JEE Mains · Physics · STD 11 - 9.2 surface tension
The surface tension of a soap bubble is \(0.03\) N/m. The work done in increasing the diameter of bubble from \(2\) cm to \(6\) cm is \(\alpha \pi \times 10^{-4}\) J. The value of \(\alpha\) is _______. (Take \(\pi = 3.14\))
- A \(0.86\)
- B \(0.64\)
- C \(1.92\)
- D \(7.68\)
Answer & Solution
Correct Answer
(C) \(1.92\)
Step-by-step Solution
Detailed explanation
Initial radius \(r_1 = \dfrac{2}{2} = 1\) cm \(= 10^{-2}\) m Final radius \(r_2 = \dfrac{6}{2} = 3\) cm \(= 3 \times 10^{-2}\) m Surface tension \(T = 0.03\) N/m A soap bubble has two free surfaces. The change in total surface area is given by:…
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