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A particle with charge Q coulomb, tied a...

A particle with charge Q coulomb, tied at the end of an inextensible string of length R metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude Q coulomb . The mass of the moving charge M is such that `Mg=Q^2/(4pivarepsilon_0R^2)`. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be

A

0

B

`2sqrt(gR))`

C

`sqrt(2gR))`

D

`sqrt(5gR))`

Text Solution

Verified by Experts

The correct Answer is:
b
Given : At the highest position of the particle, tension , T=0
`therefore` The mass of the moving charge M is such that
`Mg=(Q^(2))/(4piepsilon_(0)R^(2))`
At the highest point , centrifugal force is ,
`Mg-(kQ^(2))/(R^(2))=(Mv^(2))/(R)("here",k=(1)/(4piepsilon_(0)))`
`thereforev=0`
Gain in potential energy = Loss in kinetic energy
`rArrMg(2R)=(1)/(2)Mv_(0)^(2)`
`therefore` Velocity at the lowest point , `v_(0)=2sqrt(gR)`
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