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 equalYou need to login to perform this action.
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