Question

To increase the length of brass rod by 2% its temperature should increase by ($\alpha = 0.0002^{o}C^{- 1}$)

• A. (a) 800 °C
• A. (b) 900 °C
• A. (c) 1000 °C
• A. (d) 1100 °C

Answer 1

(c) $\frac{\Delta L}{L} = 2\% = \frac{2}{100}$

$\alpha = 0.00002{^\circ}C^{- 1}$

As $\Delta L = \alpha L\Delta T$

Or $\Delta T = \frac{\Delta T}{\alpha L} = \frac{2}{100 \times 0.00002} = \frac{1}{0.001}$

$= 10^{3} = 1000{^\circ}C$