Q126 / 160 · TS EAMCET 2015 · Physics
1000 spherical drops of water each $10^{-8} \mathrm{~m}$ in diameter coalesce to form one large spherical drop. The amount of energy liberated in this process (in joule) is (surface tension of water is $0.075 \mathrm{~N} / \mathrm{m}$ ).
- A$10.75 \pi \times 10^{-15}$
- Correct answerB$6.75 \pi \times 10^{-15}$
- C$8.65 \pi \times 10^{-15}$
- D$3.88 \pi \times 10^{-15}$
Solution
Volume of small spherical drops $=\frac{4}{3} \pi r^3$
Volume of larger spherical drops $=\frac{4}{3} \pi R^3$
$n$ small spherical drops combine to form large drops.
$$
\begin{aligned}
& \therefore \quad n\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi R^3 \\
& \Rightarrow \quad R=n^{1 / 3} r \\
&
\end{aligned}
$$
$$
\text { Here, } \quad n=1000, r=\frac{1}{2} \times 10^{-8} \mathrm{~m}
$$
Amount of energy liberated
$=$ Change in area $\times$ surface tension
$$
\begin{aligned}
& =4 \pi R^2\left(n^{1 / 3}-1\right) \times 0.075 \\
& =6.75 \pi \times 10^{-15} \mathrm{~J}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.