JEE Advanced 2026 · Paper 2 · Physics Q16

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Question Stem for Question Nos. 15 and 16

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 ( $\mathrm M_1$ and $\mathrm M_2$ ) 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 $\sqrt{10}$ 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 $\epsilon_r=15$ and the right chamber is empty $(\epsilon_r=1)$ . 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 $\epsilon_r=1$ 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 $^{-2}$ .]

Cross-section of the container between horizontal plates M₁ (top) and M₂ (bottom), 2 m apart, the plates extending beyond the walls. A thin partition divides it into two 1 m wide chambers: the left holds liquid (ε_r = 15) to a large height, the right a shallow layer (ε_r = 1 above). A small hole sits at the bottom of the partition.

The difference in the capacitance (in F) between the metal plates at $t=0$ and that at $t=500$ s is $(8-n)\epsilon_0$ , where $\epsilon_0$ is the permittivity of free space. The value of $n$ is:

Your answer

Answer and worked solution

Official answer: 1.9

The plates extend far beyond the container, but the region outside it does not change, so only the two 1 m² columns matter. A column with liquid height $h$ is two capacitors in series:
$C=\frac{\epsilon_0A}{\frac h{15}+(2-h)}.$

$t=0$ . Left ( $h=2$ ): $\frac{\epsilon_0}{2/15}=7.5\epsilon_0$ . Right ( $h=0$ ): $0.5\epsilon_0$ . Total $8\epsilon_0$ .

$t=500$ s ( $h_1=1.25$ , $h_2=0.75$ ):

left: $\dfrac{\epsilon_0}{0.0833+0.75}=1.200\epsilon_0$ ;
right: $\dfrac{\epsilon_0}{0.05+1.25}=0.769\epsilon_0$ .

Total $1.969\epsilon_0$ .

The difference is $(8-1.969)\epsilon_0=(8-n)\epsilon_0$ , so $n\approx1.97$ .

Topics:Dielectrics Series and parallel combinations

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JEE Advanced 2026 · Paper 2, Physics question 16