A spherical liquid drop of radius R is divide | vedclass

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(C) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$. Since the volume remains constant,the volume of the large dro

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$4\pi R^2 T$

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(C) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$. Since the volume remains constant,the volume of the large drop equals the volume of $8$ small droplets:$\frac{4}{3}\pi R^3 = 8 	imes \frac{4}{3}\pi r^3$$R^3 = 8r^3 \implies r = \frac{R}{2}$Work done $W$ is equal to the change in surface energy:$W = T 	imes (\Delta A) = T 	imes (A_{final} - A_{initial})$$A_{initial} = 4\pi R^2$$A_{final} = 8 	imes (4\pi r^2) = 32\pi (\frac{R}{2})^2 = 32\pi 	imes \frac{R^2}{4} = 8\pi R^2$$W = T 	imes (8\pi R^2 - 4\pi R^2) = 4\pi R^2 T$

$A$ spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$,then the work done in this process will be

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