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Question

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $$P$$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length are doubled is:


A
P
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B
3P4
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C
P2
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D
P4
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Solution

The correct option is A $$\dfrac{P}{4}$$
$$Q_{intial}=\dfrac{\pi P r^{4}}{8 \eta{L}}$$
$$Q_{final}=2Q_{initial}=\dfrac{\pi 4 P_{1}(r)^{4}}{8 \eta L}$$
$$\Rightarrow P_{1}=\dfrac{P}{4}$$

Physics

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