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Friday, 9 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 2 April 2026, Shift 1, Physics Q27

Question 27 of 75 in this shift, Physics question 2 of 25, Section A.

The diameter of a wire measured by a screw gauge of least count 0.0010.001 cm is 0.080.08 cm. The length measured by a scale of least count 0.10.1 cm is 150150 cm. When a weight of 100100 N is applied to the wire, the extension in length is 0.50.5 cm, measured by a micrometer of least count 0.0010.001 cm. The error in the measured Young's modulus is α×109 N/m2\alpha \times 10^{9}\ \mathrm{N/m^2}. The value of α\alpha is __________. (Ignore the contribution of the load to Young's modulus error calculation)
  1. (1)1.31.3
  2. (2)1.651.65Official answer
  3. (3)0.130.13
  4. (4)0.250.25

Official answer

Option 2

NTA final key.

Chapter
Units and Measurements
Topic
Elasticity
Idea tested
Error Analysis in Searle's Experiment

Same idea in other shifts

Asked 3× in all
  1. 8 Apr 2024, Shift 1 · Q48Young's modulus is determined by the equation given by Y=49000mldynecm2Y=49000\frac{m}{l}\frac{\mathrm{dyne}}{\mathrm{cm}^2} where MM is the mass and…MediumSingle correct
  2. 16 Mar 2021, Shift 2 · Q17In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.