A spherical drop of oil of radius 1, cm is br | vedclass

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(C) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.Given $R = 1\, cm$,$n = 1000$,and surface tension $T = 50\, dy

Vedclass · Accepted Answer

$1800\pi \, ergs$

Vedclass · Answer

(C) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.Given $R = 1\, cm$,$n = 1000$,and surface tension $T = 50\, dynes/cm$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $1000$ small droplets:$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$$R^3 = 1000 r^3 \implies r = \frac{R}{10} = 0.1\, cm$.The work done $W$ is equal to the increase in surface energy:$W = T 	imes \Delta A = T 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$$W = 4\pi T (n r^2 - R^2)$Substituting the values:$W = 4\pi 	imes 50 	imes (1000 	imes (0.1)^2 - 1^2)$$W = 200\pi 	imes (1000 	imes 0.01 - 1)$$W = 200\pi 	imes (10 - 1) = 200\pi 	imes 9 = 1800\pi \, ergs$.

$A$ spherical drop of oil of radius $1\, cm$ is broken into $1000$ droplets of equal radii. If the surface tension of oil is $50\, dynes/cm$,the work done is

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