JEE MAIN - Physics (2014 (Offline) - No. 23)
The pressure that has to be applied to the ends of a steel wire of length $$10$$ $$cm$$ to keep its length constant when its temperature is raised by $${100^ \circ }C$$ is:
(For steel Young's modulus is $$2 \times {10^{11}}\,\,N{m^{ - 2}}$$ and coefficient of thermal expansion is $$1.1 \times {10^{ - 5}}\,{K^{ - 1}}$$ )
$$2.2 \times {10^8}\,\,Pa$$
$$2.2 \times {10^9}\,\,Pa$$
$$2.2 \times {10^7}\,\,Pa$$
$$2.2 \times {10^6}\,\,Pa$$
Explanation
Young's modulus $$Y = {{stress} \over {strain}}$$
$$stress = Y \times strain$$
$$Stress$$ in steel wire $$=$$ Applied $$pressure$$
$$Pressure$$ $$=$$ $$stress$$ $$=$$ $$Y \times \,strain$$
$$Strain = {{\Delta L} \over L} = \alpha \Delta T$$ (As length is constant)
$$ = 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100$$
$$ = 2.2 \times {10^8}Pa$$
$$stress = Y \times strain$$
$$Stress$$ in steel wire $$=$$ Applied $$pressure$$
$$Pressure$$ $$=$$ $$stress$$ $$=$$ $$Y \times \,strain$$
$$Strain = {{\Delta L} \over L} = \alpha \Delta T$$ (As length is constant)
$$ = 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100$$
$$ = 2.2 \times {10^8}Pa$$
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