question_answer

Two small spheres of masses \[{{M}_{1}}\] and \[{{M}_{2}}\] are suspended by weightless insulating threads of length \[{{L}_{1}}\] 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 y     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 \[{{T}_{1}}\sin {{\theta }_{1}}={{F}_{1}}\,\] \[\tan {{\theta }_{1}}\,=\frac{{{F}_{1}}}{{{M}_{1}}g}\] 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