JEE Main 2025 Apr 2 Shift 1, Physics Q21: Hydrostatics

JEE Main2025Apr 2, Shift 1Physics

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 $= 10 m/s^{2}$ )

Your answer

Solution

The hinged door experiences hydrostatic pressure from both sides at depth $h = 3 m$ . The denser liquid (on one side) pushes harder than the water on the other side, so an external force is needed to hold the door shut, balancing the pressure difference times area.

Pressure on each face at depth $h$ (atmospheric pressure cancels):

Liquid side: $P_{ℓ} = P_{0} + ρ_{ℓ} g h$ .

Water side: $P_{w} = P_{0} + ρ_{w} g h$ .

Net force from the liquid pushing the door toward the water side:

$F_{ext} = (P_{ℓ} - P_{w}) A = (ρ_{ℓ} - ρ_{w}) g h A$

Substitute $ρ_{ℓ} = 1500$ , $ρ_{w} = 1000 kg/m^{3}$ ; $g = 10 m/s^{2}$ ; $h = 3 m$ ; $A = 100 cm^{2} = 100 \times 1 0^{- 4} m^{2}$ :

$F_{ext} = (1500 - 1000) \times 10 \times 3 \times 1 0^{- 2}$

$F_{ext} = 500 \times 10 \times 3 \times 1 0^{- 2} = 150 N$

$∴$ Required force $= 150 N$ .

Concept behind this question

HydrostaticsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 2 Shift 1, Physics Q21: Hydrostatics

Concept behind this question

More previous year questions on Hydrostatics

Practice more Physics