JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 6)
A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :
2.8 $$\times$$ 10$$-$$4 J
1.5 $$\times$$ 10$$-$$3 J
1.9 $$\times$$ 10$$-$$4 J
9.4 $$\times$$ 10$$-$$5 J
Explanation
$$r' = {r \over 4}$$
$$ \Rightarrow \Delta E = T(\Delta S)$$
$$ = T \times 4\pi (nr{'^2} - {r^2}),\,n = 64$$
$$ = T \times 4\pi \times (4 - 1){r^2}$$
$$ \Rightarrow \Delta E = 0.075 \times 4 \times 3.142(3) \times {10^{ - 4}}\,$$ J
$$ = 2.8 \times {10^{ - 4}}$$ J
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