MHT CET · Physics · Mechanical Properties of Fluids
The work done in splitting a water drop of radius R into 64 droplets is ( \(\mathrm{T}=\) Surface tension of water)
- A \(6 \pi \mathrm{TR}^2\)
- B \(12 \pi \mathrm{TR}^2\)
- C \(8 \pi \mathrm{TR}^2\)
- D \(\quad 24 \pi \mathrm{TR}^2\)
Answer & Solution
Correct Answer
(B) \(12 \pi \mathrm{TR}^2\)
Step-by-step Solution
Detailed explanation
Given the radius of drop is R .
Let \(r\) be the radius of smaller droplets. As the total volume remains the same,
\(\begin{array}{ll}
& \frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3 \\
\therefore \quad & r=\frac{R}{(64)^{\frac{1}{3}}}=\frac{R}{4}
\end{array}\)
Initial surface energy \(E_1=4 \pi R^2 T\)
Final surface energy \(E_2=64 \times 4 \pi \times\left(\frac{R}{4}\right)^2 \times T\)
\(=16 \pi \mathrm{R}^2 \mathrm{~T}\)
\(\therefore \quad\) Work done, \(\mathrm{W}=\mathrm{E}_2-\mathrm{E}_1\)
\(\mathrm{W}=12 \pi \mathrm{TR}^2\)
Let \(r\) be the radius of smaller droplets. As the total volume remains the same,
\(\begin{array}{ll}
& \frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3 \\
\therefore \quad & r=\frac{R}{(64)^{\frac{1}{3}}}=\frac{R}{4}
\end{array}\)
Initial surface energy \(E_1=4 \pi R^2 T\)
Final surface energy \(E_2=64 \times 4 \pi \times\left(\frac{R}{4}\right)^2 \times T\)
\(=16 \pi \mathrm{R}^2 \mathrm{~T}\)
\(\therefore \quad\) Work done, \(\mathrm{W}=\mathrm{E}_2-\mathrm{E}_1\)
\(\mathrm{W}=12 \pi \mathrm{TR}^2\)
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