A uniform circular disc of radius ' R^\prime and mass ' M^\prime 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.

Rotational Motion PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q52 with Solution

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Correct answer: B

Moment of inertia of the original disc

For a uniform disc of mass $M$ and radius $R$ , about an axis perpendicular to the plane through its centre,

$I_{\text{full}}=\frac{1}{2}MR^2$

Mass of the removed छोटे disc

The removed part is a circular disc of radius

$r=\frac{R}{2}$

Since the original disc is uniform, mass is proportional to area.

So mass of removed disc is

$m=M\cdot \frac{\pi (R/2)^2}{\pi R^2}=M\cdot \frac{1}{4}=\frac{M}{4}$

Position of the removed disc

From the figure (standard configuration), the small disc of radius $R/2$ is cut such that it is internally tangent to the outer disc. Hence the distance between the centre of the original disc and the centre of the removed part is

$d=R-\frac{R}{2}=\frac{R}{2}$

Moment of inertia of the removed part about the original centre

First, about its own centre:

$I_{\text{removed, centre}}=\frac{1}{2}m r^2$

Substitute $m=\frac{M}{4}$ and $r=\frac{R}{2}$ :

=\frac{1}{2}\cdot \frac{M}{4}\cdot \frac{R^2}{4} =\frac{MR^2}{32}$$ Now use parallel axis theorem to shift to the axis through the original centre: $$I_{\text{removed about original centre}}=I_{\text{removed, centre}}+md^2$$ $$=\frac{MR^2}{32}+\frac{M}{4}\left(\frac{R}{2}\right)^2 =\frac{MR^2}{32}+\frac{M}{4}\cdot \frac{R^2}{4}$$ $$=\frac{MR^2}{32}+\frac{MR^2}{16} =\frac{MR^2}{32}+\frac{2MR^2}{32} =\frac{3}{32}MR^2$$ 5. **Moment of inertia of the remaining part** $$I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed about original centre}}$$ $$=\frac{1}{2}MR^2-\frac{3}{32}MR^2 =\frac{16}{32}MR^2-\frac{3}{32}MR^2 =\frac{13}{32}MR^2$$ 6. **Option matching** $$\boxed{I=\frac{13}{32}MR^2}$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Moment of inertia of the original disc

For a uniform disc of mass $M$ and radius $R$ , about an axis perpendicular to the plane through its centre,

$I_{\text{full}}=\frac{1}{2}MR^2$

Mass of the removed छोटे disc

The removed part is a circular disc of radius

$r=\frac{R}{2}$

Since the original disc is uniform, mass is proportional to area.

So mass of removed disc is

$m=M\cdot \frac{\pi (R/2)^2}{\pi R^2}=M\cdot \frac{1}{4}=\frac{M}{4}$

Position of the removed disc

$d=R-\frac{R}{2}=\frac{R}{2}$

Moment of inertia of the removed part about the original centre

First, about its own centre:

$I_{\text{removed, centre}}=\frac{1}{2}m r^2$

Substitute $m=\frac{M}{4}$ and $r=\frac{R}{2}$ :

=\frac{1}{2}\cdot \frac{M}{4}\cdot \frac{R^2}{4} =\frac{MR^2}{32}$$ Now use parallel axis theorem to shift to the axis through the original centre: $$I_{\text{removed about original centre}}=I_{\text{removed, centre}}+md^2$$ $$=\frac{MR^2}{32}+\frac{M}{4}\left(\frac{R}{2}\right)^2 =\frac{MR^2}{32}+\frac{M}{4}\cdot \frac{R^2}{4}$$ $$=\frac{MR^2}{32}+\frac{MR^2}{16} =\frac{MR^2}{32}+\frac{2MR^2}{32} =\frac{3}{32}MR^2$$ 5. **Moment of inertia of the remaining part** $$I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed about original centre}}$$ $$=\frac{1}{2}MR^2-\frac{3}{32}MR^2 =\frac{16}{32}MR^2-\frac{3}{32}MR^2 =\frac{13}{32}MR^2$$ 6. **Option matching** $$\boxed{I=\frac{13}{32}MR^2}$$ So the correct option is **B**.

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