Question

# Two capillary of length $$L$$ and $$2L$$ and of radius $$R$$ and $$2R$$ are connected in series. The net rate of flow of fluid through them will be(given rate to the flow through single capillary, $$X = \dfrac {\pi P R^4}{8\eta L}$$)

A
89X
B
98X
C
57X
D
75X

Solution

## The correct option is A $$\dfrac {8}{9}X$$Fluid resistance is given by R = $$\dfrac {8\eta L}{\pi r^4}$$When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is$$\displaystyle R_s = R_1 + R_2 = \frac {8\eta L}{\pi R^4} + \frac {8\eta \times 2L}{\pi (2R)^4}= \left (\frac{8\eta L}{\pi R^4} \right ) \times \frac {9}{8}$$Rate of flow = $$\displaystyle \frac {P}{R_s} = \frac {\pi P R^4}{8\eta L}\times \frac{8}{9} = \frac{8}{9} X \left [ as X = \frac{\pi P R^4}{8\eta L} \right ]$$Physics

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