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(B) The work done $W$ in increasing the surface area is given by $W = T \Delta A$.Here, $\Delta A$ is the change in surface area, where $\Delta A = A_

Vedclass · Accepted Answer

$12$

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(B) The work done $W$ in increasing the surface area is given by $W = T \Delta A$.Here, $\Delta A$ is the change in surface area, where $\Delta A = A_{final} - A_{initial}$.Let the radius of the large drop be $R$ and the radius of each small droplet be $r$.Since the volume remains constant, $\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$.Taking the cube root on both sides, $R = 4r$, which implies $r = \frac{R}{4}$.The initial surface area $A_i = 4\pi R^2$.The final surface area $A_f = 64 	imes 4\pi r^2 = 64 	imes 4\pi (\frac{R}{4})^2 = 64 	imes 4\pi 	imes \frac{R^2}{16} = 16 	imes 4\pi R^2 = 64\pi R^2$.The change in surface area $\Delta A = A_f - A_i = 64\pi R^2 - 4\pi R^2 = 60\pi R^2$.Therefore, the work done $W = T 	imes 60\pi R^2$ is incorrect based on the provided options; let us re-evaluate: $W = T(64 	imes 4\pi (R/4)^2 - 4\pi R^2) = T(4\pi R^2(64/16 - 1)) = T(4\pi R^2(4 - 1)) = T(4\pi R^2 	imes 3) = 12\pi R^2 T$.

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T$ is the surface tension of water) (in $\pi R^2 T$)

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