A liquid drop of diameter 2 text{ mm} breaks | vedclass

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(B) Let $R$ be the radius of the original drop and $r$ be the radius of each of the $512$ small droplets.By conservation of volume,$\frac{4}{3}\pi R^3

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Let $R$ be the radius of the original drop and $r$ be the radius of each of the $512$ small droplets.By conservation of volume,$\frac{4}{3}\pi R^3 = 512 	imes \frac{4}{3}\pi r^3$,which simplifies to $R^3 = 512r^3$,so $r = \frac{R}{8}$.The change in surface energy $\Delta U = T(A_{	ext{final}} - A_{	ext{initial}}) = T(512 	imes 4\pi r^2 - 4\pi R^2)$.Substituting $r = \frac{R}{8}$,we get $\Delta U = 4\pi T (512 	imes (\frac{R}{8})^2 - R^2) = 4\pi T (8R^2 - R^2) = 28\pi T R^2$.Given $D = 2 	ext{ mm}$,so $R = 1 	ext{ mm} = 10^{-3} 	ext{ m}$ and $T = 0.08 	ext{ N/m}$.$\Delta U = 28 	imes 3.14159 	imes 0.08 	imes (10^{-3})^2 \approx 7.036 	imes 10^{-6} 	ext{ J}$.Comparing with $\alpha 	imes 10^{-6} 	ext{ J}$,the value of $\alpha$ is approximately $7$.

$A$ liquid drop of diameter $2 \text{ mm}$ breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . (Take surface tension of liquid = $0.08 \text{ N/m}$)

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