Two capillary tubes of same length but radii $r_{1}, r_{2}$ are fitted in parallel to a bottom of a vessel. The pressure head is $p$ . What should be the radius of single tube that can replace the two tubes so that the rate of flow is same as before?

r_{1} + r_{2}

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r_{1}^{2} + r_{2}^{2}

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r_{1}^{4} + r_{2}^{4}

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{(r_{1}^{2} + r_{2}^{2})}^{\frac{1}{4}}

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Solution

The correct option is D ${(r_{1}^{2} + r_{2}^{2})}^{\frac{1}{4}}$
Assuming the pipes are at equal depths, applying bernoulli's principle
efflux velocity of fluid,V is same in both tubes = ${(ρ g h)}^{\frac{1}{2}}$
Rate of flow = cross section area $\times$ velocity
= $(π {r_{1}}^{2} + π {r_{2}}^{2}) V$
Equivalent tube with same flow rate = $π R^{2} V$
$⟹$ $(π R^{2}) V$ = $(π {r_{1}}^{2} + π {r_{2}}^{2}) V$
$⟹$ R = $({r_{1}}^{2} + {r_{2}}^{2})^{\frac{1}{2}}$

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Two capillary tubes of the same length but different radii $r_{1} a n d r_{2}$ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before