149916
A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular velocity \(\omega\). Two objects, each of mass \(m\), are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity
B Initial mass of ring \(=\mathrm{M} \mathrm{kg}\) Initial angular velocity \(=\omega\) Let final angular velocity \(=\omega^{\prime}\) Final mass of the ring \(=(M+2 m) k g\) Law of conservation of angular moment \(I_{1} \omega_{1}=I_{2} \omega_{2}\) \(M^{2} \omega=(M+2 m) r^{2} \times \omega^{\prime}\) \(\omega^{\prime}=\frac{M \omega}{M+2 m}\)
AMU-2017
Rotational Motion
149917
A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 s\). The magnitude of this torque is
1 \(\frac{3 \mathrm{~A}_{0}}{4}\)
2 \(\mathrm{A}_{0}\)
3 \(4 \mathrm{~A}_{0}\)
4 \(121 \mathrm{~A}_{0}\)
Explanation:
A Given that, \(\mathrm{L}_{2}=4 \mathrm{~A}_{0}\) \(\mathrm{L}_{1}=\mathrm{A}_{0}\) \(\Delta \mathrm{t}=4 \mathrm{sec}\) Change in angular momentum \(\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}\) \(\Delta \mathrm{L}=4 \mathrm{~A}_{0}-\mathrm{A}_{0}\) \(\Delta \mathrm{L}=3 \mathrm{~A}_{0}\) Torque \((\tau)=\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}\) \(\tau=\frac{3 \mathrm{~A}_{0}}{4}\)
BCECE-2013
Rotational Motion
149918
A wheel having moment of inertia \(2 \mathrm{~kg}-\mathrm{m}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
1 \(\frac{2 \pi}{13} \mathrm{~N}-\mathrm{m}\)
2 \(\frac{\pi}{14} \mathrm{~N}-\mathrm{m}\)
3 \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4 \(\frac{\pi}{20} \mathrm{~N}-\mathrm{m}\)
Explanation:
C Given that, \(\mathrm{I}=2 \mathrm{kgm}^{2}\) \(\mathrm{~N}=60 \mathrm{rpm}\) \(t=60 \mathrm{sec}\) \(\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 60}{60}\) \(\omega=2 \pi\) Torque required to stop the wheel's rotation is- \(\tau =\mathrm{I} \cdot \alpha=\mathrm{I}\left(\frac{\omega}{\mathrm{t}}\right)\) \(=2 \times \frac{2 \pi}{60}\) \(\tau =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
AIIMS-26.05.2019 (M)
Rotational Motion
149919
A particle of mass \(m\) is moving with a constant velocity along a line parallel to the positive direction of \(x\)-axis. The magnitude of its angular momentum with respect to the origin
1 is zero
2 goes on increasing as \(x\) increases
3 goes on decreasing as \(x\) increase
4 remains constant for all positions of the particle
Explanation:
D A particle of mass \(m\) is moving with constant velocity and it is parallel to the \(\mathrm{x}\)-axis. We know that, Angular momentum \((\vec{L})=\vec{p} \times \vec{r}\) Where, \(r=\) perpendicular distance \(\mathrm{p}=\) linear momentum, \(\mathrm{L}\) is angular momentum From figure, perpendicular distance and velocity are perpendicular to each other. Now, velocity, mass and distance (r) all are constant. So, we can say that angular momentum will be constant.
AMU-2018
Rotational Motion
149920
If force \(\overrightarrow{\mathbf{F}}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) acts on a particle having position vector \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) then, the torque about the origin will be :
149916
A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular velocity \(\omega\). Two objects, each of mass \(m\), are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity
B Initial mass of ring \(=\mathrm{M} \mathrm{kg}\) Initial angular velocity \(=\omega\) Let final angular velocity \(=\omega^{\prime}\) Final mass of the ring \(=(M+2 m) k g\) Law of conservation of angular moment \(I_{1} \omega_{1}=I_{2} \omega_{2}\) \(M^{2} \omega=(M+2 m) r^{2} \times \omega^{\prime}\) \(\omega^{\prime}=\frac{M \omega}{M+2 m}\)
AMU-2017
Rotational Motion
149917
A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 s\). The magnitude of this torque is
1 \(\frac{3 \mathrm{~A}_{0}}{4}\)
2 \(\mathrm{A}_{0}\)
3 \(4 \mathrm{~A}_{0}\)
4 \(121 \mathrm{~A}_{0}\)
Explanation:
A Given that, \(\mathrm{L}_{2}=4 \mathrm{~A}_{0}\) \(\mathrm{L}_{1}=\mathrm{A}_{0}\) \(\Delta \mathrm{t}=4 \mathrm{sec}\) Change in angular momentum \(\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}\) \(\Delta \mathrm{L}=4 \mathrm{~A}_{0}-\mathrm{A}_{0}\) \(\Delta \mathrm{L}=3 \mathrm{~A}_{0}\) Torque \((\tau)=\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}\) \(\tau=\frac{3 \mathrm{~A}_{0}}{4}\)
BCECE-2013
Rotational Motion
149918
A wheel having moment of inertia \(2 \mathrm{~kg}-\mathrm{m}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
1 \(\frac{2 \pi}{13} \mathrm{~N}-\mathrm{m}\)
2 \(\frac{\pi}{14} \mathrm{~N}-\mathrm{m}\)
3 \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4 \(\frac{\pi}{20} \mathrm{~N}-\mathrm{m}\)
Explanation:
C Given that, \(\mathrm{I}=2 \mathrm{kgm}^{2}\) \(\mathrm{~N}=60 \mathrm{rpm}\) \(t=60 \mathrm{sec}\) \(\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 60}{60}\) \(\omega=2 \pi\) Torque required to stop the wheel's rotation is- \(\tau =\mathrm{I} \cdot \alpha=\mathrm{I}\left(\frac{\omega}{\mathrm{t}}\right)\) \(=2 \times \frac{2 \pi}{60}\) \(\tau =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
AIIMS-26.05.2019 (M)
Rotational Motion
149919
A particle of mass \(m\) is moving with a constant velocity along a line parallel to the positive direction of \(x\)-axis. The magnitude of its angular momentum with respect to the origin
1 is zero
2 goes on increasing as \(x\) increases
3 goes on decreasing as \(x\) increase
4 remains constant for all positions of the particle
Explanation:
D A particle of mass \(m\) is moving with constant velocity and it is parallel to the \(\mathrm{x}\)-axis. We know that, Angular momentum \((\vec{L})=\vec{p} \times \vec{r}\) Where, \(r=\) perpendicular distance \(\mathrm{p}=\) linear momentum, \(\mathrm{L}\) is angular momentum From figure, perpendicular distance and velocity are perpendicular to each other. Now, velocity, mass and distance (r) all are constant. So, we can say that angular momentum will be constant.
AMU-2018
Rotational Motion
149920
If force \(\overrightarrow{\mathbf{F}}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) acts on a particle having position vector \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) then, the torque about the origin will be :
149916
A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular velocity \(\omega\). Two objects, each of mass \(m\), are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity
B Initial mass of ring \(=\mathrm{M} \mathrm{kg}\) Initial angular velocity \(=\omega\) Let final angular velocity \(=\omega^{\prime}\) Final mass of the ring \(=(M+2 m) k g\) Law of conservation of angular moment \(I_{1} \omega_{1}=I_{2} \omega_{2}\) \(M^{2} \omega=(M+2 m) r^{2} \times \omega^{\prime}\) \(\omega^{\prime}=\frac{M \omega}{M+2 m}\)
AMU-2017
Rotational Motion
149917
A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 s\). The magnitude of this torque is
1 \(\frac{3 \mathrm{~A}_{0}}{4}\)
2 \(\mathrm{A}_{0}\)
3 \(4 \mathrm{~A}_{0}\)
4 \(121 \mathrm{~A}_{0}\)
Explanation:
A Given that, \(\mathrm{L}_{2}=4 \mathrm{~A}_{0}\) \(\mathrm{L}_{1}=\mathrm{A}_{0}\) \(\Delta \mathrm{t}=4 \mathrm{sec}\) Change in angular momentum \(\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}\) \(\Delta \mathrm{L}=4 \mathrm{~A}_{0}-\mathrm{A}_{0}\) \(\Delta \mathrm{L}=3 \mathrm{~A}_{0}\) Torque \((\tau)=\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}\) \(\tau=\frac{3 \mathrm{~A}_{0}}{4}\)
BCECE-2013
Rotational Motion
149918
A wheel having moment of inertia \(2 \mathrm{~kg}-\mathrm{m}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
1 \(\frac{2 \pi}{13} \mathrm{~N}-\mathrm{m}\)
2 \(\frac{\pi}{14} \mathrm{~N}-\mathrm{m}\)
3 \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4 \(\frac{\pi}{20} \mathrm{~N}-\mathrm{m}\)
Explanation:
C Given that, \(\mathrm{I}=2 \mathrm{kgm}^{2}\) \(\mathrm{~N}=60 \mathrm{rpm}\) \(t=60 \mathrm{sec}\) \(\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 60}{60}\) \(\omega=2 \pi\) Torque required to stop the wheel's rotation is- \(\tau =\mathrm{I} \cdot \alpha=\mathrm{I}\left(\frac{\omega}{\mathrm{t}}\right)\) \(=2 \times \frac{2 \pi}{60}\) \(\tau =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
AIIMS-26.05.2019 (M)
Rotational Motion
149919
A particle of mass \(m\) is moving with a constant velocity along a line parallel to the positive direction of \(x\)-axis. The magnitude of its angular momentum with respect to the origin
1 is zero
2 goes on increasing as \(x\) increases
3 goes on decreasing as \(x\) increase
4 remains constant for all positions of the particle
Explanation:
D A particle of mass \(m\) is moving with constant velocity and it is parallel to the \(\mathrm{x}\)-axis. We know that, Angular momentum \((\vec{L})=\vec{p} \times \vec{r}\) Where, \(r=\) perpendicular distance \(\mathrm{p}=\) linear momentum, \(\mathrm{L}\) is angular momentum From figure, perpendicular distance and velocity are perpendicular to each other. Now, velocity, mass and distance (r) all are constant. So, we can say that angular momentum will be constant.
AMU-2018
Rotational Motion
149920
If force \(\overrightarrow{\mathbf{F}}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) acts on a particle having position vector \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) then, the torque about the origin will be :
149916
A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular velocity \(\omega\). Two objects, each of mass \(m\), are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity
B Initial mass of ring \(=\mathrm{M} \mathrm{kg}\) Initial angular velocity \(=\omega\) Let final angular velocity \(=\omega^{\prime}\) Final mass of the ring \(=(M+2 m) k g\) Law of conservation of angular moment \(I_{1} \omega_{1}=I_{2} \omega_{2}\) \(M^{2} \omega=(M+2 m) r^{2} \times \omega^{\prime}\) \(\omega^{\prime}=\frac{M \omega}{M+2 m}\)
AMU-2017
Rotational Motion
149917
A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 s\). The magnitude of this torque is
1 \(\frac{3 \mathrm{~A}_{0}}{4}\)
2 \(\mathrm{A}_{0}\)
3 \(4 \mathrm{~A}_{0}\)
4 \(121 \mathrm{~A}_{0}\)
Explanation:
A Given that, \(\mathrm{L}_{2}=4 \mathrm{~A}_{0}\) \(\mathrm{L}_{1}=\mathrm{A}_{0}\) \(\Delta \mathrm{t}=4 \mathrm{sec}\) Change in angular momentum \(\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}\) \(\Delta \mathrm{L}=4 \mathrm{~A}_{0}-\mathrm{A}_{0}\) \(\Delta \mathrm{L}=3 \mathrm{~A}_{0}\) Torque \((\tau)=\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}\) \(\tau=\frac{3 \mathrm{~A}_{0}}{4}\)
BCECE-2013
Rotational Motion
149918
A wheel having moment of inertia \(2 \mathrm{~kg}-\mathrm{m}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
1 \(\frac{2 \pi}{13} \mathrm{~N}-\mathrm{m}\)
2 \(\frac{\pi}{14} \mathrm{~N}-\mathrm{m}\)
3 \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4 \(\frac{\pi}{20} \mathrm{~N}-\mathrm{m}\)
Explanation:
C Given that, \(\mathrm{I}=2 \mathrm{kgm}^{2}\) \(\mathrm{~N}=60 \mathrm{rpm}\) \(t=60 \mathrm{sec}\) \(\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 60}{60}\) \(\omega=2 \pi\) Torque required to stop the wheel's rotation is- \(\tau =\mathrm{I} \cdot \alpha=\mathrm{I}\left(\frac{\omega}{\mathrm{t}}\right)\) \(=2 \times \frac{2 \pi}{60}\) \(\tau =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
AIIMS-26.05.2019 (M)
Rotational Motion
149919
A particle of mass \(m\) is moving with a constant velocity along a line parallel to the positive direction of \(x\)-axis. The magnitude of its angular momentum with respect to the origin
1 is zero
2 goes on increasing as \(x\) increases
3 goes on decreasing as \(x\) increase
4 remains constant for all positions of the particle
Explanation:
D A particle of mass \(m\) is moving with constant velocity and it is parallel to the \(\mathrm{x}\)-axis. We know that, Angular momentum \((\vec{L})=\vec{p} \times \vec{r}\) Where, \(r=\) perpendicular distance \(\mathrm{p}=\) linear momentum, \(\mathrm{L}\) is angular momentum From figure, perpendicular distance and velocity are perpendicular to each other. Now, velocity, mass and distance (r) all are constant. So, we can say that angular momentum will be constant.
AMU-2018
Rotational Motion
149920
If force \(\overrightarrow{\mathbf{F}}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) acts on a particle having position vector \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) then, the torque about the origin will be :
149916
A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular velocity \(\omega\). Two objects, each of mass \(m\), are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity
B Initial mass of ring \(=\mathrm{M} \mathrm{kg}\) Initial angular velocity \(=\omega\) Let final angular velocity \(=\omega^{\prime}\) Final mass of the ring \(=(M+2 m) k g\) Law of conservation of angular moment \(I_{1} \omega_{1}=I_{2} \omega_{2}\) \(M^{2} \omega=(M+2 m) r^{2} \times \omega^{\prime}\) \(\omega^{\prime}=\frac{M \omega}{M+2 m}\)
AMU-2017
Rotational Motion
149917
A constant torque acting on a uniform circular wheel changes its angular momentum from \(A_{0}\) to \(4 A_{0}\) in \(4 s\). The magnitude of this torque is
1 \(\frac{3 \mathrm{~A}_{0}}{4}\)
2 \(\mathrm{A}_{0}\)
3 \(4 \mathrm{~A}_{0}\)
4 \(121 \mathrm{~A}_{0}\)
Explanation:
A Given that, \(\mathrm{L}_{2}=4 \mathrm{~A}_{0}\) \(\mathrm{L}_{1}=\mathrm{A}_{0}\) \(\Delta \mathrm{t}=4 \mathrm{sec}\) Change in angular momentum \(\Delta \mathrm{L}=\mathrm{L}_{2}-\mathrm{L}_{1}\) \(\Delta \mathrm{L}=4 \mathrm{~A}_{0}-\mathrm{A}_{0}\) \(\Delta \mathrm{L}=3 \mathrm{~A}_{0}\) Torque \((\tau)=\frac{\Delta \mathrm{L}}{\Delta \mathrm{t}}\) \(\tau=\frac{3 \mathrm{~A}_{0}}{4}\)
BCECE-2013
Rotational Motion
149918
A wheel having moment of inertia \(2 \mathrm{~kg}-\mathrm{m}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
1 \(\frac{2 \pi}{13} \mathrm{~N}-\mathrm{m}\)
2 \(\frac{\pi}{14} \mathrm{~N}-\mathrm{m}\)
3 \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4 \(\frac{\pi}{20} \mathrm{~N}-\mathrm{m}\)
Explanation:
C Given that, \(\mathrm{I}=2 \mathrm{kgm}^{2}\) \(\mathrm{~N}=60 \mathrm{rpm}\) \(t=60 \mathrm{sec}\) \(\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 60}{60}\) \(\omega=2 \pi\) Torque required to stop the wheel's rotation is- \(\tau =\mathrm{I} \cdot \alpha=\mathrm{I}\left(\frac{\omega}{\mathrm{t}}\right)\) \(=2 \times \frac{2 \pi}{60}\) \(\tau =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
AIIMS-26.05.2019 (M)
Rotational Motion
149919
A particle of mass \(m\) is moving with a constant velocity along a line parallel to the positive direction of \(x\)-axis. The magnitude of its angular momentum with respect to the origin
1 is zero
2 goes on increasing as \(x\) increases
3 goes on decreasing as \(x\) increase
4 remains constant for all positions of the particle
Explanation:
D A particle of mass \(m\) is moving with constant velocity and it is parallel to the \(\mathrm{x}\)-axis. We know that, Angular momentum \((\vec{L})=\vec{p} \times \vec{r}\) Where, \(r=\) perpendicular distance \(\mathrm{p}=\) linear momentum, \(\mathrm{L}\) is angular momentum From figure, perpendicular distance and velocity are perpendicular to each other. Now, velocity, mass and distance (r) all are constant. So, we can say that angular momentum will be constant.
AMU-2018
Rotational Motion
149920
If force \(\overrightarrow{\mathbf{F}}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) acts on a particle having position vector \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) then, the torque about the origin will be :
