JEE Main · Physics ↓ Falling

Mechanical Properties of Fluids appeared 32 times across 3 years — 3.7% of Physics. This question is from Hydrostatic Pressure.

Year 2026 2025 2024 Total
Questions 9 15 8 32
Practice all Mechanical Properties of Fluids questions → View full JEE Main weightage & trends →

A vessel with square cross-section and height of 6~m is vertically partitioned. A small window of 100~cm² 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× 10³~kg/m³. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity = 10~m/s²)

Numerical Answer Type:
Enter a numerical value Answer: 150 to 150 +4 marks

Solution & Explanation

Related Formula
P = P₀ + ρ g h F = Δ P · A
Core Logic

Let's analyze the pressure acting on the window at depth h = 3~m from both sides:

  • Water side (density ρw = 1.0 × 10³~kg/m³):
Pw = P₀ + ρw g h
  • Denser liquid side (density ρl = 1.5 × 10³~kg/m³):
Pl = P₀ + ρl g h

The net pressure difference pushing on the window is:

Δ P = Pl - Pw = (ρl - ρw) g h Δ P = (1.5 × 10³ - 1.0 × 10³) × 10 × 3 = 500 × 30 = 15000~N/m²

The window has area A = 100~cm² = 100 × 10⁻⁴~m² = 10⁻²~m².

To keep the door from opening, we must apply a force balancing the pressure difference:

F = Δ P · A = 15000 × 10⁻² = 150~N
Step 1: Final Conclusion

The required force is

Step 1: Final Conclusion

The required force is $150\mathrm{~N}.

Pattern Recognition

For a vertical interface between two fluids, the net pressure difference is simply given by

Pattern Recognition

For a vertical interface between two fluids, the net pressure difference is simply given by $\Delta P = \Delta \rho \cdot g h. The ambient atmospheric pressureP_0$ cancels out since it acts on both sides of the window.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Window partition fluid pressure diagram
Window partition fluid pressure diagram

Reference Study Guides

Mechanical Properties Of Fluids

More Mechanical Properties of Fluids Previous-Year Questions — Page 7

Q47 jee_main_2024_31_jan_evening Viscosity and Terminal Velocity
A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity:
  • A. (v)/(2)
  • B. (v)/(4)
  • C. 4v
  • D. 2v

Solution

Related Formula

At terminal velocity, downward force equals upward drag (assuming negligible buoyancy):

Mg = 6π η r v v = (Mg)/(6π η r)
Core Logic

Since the density of the medium is negligible, we ignore buoyant forces. The mass M of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower).

Step 1: Setup Proportionality

Since M, g, and η are all constants:

v ∝ (1)/(r)
Step 2: Evaluating the Ratio

For the second ball, r' = 2r. Therefore, the new terminal velocity v' is:

v' = v × ((r)/(r')) = v × ((r)/(2r)) = (v)/(2)
Pattern Recognition

Read the constraints carefully. Usually, questions keep material density uniform (v ∝ r²). However, this specifically says "same mass". This shifts the formula dependency from v ∝ r² entirely to v ∝ 1/r because M acts as a constant numerator.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2024_31_jan_morning Viscosity And Terminal Velocity
A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?
  • A.
  • B.
  • C.
  • D.

Solution

Related Formula
mg - FB - Fv = ma Fv = 6πη r v
Core Logic

Viscosity And Terminal Velocity diagram for Q46 - JEE Main 2024 Morning
Viscosity And Terminal Velocity diagram for Q46 - JEE Main 2024 Morning

When dropped, three forces act on the ball: Gravity downwards, Buoyant force upwards, and Viscous drag upwards.

mg - FB - Fv = m (dv)/(dt) (ρ (4)/(3)π r³)g - (ρL (4)/(3)π r³)g - 6πη rv = m (dv)/(dt)

Let (4π r³ g(ρ - ρL))/(3m) = K₁ and (6πη r)/(m) = K₂.

(dv)/(dt) = K₁ - K₂ v

Integrating from t=0, v=0:

∫₀^v (dv)/(K₁ - K₂ v) = ∫₀^t dt -(1)/(K₂) ln ((K₁ - K₂ v)/(K₁)) = t v = (K₁)/(K₂) ( 1 - e-K₂ t )

This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity VT = K₁ / K₂.

Chapter Mix

Class 11 Physics: Mechanical Properties Of Fluids

More Mechanical Properties of Fluids Questions — jee_main_2025_02_april_morning

Practice all Mechanical Properties of Fluids previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)