150465
A thin uniform circular ring is rolling down an inclined plane of inclination \(30^{\circ}\) without slipping. Its linear acceleration along the inclined plane will be
1 \(\frac{g}{2}\)
2 \(\frac{g}{3}\)
3 \(\frac{g}{4}\)
4 \(\frac{2 g}{3}\)
Explanation:
C We know that, acceleration on inclined plane in rolling motion- \(a=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}\) For ring, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1\) \(\mathrm{a}=\frac{\mathrm{g} \sin 30^{\circ}}{1+1}\) \(\mathrm{a}=\frac{\mathrm{g} \times \frac{1}{2}}{2}\) \(\left[\because \sin 30^{\circ}=\frac{1}{2}\right]\) \(\mathrm{a}=\frac{\mathrm{g}}{4}\)
AIPMT- 1994
Rotational Motion
150466
A very small particle rests on the top of a hemisphere of radius \(20 \mathrm{~cm}\). The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking \(\mathbf{g}=9.8 \mathbf{~ m s}^{-2}\) is
1 \(1.4 \mathrm{~ms}^{-1}\)
2 \(2.4 \mathrm{~ms}^{-1}\)
3 \(0.4 \mathrm{~ms}^{-1}\)
4 \(0.7 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\mathrm{R}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) According to question, When particle is rest on the top of hemisphere- \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv} \mathrm{v}^{2}}{\mathrm{R}}\) If particle leave the hemisphere without sliding then, \(\mathrm{N}=0\) \(\therefore \quad \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\mathrm{Rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 9.8} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=\sqrt{1.96}\) \(\mathrm{v}=1.4 \mathrm{~m} / \mathrm{s}\)
EAMCET-1993
Rotational Motion
150467
A rod of length \(l\) is held vertically stationary with its lower end located at a position \(P\) on the horizontal plane. When the rod is released to topple about \(P\), the velocity of the upper end of the rod with which it hits the ground is
1 \(\sqrt{\frac{\mathrm{g}}{l}}\)
2 \(\sqrt{3 \mathrm{~g} l}\)
3 \(3 \sqrt{\frac{\mathrm{g}}{l}}\)
4 \(\sqrt{\frac{3 \mathrm{~g}}{l}}\)
Explanation:
B The rod topples about point \(\mathrm{P}\). So, \(\mathrm{MOI}\) of \(\operatorname{rod}(\mathrm{I})=\frac{\mathrm{M} l^{2}}{3}\) By law of conservation of energy- Loss in P.E = gain in K.E \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^{2}}{3} \times \omega^{2} \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^{2}}{3}\right]\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2 \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^2}{3}\right]\) For rotation about point \(\mathrm{P}\) - \(\therefore \mathrm{v} =\omega l\) \(\text { So, } \mathrm{v} =\sqrt{\frac{3 \mathrm{~g}}{l}} \times l\) \(\mathrm{v} =\sqrt{3 \mathrm{~g} l}\)
AP EAMCET(Medical)-2009
Rotational Motion
150443
Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,
1 the heavier sphere reaches the bottom first
2 the bigger sphere reaches the bottom first
3 the two spheres reach the bottom together
4 the information given is not sufficient to tell which sphere will reach the bottom first
Explanation:
C For solid sphere- \(a=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{2}{5} \quad\) (Where \(\mathrm{K}\) is radius of gyration) So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150465
A thin uniform circular ring is rolling down an inclined plane of inclination \(30^{\circ}\) without slipping. Its linear acceleration along the inclined plane will be
1 \(\frac{g}{2}\)
2 \(\frac{g}{3}\)
3 \(\frac{g}{4}\)
4 \(\frac{2 g}{3}\)
Explanation:
C We know that, acceleration on inclined plane in rolling motion- \(a=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}\) For ring, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1\) \(\mathrm{a}=\frac{\mathrm{g} \sin 30^{\circ}}{1+1}\) \(\mathrm{a}=\frac{\mathrm{g} \times \frac{1}{2}}{2}\) \(\left[\because \sin 30^{\circ}=\frac{1}{2}\right]\) \(\mathrm{a}=\frac{\mathrm{g}}{4}\)
AIPMT- 1994
Rotational Motion
150466
A very small particle rests on the top of a hemisphere of radius \(20 \mathrm{~cm}\). The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking \(\mathbf{g}=9.8 \mathbf{~ m s}^{-2}\) is
1 \(1.4 \mathrm{~ms}^{-1}\)
2 \(2.4 \mathrm{~ms}^{-1}\)
3 \(0.4 \mathrm{~ms}^{-1}\)
4 \(0.7 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\mathrm{R}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) According to question, When particle is rest on the top of hemisphere- \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv} \mathrm{v}^{2}}{\mathrm{R}}\) If particle leave the hemisphere without sliding then, \(\mathrm{N}=0\) \(\therefore \quad \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\mathrm{Rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 9.8} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=\sqrt{1.96}\) \(\mathrm{v}=1.4 \mathrm{~m} / \mathrm{s}\)
EAMCET-1993
Rotational Motion
150467
A rod of length \(l\) is held vertically stationary with its lower end located at a position \(P\) on the horizontal plane. When the rod is released to topple about \(P\), the velocity of the upper end of the rod with which it hits the ground is
1 \(\sqrt{\frac{\mathrm{g}}{l}}\)
2 \(\sqrt{3 \mathrm{~g} l}\)
3 \(3 \sqrt{\frac{\mathrm{g}}{l}}\)
4 \(\sqrt{\frac{3 \mathrm{~g}}{l}}\)
Explanation:
B The rod topples about point \(\mathrm{P}\). So, \(\mathrm{MOI}\) of \(\operatorname{rod}(\mathrm{I})=\frac{\mathrm{M} l^{2}}{3}\) By law of conservation of energy- Loss in P.E = gain in K.E \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^{2}}{3} \times \omega^{2} \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^{2}}{3}\right]\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2 \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^2}{3}\right]\) For rotation about point \(\mathrm{P}\) - \(\therefore \mathrm{v} =\omega l\) \(\text { So, } \mathrm{v} =\sqrt{\frac{3 \mathrm{~g}}{l}} \times l\) \(\mathrm{v} =\sqrt{3 \mathrm{~g} l}\)
AP EAMCET(Medical)-2009
Rotational Motion
150443
Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,
1 the heavier sphere reaches the bottom first
2 the bigger sphere reaches the bottom first
3 the two spheres reach the bottom together
4 the information given is not sufficient to tell which sphere will reach the bottom first
Explanation:
C For solid sphere- \(a=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{2}{5} \quad\) (Where \(\mathrm{K}\) is radius of gyration) So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.
150465
A thin uniform circular ring is rolling down an inclined plane of inclination \(30^{\circ}\) without slipping. Its linear acceleration along the inclined plane will be
1 \(\frac{g}{2}\)
2 \(\frac{g}{3}\)
3 \(\frac{g}{4}\)
4 \(\frac{2 g}{3}\)
Explanation:
C We know that, acceleration on inclined plane in rolling motion- \(a=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}\) For ring, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1\) \(\mathrm{a}=\frac{\mathrm{g} \sin 30^{\circ}}{1+1}\) \(\mathrm{a}=\frac{\mathrm{g} \times \frac{1}{2}}{2}\) \(\left[\because \sin 30^{\circ}=\frac{1}{2}\right]\) \(\mathrm{a}=\frac{\mathrm{g}}{4}\)
AIPMT- 1994
Rotational Motion
150466
A very small particle rests on the top of a hemisphere of radius \(20 \mathrm{~cm}\). The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking \(\mathbf{g}=9.8 \mathbf{~ m s}^{-2}\) is
1 \(1.4 \mathrm{~ms}^{-1}\)
2 \(2.4 \mathrm{~ms}^{-1}\)
3 \(0.4 \mathrm{~ms}^{-1}\)
4 \(0.7 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\mathrm{R}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) According to question, When particle is rest on the top of hemisphere- \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv} \mathrm{v}^{2}}{\mathrm{R}}\) If particle leave the hemisphere without sliding then, \(\mathrm{N}=0\) \(\therefore \quad \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\mathrm{Rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 9.8} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=\sqrt{1.96}\) \(\mathrm{v}=1.4 \mathrm{~m} / \mathrm{s}\)
EAMCET-1993
Rotational Motion
150467
A rod of length \(l\) is held vertically stationary with its lower end located at a position \(P\) on the horizontal plane. When the rod is released to topple about \(P\), the velocity of the upper end of the rod with which it hits the ground is
1 \(\sqrt{\frac{\mathrm{g}}{l}}\)
2 \(\sqrt{3 \mathrm{~g} l}\)
3 \(3 \sqrt{\frac{\mathrm{g}}{l}}\)
4 \(\sqrt{\frac{3 \mathrm{~g}}{l}}\)
Explanation:
B The rod topples about point \(\mathrm{P}\). So, \(\mathrm{MOI}\) of \(\operatorname{rod}(\mathrm{I})=\frac{\mathrm{M} l^{2}}{3}\) By law of conservation of energy- Loss in P.E = gain in K.E \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^{2}}{3} \times \omega^{2} \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^{2}}{3}\right]\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2 \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^2}{3}\right]\) For rotation about point \(\mathrm{P}\) - \(\therefore \mathrm{v} =\omega l\) \(\text { So, } \mathrm{v} =\sqrt{\frac{3 \mathrm{~g}}{l}} \times l\) \(\mathrm{v} =\sqrt{3 \mathrm{~g} l}\)
AP EAMCET(Medical)-2009
Rotational Motion
150443
Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,
1 the heavier sphere reaches the bottom first
2 the bigger sphere reaches the bottom first
3 the two spheres reach the bottom together
4 the information given is not sufficient to tell which sphere will reach the bottom first
Explanation:
C For solid sphere- \(a=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{2}{5} \quad\) (Where \(\mathrm{K}\) is radius of gyration) So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.
150465
A thin uniform circular ring is rolling down an inclined plane of inclination \(30^{\circ}\) without slipping. Its linear acceleration along the inclined plane will be
1 \(\frac{g}{2}\)
2 \(\frac{g}{3}\)
3 \(\frac{g}{4}\)
4 \(\frac{2 g}{3}\)
Explanation:
C We know that, acceleration on inclined plane in rolling motion- \(a=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}\) For ring, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=1\) \(\mathrm{a}=\frac{\mathrm{g} \sin 30^{\circ}}{1+1}\) \(\mathrm{a}=\frac{\mathrm{g} \times \frac{1}{2}}{2}\) \(\left[\because \sin 30^{\circ}=\frac{1}{2}\right]\) \(\mathrm{a}=\frac{\mathrm{g}}{4}\)
AIPMT- 1994
Rotational Motion
150466
A very small particle rests on the top of a hemisphere of radius \(20 \mathrm{~cm}\). The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, taking \(\mathbf{g}=9.8 \mathbf{~ m s}^{-2}\) is
1 \(1.4 \mathrm{~ms}^{-1}\)
2 \(2.4 \mathrm{~ms}^{-1}\)
3 \(0.4 \mathrm{~ms}^{-1}\)
4 \(0.7 \mathrm{~ms}^{-1}\)
Explanation:
A Given, \(\mathrm{R}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) According to question, When particle is rest on the top of hemisphere- \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv} \mathrm{v}^{2}}{\mathrm{R}}\) If particle leave the hemisphere without sliding then, \(\mathrm{N}=0\) \(\therefore \quad \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\mathrm{Rg}}\) \(\mathrm{v}=\sqrt{0.2 \times 9.8} \mathrm{~m} / \mathrm{s}\) \(\mathrm{v}=\sqrt{1.96}\) \(\mathrm{v}=1.4 \mathrm{~m} / \mathrm{s}\)
EAMCET-1993
Rotational Motion
150467
A rod of length \(l\) is held vertically stationary with its lower end located at a position \(P\) on the horizontal plane. When the rod is released to topple about \(P\), the velocity of the upper end of the rod with which it hits the ground is
1 \(\sqrt{\frac{\mathrm{g}}{l}}\)
2 \(\sqrt{3 \mathrm{~g} l}\)
3 \(3 \sqrt{\frac{\mathrm{g}}{l}}\)
4 \(\sqrt{\frac{3 \mathrm{~g}}{l}}\)
Explanation:
B The rod topples about point \(\mathrm{P}\). So, \(\mathrm{MOI}\) of \(\operatorname{rod}(\mathrm{I})=\frac{\mathrm{M} l^{2}}{3}\) By law of conservation of energy- Loss in P.E = gain in K.E \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^{2}}{3} \times \omega^{2} \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^{2}}{3}\right]\) \(\mathrm{mg} \frac{l}{2}=\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2 \quad\left[\because \mathrm{I}=\frac{\mathrm{m} l^2}{3}\right]\) For rotation about point \(\mathrm{P}\) - \(\therefore \mathrm{v} =\omega l\) \(\text { So, } \mathrm{v} =\sqrt{\frac{3 \mathrm{~g}}{l}} \times l\) \(\mathrm{v} =\sqrt{3 \mathrm{~g} l}\)
AP EAMCET(Medical)-2009
Rotational Motion
150443
Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,
1 the heavier sphere reaches the bottom first
2 the bigger sphere reaches the bottom first
3 the two spheres reach the bottom together
4 the information given is not sufficient to tell which sphere will reach the bottom first
Explanation:
C For solid sphere- \(a=\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\) \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{2}{5} \quad\) (Where \(\mathrm{K}\) is radius of gyration) So acceleration is not affected in this case because it independent of mass. So, both sphere reach the bottom together.
