An thin rod of negligible mass and area of cross - section `4xx10^(-6)m^(2)`, suspended vertically from one end, has a length of 0.5 m at `100^(@)C`. The rod is cooled to `0^(@)C` but prevented from contracting by attaching a mass at the lower end. The value of this mass is
(Given, coefficient of linear expansion is `10^(-5^(@))C^(-1)`, Young's modulus is `Y=10^(11)Nm^(-2) and g=10ms^(-2))`

Answer

To solve the problem, we need to find the mass that must be attached to the lower end of a thin rod to prevent it from contracting when cooled from 100°C to 0°C. Here’s a step-by-step solution:

### Step 1: Calculate the change in length of the rod
The formula for the change in length (\(\Delta L\)) due to temperature change is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Where:
- \(L = 0.5 \, \text{m}\) (original length of the rod)
- \(\alpha = 10^{-5} \, \text{°C}^{-1}\) (coefficient of linear expansion)
- \(\Delta T = 100 \, \text{°C} - 0 \, \text{°C} = 100 \, \text{°C}\)

Substituting the values:
\[ \Delta L = 0.5 \cdot 10^{-5} \cdot 100 = 0.5 \cdot 10^{-5} \cdot 10^2 = 0.5 \cdot 10^{-3} = 5 \times 10^{-4} \, \text{m} \]

### Step 2: Relate the force to the change in length
The force (\(F\)) required to prevent the contraction of the rod can be expressed using Young's modulus (\(Y\)):
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Where:
- \(F = mg\) (force due to the mass)
- \(A = 4 \times 10^{-6} \, \text{m}^2\) (cross-sectional area)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)

Rearranging the equation to find the mass (\(m\)):
\[ F = Y \cdot \frac{\Delta L \cdot A}{L} \]
Substituting \(F = mg\):
\[ mg = Y \cdot \frac{\Delta L \cdot A}{L} \]
Thus, we can express \(m\) as:
\[ m = \frac{Y \cdot A \cdot \Delta L}{g \cdot L} \]

### Step 3: Substitute the values into the equation
Now substituting the known values:
- \(Y = 10^{11} \, \text{N/m}^2\)
- \(A = 4 \times 10^{-6} \, \text{m}^2\)
- \(\Delta L = 5 \times 10^{-4} \, \text{m}\)
- \(g = 10 \, \text{m/s}^2\)
- \(L = 0.5 \, \text{m}\)

Substituting these values into the equation for \(m\):
\[ m = \frac{10^{11} \cdot (4 \times 10^{-6}) \cdot (5 \times 10^{-4})}{10 \cdot 0.5} \]
Calculating the denominator:
\[ 10 \cdot 0.5 = 5 \]
Now substituting this back into the equation:
\[ m = \frac{10^{11} \cdot 4 \times 10^{-6} \cdot 5 \times 10^{-4}}{5} \]
The \(5\) in the numerator and denominator cancels out:
\[ m = 10^{11} \cdot 4 \times 10^{-6} \times 10^{-4} \]
\[ m = 4 \times 10^{11 - 6 - 4} = 4 \times 10^{1} = 40 \, \text{kg} \]

### Final Answer
The mass that must be attached to the lower end of the rod to prevent it from contracting is:
\[ \boxed{40 \, \text{kg}} \]