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Question #42216

Difficulty: 8/10 | 🟢 Free

The Question

A thick wire of length $L$ , cross-sectional area $A$ , and Young's modulus $Y$ is suspended vertically from a rigid support. The linear mass density of the wire is non-uniform and varies as $\lambda(x) = kx^2$ , where $x$ is the distance measured from its free lower end and $k$ is a positive constant. If the total mass of the wire is $M$ and the acceleration due to gravity is $g$ , what is the total elongation of the wire under its own weight?

Options

A.
$\frac{MgL}{4AY}$
B.
$\frac{MgL}{2AY}$
C.
$\frac{3MgL}{4AY}$
D.
$\frac{MgL}{3AY}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Wire Elongation under Self-Weight|Continuous Charge Distribution|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model continuous mass distributions and local stress-strain relationships.

A common trap is blindly applying the standard formula $\frac{MgL}{2AY}$ for a uniform wire, failing to recognize that a non-uniform density shifts the effective center of mass and creates a non-linear tension gradient.

Another trap is incorrectly calculating the center of mass distance from the suspension point, leading to $\frac{3MgL}{4AY}$ .

Students must derive the tension $T(x)$ locally by integrating the weight of the suspended segment and then integrate the differential elongation, reinforcing that macroscopic formulas are integrals of local physical laws.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Wire Elongation under Self-Weight
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free