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Question #43490

Difficulty: 8/10 | 🟢 Free

The Question

A stationary bob of mass mm is suspended from a fixed pivot by a massless string of length LL. A small bullet of mass mm moving horizontally with speed u=25gLu = 2\sqrt{5gL} strikes the bob and gets embedded in it. The string has a variable un-stretched cross-sectional area given by A(θ)=A0(2+cos⁡θ)2A(\theta) = A_0 (2 + \cos\theta)^2, where θ\theta is the angular displacement of the string from the lowest point and A0A_0 is a positive constant. If the string is designed such that it just reaches its breaking stress σb\sigma_b during the subsequent motion, find the required value of σb\sigma_b and the angle θb\theta_b at which the string breaks.

Options

  • A.

    σb=3mg2A0,  θb=π2\sigma_b = \frac{3mg}{2A_0}, \; \theta_b = \frac{\pi}{2}

  • B.

    σb=4mg3A0,  θb=0\sigma_b = \frac{4mg}{3A_0}, \; \theta_b = 0

  • C.

    σb=36mg25A0,  θb=π3\sigma_b = \frac{36mg}{25A_0}, \; \theta_b = \frac{\pi}{3}

  • D.

    σb=3mg4A0,  θb=π2\sigma_b = \frac{3mg}{4A_0}, \; \theta_b = \frac{\pi}{2}

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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

Non-Uniform Circular Motion|Breaking Stress Concept|Calculus of Variations

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of collision mechanics, non-uniform circular motion, and material stress.

A common trap (tested by Option B) is assuming that the string will break at the lowest point (θ=0\theta = 0) simply because the tension is maximum there.

However, stress is the ratio of tension to cross-sectional area.

Because the area A(θ)A(\theta) decreases as θ\theta increases, students must express stress as a continuous function of θ\theta and use calculus to find its true maximum.

Additionally, students must correctly apply conservation of angular momentum during the perfectly inelastic collision to find the correct initial velocity, and remember to use the combined mass 2m2m for the subsequent dynamics (Option D traps those who forget the mass addition).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Plane - Class 11 Part 1
ConceptNon-Uniform Circular Motion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree