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Question #42362

Difficulty: 8/10 | 🟢 Free

The Question

A small droplet of initial mass $m_0$ falls from rest under gravity through a stationary cloud. As it falls, it accumulates mass from the cloud such that its mass at a vertical distance $x$ from the release point is given by $m(x) = m_0\left(1 + \frac{x}{H}\right)$ , where $H$ is a positive constant. Assuming no resistive forces other than the momentum transfer from the accumulated stationary mass, what is the speed of the droplet after it has fallen a distance $H$ ?

Options

A.
$\sqrt{\frac{7}{6} gH}$
B.
$\sqrt{\frac{3}{2} gH}$
C.
$\sqrt{gH}$
D.
$\sqrt{2gH}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Variable Mass Systems|Gravitational Potential Energy (System)|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the correct application of Newton's Second Law in variable mass systems.

Students often fall into the trap of using the work-energy theorem ( $\Delta KE = W_{gravity}$ ) without accounting for the macroscopic kinetic energy dissipated as heat during the perfectly inelastic collisions with the stationary cloud particles.

This leads to the incorrect energy conservation equation $\frac{1}{2}m(H)v^2 = \int m(x)g dx$ , yielding option B.

Another common mistake is assuming acceleration is simply $g$ (ignoring the $v(dm/dt)$ thrust term), leading to option D.

The correct approach requires using the momentum form $F = dp/dt$ and recognizing that mechanical energy is not conserved.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Variable Mass Systems
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free