Water rises up to a height $h$ in a capillary tube immersed vertically in water. When this whole arrangement…
Water rises up to a height $h$ in a capillary tube immersed vertically in water. When this whole arrangement is taken to a depth $d$ in a mine, the water level rises up to a height $h^{\prime}$. If $R$ is the radius of the earth, then the ratio $\frac{h}{h^{\prime}}$ is
$1+\frac{d}{R}$
$1-\frac{d}{R}$
$\frac{R+d}{R-d}$
$\frac{R-d}{R+d}$
Solution
On the surface of the earth $T=\frac{r h \rho g}{2}$ for water
$\therefore h=\frac{2 T}{r \rho g}$
At a depth $d$, the value of acceleration due to gravity changes to
$g^{\prime}=g\left(1-\frac{d}{R}\right)$, therefore
$\begin{aligned}
& h^{\prime}=\frac{2 T}{r \rho g^{\prime} d}=\frac{2 T}{r \rho g\left(1-\frac{d}{R}\right)} \\
& \therefore \frac{h}{h^{\prime}}=\frac{2 T}{r \rho g} \times \frac{r \rho g\left(1-\frac{d}{R}\right)}{2 T}=1-\frac{d}{R}
\end{aligned}$