If speed (V), acceleration (A), and force (F) are taken as fundamental quantities, what is the dimensional formula for Young's modulus in terms of these quantities?

Question

Accepted Answer

V^(-4) A² F. Young's modulus Y has dimensions [M L⁻¹ T⁻²] (same as pressure). We need to express this using V=[LT⁻¹], A=[LT⁻²], F=[MLT⁻²]. From A/V = T⁻¹, so T = V/A (dimension-wise: T ~ V A⁻¹). L = V*T = V² A⁻¹. M = F/(A*L) = F/(A * V² A⁻¹) = F A⁰ V⁻²... let me redo: M = F T² L⁻¹ = F (V/A)² (V²/A)⁻¹ = F V² A⁻² * A V⁻² = F A⁻¹. Now [Y] = M L⁻¹ T⁻² = (F A⁻¹)(V² A⁻¹)⁻¹ (V A⁻¹)⁻² = F A⁻¹ * A V⁻² * A² V⁻² = F A² V⁻⁴.

If speed (V), acceleration (A), and force (F) are taken as fundamental quantities, what is the dimensional formula for Young's modulus in terms of these quantities?

Solution

Related JEE Advanced Physics questions