Start Learning Free

Before You Begin

Question #42399

Difficulty: 8/10 | 🟢 Free

The Question

An elevator accelerates vertically upwards with a constant acceleration a=ga = g. Inside the elevator, a fixed vertical circular track of radius RR is completely submerged in a stationary, incompressible ideal fluid of density σ\sigma. A small spherical shell of negligible mass and volume VV, completely filled with another incompressible fluid of density ρ=3σ\rho = 3\sigma, is projected tangentially from the lowest point of the track's interior. Assuming no viscous drag, what is the minimum initial speed v0v_0 relative to the track required for the sphere to just complete the vertical loop?

Options

  • A.

    20gR3\sqrt{\frac{20gR}{3}}

  • B.

    10gR\sqrt{10gR}

  • C.

    10gR3\sqrt{\frac{10gR}{3}}

  • D.

    5gR\sqrt{5gR}

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

Minimum Speed for Looping|Apparent Weight in Accelerating Elevators|Accelerated Frame Hydrostatics|Vector Analysis

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply non-inertial frame transformations to both solid mechanics and hydrostatics.

A common trap is assuming buoyancy remains σVg\sigma V g even in an accelerating frame, failing to realize that the hydrostatic pressure gradient scales with geff=g+ag_{eff} = g + a.

Another mistake is forgetting to divide the net effective weight by the actual mass (ρV\rho V) to find the modified kinematic acceleration g∗g^*.

By combining the looping condition with an accelerating fluid, it forces a rigorous re-derivation of the critical velocity v0=5g∗Rv_0 = \sqrt{5 g^* R} rather than relying on rote memorization of 5gR\sqrt{5gR}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Plane - Class 11 Part 1
ConceptMinimum Speed for Looping
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree