150134
A thin, uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of \((g=\) acceleration due to gravity)
1 \(\frac{L^{2} \omega^{2}}{3 g}\)
2 \(\frac{\mathrm{L}^{2} \omega^{2}}{3}\)
3 \(\frac{L^{2} \omega^{2}}{2 g}\)
4 \(\frac{L^{2} \omega^{2}}{6 g}\)
Explanation:
D According to law of conservation of energy- \(\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}\) Moment of inertia of the rod about the axis passing through its end \(\mathrm{I}=\frac{1}{3} \mathrm{ML}^{2}\) Putting the value of 'I' in equation (i), we get - \(\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \cdot \omega^{2} =\mathrm{Mgh}\) \(\mathrm{h} =\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
MHT-CET 2020
Rotational Motion
150135
Moment of inertia of a uniform solid sphere of mass ' \(M\) ' and radius ' \(R\) ' about an axis at a distance \(\left(\frac{R}{2}\right)\) from the centre is
1 \(\frac{2}{5} \mathrm{MR}^{2}\)
2 \(\frac{9}{10} \mathrm{MR}^{2}\)
3 \(\frac{13}{20} \mathrm{MR}^{2}\)
4 \(\frac{7}{5} \mathrm{MR}^{2}\)
Explanation:
C From parallel axis theorem \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{Md}^{2}\) \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2}\) \(=\frac{2}{5} \mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{4}\) \(=\frac{13}{2 \mathrm{n}} \mathrm{MR}^{2}\)
MHT-CET 2020
Rotational Motion
150136
Four spheres each of mass ' \(M\) ' and radius ' \(R\) ' are placed with their centers on the corners of a square of side ' \(L\) '. The moment of inertia of the system about any side of square is
C Given, Mass of each sphere \(=\mathrm{M}\) Radius \(=\mathrm{R}\) distance between two sphere \(=\mathrm{L}\) Now as we know moment of inertia of a sphere lying on the axis \(=\frac{2}{5} \mathrm{MR}^{2}\) So, moment of inertia of two sphere lying on the axis \(=2 \times \frac{2}{5} \mathrm{MR}^{2}=\frac{4}{5} \mathrm{MR}^{2}\) By using parallel axis theorem of moment of inertia of another two sphere lying on the line parallel to the axis at distance of \(\mathrm{L}=2\left[\frac{2}{5} \mathrm{MR}^{2}+\mathrm{ML}^{2}\right]\) \(=\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\) Total moment of inertia \(=\frac{4}{5} \mathrm{MR}^{2}+\left(\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\right)\) \(=\frac{8}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\)
MHT-CET 2020
Rotational Motion
150137
A uniform rod of length ' \(2 L\) ' has constant mass per unit length ' \(m\) '. Moment of inertia of the rod about an axis passing through its centre and perpendicular to length is
1 \(\frac{\mathrm{mL}^{3}}{3}\)
2 \(\frac{\mathrm{mL}^{2}}{4}\)
3 \(\frac{2 \mathrm{~mL}^{3}}{3}\)
4 \(\frac{\mathrm{mL}^{2}}{12}\)
Explanation:
C Given, length of \(\operatorname{rod}(l)=2 \mathrm{~L}\) Mass of \(\operatorname{rod}(\mathrm{m})=2 \mathrm{~mL}\) Moment of inertia of rod about the axis passing through its center \(\left(I_{C}\right)=\frac{1}{12} \mathrm{~m} l^{2}\) \(\mathrm{I}_{\mathrm{C}}=\frac{1}{12}(2 \mathrm{~mL}) \cdot(2 \mathrm{~L})^{2}\) \(=\frac{8}{12} \mathrm{~mL}^{3}\) \(=\frac{2}{3} \mathrm{~mL}^{3}\)
MHT-CET 2020
Rotational Motion
150138
A uniform disc of mass \(4 \mathrm{~kg}\) has radius of 0.4 m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
1 \(0.32 \mathrm{~kg}-\mathrm{m}_{2}^{2}\)
2 \(0.96 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(0.16 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(0.64 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
B Given, \(\text { Mass of disc }=4 \mathrm{~kg}\) \(\text { radius }=0.4 \mathrm{~m}\) Moment of Inertia of disc about the centroid \(\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{MR}^{2}\) Moment of inertia of disc about the axis passing through a point on its circumference, from parallel axis theorem \(I=I_{C}+M R^{2}\) \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}\) \(=\frac{3}{2} \mathrm{MR}^{2}\) \(=\frac{3}{2} \times 4 \times(0.4)^{2}\) \(=6 \times 0.16\) \(=0.96 \mathrm{~kg} \cdot \mathrm{m}^{2}\)
150134
A thin, uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of \((g=\) acceleration due to gravity)
1 \(\frac{L^{2} \omega^{2}}{3 g}\)
2 \(\frac{\mathrm{L}^{2} \omega^{2}}{3}\)
3 \(\frac{L^{2} \omega^{2}}{2 g}\)
4 \(\frac{L^{2} \omega^{2}}{6 g}\)
Explanation:
D According to law of conservation of energy- \(\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}\) Moment of inertia of the rod about the axis passing through its end \(\mathrm{I}=\frac{1}{3} \mathrm{ML}^{2}\) Putting the value of 'I' in equation (i), we get - \(\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \cdot \omega^{2} =\mathrm{Mgh}\) \(\mathrm{h} =\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
MHT-CET 2020
Rotational Motion
150135
Moment of inertia of a uniform solid sphere of mass ' \(M\) ' and radius ' \(R\) ' about an axis at a distance \(\left(\frac{R}{2}\right)\) from the centre is
1 \(\frac{2}{5} \mathrm{MR}^{2}\)
2 \(\frac{9}{10} \mathrm{MR}^{2}\)
3 \(\frac{13}{20} \mathrm{MR}^{2}\)
4 \(\frac{7}{5} \mathrm{MR}^{2}\)
Explanation:
C From parallel axis theorem \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{Md}^{2}\) \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2}\) \(=\frac{2}{5} \mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{4}\) \(=\frac{13}{2 \mathrm{n}} \mathrm{MR}^{2}\)
MHT-CET 2020
Rotational Motion
150136
Four spheres each of mass ' \(M\) ' and radius ' \(R\) ' are placed with their centers on the corners of a square of side ' \(L\) '. The moment of inertia of the system about any side of square is
C Given, Mass of each sphere \(=\mathrm{M}\) Radius \(=\mathrm{R}\) distance between two sphere \(=\mathrm{L}\) Now as we know moment of inertia of a sphere lying on the axis \(=\frac{2}{5} \mathrm{MR}^{2}\) So, moment of inertia of two sphere lying on the axis \(=2 \times \frac{2}{5} \mathrm{MR}^{2}=\frac{4}{5} \mathrm{MR}^{2}\) By using parallel axis theorem of moment of inertia of another two sphere lying on the line parallel to the axis at distance of \(\mathrm{L}=2\left[\frac{2}{5} \mathrm{MR}^{2}+\mathrm{ML}^{2}\right]\) \(=\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\) Total moment of inertia \(=\frac{4}{5} \mathrm{MR}^{2}+\left(\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\right)\) \(=\frac{8}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\)
MHT-CET 2020
Rotational Motion
150137
A uniform rod of length ' \(2 L\) ' has constant mass per unit length ' \(m\) '. Moment of inertia of the rod about an axis passing through its centre and perpendicular to length is
1 \(\frac{\mathrm{mL}^{3}}{3}\)
2 \(\frac{\mathrm{mL}^{2}}{4}\)
3 \(\frac{2 \mathrm{~mL}^{3}}{3}\)
4 \(\frac{\mathrm{mL}^{2}}{12}\)
Explanation:
C Given, length of \(\operatorname{rod}(l)=2 \mathrm{~L}\) Mass of \(\operatorname{rod}(\mathrm{m})=2 \mathrm{~mL}\) Moment of inertia of rod about the axis passing through its center \(\left(I_{C}\right)=\frac{1}{12} \mathrm{~m} l^{2}\) \(\mathrm{I}_{\mathrm{C}}=\frac{1}{12}(2 \mathrm{~mL}) \cdot(2 \mathrm{~L})^{2}\) \(=\frac{8}{12} \mathrm{~mL}^{3}\) \(=\frac{2}{3} \mathrm{~mL}^{3}\)
MHT-CET 2020
Rotational Motion
150138
A uniform disc of mass \(4 \mathrm{~kg}\) has radius of 0.4 m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
1 \(0.32 \mathrm{~kg}-\mathrm{m}_{2}^{2}\)
2 \(0.96 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(0.16 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(0.64 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
B Given, \(\text { Mass of disc }=4 \mathrm{~kg}\) \(\text { radius }=0.4 \mathrm{~m}\) Moment of Inertia of disc about the centroid \(\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{MR}^{2}\) Moment of inertia of disc about the axis passing through a point on its circumference, from parallel axis theorem \(I=I_{C}+M R^{2}\) \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}\) \(=\frac{3}{2} \mathrm{MR}^{2}\) \(=\frac{3}{2} \times 4 \times(0.4)^{2}\) \(=6 \times 0.16\) \(=0.96 \mathrm{~kg} \cdot \mathrm{m}^{2}\)
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Rotational Motion
150134
A thin, uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of \((g=\) acceleration due to gravity)
1 \(\frac{L^{2} \omega^{2}}{3 g}\)
2 \(\frac{\mathrm{L}^{2} \omega^{2}}{3}\)
3 \(\frac{L^{2} \omega^{2}}{2 g}\)
4 \(\frac{L^{2} \omega^{2}}{6 g}\)
Explanation:
D According to law of conservation of energy- \(\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}\) Moment of inertia of the rod about the axis passing through its end \(\mathrm{I}=\frac{1}{3} \mathrm{ML}^{2}\) Putting the value of 'I' in equation (i), we get - \(\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \cdot \omega^{2} =\mathrm{Mgh}\) \(\mathrm{h} =\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
MHT-CET 2020
Rotational Motion
150135
Moment of inertia of a uniform solid sphere of mass ' \(M\) ' and radius ' \(R\) ' about an axis at a distance \(\left(\frac{R}{2}\right)\) from the centre is
1 \(\frac{2}{5} \mathrm{MR}^{2}\)
2 \(\frac{9}{10} \mathrm{MR}^{2}\)
3 \(\frac{13}{20} \mathrm{MR}^{2}\)
4 \(\frac{7}{5} \mathrm{MR}^{2}\)
Explanation:
C From parallel axis theorem \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{Md}^{2}\) \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2}\) \(=\frac{2}{5} \mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{4}\) \(=\frac{13}{2 \mathrm{n}} \mathrm{MR}^{2}\)
MHT-CET 2020
Rotational Motion
150136
Four spheres each of mass ' \(M\) ' and radius ' \(R\) ' are placed with their centers on the corners of a square of side ' \(L\) '. The moment of inertia of the system about any side of square is
C Given, Mass of each sphere \(=\mathrm{M}\) Radius \(=\mathrm{R}\) distance between two sphere \(=\mathrm{L}\) Now as we know moment of inertia of a sphere lying on the axis \(=\frac{2}{5} \mathrm{MR}^{2}\) So, moment of inertia of two sphere lying on the axis \(=2 \times \frac{2}{5} \mathrm{MR}^{2}=\frac{4}{5} \mathrm{MR}^{2}\) By using parallel axis theorem of moment of inertia of another two sphere lying on the line parallel to the axis at distance of \(\mathrm{L}=2\left[\frac{2}{5} \mathrm{MR}^{2}+\mathrm{ML}^{2}\right]\) \(=\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\) Total moment of inertia \(=\frac{4}{5} \mathrm{MR}^{2}+\left(\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\right)\) \(=\frac{8}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\)
MHT-CET 2020
Rotational Motion
150137
A uniform rod of length ' \(2 L\) ' has constant mass per unit length ' \(m\) '. Moment of inertia of the rod about an axis passing through its centre and perpendicular to length is
1 \(\frac{\mathrm{mL}^{3}}{3}\)
2 \(\frac{\mathrm{mL}^{2}}{4}\)
3 \(\frac{2 \mathrm{~mL}^{3}}{3}\)
4 \(\frac{\mathrm{mL}^{2}}{12}\)
Explanation:
C Given, length of \(\operatorname{rod}(l)=2 \mathrm{~L}\) Mass of \(\operatorname{rod}(\mathrm{m})=2 \mathrm{~mL}\) Moment of inertia of rod about the axis passing through its center \(\left(I_{C}\right)=\frac{1}{12} \mathrm{~m} l^{2}\) \(\mathrm{I}_{\mathrm{C}}=\frac{1}{12}(2 \mathrm{~mL}) \cdot(2 \mathrm{~L})^{2}\) \(=\frac{8}{12} \mathrm{~mL}^{3}\) \(=\frac{2}{3} \mathrm{~mL}^{3}\)
MHT-CET 2020
Rotational Motion
150138
A uniform disc of mass \(4 \mathrm{~kg}\) has radius of 0.4 m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
1 \(0.32 \mathrm{~kg}-\mathrm{m}_{2}^{2}\)
2 \(0.96 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(0.16 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(0.64 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
B Given, \(\text { Mass of disc }=4 \mathrm{~kg}\) \(\text { radius }=0.4 \mathrm{~m}\) Moment of Inertia of disc about the centroid \(\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{MR}^{2}\) Moment of inertia of disc about the axis passing through a point on its circumference, from parallel axis theorem \(I=I_{C}+M R^{2}\) \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}\) \(=\frac{3}{2} \mathrm{MR}^{2}\) \(=\frac{3}{2} \times 4 \times(0.4)^{2}\) \(=6 \times 0.16\) \(=0.96 \mathrm{~kg} \cdot \mathrm{m}^{2}\)
150134
A thin, uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of \((g=\) acceleration due to gravity)
1 \(\frac{L^{2} \omega^{2}}{3 g}\)
2 \(\frac{\mathrm{L}^{2} \omega^{2}}{3}\)
3 \(\frac{L^{2} \omega^{2}}{2 g}\)
4 \(\frac{L^{2} \omega^{2}}{6 g}\)
Explanation:
D According to law of conservation of energy- \(\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}\) Moment of inertia of the rod about the axis passing through its end \(\mathrm{I}=\frac{1}{3} \mathrm{ML}^{2}\) Putting the value of 'I' in equation (i), we get - \(\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \cdot \omega^{2} =\mathrm{Mgh}\) \(\mathrm{h} =\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
MHT-CET 2020
Rotational Motion
150135
Moment of inertia of a uniform solid sphere of mass ' \(M\) ' and radius ' \(R\) ' about an axis at a distance \(\left(\frac{R}{2}\right)\) from the centre is
1 \(\frac{2}{5} \mathrm{MR}^{2}\)
2 \(\frac{9}{10} \mathrm{MR}^{2}\)
3 \(\frac{13}{20} \mathrm{MR}^{2}\)
4 \(\frac{7}{5} \mathrm{MR}^{2}\)
Explanation:
C From parallel axis theorem \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{Md}^{2}\) \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2}\) \(=\frac{2}{5} \mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{4}\) \(=\frac{13}{2 \mathrm{n}} \mathrm{MR}^{2}\)
MHT-CET 2020
Rotational Motion
150136
Four spheres each of mass ' \(M\) ' and radius ' \(R\) ' are placed with their centers on the corners of a square of side ' \(L\) '. The moment of inertia of the system about any side of square is
C Given, Mass of each sphere \(=\mathrm{M}\) Radius \(=\mathrm{R}\) distance between two sphere \(=\mathrm{L}\) Now as we know moment of inertia of a sphere lying on the axis \(=\frac{2}{5} \mathrm{MR}^{2}\) So, moment of inertia of two sphere lying on the axis \(=2 \times \frac{2}{5} \mathrm{MR}^{2}=\frac{4}{5} \mathrm{MR}^{2}\) By using parallel axis theorem of moment of inertia of another two sphere lying on the line parallel to the axis at distance of \(\mathrm{L}=2\left[\frac{2}{5} \mathrm{MR}^{2}+\mathrm{ML}^{2}\right]\) \(=\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\) Total moment of inertia \(=\frac{4}{5} \mathrm{MR}^{2}+\left(\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\right)\) \(=\frac{8}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\)
MHT-CET 2020
Rotational Motion
150137
A uniform rod of length ' \(2 L\) ' has constant mass per unit length ' \(m\) '. Moment of inertia of the rod about an axis passing through its centre and perpendicular to length is
1 \(\frac{\mathrm{mL}^{3}}{3}\)
2 \(\frac{\mathrm{mL}^{2}}{4}\)
3 \(\frac{2 \mathrm{~mL}^{3}}{3}\)
4 \(\frac{\mathrm{mL}^{2}}{12}\)
Explanation:
C Given, length of \(\operatorname{rod}(l)=2 \mathrm{~L}\) Mass of \(\operatorname{rod}(\mathrm{m})=2 \mathrm{~mL}\) Moment of inertia of rod about the axis passing through its center \(\left(I_{C}\right)=\frac{1}{12} \mathrm{~m} l^{2}\) \(\mathrm{I}_{\mathrm{C}}=\frac{1}{12}(2 \mathrm{~mL}) \cdot(2 \mathrm{~L})^{2}\) \(=\frac{8}{12} \mathrm{~mL}^{3}\) \(=\frac{2}{3} \mathrm{~mL}^{3}\)
MHT-CET 2020
Rotational Motion
150138
A uniform disc of mass \(4 \mathrm{~kg}\) has radius of 0.4 m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
1 \(0.32 \mathrm{~kg}-\mathrm{m}_{2}^{2}\)
2 \(0.96 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(0.16 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(0.64 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
B Given, \(\text { Mass of disc }=4 \mathrm{~kg}\) \(\text { radius }=0.4 \mathrm{~m}\) Moment of Inertia of disc about the centroid \(\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{MR}^{2}\) Moment of inertia of disc about the axis passing through a point on its circumference, from parallel axis theorem \(I=I_{C}+M R^{2}\) \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}\) \(=\frac{3}{2} \mathrm{MR}^{2}\) \(=\frac{3}{2} \times 4 \times(0.4)^{2}\) \(=6 \times 0.16\) \(=0.96 \mathrm{~kg} \cdot \mathrm{m}^{2}\)
150134
A thin, uniform metal rod of mass ' \(M\) ' and length ' \(L\) ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \(\omega\) '. Its centre of mass rises to a maximum height of \((g=\) acceleration due to gravity)
1 \(\frac{L^{2} \omega^{2}}{3 g}\)
2 \(\frac{\mathrm{L}^{2} \omega^{2}}{3}\)
3 \(\frac{L^{2} \omega^{2}}{2 g}\)
4 \(\frac{L^{2} \omega^{2}}{6 g}\)
Explanation:
D According to law of conservation of energy- \(\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{Mgh}\) Moment of inertia of the rod about the axis passing through its end \(\mathrm{I}=\frac{1}{3} \mathrm{ML}^{2}\) Putting the value of 'I' in equation (i), we get - \(\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \cdot \omega^{2} =\mathrm{Mgh}\) \(\mathrm{h} =\frac{\mathrm{L}^{2} \omega^{2}}{6 \mathrm{~g}}\)
MHT-CET 2020
Rotational Motion
150135
Moment of inertia of a uniform solid sphere of mass ' \(M\) ' and radius ' \(R\) ' about an axis at a distance \(\left(\frac{R}{2}\right)\) from the centre is
1 \(\frac{2}{5} \mathrm{MR}^{2}\)
2 \(\frac{9}{10} \mathrm{MR}^{2}\)
3 \(\frac{13}{20} \mathrm{MR}^{2}\)
4 \(\frac{7}{5} \mathrm{MR}^{2}\)
Explanation:
C From parallel axis theorem \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{Md}^{2}\) \(\mathrm{I}=\mathrm{I}_{\mathrm{C}}+\mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2}\) \(=\frac{2}{5} \mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{4}\) \(=\frac{13}{2 \mathrm{n}} \mathrm{MR}^{2}\)
MHT-CET 2020
Rotational Motion
150136
Four spheres each of mass ' \(M\) ' and radius ' \(R\) ' are placed with their centers on the corners of a square of side ' \(L\) '. The moment of inertia of the system about any side of square is
C Given, Mass of each sphere \(=\mathrm{M}\) Radius \(=\mathrm{R}\) distance between two sphere \(=\mathrm{L}\) Now as we know moment of inertia of a sphere lying on the axis \(=\frac{2}{5} \mathrm{MR}^{2}\) So, moment of inertia of two sphere lying on the axis \(=2 \times \frac{2}{5} \mathrm{MR}^{2}=\frac{4}{5} \mathrm{MR}^{2}\) By using parallel axis theorem of moment of inertia of another two sphere lying on the line parallel to the axis at distance of \(\mathrm{L}=2\left[\frac{2}{5} \mathrm{MR}^{2}+\mathrm{ML}^{2}\right]\) \(=\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\) Total moment of inertia \(=\frac{4}{5} \mathrm{MR}^{2}+\left(\frac{4}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\right)\) \(=\frac{8}{5} \mathrm{MR}^{2}+2 \mathrm{ML}^{2}\)
MHT-CET 2020
Rotational Motion
150137
A uniform rod of length ' \(2 L\) ' has constant mass per unit length ' \(m\) '. Moment of inertia of the rod about an axis passing through its centre and perpendicular to length is
1 \(\frac{\mathrm{mL}^{3}}{3}\)
2 \(\frac{\mathrm{mL}^{2}}{4}\)
3 \(\frac{2 \mathrm{~mL}^{3}}{3}\)
4 \(\frac{\mathrm{mL}^{2}}{12}\)
Explanation:
C Given, length of \(\operatorname{rod}(l)=2 \mathrm{~L}\) Mass of \(\operatorname{rod}(\mathrm{m})=2 \mathrm{~mL}\) Moment of inertia of rod about the axis passing through its center \(\left(I_{C}\right)=\frac{1}{12} \mathrm{~m} l^{2}\) \(\mathrm{I}_{\mathrm{C}}=\frac{1}{12}(2 \mathrm{~mL}) \cdot(2 \mathrm{~L})^{2}\) \(=\frac{8}{12} \mathrm{~mL}^{3}\) \(=\frac{2}{3} \mathrm{~mL}^{3}\)
MHT-CET 2020
Rotational Motion
150138
A uniform disc of mass \(4 \mathrm{~kg}\) has radius of 0.4 m. Its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane is
1 \(0.32 \mathrm{~kg}-\mathrm{m}_{2}^{2}\)
2 \(0.96 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(0.16 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(0.64 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
B Given, \(\text { Mass of disc }=4 \mathrm{~kg}\) \(\text { radius }=0.4 \mathrm{~m}\) Moment of Inertia of disc about the centroid \(\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{MR}^{2}\) Moment of inertia of disc about the axis passing through a point on its circumference, from parallel axis theorem \(I=I_{C}+M R^{2}\) \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}\) \(=\frac{3}{2} \mathrm{MR}^{2}\) \(=\frac{3}{2} \times 4 \times(0.4)^{2}\) \(=6 \times 0.16\) \(=0.96 \mathrm{~kg} \cdot \mathrm{m}^{2}\)
