JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 26 July 2022, Shift 2, Physics Q55

Question 55 of 90 in this shift, Physics question 25 of 30, Section B.

A uniform heavy rod of mass 20 kg, cross sectional area

0.4\ m^2

and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is

x \times 10^{-9}

m. The value of

x

is ___________. (Given, young modulus

Y = 2 \times 10^{11}\ \text{Nm}^{-2}

and

g = 10\ ms^{-2}

)

Official answer

NTA final key.

Chapter: Mechanical Properties of Solids
Topic: Elasticity
Idea tested: Elongation Due to Self-Weight

Same idea in other shifts

Asked 4× in all

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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 108 of the 174 JEE Main shifts held from 2017 to 2026.