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A particle with charge QC, tied at the e...

A particle with charge QC, 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 fxied charge of magnitude QC. The mass of the moving charge M is such that `Mg=(Q^(2))/(4pi epsi_(0)R^(2))` . If at the highest position of the particle, the tenstion of the string just vanishes, the horizontal velocity at the lowest point has to be

A

0

B

`2 sqrt(gR)`

C

`sqrt(gR)`

D

`sqrt(5gR)`

Text Solution

Verified by Experts

From the figure `2.116`, tension on the top of vertuicle circular path is zero.

Therefore, at that position,
`Mg-(Q^(2))/(4pi epsi_(0)R^(2))=(Mv^(2))/(R)`
But `Mg-(Q^(2))/(4pi epsi_(0) R^(2))`
`:. v=0`
Now, from the conservation of mechanical energy,
`(PE)_(B)=(Delta KE)_(A)[ :. (KE)_(B)=(1)/(2)Mv^(2)=0]`
or, `Mgxx2R=(1)/(R)Mv_(0)^(2) or, v_(0)= 2sqrt(gR)`
The option (B) is correct.
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