Question

# If the work done in blowing a soap bubble of volume $$'V'$$ is $$W$$, then the work done in blowing is soap bubble of volume $$'2V'$$ is

A
4 W
B
8 W
C
21/3W
D
41/3W

Solution

## The correct option is D $$4^{1/3}W$$$$W = T\triangle A,$$$$\triangle A =$$ change in AreaT = surface Area$$V = \dfrac{4}{3}\pi r^{3}$$$$Area = 4 \pi r^{2}$$$$\Rightarrow \boxed{ A\alpha v^{2/3}}$$$$\Rightarrow \triangle A = (\triangle v)^{2/3}$$$$\Rightarrow \dfrac{W_{1}}{W_{2}} = \dfrac{\triangle A_{1}}{\triangle A_{2}} = \left ( \dfrac{\triangle V_{1}}{\triangle V_{2}} \right )^{2/3}$$$$\Rightarrow W_{2} = W_{1}.\left ( \dfrac{2V}{V} \right )^{2/2}$$$$\Rightarrow \boxed{W_{2}. 2^{2/3}} = 4^{1/3}.w$$Physics

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