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 $V^{-4}{A}^{2}F$

$\frac{F}{A}=y\cdot \frac{\Delta l}{l}$
$\left[Y\right]=\frac{F}{A}$
Now from dimension
$F=\frac{ML}{T^2}$
$L=\frac{F}{M}\cdot T^2$
$L^2=\frac{F^2}{M^2}{\left(\frac{V}{A}\right)}^4$
$\because T=\frac{V}{A}$
$L^2=\frac{F^2}{M^2{A}^2}\frac{v^4}{A^2}$ $F=MA$
$L^2=\frac{V^4}{A^2}$
$\left[Y\right]=\frac{\left[F\right]}{\left[A\right]}=F^1{V}^{-4}{A}^2$.