JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Capacitance

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 $\sqrt{10}{\mathrm{cm}}^2$ 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 ${ϵ}_{\mathrm{r}}=15$ and the right chamber is empty $({ϵ}_{\mathrm{r}}=1)$. 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 ${ϵ}_{\mathrm{r}}=1$ and is maintained at atmospheric pressure. [Given: acceleration due to gravity is $10{\mathrm{ms}}^{-2}$.]

The difference in the capacitance (in F) between the metal plates at $t=0$ and that at $t=500\mathrm{s}$ is $(8-\mathrm{n}){ϵ}_0$, where ${ϵ}_0$ is the permittivity of free space. The value of $n$ is: