JEE Advanced 2026 — Paper 2 — PHY-IV solution

JEE Advanced2026Paper 2PHY-IV

Question Stem (Q1-Q2).
A container of height $2$ m, length $2$ m and breadth $1$ m is made of insulating vertical walls and two large area horizontal metal plates ( $M_{1}$ and $M_{2}$ ) 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 $10$ 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 $ϵ_{r} = 15$ and the right chamber is empty ( $ϵ_{r} = 1$ ). At time $t = 0$ , 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 $ϵ_{r} = 1$ and is maintained at atmospheric pressure. The schematic of the container at a time $t > 0$ is shown in the figure. [Given: acceleration due to gravity is $10$ ms $^{- 2}$ .]

Q1. The height (in m) of the liquid in left chamber at $t = 500$ s is:

Your answer

Solution

Bernoulli between the top surface of the left chamber (height $2 - h$ , point 1) and the hole (point 2, depth $h$ below the right surface but bottom of partition).

Let $h$ be the height of liquid in the right chamber. The driving head is the difference between the two surfaces, which equals $(2 - h) - h = 2 - 2 h$ . By Torricelli,

$v = 2 g (2 - 2 h) \cdot \frac{1}{2} \cdot 2 \Rightarrow v = 2 g (1 - h)$ .

Continuity at the hole. Volume rate out of left chamber equals $a v$ , and equals $A d h / d t$ for the rise in the right (cross-section $A$ ). With $A = 1 m \times 1 m = 1$ m $^{2}$ and $a = 10 \times 1 0^{- 4}$ m $^{2}$ ,

$\frac{d h}{d t} = \frac{a}{A} \cdot 2 g (1 - h)$ .

Integrate. Separating variables and integrating from $0$ to $t$ and $0$ to $h$ :

$1 - 1 - h = \frac{a}{A} g t \Rightarrow 1 - h = [1 - \frac{a}{A} g t]^{2}$ .

Substitute numbers at $t = 500$ s. $\frac{a}{A} g t = 10 \times 1 0^{- 4} \times 10 \times 500 = 10 \times 1 0^{- 4} \times 500 = 0.5$ . So $1 - h = (1 - 0.5)^{2} = 0.25 \Rightarrow h = 0.75$ m.

Left chamber height. $H_{L} = 2 - h = 2 - 0.75 = 1.25$ m.

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