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 the radius ( r )

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 the radius ( r ) of the drop, density ρ and surface tension S of the liquid. Which among the following will be a possible expression for the time period? where k is a dimensionless constant.

TestHub · Accepted Answer

According to the question,

time period, $T \propto r^{a} ρ^{b} S^{c}$

$T = k r^{a} ρ^{b} S^{c}$ ..... (i)

Thus, putting dimension, we get

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

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

Equating the dimensions of both sides, we get

$b + c = 0, a - 3 b = 0$

and $- 2 c = 1$

$∴ c = - \frac{1}{2}$

$∴ b = \frac{1}{2}$

and $a = 3 b = \frac{3}{2}$

Putting these value of $a, b$ and $c$ into Equation. (i), we get

$T = k \cdot r^{\frac{3}{2}} ρ^{\frac{1}{2}} \cdot S^{- \frac{1}{2}}$

$= k \sqrt{\frac{ρ}{S} r^{3}}$

Physics - Unit & Dimension Question with Solution | TestHub

Options:

Doubts & Discussion