Question

If the time period $\left(T\right)$ of vibrations of a liquid drop depends on the surface tension $\left(S\right)$ of the fluid over which it is kept, radius $\left(r\right)$ of the drop and density $\left(\rho \right)$ of the liquid then the equation of $T$ is
[here, $k$ is a dimensionless proportionality constant]

• A. $T=k\sqrt{\rho r^3/S}$
• B. $T=k\sqrt{{\rho }^3{r}^3/S}$
• C. $T=k\sqrt{\rho r^3/S^3}$
• D. $T=k\sqrt{\rho r^5/S}$

Answer 1

The correct option is A $T=k\sqrt{\rho r^3/S}$

Dimensional formula of,
Time period, $T$
$\left[T\right]=\left[M^0{L}^0{T}^1\right]$
Surface tension, $S$
$\left[S\right]=\left[M{L}^0{T}^{-2}\right]$
Radius, $r$
$\left[r\right]=\left[M^{0}L{T}^0\right]$
Density, $\rho$
$\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$
Let us suppose the relation is
$T=k{\rho }^a{r}^b{S}^c$
$\Rightarrow \left[T\right]=\left[k{\rho }^a{r}^b{S}^c\right]$
$\Rightarrow \left[M^0{L}^0{T}^1\right]=[M^1{L}^{-3}{T}^0{]}^a\times [M^{0}L{T}^0{]}^b\times [M{L}^0{T}^{-2}{]}^c$
$\Rightarrow \left[M^0{L}^0{T}^1\right]=\left[M^{(a+c)}{L}^{(-3a+b)}{T}^{(-2c)}\right]$
On comparing, we get,
$\begin{array}{}a+c=0& \text{(1)}\end{array}$
$\begin{array}{}-3a+b=0& \text{(2)}\end{array}$
$\begin{array}{}-2c=1& \text{(3)}\end{array}$
On solving these equations, we get,
$a=1/2,b=3/2$ and $c=-1/2$
Thus,
$T=k{\rho }^a{r}^b{S}^c\Rightarrow T=k\sqrt{\rho r^3/S}$