JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Fluid Mechanics

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of

100 cm^{2}

with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density

1.5 \times 1 0^{3} kg / m^{3}

. What force one needs to apply on the hinged door so that it does not get opened?
(Acceleration due to gravity

= 10 m / s^{2}

)

Answer: 150

Solution

The two liquids press on opposite faces of the window. With both standing at the same level, the net outward push comes only from the density difference at that depth:

F = (ρ_{ℓ} - ρ_{w}) g h A

. Here

ρ_{ℓ} - ρ_{w} = 1500 - 1000 = 500

kg/m

^{3}

h = 3

m, and

A = 100 c m^{2} = 1 0^{- 2} m^{2}

, so

F = 500 \times 10 \times 3 \times 1 0^{- 2} = 150

N must be applied to keep the door shut.

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Fluid Mechanics

Solution

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