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Chapter 8: Problem 63

A uniform rod of length \(L\) has a mass per unit length \(\lambda\) and area of cross-section \(A .\) The elongation in the rod is \(l\) due to its own weight if it is suspended from the ceiling of a room. The Young's modulus of the rod is (A) \(\frac{2 \lambda g L^{2}}{A l}\) (B) \(\frac{\lambda g L^{2}}{2 A l}\) (C) \(\frac{2 \lambda g L}{A l}\) (D) \(\frac{\lambda g l^{2}}{A L}\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution: The Young's modulus of the rod is given by: \(Y = \frac{\lambda g L^{2}}{2 A l}\).

Step by step solution

01

Determine the stress on the rod

We know that stress = force/area. In this case, the force acting on the rod is its weight, which can be calculated as the product of mass and acceleration due to gravity, g. The mass of the rod can be calculated by multiplying its mass per unit length (位) and length (L). So, stress can be given as: Stress = Force/Area = (位Lg)/A
02

Determine the strain on the rod

Strain is defined as the elongation (l) divided by the original length (L) of the rod. Strain = l/L
03

Calculate Young's modulus

Now, Young's modulus is defined as the ratio of stress to strain, so we can write: Young's modulus = Stress/Strain = (位Lg/A) / (l/L) Upon simplifying this expression, we have: Young's modulus = (位gL虏)/(Al) Comparing this expression with the given options, we find that it matches option (B). #Answer#: Hence, the Young's modulus of the rod is given by: \(Y = \frac{\lambda g L^{2}}{2 A l}\)

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