Question

If Surface tension $\left(S\right)$, Moment of Inertia $\left(I\right)$ and Planck's constant $\left(h\right)$, were to be taken as the fundamental units, then the dimensional formula for linear momentum would be:

• A. $S^{\frac{3}{2}}{I}^{\frac{1}{2}}{h}^0$
• B. $S^{\frac{1}{2}}{I}^{\frac{1}{2}}{h}^0$
• C. $S^{\frac{1}{2}}{I}^{\frac{3}{2}}{h}^{-1}$
• D. $S^{\frac{1}{2}}{I}^{\frac{1}{2}}{h}^{-1}$

Answer 1

The correct option is B $S^{\frac{1}{2}}{I}^{\frac{1}{2}}{h}^0$

Let the dimensional formula of linear momentum is,

$p=S^a{I}^b{h}^c$

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

Comparing powers of $M$, we get, $a+b+c=1$

Comparing powers of $L$, we get,
$2b+2c=1$

And, Comparing powers of $T$, we get, $-2a-c=-1$

On solving the above three equations, we get,

$a=1/2,b=1/2,c=0$

So, $p=S^{\frac{1}{2}}{I}^{\frac{1}{2}}{h}^0$

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