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JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 28 July 2022, Shift 2, Physics Q22

Question 22 of 90 in this shift, Physics question 22 of 30, Section B.

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density

\rho=\rho_0\left(1-\frac{x^2}{L^2}\right)

kg/m and length L (in meter) is

\frac{3L}{\alpha}

m. The value of

\alpha

is _______. (where

x

is the distance from end A)

Official answer

8

NTA final key.

Chapter: System of Particles and Rotational Motion
Topic: Center of Mass
Idea tested: COM of Non-Uniform Rod

Same idea in other shifts

Asked 2× in all

9 Jan 2020, Shift 2 · Q5A rod of length $L$ has non-uniform linear mass density given by $\rho(x) = a + b\left(\frac{x}{L}\right)^2$ , where $a$ and $b$ are…EasySingle 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 108 of the 174 JEE Main shifts held from 2017 to 2026.