CBSE Class 12-science Answered
Q62 given answer 2
Asked by Nishantthakre95 | 20 Aug, 2019, 08:59: AM
Expert Answer
Figure shows the various forces acting on the particle of mass m and charge q.
Tesnion T i9n the string are resolved as T cosθ in force direction and T sinθ in perpendicular direction.
Resolved component T sinθ provides the centripetal force for circular motion.
Hence, we have, T sinθ = mv2/R ....................(1)
where v is the speed of particle in circular path and R is radius of circular path.
As shown in figure F is the reultant force of weight and electtrostatic force acting on the particle.
We have, T cosθ = F .....................(2)
we eliminate tension T by dividing eqn.(1) by eqn.(2), tanθ = (1/F) mv2/R .................(3)
let us sunstitute, tanθ = qE/mg and R = L sinθ in eqn.(3, where L is length of string.
Then eqn.(3) is rewritten as, qE /( mg ) = (1/F) [ mv2 /( L sinθ ) ] ...................(4)
again by substituting F sinθ = qE , Eqn.(4) is simplified as, v2 = (qE/m)2 (L/g) or v = (qE/m) (L/g)1/2
Answered by Thiyagarajan K | 20 Aug, 2019, 03:59: PM
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