MHT CET · Physics · Mechanical Properties of Fluids
A liquid drop having surface energy ' \(E\) ' is spread into 216 droplets of the same size. The final surface energy of the droplets is
- A \(3 E\)
- B \(8 \mathrm{E}\)
- C \(2 E\)
- D \(6 \mathrm{E}\)
Answer & Solution
Correct Answer
(D) \(6 \mathrm{E}\)
Step-by-step Solution
Detailed explanation
The correct option is (D).
Concept: Surface Energy is proportional to the surface area \(\mathrm{E} \propto 4 \pi R^2\).
Considering volume conservation, the size of small droplets \(r\) is:
\(V=\frac{4}{3} \pi R^3=216 \times \frac{4}{3} \pi r^3\)
Therefore, \(\mathrm{R}=6 \mathrm{r}\)
The total surface area of the new 216 droplets is:
\(216 \times\left(4 \pi r^2\right)\)
or \(6 \times\left(4 \pi R^2\right)\)
Given \(E \propto 4 \pi R^2\)
Therefore, the total surface energy of the new droplets would be \(6 \mathrm{E}\)
Concept: Surface Energy is proportional to the surface area \(\mathrm{E} \propto 4 \pi R^2\).
Considering volume conservation, the size of small droplets \(r\) is:
\(V=\frac{4}{3} \pi R^3=216 \times \frac{4}{3} \pi r^3\)
Therefore, \(\mathrm{R}=6 \mathrm{r}\)
The total surface area of the new 216 droplets is:
\(216 \times\left(4 \pi r^2\right)\)
or \(6 \times\left(4 \pi R^2\right)\)
Given \(E \propto 4 \pi R^2\)
Therefore, the total surface energy of the new droplets would be \(6 \mathrm{E}\)
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