Question

The coefficient of linear expansion of steel and brass are $11\times {10}^{-6}{/}^{\circ }\text{C}$ and $19\times {10}^{-6}{/}^{\circ }\text{C}$, respectively. If their differences in lengths at all temperatures has to kept constant at $30\text{cm}$, their lengths at $0^{\circ }\text{C}$ should be

• A. $71.25\text{cm}$ and $41.25\text{cm}$
• B. $82\text{cm}$ and $52\text{cm}$
• C. $92\text{cm}$ and $62\text{cm}$
• D. $62.25\text{cm}$ and $32.25\text{cm}$

Answer 1

The correct option is A $71.25\text{cm}$ and $41.25\text{cm}$

Let us suppose $L_{o1}$ and $L_{o2}$ is the length of rod of steel and brass at $0^{\circ }\text{C}$ respectively and $\Delta T$ change in temperature.
We know, $L=L_o(1+\alpha \Delta T)$ [linear expansion]
For steel,
$L_1=L_{o1}(1+11\times {10}^{-6}\Delta T)$ $[{\alpha }_{\text{steel}}=11\times {10}^{-6}{/}^{\circ }\text{C},\text{given}]$
$=L_{o1}+11\times {10}^{-6}\times \Delta T\times L_{o1}$...(1)
For brass,
$L_2=L_{o2}(1+19\times {10}^{-6}\Delta T)$ $[{\alpha }_{\text{brass}}=19\times {10}^{-6}{/}^{\circ }\text{C},\text{given}]$
$=L_{o2}+19\times {10}^{-6}\times \Delta T\times L_{o2}$...(2)
On subtracting (1) from (2)
$L_2-L_1=L_{o2}-L_{o1}+19\times {10}^{-6}\times \Delta T\times L_{o2}-11\times {10}^{-6}\times \Delta T\times L_{o1}$
$\Rightarrow L_{o1}=\frac{19\times {10}^{-6}}{11\times {10}^{-6}}{L}_{o2}$
[$L_2-L_1=L_{o2}-L_{o1}$ as difference in length is constant, given]
$\Rightarrow L_{o1}=\frac{19}{11}{L}_{o2}$ ... (3)
Also, $L_{o1}-L_{o2}=30\text{cm}$ [Given]
$\Rightarrow \frac{19}{11}{L}_{o2}-L_{o2}=30$ [from (3)]
$\Rightarrow \frac{8}{11}{L}_{o2}=30$
$\Rightarrow L_{o2}=41.25\text{cm}$ ... (4)
On putting (4) in (3),
$L_{o1}=71.25\text{cm}$

$\therefore$ Lengths of the rods at $0^{\circ }\text{C}$ are $71.25\text{cm}$ and $41.25\text{cm}$ of steel and brass respectively.