JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 13)
If 1000 droplets of water of surface tension $$0.07 \mathrm{~N} / \mathrm{m}$$, having same radius $$1 \mathrm{~mm}$$ each, combine to from a single drop. In the process the released surface energy is :-
$$\left( {\mathrm{Take}\,\pi = {{22} \over 7}} \right)$$
$$7 .92 \times 10^{-4} \mathrm{~J}$$
$$7 .92 \times 10^{-6} \mathrm{~J}$$
$$8 .8 \times 10^{-5} \mathrm{~J}$$
$$9 .68 \times 10^{-4} \mathrm{~J}$$
Explanation
$1000 \times \frac{4 \pi}{3}(1)^{3}=\frac{4 \pi}{3} \mathrm{R}^{3}$
$\mathrm{R}=10 \mathrm{~mm}$
$\mathrm{T} \times 1000 \times 4 \pi\left(10^{-3}\right)^{2}-\mathrm{T} \times 4 \pi\left(10 \times 10^{-3}\right)^{2}=\Delta \mathrm{E}$
$$ \Rightarrow $$ $\Delta \mathrm{E}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$
$$ \Rightarrow $$ $\Delta \mathrm{E}=7.92 \times 10^{-4} \mathrm{~J}$
$\mathrm{R}=10 \mathrm{~mm}$
$\mathrm{T} \times 1000 \times 4 \pi\left(10^{-3}\right)^{2}-\mathrm{T} \times 4 \pi\left(10 \times 10^{-3}\right)^{2}=\Delta \mathrm{E}$
$$ \Rightarrow $$ $\Delta \mathrm{E}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$
$$ \Rightarrow $$ $\Delta \mathrm{E}=7.92 \times 10^{-4} \mathrm{~J}$
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