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(B) Let the radius of the large drop be $R = 1 \ cm = 10^{-2} \ m$. Let the number of small droplets be $n = 10^6$. Let the radius of each small dropl

Vedclass · Accepted Answer

$4356 	imes 10^{-6} \ J$

Vedclass · Answer

(B) Let the radius of the large drop be $R = 1 \ cm = 10^{-2} \ m$. Let the number of small droplets be $n = 10^6$. Let the radius of each small droplet be $r$.Since the volume remains constant,$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$.$r^3 = \frac{R^3}{n} = \frac{(10^{-2})^3}{10^6} = \frac{10^{-6}}{10^6} = 10^{-12} \ m^3$.So,$r = 10^{-4} \ m$.The change in surface energy $\Delta U$ is given by $\Delta U = T 	imes \Delta A$,where $\Delta A$ is the change in surface area.$\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2)$.$\Delta A = 4 \pi (10^6 	imes (10^{-4})^2 - (10^{-2})^2) = 4 \pi (10^6 	imes 10^{-8} - 10^{-4}) = 4 \pi (10^{-2} - 10^{-4}) = 4 \pi (0.01 - 0.0001) = 4 \pi (0.0099) \ m^2$.$\Delta U = 35 	imes 10^{-3} 	imes 4 	imes 3.1416 	imes 0.0099 \approx 4356 	imes 10^{-6} \ J$.

$A$ mercury drop of radius $1 \ cm$ is divided into $10^6$ droplets of equal size. If the surface tension of mercury is $35 \times 10^{-3} \ N/m$,then the change in surface energy in the process is:

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