Question

When a capillary tube of radius r is immersed in a liquid of density $\rho$, the liquid rises to a height h in it. If m is the mass of the liquid in the capillary tube, the P.E. of this mass of the liquid in the tube is :

• A. $\frac{mgh}{4}$
• B. $\frac{mgh}{2}$
• C. mgh
• D. 2mgh

Answer 1

The correct option is D 2mgh

$M=\pi r^{2}h\rho$

Now

Surface tension =$\frac{h\rho gr}{2}=\frac{h^{\prime }\rho g{r}^{\prime }}{2}$

Where h' is the height to which water rised in the second tube and r' its radius

Since $r^{\prime }=2r,h^{\prime }=\frac{h}{2}$

$M^{\prime }=\pi r^{\prime 2}{h}^{\prime }\rho$

$=\pi (2r{)}^2\frac{h}{2}\rho =2\pi r^{2}h\rho =2m$

So, P.E. = 2Mgh

Hence (D) option is correct