A) \[\frac{2mg}{3A}\]
B) \[\frac{2mg}{2A}\]
C) \[\frac{mg}{2A}\]
D) \[\frac{mg}{A}\]
Correct Answer: D
Solution :
For the particle to move undeflected with constant velocity, the acceleration of the particle should be zero, i.e. net force acting on the particle should be zero. \[\therefore \] \[(u\times A)+mg=0\] \[u\times A=-mg\Rightarrow |u\times A|\,=mg\] \[\Rightarrow \] \[u\,A\sin \theta =mg\] or, \[u=\frac{mg}{A\,\sin \theta }\] Now, u will be minimum, when \[\sin \theta \] will be maximum (i.e.\[\sin \theta =1\]) \[\therefore \] \[{{u}_{\min }}=mg/A\,\,\text{along}\,\,Z\text{-axis}\]You need to login to perform this action.
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