JEE Main 2022 · 29 Jun · Shift 2 · Physics Q57

← JEE Main 2022 · 29 Jun · Shift 2

PhysicsInteger+4 / -1Question 57 of 90

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_1$ . The same rod is bent into a ring and its moment of inertia about a diameter is $I_2$ . If $\dfrac{I_1}{I_2}$ is $\dfrac{x\pi^2}3$ , then the value of $x$ will be ______.

Your answer

Answer and worked solution

Official answer: 8

Rod about an end $\frac{ML^2}3$ . Ring of radius $\frac L{2\pi}$ about a diameter $\frac{ML^2}{8\pi^2}$ .

Ratio $\frac{8\pi^2}3$ , so $x=8$ .

Answer: 8

Topics:Moment of inertia

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JEE Main 2022 · 29 Jun · Shift 2, Physics question 57