Q62 given answer 2

Asked by Nishantthakre95 | 20th 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)(L/g)   or    v =  (qE/m) (L/g)1/2
 

Answered by Thiyagarajan K | 20th Aug, 2019, 03:59: PM

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