361891
A wheel is rotating at \(900\,\,{\rm{rpm}}\) about its axis. When the power is cut off, it comes to rest in \(1 \mathrm{~min}\). The angular retardation (in \(rad\,{s^{ - 2}}\) ) is
361892
A particle is moving in a radius \(R\) with constant speed \(v\). The magnitude of average acceleration after half revolution is
1 \(\frac{{2\pi }}{{Rv}}\)
2 \(\frac{{2\pi }}{{R{v^2}}}\)
3 \(\frac{{2{v^2}}}{{\pi R}}\)
4 \(\frac{{2v}}{{\pi {R^2}}}\)
Explanation:
After half revolution, change in velocity, \(\Delta v = v - ( - v) = 2v\) Time taken for half revolution, \(\Delta t = \frac{T}{2} = \frac{1}{2}\frac{{2\pi R}}{v} = \frac{{\pi R}}{v}\) \(\therefore \) Average acceleration after half revolution, \({a_{avg}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{2v}}{{\left( {\frac{{\pi R}}{v}} \right)}} = \frac{{2{v^2}}}{{\pi R}}\)
MHTCET - 2020
PHXI04:MOTION IN A PLANE
361893
Two particles of equal masses are revolving in circular paths of radii \({r_1}\) and \({r_2}\) respectively with the same speed. The ratio of their centripetal forces is
\(F = \frac{{m{v^2}}}{r}\) If \(m\) and \(n\) are constants then \(F \propto \frac{1}{r}\) \(\therefore \;\;\frac{{{F_1}}}{{{F_2}}} = \left( {\frac{{{r_2}}}{{{r_1}}}} \right)\)
PHXI04:MOTION IN A PLANE
361894
If a particle covers half the circle of radius \(R\) with constant speed \(v\) then
1 Change in momentum is \(mvr\)
2 Change in \(K\).\(E\). is \(1/2\,m{v^2}\)
3 Change in \(K\).\(E\). is \(m{v^2}\)
4 Change in \(K\).\(E\). is zero
Explanation:
As momentum is vector quantity \(\therefore \;\) Change in momentum \(\Delta P = 2mv\sin \,(\theta /2)\;\) \( = 2mv\sin \,(90) = 2mv\) But kinetic energy remains always constant so change in kinetic energy is zero.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI04:MOTION IN A PLANE
361891
A wheel is rotating at \(900\,\,{\rm{rpm}}\) about its axis. When the power is cut off, it comes to rest in \(1 \mathrm{~min}\). The angular retardation (in \(rad\,{s^{ - 2}}\) ) is
361892
A particle is moving in a radius \(R\) with constant speed \(v\). The magnitude of average acceleration after half revolution is
1 \(\frac{{2\pi }}{{Rv}}\)
2 \(\frac{{2\pi }}{{R{v^2}}}\)
3 \(\frac{{2{v^2}}}{{\pi R}}\)
4 \(\frac{{2v}}{{\pi {R^2}}}\)
Explanation:
After half revolution, change in velocity, \(\Delta v = v - ( - v) = 2v\) Time taken for half revolution, \(\Delta t = \frac{T}{2} = \frac{1}{2}\frac{{2\pi R}}{v} = \frac{{\pi R}}{v}\) \(\therefore \) Average acceleration after half revolution, \({a_{avg}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{2v}}{{\left( {\frac{{\pi R}}{v}} \right)}} = \frac{{2{v^2}}}{{\pi R}}\)
MHTCET - 2020
PHXI04:MOTION IN A PLANE
361893
Two particles of equal masses are revolving in circular paths of radii \({r_1}\) and \({r_2}\) respectively with the same speed. The ratio of their centripetal forces is
\(F = \frac{{m{v^2}}}{r}\) If \(m\) and \(n\) are constants then \(F \propto \frac{1}{r}\) \(\therefore \;\;\frac{{{F_1}}}{{{F_2}}} = \left( {\frac{{{r_2}}}{{{r_1}}}} \right)\)
PHXI04:MOTION IN A PLANE
361894
If a particle covers half the circle of radius \(R\) with constant speed \(v\) then
1 Change in momentum is \(mvr\)
2 Change in \(K\).\(E\). is \(1/2\,m{v^2}\)
3 Change in \(K\).\(E\). is \(m{v^2}\)
4 Change in \(K\).\(E\). is zero
Explanation:
As momentum is vector quantity \(\therefore \;\) Change in momentum \(\Delta P = 2mv\sin \,(\theta /2)\;\) \( = 2mv\sin \,(90) = 2mv\) But kinetic energy remains always constant so change in kinetic energy is zero.
361891
A wheel is rotating at \(900\,\,{\rm{rpm}}\) about its axis. When the power is cut off, it comes to rest in \(1 \mathrm{~min}\). The angular retardation (in \(rad\,{s^{ - 2}}\) ) is
361892
A particle is moving in a radius \(R\) with constant speed \(v\). The magnitude of average acceleration after half revolution is
1 \(\frac{{2\pi }}{{Rv}}\)
2 \(\frac{{2\pi }}{{R{v^2}}}\)
3 \(\frac{{2{v^2}}}{{\pi R}}\)
4 \(\frac{{2v}}{{\pi {R^2}}}\)
Explanation:
After half revolution, change in velocity, \(\Delta v = v - ( - v) = 2v\) Time taken for half revolution, \(\Delta t = \frac{T}{2} = \frac{1}{2}\frac{{2\pi R}}{v} = \frac{{\pi R}}{v}\) \(\therefore \) Average acceleration after half revolution, \({a_{avg}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{2v}}{{\left( {\frac{{\pi R}}{v}} \right)}} = \frac{{2{v^2}}}{{\pi R}}\)
MHTCET - 2020
PHXI04:MOTION IN A PLANE
361893
Two particles of equal masses are revolving in circular paths of radii \({r_1}\) and \({r_2}\) respectively with the same speed. The ratio of their centripetal forces is
\(F = \frac{{m{v^2}}}{r}\) If \(m\) and \(n\) are constants then \(F \propto \frac{1}{r}\) \(\therefore \;\;\frac{{{F_1}}}{{{F_2}}} = \left( {\frac{{{r_2}}}{{{r_1}}}} \right)\)
PHXI04:MOTION IN A PLANE
361894
If a particle covers half the circle of radius \(R\) with constant speed \(v\) then
1 Change in momentum is \(mvr\)
2 Change in \(K\).\(E\). is \(1/2\,m{v^2}\)
3 Change in \(K\).\(E\). is \(m{v^2}\)
4 Change in \(K\).\(E\). is zero
Explanation:
As momentum is vector quantity \(\therefore \;\) Change in momentum \(\Delta P = 2mv\sin \,(\theta /2)\;\) \( = 2mv\sin \,(90) = 2mv\) But kinetic energy remains always constant so change in kinetic energy is zero.
361891
A wheel is rotating at \(900\,\,{\rm{rpm}}\) about its axis. When the power is cut off, it comes to rest in \(1 \mathrm{~min}\). The angular retardation (in \(rad\,{s^{ - 2}}\) ) is
361892
A particle is moving in a radius \(R\) with constant speed \(v\). The magnitude of average acceleration after half revolution is
1 \(\frac{{2\pi }}{{Rv}}\)
2 \(\frac{{2\pi }}{{R{v^2}}}\)
3 \(\frac{{2{v^2}}}{{\pi R}}\)
4 \(\frac{{2v}}{{\pi {R^2}}}\)
Explanation:
After half revolution, change in velocity, \(\Delta v = v - ( - v) = 2v\) Time taken for half revolution, \(\Delta t = \frac{T}{2} = \frac{1}{2}\frac{{2\pi R}}{v} = \frac{{\pi R}}{v}\) \(\therefore \) Average acceleration after half revolution, \({a_{avg}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{2v}}{{\left( {\frac{{\pi R}}{v}} \right)}} = \frac{{2{v^2}}}{{\pi R}}\)
MHTCET - 2020
PHXI04:MOTION IN A PLANE
361893
Two particles of equal masses are revolving in circular paths of radii \({r_1}\) and \({r_2}\) respectively with the same speed. The ratio of their centripetal forces is
\(F = \frac{{m{v^2}}}{r}\) If \(m\) and \(n\) are constants then \(F \propto \frac{1}{r}\) \(\therefore \;\;\frac{{{F_1}}}{{{F_2}}} = \left( {\frac{{{r_2}}}{{{r_1}}}} \right)\)
PHXI04:MOTION IN A PLANE
361894
If a particle covers half the circle of radius \(R\) with constant speed \(v\) then
1 Change in momentum is \(mvr\)
2 Change in \(K\).\(E\). is \(1/2\,m{v^2}\)
3 Change in \(K\).\(E\). is \(m{v^2}\)
4 Change in \(K\).\(E\). is zero
Explanation:
As momentum is vector quantity \(\therefore \;\) Change in momentum \(\Delta P = 2mv\sin \,(\theta /2)\;\) \( = 2mv\sin \,(90) = 2mv\) But kinetic energy remains always constant so change in kinetic energy is zero.
