150108
The moment of inertia of a thin rod of mass ' \(M\) ' and length ' \(L\) ' about an axis passing through a point at a distance \(\frac{\mathrm{L}}{4}\) from its center and perpendicular to its length is
1 \(\frac{\mathrm{ML}^{3}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{48}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{7 \mathrm{ML}^{2}}{48}\)
Explanation:
D Given, mass of thin \(\operatorname{rod}=\mathrm{M}\) Length of thin \(\operatorname{rod}=\mathrm{L}\) Applying parallel axis theorem, Moment of inertia \((I)=I_{a}+M^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}\) \(I=\frac{M L^{2}}{12}+\frac{M L^{2}}{16}\) \(I=\frac{7}{48} M L^{2}\)
AP EAMCET-25.08.2021
Rotational Motion
150109
The moment of inertia of a hollow square cube of side length a and mass \(M\) about an axis passing through the center of the cube is a
1 \(\frac{1}{18} \mathrm{Ma}^{2}\)
2 \(\frac{1}{6} \mathrm{Ma}^{2}\)
3 \(\frac{5}{6} \mathrm{Ma}^{2}\)
4 \(\frac{5}{18} \mathrm{Ma}^{2}\)
Explanation:
D Given that, Hollow square cube of side length \(=a\) Each square plate of mass \(=\mathrm{m}\) Total mass of square cube \(M=6 \mathrm{~m}\) M.O.I of square plate about axis through C.O.M and perpendicular to its plane \(=\frac{\mathrm{ma}^{2}}{6}\) Axis of rotation passes through center of mass of two square plates (1) and (2) facing opposite to each other \(\mathrm{I}_{2}=\mathrm{I}_{1}=\frac{\mathrm{ma}^{2}}{6}\) M.O.I of a square plate about axis passing through C.O.M and parallel to its plane is. \(\mathrm{I}=\frac{\mathrm{ma}^{2}}{12}\) and the remaining 4 plates have a distance of from the cube axis. By parallel axis theorem, M.O.I of plate (3), (4), (5) and (6) about cube axis is \(\mathrm{I}_{3}=\mathrm{I}_{4}=\mathrm{I}_{5}=\mathrm{I}_{6}=\frac{\mathrm{ma}^{2}}{12}+\mathrm{m}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{ma}^{2}}{3}\) Therefore, total M.O.I about cube axis is, \(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}+\mathrm{I}_{5}+\mathrm{I}_{6}\) \(I =2 I_{1}+4 I_{3}\) \(=2 \times \frac{\mathrm{ma}^{2}}{6}+4 \times \frac{\mathrm{ma}^{2}}{3}=\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \times \mathrm{Ma}^{2}}{6 \times 3}\) \(I =\frac{5 \mathrm{Ma}^{2}}{18}\)
Assam CEE-2021
Rotational Motion
150110
Four identical rings of radius \(R\) and mass \(M\) are placed at the corner of a square in \(x-y\) plane such that each ring touches the two rings tangentially. The moment of inertia of this system about \(\mathrm{z}\)-axis passing through the centre of the square is
1 \(8 \mathrm{MR}_{2}^{2}\)
2 \(7 \mathrm{MR}^{2}\)
3 \(6 \mathrm{MR}^{2}\)
4 \(4 \mathrm{MR}^{2}\)
Explanation:
C According to question M.I. of system \(=4 \times\) M.I. of each ring apply parallel axis theorem \(\mathrm{MI})_{\text {system }}=4 \times\left[\mathrm{MR}^{2}+\mathrm{Mx}^{2}\right]\) \(=4\left[\mathrm{MR}^{2}+\mathrm{M}\left(\frac{\mathrm{R}}{\sqrt{2}}\right)^{2}\right] \quad\left[\text { Where, } \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}\right]\) \(=4\left[\mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{2}\right]\) \(\text { (M.I.) } \text { system }^{2}=6 \mathrm{MR}^{2}\)
UPSEE 2020
Rotational Motion
150111
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is ' \(I\) '. Now, the rod is bent in a ring and its moment of inertia about diameter is ' \(I_{1}\) ', Then \(I / I_{1}\) is
1 \(\frac{8 \pi^{2}}{3}\)
2 \(\frac{\pi^{2}}{3}\)
3 \(\frac{11 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Circumference of ring \(=\) Length of rod \(2 \pi \mathrm{R}=\mathrm{L}\) \(\text { or } \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of one of its ends of rod \(\mathrm{I}=\frac{\mathrm{ML}^{2}}{3}\) Moment of inertia of ring \(\left(\mathrm{I}_{1}\right)=\frac{\mathrm{MR}^{2}}{2}\) Putting the value of \(\mathrm{R}\) from equation (i), we get \(\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{8 \pi^{2}}\) Taking ratio of equation (ii) and (iii), we get \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{\mathrm{ML}^{2}}{3} \times \frac{8 \pi^{2}}{\mathrm{ML}^{2}}\) \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{8}{3} \pi^{2}\)
150108
The moment of inertia of a thin rod of mass ' \(M\) ' and length ' \(L\) ' about an axis passing through a point at a distance \(\frac{\mathrm{L}}{4}\) from its center and perpendicular to its length is
1 \(\frac{\mathrm{ML}^{3}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{48}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{7 \mathrm{ML}^{2}}{48}\)
Explanation:
D Given, mass of thin \(\operatorname{rod}=\mathrm{M}\) Length of thin \(\operatorname{rod}=\mathrm{L}\) Applying parallel axis theorem, Moment of inertia \((I)=I_{a}+M^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}\) \(I=\frac{M L^{2}}{12}+\frac{M L^{2}}{16}\) \(I=\frac{7}{48} M L^{2}\)
AP EAMCET-25.08.2021
Rotational Motion
150109
The moment of inertia of a hollow square cube of side length a and mass \(M\) about an axis passing through the center of the cube is a
1 \(\frac{1}{18} \mathrm{Ma}^{2}\)
2 \(\frac{1}{6} \mathrm{Ma}^{2}\)
3 \(\frac{5}{6} \mathrm{Ma}^{2}\)
4 \(\frac{5}{18} \mathrm{Ma}^{2}\)
Explanation:
D Given that, Hollow square cube of side length \(=a\) Each square plate of mass \(=\mathrm{m}\) Total mass of square cube \(M=6 \mathrm{~m}\) M.O.I of square plate about axis through C.O.M and perpendicular to its plane \(=\frac{\mathrm{ma}^{2}}{6}\) Axis of rotation passes through center of mass of two square plates (1) and (2) facing opposite to each other \(\mathrm{I}_{2}=\mathrm{I}_{1}=\frac{\mathrm{ma}^{2}}{6}\) M.O.I of a square plate about axis passing through C.O.M and parallel to its plane is. \(\mathrm{I}=\frac{\mathrm{ma}^{2}}{12}\) and the remaining 4 plates have a distance of from the cube axis. By parallel axis theorem, M.O.I of plate (3), (4), (5) and (6) about cube axis is \(\mathrm{I}_{3}=\mathrm{I}_{4}=\mathrm{I}_{5}=\mathrm{I}_{6}=\frac{\mathrm{ma}^{2}}{12}+\mathrm{m}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{ma}^{2}}{3}\) Therefore, total M.O.I about cube axis is, \(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}+\mathrm{I}_{5}+\mathrm{I}_{6}\) \(I =2 I_{1}+4 I_{3}\) \(=2 \times \frac{\mathrm{ma}^{2}}{6}+4 \times \frac{\mathrm{ma}^{2}}{3}=\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \times \mathrm{Ma}^{2}}{6 \times 3}\) \(I =\frac{5 \mathrm{Ma}^{2}}{18}\)
Assam CEE-2021
Rotational Motion
150110
Four identical rings of radius \(R\) and mass \(M\) are placed at the corner of a square in \(x-y\) plane such that each ring touches the two rings tangentially. The moment of inertia of this system about \(\mathrm{z}\)-axis passing through the centre of the square is
1 \(8 \mathrm{MR}_{2}^{2}\)
2 \(7 \mathrm{MR}^{2}\)
3 \(6 \mathrm{MR}^{2}\)
4 \(4 \mathrm{MR}^{2}\)
Explanation:
C According to question M.I. of system \(=4 \times\) M.I. of each ring apply parallel axis theorem \(\mathrm{MI})_{\text {system }}=4 \times\left[\mathrm{MR}^{2}+\mathrm{Mx}^{2}\right]\) \(=4\left[\mathrm{MR}^{2}+\mathrm{M}\left(\frac{\mathrm{R}}{\sqrt{2}}\right)^{2}\right] \quad\left[\text { Where, } \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}\right]\) \(=4\left[\mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{2}\right]\) \(\text { (M.I.) } \text { system }^{2}=6 \mathrm{MR}^{2}\)
UPSEE 2020
Rotational Motion
150111
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is ' \(I\) '. Now, the rod is bent in a ring and its moment of inertia about diameter is ' \(I_{1}\) ', Then \(I / I_{1}\) is
1 \(\frac{8 \pi^{2}}{3}\)
2 \(\frac{\pi^{2}}{3}\)
3 \(\frac{11 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Circumference of ring \(=\) Length of rod \(2 \pi \mathrm{R}=\mathrm{L}\) \(\text { or } \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of one of its ends of rod \(\mathrm{I}=\frac{\mathrm{ML}^{2}}{3}\) Moment of inertia of ring \(\left(\mathrm{I}_{1}\right)=\frac{\mathrm{MR}^{2}}{2}\) Putting the value of \(\mathrm{R}\) from equation (i), we get \(\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{8 \pi^{2}}\) Taking ratio of equation (ii) and (iii), we get \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{\mathrm{ML}^{2}}{3} \times \frac{8 \pi^{2}}{\mathrm{ML}^{2}}\) \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{8}{3} \pi^{2}\)
150108
The moment of inertia of a thin rod of mass ' \(M\) ' and length ' \(L\) ' about an axis passing through a point at a distance \(\frac{\mathrm{L}}{4}\) from its center and perpendicular to its length is
1 \(\frac{\mathrm{ML}^{3}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{48}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{7 \mathrm{ML}^{2}}{48}\)
Explanation:
D Given, mass of thin \(\operatorname{rod}=\mathrm{M}\) Length of thin \(\operatorname{rod}=\mathrm{L}\) Applying parallel axis theorem, Moment of inertia \((I)=I_{a}+M^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}\) \(I=\frac{M L^{2}}{12}+\frac{M L^{2}}{16}\) \(I=\frac{7}{48} M L^{2}\)
AP EAMCET-25.08.2021
Rotational Motion
150109
The moment of inertia of a hollow square cube of side length a and mass \(M\) about an axis passing through the center of the cube is a
1 \(\frac{1}{18} \mathrm{Ma}^{2}\)
2 \(\frac{1}{6} \mathrm{Ma}^{2}\)
3 \(\frac{5}{6} \mathrm{Ma}^{2}\)
4 \(\frac{5}{18} \mathrm{Ma}^{2}\)
Explanation:
D Given that, Hollow square cube of side length \(=a\) Each square plate of mass \(=\mathrm{m}\) Total mass of square cube \(M=6 \mathrm{~m}\) M.O.I of square plate about axis through C.O.M and perpendicular to its plane \(=\frac{\mathrm{ma}^{2}}{6}\) Axis of rotation passes through center of mass of two square plates (1) and (2) facing opposite to each other \(\mathrm{I}_{2}=\mathrm{I}_{1}=\frac{\mathrm{ma}^{2}}{6}\) M.O.I of a square plate about axis passing through C.O.M and parallel to its plane is. \(\mathrm{I}=\frac{\mathrm{ma}^{2}}{12}\) and the remaining 4 plates have a distance of from the cube axis. By parallel axis theorem, M.O.I of plate (3), (4), (5) and (6) about cube axis is \(\mathrm{I}_{3}=\mathrm{I}_{4}=\mathrm{I}_{5}=\mathrm{I}_{6}=\frac{\mathrm{ma}^{2}}{12}+\mathrm{m}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{ma}^{2}}{3}\) Therefore, total M.O.I about cube axis is, \(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}+\mathrm{I}_{5}+\mathrm{I}_{6}\) \(I =2 I_{1}+4 I_{3}\) \(=2 \times \frac{\mathrm{ma}^{2}}{6}+4 \times \frac{\mathrm{ma}^{2}}{3}=\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \times \mathrm{Ma}^{2}}{6 \times 3}\) \(I =\frac{5 \mathrm{Ma}^{2}}{18}\)
Assam CEE-2021
Rotational Motion
150110
Four identical rings of radius \(R\) and mass \(M\) are placed at the corner of a square in \(x-y\) plane such that each ring touches the two rings tangentially. The moment of inertia of this system about \(\mathrm{z}\)-axis passing through the centre of the square is
1 \(8 \mathrm{MR}_{2}^{2}\)
2 \(7 \mathrm{MR}^{2}\)
3 \(6 \mathrm{MR}^{2}\)
4 \(4 \mathrm{MR}^{2}\)
Explanation:
C According to question M.I. of system \(=4 \times\) M.I. of each ring apply parallel axis theorem \(\mathrm{MI})_{\text {system }}=4 \times\left[\mathrm{MR}^{2}+\mathrm{Mx}^{2}\right]\) \(=4\left[\mathrm{MR}^{2}+\mathrm{M}\left(\frac{\mathrm{R}}{\sqrt{2}}\right)^{2}\right] \quad\left[\text { Where, } \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}\right]\) \(=4\left[\mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{2}\right]\) \(\text { (M.I.) } \text { system }^{2}=6 \mathrm{MR}^{2}\)
UPSEE 2020
Rotational Motion
150111
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is ' \(I\) '. Now, the rod is bent in a ring and its moment of inertia about diameter is ' \(I_{1}\) ', Then \(I / I_{1}\) is
1 \(\frac{8 \pi^{2}}{3}\)
2 \(\frac{\pi^{2}}{3}\)
3 \(\frac{11 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Circumference of ring \(=\) Length of rod \(2 \pi \mathrm{R}=\mathrm{L}\) \(\text { or } \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of one of its ends of rod \(\mathrm{I}=\frac{\mathrm{ML}^{2}}{3}\) Moment of inertia of ring \(\left(\mathrm{I}_{1}\right)=\frac{\mathrm{MR}^{2}}{2}\) Putting the value of \(\mathrm{R}\) from equation (i), we get \(\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{8 \pi^{2}}\) Taking ratio of equation (ii) and (iii), we get \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{\mathrm{ML}^{2}}{3} \times \frac{8 \pi^{2}}{\mathrm{ML}^{2}}\) \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{8}{3} \pi^{2}\)
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Rotational Motion
150108
The moment of inertia of a thin rod of mass ' \(M\) ' and length ' \(L\) ' about an axis passing through a point at a distance \(\frac{\mathrm{L}}{4}\) from its center and perpendicular to its length is
1 \(\frac{\mathrm{ML}^{3}}{48}\)
2 \(\frac{\mathrm{ML}^{2}}{48}\)
3 \(\frac{\mathrm{ML}^{2}}{12}\)
4 \(\frac{7 \mathrm{ML}^{2}}{48}\)
Explanation:
D Given, mass of thin \(\operatorname{rod}=\mathrm{M}\) Length of thin \(\operatorname{rod}=\mathrm{L}\) Applying parallel axis theorem, Moment of inertia \((I)=I_{a}+M^{2}\) \(I=\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}\) \(I=\frac{M L^{2}}{12}+\frac{M L^{2}}{16}\) \(I=\frac{7}{48} M L^{2}\)
AP EAMCET-25.08.2021
Rotational Motion
150109
The moment of inertia of a hollow square cube of side length a and mass \(M\) about an axis passing through the center of the cube is a
1 \(\frac{1}{18} \mathrm{Ma}^{2}\)
2 \(\frac{1}{6} \mathrm{Ma}^{2}\)
3 \(\frac{5}{6} \mathrm{Ma}^{2}\)
4 \(\frac{5}{18} \mathrm{Ma}^{2}\)
Explanation:
D Given that, Hollow square cube of side length \(=a\) Each square plate of mass \(=\mathrm{m}\) Total mass of square cube \(M=6 \mathrm{~m}\) M.O.I of square plate about axis through C.O.M and perpendicular to its plane \(=\frac{\mathrm{ma}^{2}}{6}\) Axis of rotation passes through center of mass of two square plates (1) and (2) facing opposite to each other \(\mathrm{I}_{2}=\mathrm{I}_{1}=\frac{\mathrm{ma}^{2}}{6}\) M.O.I of a square plate about axis passing through C.O.M and parallel to its plane is. \(\mathrm{I}=\frac{\mathrm{ma}^{2}}{12}\) and the remaining 4 plates have a distance of from the cube axis. By parallel axis theorem, M.O.I of plate (3), (4), (5) and (6) about cube axis is \(\mathrm{I}_{3}=\mathrm{I}_{4}=\mathrm{I}_{5}=\mathrm{I}_{6}=\frac{\mathrm{ma}^{2}}{12}+\mathrm{m}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{ma}^{2}}{3}\) Therefore, total M.O.I about cube axis is, \(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}+\mathrm{I}_{5}+\mathrm{I}_{6}\) \(I =2 I_{1}+4 I_{3}\) \(=2 \times \frac{\mathrm{ma}^{2}}{6}+4 \times \frac{\mathrm{ma}^{2}}{3}=\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \mathrm{ma}^{2}}{3}\) \(I =\frac{5 \times \mathrm{Ma}^{2}}{6 \times 3}\) \(I =\frac{5 \mathrm{Ma}^{2}}{18}\)
Assam CEE-2021
Rotational Motion
150110
Four identical rings of radius \(R\) and mass \(M\) are placed at the corner of a square in \(x-y\) plane such that each ring touches the two rings tangentially. The moment of inertia of this system about \(\mathrm{z}\)-axis passing through the centre of the square is
1 \(8 \mathrm{MR}_{2}^{2}\)
2 \(7 \mathrm{MR}^{2}\)
3 \(6 \mathrm{MR}^{2}\)
4 \(4 \mathrm{MR}^{2}\)
Explanation:
C According to question M.I. of system \(=4 \times\) M.I. of each ring apply parallel axis theorem \(\mathrm{MI})_{\text {system }}=4 \times\left[\mathrm{MR}^{2}+\mathrm{Mx}^{2}\right]\) \(=4\left[\mathrm{MR}^{2}+\mathrm{M}\left(\frac{\mathrm{R}}{\sqrt{2}}\right)^{2}\right] \quad\left[\text { Where, } \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}\right]\) \(=4\left[\mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{2}\right]\) \(\text { (M.I.) } \text { system }^{2}=6 \mathrm{MR}^{2}\)
UPSEE 2020
Rotational Motion
150111
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is ' \(I\) '. Now, the rod is bent in a ring and its moment of inertia about diameter is ' \(I_{1}\) ', Then \(I / I_{1}\) is
1 \(\frac{8 \pi^{2}}{3}\)
2 \(\frac{\pi^{2}}{3}\)
3 \(\frac{11 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Circumference of ring \(=\) Length of rod \(2 \pi \mathrm{R}=\mathrm{L}\) \(\text { or } \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of one of its ends of rod \(\mathrm{I}=\frac{\mathrm{ML}^{2}}{3}\) Moment of inertia of ring \(\left(\mathrm{I}_{1}\right)=\frac{\mathrm{MR}^{2}}{2}\) Putting the value of \(\mathrm{R}\) from equation (i), we get \(\mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{8 \pi^{2}}\) Taking ratio of equation (ii) and (iii), we get \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{\mathrm{ML}^{2}}{3} \times \frac{8 \pi^{2}}{\mathrm{ML}^{2}}\) \(\frac{\mathrm{I}}{\mathrm{I}_{1}}=\frac{8}{3} \pi^{2}\)
