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 radius $\left(r\right)$ 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 $T$ (where $k$ is a dimensionless constant)? (given surface tension is force per unit length)

• 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

The correct option is C $k\sqrt{\frac{\rho r^3}{S}}$

$T=k{r}^x{\rho }^y{S}^z$
$T\to T,r\to L,\rho \to M{L}^{-3},S\to M{T}^{-2}$
$S=\frac{F}{l}=\frac{ma}{l}=\frac{\text{m}}{{\text{s}}^2}$
$T^1=K{L}^x\left(M{L}^{-3}{)}^y\right(M{T}^{-2}{)}^z$
$\Rightarrow T^1=K{L}^{x-3y}{M}^{y+z}{T}^{-2z}$
$-2z=1\Rightarrow z=-\frac{1}{2},y+z=0,y=-z=+\frac{1}{2}$
$x-3y=0\Rightarrow x=3y=\frac{3}{2}$
$T=k{L}^{3/2}{\rho }^{1/2}{s}^{-1/2}=k\sqrt{\frac{\rho r^3}{S}}$