A circular hole of radius R \over 4 is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Rotational Motion PYQ — JEE Main 2017 · 9 Apr · Shift 1 · Q47 with Solution

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Correct answer: B

Interpret the figure and given data

A thin uniform disc has:

mass $M$
radius $R$

A circular hole of radius $\dfrac{R}{4}$ is removed.

From the standard figure for this question, the small hole is tangent to the outer boundary and its center lies on the radius through $O$ . Hence the distance between the center of the original disc and the center of the hole is

$d=R-\frac{R}{4}=\frac{3R}{4}.$

We need the moment of inertia of the remaining part about an axis through $O$ (the center of the original disc) and perpendicular to the plane.

Use area-mass proportionality to find mass of removed disc

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

If $m$ is the mass of the removed small disc, then

$\frac{m}{M}=\frac{\pi (R/4)^2}{\pi R^2}=\frac{1}{16}.$

So,

$m=\frac{M}{16}.$

Moment of inertia of the original full disc about $O$

For a full disc about its center,

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

Moment of inertia of the removed small disc about $O$

First find its MOI about its own center:

Substitute $m=\dfrac{M}{16}$ :

=\frac{MR^2}{512}.$$ Now shift to axis through $O$ using parallel axis theorem: $$I_{\text{small about }O}=I_{\text{small, center}}+md^2.$$ Here, $$d=\frac{3R}{4}, \quad m=\frac{M}{16}.$$ Thus, $$I_{\text{small about }O}=\frac{MR^2}{512}+\frac{M}{16}\left(\frac{3R}{4}\right)^2.$$ $$=\frac{MR^2}{512}+\frac{M}{16}\cdot \frac{9R^2}{16}$$ $$=\frac{MR^2}{512}+\frac{9MR^2}{256}.$$ Taking LCM $512$, $$I_{\text{small about }O}=\frac{MR^2}{512}+\frac{18MR^2}{512} =\frac{19MR^2}{512}.$$ --- 5. **Moment of inertia of the remaining portion** $$I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed about }O}.$$ So, $$I_{\text{remaining}}=\frac{1}{2}MR^2-\frac{19MR^2}{512}.$$ Since $$\frac{1}{2}MR^2=\frac{256MR^2}{512},$$ we get $$I_{\text{remaining}}=\frac{256MR^2}{512}-\frac{19MR^2}{512} =\frac{237MR^2}{512}.$$ --- 6. **Match with options** $$\boxed{I=\frac{237MR^2}{512}}$$ This corresponds to **Option B**.

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View written solutionFree

Correct answer: B

Interpret the figure and given data

A thin uniform disc has:

mass $M$
radius $R$

A circular hole of radius $\dfrac{R}{4}$ is removed.

$d=R-\frac{R}{4}=\frac{3R}{4}.$

We need the moment of inertia of the remaining part about an axis through $O$ (the center of the original disc) and perpendicular to the plane.

Use area-mass proportionality to find mass of removed disc

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

If $m$ is the mass of the removed small disc, then

$\frac{m}{M}=\frac{\pi (R/4)^2}{\pi R^2}=\frac{1}{16}.$

So,

$m=\frac{M}{16}.$

Moment of inertia of the original full disc about $O$

For a full disc about its center,

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

Moment of inertia of the removed small disc about $O$

First find its MOI about its own center:

Substitute $m=\dfrac{M}{16}$ :

=\frac{MR^2}{512}.$$ Now shift to axis through $O$ using parallel axis theorem: $$I_{\text{small about }O}=I_{\text{small, center}}+md^2.$$ Here, $$d=\frac{3R}{4}, \quad m=\frac{M}{16}.$$ Thus, $$I_{\text{small about }O}=\frac{MR^2}{512}+\frac{M}{16}\left(\frac{3R}{4}\right)^2.$$ $$=\frac{MR^2}{512}+\frac{M}{16}\cdot \frac{9R^2}{16}$$ $$=\frac{MR^2}{512}+\frac{9MR^2}{256}.$$ Taking LCM $512$, $$I_{\text{small about }O}=\frac{MR^2}{512}+\frac{18MR^2}{512} =\frac{19MR^2}{512}.$$ --- 5. **Moment of inertia of the remaining portion** $$I_{\text{remaining}}=I_{\text{full}}-I_{\text{removed about }O}.$$ So, $$I_{\text{remaining}}=\frac{1}{2}MR^2-\frac{19MR^2}{512}.$$ Since $$\frac{1}{2}MR^2=\frac{256MR^2}{512},$$ we get $$I_{\text{remaining}}=\frac{256MR^2}{512}-\frac{19MR^2}{512} =\frac{237MR^2}{512}.$$ --- 6. **Match with options** $$\boxed{I=\frac{237MR^2}{512}}$$ This corresponds to **Option B**.

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