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

Answer 1

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

Given:

$Y\propto V^a{A}^b{F}^c\dots \left(i\right)$

As we know,

$Y=\frac{\Delta l}{l}=\frac{\text{Force}}{\text{Area}}=\frac{F}{a}$

Dimension of $F=ML{T}^{-2}$

Dimension of area $\left(a\right)=L^2$

Dimension of $V=L{T}^{-1}$

Dimension of acceleration $\left(A\right)=L{T}^{-2}$

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

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

By comparing the above equations, we get,

$c=1\dots \left(ii\right)$

$a+b+c=-1$

$\Rightarrow a+b=-2\dots \left(iii\right)$

$-a-2b-2c=-2$

From equation $\left(ii\right)$,

$a+2b=0\dots \left(iv\right)$

From equation $\left(iii\right)$ and $\left(iv\right)$ we get,

$a=-4$

Now put the value of $a=-4$ in equation $\left(iii\right)$ we get,

$b=2$

Hence, by putting the values of a,b,c in equation $\left(i\right)$ we get,
$Y\propto V^{-4}{A}^2{F}^1$

Final answer:(b)