Q154 / 155 · WBJEE 2016 · Physics

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
  • Correct answer
    D
    None of the above

Solution

Let mass of the drop $=m$ Weight of the drop = mg It will act downward. The force due to surface tension on the drop $=\sigma2 \pi R$ Where, $R=$ radius of tap If will act in upward direction. The drop of water will detach when $$ m g>\sigma 2 \pi R $$ $$ \begin{array}{l} \text { But, } \quad m=v \times \rho=\frac{4}{3} \pi^{3} p g \\ \therefore \quad \frac{4}{3} \pi^{3} \rho g>\sigma \times 2 \pi A \end{array} $$ $$ r>\left[\frac{3}{2} \frac{\sigma \cdot R}{\rho g}\right]^{1 / 3} $$ Hence, none of the given options is correct.

Hence, the correct answer is option: 'D'.

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