Question

Consider an insulated wire frame ABC as shown in a vertical plane. The two beads P and Q are of mass m each and carry charge $q_1$ and $q_2$ respectively and can slide down without any friction along the wires. If they are connected by a wire and are stationary,

• A. the angle, $\alpha ={60}^{\circ }$
• B. the angle, $\alpha ={30}^{\circ }$
• C. normal reaction on beads P and Q are $N_1=\sqrt{3}mg$ and $N_2=mg$ respectively
• D. normal reaction on beads P and Q are $N_1=mg$ and $N_2=\sqrt{3}mg$ respectively

Answer 1

Answer: (A), (C)

The correct options are
A the angle, $\alpha ={60}^{\circ }$
C normal reaction on beads P and Q are $N_1=\sqrt{3}mg$ and $N_2=mg$ respectively

Considering the forces acting on beads, the free body drawn will give us,

For bead P to be in equilibrium
$mg\cos {60}^{\circ }=(T-F)\cos \alpha ⟶\left(1\right)\text{and}{N}_1=mg\cos {30}^{\circ }+(T-F)\sin \alpha ⟶\left(2\right)\text{Similarly for bead Q to be in equilibrium,}mg\sin {60}^{\circ }=(T-F)\sin \alpha ⟶\left(3\right)\text{and}{N}_2=mg\cos {60}^{\circ }+(T-F)\cos \alpha ⟶\left(4\right)\text{Solving (3) and (1)},\text{we get,}\alpha ={60}^{\circ }\text{From (3),}T-F=mg\Rightarrow T=mg+F=mg+\frac{q_1{q}_2}{4\pi {ϵ}_0{l}^2}\text{From (2) and (4), we have ,}{N}_1=mg\cos {30}^{\circ }+(T-F)\sin {60}^{\circ }=mg\cos {30}^{\circ }+mg\sin {60}^{\circ }=\sqrt{3}mg\text{and}{N}_2=mg\sin {30}^{\circ }+mg\cos {60}^{\circ }=mg$