A metallic bar of Young’s modulus, 0.5 × 10¹¹ N m^–2 and coefficient of linear thermal expansion 10^–5 °C^–1 , length 1 m and area of cross-section 10^–3 m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is :

To find the compressive force developed in the metallic bar when it is heated without allowing it to expand, we can use the concepts of thermal stress and Young's modulus.

First, let us identify the given values from the problem:
Young's modulus of the bar, $Y=0.5\times {10}^{11}{\text{N m}}^{-2}$
Coefficient of linear thermal expansion, $\alpha ={10}^{-5}{\text{°C}}^{-1}$
Original length of the bar, $L=1\text{m}$
Area of cross-section, $A={10}^{-3}{\text{m}}^2$
Change in temperature, $\Delta T=100\text{°C}-0\text{°C}=100\text{°C}$

When a bar is heated, it undergoes thermal expansion. The change in length $\Delta L$ due to heating is given by the formula:
$\Delta L=L\alpha \Delta T$

Since the bar is prevented from expanding or bending, a compressive strain is developed in the bar. This thermal strain is defined as the fractional change in length that is prevented:
$\text{Strain}=\frac{\Delta L}{L}=\alpha \Delta T$

Young's modulus ($Y$) is the ratio of stress to strain:
$Y=\frac{\text{Stress}}{\text{Strain}}$

Therefore, the thermal stress developed in the bar is:
$\text{Stress}=Y\times \text{Strain}=Y\alpha \Delta T$

Since stress is also defined as force ($F$) per unit cross-sectional area ($A$), we can express the compressive force developed as:
$F=\text{Stress}\times A=YA\alpha \Delta T$

Now, let us substitute the given values into the formula to calculate the force:
$F=(0.5\times {10}^{11})\times \left({10}^{-3}\right)\times \left({10}^{-5}\right)\times 100$

Simplify the exponents:
$F=0.5\times {10}^{11-3-5}\times {10}^2$
$F=0.5\times {10}^3\times {10}^2$
$F=0.5\times {10}^5\text{N}$

Expressing this in the form matching the options:
$F=50\times {10}^3\text{N}$

Thus, the compressive force developed in the bar is $50\times {10}^3\text{N}$.