The coefficient of linear expansion of brass and steel are $\alpha_{1}$ and $\alpha_{2}$ . If we take a brass rod of length $l_{1}$ and steel rod of length $l_{2}$ at 0°C, their difference in length $\left(l_{2} - l_{1}\right)$ will remain the same at a temperature if
Text Solution
Verified by ExpertsD
$L_{2} = \ell_{2} (1 + \alpha_{2} \Delta \theta) \quad \ldots \quad (1)$
$\mathrm{L}_{1} = \ell_{1} (1 + \alpha_{1} \Delta \theta) \quad \ldots \quad (2)$
Subtract eq (2) by (1), we get
$(L_2 - L_1) = (\ell_2 - \ell_1) + \Delta \theta (\ell_2 \alpha_2 - \ell_1 \alpha_1)$
$\Delta \theta (\ell_2 \alpha_2 - \ell_1 \alpha_1) = 0$
$\alpha_2 l_2 = \alpha_1 l_1$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems