Question

Two rods one of aluminium and the other made of steel, having initial lengths and are connected together to form a single rod of length If the length of each rod increases by the same amounts on increasing the temperature by 1°C. the ratio in terms of coefficients of linear expansions is given by:

• A. $\frac{{\alpha _a }}{{\alpha _s }}$
• B. $\frac{{\alpha _s }}{{\alpha _a }}$
• C. $\frac{{\alpha _s }}{{\alpha _a + \alpha _s }}$
• D. $\frac{{\alpha _a }}{{\alpha _a + \alpha _s }}$

Answer 1

Answer: (C)

$\frac{{\alpha _s }}{{\alpha _a + \alpha _s }}$

Answer 2

Let the initial lengths be $l_1$ (aluminium) and $l_2$ (steel) with coefficients of linear expansion $\alpha_a$ and $\alpha_s$, respectively. When the temperature increases by 1°C, the increases in lengths are:

$$\Delta l_1 = l_1 \alpha_a, \quad \Delta l_2 = l_2 \alpha_s.$$

Since the increases are equal,

$$l_1 \alpha_a = l_2 \alpha_s \quad \Rightarrow \quad \frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}.$$

The total length is $l_1 + l_2$. Express $l_2$ as:

$$l_2 = \frac{l_1 \alpha_a}{\alpha_s}.$$

Thus,

$$l_1 + l_2 = l_1\left(1 + \frac{\alpha_a}{\alpha_s}\right) = l_1 \cdot \frac{\alpha_s+\alpha_a}{\alpha_s}.$$

So,

$$\frac{l_1}{l_1+l_2} = \frac{l_1}{l_1 \cdot \frac{\alpha_a+\alpha_s}{\alpha_s}} = \frac{\alpha_s}{\alpha_a+\alpha_s}.$$