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Physics · Mechanical Properties of Matter

JEE Main 2026 — 4 April, Evening Shift — Question 6

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli (YA/YB)(Y_A/Y_B) is 0.5. The ratio of elongations in A to B is

  1. Option A:

    1

  2. Option B:

    4

  3. Option C:

    8

  4. Option D:

    6

    Correct

Answer: D

Step-by-step solution

Using ΔL=FLAY\Delta L = \frac{FL}{AY}. For string A, tension = 3Mg3Mg, length = 2L2L, YA=0.5YBY_A = 0.5 Y_B. For string B, tension = 2Mg2Mg, length = LL, YBY_B. Ratio ΔLAΔLB=3Mg⋅2L/(A⋅0.5YB)2Mg⋅L/(AYB)=61=6\frac{\Delta L_A}{\Delta L_B} = \frac{3Mg \cdot 2L / (A \cdot 0.5Y_B)}{2Mg \cdot L / (A Y_B)} = \frac{6}{1} = 6.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity