JEE Main 2025 Apr 2 Shift 1, Physics Q16: Moment of Inertia

JEE Main2025Apr 2, Shift 1Physics

Moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its center and normal to its length is $α$ . Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

A
$α$
B
$α /4$
C
$α /8$
D
$α /2$

Solution

Original rod:

$α = \frac{M L ^{2}}{12}$

After cutting into two equal halves: each half has mass $M /2$ and length $L /2$ . The two halves are joined at their midpoints to form a perpendicular cross, with the axis perpendicular to the plane through the centre.

For one half rod about its own midpoint (which is also the cross's centre), perpendicular to its length:

$I_{half} = \frac{( M /2 ) ( L /2 ) ^{2}}{12} = \frac{M L ^{2}}{96}$

The axis is perpendicular to the plane of the cross, which lies along the length of each rod (perpendicular distance from each rod's length is the value of $I$ about an axis perpendicular to its length through midpoint). Total for two perpendicular rods:

$I_{cross} = 2 \times \frac{M L ^{2}}{96} = \frac{M L ^{2}}{48}$

Comparing with $α = \frac{M L ^{2}}{12}$ :

$I_{cross} = \frac{M L ^{2}}{48} = \frac{α}{4}$

$∴$ Option (2) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q16: Moment of Inertia

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