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
B
3P4
C
P2
D
P4

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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