Question

Coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$. Lengths of brass and steel rods are $ l_1$ and $l_2$ respectively. If $(l_2 - l_1)$ is maintained same at all temperatures, which one of the following relations holds good ?

• A. $\alpha_1 l^2_2 = \alpha_2 l_1^2$
• B. $\alpha_1^2 l_2 = \alpha_2^2 l_1$
• C. $\alpha_1 l_1 = \alpha_2 l_2$
• D. $\alpha_1 l_2 = \alpha_2 l_1$

Answer 1

The Correct Option is C

To solve this problem, we need to understand the principle of linear expansion of materials due to temperature changes. The coefficient of linear expansion, $\alpha$, represents how much a material expands per degree change in temperature. If the initial length of a rod is $l_0$ and the temperature changes by $\Delta T$, the change in length $\Delta l$ can be calculated by:

$$ \Delta l = \alpha \cdot l_0 \cdot \Delta T $$

Given:

• Coefficient of linear expansion for brass, $\alpha_1$
• Coefficient of linear expansion for steel, $\alpha_2$
• Initial length of brass rod, $l_1$
• Initial length of steel rod, $l_2$
• Difference in lengths $(l_2 - l_1)$ remains the same at all temperatures

The condition implies that even after expansion due to temperature changes, the difference in lengths should remain the same:

$$ (l_2 + \Delta l_2) - (l_1 + \Delta l_1) = l_2 - l_1 $$

Substituting the formulas for linear expansion, we have:

$$ (l_2 + \alpha_2 l_2 \Delta T) - (l_1 + \alpha_1 l_1 \Delta T) = l_2 - l_1 $$

Simplifying the equation:

$$ l_2 + \alpha_2 l_2 \Delta T - l_1 - \alpha_1 l_1 \Delta T = l_2 - l_1 $$

Cancelling $(l_2 - l_1)$ from both sides, we get:

$$ \alpha_2 l_2 \Delta T = \alpha_1 l_1 \Delta T $$

Further simplifying, we find:

$$ \alpha_1 l_1 = \alpha_2 l_2 $$

Therefore, the correct relation that holds true is:

$$ \alpha_1 l_1 = \alpha_2 l_2 $$

This equation ensures that the expansion is such that the difference in lengths remains constant.