Right answer is (c) v = \(\frac{hγ_w}{4ηL}(R^2-r^2)\)
The explanation: Consider a capillary tube of length L and radius R.
In equilibrium the sum of the forces is zero,
πr^2h1γw– πr^2h1γw-τ(2πrL)=0
on simplification, \(dv=-\frac{hγ_w}{2ηL}rdr\)
on integration,
\(v =- \frac{hγ_w}{4ηL}r^2 + C \,where\, C=\frac{hγ_w}{4ηL} R^2\)
∴ \(v = \frac{hγ_w}{4ηL}(R^2-r^2).\)