Question

The lower end of a capillary tube of radius r is placed vertically in water. With rise of water in the capillary, the heat evolved will be:

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

Answer 1

The correct option is A $+\frac{\pi r^2{h}^{2}dg}{2J}$

When the tube is placed vertically in water, water rises through height h given by $h=\frac{2Tcos\theta }{dgr}$
Upward force = $2\pi rTcos\theta$
work done by this force in raising water column through height h is given by
$\Delta W=\left(2\pi rTcos\theta \right)h=\left(2\pi rhcos\theta \right)\left(\frac{rhdg}{2cos\theta }\right)=\pi r^2{h}^{2}dg$
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
So, $\Delta E_p=\left(\pi r^{2}hd\right)\left(\frac{hg}{2}\right)=\frac{\pi r^2{h}^{2}dg}{2}$ (since m = $\pi r^{2}hd)$
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 force and frictional forces between water and glass surface and appears as heat.
So, heat released = $\frac{\Delta W-\Delta E_p}{J}=\frac{\pi r^2{h}^{2}dg}{2J}$