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(D) Since the volume of the mercury remains constant during the splitting process:$V_{initial} = V_{final}$$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) Since the volume of the mercury remains constant during the splitting process:$V_{initial} = V_{final}$$\frac{4}{3} \pi R^3 = 1000 	imes \frac{4}{3} \pi r^3$$R^3 = 1000 r^3$Taking the cube root on both sides,we get $R = 10r$,or $r = \frac{R}{10}$.The change in surface area $\Delta A$ is given by:$\Delta A = A_{final} - A_{initial}$$\Delta A = (1000 	imes 4 \pi r^2) - 4 \pi R^2$Substitute $r = \frac{R}{10}$ into the equation:$\Delta A = 4 \pi \left( 1000 	imes \left( \frac{R}{10} ight)^2 - R^2 ight)$$\Delta A = 4 \pi \left( 1000 	imes \frac{R^2}{100} - R^2 ight)$$\Delta A = 4 \pi (10 R^2 - R^2) = 4 \pi (9 R^2) = 36 \pi R^2$The change in surface energy is given by $\Delta U = T 	imes \Delta A$:$\Delta U = T 	imes 36 \pi R^2 = 36 \pi R^2 T$.

When a mercury drop of radius $R$ splits up into $1000$ droplets of radius $r$,the change in surface energy is ($\pi R^2 T$). ($T=$ surface tension of mercury).

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