JEE Challenger
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PhyHardAdv 2026 • Paper 2

Calculate Height of Liquid in Container Left Chamber at Time 500 Seconds

Comprehension Passage

A container of height 2 m2\text{ m}, length 2 m2\text{ m} and breadth 1 m1\text{ m} is made of insulating vertical walls and two large area horizontal metal plates (M1\text{M}_1 and M2\text{M}_2) which extend far beyond the vertical walls in all directions. The container is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of cross-sectional area 10 cm2\sqrt{10}\text{ cm}^2 near its bottom edge. Initially the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant ϵr=15\epsilon_r = 15 and the right chamber is empty (ϵr=1\epsilon_r = 1). At time t=0t = 0, the hole is opened and the liquid flows from the left chamber to the right chamber. In both the chambers, the space above the liquid has ϵr=1\epsilon_r = 1 and is maintained at atmospheric pressure. The schematic of the container at a time t>0t > 0 is shown in the figure.

[Given: acceleration due to gravity is 10 ms−210\text{ ms}^{-2}.]

The height (in m\text{m}) of the liquid in left chamber at t=500 st = 500\text{ s} is:

Question Diagram 1
Official Numerical Answer1.25

Topics & Concepts

Electrostatic Potential and CapacitanceCapacitorsDielectricsFluid Dynamics

Step-by-Step Solution

To find the height of the liquid in the left chamber at time t=500 st = 500\text{ s}, we apply fluid dynamics principles (specifically Torricelli's law and conservation of mass).

1. Geometric Parameters and Initial Conditions

  • Chamber Dimensions:

    • Total length of container =2 m= 2\text{ m} divided equally into two chambers, so length of left chamber L=1 mL = 1\text{ m}.
    • Breadth of chamber B=1 mB = 1\text{ m}.
    • Cross-sectional area of each chamber A=L×B=1 m2A = L \times B = 1\text{ m}^2.

  • Hole Parameters:

    • Area of hole near the bottom a=10 cm2=10×10−4 m2a = \sqrt{10}\text{ cm}^2 = \sqrt{10} \times 10^{-4}\text{ m}^2.
    • Acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2.

  • Initial State (t=0t = 0):

    • Height of liquid in left chamber, y1(0)=2 my_1(0) = 2\text{ m}.
    • Height of liquid in right chamber, y2(0)=0 my_2(0) = 0\text{ m}.

2. Conservation of Volume

Since the liquid is incompressible and total volume is conserved: y1(t)+y2(t)=2 m  ⟹  y2(t)=2−y1(t)y_1(t) + y_2(t) = 2\text{ m} \implies y_2(t) = 2 - y_1(t)

The height difference between the liquid levels in the left and right chambers at any time tt is: Δh=y1−y2=y1−(2−y1)=2y1−2\Delta h = y_1 - y_2 = y_1 - (2 - y_1) = 2y_1 - 2

3. Rate of Flow (Torricelli's Law)

The velocity of liquid efflux through the small hole is: v=2gΔh=2g(2y1−2)v = \sqrt{2g \Delta h} = \sqrt{2g(2y_1 - 2)}

The rate of decrease of liquid level in the left chamber is governed by the continuity equation: −Ady1dt=av-A \frac{dy_1}{dt} = a v

Substituting the given values into the equation: −dy1dt=(10×10−4)2(10)(2y1−2)-\frac{dy_1}{dt} = \left(\sqrt{10} \times 10^{-4}\right) \sqrt{2(10)(2y_1 - 2)}

−dy1dt=10×10−4×40(y1−1)-\frac{dy_1}{dt} = \sqrt{10} \times 10^{-4} \times \sqrt{40(y_1 - 1)}

−dy1dt=10×10−4×210y1−1-\frac{dy_1}{dt} = \sqrt{10} \times 10^{-4} \times 2\sqrt{10}\sqrt{y_1 - 1}

−dy1dt=20×10−4y1−1=2×10−3y1−1-\frac{dy_1}{dt} = 20 \times 10^{-4} \sqrt{y_1 - 1} = 2 \times 10^{-3} \sqrt{y_1 - 1}

4. Integration

Separating variables and integrating from t=0t = 0 (where y1=2 my_1 = 2\text{ m}) to t=500 st = 500\text{ s} (where y1=hy_1 = h):

∫2hdy1y1−1=−2×10−3∫0500dt\int_{2}^{h} \frac{dy_1}{\sqrt{y_1 - 1}} = -2 \times 10^{-3} \int_{0}^{500} dt

[2y1−1]2h=−2×10−3×500\left[ 2\sqrt{y_1 - 1} \right]_{2}^{h} = -2 \times 10^{-3} \times 500

2h−1−22−1=−12\sqrt{h - 1} - 2\sqrt{2 - 1} = -1

2h−1−2=−12\sqrt{h - 1} - 2 = -1

2h−1=12\sqrt{h - 1} = 1

h−1=0.5\sqrt{h - 1} = 0.5

h−1=0.25  ⟹  h=1.25 mh - 1 = 0.25 \implies h = 1.25\text{ m}

Final Answer

The height of the liquid in the left chamber at t=500 st = 500\text{ s} is 1.25 m.

Calculate Height of Liquid in Container Left Chamber at Time 500 Seconds | Physics PYQ Solution - JEE Challenger