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Physics · MECHANICAL PROPERTIES OF FLUIDS

हिन्दी

An open-tube mercury manometer (\( \rho = 13.6 \times 10^3\text{ kg/m}^3 \)) is connected to a gas container. Initially, the mercury level in the atmospheric arm is \( 15\text{ cm} \) higher than that in the arm connected to the gas. Atmospheric pressure is \( 1.010 \times 10^5\text{ Pa} \). The gas is then heated until its absolute pressure increases to \( 1.418 \times 10^5\text{ Pa} \). Taking \( g = 10\text{ m/s}^2 \), the new difference in heights between the mercury levels in the two arms is:

  1. A
    \( 30\text{ cm} \)
  2. B
    \( 25\text{ cm} \)
  3. C
    \( 45\text{ cm} \)
  4. D
    \( 20\text{ cm} \)
View Answer & Solution

Correct answer · Option A

\( 30\text{ cm} \)

Explanation

Final gauge pressure \( P_g = P - P_a = 1.418 \times 10^5 - 1.010 \times 10^5 = 0.408 \times 10^5\text{ Pa} = 40800\text{ Pa} \). From \( P_g = \rho g h \), \( h = \frac{40800}{13600 \times 10} = 0.30\text{ m} = 30\text{ cm} \).
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