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 `theta_(1)` and `theta_(2)`respectively . Which one of the following conditions is essential for `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

2

Explanation

There are three forces acting on each sphere are (i) tension (ii) weight (iii) eelctrostatic force of repulsion for sphere
In equilibrium from figure
`tantheta_(1)=F_(1)//Mg`
From sphere `(2)` in equilibrium form figure
`tantheta_(2)//M_(2)g`
For `F_(1)=F_(2)`
`or theta_(1)=theta_(2)` only for `(F_(1))/(M_(1)g)=(F_(2))/(M_(2)g)`
But ,`F_(1)=F_(2)` and then `M_(1)=M_(2)`