Question

If speed $V$, area $A$ and force $F$ are chosen as fundamental units, then the dimension of Young's modulus will be :

• A. $FA ^{-1} V ^{0}$
• B. $FA ^{2} V ^{-1}$
• C. $FA ^{2} V ^{-3}$
• D. $FA ^{2} V ^{-2}$

Answer 1

The Correct Option is A

 To find the dimension of Young's modulus with given fundamental units, we need to express Young's modulus in terms of force \(F\), area \(A\), and speed \(V\).

Young's modulus \(E\) is defined as the ratio of stress to strain. Mathematically, stress is force per unit area, and strain is dimensionless. The dimensional formula for Young's modulus becomes:

\(E = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}\)

So, its dimensional formula according to the traditional fundamental units (mass \(M\), length \(L\), time \(T\)) is:

\([E] = [M^1 L^{-1} T^{-2}]\)

We now need to express this dimensional formula in terms of \(F\), \(A\), and \(V\).

Given:

• Force \([F] = [M^1 L^1 T^{-2}]\)
• Area \([A] = [L^2]\)
• Speed \([V] = [L^1 T^{-1}]\)

Express other dimensions using \(F\), \(A\), and \(V\):

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

Plug these into the Young's modulus formula:

\([E] = [\frac{F}{A}] = [F] [A^{-1}] = [F] [L^{-2}]\\)

Now substitute \([L]\) in terms of \(F, A, V\):

\([L] = [A^{1/2}]\\)

Thus,

\([E] = [F A^{-1} V^0]\\)This matches the option \(FA^{-1} V^{0}\), which is the correct answer.

Therefore, the dimension of Young's modulus in terms of \(F\), \(A\), and \(V\) is \(FA^{-1} V^{0}\).