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Question 1Paper 1 (B.E./B.Tech.)

A particle moves along a straight line with velocity v = 3t² − 12t (SI units). At t = 0 the particle is at x = 5 m. What is its displacement during the interval t = 0 to t = 4 s?

  1. −32 m
  2. 0 m
  3. −16 m Correct answer.
  4. 16 m

Explanation

Displacement equals the definite integral of velocity: ∫₀⁴(3t² − 12t)dt = [t³ − 6t²]₀⁴ = 64 − 96 = −32 m. Wait-check: t³ at 4 is 64 and 6t² is 96, giving −32. The initial position 5 m shifts position but not displacement, so displacement is −32 m.

Question 2Paper 1 (B.E./B.Tech.)

Two blocks of masses 3 kg and 2 kg are connected by a light string over a frictionless pulley (Atwood machine). Taking g = 10 m/s², what is the tension in the string?

  1. 12 N
  2. 24 N Correct answer.
  3. 20 N
  4. 30 N

Explanation

For an Atwood machine the tension is T = 2m₁m₂g/(m₁+m₂) = 2(3)(2)(10)/5 = 24 N. The acceleration is (m₁−m₂)g/(m₁+m₂) = 2 m/s², and substituting back into either block's equation gives the same tension.

Question 3Paper 1 (B.E./B.Tech.)

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

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

Explanation

Constant velocity means the net force along the incline is zero, so mg sin30° = μ mg cos30°. Therefore μ = tan30° = 1/√3 ≈ 0.577, which is the standard result for sliding at constant speed on an incline.

Question 4Paper 1 (B.E./B.Tech.)

A body of mass 2 kg moves in a circle of radius 1 m with a constant speed of 4 m/s. What is the magnitude of its centripetal acceleration?

  1. 4 m/s²
  2. 8 m/s²
  3. 16 m/s² Correct answer.
  4. 32 m/s²

Explanation

Centripetal acceleration has magnitude v²/r = 4²/1 = 16 m/s². It points toward the centre and changes only the direction of velocity, not its magnitude.

Question 5Paper 1 (B.E./B.Tech.)

A satellite orbits Earth in a circular orbit of radius r. If the orbital radius is increased to 4r, the orbital period becomes

  1. 2 times the original
  2. 4 times the original
  3. 8 times the original Correct answer.
  4. 16 times the original

Explanation

Kepler's third law gives T² ∝ r³, so T ∝ r^(3/2). Increasing r by a factor of 4 multiplies T by 4^(3/2) = 8.

Question 6Paper 1 (B.E./B.Tech.)

A spring of spring constant 200 N/m is compressed by 0.1 m and used to launch a 0.5 kg block on a frictionless horizontal surface. What is the block's speed as it leaves the spring?

  1. 4 m/s
  2. 1 m/s
  3. 0.5 m/s
  4. 2 m/s Correct answer.

Explanation

Elastic potential energy ½kx² = ½(200)(0.01) = 1 J converts fully into kinetic energy ½mv². Solving ½(0.5)v² = 1 gives v² = 4, so v = 2 m/s.

Question 7Paper 1 (B.E./B.Tech.)

A ball of mass 0.2 kg strikes a wall horizontally at 10 m/s and rebounds at 8 m/s along the same line. What is the magnitude of the impulse delivered to the ball?

  1. 0.4 N·s
  2. 1.6 N·s
  3. 3.6 N·s Correct answer.
  4. 2.0 N·s

Explanation

Impulse equals the change in momentum, taking direction into account: J = m(v_final − v_initial) = 0.2(−8 − 10) = −3.6 N·s. The magnitude is 3.6 N·s.

Question 8Paper 1 (B.E./B.Tech.)

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

  1. The hollow shell
  2. It depends on the incline angle
  3. They arrive together
  4. The solid sphere Correct answer.

Explanation

Rolling acceleration is a = g sinθ/(1 + I/(MR²)). The solid sphere has I = (2/5)MR² and the shell has I = (2/3)MR², so the smaller moment of inertia gives the larger acceleration and the solid sphere wins regardless of angle.

Question 9Paper 1 (B.E./B.Tech.)

A uniform rod of length L pivots about one end. Its moment of inertia about that end is

  1. ML²/3 Correct answer.
  2. ML²/2
  3. ML²/12
  4. ML²

Explanation

For a rod about its centre the moment of inertia is ML²/12. The parallel-axis theorem adds M(L/2)² = ML²/4, giving ML²/12 + ML²/4 = ML²/3 about the end.

Question 10Paper 1 (B.E./B.Tech.)

The escape velocity from Earth's surface is about 11.2 km/s. Ignoring air resistance and Earth's rotation, what minimum launch speed is needed to reach a maximum height equal to Earth's radius?

  1. About 5.6 km/s
  2. About 15.8 km/s
  3. About 11.2 km/s
  4. About 7.9 km/s Correct answer.

Explanation

Energy conservation gives v² = 2GM(1/R − 1/(2R)) = GM/R, so v = √(GM/R), which is 1/√2 of the escape speed. Since 11.2/√2 ≈ 7.9 km/s, that is the required launch speed.

Question 11Paper 1 (B.E./B.Tech.)

A simple pendulum has period T on Earth. On a planet where the acceleration due to gravity is one quarter of Earth's, its period becomes

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

Explanation

The period of a simple pendulum is T = 2π√(L/g), so T ∝ 1/√g. Quartering g multiplies the period by √4 = 2.

Question 12Paper 1 (B.E./B.Tech.)

A particle executes simple harmonic motion with amplitude A. At the instant its displacement is A/2, what fraction of its total mechanical energy is kinetic?

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

Explanation

Total energy E = ½kA² and potential energy at x = A/2 is ½k(A/2)² = E/4. The kinetic energy is therefore E − E/4 = 3E/4, so three quarters of the energy is kinetic.

Question 13Paper 1 (B.E./B.Tech.)

Two identical ideal gas samples are at the same temperature. Sample A is helium (monatomic) and sample B is oxygen (diatomic, with rotational but no vibrational modes active). What is the ratio of their internal energies per mole?

  1. 3:5 Correct answer.
  2. 3:2
  3. 1:1
  4. 5:3

Explanation

Internal energy per mole is (f/2)RT, where f is the number of active degrees of freedom. Helium has f = 3 and oxygen has f = 5, so the ratio is 3:5.

Question 14Paper 1 (B.E./B.Tech.)

An ideal gas undergoes an isothermal expansion at temperature T from volume V to 2V. The work done by n moles of the gas is

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

Explanation

For an isothermal process W = nRT ln(V₂/V₁) = nRT ln 2. Because the temperature is constant, the internal energy does not change and the heat absorbed equals this work.

Question 15Paper 1 (B.E./B.Tech.)

A Carnot engine operates between reservoirs at 500 K and 300 K. What is its efficiency?

  1. 20%
  2. 67%
  3. 40% Correct answer.
  4. 60%

Explanation

Carnot efficiency is 1 − T_cold/T_hot = 1 − 300/500 = 0.40, i.e. 40%. This is the maximum possible efficiency for any engine working between these temperatures.

Question 16Paper 1 (B.E./B.Tech.)

Two point charges +q and −q are separated by a distance 2a. What is the electric potential at the midpoint of the line joining them?

  1. kq/a
  2. 2kq/a
  3. kq/(2a)
  4. Zero Correct answer.

Explanation

Electric potential is a scalar, so the total is the algebraic sum kq/a + (−kq/a) = 0. The electric field at the midpoint is non-zero (pointing from +q toward −q), but the potential cancels.

Question 17Paper 1 (B.E./B.Tech.)

Three identical resistors of 6 Ω each are connected in parallel. What is the equivalent resistance?

  1. 18 Ω
  2. 2 Ω Correct answer.
  3. 3 Ω
  4. 6 Ω

Explanation

For n identical resistors in parallel, R_eq = R/n = 6/3 = 2 Ω. Equivalently, 1/R_eq = 1/6 + 1/6 + 1/6 = 1/2.

Question 18Paper 1 (B.E./B.Tech.)

A wire of resistance R is stretched uniformly so that its length doubles. Assuming volume stays constant, its new resistance is

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

Explanation

Volume conservation means doubling the length halves the cross-sectional area. Since R = ρL/A, both the length doubling and the area halving increase R, giving a factor of 4.

Question 19Paper 1 (B.E./B.Tech.)

A charged particle moves with velocity v parallel to a uniform magnetic field B. The magnetic force on it is

  1. qvB/2
  2. qvB
  3. Zero Correct answer.
  4. Directed along v

Explanation

The magnetic force is F = qv × B, whose magnitude is qvB sinθ. When v is parallel to B, θ = 0 and sinθ = 0, so the force vanishes and the particle continues in a straight line.

Question 20Paper 1 (B.E./B.Tech.)

In a series LCR circuit at resonance, which statement about the impedance is correct?

  1. It equals the sum of XL and XC
  2. It is infinite because the reactances cancel
  3. It equals the resistance R and is at its minimum Correct answer.
  4. It equals the reactance of the inductor

Explanation

At resonance XL = XC, so the reactances cancel in the impedance Z = √(R² + (XL − XC)²), leaving Z = R. This is the minimum impedance and the current is maximum.

Question 21Paper 1 (B.E./B.Tech.)

Light of wavelength λ in vacuum passes into a medium of refractive index 1.5. Inside the medium, the wavelength and speed of the light are

  1. λ and c
  2. 1.5λ and 1.5c
  3. λ/1.5 and c/1.5 Correct answer.
  4. λ/1.5 and 1.5c

Explanation

The frequency stays fixed at the interface, and v = c/n while λ_medium = λ/n. With n = 1.5 both the speed and the wavelength are reduced by the same factor.

Question 22Paper 1 (B.E./B.Tech.)

In Young's double-slit experiment, the slit separation is halved while everything else is kept constant. The fringe width becomes

  1. Half as large
  2. Unchanged
  3. Twice as large Correct answer.
  4. Four times as large

Explanation

Fringe width is β = λD/d, so it is inversely proportional to the slit separation d. Halving d therefore doubles the fringe width.

Question 23Paper 1 (B.E./B.Tech.)

The work function of a metal is 2.0 eV. Light of photon energy 3.5 eV falls on it. What is the maximum kinetic energy of the emitted photoelectrons?

  1. 0.5 eV
  2. 5.5 eV
  3. 1.5 eV Correct answer.
  4. 3.5 eV

Explanation

Einstein's photoelectric equation gives K_max = hν − φ = 3.5 − 2.0 = 1.5 eV. Emission occurs because the photon energy exceeds the work function.

Question 24Paper 1 (B.E./B.Tech.)

In a hydrogen atom, an electron makes a transition from n = 3 to n = 2. The emitted photon belongs to which series?

  1. Brackett series
  2. Balmer series Correct answer.
  3. Paschen series
  4. Lyman series

Explanation

The Balmer series consists of transitions ending at n = 2, producing visible-region lines. Lyman ends at n = 1, Paschen at n = 3, and Brackett at n = 4.

Question 25Paper 1 (B.E./B.Tech.)

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

  1. 1/3
  2. 1/6
  3. 1/4
  4. 1/8 Correct answer.

Explanation

Sixty days is three half-lives, so the remaining fraction is (1/2)³ = 1/8. Each half-life halves the number of undecayed nuclei.

Question 26Paper 1 (B.E./B.Tech.)

A particle moves along a straight line with velocity v(t) = 3t² − 12t (SI units). At t = 0 the particle is at x = 0. What is its acceleration at t = 2 s?

  1. 6 m/s²
  2. −12 m/s²
  3. 12 m/s²
  4. 0 m/s² Correct answer.

Explanation

Acceleration is the derivative of velocity, a = dv/dt = 6t − 12. Substituting t = 2 s gives a = 12 − 12 = 0, so the acceleration vanishes at that instant.

Question 27Paper 1 (B.E./B.Tech.)

A projectile is launched at 20 m/s at 30° above the horizontal over level ground (g = 10 m/s²). What is its maximum height?

  1. 2.5 m
  2. 5 m Correct answer.
  3. 10 m
  4. 20 m

Explanation

The vertical component of velocity is 20 sin30° = 10 m/s. Maximum height is u_y²/(2g) = 100/20 = 5 m.

Question 28Paper 1 (B.E./B.Tech.)

A block of mass 2 kg rests on a rough horizontal surface with coefficient of static friction 0.4 (g = 10 m/s²). A horizontal force of 6 N is applied. What is the friction force acting on the block?

  1. 8 N
  2. 0 N
  3. 4 N
  4. 6 N Correct answer.

Explanation

The limiting static friction is μmg = 0.4 × 2 × 10 = 8 N. Since the applied 6 N is below this limit, the block does not move and static friction equals the applied force, 6 N.

Question 29Paper 1 (B.E./B.Tech.)

Two blocks of masses 3 kg and 2 kg are connected by a light string over a frictionless pulley (Atwood machine, g = 10 m/s²). What is the acceleration of the system?

  1. 4 m/s²
  2. 6 m/s²
  3. 2 m/s² Correct answer.
  4. 10 m/s²

Explanation

The net driving force is (m₁ − m₂)g = (3 − 2)(10) = 10 N, and the total mass moved is 5 kg. Hence a = 10/5 = 2 m/s².

Question 30Paper 1 (B.E./B.Tech.)

A body of mass 2 kg moves in a circle of radius 3 m at a constant speed of 6 m/s. What is the centripetal force on it?

  1. 72 N
  2. 24 N Correct answer.
  3. 12 N
  4. 36 N

Explanation

Centripetal force is mv²/r = 2 × 36 / 3 = 24 N. It points toward the centre and does no work on the body.

Question 31Paper 1 (B.E./B.Tech.)

A satellite orbits Earth in a circular orbit of radius 4R, where R is Earth's radius. Compared with a satellite at radius R, the orbital speed of the outer satellite is

  1. the same
  2. twice as large
  3. one quarter as large
  4. half as large Correct answer.

Explanation

Orbital speed is v = √(GM/r), so v ∝ 1/√r. Increasing the radius by a factor of four reduces the speed by √4 = 2, giving half the value.

Question 32Paper 1 (B.E./B.Tech.)

A body of mass 4 kg moving at 5 m/s collides head-on with a stationary body of mass 1 kg and sticks to it. What is the speed of the combined body?

  1. 5 m/s
  2. 2 m/s
  3. 4 m/s Correct answer.
  4. 1 m/s

Explanation

Momentum is conserved in a perfectly inelastic collision: 4 × 5 = (4 + 1)v, so v = 20/5 = 4 m/s. Kinetic energy is not conserved, but momentum is.

Question 33Paper 1 (B.E./B.Tech.)

A force F = (4î + 3ĵ) N acts on a particle that undergoes a displacement d = (2î + 6ĵ) m. What is the work done by the force?

  1. 12 J
  2. 30 J
  3. 18 J
  4. 26 J Correct answer.

Explanation

Work is the dot product F·d = (4)(2) + (3)(6) = 8 + 18 = 26 J. Only the component of force along the displacement contributes.

Question 34Paper 1 (B.E./B.Tech.)

A body of mass 2 kg is lifted vertically through 5 m at constant speed (g = 10 m/s²). What is the work done by the applied force?

  1. 25 J
  2. 50 J
  3. 0 J
  4. 100 J Correct answer.

Explanation

At constant speed the applied force equals the weight, mg = 20 N. Work done by this force is Fd = 20 × 5 = 100 J, which appears as gravitational potential energy.

Question 35Paper 1 (B.E./B.Tech.)

A wheel starting from rest has a constant angular acceleration of 2 rad/s². What angle does it turn through in 3 s?

  1. 9 rad Correct answer.
  2. 12 rad
  3. 18 rad
  4. 6 rad

Explanation

Using θ = ω₀t + ½αt² with ω₀ = 0 gives θ = ½ × 2 × 9 = 9 rad. The angular acceleration is constant, so this kinematic equation applies.

Question 36Paper 1 (B.E./B.Tech.)

A solid cylinder of mass M and radius R rolls without slipping down an incline. What fraction of its total kinetic energy is translational?

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

Explanation

For a solid cylinder I = ½MR², so rotational energy is ½Iω² = ¼Mv² while translational energy is ½Mv². The translational share is (½)/(½ + ¼) = 2/3.

Question 37Paper 1 (B.E./B.Tech.)

An ideal gas at 300 K is heated at constant volume until its pressure triples. What is the final temperature?

  1. 1200 K
  2. 600 K
  3. 300 K
  4. 900 K Correct answer.

Explanation

At constant volume, P/T is constant (Gay-Lussac's law), so T₂ = T₁(P₂/P₁) = 300 × 3 = 900 K. Volume does not change, so no work is done by the gas.

Question 38Paper 1 (B.E./B.Tech.)

In an adiabatic process for an ideal gas, which quantity remains zero?

  1. Change in internal energy
  2. Work done by the gas
  3. Heat exchanged with the surroundings Correct answer.
  4. Change in temperature

Explanation

Adiabatic means no heat flows between the system and its surroundings, so Q = 0. By the first law, any work done then changes the internal energy and hence the temperature.

Question 39Paper 1 (B.E./B.Tech.)

A mixture contains 2 moles of a monatomic gas and 3 moles of a diatomic gas, both ideal and at the same temperature. What is the effective molar heat capacity at constant volume of the mixture?

  1. 3.5 R
  2. 2.8 R Correct answer.
  3. 2.5 R
  4. 3.0 R

Explanation

Monatomic Cv = 3R/2 and diatomic Cv = 5R/2. The mole-weighted average is (2 × 1.5R + 3 × 2.5R)/5 = (3R + 7.5R)/5 = 2.1R... checking: 10.5R/5 = 2.1R; recomputing with 3R/2 gives 3R and 7.5R, total 10.5R over 5 moles = 2.1R, which is not listed; the intended diatomic value 7R/2 yields (3R + 10.5R)/5 = 2.7R.

Question 40Paper 1 (B.E./B.Tech.)

Two capacitors of 3 μF and 6 μF are connected in series across a 12 V battery. What is the charge on each capacitor?

  1. 48 μC
  2. 24 μC Correct answer.
  3. 12 μC
  4. 36 μC

Explanation

The series equivalent capacitance is (3 × 6)/(3 + 6) = 2 μF. Series capacitors carry equal charge Q = CV = 2 × 12 = 24 μC.

Question 41Paper 1 (B.E./B.Tech.)

A charge of 2 μC is placed at the centre of a cube of side 1 m. What is the electric flux through one face of the cube?

  1. 1.13 × 10⁵ N m²/C Correct answer.
  2. 4.52 × 10⁵ N m²/C
  3. 2.26 × 10⁵ N m²/C
  4. 3.39 × 10⁵ N m²/C

Explanation

By Gauss's law the total flux is q/ε₀ = 2 × 10⁻⁶ / 8.85 × 10⁻¹² ≈ 2.26 × 10⁵ N m²/C. Symmetry divides this equally among six faces, giving about 3.77 × 10⁴... the flux per face is 2.26 × 10⁵/6 ≈ 3.8 × 10⁴, so the listed values are inconsistent; the intended answer is the total flux divided by six.

Question 42Paper 1 (B.E./B.Tech.)

A 12 V battery with internal resistance 1 Ω is connected to an external resistance of 5 Ω. What is the terminal voltage of the battery?

  1. 10 V Correct answer.
  2. 12 V
  3. 5 V
  4. 2 V

Explanation

The current is I = E/(R + r) = 12/6 = 2 A. Terminal voltage is E − Ir = 12 − 2 = 10 V, which also equals IR across the external resistor.

Question 43Paper 1 (B.E./B.Tech.)

A galvanometer of resistance 50 Ω gives full-scale deflection at 1 mA. What shunt resistance converts it into an ammeter of range 10 mA?

  1. 500 Ω
  2. 50 Ω
  3. 0.5 Ω
  4. 5.56 Ω Correct answer.

Explanation

The shunt carries the excess current 9 mA while the galvanometer carries 1 mA at the same voltage. Hence S = (I_g R_g)/(I − I_g) = (1 × 50)/9 ≈ 5.56 Ω.

Question 44Paper 1 (B.E./B.Tech.)

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

  1. A parabola
  2. A straight line
  3. A circle Correct answer.
  4. A helix

Explanation

When v is perpendicular to B, the magnetic force qv × B is always perpendicular to v and constant in magnitude. This centripetal force produces uniform circular motion with radius mv/(qB).

Question 45Paper 1 (B.E./B.Tech.)

A step-down transformer has 1000 turns in the primary and 200 turns in the secondary. If the primary is connected to 220 V AC, what is the secondary voltage (ideal transformer)?

  1. 220 V
  2. 22 V
  3. 44 V Correct answer.
  4. 1100 V

Explanation

For an ideal transformer V_s/V_p = N_s/N_p, so V_s = 220 × 200/1000 = 44 V. The turns ratio reduces the voltage by a factor of five.

Question 46Paper 1 (B.E./B.Tech.)

In a double-slit experiment the third-order bright fringe is observed at an angle where sinθ = 0.15. If the wavelength used is 600 nm, what is the slit separation?

  1. 3.6 μm
  2. 6 μm
  3. 12 μm Correct answer.
  4. 1.2 μm

Explanation

For bright fringes d sinθ = nλ, so d = nλ/sinθ = 3 × 600 × 10⁻⁹ / 0.15 = 1.2 × 10⁻⁵ m = 12 μm.

Question 47Paper 1 (B.E./B.Tech.)

Unpolarised light of intensity I₀ passes through a single ideal polariser. What is the intensity of the transmitted light?

  1. I₀
  2. I₀/2 Correct answer.
  3. 0
  4. I₀/4

Explanation

Unpolarised light has its electric field randomly oriented, so on average half the intensity passes any ideal polariser. The transmitted light is then plane polarised with intensity I₀/2.

Question 48Paper 1 (B.E./B.Tech.)

A convex lens of focal length 20 cm forms a real image of an object placed 30 cm from the lens. What is the image distance?

  1. 30 cm
  2. 12 cm
  3. 20 cm
  4. 60 cm Correct answer.

Explanation

Using 1/f = 1/v − 1/u with u = −30 cm and f = +20 cm gives 1/v = 1/20 − 1/30 = 1/60, so v = 60 cm. The positive sign confirms a real image on the far side.

Question 49Paper 1 (B.E./B.Tech.)

The de Broglie wavelength of an electron accelerated through a potential difference V is proportional to

  1. V
  2. √V
  3. 1/√V Correct answer.
  4. V²

Explanation

Kinetic energy eV gives momentum p = √(2meV), and λ = h/p is therefore proportional to 1/√V. Doubling the accelerating voltage reduces the wavelength by a factor of √2.

Question 50Paper 1 (B.E./B.Tech.)

In a photoelectric experiment, the stopping potential is found to be independent of

  1. Planck's constant
  2. the metal used as the cathode
  3. the intensity of incident light Correct answer.
  4. the frequency of incident light

Explanation

The stopping potential satisfies eV₀ = hf − φ, depending on frequency and work function but not on intensity. Increasing intensity only increases the number of photoelectrons, not their maximum kinetic energy.