Correct Answer - D
`T sin theta = (kq^(2))/(x^(2))`
`T cos theta = mg`
Dividing the equations
`tan theta = (kq^(2))/(mgx^(2))` here `tan theta ~~ theta = (x)/(2l)`
`rArr (x)/(2l) = (kq^(2))/(x^(2))`
`rArr q prop c^(3//2)`
`rArr (dq)/(dt) prop (3)/(2) x^(1//2) ((dx)/(dt)) rArr (dx)/(dt) prop x^(-1//2)`