MHT CET · Physics · Mechanical Properties of Fluids
Twenty seven droplets of water each of radius \(0.1 \mathrm{~mm}\) merge to form a single drop then the energy released is
- A \(1.6 \times 10^{-3} \mathrm{~J}\)
- B \(1.6 \mathrm{~J}\)
- C \(1600 \mathrm{~J}\)
- D \(1.6 \times 10^{-7} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(1.6 \times 10^{-7} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
From
\(
\begin{aligned}
& \mathrm{W}=4 \pi \mathrm{r}^2 \mathrm{~T}\left(\mathrm{n}-\mathrm{n}^{2 / 3}\right) \\
& \mathrm{W}=4 \pi \times\left(0.1 \times 10^{-3}\right)^2 \times 0.072\left[27-(3)^{2 / 3}\right] \\
& =1.627 \times 10^{-7} \quad\left(\mathrm{~T}_{\text {water }}=0.075 \mathrm{~N} / \mathrm{m}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{W}=4 \pi \mathrm{r}^2 \mathrm{~T}\left(\mathrm{n}-\mathrm{n}^{2 / 3}\right) \\
& \mathrm{W}=4 \pi \times\left(0.1 \times 10^{-3}\right)^2 \times 0.072\left[27-(3)^{2 / 3}\right] \\
& =1.627 \times 10^{-7} \quad\left(\mathrm{~T}_{\text {water }}=0.075 \mathrm{~N} / \mathrm{m}\right)
\end{aligned}
\)
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