Question

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

• A. $V^{-2} A^2 F^2$
• B. $V^{-4} A^2 F$
• C. $V^{-4} A^{-2} F$
• D. $V^{-2} A^2 F^{-2}$

Answer 1

The Correct Option is B

To determine the dimension of Young's modulus when speed V, acceleration A, and force F are considered as fundamental units, we start by recalling the dimensional formula of Young's modulus.

Young's modulus, Y, is defined as the ratio of stress to strain. The dimensional formula for stress is the same as the dimensional formula for pressure, which is force per unit area.

Given the dimensional formula of force as [F] and area as having the dimensional formula of [L^2], the formula for stress is:

[\text{Stress}] = [F][L^{-2}]

Strain is dimensionless, as it is the ratio of two lengths.

Hence, the dimensional formula for Young's modulus, which is stress/strain, remains:

[\text{Y}] = [F][L^{-2}]

We know that:

• [V] = [L][T^{-1}]
• [A] = [L][T^{-2}]
• [F] = [M][L][T^{-2}]

We need to express the dimension of Young's modulus [\text{Y}] in terms of [V], [A], and [F].

Since Young's modulus is [F][L^{-2}], we substitute:

[F] = [M][L][T^{-2}] \Rightarrow [L^2] = (A^2/V^2)

In the given scenario, [L^2] = [A^2V^{-2}] (since acceleration A = [L][T^{-2}], and speed V = [L][T^{-1}]).

Thus, we can write:

[\text{Y}] = [F][L^{-2}] = [F] \times [A^2V^{-2}]^{-1}

Simplifying further:

[\text{Y}] = [F] \times V^{-2} \times A^{-2}

Therefore, the dimension of Young's modulus in terms of given units is:

V^{-4} A^2 F

Thus, the correct answer is V^{-4} A^2 F.