JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Fluid Mechanics

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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. [Given: acceleration due to gravity is

10 ms^{- 2}

.]

The height (in m) of the liquid in left chamber at

t = 500 s

is:

Answer: 1.25

Solution

Using Bernoulli's theorem between the top surface (point 1) and the hole (point 2):

ρ g (2 - h) = \frac{1}{2} ρ v^{2}

, giving

v = 2 g (1 - h)

.

Continuity equation:

a v = A \frac{d h}{d t}

, where

a = 10 \times 1 0^{- 4}

m² and

A = 2

m².

Integrating:

\frac{a}{A} g t = - 2 (1 - h - 1)

.

At

t = 500

1 - h = [1 - \frac{a}{A} g t]^{2} = [1 - \frac{10 \times 1 0 ^{- 4}}{2} \times 10 \times 500]^{2} = [1 - 0.5]^{2} = 0.25

.

So

h = 0.75

m is the height dropped. Height in left chamber =

2 - 0.75 = 1.25

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Fluid Mechanics

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