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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 $)$
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Verified Answer
The correct answers are:
None of the above
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.
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.
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