Units and Dimensions: 17 MCQs with answers

D · 9.8 m/s²

A physical quantity has both a numerical value and a unit. 9.8 m/s² has both; the others are pure numbers.

A · Seven

The SI system has exactly seven base units: metre, kilogram, second, ampere, kelvin, mole, and candela.

A · [M L T⁻²]

Force = mass × acceleration, so [M] × [L T⁻²] = [M L T⁻²].

B · newton

Newton is formed from base units: kg·m/s². The others are base units.

D · A unit

A physical quantity needs a unit. '60' alone does not specify km/h, m/s, or mph.

A · True

The SI system has exactly seven base units, and all other units are derived from them.

D · Work and torque

Work and torque both have dimensions [M L² T⁻²], though they are different physical quantities.

B · The dimensions of all terms on both sides must be equal.

Homogeneity means dimensional consistency: every term in a valid physical equation must have the same dimensions.

D · The value of a dimensionless constant like ½

Dimensional analysis cannot find dimensionless numerical constants; it only checks dimensions.

D · 10⁵ dyne

1 N = 1 kg·m/s² = 1000 g × 100 cm/s² = 10⁵ g·cm/s² = 10⁵ dyne.

A · [M L⁻¹ T⁻²]

Pressure = force/area = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²].

B · Strain

Strain is a ratio of lengths, so it has no dimensions. Angle, refractive index, and relative density are also dimensionless.

A · All three terms

By homogeneity, each term must have dimension [L]: $s$ is length, $ut$ is velocity×time, and $at^2$ is acceleration×time².

C · Power

Power = work/time = [M L² T⁻²] / [T] = [M L² T⁻³].

A · The dimension of force is the same in SI and CGS.

Dimensions are independent of the unit system. Force has dimension [M L T⁻²] in both SI and CGS, though units differ (newton vs dyne).

B · The student is wrong; same dimensions do not imply same physical quantity.

Dimensional analysis cannot distinguish between quantities with the same dimensions. Work and torque are different concepts despite sharing [M L² T⁻²].

C · [M L^2 T^{-2}]

The dimensions of the rate of change of momentum with respect to time can be found by dividing the dimensions of momentum \( [M L T^{-1}] \) by the dimensions of time \( [T] \). Thus, \( [F] = [M L T^{-1}] / [T] = [M L T^{-2}] \), which corresponds to the dimensions of force.