A vessel of square cross-section and height 6 m is vertically partitioned. A small window of area 100 cm² with a hinged door is fitted at a depth of 3 m in the partition wall. One side is filled completely with water and the other side with a liquid of density 1.5×10³ kg/m³. Take g = 10 m/s². Find the force in newtons required to keep the hinged door from opening.
Physics · Fluid Mechanics · question based on variation of pressure
JEE Mains 2025 2 April Morning Shift

Question

A vessel of square cross-section and height 6 m is vertically partitioned. A small window of area 100 cm² with a hinged door is fitted at a depth of 3 m in the partition wall. One side is filled completely with water and the other side with a liquid of density 1.5×10³ kg/m³. Take g = 10 m/s². Find the force in newtons required to keep the hinged door from opening.

Physics · Fluid Mechanics · question based on variation of pressure

JEE Mains 2025 2 April Morning Shift

G1 Physics · Accepted Answer

At the same depth h=3 m, the pressure difference across the small door is due to the density difference: ΔP = (ρ₂-ρ₁)gh. Here, ρ₂ = 1.5×10³ kg/m³, ρ₁ = 1.0×10³ kg/m³, g = 10 m/s², h = 3 m. Thus ΔP = (500)(10)(3) = 15000 Pa. Area: A = 100 cm² = 100×10⁻⁴ = 0.01 m². Required balancing force: F = ΔP·A = 15000×0.01 = 150 N. Answer: 150.