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Question #42285

Difficulty: 8/10 | 🟢 Free

The Question

A smooth circular wire of radius $r$ is fixed concentrically around a solid non-conducting sphere of mass $M$ and uniform charge $+Q$ . A small bead of mass $m$ and charge $-q$ (where $Q, q > 0$ ) slides along the wire in uniform circular motion. The system is in deep space, isolated from other external fields. If the bead experiences a state of 'weightlessness' (i.e., the normal force exerted by the wire on the bead is exactly zero), which of the following expressions correctly gives the orbital speed $v$ of the bead?

Options

A.
$\sqrt{ \frac{G M}{r} + \frac{Q q}{4\pi\epsilon_0 m r} }$
B.
$\sqrt{ \frac{G M}{r} }$
C.
$\sqrt{ \frac{G M}{r} - \frac{Q q}{4\pi\epsilon_0 m r} }$
D.
$\sqrt{ \frac{G M}{r} + \frac{Q q}{4\pi\epsilon_0 r} }$

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

Weightlessness in Orbit|Non-conducting Charged Sphere|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's fundamental understanding of 'weightlessness' and orbital dynamics.

A common misconception is treating orbital velocity as a fixed geometric constant ( $v = \sqrt{GM/r}$ ) derived solely from gravity.

By introducing a central electric field via a non-conducting charged sphere, the student is forced to re-evaluate the radial force balance.

The trap lies in assuming 'weightlessness' implies only gravitational free-fall, whereas it physically means the normal force is zero ( $N=0$ ), requiring the centripetal force to be the vector sum of all active central forces (gravitational and electrostatic).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Weightlessness in Orbit
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free