JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 13)
A water drop of radius $$1 \mathrm{~cm}$$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $$\mathrm{cm}$$, then the gain in surface energy upto first decimal place will be :
(Given $$\pi=3.14$$ )
$$8.5 \times 10^{-4} \mathrm{~J}$$
$$8.2 \times 10^{-4} \mathrm{~J}$$
$$7.5 \times 10^{-4} \mathrm{~J}$$
$$5.3 \times 10^{-4} \mathrm{~J}$$
Explanation
$$729 \times {4 \over 3}\pi {r^3} = {4 \over 3}\pi {R^3}$$
$$ \Rightarrow R = 9r$$ ........ (1)
$$\Delta U = S \times \Delta A$$ ..... (2)
$$ \Rightarrow \Delta U = S \times \{ - 4\pi {R^2} + 729 \times 4\pi {r^2}\} $$
$$ = S \times 4\pi \{ 729{r^2} - 81{r^2}\} $$
$$ = 7.5 \times {10^{ - 4}}\,J$$
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