TS EAMCET · Physics · Mechanical Properties of Fluids
A water drop breaks into 64 identical droplets of each surface area \(10^{-7} \mathrm{~m}^2\). If the surface tension of water is \(0.07 \mathrm{Nm}^{-1}\), the increase in the surface energy in the process is
- A \(158 \times 10^{-9} \mathrm{~J}\)
- B \(432 \times 10^{-9} \mathrm{~J}\)
- C \(216 \times 10^{-9} \mathrm{~J}\)
- D \(336 \times 10^{-9} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(336 \times 10^{-9} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(A_{initial} = 64^{2/3} \times 10^{-7} \mathrm{~m}^2 = 16 \times 10^{-7} \mathrm{~m}^2\) \(A_{final} = 64 \times 10^{-7} \mathrm{~m}^2\) \(\Delta A = A_{final} - A_{initial} = (64 - 16) \times 10^{-7} \mathrm{~m}^2 = 48 \times 10^{-7} \mathrm{~m}^2\)…
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