NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150317
A particle executes uniform circular motion with angular momentum ' \(L\) '. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{\mathrm{L}}{4}\)
3 \(2 \mathrm{~L}\)
4 \(\frac{\mathrm{L}}{2}\)
Explanation:
B Suppose that, Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) And \(\text { Kinetic energy }(\mathrm{K}) =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K} =\frac{1}{2} \mathrm{I} \omega . \omega\) \(\mathrm{K} =\frac{\omega \mathrm{L}}{2} \quad(\mathrm{~L}=\mathrm{I} \omega)\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) When angular frequency is doubled and its rotational kinetic energy be comes half. \(\text { Frequency }\left(\omega^{\prime}\right)=2 \omega \text { and } K^{\prime}=\frac{K}{2}\) Then, new angular momentum \(\mathrm{L}^{\prime}=\frac{2 \times\left(\frac{\mathrm{K}}{2}\right)}{2 \omega}=\frac{2 \mathrm{~K}}{4 \omega}\) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
MHT-CET 2020
Rotational Motion
150318
Two bodies have their moments of inertia I and \(2 \mathrm{I}\) respectively about their axes of rotation. If their kinetic energies of rotation are equal, their angular moment will be in the ratio
150319
A rotating body has angular momentum ' \(L\) '. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
150320
A solid cylinder of mass ' \(M\) ' ' and radius ' \(R\) ' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(v\) '. The height of the inclined plane is \(\mathrm{g}=\) acceleration due to gravity)
1 \(\frac{2 v^{2}}{3 g}\)
2 \(\frac{7 v^{2}}{9 g}\)
3 \(\frac{4 \mathrm{v}^{2}}{5 \mathrm{~g}}\)
4 \(\frac{3 v^{2}}{4 g}\)
Explanation:
D Potential energy of the solid cylinder at height \(\mathrm{h})=\mathrm{Mgh}\) K.E. of center of mass when reached at bottom, \(\text { K.E. }=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} {\left[\mathrm{I}=\mathrm{MK}^{2}\right]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{MK}^{2} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}} {[\mathrm{v}=\mathrm{R} \omega]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right) \quad \text {; where } \mathrm{K} \text { is gyration }\) radius For a solid cylinder, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2}\) K.E. \(=\frac{3}{4} \mathrm{Mv}^{2}\) Now, from conservation of energy theorem, \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\)
150317
A particle executes uniform circular motion with angular momentum ' \(L\) '. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{\mathrm{L}}{4}\)
3 \(2 \mathrm{~L}\)
4 \(\frac{\mathrm{L}}{2}\)
Explanation:
B Suppose that, Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) And \(\text { Kinetic energy }(\mathrm{K}) =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K} =\frac{1}{2} \mathrm{I} \omega . \omega\) \(\mathrm{K} =\frac{\omega \mathrm{L}}{2} \quad(\mathrm{~L}=\mathrm{I} \omega)\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) When angular frequency is doubled and its rotational kinetic energy be comes half. \(\text { Frequency }\left(\omega^{\prime}\right)=2 \omega \text { and } K^{\prime}=\frac{K}{2}\) Then, new angular momentum \(\mathrm{L}^{\prime}=\frac{2 \times\left(\frac{\mathrm{K}}{2}\right)}{2 \omega}=\frac{2 \mathrm{~K}}{4 \omega}\) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
MHT-CET 2020
Rotational Motion
150318
Two bodies have their moments of inertia I and \(2 \mathrm{I}\) respectively about their axes of rotation. If their kinetic energies of rotation are equal, their angular moment will be in the ratio
150319
A rotating body has angular momentum ' \(L\) '. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
150320
A solid cylinder of mass ' \(M\) ' ' and radius ' \(R\) ' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(v\) '. The height of the inclined plane is \(\mathrm{g}=\) acceleration due to gravity)
1 \(\frac{2 v^{2}}{3 g}\)
2 \(\frac{7 v^{2}}{9 g}\)
3 \(\frac{4 \mathrm{v}^{2}}{5 \mathrm{~g}}\)
4 \(\frac{3 v^{2}}{4 g}\)
Explanation:
D Potential energy of the solid cylinder at height \(\mathrm{h})=\mathrm{Mgh}\) K.E. of center of mass when reached at bottom, \(\text { K.E. }=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} {\left[\mathrm{I}=\mathrm{MK}^{2}\right]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{MK}^{2} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}} {[\mathrm{v}=\mathrm{R} \omega]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right) \quad \text {; where } \mathrm{K} \text { is gyration }\) radius For a solid cylinder, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2}\) K.E. \(=\frac{3}{4} \mathrm{Mv}^{2}\) Now, from conservation of energy theorem, \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\)
150317
A particle executes uniform circular motion with angular momentum ' \(L\) '. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{\mathrm{L}}{4}\)
3 \(2 \mathrm{~L}\)
4 \(\frac{\mathrm{L}}{2}\)
Explanation:
B Suppose that, Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) And \(\text { Kinetic energy }(\mathrm{K}) =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K} =\frac{1}{2} \mathrm{I} \omega . \omega\) \(\mathrm{K} =\frac{\omega \mathrm{L}}{2} \quad(\mathrm{~L}=\mathrm{I} \omega)\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) When angular frequency is doubled and its rotational kinetic energy be comes half. \(\text { Frequency }\left(\omega^{\prime}\right)=2 \omega \text { and } K^{\prime}=\frac{K}{2}\) Then, new angular momentum \(\mathrm{L}^{\prime}=\frac{2 \times\left(\frac{\mathrm{K}}{2}\right)}{2 \omega}=\frac{2 \mathrm{~K}}{4 \omega}\) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
MHT-CET 2020
Rotational Motion
150318
Two bodies have their moments of inertia I and \(2 \mathrm{I}\) respectively about their axes of rotation. If their kinetic energies of rotation are equal, their angular moment will be in the ratio
150319
A rotating body has angular momentum ' \(L\) '. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
150320
A solid cylinder of mass ' \(M\) ' ' and radius ' \(R\) ' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(v\) '. The height of the inclined plane is \(\mathrm{g}=\) acceleration due to gravity)
1 \(\frac{2 v^{2}}{3 g}\)
2 \(\frac{7 v^{2}}{9 g}\)
3 \(\frac{4 \mathrm{v}^{2}}{5 \mathrm{~g}}\)
4 \(\frac{3 v^{2}}{4 g}\)
Explanation:
D Potential energy of the solid cylinder at height \(\mathrm{h})=\mathrm{Mgh}\) K.E. of center of mass when reached at bottom, \(\text { K.E. }=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} {\left[\mathrm{I}=\mathrm{MK}^{2}\right]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{MK}^{2} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}} {[\mathrm{v}=\mathrm{R} \omega]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right) \quad \text {; where } \mathrm{K} \text { is gyration }\) radius For a solid cylinder, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2}\) K.E. \(=\frac{3}{4} \mathrm{Mv}^{2}\) Now, from conservation of energy theorem, \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Rotational Motion
150317
A particle executes uniform circular motion with angular momentum ' \(L\) '. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{\mathrm{L}}{4}\)
3 \(2 \mathrm{~L}\)
4 \(\frac{\mathrm{L}}{2}\)
Explanation:
B Suppose that, Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) And \(\text { Kinetic energy }(\mathrm{K}) =\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{~K} =\frac{1}{2} \mathrm{I} \omega . \omega\) \(\mathrm{K} =\frac{\omega \mathrm{L}}{2} \quad(\mathrm{~L}=\mathrm{I} \omega)\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) When angular frequency is doubled and its rotational kinetic energy be comes half. \(\text { Frequency }\left(\omega^{\prime}\right)=2 \omega \text { and } K^{\prime}=\frac{K}{2}\) Then, new angular momentum \(\mathrm{L}^{\prime}=\frac{2 \times\left(\frac{\mathrm{K}}{2}\right)}{2 \omega}=\frac{2 \mathrm{~K}}{4 \omega}\) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
MHT-CET 2020
Rotational Motion
150318
Two bodies have their moments of inertia I and \(2 \mathrm{I}\) respectively about their axes of rotation. If their kinetic energies of rotation are equal, their angular moment will be in the ratio
150319
A rotating body has angular momentum ' \(L\) '. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
150320
A solid cylinder of mass ' \(M\) ' ' and radius ' \(R\) ' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(v\) '. The height of the inclined plane is \(\mathrm{g}=\) acceleration due to gravity)
1 \(\frac{2 v^{2}}{3 g}\)
2 \(\frac{7 v^{2}}{9 g}\)
3 \(\frac{4 \mathrm{v}^{2}}{5 \mathrm{~g}}\)
4 \(\frac{3 v^{2}}{4 g}\)
Explanation:
D Potential energy of the solid cylinder at height \(\mathrm{h})=\mathrm{Mgh}\) K.E. of center of mass when reached at bottom, \(\text { K.E. }=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} {\left[\mathrm{I}=\mathrm{MK}^{2}\right]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{MK}^{2} \frac{\mathrm{v}^{2}}{\mathrm{R}^{2}} {[\mathrm{v}=\mathrm{R} \omega]}\) \(=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right) \quad \text {; where } \mathrm{K} \text { is gyration }\) radius For a solid cylinder, \(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2}\) K.E. \(=\frac{3}{4} \mathrm{Mv}^{2}\) Now, from conservation of energy theorem, \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\)