An iron ring of radius 50 cm and rhickness 1 cm is heated sufficiently and then put on a wooden carrrrt whell of the same radius at `40^(@)C` WHAT IS the force per unit area wih which the ring grips over the whell cooled to `27^(@)C` ? Toung's modulus of iron`=2xx10^(11)` `N m^(-2)` and liner expansivity of iron `=11xx10^(-6) K^(-1)` Neglect thermal contraction of the wooden wheel.

Answer

To solve the problem, we need to determine the force per unit area (pressure) with which the iron ring grips the wooden wheel after cooling. We will follow these steps:

### Step 1: Determine the change in temperature
The initial temperature of the iron ring is \( T_i = 40^\circ C \) and the final temperature after cooling is \( T_f = 27^\circ C \).

\[ \Delta T = T_f - T_i = 27^\circ C - 40^\circ C = -13^\circ C \]

### Step 2: Calculate the change in radius of the iron ring
Using the linear expansivity of iron (\( \alpha = 11 \times 10^{-6} \, K^{-1} \)), we can calculate the change in radius (\( \Delta R \)) due to the temperature change.

\[ \Delta R = R \cdot \alpha \cdot \Delta T \]

Where:
- \( R = 0.5 \, m \) (radius in meters)
- \( \Delta T = -13 \, K \)

Substituting the values:

\[ \Delta R = 0.5 \cdot (11 \times 10^{-6}) \cdot (-13) = -7.15 \times 10^{-6} \, m \]

### Step 3: Calculate the stress in the iron ring
The stress (\( \sigma \)) in the iron ring can be calculated using Young's modulus (\( Y = 2 \times 10^{11} \, N/m^2 \)) and the change in radius.

\[ \sigma = Y \cdot \text{strain} = Y \cdot \frac{\Delta R}{R} \]

Substituting the values:

\[ \sigma = 2 \times 10^{11} \cdot \frac{-7.15 \times 10^{-6}}{0.5} = -2.86 \times 10^{6} \, N/m^2 \]

### Step 4: Calculate the force per unit area (pressure)
The force per unit area (pressure, \( P \)) is equal to the stress in the ring:

\[ P = \sigma = -2.86 \times 10^{6} \, N/m^2 \]

Since we are interested in the magnitude of the force per unit area, we take the absolute value:

\[ P = 2.86 \times 10^{6} \, N/m^2 \]

### Final Result
The force per unit area with which the ring grips over the wheel is approximately:

\[ P \approx 2.86 \times 10^{6} \, N/m^2 \]