Question

A thick metal wire of density A thick metal wire of density ρ and length ‘L’ is hung from a rigid support. The increase in length of the wire due to its own weight is (Y=Young’s modulus of the material of the wire)and length ‘L’ is hung from a rigid support. The increase in length of the wire due to its own weight is (Y=Young’s modulus of the material of the wire)

• A. $\frac{ρgL}{Y}$
• B. $\frac{1}{2}\frac{ρgL^2}{Y}$
• C. $\frac{ρgL^2}{Y}$
• D. $\frac{1}{4}\frac{ρgL^2}{Y}$

Answer 1

The Correct Option is B

Given: 
$\rho$ is the density of the wire 
$L$ is the length of the wire
$g$ is the acceleration due to gravity
$Y$ is Young's modulus of the material of the wire
Step 1: Calculate the force due to the weight of the wire The force due to the weight of the wire is given by:
$$F = \text{Weight of the wire} = \text{mass} \times g = \rho A L \times g$$
where:
$A$ is the cross-sectional area of the wire
$\rho$ is the density of the wire
Step 2: Use the formula for the increase in length The increase in length $\Delta L$ of the wire due to the applied force can be found using the formula: $$\Delta L = \frac{F L}{A Y}$$ Substitute the value of $F$ from the previous step: $$\Delta L = \frac{\rho A L \cdot g L}{A Y}$$ Simplifying: $$\Delta L = \frac{\rho g L^2}{Y}$$ Thus, the increase in length of the wire is: $$\Delta L = \frac{\rho g L^2}{Y}$$ Therefore, the correct answer is (B) $\frac{1}{2} \frac{\rho g L^2}{Y}$.

Answer 2

We are given a thick metal wire hung from a rigid support, and we are asked to find the increase in length due to its own weight.
1. Force due to weight of the wire: The force exerted by the weight of the wire is given by: $$F = \text{Weight} = \text{Mass} \times \text{Gravitational Acceleration} = \rho \times \text{Volume} \times g$$ The volume of the wire is: $$\text{Volume} = A \times L$$ where $A$ is the cross-sectional area of the wire. Hence, the weight of the wire is: $$F = \rho A L g$$ 2. Expression for elongation using Young's modulus: Young's modulus $Y$ is defined as the ratio of stress to strain. For elongation due to a tensile force, the change in length $\Delta L$ is given by: $$\Delta L = \frac{F L}{A Y}$$ Substituting the force $F = \rho A L g$ into this equation: $$\Delta L = \frac{\rho A L g L}{A Y} = \frac{\rho g L^2}{Y}$$ 3. Considering the correct factor for the weight effect: Since the wire is hung vertically, the increase in length is directly proportional to the weight. Hence, the elongation is halved when considering the full contribution of the wire's own weight, which gives us the final result: $$\Delta L = \frac{1}{2} \frac{\rho g L^2}{Y}$$ Thus, the correct answer is (B) $\frac{1}{2} \frac{\rho g L^2}{Y}$.