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MHT CET 2026 (20 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2026 (20 May, Shift 1)

Q9MHT CET 2026 · 20 MAY · SHIFT 1Medium

Acceleration Due to Gravity

If a capillary tube is immersed in water, rise of water in capillary is h1h_1. When the whole arrangement is taken to a depth dd in a mine, the water level rises to h2h_2. The ratio h2h1\frac{h_2}{h_1} is (R=radius of earth)

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Answer: (d)

  1. Capillary rise h=2Tcos⁡θrρgh = \frac{2T\cos\theta}{r\rho g}, so h∝1gh \propto \frac{1}{g}
  2. At depth dd: gd=g(1−dR)=g R−dRg_d = g\left(1 - \frac{d}{R}\right) = g\,\frac{R-d}{R}
  3. h2h1=ggd=RR−d\frac{h_2}{h_1} = \frac{g}{g_d} = \frac{R}{R-d}
  4. So the answer is (d) RR−d\frac{R}{R-d}.
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