A certain number of spherical drops of a liqu | vedclass

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(C) When small drops coalesce into a larger drop,the total surface area decreases,which results in the release of energy.Let $n$ be the number of smal

Vedclass · Accepted Answer

Energy $= 3VT \left( \frac{1}{r} - \frac{1}{R} \right)$ is released.

Vedclass · Answer

(C) When small drops coalesce into a larger drop,the total surface area decreases,which results in the release of energy.Let $n$ be the number of small drops of radius $r$. The volume is conserved,so $V = n \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi R^3$,which implies $n = \frac{R^3}{r^3}$.The initial surface area $A_i = n (4 \pi r^2) = \left( \frac{R^3}{r^3} \right) (4 \pi r^2) = 4 \pi R^3 / r$.The final surface area $A_f = 4 \pi R^2$.The change in surface area $\Delta A = A_i - A_f = 4 \pi R^3 / r - 4 \pi R^2 = 4 \pi R^2 \left( \frac{R}{r} - 1 \right)$.Energy released $\Delta E = T \Delta A = T (4 \pi R^2) \left( \frac{R}{r} - 1 \right) = T (4 \pi R^3) \left( \frac{1}{r} - \frac{1}{R} \right)$.Since $V = \frac{4}{3} \pi R^3$,we have $4 \pi R^3 = 3V$.Substituting this,$\Delta E = 3VT \left( \frac{1}{r} - \frac{1}{R} \right)$.

$A$ certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,then:

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