A stone of mass `m` is tied to one end of a wire of length L. The diameter of the wire is D and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle θ with the vertical. If Young's modulus of the wire is `Y`, Then the increase in the length of the wire is

Options

A. `(4mgl)/(piD^(2)Y)`
B. `(4mgl)/(piD^(2)Ysin theta)`
C. `(4mgl)/(piD^(2)Y cos theta)`
D. `(4mgl)/(piD^(2)Y tan theta)`

Explanation

To find the increase in the length of the wire when a stone of mass `m` is rotated in a horizontal plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Stone:**
The stone is subjected to two forces: the gravitational force (weight) acting downward, which is `mg`, and the tension `T` in the wire acting along the wire.

2. **Resolve the Tension Force:**
When the stone is rotated, it makes an angle `θ` with the vertical. The vertical component of the tension can be expressed as:
\[ T \cos \theta \]
Since the stone is in equilibrium in the vertical direction (no vertical motion), we can set the vertical component of the tension equal to the weight of the stone:
\[ T \cos \theta = mg \]
From this, we can solve for the tension `T`:
\[ T = \frac{mg}{\cos \theta} \]

3. **Calculate the Cross-Sectional Area of the Wire:**
The area `A` of the wire can be calculated using its diameter `D`:
\[ A = \frac{\pi D^2}{4} \]

4. **Use Young's Modulus Formula:**
Young's modulus `Y` is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} \]
Rearranging this gives us the increase in length `ΔL`:
\[ \Delta L = \frac{F L}{A Y} \]

5. **Substitute the Values:**
Substitute the tension `T` for `F`, the length `L`, and the area `A` into the equation for `ΔL`:
\[ \Delta L = \frac{T L}{A Y} = \frac{\left(\frac{mg}{\cos \theta}\right) L}{\frac{\pi D^2}{4} Y} \]

6. **Simplify the Expression:**
Simplifying the expression, we get:
\[ \Delta L = \frac{4mgL}{\pi D^2 Y \cos \theta} \]

### Final Result:
Thus, the increase in the length of the wire is given by:
\[ \Delta L = \frac{4mgL}{\pi D^2 Y \cos \theta} \]