IMO Practice Test: Units and Measurements | Class 11 Physics

Question 1 of 13

If force F, velocity v and time T are chosen as fundamental quantities, the dimensions of mass are:

[F v^-1 T]

[F v T^-1]

[F v T]

[F v^-1 T^-1]

Explanation: Since F = m a = m v / T, mass m = F T / v = [F v^-1 T^1].

Question 2 of 13

The quantity that has the same dimensions as Planck's constant is:

energy

power

angular momentum

force

Explanation: Planck's constant [M L^2 T^-1] matches angular momentum (L = m v r).

Question 3 of 13

If the percentage errors in measuring mass and speed are 2% and 3%, the maximum percentage error in kinetic energy (1/2 m v^2) is:

5%

7%

8%

11%

Explanation: ΔE/E = Δm/m + 2 Δv/v = 2% + 2(3%) = 8%.

Question 4 of 13

The dimensions of the coefficient of viscosity are:

[M L^-1 T^-1]

[M L T^-1]

[M L^-1 T^-2]

[M L^-2 T^-1]

Explanation: From F = η A (dv/dx), η = F /(A · velocity gradient) = [M L^-1 T^-1].

Question 5 of 13

Which expression is dimensionally NOT possible as a single physical equation?

v = u + at

s = ut + (1/2) a t^2

v = u + (1/2) a t^2

v^2 = u^2 + 2as

Explanation: In v = u + (1/2)at^2, the term (1/2)at^2 has dimensions [L], which cannot add to a velocity [L T^-1].

Question 6 of 13

The number 0.0006070 has how many significant figures?

3

4

5

7

Explanation: Leading zeros don't count; 6, 0, 7 and the trailing 0 do — four significant figures.

Question 7 of 13

The Reynolds number (ρ v D / η) is:

dimensionless

[L T^-1]

[M L^-1 T^-1]

[M L T^-2]

Explanation: All dimensions cancel, so the Reynolds number is a pure (dimensionless) number.

Question 8 of 13

If energy E, velocity v and time T are fundamental, the dimensions of surface tension are:

[E v^-2 T^-2]

[E v^-2 T^-1]

[E v^-1 T^-2]

[E v^2 T^-2]

Explanation: Surface tension [M T^-2]; with M = E v^-2 (from E = m v^2), it becomes [E v^-2 T^-2].

Question 9 of 13

A calorie is 4.2 J (energy has dimensions [M L^2 T^-2]). In a new system the unit of mass is 100 g, of length 10 cm and of time 1 minute. The magnitude of 1 calorie in this system is:

3.6 × 10^3

1.5 × 10^4

1.5 × 10^7

4.2 × 10^4

Explanation: n2 = n1 (M1/M2)(L1/L2)^2 (T1/T2)^-2 = 4.2 × (1000/100) × (100/10)^2 × (1/60)^-2 = 4.2 × 10 × 100 × 3600 = 1.512 × 10^7.

Question 10 of 13

Two resistances are R1 = (100 ± 3) Ω and R2 = (200 ± 4) Ω in series. The equivalent resistance is:

(300 ± 1) Ω

(300 ± 5) Ω

(300 ± 7) Ω

(300 ± 12) Ω

Explanation: In series R = R1 + R2 = 300 Ω; absolute errors add: 3 + 4 = 7 Ω.

Question 11 of 13

The dimensional formula [M L^2 T^-2 K^-1] corresponds to:

heat capacity

specific heat

thermal conductivity

latent heat

Explanation: Heat capacity = energy per kelvin = [M L^2 T^-2 K^-1]; specific heat carries an extra M^-1.

Question 12 of 13

The radius of a wire is measured as 0.26 cm with least count 0.01 cm. The percentage error in the cross-sectional area (π r^2) is about:

3.8%

5.0%

7.7%

8.0%

Explanation: ΔA/A = 2 Δr/r = 2 × (0.01/0.26) ≈ 2 × 3.85% ≈ 7.7%.

Question 13 of 13

Which set of quantities can be used as fundamental to express all mechanical quantities?

force, length, time

velocity, acceleration, force

energy, velocity, force

momentum, energy, power

Explanation: Force, length and time are dimensionally independent and span [M], [L], [T], so any mechanical quantity can be built from them.

IMO Practice Test — Units and Measurements