JEE Advanced 2020 — Paper 2 — Question 5

Given: Height of vessel $= 8\,\mathrm{m}$ Mass of partition $M = 8.3\,\mathrm{kg}$ Moles of gas in each part $= 0.1$ Temperature $T = 300\,\mathrm{K}$ Acceleration due to gravity $g = 10\,\mathrm{m/s^2}$ Universal gas constant $R = 8.3\,\mathrm{J\,mol^{-1}K^{-1}}$

Let $h$ be the distance of the partition from the top at equilibrium.

From ideal gas law, pressure on top part: $P_1 = \frac{nRT}{V_1} = \frac{0.1RT}{A h}$

Pressure on bottom part: $P_2 = \frac{nRT}{V_2} = \frac{0.1RT}{A (8 - h)}$

Mechanical equilibrium of the partition: $P_2 - P_1 = \frac{M g}{A}$

Substitute pressures: $0.1 R T \left(\frac{1}{8 - h} - \frac{1}{h} \right) = M g$

Simplify difference of fractions: $0.1 R T \cdot \frac{2(h - 4)}{h (8 - h)} = M g$

Substitute numerical values: $0.1 \times 8.3 \times 300 \times \frac{2(h - 4)}{h (8 - h)} = 8.3 \times 10$

Calculate constants: $249 \times \frac{2(h - 4)}{h (8 - h)} = 83$

Rearranged: $2(h - 4) = \frac{83 h (8 - h)}{249}$

Approximate fraction: $2(h - 4) = 0.333 \times h (8 - h)$

Expand: $2h - 8 = 2.664 h - 0.333 h^2$

Bring all terms to one side: $0.333 h^2 - 0.664 h - 8 = 0$

Multiply entire equation by 3: $h^2 - 1.992 h - 24 = 0$

Use quadratic formula: $h = \frac{1.992 \pm \sqrt{(1.992)^2 + 4 \times 24}}{2}$

Calculate discriminant: $\sqrt{3.968 + 96} = \sqrt{99.968} \approx 9.998$

Calculate roots: $h = \frac{1.992 \pm 9.998}{2}$

Valid root: $h = \frac{1.992 + 9.998}{2} = 6$

Discard invalid negative root.

Final answer: $\boxed{6\, \mathrm{m}}$