Question

If the velocity $(V)$, acceleration $(A)$, and force $(F)$ are taken as fundamental quantities instead of mass $(M)$, length $(L)$, and time $(T)$, the dimensions of Young's modulus $(Y)$ would be

• A. $F A^{2} V^{-4}$
• B. $F A^{2} V^{-5}$
• C. $F A^{2} V^{-3}$
• D. $F A^{2} V^{-2}$

Answer 1

Answer: (A)

Let $Y=\left[V^{a} A^{b} F^{c}\right]$
$\left[M L^{-1} T^{-2}\right]=\left[L T^{-1}\right]^{a}\left[L T^{-2}\right]^{b}\left[M L T^{-2}\right]^{c}$
$M L^{-1} T^{-2}=M^{c} L^{a+b+c} T^{-a-2 b-2 c}$
$\therefore c=1, a+b+c=-1,-a-2 b-2 c=-2$
On solving, we get $a=-4, b=2$, and $c=1$.