A drop of water detaches itself from the exit | vedclass

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Question

(D) Let the mass of the drop be $m$. The weight of the drop is $W = mg$,which acts in the downward direction. The force due to surface tension acting

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) Let the mass of the drop be $m$. The weight of the drop is $W = mg$,which acts in the downward direction. The force due to surface tension acting on the drop at the exit of the tap is $F_s = \sigma \cdot 2\pi R$,where $R$ is the radius of the tap exit. This force acts in the upward direction. The drop of water will detach when the weight of the drop exceeds the upward force due to surface tension: $mg > \sigma \cdot 2\pi R$ Since the mass $m$ of the spherical drop is given by $m = V \cdot \rho = \frac{4}{3} \pi r^3 \rho$,where $r$ is the radius of the drop,we substitute this into the inequality: $\frac{4}{3} \pi r^3 \rho g > \sigma \cdot 2\pi R$ Solving for $r$: $r^3 > \frac{2\pi R \sigma \cdot 3}{4\pi \rho g}$ $r^3 > \frac{3 R \sigma}{2 \rho g}$ $r > \left( \frac{3 R \sigma}{2 \rho g} \right)^{1/3}$ Comparing this result with the given options,none of the options $A$,$B$,or $C$ match the derived expression. Therefore,the correct option is $D$.

$A$ 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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