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A cylindrical vessel of area of cross-section A is filled with water to a height H. It has capillary tube of length $l$ and radius r fitted horizontally at its bottom. If the coefficient of viscosity of water is $η$ , then time required in which level will fall to a height $\frac{H}{2}$ is (density of water is $ρ$ )

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$\frac{η l r^{2}}{4 Aρ} l n (2)$

$\frac{4 η l r^{4}}{πgAρ} l n (2)$

$\frac{8 η l A}{{ρπgr}^{4}} l n (2)$

$\frac{4 Hη l ρ}{{πgr}^{4}} l n (2)$

answer is C.

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Detailed Solution

$- \frac{Adh}{dt} = \frac{π (hρg) r^{4}}{8 η l} \Rightarrow \int_{0}^{t} dt = \int_{H}^{H / 2} - \frac{8 η l A}{{πρgr}^{4}} (\frac{1}{h}) d h or t = \frac{8 η l A}{{πρgr}^{4}} l n (2)$

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