Question

The length of a steel rod is $5cm$ greater than that of a brass rod. If this difference in their lengths is to remain the same at all temperatures, then the length of brass rod will be:

[Coefficient of linear expansion for steel and brass are $12\times {10}^{-6}{/}^{\circ }C$ and $18\times {10}^{-6}{/}^{\circ }C$ respectively]

• A. $20cm$
• B. $15cm$
• C. $5cm$
• D. $10cm$

Answer 1

The correct option is D $10cm$

Given : ${\alpha }_s=12\times {10}^{-6}{/}^{o}C$ ${\alpha }_B=18\times {10}^{-6}{/}^{o}C$

Let the initial length of steel rod and brass rod be $L_s$ and $L_B$ respectively.

$\therefore$ $L_s-L_B=5$ cm ............(1)

Let the temperature of both the rods is increased by $\Delta T$.

$\therefore$ New length of steel rod $L_s^{\prime }=L_s(1+{\alpha }_s\Delta T)$

Also new length of brass rod $L_B^{\prime }=L_B(1+{\alpha }_B\Delta T)$

According to the question, $L_s^{\prime }-L_B^{\prime }=5$

$\therefore$ $L_s(1+{\alpha }_s\Delta T)-$ $L_B(1+{\alpha }_B\Delta T)=5$

OR $L_s-L_B+(L_s{\alpha }_s-L_B{\alpha }_B)\Delta T=5$ ..........(2)

Using (1) in (2) we get, $(L_s{\alpha }_s-L_B{\alpha }_B)\Delta T=0$

$⟹$ $\frac{L_s}{L_B}=\frac{{\alpha }_B}{{\alpha }_s}=\frac{18\times {10}^{-6}}{12\times {10}^{-6}}$ $⟹L_s=1.5L_B$ ..........(3)

From (1) and (3), $1.5L_B-L_B=5$

OR $0.5L_B=5$ $⟹L_B=10$ cm