Question

The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the capillary, heat evolved is

• A. $+ \frac{\pi^{2}r^{2}h^{2}}{J}dg$
• B. $+ \frac{\pi r^{2}h^{2}dg}{2J}$
• C. $- \frac{\pi r^{2}h^{2}dg}{2J}$
• D. $- \frac{\pi r^{2}h^{2}dg}{J}$

Answer 1

Answer: (B)

$+ \frac{\pi r^{2}h^{2}dg}{2J}$

Answer 2

When the tube is placed vertically in water, water rises through height h given by $h = \frac{2T\cos\theta}{rdg}$

Upward force $= 2\pi r \times T\cos\theta$

Work done by this force in raising water column through height h is given by

$\Delta W = (2\pi rT\cos\theta)h = (2\pi rh\cos\theta)T$

$= (2\pi rh\cos\theta)\left( \frac{rhdg}{2\cos\theta} \right) = \pi r^{2}h^{2}dg$

However, the increase in potential energy $\Delta E_{p}$ of the raised water column $= mg\frac{h}{2}$

where m is the mass of the raised column of water $\therefore m = \pi r^{2}hd$

So,$\Delta E_{P} = (\pi r^{2}hd)\left( \frac{hg}{2} \right) = \frac{\pi r^{2}h^{2}dg}{2}$Further,

$\Delta W - \Delta E_{p} = \frac{\pi r^{2}h^{2}dg}{2}$

The part $(\Delta W - \Delta E_{P})$ is used in doing work against viscous forces and frictional forces between water and glass surface and appears as heat. So heat released

=$\frac{\Delta W - \Delta E_{p}}{J} = \frac{\pi r^{2}h^{2}dg}{2J}$