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JEE PYQ: Rotational Motion Question 51

Question 51 - 2019 (10 Apr Shift 1)

A thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma(r) = kr^2$ where $r$ is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

(1) $\frac{MR^2}{3}$

(2) $\frac{2MR^2}{3}$

(3) $\frac{MR^2}{6}$

(4) $\frac{MR^2}{2}$

Show Answer

Answer: (2)


Learning Progress: Step 51 of 70 in this series