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MHT-CET practice questions

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Question 1Physics

A body starts from rest and moves along a straight line with uniform acceleration of 4 m/s². The distance covered by the body during the third second of its motion is

  1. 10 m Correct answer.
  2. 20 m
  3. 18 m
  4. 8 m

Explanation

For uniformly accelerated motion starting from rest, the distance covered in the nth second is (a/2)(2n − 1). Substituting a = 4 m/s² and n = 3 gives 2 × 5 = 10 m, which is smaller than the 18 m covered in the first three seconds combined.

Question 2Physics

The dimensional formula of Planck's constant is

  1. [ML²T⁻³]
  2. [ML²T⁻²]
  3. [MLT⁻¹]
  4. [ML²T⁻¹] Correct answer.

Explanation

From E = hν, Planck's constant equals energy divided by frequency, so its dimensions are ML²T⁻² divided by T⁻¹, which gives ML²T⁻¹. The same dimensions are shared by angular momentum, which is why h is also quoted in joule-seconds.

Question 3Physics

A metal surface has a work function of 2.0 eV. Monochromatic light of photon energy 3.5 eV is incident on it. The maximum kinetic energy of the emitted photoelectrons is

  1. 1.5 eV Correct answer.
  2. 5.5 eV
  3. 3.5 eV
  4. 7.0 eV

Explanation

Einstein's photoelectric equation states that KEmax = hν − φ, where φ is the work function. Subtracting the 2.0 eV needed to escape the metal from the 3.5 eV photon energy leaves 1.5 eV of kinetic energy for the most energetic photoelectrons.

Question 4Physics

A solid cylinder rolls without slipping along a horizontal surface. The fraction of its total kinetic energy that is rotational is

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

Explanation

For a solid cylinder I = ½MR², so the rotational energy is ¼Mv² while the translational energy is ½Mv². The rotational share is therefore (¼)/(¾) = 1/3, and the remaining 2/3 is translational.

Question 5Physics

Two capacitors of 2 μF and 3 μF are joined in series and the combination is connected across a 10 V battery. The charge on each capacitor is

  1. 20 μC
  2. 6 μC
  3. 30 μC
  4. 12 μC Correct answer.

Explanation

Capacitors in series store the same charge, and the equivalent capacitance is C₁C₂/(C₁ + C₂) = 6/5 = 1.2 μF. Using Q = CV with V = 10 V gives 12 μC on each capacitor, so the p.d. across them is 6 V and 4 V respectively.

Question 6Physics

A block of mass 0.2 kg rests on a frictionless horizontal surface and is attached to a spring of force constant 80 N/m. When displaced and released, the block oscillates with a period of

  1. 0.4π s
  2. 0.1π s Correct answer.
  3. 0.05π s
  4. 0.2π s

Explanation

A mass-spring oscillator has period T = 2π√(m/k), independent of amplitude. Here m/k = 0.2/80 = 2.5 × 10⁻³ s², whose square root is 0.05 s, so T = 2π × 0.05 = 0.1π s, about 0.31 s.

Question 7Physics

A uniform wire of resistance R is stretched uniformly until its length becomes double the original length. The resistance of the stretched wire is

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

Explanation

Stretching keeps the volume of the wire constant, so doubling the length halves its cross-sectional area. Since R = ρl/A, the length doubles while the area halves, making the new resistance 2 × 2 = 4 times the original value.

Question 8Physics

In Young's double slit experiment, the entire apparatus is immersed in a transparent liquid of refractive index 4/3. The fringe width becomes

  1. 3β/4 Correct answer.
  2. β
  3. 16β/9
  4. 4β/3

Explanation

Fringe width is β = λD/d, so it is directly proportional to the wavelength in the medium between the slits and the screen. In a liquid of refractive index n the wavelength falls to λ/n, so the fringe width shrinks by the same factor and becomes (3/4)β.

Question 9Physics

A body weighs 100 N on the surface of the Earth. At a height above the surface equal to half the Earth's radius, its weight is nearly

  1. 25.0 N
  2. 44.4 N Correct answer.
  3. 66.7 N
  4. 50.0 N

Explanation

Above the surface, acceleration due to gravity varies as g' = g/(1 + h/R)². With h = R/2 the denominator factor is (1.5)² = 2.25, so the weight becomes 100/2.25 ≈ 44.4 N, roughly four-ninths of the surface value.

Question 10Physics

A thin equiconvex lens made of glass of refractive index 1.5 has focal length f in air. When the lens is completely immersed in water of refractive index 4/3, its focal length becomes

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

Explanation

The lens maker's formula gives 1/f = (n_glass/n_medium − 1)(1/R₁ − 1/R₂). In air the factor is 0.5, while in water it is 1.5 ÷ (4/3) − 1 = 1/8, which is one quarter as large; a quarter of the power means four times the focal length, so f' = 4f.

Question 11Physics

The dimensional formula of the universal gravitational constant G is

  1. [M⁻¹ L³ T⁻²] Correct answer.
  2. [M L³ T⁻²]
  3. [M⁻¹ L² T⁻²]
  4. [M⁻¹ L³ T⁻¹]

Explanation

Using Newton's law of gravitation, F = Gm₁m₂/r², so G = Fr²/(m₁m₂). Substituting dimensions gives [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²].

Question 12Physics

An ideal gas absorbs 300 J of heat and does 120 J of work on its surroundings. The change in its internal energy is

  1. 420 J
  2. 180 J Correct answer.
  3. −180 J
  4. 120 J

Explanation

From the first law of thermodynamics, ΔU = Q − W. Here Q = +300 J because heat is absorbed and W = +120 J because work is done by the gas, so ΔU = 300 − 120 = 180 J.

Question 13Physics

A steady current of 2.0 A flows through a metallic conductor. The number of conduction electrons crossing any cross-section of the conductor per second is (e = 1.6 × 10⁻¹⁹ C)

  1. 1.6 × 10¹⁹
  2. 6.25 × 10¹⁸
  3. 1.25 × 10¹⁹ Correct answer.
  4. 3.2 × 10¹⁹

Explanation

Current is charge per unit time, so I = ne and n = I/e. Thus n = 2.0/(1.6 × 10⁻¹⁹) = 1.25 × 10¹⁹ electrons per second.

Question 14Physics

The moment of inertia of a thin circular ring of mass M and radius R about a tangent lying in its plane is

  1. ½MR²
  2. 3/2 MR² Correct answer.
  3. MR²
  4. 2MR²

Explanation

For a ring, the moment of inertia about a diameter is MR²/2. By the parallel-axis theorem, about a tangent in its plane, I = MR²/2 + MR² = 3MR²/2.

Question 15Physics

Two point charges +2 μC and −2 μC are placed 0.10 m apart in air. The magnitude of the electric field at the midpoint of the line joining them is (k = 9 × 10⁹ N m² C⁻²)

  1. 7.2 × 10⁶ N/C
  2. zero
  3. 1.44 × 10⁸ N/C
  4. 1.44 × 10⁷ N/C Correct answer.

Explanation

At the midpoint each charge is 0.05 m away, and each field has magnitude kq/r² = 9 × 10⁹ × 2 × 10⁻⁶/(0.05)² = 7.2 × 10⁶ N/C. The two fields point in the same direction, from the positive charge toward the negative charge, so the resultant is 1.44 × 10⁷ N/C.

Question 16Physics

Two long straight parallel wires separated by 0.20 m carry currents of 5.0 A and 10 A in the same direction. The force per unit length on the wires is

  1. 2.5 × 10⁻⁵ N m⁻¹, repulsive
  2. 5.0 × 10⁻⁵ N m⁻¹, attractive Correct answer.
  3. 1.0 × 10⁻⁴ N m⁻¹, repulsive
  4. 5.0 × 10⁻⁴ N m⁻¹, attractive

Explanation

The force per unit length between long parallel currents is μ₀I₁I₂/(2πd). Substitution gives (2 × 10⁻⁷)(5.0)(10)/0.20 = 5.0 × 10⁻⁵ N/m, and currents in the same direction attract.

Question 17Physics

A series LCR circuit has L = 1.0 H and C = 1.0 μF. Its resonant angular frequency is

  1. 1.0 × 10² rad/s
  2. 1.0 × 10³ rad/s Correct answer.
  3. 1.0 × 10⁴ rad/s
  4. 1.0 × 10⁶ rad/s

Explanation

At resonance the reactances cancel, and the natural angular frequency is ω₀ = 1/√(LC). With L = 1.0 H and C = 1.0 × 10⁻⁶ F, ω₀ = 1/√(1.0 × 10⁻⁶) = 1.0 × 10³ rad/s.

Question 18Physics

A transparent medium has refractive index 1.5. The polarising angle for light reflected from its surface is

  1. 33.7°
  2. 48.2°
  3. 41.8°
  4. 56.3° Correct answer.

Explanation

Brewster's law gives tan θp = n. For n = 1.5, θp = tan⁻¹(1.5) ≈ 56.3°, the angle at which the reflected light is completely plane-polarised.

Question 19Physics

In a hydrogen atom, an electron makes a transition from n = 4 to n = 2. The wavelength of the emitted photon is closest to

  1. 103 nm
  2. 122 nm
  3. 656 nm
  4. 486 nm Correct answer.

Explanation

The photon energy is ΔE = 13.6(1/2² − 1/4²) eV = 13.6(1/4 − 1/16) = 2.55 eV. Using λ = 1240 eV·nm/ΔE gives λ ≈ 1240/2.55 ≈ 486 nm.

Question 20Physics

A rectangular coil of 100 turns and area 0.020 m² rotates at 50 revolutions per second in a uniform magnetic field of 0.10 T. The peak emf induced in the coil is

  1. 6.3 V
  2. 31 V
  3. 63 V Correct answer.
  4. 126 V

Explanation

The angular frequency is ω = 2πf = 2π(50) ≈ 314 rad/s. The peak emf is e₀ = NABω = 100 × 0.020 × 0.10 × 314 ≈ 63 V.

Question 21Physics

A train moving at 20 m/s slows down uniformly to 10 m/s while covering a distance of 75 m. What is the magnitude of its retardation?

  1. 0.5 m/s²
  2. 4.0 m/s²
  3. 2.0 m/s² Correct answer.
  4. 1.0 m/s²

Explanation

Use v² = u² − 2as with u = 20 m/s, v = 10 m/s and s = 75 m. This gives 100 = 400 − 150a, so a = 2.0 m/s². The retardation magnitude is therefore 2.0 m/s².

Question 22Physics

A constant force F = (3i + 4j) N acts on a particle that undergoes a displacement from (0, 0) to (2, 3) m. The work done by the force is

  1. 12 J
  2. 21 J
  3. 24 J
  4. 18 J Correct answer.

Explanation

Work done by a constant force is W = F · s. Here F = 3i + 4j N and s = 2i + 3j m, so W = (3)(2) + (4)(3) = 18 J. The perpendicular components do not contribute because the dot product uses only matching components.

Question 23Physics

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

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

Explanation

For a simple pendulum, T = 2π√(L/g), so T is inversely proportional to √g. If g becomes g/6, the new period is √6 times the original period. The length of the pendulum is unchanged.

Question 24Physics

A thin uniform disc of mass M and radius R has moment of inertia MR²/4 about any diameter. Its moment of inertia about an axis tangent to its rim and perpendicular to its plane is

  1. 3/2 MR² Correct answer.
  2. 5/4 MR²
  3. MR²
  4. 3/4 MR²

Explanation

For a disc, moment of inertia about a diameter is MR²/4, so about the central perpendicular axis it is MR²/2 by the perpendicular-axis theorem. The parallel-axis theorem then adds MR² for a tangent axis perpendicular to the plane. Hence I = MR²/2 + MR² = 3MR²/2.

Question 25Physics

Four identical point charges +q are placed at the corners of a square of side a. The electric potential at the centre of the square is

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

Explanation

Each corner charge is at distance a/√2 from the square's centre. The potential from one charge is kq/(a/√2) = √2 kq/a. Four identical charges add algebraically, giving 4√2 kq/a.

Question 26Physics

A galvanometer of resistance 50 Ω shows full-scale deflection for a current of 1.0 mA. To convert it into a voltmeter of range 0–10 V, the resistance to be connected in series is

  1. 10050 Ω
  2. 10000 Ω
  3. 50 Ω
  4. 9950 Ω Correct answer.

Explanation

For a voltmeter, a high resistance is connected in series with the galvanometer. The required series resistance is R = V/Ig − G = 10/(1.0×10⁻³) − 50 = 9950 Ω. This limits the full-scale current to 1.0 mA at 10 V.

Question 27Physics

An object is placed 15 cm in front of a concave mirror of focal length 10 cm. The magnification produced by the mirror is

  1. −2 Correct answer.
  2. +0.5
  3. −0.5
  4. +2

Explanation

Using the mirror formula with f = −10 cm and u = −15 cm gives 1/v = 1/f − 1/u = −1/30, so v = −30 cm. The magnification is m = −v/u = −(−30)/(−15) = −2. The image is real, inverted and twice the object size.

Question 28Physics

An electron is accelerated from rest through a potential difference of 100 V. Its de Broglie wavelength is about

  1. 12.3 nm
  2. 0.0123 nm
  3. 1.23 nm
  4. 0.123 nm Correct answer.

Explanation

The de Broglie wavelength of an electron accelerated through V volts is λ ≈ 1.227/√V nm. For V = 100 V, λ ≈ 1.227/10 = 0.123 nm. This is a standard non-relativistic estimate valid at 100 V.

Question 29Physics

In a series LCR circuit, R = 30 Ω, X_L = 40 Ω and X_C = 80 Ω. The power factor of the circuit is

  1. 0.6 leading Correct answer.
  2. 0.8 leading
  3. 0.8 lagging
  4. 0.6 lagging

Explanation

The impedance is Z = √(R² + (X_L − X_C)²) = √(30² + (−40)²) = 50 Ω. The power factor is cos φ = R/Z = 30/50 = 0.6. Because X_C > X_L, the circuit is capacitive and the current leads the voltage.

Question 30Physics

A satellite of mass m is in a circular orbit of radius r around the Earth of mass M. The minimum additional energy required to transfer it to a circular orbit of radius 2r is

  1. GMm/r
  2. GMm/(4r) Correct answer.
  3. 3GMm/(4r)
  4. GMm/(2r)

Explanation

The total energy in a circular orbit of radius r is −GMm/(2r). At radius 2r it is −GMm/(4r). The additional energy required is the difference, ΔE = −GMm/(4r) − (−GMm/(2r)) = GMm/(4r).

Question 31Physics

A vernier caliper has a main scale division of 0.5 mm. Ten vernier scale divisions coincide with nine main scale divisions. Its least count is

  1. 5.00 mm
  2. 0.50 mm
  3. 0.10 mm
  4. 0.05 mm Correct answer.

Explanation

The least count equals one main-scale division minus one vernier-scale division. Here one vernier division is (9/10)(0.5 mm) = 0.45 mm, so LC = 0.50 - 0.45 = 0.05 mm.

Question 32Physics

A metallic wire of length L and cross-sectional area A extends by e under a load F. Another wire of the same material has length 2L and area 2A. Under the same load, its extension will be

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

Explanation

For a wire under tension, extension is e = FL/(AY). Replacing L by 2L and A by 2A multiplies the numerator and denominator by 2, so the extension remains e.

Question 33Physics

A Carnot engine works between a source at 500 K and a sink at 300 K. Its efficiency is

  1. 20%
  2. 30%
  3. 60%
  4. 40% Correct answer.

Explanation

For a Carnot engine, efficiency is η = 1 - T_c/T_h. Thus η = 1 - 300/500 = 0.40, or 40%.

Question 34Physics

A projectile is launched with speed 20 m/s at 30° above the horizontal on level ground. Taking g = 10 m/s², its horizontal range is about

  1. 20.0 m
  2. 40.0 m
  3. 34.6 m Correct answer.
  4. 69.3 m

Explanation

The range on level ground is R = u² sin 2θ / g. With u = 20 m/s, θ = 30°, and g = 10 m/s², R = (400 × sin 60°)/10 ≈ 34.6 m.

Question 35Physics

A circular road curve of radius 100 m is banked for a design speed with no friction. If the banking angle is 45°, the design speed is (g = 10 m/s²)

  1. 20.0 m/s
  2. 31.6 m/s Correct answer.
  3. 100 m/s
  4. 10.0 m/s

Explanation

For frictionless banking, tan θ = v²/(rg). Therefore v = √(rg tan θ) = √(100 × 10 × 1) ≈ 31.6 m/s.

Question 36Physics

A particle executes simple harmonic motion with amplitude A. At which displacement from the mean position is its kinetic energy equal to its potential energy?

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

Explanation

In SHM, total energy E = ½kA² and potential energy U = ½kx². KE equals PE when U = E/2, giving x² = A²/2, so x = A/√2.

Question 37Physics

A 2 μF capacitor charged to 100 V is disconnected from the battery and then connected in parallel across an uncharged 2 μF capacitor. The common potential difference is

  1. 200 V
  2. 25 V
  3. 50 V Correct answer.
  4. 100 V

Explanation

The initial charge is Q = CV = 2 μF × 100 V = 200 μC. When shared between two 2 μF capacitors, total capacitance is 4 μF, so V = Q/C = 200 μC / 4 μF = 50 V.

Question 38Physics

A potentiometer wire is 10 m long and has resistance 20 Ω. It is connected in series with a 1.5 V cell and an 80 Ω resistor. The potential gradient along the wire is

  1. 0.015 V/m
  2. 0.15 V/m
  3. 0.30 V/m
  4. 0.030 V/m Correct answer.

Explanation

The current is I = 1.5 V/(20 Ω + 80 Ω) = 0.015 A. The potential drop across the 10 m wire is IR = 0.015 × 20 = 0.30 V, so the gradient is 0.30 V/10 m = 0.030 V/m.

Question 39Physics

A proton and an alpha particle, both accelerated from rest through the same potential difference, enter a uniform magnetic field perpendicular to their velocities. The ratio of the radius of the alpha particle's path to that of the proton is

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

Explanation

The radius is r = mv/(qB), and after acceleration through potential V, r = √(2mV)/(qB). Thus r ∝ √(m/q); for the alpha particle m = 4m_p and q = 2q_p, giving r_α/r_p = √(4/2) = √2.

Question 40Physics

A nucleus of mass number 240 and binding energy per nucleon 7.6 MeV undergoes fission into two fragments of mass number 120 each, with binding energy per nucleon 8.4 MeV. The energy released is

  1. 96 MeV
  2. 0.8 MeV
  3. 2016 MeV
  4. 192 MeV Correct answer.

Explanation

Energy released equals final total binding energy minus initial total binding energy. Initial BE = 240 × 7.6 = 1824 MeV; final BE = 2 × 120 × 8.4 = 2016 MeV; released energy = 192 MeV.

Question 41Physics

The dimensional formula of Planck’s constant is identical to that of which physical quantity?

  1. Power
  2. Angular momentum Correct answer.
  3. Energy
  4. Linear momentum

Explanation

Planck’s constant has dimensions of energy divided by frequency, which gives ML²T⁻¹. Angular momentum also has dimensions ML²T⁻¹. Energy, power, and linear momentum have different time or length powers.

Question 42Physics

In a perfectly inelastic collision between two isolated bodies, which quantity remains conserved during the collision?

  1. Neither kinetic energy nor linear momentum
  2. Both kinetic energy and linear momentum
  3. Kinetic energy only
  4. Linear momentum only Correct answer.

Explanation

For an isolated system, no external force acts, so total linear momentum is conserved. In a perfectly inelastic collision the bodies stick together and some kinetic energy is converted into deformation, heat, or sound.

Question 43Physics

The escape speed from the Earth’s surface is 11.2 km/s. For a planet whose mass and radius are both twice those of the Earth, the escape speed at its surface is

  1. 15.8 km/s
  2. 11.2 km/s Correct answer.
  3. 5.6 km/s
  4. 22.4 km/s

Explanation

Escape speed is v = √(2GM/R). If both M and R are doubled, the ratio inside the square root becomes (2M)/(2R) = M/R, so the escape speed is unchanged at 11.2 km/s.

Question 44Physics

A solid cylinder rolls without slipping down an inclined plane. What fraction of its total kinetic energy is rotational kinetic energy?

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

Explanation

For a solid cylinder, rotational kinetic energy is ½Iω² = ¼Mv², while translational kinetic energy is ½Mv². The total is ¾Mv², so the rotational fraction is (¼)/(¾) = 1/3.

Question 45Physics

A bar magnet is moved with its north pole towards a closed circular coil. The face of the coil nearer the magnet acquires

  1. North polarity Correct answer.
  2. No magnetic polarity
  3. Alternating polarity
  4. South polarity

Explanation

By Lenz’s law, the induced current opposes the increase in magnetic flux caused by the approaching north pole. The nearer face therefore behaves like a north pole and repels the magnet.

Question 46Physics

In Young’s double-slit experiment, the separation between the slits is doubled while the screen distance and wavelength remain unchanged. The fringe width becomes

  1. Double its original value
  2. Half its original value Correct answer.
  3. Unchanged
  4. Four times its original value

Explanation

Fringe width is given by β = λD/d. Doubling the slit separation d while keeping λ and D constant makes the fringe width half of its original value.

Question 47Physics

A metallic wire of resistance R is stretched uniformly until its length becomes twice the original length. Its new resistance is

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

Explanation

Stretching keeps the volume constant, so doubling the length halves the cross-sectional area. Since R = ρL/A, the resistance becomes 2L divided by A/2, which gives 4R.

Question 48Physics

An ideal gas expands isothermally at temperature T from volume V to volume 2V. The work done by n moles of the gas is

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

Explanation

For a reversible isothermal expansion of an ideal gas, W = nRT ln(V_f/V_i). With V_f/V_i = 2, the work done by the gas is nRT ln 2.

Question 49Physics

An air-filled parallel-plate capacitor has capacitance C. A dielectric slab of dielectric constant K and thickness equal to half the plate separation is inserted between the plates. The new capacitance is

  1. (K + 1)C/(2K)
  2. C
  3. 2K C/(K + 1) Correct answer.
  4. K C

Explanation

The air gap and the dielectric slab act as two capacitors in series. Their capacitances are 2C and 2KC, so the equivalent capacitance is (2C)(2KC)/(2C + 2KC), which simplifies to 2KC/(K + 1).

Question 50Physics

In a photoelectric experiment, light of frequency ν gives a stopping potential V₀. If the frequency is increased to 2ν while the same cathode is used, the new stopping potential is

  1. V₀
  2. 2V₀
  3. V₀ + hν/e Correct answer.
  4. V₀ − hν/e

Explanation

Einstein’s photoelectric equation gives eV₀ = hν − φ, where φ is the work function. For frequency 2ν, eV′ = 2hν − φ = hν + (hν − φ) = hν + eV₀, so V′ = V₀ + hν/e.