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Question #42230

Difficulty: 8/10 | 🟢 Free

The Question

A spherical planet of radius $R$ consists of a solid core of radius $R/3$ and uniform density $4\rho$ , surrounded by an incompressible fluid of uniform density $\rho$ up to the surface $r=R$ . The pressure at the surface of the planet is zero. What is the hydrostatic pressure at the surface of the solid core? (Assume $G$ is the universal gravitational constant)

Options

A.
$\frac{8}{9}\pi G \rho^2 R^2$
B.
$\frac{80}{81}\pi G \rho^2 R^2$
C.
$\frac{32}{27}\pi G \rho^2 R^2$
D.
$\frac{16}{27}\pi G \rho^2 R^2$

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

Pascal’s Law Limitations|Gravitational Field Variation (Depth vs Height)|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the application of hydrostatic equilibrium in a non-uniform, self-gravitating fluid body.

A common lethal mistake is assuming Pascal's Law in the form $P = \rho g h$ applies, using either the surface gravity or core gravity, which ignores the spatial variation of $g(r)$ .

Another frequent trap is incorrectly applying the Shell Theorem by superimposing the core's mass without subtracting the displaced fluid mass, which coincidentally yields the exact same incorrect result as using the surface gravity linear approximation.

Students must rigorously integrate the differential equation $dP/dr = -\rho g(r)$ using the correct enclosed mass function.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Pascal’s Law Limitations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free