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JEE Advanced 2026Paper 2
Exam
JEE Advanced
Year
2026
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2026 · Paper 2 · Physics · Mechanical Properties of Fluids

Passage: A container of height 2 m, length 2 m and breadth 1 m is...

Passage: A container of height 22 m, length 22 m and breadth 11 m is made of insulating vertical walls and two large area horizontal metal plates (M1M_1 and M2M_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\sqrt{10} cm2^2 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 1010 ms−2^{-2}.]
Passage: A container of height 2 m, length 2 m and breadth 1 m is made of insulating vertical walls and two large area

Question: The height (in m) of the liquid in left chamber at t=500t = 500 s is:

Correct answer

1.25

Step-by-step solution

Let AA be the cross-sectional area of each chamber and aa be the cross-sectional area of the hole.
From the given dimensions, the area of each chamber is A=1 m×1 m=1 m2A = 1 \text{ m} \times 1 \text{ m} = 1 \text{ m}^2.
The area of the hole is a=10 cm2=10×10−4 m2a = \sqrt{10} \text{ cm}^2 = \sqrt{10} \times 10^{-4} \text{ m}^2.
Let h1h_1 and h2h_2 be the heights of the liquid in the left and right chambers at time tt, respectively.
By conservation of volume, Ah1+Ah2=AHA h_1 + A h_2 = A H, where H=2H = 2 m is the initial height of the liquid in the left chamber.
⇒h1+h2=2\Rightarrow h_1 + h_2 = 2

The velocity of efflux through the hole is given by Torricelli's law:
v=2g(h1−h2)v = \sqrt{2g(h_1 - h_2)}

The rate of flow of the liquid is:
−Adh1dt=a2g(h1−h2)-A \dfrac{dh_1}{dt} = a \sqrt{2g(h_1 - h_2)}

Substitute h2=2−h1h_2 = 2 - h_1 into the equation:
−Adh1dt=a2g(2h1−2)-A \dfrac{dh_1}{dt} = a \sqrt{2g(2h_1 - 2)}

Let y=2h1−2y = 2h_1 - 2. Then dy=2dh1dy = 2 dh_1, which gives dh1=dy2dh_1 = \dfrac{dy}{2}.
The differential equation becomes:
−A2dydt=a2gy-\dfrac{A}{2} \dfrac{dy}{dt} = a \sqrt{2gy}
⇒dyy=−2a2gAdt\Rightarrow \dfrac{dy}{\sqrt{y}} = -\dfrac{2a\sqrt{2g}}{A} dt

Integrating both sides from t=0t = 0 to t=500t = 500 s. At t=0t = 0, h1=2h_1 = 2 m, so y(0)=2(2)−2=2y(0) = 2(2) - 2 = 2 m.
∫2yy−1/2dy=−2a2gA∫0500dt\int_{2}^{y} y^{-1/2} dy = -\dfrac{2a\sqrt{2g}}{A} \int_{0}^{500} dt
[2y]2y=−2×(10×10−4)×201×500[2\sqrt{y}]_{2}^{y} = -\dfrac{2 \times (\sqrt{10} \times 10^{-4}) \times \sqrt{20}}{1} \times 500
2y−22=−1000×10−4×2002\sqrt{y} - 2\sqrt{2} = -1000 \times 10^{-4} \times \sqrt{200}
2y−22=−0.1×1022\sqrt{y} - 2\sqrt{2} = -0.1 \times 10\sqrt{2}
2y−22=−22\sqrt{y} - 2\sqrt{2} = -\sqrt{2}
2y=22\sqrt{y} = \sqrt{2}
y=22\sqrt{y} = \dfrac{\sqrt{2}}{2}

Squaring both sides, we get:
y=0.5y = 0.5

Since y=2h1−2y = 2h_1 - 2, we have:
2h1−2=0.52h_1 - 2 = 0.5
2h1=2.52h_1 = 2.5
h1=1.25 mh_1 = 1.25 \text{ m}

Answer: 1.251.25
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