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Question

Physics Question on Youngs double slit experiment

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 :

A

5 × 103 N

B

50 × 103 N

C

100 × 103 N

D

2 × 103 N

Answer

50 × 103 N

Explanation

Solution

Thermal stress is:
Stress=YαΔT\text{Stress} = Y \cdot \alpha \cdot \Delta T.
The force is:
F=StressA=YαΔTAF = \text{Stress} \cdot A = Y \cdot \alpha \cdot \Delta T \cdot A.
Substituting values:
F=0.5×1011105100103=50×103NF = 0.5 \times 10^{11} \cdot 10^{-5} \cdot 100 \cdot 10^{-3} = 50 \times 10^3 \, N.