Question

The design of a physical instrument requires that there be a constant difference in length of 10 cm between an iron rod and a copper cylinder laid side by side at all temperatures. If ${\alpha }_{Fe}=11\times {{10}^{-6}}^{\circ }{C}^{-1}$ and ${\alpha }_{cu}=17\times {10}^{-6}{}^{\circ }{C}^{-1}$ their lengths are

• A. 28.3 cm, 18.3 cm
• B. 23.8 cm, 13.8 cm
• C. 23.9 cm, 13.9 cm
• D. 27.5 cm, 17.5 cm

Answer 1

The correct option is A 28.3 cm, 18.3 cm

Since a constant difference in length of 10 cm between an iron rod and a copper cylinder is required therefore
$L_{Fe}-L_{Cu}=10cm\dots \dots \left(i\right)$
or $\Delta L_{Fe}-\Delta L_{Cu}=O$ $\therefore$ $\Delta L_{Fe}=\Delta L_{Cu}$
i.e., Linear expansion of iron rod = Linear expansion of copper cylinder
$\Rightarrow$ $L_{Fe}\times {\alpha }_{Fe}\times \Delta T=L_{Cu}\times {\alpha }_{Cu}\times \Delta T$ $\Rightarrow$ $\frac{L_{Fe}}{L_{Cu}}=\frac{{\alpha }_{Cu}}{{\alpha }_{Fe}}=\frac{17}{11}$$\therefore$ $\frac{L_{Fe}}{L_{Cu}}=\frac{17}{11}\dots \dots \left(ii\right)$
From (i) and (ii) $L_{Fe}=28.3cm,L_{Cu}=18.3cm$.