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Standard IX

Physics

Third Law of Motion

Two small sph...

Question

Two small spheres have mass $m_{1}$ and $m_{2}$ and hanging from massless insulating threads of lengths $l_{1}$ and $l_{2}$ . Two spheres carry charges $q_{1}$ and $q_{2}$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle $θ_{1}$ and $θ_{2}$ respectively. If $F_{1} = F_{2}$ , then:

θ_{1} = θ_{2}

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M_{1} = M_{2}

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\frac{l_{1}}{tan θ_{1}} = \frac{l_{2}}{tan θ_{2}}

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\frac{q_{1}}{tan θ_{1}} = \frac{q_{2}}{tan θ_{2}}

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Solution

The correct option is B $M_{1} = M_{2}$
For sphere $1$
In equilibrium, from figure.
$\begin{matrix} T_{1} cos θ_{1} = M_{1} g; T_{1} sin θ_{1} = F_{1} ∴ tan θ_{1} = \frac{F_{1}}{M_{1} g} . \end{matrix}$
For sphere $2$
In equilibrium, from figure.
$\begin{matrix} T_{2} cos θ_{2} = M_{2} g; T_{2} sin θ_{2} = F_{2} ∴ tan θ_{2} = \frac{F_{2}}{M_{2} g} . \end{matrix}$
Force of repulsion between two charges are same
$∴ F_{1} = F_{2}$
$θ_{1} = θ_{2}$ only if $\frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g} .$
But $F_{1} = F_{2}$ , then $M_{1} = M_{2}$ .

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Two positively charged spheres of masses $m_{1}$ and $m_{2}$ , are suspended from a common point at the ceiling by identical insulating massless strings of length l. Charges on the two spheres are $q_{1}$ and $q_{2}$ , respectively. At equilibrium both strings make the same angle $θ$ with the vertical. Then

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The correct option is B M1=M2For sphere 1In equilibrium, from figure.T1cosθ1=M1g;T1sinθ1=F1∴tanθ1=F1M1g.For sphere 2In equilibrium, from figure.T2cosθ2=M2g;T2sinθ2=F2∴tanθ2=F2M2g.Force of repulsion between two charges are same∴F1=F2θ1=θ2 only if F1M1g=F2M2g.But F1=F2, then M1=M2.