Question

A brass rod and a steel rod are both measured at $0^{\circ }$C. Their lengths are found to be 150 cm and 150.2 cm respectively. At what common temperature will their lengths be equal?
(${\alpha }_{steel}=12\times {10}^{-6}{/}^{\circ }$ C, ${\alpha }_{brass}=18\times {10}^{-6}{/}^{\circ }$ C)

• A. ${111.4}^{\circ }$C
• B. ${167.2}^{\circ }$C
• C. ${222.8}^{\circ }$C
• D. ${278.3}^{\circ }$C

Answer 1

Answer: (C)

${222.8}^{\circ }$C

We know, $l_2-l_1=\alpha \times l_1\times t$, where $l_1$ $=$ initial length, and $l_2$ $=$ final length; $t=$ temperature difference (as initial temperature $=$ 0)

Now final length for steel and brass rod will be same.

Generally written, $l_2=l_1(1+\alpha t)$

$l_2$ for brass $=$ $l_2$ for steel

$150(1+18\times {10}^{-6}\times t)=150.2(1+12\times {10}^{-6}\times t)$

Solving we get, $t=222.8163$ degree centigrade