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Step-by-Step Solution
Step 1: Understand the Physical Situation
We have a conical wire of original (unextended) length $L$. The wire’s top radius is $R$, and the bottom radius is $3R$. The wire is fixed at the top to a rigid support, and a mass $M$ is hung from its bottom end.
When the mass $M$ is attached, the wire stretches due to the tensile force $F = Mg$.
Step 2: Express the Elemental Elongation
To find the total change in length, we consider a thin cross-sectional element of thickness $dx$ at a distance $x$ from the top of the wire. The formula for the small extension $d(\Delta L)$ in that element is given by:
$$
d(\Delta L) = \frac{F \, dx}{A(x) \, Y}
$$
where
• $F = Mg$ is the tensile force (constant throughout the wire),
• $A(x)$ is the cross-sectional area at distance $x$,
• $Y$ is Young’s modulus of the wire.
Step 3: Find the Radius and Cross-Sectional Area at a General Position
Because the wire tapers uniformly from radius $R$ at $x=0$ to $3R$ at $x=L$, the radius $r(x)$ increases linearly with $x$. The slope of increase is:
$$
\frac{3R - R}{L} = \frac{2R}{L}.
$$
Thus, at a distance $x$ from the top,
$$
r(x) = R + \left(\frac{2R}{L}\right) x \;=\; R\left(1 + \frac{2x}{L}\right).
$$
Consequently, the cross-sectional area at $x$ is:
$$
A(x) = \pi \,[r(x)]^2 = \pi\,R^2 \left(1 + \frac{2x}{L}\right)^2.
$$
Step 4: Set Up the Integral for Total Elongation
The total extension $\Delta L$ is found by integrating the expression for $d(\Delta L)$ from $x=0$ to $x=L$:
$$
\Delta L = \int_{0}^{L} \frac{F \, dx}{A(x) \, Y}
= \int_{0}^{L} \frac{Mg \, dx}{\pi R^2 \left(1 + \frac{2x}{L}\right)^2 \, Y}.
$$
Factor out the constants:
$$
\Delta L
= \frac{Mg}{\pi R^2 Y} \int_{0}^{L} \frac{dx}{\left(1 + \frac{2x}{L}\right)^2}.
$$
Step 5: Evaluate the Integral
Use the substitution $u = 1 + \tfrac{2x}{L}$, which gives $du = \tfrac{2}{L} dx$ or $dx = \tfrac{L}{2}\,du$. When $x=0$, $u=1$. When $x=L$, $u=3$. Hence the integral becomes:
$$
\Delta L = \frac{Mg}{\pi R^2 Y}
\int_{u=1}^{u=3} \frac{1}{u^2} \cdot \frac{L}{2} \, du.
$$
This simplifies to:
$$
\Delta L
= \frac{Mg L}{2 \pi R^2 Y} \int_{1}^{3} u^{-2}\,du
= \frac{Mg L}{2 \pi R^2 Y}\left[ -\,u^{-1} \right]_{1}^{3}.
$$
Evaluate the definite integral:
$$
-\,u^{-1}\Bigg|_{1}^{3} = -\frac{1}{3} - \left(-1\right)
= 1 - \frac{1}{3}
= \frac{2}{3}.
$$
Thus,
$$
\Delta L = \frac{Mg L}{2 \pi R^2 Y} \cdot \frac{2}{3}
= \frac{Mg L}{3 \pi R^2 Y}.
$$
Step 6: Find the Final Length
The final, extended length of the wire is its original length plus the extension:
$$
L_{\text{final}}
= L + \Delta L
= L + \frac{Mg L}{3 \pi R^2 Y}
= L \left( 1 + \frac{Mg}{3 \pi R^2 Y} \right).
$$
Therefore, the equilibrium extended length of the conical wire is:
$$
\boxed{
L \left( 1 + \frac{1}{3}\frac{Mg}{\pi Y R^2} \right).
}
$$
This matches Option 2 from the given choices.
Step 7: Provided Diagram
