JEE Main · Physics ↓ Falling

Mechanical Properties of Fluids appeared 32 times across 3 years — 3.7% of Physics. This question is from Work Done by Gravity in Connecting Vessels.

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Two cylindrical vessels of equal cross sectional area of 2~m² contain water upto height 10~m and 6~m, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is : (Density of water is 10³~kg/m³ and g=10~m/s²)

Solution & Explanation

Related Formula

The gravitational potential energy U of a liquid column of mass m and height h is evaluated relative to its bottom by placing its total mass at its center of mass (h/2):

U = m g ((h)/(2)) = (ρ A h) g ((h)/(2)) = (1)/(2) ρ A g h²

Work done by the force of gravity (W) equals the negative change in potential energy:

W = -Δ U = Uᵢ - Uf
Core Logic

Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at:

hf = (10 + 6)/(2) = 8~m
Step 1: Calculate Initial Potential Energy (Uᵢ)

Let the reference level U=0 be at the bottom:

Uᵢ = U₁ + U₂ = (1)/(2) ρ A g h₁² + (1)/(2) ρ A g h₂² Uᵢ = (1)/(2) ρ A g (10² + 6²) = (1)/(2) ρ A g (100 + 36) = 68 ρ A g

Work Done by Gravity in Connecting Vessels
Work Done by Gravity in Connecting Vessels

Step 2: Calculate Final Potential Energy (Uf)

Both vessels equalize to hf = 8~m:

Uf = 2 × [ (1)/(2) ρ A g hf² ] = ρ A g (8²) = 64 ρ A g
Step 3: Work Done by Gravity (W)
W = Uᵢ - Uf = 68 ρ A g - 64 ρ A g = 4 ρ A g

Substitute the given values (ho = 10³~kg/m³, A = 2~m², g = 10~m/s²):

W = 4 × 10³ × 2 × 10 = 8 × 10⁴~J
Pattern Recognition

For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by

Pattern Recognition

For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2. Applying this directly:

Δ U = (1)/(4) × 10³ × 2 × 10 × (10 - 6)² = 5000 × 16 = 8 × 10⁴~J$
Chapter Mix

Class 11 Physics: 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_03_april_evening

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