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Physics · Mechanical Properties of Matter

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

A lift of mass 1600kg1600\mathrm{kg} is supported by thick iron wire. If the maximum stress which the wire can withstand is 4×108N/m24\times 10^{8}\mathrm{N/m}^2 and its radius is 4mm4\mathrm{mm} then maximum acceleration the lift can take is m/s2\mathrm{m/s}^2 (take g=10m/s2g = 10\mathrm{m/s}^2 and π=3.14\pi = 3.14)

  1. Option A:

    2.56

    Correct
  2. Option B:

    3.89

  3. Option C:

    4.32

  4. Option D:

    5.16

Answer: A

Step-by-step solution

Maximum tension Tmax=stress×area=4×108×π(0.004)2=4×108×3.14×16×10−6=20096T_{max} = \text{stress} \times \text{area} = 4\times10^8 \times \pi (0.004)^2 = 4\times10^8 \times 3.14 \times 16\times10^{-6} = 20096 N. For lift moving upward with acceleration a: T−mg=ma⇒a=T/m−g=20096/1600−10=12.56−10=2.56T - mg = ma \Rightarrow a = T/m - g = 20096/1600 - 10 = 12.56 - 10 = 2.56 m/s².

Solution figure

Answer key and solution verified before publishing.

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

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