269419
For a body rolling along a level surface, the translational and rotational K.E. areequal.The body is
1 Solid cylinder
2 disc
3 ring
4 hollow sphere
Explanation:
\(\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269420
A ring and a disc ofsame mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is \(8 \mathrm{~J}\), then that of disc is
1 \(2 J\)
2 \(4 \mathrm{~J}\)
3 \(6 \mathrm{~J}\)
4 \(16 \mathrm{~J}\)
Explanation:
\(\left.K E=\frac{1}{2} m v^{2} \frac{\square}{\square}+\frac{k^{2}}{R^{2}}\right]\)
Rotational Motion
269421
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E.which is translational is
269422
If a sphere of mass\(2 \mathrm{~kg}\) and diameter \(10 \mathrm{~cm}\) is rolling at speed of \(5 \mathrm{~ms}^{-1}\). Its rotational kinetic energy is
1 10J
2 \(30 \mathrm{~J}\)
3 50J
4 \(70 \mathrm{~J}\)
Explanation:
K E_{\text {rot }}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269475
A ringis allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its center of mass is
269419
For a body rolling along a level surface, the translational and rotational K.E. areequal.The body is
1 Solid cylinder
2 disc
3 ring
4 hollow sphere
Explanation:
\(\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269420
A ring and a disc ofsame mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is \(8 \mathrm{~J}\), then that of disc is
1 \(2 J\)
2 \(4 \mathrm{~J}\)
3 \(6 \mathrm{~J}\)
4 \(16 \mathrm{~J}\)
Explanation:
\(\left.K E=\frac{1}{2} m v^{2} \frac{\square}{\square}+\frac{k^{2}}{R^{2}}\right]\)
Rotational Motion
269421
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E.which is translational is
269422
If a sphere of mass\(2 \mathrm{~kg}\) and diameter \(10 \mathrm{~cm}\) is rolling at speed of \(5 \mathrm{~ms}^{-1}\). Its rotational kinetic energy is
1 10J
2 \(30 \mathrm{~J}\)
3 50J
4 \(70 \mathrm{~J}\)
Explanation:
K E_{\text {rot }}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269475
A ringis allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its center of mass is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
269419
For a body rolling along a level surface, the translational and rotational K.E. areequal.The body is
1 Solid cylinder
2 disc
3 ring
4 hollow sphere
Explanation:
\(\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269420
A ring and a disc ofsame mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is \(8 \mathrm{~J}\), then that of disc is
1 \(2 J\)
2 \(4 \mathrm{~J}\)
3 \(6 \mathrm{~J}\)
4 \(16 \mathrm{~J}\)
Explanation:
\(\left.K E=\frac{1}{2} m v^{2} \frac{\square}{\square}+\frac{k^{2}}{R^{2}}\right]\)
Rotational Motion
269421
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E.which is translational is
269422
If a sphere of mass\(2 \mathrm{~kg}\) and diameter \(10 \mathrm{~cm}\) is rolling at speed of \(5 \mathrm{~ms}^{-1}\). Its rotational kinetic energy is
1 10J
2 \(30 \mathrm{~J}\)
3 50J
4 \(70 \mathrm{~J}\)
Explanation:
K E_{\text {rot }}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269475
A ringis allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its center of mass is
269419
For a body rolling along a level surface, the translational and rotational K.E. areequal.The body is
1 Solid cylinder
2 disc
3 ring
4 hollow sphere
Explanation:
\(\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269420
A ring and a disc ofsame mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is \(8 \mathrm{~J}\), then that of disc is
1 \(2 J\)
2 \(4 \mathrm{~J}\)
3 \(6 \mathrm{~J}\)
4 \(16 \mathrm{~J}\)
Explanation:
\(\left.K E=\frac{1}{2} m v^{2} \frac{\square}{\square}+\frac{k^{2}}{R^{2}}\right]\)
Rotational Motion
269421
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E.which is translational is
269422
If a sphere of mass\(2 \mathrm{~kg}\) and diameter \(10 \mathrm{~cm}\) is rolling at speed of \(5 \mathrm{~ms}^{-1}\). Its rotational kinetic energy is
1 10J
2 \(30 \mathrm{~J}\)
3 50J
4 \(70 \mathrm{~J}\)
Explanation:
K E_{\text {rot }}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269475
A ringis allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its center of mass is
269419
For a body rolling along a level surface, the translational and rotational K.E. areequal.The body is
1 Solid cylinder
2 disc
3 ring
4 hollow sphere
Explanation:
\(\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269420
A ring and a disc ofsame mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is \(8 \mathrm{~J}\), then that of disc is
1 \(2 J\)
2 \(4 \mathrm{~J}\)
3 \(6 \mathrm{~J}\)
4 \(16 \mathrm{~J}\)
Explanation:
\(\left.K E=\frac{1}{2} m v^{2} \frac{\square}{\square}+\frac{k^{2}}{R^{2}}\right]\)
Rotational Motion
269421
When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E.which is translational is
269422
If a sphere of mass\(2 \mathrm{~kg}\) and diameter \(10 \mathrm{~cm}\) is rolling at speed of \(5 \mathrm{~ms}^{-1}\). Its rotational kinetic energy is
1 10J
2 \(30 \mathrm{~J}\)
3 50J
4 \(70 \mathrm{~J}\)
Explanation:
K E_{\text {rot }}=\frac{1}{2} I \omega^{2}\)
Rotational Motion
269475
A ringis allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its center of mass is