Applying dimensional method, dimension of critical velocity, V₀ = [LT⁻¹], co-efficient of viscosity, η = F/6πrV, so dimension of η = [ML⁻²]/[L][LT⁻¹] = [ML⁻¹T⁻¹]. Dimension of density, ρ = [M]/[L³] = [ML⁻³]. Dimension of radius, r = [L]. Put these values in equation (i), [M⁰L⁻¹T⁻¹] = [ML⁻³

Question

Applying dimensional method, dimension of critical velocity, V₀ = [LT⁻¹], co-efficient of viscosity, η = F/6πrV, so dimension of η = [ML⁻²]/[L][LT⁻¹] = [ML⁻¹T⁻¹]. Dimension of density, ρ = [M]/[L³] = [ML⁻³]. Dimension of radius, r = [L]. Put these values in equation (i), [M⁰L⁻¹T⁻¹] = [ML⁻³T⁰ρⁿ][M⁰L⁰T⁰]ᵉ. Equating powers both sides, x + y = 0; x = -1; x = 1, -x + 3y + z = 1, -1 - 3(-1) + z = 1, -1 + 3 + z = 1, z = -1.

Accepted Answer

Torque = Force × force arm = mass × acceleration × length = [M] × [LT⁻²] × [L] = [ML²T⁻²]. The dimension of torque is correctly derived as [ML²T⁻²], which matches the given explanation in option D. The other options either have incorrect dimensional analysis or are irrelevant to the question.

Solution

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