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Question #42291

Difficulty: 8/10 | 🟢 Free

The Question

A spherical planet of radius $R$ rotates about its axis with a constant angular velocity $\omega$ . The density of the planet varies with the radial distance $r$ from its center as $\rho(r) = \rho_0 \left(1 - \frac{r}{R}\right)$ , where $\rho_0$ is a positive constant. What is the effective acceleration due to gravity (radial component) measured by an observer located inside the planet at a radial distance $r = \frac{R}{2}$ and at a latitude $\lambda = 60^\circ$ ?

Options

A.
$\frac{5 \pi G \rho_0 R}{12} - \frac{\omega^2 R}{8}$
B.
$\frac{\pi G \rho_0 R}{6} - \frac{\omega^2 R}{8}$
C.
$\frac{5 \pi G \rho_0 R}{12} - \frac{\omega^2 R}{2}$
D.
$\frac{\pi G \rho_0 R}{6} - \frac{\omega^2 R}{2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Effect of Earth's Rotation on g|Superposition Principle|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Gauss's Law of Gravity in non-uniform mass distributions combined with non-inertial reference frames.

A common trap is assuming the internal gravitational field scales linearly with radius ( $g \propto r$ ), which is only true for uniform density.

Students falling for this will calculate the surface gravity and halve it, leading to an incorrect first term.

Furthermore, the problem requires correctly projecting the centrifugal acceleration onto the radial vector at a specific latitude and internal depth, testing the precise geometric understanding of the rotational correction $g_{eff} = g - \omega^2 r \cos^2\lambda$ rather than blindly applying the surface formula.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Effect of Earth's Rotation on g
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free