If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be

Engineering

Physics

Dimensions

SI Units

Question

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be

V^–2 A² F^–2

V^–2 A² F²

V^–2 A^–2 F

V^–4 A² F

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{F}{A} = y . \frac{Δℓ}{ℓ}; [Y] = \frac{F}{A}$

Now from dimension

$\frac{F}{A} = \frac{ML}{T^{2}}; L = \frac{F}{M} \cdot T^{2}$

$L^{2} = \frac{F^{2}}{M^{2}} {(\frac{V}{A})}^{4} ∵ T = \frac{V}{A}$

$L^{2} = \frac{F^{2}}{M^{2} A^{2}} \frac{v^{2}}{A^{2}} F = MA$

$L^{2} = \frac{V^{4}}{A^{2}}$

$[Y] = \frac{[F]}{[A]} = F^{1} V^{- 4} A^{2}$