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A mercury drop of radius $1 \mathrm{~cm}$ is sprayed into $10^6$ drops of equal size. The energy expended in joules is
(Surface tension of mercury is $460 \times 10^{-3}$ $\mathrm{N} / \mathrm{m})$
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(Surface tension of mercury is $460 \times 10^{-3}$ $\mathrm{N} / \mathrm{m})$
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Verified Answer
The correct answer is:
$0.057$
$R=1 \mathrm{~cm}, n=10^6$
Total volume of small drops = volume of big drop
$\begin{aligned} 10^6 \times \frac{4}{3} \pi r^3 & =\frac{4}{3} \pi R^3 \\ 10^6 r^3 & =R^3 \\ r & =\frac{R}{10^2}\end{aligned}$
Energy expended $=$ work done
$\begin{aligned} & =\Delta A \times T \\ & =\left(n 4 \pi r^2-4 \pi R^2\right) T \\ & =4 \pi\left(n r^2-R^2\right) T \\ & =4 \pi\left(n\left(\frac{R}{10^2}\right)^2-R^2\right) T \\ & =4 \pi R^2\left[n \frac{1}{10^4}-1\right] T \\ & =4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \\ & \quad \times\left[10^6 \times \frac{1}{10^4}-1\right] \times 460 \times 10^{-3} \\ & =4 \times 3.14 \times 10^{-4} \times\left(10^2-1\right) \times 460 \times 10^{-3} \\ & =4 \times 3.14 \times 10^{-4} \times 99 \times 460 \times 10^{-3} \\ & =0.057 \mathrm{~J}\end{aligned}$
Total volume of small drops = volume of big drop
$\begin{aligned} 10^6 \times \frac{4}{3} \pi r^3 & =\frac{4}{3} \pi R^3 \\ 10^6 r^3 & =R^3 \\ r & =\frac{R}{10^2}\end{aligned}$
Energy expended $=$ work done
$\begin{aligned} & =\Delta A \times T \\ & =\left(n 4 \pi r^2-4 \pi R^2\right) T \\ & =4 \pi\left(n r^2-R^2\right) T \\ & =4 \pi\left(n\left(\frac{R}{10^2}\right)^2-R^2\right) T \\ & =4 \pi R^2\left[n \frac{1}{10^4}-1\right] T \\ & =4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \\ & \quad \times\left[10^6 \times \frac{1}{10^4}-1\right] \times 460 \times 10^{-3} \\ & =4 \times 3.14 \times 10^{-4} \times\left(10^2-1\right) \times 460 \times 10^{-3} \\ & =4 \times 3.14 \times 10^{-4} \times 99 \times 460 \times 10^{-3} \\ & =0.057 \mathrm{~J}\end{aligned}$
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