A wire longates by 9 mm when a load of 10kg is suspended from it. What is the elongation when its radius is doubled, iff all other quantities are same as before ?

Explanation

To solve the problem of elongation of a wire when its radius is doubled, we can follow these steps:

### Step 1: Understand the Given Information
We know that a wire elongates by 9 mm when a load of 10 kg is suspended from it. We need to find the elongation when the radius of the wire is doubled, keeping all other quantities the same.

### Step 2: Identify the Relevant Formula
The elongation (ΔL) of a wire under a load can be calculated using the formula:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Where:
- \( F \) = Force applied (weight of the load)
- \( L \) = Original length of the wire
- \( A \) = Cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

### Step 3: Calculate the Force
The force \( F \) due to the weight of the load can be calculated as:

\[ F = m \cdot g \]

Where \( m = 10 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \).

### Step 4: Calculate the Cross-Sectional Area
The cross-sectional area \( A \) of the wire is given by:

\[ A = \pi r^2 \]

Where \( r \) is the radius of the wire.

### Step 5: Set Up the Equations for Both Cases
1. **Case 1** (Original radius \( r \)):
\[ \Delta L_1 = \frac{F \cdot L}{\pi r^2 \cdot Y} \]
Given that \( \Delta L_1 = 9 \, \text{mm} \).

2. **Case 2** (Doubled radius \( 2r \)):
\[ \Delta L_2 = \frac{F \cdot L}{\pi (2r)^2 \cdot Y} = \frac{F \cdot L}{\pi \cdot 4r^2 \cdot Y} \]

### Step 6: Relate the Two Cases
To find the relationship between \( \Delta L_1 \) and \( \Delta L_2 \), we can divide the two equations:

\[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{F \cdot L}{\pi r^2 \cdot Y}}{\frac{F \cdot L}{\pi \cdot 4r^2 \cdot Y}} = \frac{4}{1} \]

This implies:

\[ \Delta L_2 = \frac{\Delta L_1}{4} \]

### Step 7: Substitute the Known Value
Now, substituting \( \Delta L_1 = 9 \, \text{mm} \):

\[ \Delta L_2 = \frac{9 \, \text{mm}}{4} = 2.25 \, \text{mm} \]

### Conclusion
The elongation when the radius is doubled is:

\[ \Delta L_2 = 2.25 \, \text{mm} \]