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(D) Let the radius of each small droplet be $r = 0.1 \,mm = 10^{-4} \,m$ and the number of droplets be $n = 27$.When $n$ droplets merge to form a sing

Vedclass · Accepted Answer

$1.6 	imes 10^{-7} \,J$

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(D) Let the radius of each small droplet be $r = 0.1 \,mm = 10^{-4} \,m$ and the number of droplets be $n = 27$.When $n$ droplets merge to form a single large drop of radius $R$, the volume remains constant: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$.Thus, $R = n^{1/3} r = (27)^{1/3} 	imes r = 3r$.The energy released is equal to the decrease in surface area multiplied by the surface tension $T$:$\Delta E = T 	imes (A_{initial} - A_{final}) = T 	imes (n 	imes 4 \pi r^2 - 4 \pi R^2)$.Substituting $R = 3r$:$\Delta E = 4 \pi T r^2 (n - 9)$.Given $n = 27$, $\Delta E = 4 \pi T r^2 (27 - 9) = 4 \pi T r^2 (18) = 72 \pi T r^2$.Substituting values: $\Delta E = 72 	imes 3.14 	imes 0.072 	imes (10^{-4})^2$.$\Delta E \approx 1.627 	imes 10^{-7} \,J$.

Twenty-seven droplets of water, each of radius $0.1 \,mm$, merge to form a single drop. Calculate the energy released during this process. (Take surface tension of water $T = 0.072 \,N/m$)

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