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(D) Let the radius of the large drop be $R = 1\, mm = 10^{-3}\, m$ and the number of small droplets be $n = 10^6$.Let the radius of each small droplet

Vedclass · Accepted Answer

$8.95 	imes 10^{-5}\, J$

Vedclass · Answer

(D) Let the radius of the large drop be $R = 1\, mm = 10^{-3}\, m$ and the number of small droplets be $n = 10^6$.Let the radius of each small droplet be $r$.Since the total volume remains constant:$n 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$10^6 	imes r^3 = (10^{-3})^3$$r^3 = 10^{-15} \Rightarrow r = 10^{-5}\, m$.Increase in surface area $\Delta A = n(4 \pi r^2) - 4 \pi R^2$$\Delta A = 10^6 	imes 4 \pi (10^{-5})^2 - 4 \pi (10^{-3})^2$$\Delta A = 4 \pi (10^{-4} - 10^{-6}) = 4 \pi 	imes 10^{-6} (100 - 1) = 4 \pi 	imes 99 	imes 10^{-6}\, m^2$.Work done $W = T 	imes \Delta A$$W = 72 	imes 10^{-3} 	imes 4 \pi 	imes 99 	imes 10^{-6}$$W = 72 	imes 4 	imes 3.14159 	imes 99 	imes 10^{-9} \approx 8.95 	imes 10^{-5}\, J$.

The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $T = 72 \times 10^{-3}\, N/m$):

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