Question

The time period $T$ of a small drop of liquid (due to surface tension) depends on density $\rho$, radius $r$ and surface tension $S$. The relation is

• A. $T\propto (\rho r^3/S{)}^{1/2}$
• B. $T\propto \rho rS$
• C. $T\propto \rho r/S$
• D. $T\propto S/\rho r$

Answer 1

Answer: (A)

$T\propto (\rho r^3/S{)}^{1/2}$

Let $T\propto {\rho }^{\alpha }{r}^{\beta }{S}^{\gamma }$
$\Rightarrow \left[T\right]=\left[\rho {]}^{\alpha }\right[r{]}^{\beta }[S{]}^{\gamma }$
$=\left[M^1{L}^{-3}{T}^0{]}^{\alpha }\right[M^0{L}^1{T}^0{]}^{\beta }[M^1{L}^0{T}^{-2}{]}^{\gamma }$
$\left[M^0{L}^0{T}^1\right]=\left[M^{\alpha +\gamma }{L}^{-3\alpha -\beta }{T}^{-2\gamma }\right]$
Comparing the exponents on both sides
$\alpha +\gamma =0\Rightarrow \alpha =-\gamma$
$-3\alpha +\beta =0\Rightarrow \beta =3\alpha$
$-2\gamma =1\Rightarrow \gamma =-\frac{1}{2}$
$\Rightarrow \gamma =-\frac{1}{2},\alpha =\frac{1}{2},\beta =\frac{3}{2}$
Required relation is $T\propto {\left(\frac{\rho r^3}{S}\right)}^{\frac{1}{2}}$.