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Two pitch balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become

A

`((2r)/3)`

B

`(1/sqrt2)^2`

C

`(r/(root(3)(2)))`

D

`((2r)/sqrt3)`

Text Solution

Verified by Experts

The correct Answer is:
C

`tantheta=(r//2)/y=r/(2y)`
`T cos theta = mg `
`T sin theta =F`
`tan theta =F/(mg)`

By eq. (i) and (ii)
`r/(2y)=F/(mg) =(Kq^2)/(r^2mg)impliesy =(r^3mg)/(2Kq^2)`
Since, `y prop r^3`
Now ,`y/(y//2) =(r/r_1)^3`
`implies r/r_1=(2)^(1//3)implies r_1 =r (1/2)^(1//3)`
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