AP EAMCET · PHYSICS · Mechanical Properties of Fluids
A mercury drop of radius 1 cm is divided into \(10^6\) droplets of equal size. If surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nm}^{-1}\), then the change in suface energy in the process is
- A \(8712 \mu \mathrm{~J}\)
- B \(8712 erg\)
- C \(4356 \mu \mathrm{~J}\)
- D \(4356 erg\)
Answer & Solution
Correct Answer
(C) \(4356 \mu \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(R = 1 \mathrm{~cm} = 0.01 \mathrm{~m}\) \(\Delta A = 4\pi R^2 (n^{1/3} - 1)\) \(\Delta A = 4 \times \frac{22}{7} \times (0.01)^2 ((10^6)^{1/3} - 1)\) \(\Delta A = 4 \times \frac{22}{7} \times 10^{-4} (100 - 1)\)…
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