Question

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

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

Answer 1

The correct option is D $V^{-4}{A}^{2}F$

Let $\left[Y\right]=\left[V{]}^a\right[F{]}^b[A{]}^c$

$\left[M{L}^{-1}{T}^{-2}\right]=\left[L{T}^{-1}{]}^a\right[ML{T}^{-2}{]}^b[L{T}^{-2}{]}^c$

$\left[M{L}^{-1}{T}^{-2}\right]=\left[M^b{L}^{a+b+c}{T}^{-a-2b-2c}\right]$

Comparing powers on both sides, of similar terms, we get,

$b=1,a+b+c=-1,-a-2b-2c=-2$

Solving above equations, we get,
$a=-4,b=1,c=2$

$\therefore \left[Y\right]=\left[V^{-4}F{A}^2\right]=\left[V^{-4}{A}^{2}F\right]$

Hence, $\left(D\right)$ is the correct answer.