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:

Given data:

Young’s modulus $E$ = 0.5 × 10¹¹ N·m^–2

Coefficient of linear thermal expansion $\alpha$ = 10^–5 °C^–1

Original length $L$ = 1 m

Cross‑sectional area $A$ = 10^–3 m²

Temperature increase $\mathrm{\Delta T}$ = 100 °C

Step 1: Thermal strain that would occur if the bar were free to expand

\epsilon =\alpha \mathrm{\Delta T}

Substituting the values:

\epsilon =10–5·100=10–3

Step 2: Because the bar is prevented from expanding, the actual strain is zero. The required compressive stress is therefore

\sigma =–E\epsilon

Negative sign indicates compression.

Compute the stress:

\sigma =–(0.5 \times 1011)·(10–3)

\sigma =–0.5 \times 108 N·m–2

Since $\mathrm{0.5 \times 10⁸}$ = 5 × 10⁷, we have

\sigma =–5 \times 107 N·m–2

Step 3: Compressive force $F$ is stress multiplied by the cross‑sectional area:

F=\sigma ·A

Substitute $\sigma$ and $A$:

F=(5 \times 107)·(10–3)

F=5 \times 104 N

Step 4: Express the result in the form requested in the options:

F=50 \times 103 N

Therefore, the compressive force developed in the bar is 50 × 10³ N.