Question

A drop of water detaches itself from the exit of a tap when ($\sigma =$ surface tension of water. $\rho =$ density of water, $R=$ radius of the tap exit, $r=$ radius of the drop)

• A. $r>{\left(\frac{2}{3}\frac{R\sigma }{\rho g}\right)}^{1/3}$
• B. $r>\frac{2}{3}\frac{\sigma }{\rho g}$
• C. $\frac{2\sigma }{r}>$ atmospheric pressure
• D. $r>{\left(\frac{2}{3}\frac{R\sigma }{\rho g}\right)}^{2/3}$
• E. None of the above

Answer 1

The correct option is E None of the above

When the drop of water detaches itself from the exit of tap, the surface tension acting on the drop upwards is less than the weight of the drop downwards.

Hence $\sigma \left(2\pi r\right)cos\theta <\rho \times \frac{4}{3}\pi R^{3}g$

$⟹\sigma \left(2\pi r\right)\frac{r}{R}<\rho \times \frac{4}{3}\pi R^{3}g$

$⟹r<(\frac{2\rho g}{3\sigma }{)}^{1/2}{R}^2$