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(T) of the liquid drop depends on radius (r) of the drop, density $\left(\rho \right)$ and surface tension(s) of the liquid. Which among the following will be a possible expression for T(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)

$k\sqrt{\frac{\rho r^3}{s}}$

$T=k{r}^x{\rho }^y{s}^z$
$T\to T,r\to L,\rho \to M{L}^{-3},S\to M{T}^{-2}$
$S=\frac{F}{l}=\frac{ma}{l}=\frac{m}{S^2}$
$T^1=K{L}^x\left(M{L}^{-2}{)}^y\right(M{T}^{-2}{)}^z$
$\Rightarrow T^1=K{L}^{x-3y}{M}^{y+z}{T}^{-2z}$
$-2z=1$ $\Rightarrow z=-\frac{1}{2},y+z=0$ $y=-z=+\frac{1}{2}$
$x-3y=0$ $\Rightarrow x=3y=\frac{3}{2}$
$T=K{6}^{3/2}{\rho }^{1/2}{s}^{-1/2}=K\sqrt{\frac{\rho r^3}{S}}$