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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2026 · 4 April · Evening Shift · Physics · Mechanical Properties of Solids

A metal string A is suspended from a rigid support and its free end...

A metal string AA is suspended from a rigid support and its free end is attached to a block of mass MM. Second block having mass 2M2M is suspended at the bottom of the first block using a string BB. The area of cross sections of strings AA and BB are same. The ratio of lengths of strings of AA to BB is 22 and the ratio of their Young's moduli (YA/YB)(Y_A/Y_B) is 0.50.5. The ratio of elongations in AA to BB is ______.

Options

  1. A11
  2. B44
  3. C88
  4. D66

Correct answer

D. 66

Step-by-step solution

Tension in string BB is due to the block of mass 2M2M:

TB=2MgT_B = 2Mg

Tension in string AA is due to both blocks of mass MM and 2M2M:

TA=Mg+2Mg=3MgT_A = Mg + 2Mg = 3Mg

The elongation in a string is given by Hooke's law as ΔL=TLAY\Delta L = \dfrac{TL}{AY}.

The ratio of elongations in string AA to string BB is:

ΔLAΔLB=(TATB)(LALB)(ABAA)(YBYA)\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{T_A}{T_B}\right) \left(\dfrac{L_A}{L_B}\right) \left(\dfrac{A_B}{A_A}\right) \left(\dfrac{Y_B}{Y_A}\right)

Given that LALB=2\dfrac{L_A}{L_B} = 2, ABAA=1\dfrac{A_B}{A_A} = 1, and YAYB=0.5⇒YBYA=2\dfrac{Y_A}{Y_B} = 0.5 \Rightarrow \dfrac{Y_B}{Y_A} = 2.

Substituting the values into the ratio equation:

ΔLAΔLB=(3Mg2Mg)×(2)×(1)×(2)\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{3Mg}{2Mg}\right) \times (2) \times (1) \times (2)

ΔLAΔLB=32×4=6\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{3}{2} \times 4 = 6

Answer: 66
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