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Question #42384

Difficulty: 8/10 | 🟢 Free

The Question

A solid sphere of radius $R$ has a non-uniform mass density given by $\rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right)$ , where $r$ is the radial distance from its centre. The sphere is released from rest at a height $H = 1.8 \text{ m}$ on a rough inclined plane. It rolls down without slipping and smoothly transitions to a horizontal ledge at height $h = 5.0 \text{ m}$ above the ground. It then leaves the ledge horizontally and falls to the ground. Assuming acceleration due to gravity $g = 10 \text{ m/s}^2$ and no energy loss at the transition, what is the horizontal distance from the edge of the ledge to the point where the sphere hits the ground?

Options

A.
$2\sqrt{7} \text{ m}$
B.
$6\sqrt{\frac{5}{7}} \text{ m}$
C.
$6 \text{ m}$
D.
$\frac{6\sqrt{15}}{5} \text{ m}$

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

Rolling on Inclines|Projectile on Inclined Plane|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple variable mass density integration with rigid body dynamics and projectile motion.

A common trap is blindly assuming the standard moment of inertia for a uniform solid sphere ( $I = \frac{2}{5}MR^2$ ), which leads to an incorrect rotational kinetic energy partition.

By forcing the derivation of $I = \frac{2}{7}MR^2$ from $\rho(r)$ , it invalidates formula-based recall.

Furthermore, it verifies the understanding that rotational kinetic energy remains constant during the projectile phase, meaning only the translational velocity contributes to the horizontal range.

Quick Facts About This Question

Question Information

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