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. $\left[V^{- 2} A^{2} F^{2}\right]$
• B. $\left[V^{- 2} A^{2} F^{- 2}\right]$
• C. $\left[V^{- 4} A^{- 2} F\right]$
• D. $\left[V^{- 4} A^{2} F\right]$

Answer 1

Answer: (D)

Let $Y=f\left(V , \, F , \, A\right)$
$Y=KV^{x}F^{y}A^{z}, \, K \rightarrow $ Unit less
$\left[Y\right]=\left[V\right]^{x}\left[F\right]^{y}\left[A\right]^{z}$
$\left[ML^{- 1} T^{- 2}\right]=\left[LT^{- 1}\right]^{x}\left[MLT^{- 2}\right]^{y}\left[LT^{- 2}\right]^{z}$
$\left[M\right]\left[L^{- 1}\right]\left[T^{- 2}\right]=\left[M\right]^{y}\left[L^{x + y + z}\right]\left[T^{- x - 2 y - 2 z}\right]$
$\Rightarrow \, \, y=1, \, x+y+z=-1; \, -x-2y-2z=-2$
$x+z=-2 \, \, x+2y+2z=2$
$z=2, \, x=-4 \, \, x+2z=0$
$\left|Y\right|=\left[V^{- 4} A^{2} F\right]$