Question

What would be the temperature rise of a brass rod $(\alpha =2\times {10}^{-5}{}^{\circ }{C}^{-1})$ so as to increase its length by 1$\%$?

• A. 1000 ⁰C
• B. 500 ⁰C
• C. 200 ⁰C
• D. 100 ⁰C

Answer 1

The correct option is B

500 ⁰C

$\alpha$ = coefficient of linear expansion

Initial length = $l_0$

Final length = $l$

Change in temperature = $t$

Change in length = $l-l_0=0.01l_0$

= $\alpha \times l_0\times t=0.01l_0$

= $\alpha \times t=0.01$

= $t=\frac{0.01}{2}\times {10}^5$

= $t=\frac{1000}{2}=500{}^{\circ }C$