question_answer

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 \[{{\theta }_{1}}={{\theta }_{2}}\]?

A)  \[{{M}_{1}}\ne {{M}_{2}},but{{Q}_{1}}={{Q}_{2}}\]

B)  \[{{M}_{1}}={{M}_{2}}\]

C)  \[{{Q}_{1}}={{Q}_{2}}\]

D)  \[{{L}_{1}}={{L}_{2}}\]

Correct Answer: B

Solution :

For sphere 1, in equilibrium \[{{T}_{1}}\cos {{\theta }_{1}}={{M}_{1}}g\]and\[{{T}_{1}}\sin {{\theta }_{1}}={{F}_{1}}\] \[\therefore \] \[{{\theta }_{1}}=\frac{{{F}_{1}}}{{{M}_{1}}g}\] Similarly for sphere \[2,\tan {{\theta }_{2}}=\frac{{{F}_{2}}}{{{M}_{2}}g}\]F is same on both the charges, \[\theta \]will be same only if their masses M are equal.