365628
Assertion : Torque is equal to rate of change of angular momentum. Reason : Angular momentum depends on moment of inertia.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Torque \(\tau=I \alpha\) But \(\alpha=\dfrac{d \omega}{d t}\) and \(L=I \omega\) \(\Rightarrow \tau=\dfrac{d L}{d t}=\dfrac{d(I \omega)}{d t}=I \alpha\) So correct option is (2).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365629
Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is \(\,8 \times {10^{ - 7}}kg{m^2}\)
365630
A ring of mass \(10\;kg\) and diameter \(0.4\;m\) is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
365631
A solid cylinder of mass \(20\;kg\) and radius \(20\;cm\) rotates about its axis with an angular speed of \(100{\mkern 1mu} \,rad\,{s^{ - 1}}\). The angular momentum of the cylinder about its axis is
1 \(40\;J\;s\)
2 \(400\;J\;s\)
3 \(20\;J\;s\)
4 \(200\;J\;s\)
Explanation:
Here, \(M = 20\;kg\) \(R = 20\;cm = 20 \times {10^{ - 2}}\;m,\omega = 100\,{\mkern 1mu} rad{s^{ - 1}}\) Moment of inertia of the solid cylinder about its axis is \(I = \frac{{M{R^2}}}{2} = \frac{{(20\;kg){{\left( {20 \times {{10}^{ - 2}}\;m} \right)}^2}}}{2} = 0.4\;kg\;{m^2}\) Angular momentum of the cylinder about its axis is \(\left. {L = I\omega = \left( {0.4\;kg\;{m^2}} \right)100{\mkern 1mu} \,rad{s^{ - 1}}} \right) = 40\;Js\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365632
If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(T\) is
1 \(\dfrac{4 \pi M R^{2}}{5 T}\)
2 \(\dfrac{\pi M R^{2}}{T}\)
3 \(\dfrac{M R^{2} \pi}{T}\)
4 \(\dfrac{2 \pi M R^{2}}{5 T}\)
Explanation:
Angular momentum of earth about its axis of rotation, \(L=I \omega=\dfrac{2}{5} M R^{2} \times \dfrac{2 \pi}{T}=\dfrac{4 \pi M R^{2}}{5 T}\)
365628
Assertion : Torque is equal to rate of change of angular momentum. Reason : Angular momentum depends on moment of inertia.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Torque \(\tau=I \alpha\) But \(\alpha=\dfrac{d \omega}{d t}\) and \(L=I \omega\) \(\Rightarrow \tau=\dfrac{d L}{d t}=\dfrac{d(I \omega)}{d t}=I \alpha\) So correct option is (2).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365629
Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is \(\,8 \times {10^{ - 7}}kg{m^2}\)
365630
A ring of mass \(10\;kg\) and diameter \(0.4\;m\) is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
365631
A solid cylinder of mass \(20\;kg\) and radius \(20\;cm\) rotates about its axis with an angular speed of \(100{\mkern 1mu} \,rad\,{s^{ - 1}}\). The angular momentum of the cylinder about its axis is
1 \(40\;J\;s\)
2 \(400\;J\;s\)
3 \(20\;J\;s\)
4 \(200\;J\;s\)
Explanation:
Here, \(M = 20\;kg\) \(R = 20\;cm = 20 \times {10^{ - 2}}\;m,\omega = 100\,{\mkern 1mu} rad{s^{ - 1}}\) Moment of inertia of the solid cylinder about its axis is \(I = \frac{{M{R^2}}}{2} = \frac{{(20\;kg){{\left( {20 \times {{10}^{ - 2}}\;m} \right)}^2}}}{2} = 0.4\;kg\;{m^2}\) Angular momentum of the cylinder about its axis is \(\left. {L = I\omega = \left( {0.4\;kg\;{m^2}} \right)100{\mkern 1mu} \,rad{s^{ - 1}}} \right) = 40\;Js\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365632
If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(T\) is
1 \(\dfrac{4 \pi M R^{2}}{5 T}\)
2 \(\dfrac{\pi M R^{2}}{T}\)
3 \(\dfrac{M R^{2} \pi}{T}\)
4 \(\dfrac{2 \pi M R^{2}}{5 T}\)
Explanation:
Angular momentum of earth about its axis of rotation, \(L=I \omega=\dfrac{2}{5} M R^{2} \times \dfrac{2 \pi}{T}=\dfrac{4 \pi M R^{2}}{5 T}\)
365628
Assertion : Torque is equal to rate of change of angular momentum. Reason : Angular momentum depends on moment of inertia.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Torque \(\tau=I \alpha\) But \(\alpha=\dfrac{d \omega}{d t}\) and \(L=I \omega\) \(\Rightarrow \tau=\dfrac{d L}{d t}=\dfrac{d(I \omega)}{d t}=I \alpha\) So correct option is (2).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365629
Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is \(\,8 \times {10^{ - 7}}kg{m^2}\)
365630
A ring of mass \(10\;kg\) and diameter \(0.4\;m\) is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
365631
A solid cylinder of mass \(20\;kg\) and radius \(20\;cm\) rotates about its axis with an angular speed of \(100{\mkern 1mu} \,rad\,{s^{ - 1}}\). The angular momentum of the cylinder about its axis is
1 \(40\;J\;s\)
2 \(400\;J\;s\)
3 \(20\;J\;s\)
4 \(200\;J\;s\)
Explanation:
Here, \(M = 20\;kg\) \(R = 20\;cm = 20 \times {10^{ - 2}}\;m,\omega = 100\,{\mkern 1mu} rad{s^{ - 1}}\) Moment of inertia of the solid cylinder about its axis is \(I = \frac{{M{R^2}}}{2} = \frac{{(20\;kg){{\left( {20 \times {{10}^{ - 2}}\;m} \right)}^2}}}{2} = 0.4\;kg\;{m^2}\) Angular momentum of the cylinder about its axis is \(\left. {L = I\omega = \left( {0.4\;kg\;{m^2}} \right)100{\mkern 1mu} \,rad{s^{ - 1}}} \right) = 40\;Js\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365632
If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(T\) is
1 \(\dfrac{4 \pi M R^{2}}{5 T}\)
2 \(\dfrac{\pi M R^{2}}{T}\)
3 \(\dfrac{M R^{2} \pi}{T}\)
4 \(\dfrac{2 \pi M R^{2}}{5 T}\)
Explanation:
Angular momentum of earth about its axis of rotation, \(L=I \omega=\dfrac{2}{5} M R^{2} \times \dfrac{2 \pi}{T}=\dfrac{4 \pi M R^{2}}{5 T}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365628
Assertion : Torque is equal to rate of change of angular momentum. Reason : Angular momentum depends on moment of inertia.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Torque \(\tau=I \alpha\) But \(\alpha=\dfrac{d \omega}{d t}\) and \(L=I \omega\) \(\Rightarrow \tau=\dfrac{d L}{d t}=\dfrac{d(I \omega)}{d t}=I \alpha\) So correct option is (2).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365629
Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is \(\,8 \times {10^{ - 7}}kg{m^2}\)
365630
A ring of mass \(10\;kg\) and diameter \(0.4\;m\) is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
365631
A solid cylinder of mass \(20\;kg\) and radius \(20\;cm\) rotates about its axis with an angular speed of \(100{\mkern 1mu} \,rad\,{s^{ - 1}}\). The angular momentum of the cylinder about its axis is
1 \(40\;J\;s\)
2 \(400\;J\;s\)
3 \(20\;J\;s\)
4 \(200\;J\;s\)
Explanation:
Here, \(M = 20\;kg\) \(R = 20\;cm = 20 \times {10^{ - 2}}\;m,\omega = 100\,{\mkern 1mu} rad{s^{ - 1}}\) Moment of inertia of the solid cylinder about its axis is \(I = \frac{{M{R^2}}}{2} = \frac{{(20\;kg){{\left( {20 \times {{10}^{ - 2}}\;m} \right)}^2}}}{2} = 0.4\;kg\;{m^2}\) Angular momentum of the cylinder about its axis is \(\left. {L = I\omega = \left( {0.4\;kg\;{m^2}} \right)100{\mkern 1mu} \,rad{s^{ - 1}}} \right) = 40\;Js\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365632
If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(T\) is
1 \(\dfrac{4 \pi M R^{2}}{5 T}\)
2 \(\dfrac{\pi M R^{2}}{T}\)
3 \(\dfrac{M R^{2} \pi}{T}\)
4 \(\dfrac{2 \pi M R^{2}}{5 T}\)
Explanation:
Angular momentum of earth about its axis of rotation, \(L=I \omega=\dfrac{2}{5} M R^{2} \times \dfrac{2 \pi}{T}=\dfrac{4 \pi M R^{2}}{5 T}\)
365628
Assertion : Torque is equal to rate of change of angular momentum. Reason : Angular momentum depends on moment of inertia.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Torque \(\tau=I \alpha\) But \(\alpha=\dfrac{d \omega}{d t}\) and \(L=I \omega\) \(\Rightarrow \tau=\dfrac{d L}{d t}=\dfrac{d(I \omega)}{d t}=I \alpha\) So correct option is (2).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365629
Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is \(\,8 \times {10^{ - 7}}kg{m^2}\)
365630
A ring of mass \(10\;kg\) and diameter \(0.4\;m\) is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
365631
A solid cylinder of mass \(20\;kg\) and radius \(20\;cm\) rotates about its axis with an angular speed of \(100{\mkern 1mu} \,rad\,{s^{ - 1}}\). The angular momentum of the cylinder about its axis is
1 \(40\;J\;s\)
2 \(400\;J\;s\)
3 \(20\;J\;s\)
4 \(200\;J\;s\)
Explanation:
Here, \(M = 20\;kg\) \(R = 20\;cm = 20 \times {10^{ - 2}}\;m,\omega = 100\,{\mkern 1mu} rad{s^{ - 1}}\) Moment of inertia of the solid cylinder about its axis is \(I = \frac{{M{R^2}}}{2} = \frac{{(20\;kg){{\left( {20 \times {{10}^{ - 2}}\;m} \right)}^2}}}{2} = 0.4\;kg\;{m^2}\) Angular momentum of the cylinder about its axis is \(\left. {L = I\omega = \left( {0.4\;kg\;{m^2}} \right)100{\mkern 1mu} \,rad{s^{ - 1}}} \right) = 40\;Js\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365632
If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(T\) is
1 \(\dfrac{4 \pi M R^{2}}{5 T}\)
2 \(\dfrac{\pi M R^{2}}{T}\)
3 \(\dfrac{M R^{2} \pi}{T}\)
4 \(\dfrac{2 \pi M R^{2}}{5 T}\)
Explanation:
Angular momentum of earth about its axis of rotation, \(L=I \omega=\dfrac{2}{5} M R^{2} \times \dfrac{2 \pi}{T}=\dfrac{4 \pi M R^{2}}{5 T}\)