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Question #42411

Difficulty: 8/10 | 🟢 Free

The Question

A solid planet of radius $R$ has a spherically symmetric density distribution given by $\rho(r) = \rho_0 \left(\frac{r}{R}\right)$ , where $r$ is the distance from the center of the planet and $\rho_0$ is a positive constant. A small particle is projected radially outward from the center of the planet through a narrow, smooth tunnel. If $g_s$ is the acceleration due to gravity at the surface of the planet, what is the minimum initial speed $v_0$ required for the particle to just reach the surface?

Options

A.
$\sqrt{\frac{2}{3} g_s R}$
B.
$\sqrt{g_s R}$
C.
$\sqrt{2 g_s R}$
D.
$\sqrt{\frac{1}{3} g_s R}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Maximum Height Trap|Gravitational Potential Energy (System)|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply the work-energy theorem in a non-uniform gravitational field.

A common trap is reflexively using the constant acceleration formula $v^2 = 2gH$ (yielding $\sqrt{2g_sR}$ ) or assuming the standard uniform density result where $g(r)$ is linear (yielding $\sqrt{g_sR}$ ).

The student must recognize that the variable density $\rho \propto r$ requires integrating the mass distribution to find $M(r) \propto r^4$ , leading to a quadratic gravitational field $g(r) \propto r^2$ .

Integrating this field from the center to the surface correctly evaluates the change in potential energy required for energy conservation.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Maximum Height Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free