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Question #43535

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $m$ is suspended in equilibrium from a rigid support by a vertical, uniform, massless wire of length $L$ , cross-sectional area $A$ , and Young's modulus $Y$ . The breaking stress of the wire is $\sigma_0$ (where $\sigma_0 > \frac{mg}{A}$ ). A sudden downward impulse $J$ is imparted to the block. Assuming the wire behaves as a linear spring and does not slacken, what is the maximum value of $J$ such that the wire does not break during the subsequent oscillation?

Options

A.
$(A \sigma_0 - mg) \sqrt{\frac{mL}{YA}}$
B.
$A \sigma_0 \sqrt{\frac{mL}{YA}}$
C.
$(A \sigma_0 + mg) \sqrt{\frac{mL}{YA}}$
D.
$(A \sigma_0 - mg) \frac{mL}{YA}$

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

Breaking Stress Concept|Effective Spring Constant|Vertical Oscillations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to differentiate between static equilibrium loading and dynamic loading in an oscillating system.

The primary trap is calculating the breaking condition based solely on the dynamic force (the impulse-induced tension) while neglecting the pre-existing static tension ( $mg$ ) required for equilibrium.

Alternatively, students might incorrectly apply the breaking stress to the equilibrium position rather than the lowest point of the oscillation where the upward acceleration (and thus the tension) is maximized ( $T_{max} = mg + m\omega^2 x_0$ ).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Breaking Stress Concept
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free