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Q.

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

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a

6W

b

12W

c

4W

d

2W

answer is C.

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Detailed Solution

$\begin{array}{l} W = T \times 4 π r^{2} \\ W \propto r^{2} \\ \frac{W_{1}}{W_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} \\ W_{2} = W_{1} \times {(\frac{2 r}{r})}^{2} \\ W_{2} = 4 W \end{array}$

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