Question

Magnitude of force developed by raising the temperature from $0^{\circ }C$ to ${100}^{\circ }C$ of an iron bar $100cm$ long and $1c{m}^2$ cross-section when it is held so that it is not permitted to expand or bend is($\alpha =12\times {10}^{-6}$ per ${}^{\circ }C$ and $Y=20\times {10}^{11}{\text{dyne/cm}}^2$)

• A. $10N$
• B. ${10}^{2}N$
• C. $24\times {10}^{3}N$
• D. ${10}^{4}N$

Answer 1

Answer: (D)

${10}^{4}N$

Expression of force is $F=\times Y\times A\times \alpha \times \Delta T$ where $Y$ is the Young's Modulus.

$\therefore F={10}^{11}]cfracN{m}^2\times 1\times {10}^{-4}{m}^2\times ({10}^{-5})\times 100°CF={10}^{4}N$

The magnitude of force $={10}^{4}N$