A uniformly tapering conical wire is mae 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 to:

Options

A. `L(1+(1)/(9)(Mg)/(piRM^(3)))`
B. `L(1+(2)/(9)(Mg)/(piYR^(2)))`
C. `L(1+(2)/(3)(Mg)/(piYR^(2)))`
D. `L(1+(1)/(3)(Mg)/(piYR^(2)))`

Answer

To solve the problem of finding the equilibrium extended length of a uniformly tapering conical wire, we can follow these steps:

### Step 1: Understand the Geometry of the Wire
The wire is conical, with the upper radius \( R \) and the lower radius \( 3R \). The unextended length of the wire is \( L \).

### Step 2: Determine the Radius as a Function of Length
Since the wire tapers, we need to express the radius \( r \) at a distance \( x \) from the upper end. The relationship can be derived from similar triangles.

Using the geometry:
\[ \frac{3R - r}{L} = \frac{3R - R}{L} = \frac{2R}{L} \]
From this, we can express \( r \):
\[ r = 3R - \frac{2R}{L} x \]

### Step 3: Calculate the Extension of a Small Element
Consider a small element of length \( dx \) at a distance \( x \) from the upper end. The force acting on this element due to the weight \( M \) is \( mg \).

The extension \( d\delta l \) of this small element is given by:
\[ d\delta l = \frac{F \cdot dx}{A \cdot Y} \]
Where:
- \( F = mg \)
- \( A = \pi r^2 = \pi \left(3R - \frac{2R}{L} x\right)^2 \)
- \( Y \) is the Young's modulus.

Thus, we have:
\[ d\delta l = \frac{mg \cdot dx}{\pi \left(3R - \frac{2R}{L} x\right)^2 Y} \]

### Step 4: Integrate to Find Total Extension
To find the total extension \( \delta l \), we integrate from \( x = 0 \) to \( x = L \):
\[ \delta l = \int_0^L \frac{mg \cdot dx}{\pi \left(3R - \frac{2R}{L} x\right)^2 Y} \]

### Step 5: Solve the Integral
Let \( u = 3R - \frac{2R}{L} x \), then \( du = -\frac{2R}{L} dx \) or \( dx = -\frac{L}{2R} du \). The limits change accordingly:
- When \( x = 0 \), \( u = 3R \)
- When \( x = L \), \( u = R \)

Now substituting:
\[ \delta l = \int_{3R}^{R} \frac{mg \cdot \left(-\frac{L}{2R}\right) du}{\pi u^2 Y} \]
This simplifies to:
\[ \delta l = \frac{mgL}{2\pi Y} \int_{R}^{3R} \frac{du}{u^2} \]
Calculating the integral:
\[ \int \frac{du}{u^2} = -\frac{1}{u} \]
Thus:
\[ \delta l = \frac{mgL}{2\pi Y} \left[-\frac{1}{u}\right]_{R}^{3R} = \frac{mgL}{2\pi Y} \left(-\frac{1}{3R} + \frac{1}{R}\right) = \frac{mgL}{2\pi Y} \left(\frac{2}{3R}\right) \]
So:
\[ \delta l = \frac{mgL}{3\pi Y R} \]

### Step 6: Find the Total Extended Length
The total extended length \( L' \) is the original length plus the extension:
\[ L' = L + \delta l = L + \frac{mgL}{3\pi Y R} = L \left(1 + \frac{mg}{3\pi Y R}\right) \]

### Final Answer
The equilibrium extended length of the wire is:
\[ L' = L \left(1 + \frac{mg}{3\pi Y R}\right) \]