JEE Main 2025 Jan 22 Shift 1, Physics Q12: Moment of Inertia

JEE Main2025Jan 22, Shift 1Physics

A uniform circular disc of radius 'R' and mass 'M' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

A
$\frac{7}{32} M R^{2}$
B
$\frac{9}{32} M R^{2}$
C
$\frac{17}{32} M R^{2}$
D
$\frac{13}{32} M R^{2}$

Solution

Mass per unit area $σ = \frac{M}{π R ^{2}}$ . The removed disc of radius $R /2$ has mass $m = σ \cdot π (R /2)^{2} = M /4$ , with its centre at distance $R /2$ from the centre of the big disc.

Moment of inertia of full disc about the centre: $I_{f u l l} = \frac{1}{2} M R^{2}$ .

Moment of inertia of the removed piece about the centre of the big disc (parallel-axis):

$I_{r e m} = \frac{1}{2} (\frac{M}{4}) (\frac{R}{2})^{2} + \frac{M}{4} (\frac{R}{2})^{2} = \frac{M R ^{2}}{32} + \frac{M R ^{2}}{16} = \frac{3 M R ^{2}}{32}$ .

Remaining: $I = \frac{M R ^{2}}{2} - \frac{3 M R ^{2}}{32} = \frac{16 - 3}{32} M R^{2} = \frac{13}{32} M R^{2}$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q12: Moment of Inertia

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