Question

If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop, and density $\left(\rho \right)$ of the liquid, then find the expression of T.

Answer 1

$T=K\left(s^a{r}^b{\rho }^c\right)$

$M^0{L}^0{T}^1=R\left(M{T}^{-2}{)}^{-2}\right(L^b\left)\right(M{L}^{-3}{)}^c$

Comparing powers

of mi

$0=a+c...\left(1\right)$

Of Li

$0=b-3c\Rightarrow b=3e...\left(2\right)$

Of Ti

$1=-2a\Rightarrow a=-1/2$

Solving (1) & (2)

$c=1/2$

$b=3/2$

$\therefore T=K\frac{r^{3/2}{\rho }^{1/2}}{s^{r2}}$

$T=K\sqrt{\frac{r^3\rho }{s}}$