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NEET PHYSICSThermal Properties of MatterMedium

Question

The coefficient of linear expansion of brass and steel rods are α1\alpha_1 and α2\alpha_2. Lengths of brass and steel rods are L1L_1 and L2L_2 respectively. If (L2L1)(L_2-L_1) remains the same at all temperatures, which one of the following relations holds good?

A

α1L22=α2L12\alpha_1L_2^2=\alpha_2L_1^2

B

α12L2=α22L1\alpha_1^2L_2=\alpha_2^2L_1

C

α1L1=α2L2\alpha_1L_1=\alpha_2L_2

D

α1L2=α2L1\alpha_1L_2=\alpha_2L_1

Step-by-Step Solution

Let the change in temperature be ΔT\Delta T. The new lengths of the brass and steel rods will be: L1=L1(1+α1ΔT)L_1' = L_1(1 + \alpha_1\Delta T) L2=L2(1+α2ΔT)L_2' = L_2(1 + \alpha_2\Delta T) Given that the difference in lengths (L2L1)(L_2 - L_1) remains the same at all temperatures: L2L1=L2L1L_2' - L_1' = L_2 - L_1 L2(1+α2ΔT)L1(1+α1ΔT)=L2L1L_2(1 + \alpha_2\Delta T) - L_1(1 + \alpha_1\Delta T) = L_2 - L_1 L2+L2α2ΔTL1L1α1ΔT=L2L1L_2 + L_2\alpha_2\Delta T - L_1 - L_1\alpha_1\Delta T = L_2 - L_1 L2α2ΔTL1α1ΔT=0L_2\alpha_2\Delta T - L_1\alpha_1\Delta T = 0     L2α2=L1α1\implies L_2\alpha_2 = L_1\alpha_1     α1L1=α2L2\implies \alpha_1L_1 = \alpha_2L_2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Mattercoefficientlinearexpansionlengthsrespectively

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