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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 10mB. the time of flight of the particle is 2 s.C. the maximum height reached is 5 m.D. the horizontal range of the particle is 0. |
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Answer» Correct Answer - A::B::C a.,b.,c. Time of flight `1(t)=(2u)/(g)=(2xx10)/(10)=2s` `H=(u^(2))/(2g)=(10^(2))/(2xx10)=5m` `R=0+(1)/(2)((qE)/(m))t^(2)=(1)/(2)xx(10^(-3)xx10^(4)xx2xx2)/(2)=10m` |
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