A stone of mass 1.3 kg is being rotated in a horizontal plane as a conical pendulum with the help of a 140 cm long aluminium wire of cross - sectional area `1mm^(2)`. The wire makes an angle `theta=75^(@)` with the vertical. What is the increment in the length (in mm) of the wire? `["Young's modulus of aluminium " Y_(Al)=7xx10^(10)"N m"^(-2), `
` sin 75^(@)~~0.97, cos 75^(@)~~0.26, g=10ms^(-2)]`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the stone
The stone of mass \( m = 1.3 \, \text{kg} \) is subjected to two main forces:
1. The gravitational force \( F_g = mg \).
2. The tension \( T \) in the wire.

### Step 2: Calculate the gravitational force
Using the formula for gravitational force:
\[ F_g = mg = 1.3 \, \text{kg} \times 10 \, \text{m/s}^2 = 13 \, \text{N} \]

### Step 3: Relate the tension to the gravitational force
Since the wire makes an angle \( \theta = 75^\circ \) with the vertical, we can resolve the tension into its vertical and horizontal components. The vertical component of the tension balances the weight of the stone:
\[ T \cos(75^\circ) = mg \]
Thus, we can express the tension \( T \) as:
\[ T = \frac{mg}{\cos(75^\circ)} = \frac{13 \, \text{N}}{\cos(75^\circ)} \]
Using \( \cos(75^\circ) \approx 0.26 \):
\[ T = \frac{13}{0.26} \approx 50 \, \text{N} \]

### Step 4: Use Young's modulus to find the increment in length
Young's modulus \( Y \) is given by the formula:
\[ Y = \frac{F L}{A \Delta L} \]
Rearranging this gives:
\[ \Delta L = \frac{F L}{A Y} \]
Where:
- \( F = T \approx 50 \, \text{N} \)
- \( L = 140 \, \text{cm} = 1.4 \, \text{m} \)
- \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- \( Y = 7 \times 10^{10} \, \text{N/m}^2 \)

Substituting the values into the equation:
\[ \Delta L = \frac{50 \times 1.4}{1 \times 10^{-6} \times 7 \times 10^{10}} \]

### Step 5: Calculate the increment in length
Calculating the numerator:
\[ 50 \times 1.4 = 70 \]
Calculating the denominator:
\[ 1 \times 10^{-6} \times 7 \times 10^{10} = 7 \times 10^{4} \]
Thus:
\[ \Delta L = \frac{70}{7 \times 10^{4}} = 1 \times 10^{-3} \, \text{m} \]
Converting to millimeters:
\[ \Delta L = 1 \, \text{mm} \]

### Final Answer
The increment in the length of the wire is \( 1 \, \text{mm} \).

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