Water rises up to height ' $X$ ' in a capillary tube immersed vertically in water. When the whole…

Water rises up to height ' $X$ ' in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth ' $d$ ' in a mine, the water level rises up to height ' $Y$ '. If ' $R$ ' is the radius of earth then the ratio $\frac{Y}{X}$ is
  1. $\left(1-\frac{d}{R}\right)^{-1}$
  2. $\left(1-\frac{d}{R}\right)$
  3. $\left(1+\frac{d}{R}\right)^{-1}$
  4. $\left(1+\frac{d}{R}\right)$

Solution

Rise in capillary tube is given as, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rgg}}$
As all the other quantities are kept constant, in mine of depth d, $\mathrm{h} \propto \frac{1}{\mathrm{~g}}$ At à depth $\mathrm{d}, \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(\frac{1-\mathrm{d}}{\mathrm{R}}\right)$ Also, given that, $h=x$ and $h_d=y$ $\therefore \quad \frac{x}{y}=\frac{g_d}{g}=\left(1-\frac{d}{R}\right) \Rightarrow \frac{y}{x}=\left(1-\frac{d}{R}\right)^{-1}$

Asked in: MHT CET 2024 (04 May Shift 1)

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