Consider the following charged suspended ball | vedclass

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(D) For each ball,the forces acting are the tension $T$,gravitational force $mg$ downwards,and the electrostatic repulsive force $F_e$ horizontally.At

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$(b, d)$

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(D) For each ball,the forces acting are the tension $T$,gravitational force $mg$ downwards,and the electrostatic repulsive force $F_e$ horizontally.At equilibrium,the horizontal force $F_e = T \sin 	heta$ and the vertical force $mg = T \cos 	heta$.Dividing these gives $	an 	heta = \frac{F_e}{mg}$.Since both balls experience the same electrostatic force $F_e = \frac{k Q_1 Q_2}{r^2}$,the ratio $\frac{F_e}{mg}$ determines the angle.For the left ball,$	an \alpha = \frac{F_e}{m_1 g}$. For the right ball,$	an \beta = \frac{F_e}{m_2 g}$.Given $\alpha This simplifies to $m_1 > m_2$.Since the problem does not provide a specific relationship between $Q_1$ and $Q_2$,they can be related in any way as long as the electrostatic force is equal and opposite. Thus,$m_1 > m_2$ must hold,while $Q_1$ and $Q_2$ can have any relative values.Looking at the options,both $(b)$ and $(d)$ satisfy the condition $m_1 > m_2$.

Consider the following charged suspended ball system. If $\alpha < \beta$,then out of the following statements,which can be true at equilibrium? :-
$(a) Q_1 > Q_2, m_1 < m_2$
$(b) Q_1 > Q_2, m_1 > m_2$
$(c) Q_1 < Q_2, m_1 = m_2$
$(d) Q_1 < Q_2, m_1 > m_2$

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Consider the following charged suspended ball system. If $\alpha < \beta$,then out of the following statements,which can be true at equilibrium? :- $(a) Q_1 > Q_2, m_1 < m_2$ $(b) Q_1 > Q_2, m_1 > m_2$ $(c) Q_1 < Q_2, m_1 = m_2$ $(d) Q_1 < Q_2, m_1 > m_2$