Question: 44. A metallic bar of Young’s modulus, 0.5 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} and coefficient of linear thermal expansion 10^{-5}{ }^{\circ} \mathrm{C}^{-1}, length 1 m and area of cross-section 10^{-3} \mathrm{~m}^{2} is heated from 0^{\circ} \mathrm{C} to 100^{\circ} \mathrm{C} without expansion or bending. The compressive force developed in it is:
(1) 5 \times 10^{3} \mathrm{~N}
(2) 50 \times 10^{3} \mathrm{~N}
(3) 100 \times 10^{3} \mathrm{~N}
(4) 2 \times 10^{3} \mathrm{~N}
Answer: Option (2)
Explanation:
When a metallic bar is heated and its expansion is completely prevented,
thermal stress is developed in the bar.
The thermal stress produced is given by
\text{Stress} = Y \alpha \Delta T,
where Y is Young’s modulus, \alpha is the coefficient of linear expansion,
and \Delta T is the rise in temperature.
Given Y = 0.5 \times 10^{11} \mathrm{~N\,m^{-2}},
\alpha = 10^{-5} \,^{\circ}\mathrm{C}^{-1},
and \Delta T = 100^{\circ}\mathrm{C}.
So,
\text{Stress} = 0.5 \times 10^{11} \times 10^{-5} \times 100 = 0.5 \times 10^{8} \mathrm{~N\,m^{-2}}.
The compressive force developed is given by
F = \text{Stress} \times A,
where A is the area of cross-section.
Given A = 10^{-3} \mathrm{~m^{2}},
F = 0.5 \times 10^{8} \times 10^{-3} = 0.5 \times 10^{5} \mathrm{~N}.
F = 50 \times 10^{3} \mathrm{~N}.
Hence, the compressive force developed in the bar is 50 \times 10^{3} \mathrm{~N},
which corresponds to option (2).