PhysicsNEET 2024Moment of inertia, theorems of axes

System of Particles and Rotational Motion

The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is $2400 g c m^{2}$ . The length of the 400 g rod is nearly:

A.17.5 cm
B.20.7 cm
C.72.0 cm
D.8.5 cm✓

Solution

$I = \frac{M L ^{2}}{12} \Rightarrow L^{2} = \frac{12 \times 2400}{400} = 72 \Rightarrow L \approx 8.5$ cm.

Save for spaced revision (free account)

More from this chapter in past papers

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the…NEET 2025
The ratio of the radius of gyration of a thin uniform disc about an axis passing through…NEET 2022
From a circular ring of mass M and radius R an arc corresponding to a 90° sector is…NEET 2021

From Physics › System of Particles and Rotational Motion › Moment of inertia, theorems of axes — lectures, notes and every past question on it.

Last revised 7 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsApp Telegram