Radial & Tangential Acceleration for Non Uniform Circular Motion
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A proton, a deuteron, and an -particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is
- 640 m/s2
- 40 m/s2
- 40π m/s2
- 160 m/s2
- Zero
- √gl
- √1.5gl
- √2gl
- 4 m/s2
- 2 m/s2
- 8 m/s2
- 13 m/s2
- √98 m/s
- 7 m/s
- √490 m/s
- √4.9
Give an example of an accelerated body, moving with a uniform speed.
- 4:3
- 3:2
- 5:4
- 1:4
- 14 m
- 28 m
- 35 m
- 7 m
- 5 rad/s
- 5√2 rad/s
- 2√2 rad/s
- 2 rad/s
- θ=cos−1(13)
- θ=60∘
- θ=cos−1(23)
- θ=0∘
- Mg√1−(R−aR)2
- Mg√(RR−a)2−1
- MgaR
- Mg√1−a2R2
- g√10
- g√6
- g
- g√2
- v=v0e−SR
- v=v0e−SR
- v=v0eRS
- v=v0e−RS
- 4 m/s2
- 8 m/s2
4√2 m/s2- 4√3 m/s2
Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension inthe string ocnecting the 1.0 kg blocks and (c) the tensionin th tring attached to 0.50 kg.
- For h=5R, the resultant force acting on the block at Q is √75mg
- If the force exerted by the block against the track at the top of the loop equals its weight, then h=3R
- For h=5R, the resultant force acting on the block at Q is √65mg
- If the block should not fall off at the top of the circular track, it must be released from a minimum height of h=5R2
Is acceleration possible in uniform motion? Give example.
- 20 rad, 4√10 s
- 40 rad, √10 s
- 40 rad, 4√10 s
- 20 rad, √10 s
- At the top of the circle.
- At the bottom of the circle.
- Half way down
- None of above
- 2as2R
- 2as(1+s2R2)12
- 2as
- 2aR2s
- Rv0
- Rv0e−4π
- Rv0(1−e−4π)
- Rv0(1+e−4π)
- 2 mg
- mg
- 3 mg
- √3mg
A proton of mass 1.6×10−27 kg goes round in a circular orbit of radius 0.10 m under a centripetal force of 4×10−13N. then the frequency of revolution of the proton is about
0.08×108 cycles per sec
12×108 cycles per sec
8×108 cycles per sec
4×108 cycles per sec
- T+mg
- Tmg√(mg)2+T2
- mgT√T2−(mg)2
- mgT√(mg)2+T2
- 11√5 m/s2
- 11√10 m/s2
- √5 m/s2
- 11√5 m/s2
- 8mg
- 4mg
- 6mg
- 2mg
- T=mg(2cosθ−3cosθ0), θmax=30∘
- T=mg(3cosθ−2cosθ0), θmax=60∘
- T=mg(2cosθ−3cosθ0), θmax=60∘
- T=mg(3cosθ−2cosθ0), θmax=30∘
- 1400 N, 700 N, 2100 N
- 2100 N, 700 N, 1400 N
- 2100 N, 700 N, 700√5 N
- 700 N, 700√5 N, 2100 N
- 30 μN
- 50 μN
- 40 μN
- 60 μN