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NUST NET Physics MCQs with answers

Practise NUST NET Physics (Engineering) with 151 exam-style MCQs, each with the answer and a short explanation. Every test is marked the moment you finish and shows your score chapter by chapter, so you know what to revise next. It is free and needs no sign-up.

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NUST NET Physics MCQs with answers

Practise NUST NET Physics (Engineering) with 151 exam-style MCQs, each with the answer and a short explanation. Every test is marked the moment you finish and shows your score chapter by chapter, so you know what to revise next. It is free and needs no sign-up.

151 questions · 8 chapters

What each chapter covers 8 chapters

The questions follow the syllabus chapter by chapter. You can test the whole subject or one chapter at a time.

  1. Physics · Measurement, vectors and equilibrium 14 questionsMeasurement, Vectors and equilibrium
  2. Physics · Mechanics 35 questionsMotion and force, Work, energy and power, Circular and rotational motion, Fluid dynamics
  3. Physics · Oscillations, waves and light 23 questionsOscillations, Waves and sound, Physical optics and optical instruments
  4. Physics · Heat and thermodynamics 8 questionsKinetic theory and thermodynamics
  5. Physics · Electricity 18 questionsElectrostatics, Current electricity
  6. Physics · Magnetism and alternating current 20 questionsElectromagnetism, Electromagnetic induction, Alternating current
  7. Physics · Solids and electronics 13 questionsPhysics of solids, Electronics
  8. Physics · Modern and nuclear physics 20 questionsDawn of modern physics, Atomic spectra, Nuclear physics
Sample NUST NET Physics MCQs with answers 12 questions

12 questions from the test, one or two from each chapter. Try each one, then open the answer.

1. The mass and speed of a body are measured with uncertainties of 2% and 3%. The percentage uncertainty in its kinetic energy ½mv2 is

  1. A
    5%
  2. B
    6%
  3. C
    8%
  4. D
    11%
Show answer

Answer: C. Percentage uncertainties add, each multiplied by its power: 2% + 2 × 3% = 8%. The ½ is exact and adds nothing. Forgetting the square on v gives 5%.

2. A 2.0 kg mass is raised through a height of 3.0 m. Its gain in gravitational potential energy is

  1. A
    6.0 J
  2. B
    29.4 J
  3. C
    58.8 J
  4. D
    5.88 J
Show answer

Answer: C. ΔPE = mgh = 2.0 × 9.8 × 3.0 = 58.8 J. 6.0 J leaves out g, and 29.4 J brings in a stray ½.

3. Light of wavelength 500 nm falls normally on a diffraction grating. For a first-order maximum to be formed at all, the number of lines per centimetre must be less than

  1. A
    2.0 × 106
  2. B
    2.0 × 104
  3. C
    1.0 × 104
  4. D
    5.0 × 10−5
Show answer

Answer: B. d sin θ = λ needs sin θ = λ/d to be less than 1, so d must exceed λ = 5.0 × 10−5 cm. Lines per cm = 1/d, which must be less than 1/(5.0 × 10−5) = 2.0 × 104. 2.0 × 106 is the limit per metre, and 1.0 × 104 is the limit for a second-order maximum.

4. For a monatomic ideal gas Cv = 3R/2. The ratio γ = Cp/Cv is

  1. A
    3/2
  2. B
    7/5
  3. C
    4/3
  4. D
    5/3
Show answer

Answer: D. Cp = Cv + R = 5R/2, so γ = (5R/2) ÷ (3R/2) = 5/3. 7/5 is the value for a diatomic gas.

5. Three identical cells, each of EMF 2.0 V and internal resistance 0.50 Ω, are connected in series to a 4.5 Ω resistor. The current in the resistor is

  1. A
    1.3 A
  2. B
    1.0 A
  3. C
    0.33 A
  4. D
    1.2 A
Show answer

Answer: B. In series the EMFs and the internal resistances both add: E = 6.0 V and r = 1.5 Ω. I = E/(R + r) = 6.0/6.0 = 1.0 A. Ignoring the internal resistances gives 1.3 A, and counting only one of them gives 1.2 A.

6. The coil of an AC generator has 100 turns of area 0.010 m2 and rotates at 50 revolutions per second in a uniform field of 0.20 T. The peak EMF is

  1. A
    20π V
  2. B
    10 V
  3. C
    10π V
  4. D
    0.2π V
Show answer

Answer: A. ω = 2πf = 100π rad s−1, so ε0 = NABω = 100 × 0.010 × 0.20 × 100π = 20π V (about 63 V). Using f in place of ω gives 10 V, and leaving out N gives 0.2π V.

7. A wire of Young's modulus 1.0 × 1011 Pa is under a tensile stress of 5.0 × 107 Pa. Its tensile strain is

  1. A
    2.0 × 103
  2. B
    5.0 × 10−4
  3. C
    5.0 × 1018
  4. D
    5.0 × 10−2
Show answer

Answer: B. Strain = stress/Y = 5.0 × 107 ÷ 1.0 × 1011 = 5.0 × 10−4 (0.05%); it has no unit. 2.0 × 103 divides the wrong way round, and 5.0 × 10−2 writes 0.05% as if it were 0.05.

8. Energy is released both when a heavy nucleus splits (fission) and when light nuclei join (fusion), because in each case

  1. A
    the total number of nucleons decreases
  2. B
    the products have a greater binding energy per nucleon
  3. C
    the products have a smaller binding energy per nucleon
  4. D
    neutrons are turned completely into energy
Show answer

Answer: B. Binding energy per nucleon is greatest near iron-56. Splitting heavy nuclei or fusing light ones moves towards this peak, so the products are more tightly bound; the mass lost is released as energy. The number of nucleons does not change.

9. A force F = (i + 3j) N acts at the point r = (2i + j) m. The torque τ = r × F about the origin is

  1. A
    5k N m
  2. B
    7k N m
  3. C
    −5k N m
  4. D
    6k N m
Show answer

Answer: A. τ = (2i + j) × (i + 3j) = 6(i × j) + 1(j × i) = 6k − k = 5k N m. Adding the two products gives 7k; working out F × r instead of r × F gives −5k.

10. A 0.20 kg ball is dropped from a height of 5.0 m and rebounds to 3.2 m. Taking g = 10 m s−2, the energy lost in the bounce is

  1. A
    6.4 J
  2. B
    3.6 J
  3. C
    1.8 J
  4. D
    10 J
Show answer

Answer: B. Loss = mg(h1 − h2) = 0.20 × 10 × 1.8 = 3.6 J. 10 J is the whole starting energy and 6.4 J is the energy kept after the bounce.

11. An astronomical telescope in normal adjustment is 44 cm long and has a magnifying power of 10. The focal lengths of its objective and eyepiece are

  1. A
    4 cm and 40 cm
  2. B
    40 cm and 4 cm
  3. C
    22 cm and 22 cm
  4. D
    44 cm and 4.4 cm
Show answer

Answer: B. In normal adjustment the length is fo + fe = 44 cm and M = fo/fe = 10. So fe = 4 cm and fo = 40 cm: the objective has the longer focal length.

12. When a gas is heated at constant volume,

  1. A
    all the heat supplied increases its internal energy
  2. B
    all the heat supplied is used to do external work
  3. C
    its internal energy stays constant
  4. D
    its temperature stays constant
Show answer

Answer: A. With no change in volume the gas does no work (W = PΔV = 0), so the first law gives Q = ΔU: all the heat raises the internal energy and hence the temperature.

Try 139 more questions

Ways to practise 5 ways

When you get a question wrong, a short note called "The idea behind this" explains the topic it belongs to, with the rules to remember and a worked example.

Common questions

Is this NUST NET Physics test free?

Yes. Every test and every explanation is free, and you do not need an account. If you sign in, your results are kept across your devices.

Are these past paper questions?

The 151 MCQs are our own, written to the syllabus in the style of the exam.

How is the test marked?

Instantly. You get your score, the right answer and an explanation for every question, and a list of the topics to work on.

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