Question

A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation $\left(T\right)$ of the liquid drop depends on the radius $(r)$ of the drop, density $\left(\rho \right)$ and surface tension $\left(S\right)$ of the liquid. Which among the following will be a possible expression for the time period? where $k$ is a dimensionless constant.

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

Answer 1

Answer: (C)

According to the question,
time period, $T\propto r^a{\rho }^b{S}^c$
$T=k{r}^a{\rho }^b{S}^c$ ..... (i)
Thus, putting dimension, we get
$[T]=[L{]}^a[M{L}^{-3}{]}^b[M{T}^{-2}{]}^c$
$\left[T\right]={\left[M\right]}^{b+c}\cdot {\left[L\right]}^{a-3b}\cdot {\left[T\right]}^{-2c}$
Equating the dimensions of both sides, we get
$b+c=0,a-3b=0$
and $-2c=1$
$\therefore c=-\frac{1}{2}$
$\therefore b=\frac{1}{2}$
and $a=3b=\frac{3}{2}$
Putting these value of $a,b$ and $c$ into Equation. (i), we get
$T=k\cdot r^{\frac{3}{2}}{\rho }^{\frac{1}{2}}\cdot S^{-\frac{1}{2}}$
$=k\sqrt{\frac{\rho }{S}{r}^3}$