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An iron bar is heated from $\;0^{0}C\;$ to $\;50^{0}C\;$ . If $\;\gamma\;$ of iron = $\;2 \times 10^{12} \;dyne/cm^{2}\;$ and $\;\alpha\;$ of iron is $\;12 \times 10^{-6}/^{0}C\;$ . The stress developed in the rod if it is prevented from expanding is

$(a)\;12 \times 10^{5}\;N/m^2\qquad(b)\;12 \times 10^{6}\;N/m^2\qquad(c)\;12 \times 10^{7}\;N/m^2\qquad(d)\;12 \times 10^{8}\;N/m^2$

1 Answer

Answer : $\;12 \times 10^{7}\;N/m^2$
Explanation :
$\bigtriangleup l = l \alpha \bigtriangleup t$
$stress = \gamma \large\frac{\bigtriangleup l}{l}$
$=\gamma \alpha \bigtriangleup t$
answered Mar 24, 2014 by yamini.v

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