Two small spheres of each charge `q`, mass m and material density `d` are suspended from a fixed point with the help of inextensible light thread. When the spheres are in air, the angle between the threads is `90^(@)`. When the spheres are suspended in a liquid of density `2/3 d`, the angle between the threads is `60^(@)` . The value of dielectric constant of the liquid is

Options

A. `6sqrt(3)`
B. `2 sqrt(5)`
C. `5 sqrt(3)`
D. `7sqrt(2)`

Answer

A

Explanation

Key Idea At any instant,
`T cos theta = ` mg and `T sin theta = F_(e) rArr tan theta mg = F_( e)`

`F_(e) =(kq^(2))/r^(2)`
` r = 2lcos theta" " [ "from Fig."]`
`tan theta = F_(e)/(mg) `
So, for ` theta = 45^(@) and phi = 45^(@)`
`r = (2l)/v^(2) = sqrt(2)l rArr F_(e) = (Kq^(2))/(2l)`
`tan 45^(@) = (kq^(2))/(2l" " mg) " " (because tan 45^(2) = 1)`
`mg = (kq^(2))/(2l)" "...(i)`
For, ` theta = 30 ^(@) and phi = 60^(@)`
`r = 2l xx 1/2 = l`
So `F_(e) = (kq^(2))/l^(2) rArrtan 30^(@) = (k.q^(2))/(l^(2) - m.g)` As, the sphere is suspended in a liquid of density `2/3 d` then the observed weight of the body,
`m. = V ( d - (2d)/3) = m/3" " [ because m = V cdot d]`
`1/sqrt(3) = (3k.q^(2))/(l^(2) mg) rArr mg = - (3sqrt(3)k.q^(2))/l^(2)" "...(ii)`
So, from Eq. (i) and (ii), we get
`3sqrt(3) k. = k/2 rArr k. = 1/ (4 pivarepsilon _(0) varepsilon_(r)), K = 1/ (4pi varepsilon_(0))`
`rArr varepsilon_(r) = 6 sqrt(3)`