Figure shows graph between stress and strain for a uniform wire at two different temperatures. Then

T_{1} > T_{2}

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T_{2} > T_{1}

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T_{1} = T_{2}

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None of these

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Solution

The correct option is B $T_{1} > T_{2}$
The more elastic wire then the deformation is less, so the young modulus is more.
From graph, The young modulus of $T_{2}$ is more than the $T_{2}$
$Y_{2} > Y_{1}$
The temperature of the wire if inversely proportional to the Young's modulus,
$T \propto \frac{1}{Y}$
So, $T_{1} > T_{2}$
The correct option is A.

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Figure shows graph between stress and strain for a uniform wire at two different temperatures. Then

The correct option is B T1>T2The more elastic wire then the deformation is less, so the young modulus is more.From graph, The young modulus of T2 is more than the T2Y2>Y1The temperature of the wire if inversely proportional to the Young's modulus, T∝1YSo, T1>T2The correct option is A.