Question

A metal rod has length of 1m at 30 C. '$\alpha$' of metal is $2.5\times {10}^5/C$.The temeperature at which it will be shortened by 1 mm is

• A. -30°C
• B. -40°C
• C. -10°C
• D. 10°C

Answer 1

The correct option is C -10°C

Coefficient of linear expansion:

$\alpha =\frac{\Delta L}{L\Delta T}$

$\Delta T=\frac{\Delta T}{L\alpha }$

$=\frac{{10}^{-3}}{1\times 2.5\times {10}^{-5}}$

$\Delta T=T-T^{\prime }$

$T^{\prime }=T-\Delta T=30-40=-10°C$