Physics · Mechanical Properties of Matter

JEE Main 2026 — 5 April, Morning Shift — Question 16

Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 10^5$ N/m². Area of cross-section of upper wire is $0.008$ cm² and of lower wire is $0.004$ cm². The maximum mass that can be added to pan without breaking any wire is ______ kg (take g = 10 m/s²).

Option A:
56
Option B:
38
Correct
Option C:
96
Option D:
5.6

Answer: B

Step-by-step solution

Upper wire tension = (40+m)g, lower wire tension = (10+m)g. Stress upper = (40+m)g/(0.008×10^{-4}) ≤ 12×10^5 ⇒ m ≤ 38 kg. Stress lower = (10+m)g/(0.004×10^{-4}) ≤ 12×10^5 ⇒ m ≤ 38 kg as well. So m_max = 38 kg.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Stress,Strain and Modulus of Elasticity

Answer: B

Step-by-step solution

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