MyCompetitiveExams Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.
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:
F = Y · A · α · ΔT = (0.5 × 10¹¹) × (10⁻³) × (10⁻⁵) × (100) = 0.5 × 10⁵ = 50 × 10³ N!
When a metallic bar of length L is heated through temperature difference ΔT, its natural unrestricted thermal expansion would be ΔL = L α ΔT. Because the bar is rigidly held between immovable walls with no expansion allowed, a compressive mechanical strain is developed:
From Hooke’s law, compressive thermal stress σ is the product of Young’s modulus Y and thermal strain:
The compressive force F developed in the cross-section is the stress multiplied by the cross-sectional area A:
Substitute Y = 0.5 × 10¹¹ N m⁻², A = 10⁻³ m², α = 10⁻⁵ °C⁻¹, and ΔT = 100°C:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Fails to multiply by 0.5 in Young’s modulus.
Arithmetic power error.
Off by a factor of 10.
Exact evaluation: 0.5 × 10⁵ N = 50 × 10³ N.
Students search for where to multiply length L = 1 m, sometimes dividing or multiplying erroneously.
"Thermal stress in a rigidly clamped rod is independent of its length and depends solely on Young’s modulus, coefficient of expansion, and temperature rise (σ = Y α ΔT)."
Test your speed and accuracy in the authentic 200-minute timed NTA simulator.
Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)
Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.