- Home
- Standard 11
- Physics
A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let $R$ be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency $v$ about its equilibrium shape. By dimensional analysis, the ratio $\frac{v}{\sqrt{\sigma / \rho R^3}}$ can be (Here, $\sigma$ is surface tension, $\rho$ is density, $g$ is acceleration due to gravity and $k$ is an arbitrary dimensionless constant)
$k \rho g R^2 / \sigma$
$k \rho R^3 / g_\sigma$
$k \rho R^2 / g \sigma$
$k \rho / g \sigma$
Solution
(a)
Dimensions of
$\frac{v}{\sqrt{\sigma}}=\frac{[v]}{\sqrt{\rho \cdot h^3}}=\frac{\left[ T ^{-1}\right]}{\sqrt{\left[\rho R ^3\right]}}=\frac{\left[ MT ^{-2}\right]}{\sqrt{\left[\frac{ M }{ L ^3} \cdot L ^3\right]}}$
$=\left[ M ^0 L ^0 T ^0\right]$
Now, from option $(a)$,
Dimensions of $\frac{k \rho g R^2}{\sigma}=\frac{\left[k \rho g R^2\right]}{[\sigma]}$
$=\frac{\left[\frac{ M }{ L ^3}\right] \cdot\left[ LT ^{-2}\right] \cdot\left[ L ^2\right]}{\left[ MT ^{-2}\right]}$
$=\left[ M ^0 L ^0 T ^0\right]$
So, by dimensional analysis, option $(a)$ is correct.
