150093
Five particles of mass \(2 \mathrm{~kg}\) are attached to the rim of a circular disc of radius \(0.1 \mathrm{~m}\) and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
1 \(1 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.1 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(2 \mathrm{~kg} \mathrm{~m}^{2}\)
4 \(0.2 \mathrm{~kg} \mathrm{~m}^{2}\)
Explanation:
B Given, Mass of particle \(m=2 \mathrm{~kg}\) Radius of circular disc \(=0.1 \mathrm{~m}\) According to question, As we know that, Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc \(\mathrm{I}_{\mathrm{c.m}} =\mathrm{MR}^{2}\) \(\mathrm{I}_{\text {net }} =5 \times \mathrm{I}_{\text {each }}\) \(=5 \mathrm{mr}^{2}\) \(=5 \mathrm{mr}^{2}=5 \times 2 \times(0.1)^{2}\) \(=10 \times 1 / 100=0.1 \mathrm{~kg} \cdot \mathrm{m}^{2}\) \(\mathrm{I}_{\text {net }} =0.1 \mathrm{Kg} \cdot \mathrm{m}^{2}\)
RPMT-2006
Rotational Motion
150094
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is
1 \(\sqrt{3}: \sqrt{2}\)
2 \(1: \sqrt{2}\)
3 \(\sqrt{3}: 1\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B Radius of gyration, \(\because\) Moment of inertia, \(\mathrm{I}=\mathrm{MK}^{2}\) \(\mathrm{K}^{2} =\frac{\mathrm{I}}{\mathrm{M}}\) \(\text { Radius of gyration, } \mathrm{K} =\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}\) Moment of inertia of disc, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{\frac{1}{2} \mathrm{MR}^{2}}{\mathrm{M}}}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{R}^{2}}{2}}\) Moment of inertia of ring, \(\mathrm{I}=\mathrm{MR}^{2}\) \(K_{\text {ring }}=\sqrt{\frac{M R^{2}}{M}}\) \(K_{\text {ring }}=\sqrt{R^{2}}\) Formeq.(i) \(\div\) (ii), \(\because \frac{\mathrm{K}_{\text {disc }}}{\mathrm{K}_{\text {ring }}}=\frac{\sqrt{\frac{1}{2} \mathrm{R}^{2}}}{\sqrt{\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}}\)
JCECE-2014
Rotational Motion
150095
The moment of inertia of a uniform thin rod of length \(L\) and mass \(M\) about an axis passing through a point at a distance \(\frac{L}{3}\) from one of its ends and perpendicular to the rod is
1 \(\frac{7 \mathrm{ML}^{2}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{9}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{\mathrm{ML}^{2}}{3}\)
Explanation:
B According to parallel axis theorem- \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{M}\left(\frac{\mathrm{L}}{6}\right)^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36}\) \(=\frac{4 \mathrm{ML}^{2}}{36}\) \(=\frac{\mathrm{ML}^{2}}{}\)
AMU-2010
Rotational Motion
150096
From a disc of mass ' \(M\) ' and radius ' \(R\) ', a circular hole of a diameter ' \(R\) ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
1 \(\frac{9 \mathrm{MR}^{2}}{32}\)
2 \(\frac{13 \mathrm{MR}^{2}}{32}\)
3 \(\frac{11 \mathrm{MR}^{2}}{32}\)
4 \(\frac{7 \mathrm{MR}^{2}}{32}\)
Explanation:
B Given that, mass of disc \(=\) M, radius of disc \(=\) \(\mathrm{R}\) Moment of inertia \(\left(\mathrm{I}_{\text {big disc }}\right)=\frac{1}{2} \mathrm{MR}^{2}\) Let, moment of inertia of small disc is \(I_{1}\) which is cut from big disc. So, \(\mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) From parallel axis theorem, moment of inertia of the small disc \(\left(I_{1}\right)=I_{c}+m_{1} R_{1}^{2}\) \(\mathrm{I}_1=\frac{1}{2} \cdot \frac{\mathrm{M}}{4} \cdot\left(\frac{\mathrm{R}}{2}\right)^2+\frac{\mathrm{M}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^2 \quad\left[\begin{array}{l}\because \mathrm{m}_1=\mathrm{M} / 4 \\ \mathrm{R}_1=\mathrm{R} / 2\end{array} \& \mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{R}_1^2\right]\) \(=\frac{M^{2}}{32}+\frac{M R^{2}}{16}\) \(I_{1} =\frac{3 M^{2}}{32}\) Putting the value of these in equation (ii), we get- \(\therefore \quad \mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) \(\mathrm{I}_{\text {remaining }}=\frac{1}{2} \mathrm{MR}^{2}-\frac{3 \mathrm{MR}^{2}}{32}\) \(\mathrm{I}_{\text {remaining }}=\frac{13 \mathrm{MR}^{2}}{32}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150093
Five particles of mass \(2 \mathrm{~kg}\) are attached to the rim of a circular disc of radius \(0.1 \mathrm{~m}\) and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
1 \(1 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.1 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(2 \mathrm{~kg} \mathrm{~m}^{2}\)
4 \(0.2 \mathrm{~kg} \mathrm{~m}^{2}\)
Explanation:
B Given, Mass of particle \(m=2 \mathrm{~kg}\) Radius of circular disc \(=0.1 \mathrm{~m}\) According to question, As we know that, Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc \(\mathrm{I}_{\mathrm{c.m}} =\mathrm{MR}^{2}\) \(\mathrm{I}_{\text {net }} =5 \times \mathrm{I}_{\text {each }}\) \(=5 \mathrm{mr}^{2}\) \(=5 \mathrm{mr}^{2}=5 \times 2 \times(0.1)^{2}\) \(=10 \times 1 / 100=0.1 \mathrm{~kg} \cdot \mathrm{m}^{2}\) \(\mathrm{I}_{\text {net }} =0.1 \mathrm{Kg} \cdot \mathrm{m}^{2}\)
RPMT-2006
Rotational Motion
150094
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is
1 \(\sqrt{3}: \sqrt{2}\)
2 \(1: \sqrt{2}\)
3 \(\sqrt{3}: 1\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B Radius of gyration, \(\because\) Moment of inertia, \(\mathrm{I}=\mathrm{MK}^{2}\) \(\mathrm{K}^{2} =\frac{\mathrm{I}}{\mathrm{M}}\) \(\text { Radius of gyration, } \mathrm{K} =\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}\) Moment of inertia of disc, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{\frac{1}{2} \mathrm{MR}^{2}}{\mathrm{M}}}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{R}^{2}}{2}}\) Moment of inertia of ring, \(\mathrm{I}=\mathrm{MR}^{2}\) \(K_{\text {ring }}=\sqrt{\frac{M R^{2}}{M}}\) \(K_{\text {ring }}=\sqrt{R^{2}}\) Formeq.(i) \(\div\) (ii), \(\because \frac{\mathrm{K}_{\text {disc }}}{\mathrm{K}_{\text {ring }}}=\frac{\sqrt{\frac{1}{2} \mathrm{R}^{2}}}{\sqrt{\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}}\)
JCECE-2014
Rotational Motion
150095
The moment of inertia of a uniform thin rod of length \(L\) and mass \(M\) about an axis passing through a point at a distance \(\frac{L}{3}\) from one of its ends and perpendicular to the rod is
1 \(\frac{7 \mathrm{ML}^{2}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{9}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{\mathrm{ML}^{2}}{3}\)
Explanation:
B According to parallel axis theorem- \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{M}\left(\frac{\mathrm{L}}{6}\right)^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36}\) \(=\frac{4 \mathrm{ML}^{2}}{36}\) \(=\frac{\mathrm{ML}^{2}}{}\)
AMU-2010
Rotational Motion
150096
From a disc of mass ' \(M\) ' and radius ' \(R\) ', a circular hole of a diameter ' \(R\) ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
1 \(\frac{9 \mathrm{MR}^{2}}{32}\)
2 \(\frac{13 \mathrm{MR}^{2}}{32}\)
3 \(\frac{11 \mathrm{MR}^{2}}{32}\)
4 \(\frac{7 \mathrm{MR}^{2}}{32}\)
Explanation:
B Given that, mass of disc \(=\) M, radius of disc \(=\) \(\mathrm{R}\) Moment of inertia \(\left(\mathrm{I}_{\text {big disc }}\right)=\frac{1}{2} \mathrm{MR}^{2}\) Let, moment of inertia of small disc is \(I_{1}\) which is cut from big disc. So, \(\mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) From parallel axis theorem, moment of inertia of the small disc \(\left(I_{1}\right)=I_{c}+m_{1} R_{1}^{2}\) \(\mathrm{I}_1=\frac{1}{2} \cdot \frac{\mathrm{M}}{4} \cdot\left(\frac{\mathrm{R}}{2}\right)^2+\frac{\mathrm{M}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^2 \quad\left[\begin{array}{l}\because \mathrm{m}_1=\mathrm{M} / 4 \\ \mathrm{R}_1=\mathrm{R} / 2\end{array} \& \mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{R}_1^2\right]\) \(=\frac{M^{2}}{32}+\frac{M R^{2}}{16}\) \(I_{1} =\frac{3 M^{2}}{32}\) Putting the value of these in equation (ii), we get- \(\therefore \quad \mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) \(\mathrm{I}_{\text {remaining }}=\frac{1}{2} \mathrm{MR}^{2}-\frac{3 \mathrm{MR}^{2}}{32}\) \(\mathrm{I}_{\text {remaining }}=\frac{13 \mathrm{MR}^{2}}{32}\)
150093
Five particles of mass \(2 \mathrm{~kg}\) are attached to the rim of a circular disc of radius \(0.1 \mathrm{~m}\) and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
1 \(1 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.1 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(2 \mathrm{~kg} \mathrm{~m}^{2}\)
4 \(0.2 \mathrm{~kg} \mathrm{~m}^{2}\)
Explanation:
B Given, Mass of particle \(m=2 \mathrm{~kg}\) Radius of circular disc \(=0.1 \mathrm{~m}\) According to question, As we know that, Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc \(\mathrm{I}_{\mathrm{c.m}} =\mathrm{MR}^{2}\) \(\mathrm{I}_{\text {net }} =5 \times \mathrm{I}_{\text {each }}\) \(=5 \mathrm{mr}^{2}\) \(=5 \mathrm{mr}^{2}=5 \times 2 \times(0.1)^{2}\) \(=10 \times 1 / 100=0.1 \mathrm{~kg} \cdot \mathrm{m}^{2}\) \(\mathrm{I}_{\text {net }} =0.1 \mathrm{Kg} \cdot \mathrm{m}^{2}\)
RPMT-2006
Rotational Motion
150094
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is
1 \(\sqrt{3}: \sqrt{2}\)
2 \(1: \sqrt{2}\)
3 \(\sqrt{3}: 1\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B Radius of gyration, \(\because\) Moment of inertia, \(\mathrm{I}=\mathrm{MK}^{2}\) \(\mathrm{K}^{2} =\frac{\mathrm{I}}{\mathrm{M}}\) \(\text { Radius of gyration, } \mathrm{K} =\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}\) Moment of inertia of disc, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{\frac{1}{2} \mathrm{MR}^{2}}{\mathrm{M}}}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{R}^{2}}{2}}\) Moment of inertia of ring, \(\mathrm{I}=\mathrm{MR}^{2}\) \(K_{\text {ring }}=\sqrt{\frac{M R^{2}}{M}}\) \(K_{\text {ring }}=\sqrt{R^{2}}\) Formeq.(i) \(\div\) (ii), \(\because \frac{\mathrm{K}_{\text {disc }}}{\mathrm{K}_{\text {ring }}}=\frac{\sqrt{\frac{1}{2} \mathrm{R}^{2}}}{\sqrt{\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}}\)
JCECE-2014
Rotational Motion
150095
The moment of inertia of a uniform thin rod of length \(L\) and mass \(M\) about an axis passing through a point at a distance \(\frac{L}{3}\) from one of its ends and perpendicular to the rod is
1 \(\frac{7 \mathrm{ML}^{2}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{9}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{\mathrm{ML}^{2}}{3}\)
Explanation:
B According to parallel axis theorem- \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{M}\left(\frac{\mathrm{L}}{6}\right)^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36}\) \(=\frac{4 \mathrm{ML}^{2}}{36}\) \(=\frac{\mathrm{ML}^{2}}{}\)
AMU-2010
Rotational Motion
150096
From a disc of mass ' \(M\) ' and radius ' \(R\) ', a circular hole of a diameter ' \(R\) ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
1 \(\frac{9 \mathrm{MR}^{2}}{32}\)
2 \(\frac{13 \mathrm{MR}^{2}}{32}\)
3 \(\frac{11 \mathrm{MR}^{2}}{32}\)
4 \(\frac{7 \mathrm{MR}^{2}}{32}\)
Explanation:
B Given that, mass of disc \(=\) M, radius of disc \(=\) \(\mathrm{R}\) Moment of inertia \(\left(\mathrm{I}_{\text {big disc }}\right)=\frac{1}{2} \mathrm{MR}^{2}\) Let, moment of inertia of small disc is \(I_{1}\) which is cut from big disc. So, \(\mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) From parallel axis theorem, moment of inertia of the small disc \(\left(I_{1}\right)=I_{c}+m_{1} R_{1}^{2}\) \(\mathrm{I}_1=\frac{1}{2} \cdot \frac{\mathrm{M}}{4} \cdot\left(\frac{\mathrm{R}}{2}\right)^2+\frac{\mathrm{M}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^2 \quad\left[\begin{array}{l}\because \mathrm{m}_1=\mathrm{M} / 4 \\ \mathrm{R}_1=\mathrm{R} / 2\end{array} \& \mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{R}_1^2\right]\) \(=\frac{M^{2}}{32}+\frac{M R^{2}}{16}\) \(I_{1} =\frac{3 M^{2}}{32}\) Putting the value of these in equation (ii), we get- \(\therefore \quad \mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) \(\mathrm{I}_{\text {remaining }}=\frac{1}{2} \mathrm{MR}^{2}-\frac{3 \mathrm{MR}^{2}}{32}\) \(\mathrm{I}_{\text {remaining }}=\frac{13 \mathrm{MR}^{2}}{32}\)
150093
Five particles of mass \(2 \mathrm{~kg}\) are attached to the rim of a circular disc of radius \(0.1 \mathrm{~m}\) and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
1 \(1 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.1 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(2 \mathrm{~kg} \mathrm{~m}^{2}\)
4 \(0.2 \mathrm{~kg} \mathrm{~m}^{2}\)
Explanation:
B Given, Mass of particle \(m=2 \mathrm{~kg}\) Radius of circular disc \(=0.1 \mathrm{~m}\) According to question, As we know that, Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc \(\mathrm{I}_{\mathrm{c.m}} =\mathrm{MR}^{2}\) \(\mathrm{I}_{\text {net }} =5 \times \mathrm{I}_{\text {each }}\) \(=5 \mathrm{mr}^{2}\) \(=5 \mathrm{mr}^{2}=5 \times 2 \times(0.1)^{2}\) \(=10 \times 1 / 100=0.1 \mathrm{~kg} \cdot \mathrm{m}^{2}\) \(\mathrm{I}_{\text {net }} =0.1 \mathrm{Kg} \cdot \mathrm{m}^{2}\)
RPMT-2006
Rotational Motion
150094
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is
1 \(\sqrt{3}: \sqrt{2}\)
2 \(1: \sqrt{2}\)
3 \(\sqrt{3}: 1\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B Radius of gyration, \(\because\) Moment of inertia, \(\mathrm{I}=\mathrm{MK}^{2}\) \(\mathrm{K}^{2} =\frac{\mathrm{I}}{\mathrm{M}}\) \(\text { Radius of gyration, } \mathrm{K} =\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}\) Moment of inertia of disc, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{\frac{1}{2} \mathrm{MR}^{2}}{\mathrm{M}}}\) \(\mathrm{K}_{\text {disc }}=\sqrt{\frac{\mathrm{R}^{2}}{2}}\) Moment of inertia of ring, \(\mathrm{I}=\mathrm{MR}^{2}\) \(K_{\text {ring }}=\sqrt{\frac{M R^{2}}{M}}\) \(K_{\text {ring }}=\sqrt{R^{2}}\) Formeq.(i) \(\div\) (ii), \(\because \frac{\mathrm{K}_{\text {disc }}}{\mathrm{K}_{\text {ring }}}=\frac{\sqrt{\frac{1}{2} \mathrm{R}^{2}}}{\sqrt{\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}}\)
JCECE-2014
Rotational Motion
150095
The moment of inertia of a uniform thin rod of length \(L\) and mass \(M\) about an axis passing through a point at a distance \(\frac{L}{3}\) from one of its ends and perpendicular to the rod is
1 \(\frac{7 \mathrm{ML}^{2}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{9}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{\mathrm{ML}^{2}}{3}\)
Explanation:
B According to parallel axis theorem- \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{M}\left(\frac{\mathrm{L}}{6}\right)^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\frac{\mathrm{ML}^{2}}{36}\) \(=\frac{4 \mathrm{ML}^{2}}{36}\) \(=\frac{\mathrm{ML}^{2}}{}\)
AMU-2010
Rotational Motion
150096
From a disc of mass ' \(M\) ' and radius ' \(R\) ', a circular hole of a diameter ' \(R\) ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
1 \(\frac{9 \mathrm{MR}^{2}}{32}\)
2 \(\frac{13 \mathrm{MR}^{2}}{32}\)
3 \(\frac{11 \mathrm{MR}^{2}}{32}\)
4 \(\frac{7 \mathrm{MR}^{2}}{32}\)
Explanation:
B Given that, mass of disc \(=\) M, radius of disc \(=\) \(\mathrm{R}\) Moment of inertia \(\left(\mathrm{I}_{\text {big disc }}\right)=\frac{1}{2} \mathrm{MR}^{2}\) Let, moment of inertia of small disc is \(I_{1}\) which is cut from big disc. So, \(\mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) From parallel axis theorem, moment of inertia of the small disc \(\left(I_{1}\right)=I_{c}+m_{1} R_{1}^{2}\) \(\mathrm{I}_1=\frac{1}{2} \cdot \frac{\mathrm{M}}{4} \cdot\left(\frac{\mathrm{R}}{2}\right)^2+\frac{\mathrm{M}}{4} \times\left(\frac{\mathrm{R}}{2}\right)^2 \quad\left[\begin{array}{l}\because \mathrm{m}_1=\mathrm{M} / 4 \\ \mathrm{R}_1=\mathrm{R} / 2\end{array} \& \mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{~m}_1 \mathrm{R}_1^2\right]\) \(=\frac{M^{2}}{32}+\frac{M R^{2}}{16}\) \(I_{1} =\frac{3 M^{2}}{32}\) Putting the value of these in equation (ii), we get- \(\therefore \quad \mathrm{I}_{\text {remaining }}=\mathrm{I}_{\text {big disc }}-\mathrm{I}_{1}\) \(\mathrm{I}_{\text {remaining }}=\frac{1}{2} \mathrm{MR}^{2}-\frac{3 \mathrm{MR}^{2}}{32}\) \(\mathrm{I}_{\text {remaining }}=\frac{13 \mathrm{MR}^{2}}{32}\)
