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9.1 , fluid mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics9.1 , fluid mechanics1 Mark
MCQ
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 of the flow through single capillary, $X = \pi P{R^4}/8\eta L)$
  • ✓
    $\frac{8}{9}X$
  • B
    $\frac{9}{8}X$
  • C
    $\frac{5}{7}X$
  • D
    $\frac{7}{5}X$
✓

Answer

Correct option: A.
$\frac{8}{9}X$
a
(a)Fluid resistance is given by $R = \frac{{8\eta l}}{{\pi {r^4}}}.$
When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is
${R_e} = {R_1} + {R_2} = \frac{{8\eta L}}{{\pi {r^4}}} + \frac{{8\eta \times 2L}}{{\pi {{(2R)}^4}}} = \left( {\frac{{8\eta L}}{{\pi {r^4}}}} \right) \times \frac{9}{8}$
Equivalent resistance becomes $\frac{9}{8}$times so rate of flow will be $\frac{8}{9}X$

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