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Class XI · Physics

Physical World and Measurement: NEET MCQ practice

231 practice questions on Physical World and Measurement, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

231

Practice questions

175

Easy

55

Medium

1

Hard

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What Physical World and Measurement covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Units of measurement, System of Units, SI units, fundamental and derived units
  • Dimensions of physical quantities, dimensional analysis and its applications
  • Errors in measurement, significant figures

Sample questions on Physical World and Measurement

Try each one, then open the answer and explanation.

Question 1 · Practice question

Checking the correctness of equations using the method of dimensions is based on

  1. A.the type of system
  2. B.equality of inertial frames of references
  3. C.principle of homogeneity of dimensions
  4. D.none of these
Show answer

Answer: C. principle of homogeneity of dimensions

Dimensional analysis checks whether an equation is dimensionally consistent by ensuring all terms have the same dimensions — this is the principle of homogeneity of dimensions. The other options are unrelated: the 'type of system' doesn’t affect dimensional checks, and 'equality of inertial frames' is from relativity/mechanics, not dimensional analysis.

Question 2 · Practice question

Which of the following relations is dimensionally incorrect?

  1. A.1 u = 931.5 MeV
  2. B.1 u = 931.5 MeV/c 2
  3. C.1 u = 1.67 x 10 -27 kg
  4. D.None of these
Show answer

Answer: A. 1 u = 931.5 MeV

Atomic mass unit (u) has dimension of mass, while MeV is energy. Dimensional analysis requires both sides to have the same dimensions, so 1 u = 931.5 MeV is dimensionally incorrect. Option (b) is correct because dividing MeV by c² gives it mass dimension, and (c) is straightforward mass equivalence, so only (a) is wrong.

Question 3 · Practice question

A new system of units is proposed in which unit of mass is α kg, unit of length is β m and unit of time is γ s. What will be value of 5 J in this new system?

  1. A.5αβ 2 γ −2
  2. B.5α −1 β -2 γ 2
  3. C.5α −2 β −1 γ −2
  4. D.5α −1 β 2 γ 2
Show answer

Answer: B. 5α −1 β -2 γ 2

Energy has dimensions [M¹L²T⁻²]. When units are scaled (mass by factor α, length by β, time by γ), the numerical value scales inversely with each dimension's power: multiply by α⁻¹ (for mass), β⁻² (for length squared), and γ² (for time⁻²). So 5 J becomes 5 α⁻¹ β⁻² γ² in the new system. Option (a) wrongly puts α and β² in the numerator, which would imply a larger unit giving a smaller number—incorrect because larger mass/length units should decrease the numerical value of energy.

Question 4 · Practice question

If the energy, E = G p h q c r where G is the universal gravitational constant, h is the Planck's constant and c is the velocity of light, then the values o f p, q and r are, respectively

  1. A.- 1/2, 1/2 and 5/2
  2. B.1/2, - 1/2 and - 5/2
  3. C.- 1/2, 1/2 and 3/2
  4. D.1/2, - 1/2 and - 3
Show answer

Answer: A. - 1/2, 1/2 and 5/2

Using dimensional analysis, we equate dimensions of E ([M L² T⁻²]) to those of G^(p) h^(q) c^(r), solve three simultaneous equations, and get p = -1/2, q = 1/2, r = 5/2. Option (a) matches exactly. The common trap is misremembering the dimension of G, which could lead to picking (c) or (d), but careful solving gives (a).

Question 5 · Practice question

The velocity of a particle (v) at an instant t is given by v = at + bt 2. The dimension of b is the

  1. A.[L]
  2. B.[LT -1 ]
  3. C.[LT -2 ]
  4. D.[LT -3 ]
Show answer

Answer: D. [LT -3 ]

The dimensions of each term in an equation must match. Here, the term b t² must have the same dimensions as velocity [LT⁻¹]. Dividing [LT⁻¹] by [T²] gives [LT⁻³]. Distractor [LT⁻²] might tempt students if they mistakenly think b is an acceleration, but t² is already present.

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