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Two small spheres of masses M(1)and M(2)...

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)` ?

A

`m_1 ne m_2 ` but `Q_1 = Q_2`

B

`m_1 =m_2`

C

`Q_1 =Q_2`

D

`L_1 =L_2`

Text Solution

Verified by Experts

The correct Answer is:
B
Force of repulsion between charges are same
`:. F_1 =F_2 =F`

`{:(T_1cos theta_1=m_1g,T_2sin theta_2=F),(T_1 sintheta_1=F,T_2 costheta_2=m_2g),(tan theta_1=F/(m_1g),tan theta_2=F/(m_2g)):}`
Since, `theta_1=theta_2`
`impliesF/(m_1g)=F/(m_2g)implies m_1=m_2`
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