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 $^{- 2}$ .]

Q2. The difference in the capacitance (in F) between the metal plates at $t = 0$ and that at $t = 500$ s is $(8 - n) ϵ_{0}$ , where $ϵ_{0}$ is the permittivity of free space. The value of $n$ is:

Your answer

Solution

Setup. The two large plates $M_{1}$ and $M_{2}$ separated by height $2$ m form a parallel-plate capacitor. The region between the plates contains the container (with dielectrics in the two chambers) plus a much larger area outside the container (with $ϵ_{r} = 1$ ). Let $C_{0}$ be the capacitance of the outside region. The container contributes two parallel capacitors of base area $1 m \times 1 m = 1 m^{2}$ each.

At $t = 0$ : Left chamber is fully filled with $ϵ_{r} = 15$ over height $2$ m, right chamber is empty (air, $ϵ_{r} = 1$ ) over height $2$ m.

$C_{L}^{(0)} = \frac{15 ϵ _{0} \cdot 1}{2}, C_{R}^{(0)} = \frac{ϵ _{0} \cdot 1}{2}$ . Sum $= \frac{15 ϵ _{0}}{2} + \frac{ϵ _{0}}{2} = 8 ϵ_{0}$ .

So $C_{i} = C_{0} + 8 ϵ_{0}$ .

At $t = 500$ s: Left chamber has liquid of height $1.25$ m (with $ϵ_{r} = 15$ ) and air of height $0.75$ m on top. Right chamber has liquid of height $0.75$ m and air of height $1.25$ m on top. Each chamber is then a series combination of two dielectrics.

For the left chamber, the two dielectric slabs (each of base area $1$ m $^{2}$ ) in series:

$\frac{1}{C _{L}^{(f)}} = \frac{0.75}{ϵ _{0} \cdot 1} + \frac{1.25}{15 ϵ _{0} \cdot 1} \Rightarrow C_{L}^{(f)} = \frac{ϵ _{0}}{0.75 + 1.25/15}$ .

For the right chamber:

$\frac{1}{C _{R}^{(f)}} = \frac{0.75}{15 ϵ _{0}} + \frac{1.25}{ϵ _{0}} \Rightarrow C_{R}^{(f)} = \frac{ϵ _{0}}{0.75/15 + 1.25}$ .

Sum: $C_{L}^{(f)} + C_{R}^{(f)} = \frac{128}{65} ϵ_{0}$ .

So $C_{f} = C_{0} + \frac{128}{65} ϵ_{0}$ .

Difference. $C_{i} - C_{f} = (8 - \frac{128}{65}) ϵ_{0}$ , so $n = \frac{128}{65} \approx 1.97$ .

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