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 ( `rho`) and surface tension (s) of the liquid. Which among the following will we be a possible expression for T (where k is a dimensionless constant )?

Options

A. `ksqrt((rhor)/(s))`
B. `ksqrt((rho^(2)r)/(s))`
C. `ksqrt((rhor^(3))/(s))`
D. `ksqrt((rhor^(3))/(s^(2)))`

Answer

C

Explanation

Let `T prop r^(a)rho^(b)s^( c)`
or, T= `kr^(a)rho^(b)s^( c)`
From dimensional analysis,
`T^(1)= L^(a)(ML^(-3))^(b)(MT^(-2))^(c)`
`=L^(a-3b).M^(b+c).T^(-2c)`
Equating power of both sides ,
`c= -(1)/(2),b = (1)/(2), a = (3)/(2)`
`:.` From equation (1), T= `kr^((+3)/(2)).rho^((1)/2).s^((1)/(2))`
or,` T= ksqrt((rhor^(3))/(s))`