Class JEE Physics ALL Q #1183
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
Two rods of equal length 60 cm each are joined together end to end. The coefficients of linear expansion are $24\times10^{-6\circ}C^{-1}$ and $1.2\times10^{-5\circ}C^{-1}$. Initial temperature $30^{\circ}C$ is increased to $100^{\circ}C$. Find final length (in cm).
(A) $120.1321$
(B) $120.1123$
(C) $120.1512$
(D) $120.1084$
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AI Tutor Explanation

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Step-by-Step Solution

  1. Step 1: Calculate the change in temperature.

    ΔT = Final temperature - Initial temperature = 100°C - 30°C = 70°C

  2. Step 2: Calculate the change in length for the first rod.

    ΔL1 = L1 * α1 * ΔT = 60 cm * (24 × 10-6 °C-1) * 70°C = 0.1008 cm

  3. Step 3: Calculate the change in length for the second rod.

    ΔL2 = L2 * α2 * ΔT = 60 cm * (1.2 × 10-5 °C-1) * 70°C = 0.0504 cm

  4. Step 4: Calculate the total change in length.

    ΔLtotal = ΔL1 + ΔL2 = 0.1008 cm + 0.0504 cm = 0.1512 cm

  5. Step 5: Calculate the final length.

    Final Length = Initial Length + Total Change in Length = (60 cm + 60 cm) + 0.1512 cm = 120 cm + 0.1512 cm = 120.1512 cm

Correct Answer: 120.1512

AI Suggestion: Option C

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the formula for linear expansion to calculate the final length of the combined rods.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure (applying the formula for linear expansion) to arrive at the solution.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires application of the concept of thermal expansion in a problem-solving scenario, going beyond rote memorization of formulas.

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