Two small spheres of masses `M_(1)` and `M_(2)` are suspended by weightless insulating threads of lengths `L_(1)` and `L_(2)` The spheres carry charges `Q_(1)` and `Q_(2)` respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of `theta_(1)` and `theta_(2)` as shown. Which one of the following conditions is essential, if `theta_(1)=theta_(2)` ?

Options

A. `M_(1)!=M_(2)" but "Q_(1)=Q_(2)`
B. `M_(1)=M_(2)`
C. `Q_(1)=Q_(2)`
D. `L_(1)=L_(2)`

Answer

B

Explanation

The three forces acting on each sphere are :
(i) Tension (ii) Weight
(iii) electrostatic force of repulsion

For sphere 1
In equilibrium, from figure
`T_(1)costheta_(1)=M_(1)g,T_(!)sintheta_(1)=F_(1)`
`:.tantheta_(1)=(F_(1))/(M_(1)g)`
For sphere 2
In equilibrium, from figure
`T_(1)costheta_(1)=M_(1)g,T_(!)sintheta_(1)=F_(1)`
` :.tan theta_(1)=(F_(1))/(M_(1)g)`
For sphere 2
In equilibrium, from figure
`T_(1)costheta_(1)=M_(1)g,T_(1)sintheta_(1)sintheta_(1)=F_(1)`
`:." "tantheta_(1)=(F_(1))/(M_(1)g)`
For sphere 2
In equilibrium, from figure
`T_(2)costheta_(2)=M_(2)g,T_(2)sintheta_(2)=F_(2):.tantheta_(2)=(F_(2))/(M_(2)g)`
Foirce of repulsion between two charges are same
`:." " F_(1)=F_(2)`
`theta_(1)=theta_(2)` only if
`(rho_(1))/(M_(1)g)=(F_(2))/(M_(2)g)`
But `F_(1)=F_(2),` then `M_(1)=M_(2)`