InterviewSolution
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InterviewSolution
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A particle of mass 2 kg chrge 1 mC is projected vertially with velocity k`10 ms^-1`. There is as uniform horizontal electric field of `10^4N//C,` thenA. the horizontal range of the particle is `10 m`B. the time of flight of the particle is `2s`C. the maximum heighty reached is `5m`D. the horizontal range of the particle is `5m` |
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Answer» Correct Answer - A::B::C `T=(2u_y)/g=(2xx10)/10=2s` `H=u_y^2/(2g)=((10)^2)/20=5m` `R=1/2a_xT^2=1/2((qE)/m)T^2` `=1/2((10^-3xx10^4)/2)(2)^2` `=10m` |
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