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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
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B
8 W
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C
21/3W
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D
41/3W
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Solution

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

1201326_1069626_ans_01ef044188ce466fb0c97dffb482c580.jpg

Physics

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