Units and Measurements – Physical Quantities and Their Units: NEET MCQs with Answers

Explanation: Since 1 eV = 1.6 × 10⁻¹⁹ J, the energy in eV is 10⁻²⁰/(1.6 × 10⁻¹⁹) = 0.0625 ≈ 0.06 eV. The electron volt is the convenient unit for atomic and molecular energies.

Explanation: By the principle of homogeneity of dimensions, every term on the right-hand side must have the same unit as x, i.e. kilometre. Hence unit of (b t²) = km ⇒ unit of b = km/(unit of t)² = km/s². (For completeness, the same reasoning gives the unit of a as km/s.)

Explanation: Force has the dimensional formula [F] = [MLT⁻²], so 1 unit of force = (unit of mass)(unit of length)(unit of time)⁻². Converting each quantity to SI: 10 g = 10⁻² kg, 10 cm = 10⁻¹ m, 0.1 s = 10⁻¹ s. 1 unit of force = (10⁻² kg)(10⁻¹ m)(10⁻¹ s)⁻² = 10⁻² × 10⁻¹ × 10² N = 10⁻¹ N = 0.1 N.

Explanation: By Stefan's law, the energy radiated per unit area per unit time by a perfect black body at absolute temperature T is E/(A t) = σ T⁴. Hence σ = (E/A t)/T⁴ = (power/area)/T⁴. Unit of σ = (watt/metre²)/kelvin⁴ = W m⁻² K⁻⁴. Its SI value is σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

Explanation: From E = hν, h = E/ν has the unit joule-second (J s) and dimensions [ML²T⁻¹]. Its accepted SI value is h = 6.63 × 10⁻³⁴ J s.

Explanation: From Coulomb's law, F = (1/4πε₀) q₁q₂/r², so ε₀ = q₁q₂/(4πFr²). The unit of ε₀ is therefore coulomb²/(newton × metre²), i.e. C² N⁻¹ m⁻²; only the metre is squared in the denominator, not the newton.

Explanation: The magnitude of a physical quantity is invariant, so n₁u₁ = n₂u₂. In the new system, the unit of density is u₂ = (unit of mass)/(unit of length)³ = 100 g/(10 cm)³ = 100 g/1000 cm³ = 0.1 g cm⁻³. Hence n₂ = n₁u₁/u₂ = (4 g cm⁻³)/(0.1 g cm⁻³) = 40. The density therefore reads 40 units in the new system. (The smaller the unit chosen, the larger the numerical value — here the density unit shrank, so the number grew.)

Explanation: Since the damping force is proportional to velocity, F ∝ v, i.e. F = k v, where k is the damping constant. Therefore k = F/v. Unit of k = newton/(metre per second) = (kg m s⁻²)/(m s⁻¹) = kg s⁻¹. Dimensionally, [k] = [MLT⁻²]/[LT⁻¹] = [MT⁻¹], which is consistent with the unit kg s⁻¹.

Explanation: For steady conduction through a slab of thickness L and cross-sectional area A, dQ/dt = K A (ΔT/L) Rearranging, K = (dQ/dt) × L /(A ΔT). Unit of K = (joule per second × metre)/(metre² × kelvin) = (watt × metre)/(metre² × kelvin) = W m⁻¹ K⁻¹. Since 1 W = 1 J s⁻¹, the unit may also be written as J s⁻¹ m⁻¹ K⁻¹, but never as J m⁻¹ K⁻¹ (the time factor cannot be dropped).

Explanation: One minute of arc is 1′ = (1/60)°, and 1° = π/180 rad. So 1′ = (1/60) × (π/180) rad = π/10800 rad = 2.91 × 10⁻⁴ rad.

Explanation: Plane angle is measured in radian (rad) and solid angle in steradian (sr); both are recognised units in the SI system. However, plane angle = arc/radius = [L]/[L] and solid angle = area/radius² = [L²]/[L²], so both reduce to [M⁰L⁰T⁰]. Thus these two quantities possess units but are dimensionless. In the NCERT scheme they were earlier called supplementary quantities and are now treated as dimensionless derived quantities.

Explanation: A solid angle is defined as (area intercepted)/(radius)², so its dimensional formula is [M⁰L⁰T⁰] — it is dimensionless. Strain = (change in dimension)/(original dimension), which is a ratio of two lengths and is therefore dimensionless. Plane angle = (arc)/(radius), again a ratio of two lengths and dimensionless. Among the remaining choices, stress = force/area = [ML⁻¹T⁻²], length of arc = [L], and angular speed = [T⁻¹] — none of these is dimensionless. Hence strain and plane angle form the required pair.

Explanation: bt² must have the unit of distance (km), so b = km/s².

Explanation: New unit of mass = 100 g, new unit of length = 10 cm ⇒ new unit of volume = 1000 cm³. Density = 4 g cm⁻³ = 4 × (1000 cm³ / 100 g) new units = 4 × 10 = 40 in the new system.

Explanation: From Coulomb's law F = q₁q₂/(4πε₀r²), ε₀ = q₁q₂/(4πF r²). Unit = C²/(N·m²) = coulomb²/newton-metre².