A mercury drop of radius 1, cm is sprayed int | vedclass

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(A) The change in surface energy $W$ is given by the formula $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface

Vedclass · Accepted Answer

$0.057$

Vedclass · Answer

(A) The change in surface energy $W$ is given by the formula $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.When a large drop of radius $R$ is broken into $n$ smaller drops of radius $r$,the change in area is $\Delta A = n(4\pi r^2) - 4\pi R^2$.Since the volume remains constant,$\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,which implies $r = R / n^{1/3}$.Substituting this into the area formula: $\Delta A = 4\pi R^2 (n^{1/3} - 1)$.Given $R = 1\, cm = 10^{-2}\, m$,$n = 10^6$,and $T = 460 	imes 10^{-3}\, N/m$.$W = 4 \pi (10^{-2})^2 (460 	imes 10^{-3}) ((10^6)^{1/3} - 1)$.$W = 4 	imes 3.14159 	imes 10^{-4} 	imes 0.46 	imes (100 - 1)$.$W = 4 	imes 3.14159 	imes 10^{-4} 	imes 0.46 	imes 99$.$W \approx 0.057\, J$.

$A$ mercury drop of radius $1\, cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of mercury is $460 \times 10^{-3}\, N/m$)

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