Question

The time of oscillation $T$ of a small drop of liquid depends on radius $r$, density $\rho$ and surface tension $S$. The relation between them is given by

• A. $T\propto \sqrt{\frac{S}{\rho r^3}}$
• B. $T\propto \sqrt{\frac{\rho r^3}{S}}$
• C. $T\propto \sqrt{\frac{S^2{r}^3}{\rho }}$
• D. $T\propto \sqrt{\frac{\rho r^3}{2S}}$

Answer 1

The correct option is B $T\propto \sqrt{\frac{\rho r^3}{S}}$

Let $T=S^a{r}^b{\rho }^c$
So, $\left[T\right]=\left[M^0{L}^0{T}^1\right]$

$=\left[M^1{L}^{-1}{T}^{-2}{]}^a\right[L^1{]}^b[M{L}^{-3}{]}^c$

$=\left[M^{a+c}{L}^{b-3c}{T}^{-2a}\right]$

Applying principle of homogeneity,

$a+c=0;b-3c=0;-2a=1$
On solving, we get $a=-1/2.b=3/2,c=1/2$
$T\propto \sqrt{\frac{\rho r^3}{S}}$