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A steel wire of uniform cross- sectional area $2\;mm^2$ in heated upto $50^{\circ}C$ and is stretched by clamping its two ends rigidly. The change in tension when the temperature falls from $50^{\circ}C$ to $30^{\circ}C$ is $( \alpha =1.1 \times 10^{-5} C)\;(Y=2.0 \times 10^{11} N/ m^2)$

\[\begin {array} {1 1} (a)\;88\;N \\ (b)\;5\;N \\ (c)\;1.5 \times 10^{10}\;N \\ (d)\;2.5 \times 2.5 \times 10^{10} \;N \end {array}\]

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1 Answer

Change in the tension $F= YA \;\alpha\; \Delta T$
$Y$- Young's modulus
$\Delta T$- change in temperature
$A$- area of cross - section
$F=(2.0 \times 10^{11})(1.1 \times 10^{-5}) \times 2 \times 10^{-6} \times 20$
Hence a is the correct answer


answered Oct 3, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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