Question

The radii of the two columns in $U$ tube are $r_{1}$ and $r_{2}$. When a liquid of density $ \rho$ (angle of contact is $0^{\circ}$) is filled in it the level difference of liquid in two arms is $h$. The surface tension of liquid is ($g$ = acceleration due to gravity)

• A. $\frac{\rho ghr_{1}r_{2}}{2\left(r_{2}-r_{1}\right)}$
• B. $\frac{\rho gh\left(r_{1}-r_{2}\right)}{2r_{1}r_{2}}$
• C. $\frac{2\left(r_{2}-r_{1}\right)}{\rho ghr_{1}r_{2}}$
• D. $\frac{\rho gh}{2 \left(r_{2}-r_{1}\right)}$

Answer 1

Answer: (A)

Let $h_{1}$, $h_{2}$ be the heights to which liquid rises in two columns of radii $r_{1}$ and $r_{2}$ respectively. Then
$h_{1}=\frac{2S \, cos\,0^{\circ}}{r_{1}\rho g}=\frac{2S}{r_{1}\rho g}$
where $s$ is the surface tension of liquid.
and $h_{2}=\frac{2S\,cos \,0^{\circ}}{r_{2}\,\rho g}=\frac{2S}{r_{2}\,\rho g}$
$\therefore $ Difference in levels of liquid in two arms of $U$ tube is
$h=h_{1}-h_{2}=\frac{2S}{\rho g} \left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]$
$=2S \frac{\left(r_{2}-r_{1}\right)}{r_{1}r_{2}\rho g}$
$S=\frac{r_{1}r_{2}\,\rho gh}{2 \left(r_{2}-r_{1}\right)}$