Q138 / 155 · WBJEE 2016 · Physics
1000 droplets of water having $2 \mathrm{mm}$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 \mathrm{Nm}^{-1}$. The energy loss in the process is
- Correct answerA$8.146 \times 10^{-4} \mathrm{J}$
- B$4.4 \times 10^{-4} \mathrm{J}$
- C$2108 \times 10^{-5} \mathrm{J}$
- D$4.7 \times 10^{-1} \mathrm{J}$
Solution
Let the radius of single drop = $r$
Radius of small drop $=\frac{2 m m}{2}$
$$
\begin{array}{l}
=1 \mathrm{mm} \\
=1 \times 10^{-3} \mathrm{m}
\end{array}
$$
Surface tension of water $=0.072 \mathrm{N} / \mathrm{m}$
The volume of large drop must be equal volume of all small drops.
$$
\frac{4}{3} \pi R^{3}=1000\left(\frac{4}{3} \pi r^{3}\right)
$$
$\therefore$
$$
A=(1000)^{\frac{1}{3}} \cdot r=10 r
$$
The initial potential energy of drops system $U_{i}=s \times n \times 4 \pi r^{2}$
$$
\begin{array}{l}
=s \times 1000 \times 4 \pi^{2} \\
=1000 \mathrm{s} \cdot 4 \pi^{2}
\end{array}
$$
When the droplets coalesce, the final potential energy of the system $U_{1}=s \cdot 4 \pi R^{2}$
$$
=s \cdot 4 \pi \times 100 r^{2}
$$
Energy loss in the formation of large drop
$$
\begin{aligned}
\Delta E &=U_{t}-U_{i} \\
&=s \cdot 4 \pi 100 r^{2}-1000 \mathrm{s} \cdot 4 \pi r^{2} \\
&=s \cdot \pi r^{2}(400-4000) \\
&=-3600 \mathrm{s} \cdot \pi^{2} \\
&=-3600 \times 0.72 \times 3.14 \times\left(1 \times 10^{-3}\right)^{2}
\end{aligned}
$$
From catenation
$$
\begin{array}{l}
=813.888 \times 10^{-6} \\
=8.1388 \times 10^{-4} \mathrm{J} \\
=8.146 \times 10^{-4} \mathrm{J}
\end{array}
$$
Hence, the correct answer is option: 'A'.