1000 spherical drops of water each 10^{-8} m | vedclass

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(B) Let $n = 1000$ be the number of small drops and $r$ be the radius of each small drop. The diameter is $10^{-8} \ m$, so $r = 0.5 	imes 10^{-8} \

Vedclass · Accepted Answer

$6.75 \pi 	imes 10^{-15}$

Vedclass · Answer

(B) Let $n = 1000$ be the number of small drops and $r$ be the radius of each small drop. The diameter is $10^{-8} \ m$, so $r = 0.5 	imes 10^{-8} \ m$.Volume of $n$ small drops = Volume of one large drop of radius $R$.$n 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3 \implies R = n^{1/3} r$.$R = (1000)^{1/3} 	imes (0.5 	imes 10^{-8} \ m) = 10 	imes 0.5 	imes 10^{-8} \ m = 5 	imes 10^{-8} \ m$.Energy liberated $\Delta U = T 	imes \Delta A$, where $\Delta A = (n 	imes 4 \pi r^2) - (4 \pi R^2)$.$\Delta A = 4 \pi (n r^2 - R^2) = 4 \pi (1000 	imes (0.5 	imes 10^{-8})^2 - (5 	imes 10^{-8})^2)$.$\Delta A = 4 \pi (1000 	imes 0.25 	imes 10^{-16} - 25 	imes 10^{-16}) = 4 \pi (250 - 25) 	imes 10^{-16} = 4 \pi 	imes 225 	imes 10^{-16} = 900 \pi 	imes 10^{-16} = 9 \pi 	imes 10^{-14} \ m^2$.Energy liberated $\Delta U = 0.075 	imes 9 \pi 	imes 10^{-14} = 0.675 \pi 	imes 10^{-14} = 6.75 \pi 	imes 10^{-15} \ J$.

$1000$ spherical drops of water each $10^{-8} \ m$ in diameter coalesce to form one large spherical drop. The amount of energy liberated in this process (in joule) is (surface tension of water is $0.075 \ N/m$).

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