If work W is done in blowing a bubble of radius R from a soap solution, then the work done in blowing a bubble of radius 2R from the same solution is

W/2

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2(1/3)W

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Solution

The correct option is C
4W

$W = 8 π R^{2} T \Rightarrow W α R^{2} \Rightarrow \frac{W_{2}}{W_{1}} = \frac{R_{2}^{2}}{R_{1}^{2}} \Rightarrow W_{2} = \frac{4 R_{1}^{2}}{R_{1}^{2}} W \Rightarrow W_{2} = 4 W$

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