Two small spheres of masses `M_(1)`and `M_(2)` are suspended by weightless insulating threads of lengths `L_(1)` and `L_(2)`. The speres carry charges `Q_(1)` and `Q_(2)` respectively. The spheres are suspended such that they are in level with one another adn the threads are inclined to the verticle 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. `Q_(1) = Q_(2)`
C. `L_(1) = L_(2)`
D. `M_(1) = M_(2)`

Answer

D

Explanation

Refer to Fig.

Sphere `1` will be in equilibrium, when `T_(1) cos theta_(1) = M_(1) g` and `T_(1) sin theta_(1) = F_(1)`
`:. tan theta_(1) = (F_(1))/(M_(1)g)`
Similarly, sphere 2 will be in equilibrium, when
`T_(2) cos theta_(2) = M_(2) g` and `T_(2) sin theta_(2) = F_(2)`
`:. tan theta_(1) = (F_(2))/(M_(1) g)`
As charges repel eachother with the same force, therefore `F_(1) = F_(2)` For `theta_(1) = theta_(2), M_(1) = M_(2)`.