The surface energy of a liquid drop is V. It | vedclass

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(B) Let the radius of the initial large drop be $R$ and the radius of each small droplet be $r$.The volume of the large drop is equal to the total vol

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$10 V$

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(B) Let the radius of the initial large drop be $R$ and the radius of each small droplet be $r$.The volume of the large drop is equal to the total volume of the $1000$ small droplets:$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}{3} \pi r^3$$R^3 = 1000 r^3 \implies R = 10r$ or $r = \frac{R}{10}$.The initial surface energy is $V = T 	imes A = T 	imes 4 \pi R^2$,where $T$ is the surface tension.The total surface area of $1000$ small droplets is $A' = 1000 	imes 4 \pi r^2$.Substituting $r = \frac{R}{10}$:$A' = 1000 	imes 4 \pi \left(\frac{R}{10}ight)^2 = 1000 	imes 4 \pi 	imes \frac{R^2}{100} = 10 	imes 4 \pi R^2$.The new surface energy $V'$ is $T 	imes A' = T 	imes 10 	imes 4 \pi R^2 = 10 V$.

The surface energy of a liquid drop is $V$. It is sprayed into $1000$ equal droplets. The surface energy of all the droplets is

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