TS EAMCET · Physics · Mechanical Properties of Fluids
A thin film of water is formed between two straight parallel wires each of length 8 cm separated by distance of 0.6 cm. The work done to increase the distance between the wires to 0.8 cm is
(Surface tension of water \(=0.07 \mathrm{Nm}^{-1}\) )
- A \(33.6 \mu \mathrm{~J}\)
- B \(22.4 \mu \mathrm{~J}\)
- C \(11.2 \mu \mathrm{~J}\)
- D \(44.8 \mu \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(22.4 \mu \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(W = T \cdot \Delta A\) \(\Delta A = 2 L (x_2 - x_1) = 2(0.08 \text{ m})(0.008 \text{ m} - 0.006 \text{ m})\) \(\Delta A = 2(0.08 \text{ m})(0.002 \text{ m}) = 0.00032 \text{ m}^2\) \(W = (0.07 \text{ Nm}^{-1})(0.00032 \text{ m}^2) = 0.0000224 \text{ J}\)…
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