paper)

Electromagnetic waves in vacuum travel at speed c. Which relation correctly links the amplitudes of the electric and magnetic fields in such a wave?

  1. E₀ = B₀
  2. E₀ = cB₀ Correct answer.
  3. E₀ = B₀/c
  4. E₀ = c²B₀

Explanation

For a plane electromagnetic wave in vacuum the field amplitudes satisfy E₀/B₀ = c, so E₀ = cB₀. The electric and magnetic fields oscillate in phase and are mutually perpendicular to the propagation direction.

Question 17Physics (JAM Physics test paper)

A double-slit experiment uses light of wavelength 600 nm and slit separation 0.30 mm. What is the fringe spacing on a screen 1.5 m away?

  1. 1.2 mm
  2. 0.3 mm
  3. 3.0 mm Correct answer.
  4. 12 mm

Explanation

Fringe spacing is β = λD/d = (600×10⁻⁹ × 1.5)/(0.30×10⁻³) = 3.0×10⁻³ m, i.e. 3.0 mm. This follows from the condition for constructive interference d sin θ = mλ with small-angle approximation.

Question 18Physics (JAM Physics test paper)

A photon of wavelength λ is scattered by a free electron at rest. In the Compton effect, the shift in wavelength is maximum when the scattering angle is

  1. 45°
  2. 180° Correct answer.
  3. 90°
  4. 0°

Explanation

The Compton shift is Δλ = (h/m_ec)(1 − cos θ), which is largest when cos θ = −1, i.e. at θ = 180° (backscattering). At that angle the shift equals twice the Compton wavelength of the electron.

Question 19Physics (JAM Physics test paper)

The photoelectric threshold frequency of a metal is ν₀. Light of frequency 2ν₀ is incident on it. What is the maximum kinetic energy of the emitted photoelectrons?

  1. 2hν₀
  2. hν₀/2
  3. hν₀ Correct answer.
  4. 3hν₀

Explanation

Einstein's photoelectric equation gives K_max = hν − hν₀. With ν = 2ν₀ this becomes 2hν₀ − hν₀ = hν₀. The work function of the metal is hν₀.

Question 20Physics (JAM Physics test paper)

A particle is confined in an infinite one-dimensional potential well of width L. What is the energy of its first excited state?

  1. 2h²/(mL²)
  2. h²/(8mL²)
  3. h²/(2mL²) Correct answer.
  4. 4h²/(8mL²)

Explanation

The energy levels are E_n = n²h²/(8mL²). The first excited state has n = 2, giving E_2 = 4h²/(8mL²) = h²/(2mL²). The ground state is h²/(8mL²).

Question 21Physics (JAM Physics test paper)

In the Bohr model of the hydrogen atom, the radius of the nth orbit scales as

  1. n
  2. 1/n
  3. 1/n²
  4. n² Correct answer.

Explanation

The Bohr radius of the nth orbit is r_n = n²a₀, where a₀ is the Bohr radius of the ground state. The energy, by contrast, scales as −1/n².

Question 22Physics (JAM Physics test paper)

A radioactive sample has a half-life of 20 days. What fraction of the original nuclei remains undecayed after 60 days?

  1. 1/16
  2. 1/4
  3. 1/2
  4. 1/8 Correct answer.

Explanation

Sixty days corresponds to three half-lives. Each half-life halves the remaining number, so the fraction left is (1/2)³ = 1/8. The decay is exponential, N = N₀(1/2)^(t/T).

Question 23Physics (JAM Physics test paper)

A rigid body rotates about a fixed axis with angular velocity ω and angular acceleration α. Which statement about a point at distance r from the axis is correct at that instant?

  1. Its tangential acceleration is ω²r and its centripetal acceleration is αr.
  2. Its tangential acceleration is αr and its centripetal acceleration is ω²r. Correct answer.
  3. Its total acceleration equals αr only.
  4. Both accelerations equal αr.

Explanation

The tangential component of acceleration is d(ωr)/dt = αr, while the centripetal component directed toward the axis is ω²r. The total acceleration is the vector sum of these two perpendicular components.

Question 24Physics (JAM Physics test paper)

A uniform rod of length L and mass M is pivoted at one end and released from rest in a horizontal position. Using energy conservation, what is the angular speed when it becomes vertical? The moment of inertia about the pivot is ML²/3.

  1. √(2g/L)
  2. √(3g/L) Correct answer.
  3. √(g/L)
  4. √(6g/L)

Explanation

The centre of mass falls a distance L/2, so the lost potential energy is MgL/2. Equating this to (1/2)Iω² = (1/2)(ML²/3)ω² gives ω² = 3g/L, hence ω = √(3g/L).

Question 25Physics (JAM Physics test paper)

A particle of mass m moves in a one-dimensional potential V(x) = (1/2)kx². It oscillates with amplitude A. Using the Bohr–Sommerfeld quantization condition ∮p dx = nh, what is the quantized energy spectrum?

  1. E_n = n²ℏω
  2. E_n = (n + 1)ℏω
  3. E_n = (n + 1/2)ℏω Correct answer.
  4. E_n = nℏω

Explanation

For a harmonic oscillator the phase-space area enclosed by an orbit of energy E is 2πE/ω, so the Bohr–Sommerfeld condition gives E_n = (n + 1/2)ℏω with ω = √(k/m). This matches the exact quantum result, including the zero-point energy ℏω/2.

Question 26Physics (JAM Physics test paper)

A particle moves in a plane with position vector r = (3t²) i + (4t) j, where t is in seconds and distances in metres. What is the magnitude of its acceleration at t = 1 s?

  1. 6 m s⁻² Correct answer.
  2. 10 m s⁻²
  3. 4 m s⁻²
  4. 0 m s⁻²

Explanation

Differentiate the position twice: velocity is (6t) i + 4 j and acceleration is a constant (6) i, since the j-component is linear in t. Hence the acceleration magnitude is 6 m s⁻² at every instant, including t = 1 s.

Question 27Physics (JAM Physics test paper)

A block slides down a rough incline of angle 30° with constant velocity. What is the coefficient of kinetic friction between block and incline?

  1. 1/2
  2. √3
  3. zero
  4. 1/√3 Correct answer.

Explanation

Constant velocity means the net force along the incline vanishes, so mg sin30° = μ mg cos30°. Therefore μ = tan30° = 1/√3 ≈ 0.577.

Question 28Physics (JAM Physics test paper)

A particle of mass m moves in a circle of radius R with constant speed v. What is the magnitude of the angular momentum of the particle about the centre of the circle?

  1. m v R²
  2. m v / R
  3. m v R Correct answer.
  4. m v² R

Explanation

For circular motion the velocity is perpendicular to the radius vector, so L = r × p has magnitude R·mv = mvR. The direction is along the axis of the circle, given by the right-hand rule.

Question 29Physics (JAM Physics test paper)

A solid sphere and a hollow spherical shell have the same mass and the same radius. Both roll without slipping down an incline from rest. Which reaches the bottom first?

  1. The hollow shell, because its mass is farther from the axis.
  2. They arrive together because both have the same total energy.
  3. The solid sphere, because it has the smaller moment of inertia. Correct answer.
  4. They arrive together because masses and radii are equal.

Explanation

Rolling time depends on the fraction of energy stored in rotation, set by I/(MR²). The solid sphere has I = (2/5)MR², smaller than (2/3)MR² for the shell, so more energy goes into translation and it accelerates faster.

Question 30Physics (JAM Physics test paper)

A simple pendulum of length L is taken to the Moon, where the acceleration due to gravity is about g/6. By what factor does its period change relative to Earth?

  1. 6
  2. √6 Correct answer.
  3. 36
  4. 1/√6

Explanation

The period of a simple pendulum is T = 2π√(L/g), so T ∝ 1/√g. Replacing g by g/6 multiplies the period by √6 ≈ 2.45.

Question 31Physics (JAM Physics test paper)

A particle executes simple harmonic motion with amplitude A. At what displacement from equilibrium is its kinetic energy equal to its potential energy?

  1. A
  2. A/√2 Correct answer.
  3. A/2
  4. A/4

Explanation

Total energy is (1/2)kA² and potential energy is (1/2)kx². Setting (1/2)kx² = (1/2)(1/2)kA² gives x² = A²/2, so x = A/√2.

Question 32Physics (JAM Physics test paper)

An ideal gas undergoes an adiabatic free expansion into a vacuum inside an insulated container. Which quantity remains unchanged?

  1. Temperature Correct answer.
  2. Pressure
  3. Entropy
  4. Volume

Explanation

For free expansion no work is done (vacuum) and no heat is exchanged (insulated), so internal energy is constant. For an ideal gas internal energy depends only on temperature, so temperature stays the same.

Question 33Physics (JAM Physics test paper)

A wire of resistance R is stretched uniformly so that its length doubles while its volume stays constant. What is its new resistance?

  1. 2R
  2. R/2
  3. R/4
  4. 4R Correct answer.

Explanation

Constant volume with doubled length halves the cross-sectional area. Since R = ρL/A, doubling L and halving A multiplies the resistance by 4.

Question 34Physics (JAM Physics test paper)

A galvanometer of resistance G gives full-scale deflection for current I_g. To convert it into a voltmeter of range V, what series resistance is required?

  1. V/I_g − G Correct answer.
  2. V·I_g − G
  3. V/I_g + G
  4. G − V/I_g

Explanation

The full-scale current I_g must flow when voltage V is applied across the series combination of G and R. Thus V = I_g(G + R), giving R = V/I_g − G.

Question 35Physics (JAM Physics test paper)

A charged particle enters a region of uniform magnetic field with its velocity parallel to the field. What is the path of the particle?

  1. A circle in the plane perpendicular to the field
  2. A parabola bending away from the field
  3. A straight line along the field direction Correct answer.
  4. A helix along the field direction

Explanation

The magnetic force is qv × B, which vanishes when v is parallel to B. With no force (ignoring other fields) the particle continues in a straight line at constant speed.

Question 36Physics (JAM Physics test paper)

A small bar magnet is moved quickly into a coil and then held stationary inside it. Which statement describes the induced emf in the coil?

  1. The emf reverses sign and remains constant after the magnet stops.
  2. An emf appears only while the magnet moves, then falls to zero. Correct answer.
  3. The emf is largest when the magnet is at rest inside the coil.
  4. A constant emf persists while the magnet is stationary inside the coil.

Explanation

Faraday's law states the induced emf equals the negative rate of change of magnetic flux. Flux changes only while the magnet moves, so the emf exists during motion and is zero once the magnet is stationary.

Question 37Physics (JAM Physics test paper)

Unpolarised light of intensity I₀ passes through two ideal polarisers whose transmission axes are at 60° to each other. What is the intensity of the emerging light?

  1. I₀/8 Correct answer.
  2. I₀/2
  3. I₀/4
  4. 3I₀/8

Explanation

The first polariser halves the intensity to I₀/2. Malus's law then gives (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8.

Question 38Physics (JAM Physics test paper)

A convex lens of focal length f forms a real image of an object placed at distance 2f. Where is the image formed and what is its magnification?

  1. At 2f on the other side, magnification +1
  2. At f on the other side, magnification −1/2
  3. At 2f on the other side, magnification −1 Correct answer.
  4. At infinity, magnification undefined

Explanation

Using 1/v − 1/u = 1/f with u = −2f gives v = 2f, so the image is real and at distance 2f on the far side. Magnification m = v/u = −1, so the image is inverted and the same size.

Question 39Physics (JAM Physics test paper)

In a hydrogen atom, an electron makes a transition from n = 3 to n = 2. What is the wavelength of the emitted photon in terms of the Rydberg constant R?

  1. 1/(5R)
  2. 36/(5R) Correct answer.
  3. 5R/36
  4. 5/(36R)

Explanation

The Rydberg formula gives 1/λ = R(1/2² − 1/3²) = R(1/4 − 1/9) = 5R/36. Inverting this yields λ = 36/(5R), the first Balmer line (Hα).

Question 40Physics (JAM Physics test paper)

A nucleus of mass number A has a radius given by R = R₀A^(1/3). If the mass number is increased by a factor of 8, by what factor does the nuclear volume increase?

  1. 64
  2. 2
  3. 4
  4. 8 Correct answer.

Explanation

Volume scales as R³ ∝ (A^(1/3))³ = A. Increasing A by a factor 8 therefore increases the volume by the same factor 8, reflecting roughly constant nuclear density.

Question 41Physics (JAM Physics test paper)

A particle is described by a normalised wavefunction. What is the physical significance of the squared modulus |ψ(x)|²?

  1. The energy density at position x
  2. The phase of the wavefunction at x
  3. The probability density of finding the particle at x Correct answer.
  4. The momentum of the particle at x

Explanation

Born's interpretation states that |ψ(x)|² dx gives the probability of finding the particle in the interval between x and x + dx. This is the probability density, and it must integrate to unity for a normalised state.

Question 42Physics (JAM Physics test paper)

A particle of mass m is confined in a one-dimensional box of width L. What is the expectation value of its position in the ground state?

  1. L/2 Correct answer.
  2. L/4
  3. 2L/3
  4. L

Explanation

The ground-state probability density sin²(πx/L) is symmetric about the centre of the box. Hence the expectation value of x is L/2, the midpoint, even though the particle is never found exactly at the walls.

Question 43Physics (JAM Physics test paper)

Two identical resistors are connected first in series and then in parallel across the same battery. What is the ratio of power dissipated in the series combination to that in the parallel combination?

  1. 1:4 Correct answer.
  2. 2:1
  3. 4:1
  4. 1:2

Explanation

Series resistance is 2R and parallel resistance is R/2, giving a ratio of 4. Since P = V²/R_eq for the same V, the power ratio is the inverse, 1:4.

Question 44Physics (JAM Physics test paper)

A parallel-plate capacitor is charged and then isolated. A dielectric slab of relative permittivity 4 is inserted to fill the gap. What happens to the stored energy?

  1. It increases by a factor of 4
  2. It decreases to one half
  3. It stays the same
  4. It decreases to one quarter Correct answer.

Explanation

With the capacitor isolated, charge Q is constant and energy is U = Q²/(2C). Inserting the dielectric multiplies C by 4, so U falls to one quarter of its initial value.

Question 45Physics (JAM Physics test paper)

A sound source moves toward a stationary observer with speed equal to one tenth of the speed of sound. What is the observed frequency in terms of the source frequency f₀?

  1. (10/9) f₀ Correct answer.
  2. (9/10) f₀
  3. (11/10) f₀
  4. (10/11) f₀

Explanation

For a moving source and stationary observer, f = f₀ v/(v − v_s). With v_s = v/10 this becomes f₀ v/(0.9v) = (10/9)f₀, a higher pitch as the source approaches.

Question 46Physics (JAM Physics test paper)

A particle moves in a central potential V(r) = −k/r with k > 0. Which quantity is conserved because of the spherical symmetry of the potential?

  1. Angular momentum Correct answer.
  2. Linear momentum
  3. Kinetic energy alone
  4. Total mechanical energy only

Explanation

A central force exerts no torque about the origin, so angular momentum L = r × p is conserved. Total energy is also conserved, but angular momentum is the quantity tied specifically to the rotational symmetry.

Question 47Physics (JAM Physics test paper)

An ideal gas of γ = 5/3 undergoes an adiabatic process. If its volume is halved reversibly, by what factor does its pressure change?

  1. 2^(3/5)
  2. 2
  3. 2^(2/3)
  4. 2^(5/3) Correct answer.

Explanation

For a reversible adiabatic process pV^γ is constant, so p ∝ V^(−γ). Halving the volume multiplies the pressure by 2^(5/3) ≈ 3.17.

Question 48Physics (JAM Physics test paper)

A particle of mass m is bound in a finite square well. Compared with the infinite well of the same width, how do the bound-state energies behave?

  1. They are identical for all levels below the well depth.
  2. They are higher because the walls are softer.
  3. They are lower, and only finitely many bound states exist. Correct answer.
  4. They are lower, and infinitely many bound states exist.

Explanation

A finite well allows the wavefunction to penetrate the walls, which lowers the kinetic energy and hence the energy levels compared with the infinite well. Because the well depth is finite, only a finite number of bound states fit inside it.

Question 49Physics (JAM Physics test paper)

A sample contains two radioactive isotopes A and B with half-lives in the ratio 1:2 and equal initial activities. After a time equal to one half-life of A, what is the ratio of the activity of A to that of B?

  1. 1:1
  2. 1:2
  3. 2:1
  4. 1:√2 Correct answer.

Explanation

Activity is A₀(1/2)^(t/T). After t = T_A, isotope A has halved while B has decayed by (1/2)^(1/2) = 1/√2. The ratio of activities is therefore (1/2):(1/√2) = 1:√2.

Question 50Physics (JAM Physics test paper)

A diffraction grating with N slits per unit length is illuminated at normal incidence by light of wavelength λ. What is the condition for the principal maxima, and how does the number of slits affect the sharpness of these maxima?

  1. d sinθ = (m + 1/2)λ, with maxima becoming sharper as N increases
  2. d cosθ = mλ, with maxima becoming sharper as N increases
  3. d sinθ = mλ, with maxima becoming broader as N increases
  4. d sinθ = mλ, with maxima becoming sharper as N increases Correct answer.

Explanation

Constructive interference from a grating occurs when the path difference between adjacent slits, d sinθ, equals an integer number of wavelengths. Increasing the number of illuminated slits narrows each principal maximum, making the lines sharper and better resolved.