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JEE Mains · Physics · STD 11 - 9.2 surface tension

The surface tension of a soap solution is \(3.5 \times 10^{-2}\) N/m. The work required to increase the radius of a soap bubble from \(1\) cm to \(2\) cm is \(\alpha \times 10^{-6}\) J. The value of \(\alpha\) is _____. (\(\pi = 22/7\))

  1. A 262
  2. B 263
  3. C 264
  4. D 265
Verified Solution

Answer & Solution

Correct Answer

(C) 264

Step-by-step Solution

Detailed explanation

The work done to increase the radius of a soap bubble is given by \(W = T \times \Delta A\) Since a soap bubble has two free surfaces, the change in surface area is \(\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)\) Substituting the given values:…
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