Question

Two rods, one of aluminium and the other of steel, having initial length $l_1$ and $l_2$ respectively are connected together to form a single rod of length $l_1+l_2$. The coefficients of linear expansion for aluminium and steel are ${\alpha }_a$ and ${\alpha }_s$ respectively. If the length of each rod increases by the same amount when their temperatures are raised by $t{}^{\circ }\text{C}$, then find the ratio $l_1/(l_1+l_2)$

• A. ${\alpha }_s/{\alpha }_a$
• B. ${\alpha }_a/{\alpha }_s$
• C. ${\alpha }_s/({\alpha }_a+{\alpha }_s)$
• D. ${\alpha }_a/({\alpha }_a+{\alpha }_s)$

Answer 1

The correct option is C ${\alpha }_s/({\alpha }_a+{\alpha }_s)$

Given:
For aluminium rod,
Initial length $=l_1$
Co-efficient of linear expansion $={\alpha }_a$
For steel rod,
Initial length $=l_2$
Co-efficient of linear expansion $={\alpha }_s$

To find:
Ratio, $\frac{l_1}{l_1+l_2}=?$

Given, change in length of both the rods are equal when the temperature of both the rods are raised by $t{}^{\circ }\text{C}$
$\Rightarrow \Delta l_a=\Delta l_s$
$\Rightarrow l_1{\alpha }_{a}t=l_2{\alpha }_{s}t$
[ from formula of linear expansion ]
$\Rightarrow \frac{l_2}{l_1}=\frac{{\alpha }_a}{{\alpha }_s}$
$\Rightarrow \frac{l_2}{l_1}+1=\frac{{\alpha }_a}{{\alpha }_s}+1$
$\Rightarrow \frac{l_2+l_1}{l_1}=\frac{{\alpha }_a+{\alpha }_s}{{\alpha }_s}$
$\Rightarrow \frac{l_1}{l_1+l_2}=\frac{{\alpha }_s}{{\alpha }_a+{\alpha }_s}$