Question

A uniformly tapering conical wire is made from a material of Young?s modulus $Y$ and has a normal, unextended length $L$. The radii, at the upper and lower ends of this conical wire, have values $R$ and $3R$, respectively. The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal :

• A. $L \left(1 + \frac{2}{9} \frac{Mg}{\pi YR^{2}} \right) $
• B. $L \left(1 + \frac{1}{3} \frac{Mg}{\pi YR^{2}} \right) $
• C. $L \left(1 + \frac{1}{9} \frac{Mg}{\pi YR^{2}} \right) $
• D. $L \left(1 + \frac{2}{3} \frac{Mg}{\pi YR^{2}} \right) $

Answer 1

The Correct Option is B

To solve this problem, we need to find the equilibrium extended length of a uniformly tapering conical wire under the load of a mass \( M \). Here's a step-by-step explanation:

1. Understand the geometry and properties of the conical wire:
   • The wire is tapering, with the upper end radius \( R \) and lower end radius \( 3R \).
   • The unextended length of the wire is \( L \).
   • The mass \( M \) is attached to the lower end.
2. Since the wire is conical, its cross-sectional area varies linearly from top to bottom. The area \( A(x) \) at a distance \( x \) from the top is given by: A(x) = \pi \left(R + \frac{2R}{L}x \right)^{2}.
3. The force acting due to the mass \( M \) is \( Mg \), where \( g \) is the acceleration due to gravity.
4. Young’s modulus \( Y \) relates stress to strain as: \[ \text{Stress} = \frac{\text{Force}}{\text{Area}}, \quad \text{Strain} = \frac{\Delta L}{L} \] For a small element \( dx \) at a distance \( x \), the change in length \( d(\Delta L) \) can be expressed as: \[ d(\Delta L) = \frac{Mg \cdot dx}{\pi \left(R + \frac{2R}{L}x \right)^{2} \cdot Y} \]
5. To find the total change in length \( \Delta L \), integrate \( d(\Delta L) \) from \( 0 \) to \( L \): \[ \Delta L = \int_{0}^{L} \frac{Mg \cdot dx}{\pi \left(R + \frac{2R}{L}x \right)^{2} \cdot Y} \] This integral involves evaluating: \[ \Delta L = \int_{0}^{L} \frac{Mg}{\pi Y\left(R^{2} + \frac{4R^{2}x}{L} + \frac{4R^{2}x^{2}}{L^{2}} \right)} \, dx \]
6. Calculate this integral. Given the linear variation of area, simplifying it leads to: \[ \Delta L = \frac{Mg}{\pi YR^{2}} \cdot \frac{1}{3} \]
7. The equilibrium extended length of the wire is the original length plus the change in length: \[ L_{\text{extended}} = L + \Delta L = L \left(1 + \frac{1}{3} \frac{Mg}{\pi YR^{2}} \right) \]

Therefore, the correct option is $L \left(1 + \frac{1}{3} \frac{Mg}{\pi YR^{2}} \right) $.