Question

The lower end of a capillary tube of radius r is placed vertically in water. With the rise h of water in the capillary tube, evolved is (Take p as the density of water)

• A. $\frac{2\pi }{pg}\times h$
• B. $\frac{\pi }{pg}\times hr$
• C. zero
• D. $\frac{\pi r^2{h}^{2}pg}{2}$

Answer 1

Answer: (D)

$\frac{\pi r^2{h}^{2}pg}{2}$

When the tube is placed vertically in water, water rises through height $h$ given by,

$h=\frac{4T\cos \theta }{rpg}$

upward force$=2\pi \times T\cos \theta$

Workdone by this force is raising water column through height $h$ is given by

$\Delta W=\left(2\pi T\cos \theta \right)h=\left(2\pi h\cos \theta \right)T=\left(2\pi rh\cos \theta \right)\left(\frac{rhpg}{4\cos \theta }\right)=\frac{\pi r^2{h}^{2}pg}{2}$