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A metal rod has a length of 1m at 30°C. ‘ $α$ ’ of metal is 2.5 x 10^-5/°C. The temperature at which it will be shortened by 1mm is

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– 30°C

– 10°C

– 40°C

10°C

answer is C.

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Detailed Solution

$α = \frac{l_{2} - l_{1}}{l_{1} (t_{2} - t_{1})} . t_{2} - t_{1} = \frac{l_{2} - l_{1}}{l_{1} α}$
$t_{2} - 30 = \frac{- 1 \times 10^{- 3}}{1 \times 2 .5 \times 10^{- 5}}$ =-40^oC
$t_{2} = - 10 ° C$

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