A capillary tube is vertically immersed in water, water rises upto a height 'h $_{1}$ '. When the whole…
A capillary tube is vertically immersed in water, water rises upto a height 'h $_{1}$ '. When
the whole arrangement is taken to a depth 'd' in a mine, the water level rises upto
height ${ }^{\prime} h_{2}$ '. The ratio $\mathrm{h}_{1} / \mathrm{h}_{2}$ is
$(\mathrm{R}=$ radius of earth $)$
$\left(1+\frac{2 \mathrm{~d}}{\mathrm{R}}\right)$
$\left(1-\frac{d}{R}\right)$
$\left(1+\frac{\mathrm{d}}{\mathrm{R}}\right)$
$\left(1-\frac{2 d}{R}\right)$
Solution
$\begin{aligned} & h=\frac{2 T \cos \theta}{r \rho g} \\ \therefore & h \alpha \frac{1}{g} \\ \therefore & \frac{h_{1}}{h_{2}}=\frac{g_{2}}{g_{1}}=1-\frac{d}{R} \end{aligned}$