MarksRiser
MarksRiser
NEET2022Physics-Mechanics

NEET 2022 Physics Fluid Dynamics MCQ Question

Type: MCQ-conceptual-Hard-Class 11

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is:

Question diagram
A

C

B

D

C

A

D

B

Correct Answer

Option D

Detailed Explanation

✅ Correct answer: D — B

NCERT states that for a sphere falling through a viscous medium:

  • Initially, it accelerates due to gravity.

  • As velocity increases, the retarding viscous force increases because

    Fv=6πηavF_v=6\pi\eta av
  • Eventually, viscous force + buoyant force = gravitational force.

  • Net force becomes zero, so acceleration becomes zero.

  • The sphere then moves with constant terminal velocity. (Sathee)

Therefore the v−tv-t curve must:

start from 0→increase→become horizontal\boxed{\text{start from }0\rightarrow\text{increase}\rightarrow\text{become horizontal}}

That corresponds to curve B in your diagram.

❌ Why other options are wrong

  • A: Straight line → constant acceleration. ❌ But acceleration decreases as viscous force increases.
  • C: Velocity eventually decreases → not possible here. ❌ The ball approaches a constant terminal velocity.
  • D: Velocity falls toward zero → incorrect. ❌ It continues falling at terminal velocity.

📌 NCERT line/concept

As velocity increases, the retarding force also increases. Finally, viscous force plus buoyant force becomes equal to gravity; net force and acceleration become zero, and the sphere descends with constant velocity.

🧠 NEET Trick

Viscous falling sphere:

v↑,a↓,finally v=constant\boxed{v\uparrow,\quad a\downarrow,\quad \text{finally }v=\text{constant}}

So:

Ans = D (B)\boxed{\text{Ans = D (B)}}

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