Dimensions of Physics quantities, Dimensional Analysis and Its Applications: NEET MCQs with Answers

Explanation: Stress is the restoring force per unit area, [MLT⁻²]/[L²] = [ML⁻¹T⁻²]. It therefore shares its dimensional formula with pressure and with energy density, while strain is dimensionless.

Explanation: The combination e²/(4πε₀) has the dimensions of energy × length, [ML³T⁻²]. Multiplying by G = [M⁻¹L³T⁻²] gives [L⁶T⁻⁴]; taking the square root gives [L³T⁻²], and dividing by c² = [L²T⁻²] leaves [L]. Hence the correct combination is (1/c²)[Ge²/4πε₀]¹ᐟ².

Explanation: Take L = h^a c^b G^d. Using h = [ML²T⁻¹], c = [LT⁻¹], G = [M⁻¹L³T⁻²] and comparing powers gives a = 1/2, d = 1/2 and b = −3/2. Hence L = √(hG)/c³ᐟ², the Planck length (≈ 1.6 × 10⁻³⁵ m).

Explanation: Substituting η = [ML⁻¹T⁻¹], ρ = [ML⁻³] and r = [L] into [LT⁻¹]: from T, −x = −1 → x = 1; from M, x + y = 0 → y = −1; from L, −x − 3y + z = 1 → −1 + 3 + z = 1 → z = −1. Thus v𝒸 = kη/(ρr), the familiar Reynolds-number result.

Explanation: Surface tension is force per unit length, [MT⁻²]. Writing S = E^a V^b T^c with E = [ML²T⁻²], V = [LT⁻¹]: M gives a = 1; L gives 2a + b = 0 → b = −2; T gives −2a − b + c = −2 → −2 + 2 + c = −2 → c = −2. Hence [EV⁻²T⁻²].

Explanation: Mass = force/acceleration, and acceleration = velocity/time, so m = F/(V/T) = FV⁻¹T. Checking with SI: N/(m s⁻¹) × s = kg, as required.

Explanation: Density, ρ = m/V, so the maximum fractional error is Δρ/ρ = Δm/m + ΔV/V = 0.01/22.42 + 0.1/4.7 = 0.000446 + 0.02128 = 0.02172 Percentage error = 0.02172 × 100 ≈ 2.17% ≈ 2%. Notice that almost the whole of the error comes from the volume measurement, because 4.7 cc is quoted to far fewer significant figures than 22.42 g.

Explanation: Kinetic energy, K = ½mv². Hence (ΔK/K)×100 = (Δm/m)×100 + 2(Δv/v)×100 = 2% + (2 × 3%) = 2% + 6% = 8% The constant factor ½ contributes nothing to the error, since it is exactly known. The speed contributes twice its own error because it appears squared.

Explanation: For a cube of side l, ρ = m/V = m/l³. Therefore (Δρ/ρ)×100 = (Δm/m)×100 + 3(Δl/l)×100 = 4% + (3 × 3%) = 4% + 9% = 13% The length contributes three times its own percentage error because volume varies as the cube of the side.

Explanation: The volume of a sphere is V = (4/3)πr³, so V ∝ r³. Hence (ΔV/V)×100 = 3 × (Δr/r)×100 = 3 × 2% = 6%. (For comparison, the surface area varies as r², so its error would have been 2 × 2% = 4%.)

Explanation: For free fall from rest, h = ½gt², hence g = 2h/t². Taking logarithms: ln g = ln 2 + ln h − 2 ln t. Differentiating and taking the magnitudes of all terms for the maximum possible error: (Δg/g)×100 = (Δh/h)×100 + 2(Δt/t)×100 = e₁ + 2e₂ The time contributes twice because it enters the formula as t². Always take the absolute value of each term when computing the maximum error, even where the exponent is negative.

Explanation: The maximum percentage error in P is obtained by adding the percentage errors of all factors, each multiplied by the magnitude of its exponent: (ΔP/P)×100 = 3(Δa/a)×100 + 2(Δb/b)×100 + (Δc/c)×100 + (Δd/d)×100 = (3 × 1) + (2 × 2) + (1 × 3) + (1 × 4) = 3 + 4 + 3 + 4 = 14% Quantities appearing in the denominator contribute exactly as much as those in the numerator, because errors are always combined additively for the worst case.

Explanation: The maximum percentage error is the sum of each percentage error multiplied by the magnitude of its power: 2(1%) + ½(2%) + ⅓(3%) + 3(4%) = 2 + 1 + 1 + 12 = 16%. Errors always add in the worst case, so a maximum error cannot be negative.

Explanation: Arithmetically, 9.99 − 0.0099 = 9.9801 m. In addition and subtraction the result must be rounded off so as to retain as many decimal places as are present in the quantity with the fewest decimal places. Here 9.99 m has two decimal places and 0.0099 m has four, so the answer must be reported to two decimal places.

Explanation: Mean value: T̄ = (1.25 + 1.24 + 1.27 + 1.21 + 1.28)/5 = 6.25/5 = 1.25 s Absolute deviations from the mean: |1.25 − 1.25| = 0.00; |1.25 − 1.24| = 0.01; |1.25 − 1.27| = 0.02; |1.25 − 1.21| = 0.04; |1.25 − 1.28| = 0.03 Mean absolute error: ΔT̄ = (0.00 + 0.01 + 0.02 + 0.04 + 0.03)/5 = 0.10/5 = 0.02 s Percentage relative error: = (ΔT̄/T̄) × 100 = (0.02/1.25) × 100 = 1.6%