MHT CET · Physics · Mechanical Properties of Fluids
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
- A \(\left(1-\frac{d}{R}\right)^{-1}\)
- B \(\left(1-\frac{d}{R}\right)\)
- C \(\left(1+\frac{d}{R}\right)^{-1}\)
- D \(\left(1+\frac{d}{R}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(1-\frac{d}{R}\right)^{-1}\)
Step-by-step Solution
Detailed explanation
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 \(h_d=y\)
\(\therefore \quad \frac{x}{y}=\frac{g_d}{g}=\left(1-\frac{d}{R}\right) \Rightarrow \frac{y}{x}=\left(1-\frac{d}{R}\right)^{-1}\)
\(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{rgg}}\)
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 \(h_d=y\)
\(\therefore \quad \frac{x}{y}=\frac{g_d}{g}=\left(1-\frac{d}{R}\right) \Rightarrow \frac{y}{x}=\left(1-\frac{d}{R}\right)^{-1}\)
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