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 $\left(\right.r\left.\right)$ 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, $T \propto r^{a}\rho ^{b}S^{c}$
$T=kr^{a}\rho ^{b}S^{c}$ ..... (i)
Thus, putting dimension, we get
$\left[\right. T \left]\right. = \left[\right. L \left]\right.^{a} \left[\right. M L^{- 3} \left]\right.^{b} \left[\right. M T^{- 2} \left]\right.^{c}$
$\left[T\right]=\left[M\right]^{b + c}\cdot \left[L\right]^{a - 3 b}\cdot \left[T\right]^{- 2 c}$
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
$T=k\cdot r^{\frac{3}{2}}\rho ^{\frac{1}{2}}\cdot S^{- \frac{1}{2} \, }$
$=k\sqrt{\frac{\rho }{S} r^{3}}$