A slightly conical wire of length \(L\) and radii a and \(b\) is stretched by two forces each of magnitude \(F\), applied parallel to its length in opposite directions and normal to end faces. If \(Y\) denotes the Young's modulus, then the extension produced is : (1) \(\frac{F L}{\pi\left(a^{2}+b^{2}\right) Y}\) (2) \(\frac{F L}{\pi\left(a^{2}-b^{2}\right) Y}\) (3) \(\frac{\mathrm{FL}}{\pi \mathrm{abY}}\) (4) None of these

A conical wire is a type of rod whose cross-sectional area changes along its length. Typically, it tapers from a wider end to a narrower end. This variation in radius means its properties and behavior under stress vary depending on the segment analyzed.
In this exercise, the conical wire has radii at the ends of `a` and `b` and length `L`. When forces are applied parallel to the wire length and in opposite directions, the material stretches, making it crucial to understand these variations.
The radius at any point `x` from the narrower end can be calculated with the formula:
\( r(x) = a + \frac{(b - a)x}{L} \).
This formula shows a linear change from radius `a` to `b` along the length `L` of the wire.
As the radius changes, the cross-sectional area at any point `x` is \(Area = \pi \big(r(x)\big)^2\). Since area affects stretch under force, accurately accounting for this change is essential for precise calculations.

Material deformation refers to the change in shape or size of a material under stress. In this context, it's about stretching a conical wire under tensile forces. These forces act parallel and in opposite directions, causing the wire to elongate.
The extension in a uniform rod can be simplified, but for a conical wire, the changing area complicates things. The infinitesimal extension for a small segment `dx` must be calculated using:
\( d(\triangle L) = \frac{F dx}{\text{Area} \cdot Y} \),
with Area being \pi \big(r(x)\big)^2\.
Here, `Y` denotes Young's modulus—a measure of stiffness of the material. It relates stress (force per unit area) to strain (proportional change in length). To find total deformation, you need to integrate these infinitesimal extensions across the entire wire length.

Integral calculus is essential for dealing with the varying cross-sectional area in our conical wire problem. Integration allows us to sum up infinitesimal changes in stretch over the wire's length.
We start with the integral for infinitesimal extension:
\( \triangle L = \frac{F}{\pi Y} \int_{0}^{L} \frac{dx}{(a + \frac{(b - a)x}{L})^2}\).
Applying substitution, where \u = a + \frac{(b - a)x}{L}\ and \du = \frac{(b - a)}{L}dx\, transforms the integral into a simpler form:

\( dx = \frac{L}{(b - a)} du \).
Integrating between limits `a` to `b`, we derive:
\( \triangle L = \frac{F L}{\pi Y (b - a)} \int_{a}^{b} \frac{du}{u^2} = \frac{F L}{\pi Y (b - a)} \left[ -\frac{1}{u} \right]_{a}^{b} \).
Solving gives the extension of the wire:

\( \triangle L = \frac{F L}{\pi Y (b - a)} \left( \frac{1}{a} - \frac{1}{b} \right)\).
Combining terms into a single fraction results in the final formula:
\( \triangle L = \frac{F L}{\pi ab Y} \).
This step-by-step integration helps understand how the changing radius affects total elongation and highlights the power of integral calculus in solving complex deformation problems.