AIIMS 2003 Physics Fluid Dynamics MCQ Question

To analyze the motion of a lead shot of 1 mm diameter falling through a long column of glycerine, we need to understand the dynamics of an object falling through a viscous fluid. This is a classic problem in fluid dynamics, illustrating the effects of drag force on motion.

Explanation of the Correct Answer (C)

When an object falls through a fluid, it experiences three primary forces:

1. Weight (Gravitational Force): This force pulls the object downward and is given by: $F_g = mg$ where $m$ is the mass of the object and $g$ is the acceleration due to gravity.

2. Buoyant Force: This force acts upward and is equal to the weight of the fluid displaced by the object, given by Archimedes’ principle: $F_b = \rho_f V g$ where $\rho_f$ is the density of the fluid, $V$ is the volume of the object, and $g$ is the acceleration due to gravity.

3. Drag Force: As the object moves through the fluid, it experiences a drag force that opposes its motion. For small spheres in a viscous fluid, the drag force can be described by Stokes' Law: $F_d = 6\pi \eta r v$ where $\eta$ is the dynamic viscosity of the fluid, $r$ is the radius of the sphere, and $v$ is the velocity of the object.

Motion Analysis

Initially, when the lead shot is dropped, it accelerates due to gravity until the drag force balances the gravitational force. At terminal velocity $v_t$, the net force acting on the lead shot becomes zero. This condition can be expressed as: $F_g - F_b - F_d = 0$

Substituting the forces into the equation: $mg - \rho_f V g - 6\pi \eta r v_t = 0$

By solving this equation, we can find the terminal velocity $v_t$:

1. The volume $V$ of the lead shot (spherical) is given by: $V = \frac{4}{3} \pi r^3$ where $r = 0.5 \text{ mm} = 0.0005 \text{ m}$.

2. Rearranging the balance of forces gives: $v_t = \frac{(mg - \rho_f V g)}{6\pi \eta r}$

As the lead shot falls through the glycerine, the velocity $v$ initially increases until it reaches a steady state (terminal velocity) where it no longer accelerates. This results in a characteristic curve where velocity increases with distance initially and then levels off as it approaches terminal velocity.

Why Other Options Are Incorrect

Since the question specifically states that option C is the correct representation of the velocity versus distance graph for the lead shot in glycerine, options D (and any other options) must not accurately represent this behavior. The common characteristics of the options could be:

• They may suggest an incorrect linear relationship indicating constant acceleration or deceleration rather than leveling off, which does not occur in viscous drag scenarios.
• They might not reflect the asymptotic nature of the terminal velocity.

Conclusion

In summary, the falling lead shot's velocity increases rapidly at first but then approaches a maximum constant value (terminal velocity) due to the balance of forces acting on it. This is characteristic of motion in a viscous fluid like glycerine, confirming that option C accurately depicts the situation described in the problem.

Understanding these dynamics is crucial for mastering concepts in fluid mechanics and preparing for NEET/JEE exams.