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Standard XII

Physics

Displacement of COM:Application

Two small sph...

Question

Two small spheres with mass $m_{1}$ and $m_{2}$ hang from massless insulating threads of length $l_{1}$ and $l_{2}$ . The 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 angles $θ_{1}$ and $θ_{2}$ . Which of the conditions is required if $θ_{1} = θ_{2}$ ?

m_{1} = m_{2}

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| q_{1} | = | q_{2} |

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l_{1} = l_{2}

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\frac{q_{1}}{m_{1}} = \frac{q_{2}}{m_{2}}

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Solution

The correct option is A $m_{1} = m_{2}$
Suppose $F$ is the electrostatic force between the spheres.

From equilibrium of forces,
$T_{1} sin θ_{1} = F$ and $T_{1} cos θ_{1} = m_{1} g$
So, $tan θ_{1} = \frac{F}{m_{1} g} - - - - - (1)$
Similarly, $tan θ_{2} = \frac{F}{m_{2} g} - - - - - (2)$
For $θ_{1} = θ_{2}$ ,
from (1) and (2), $m_{1} = m_{2}$

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Displacement of COM:Application

Standard XII Physics

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The correct option is A m1=m2Suppose F is the electrostatic force between the spheres. From equilibrium of forces, T1sinθ1=F and T1cosθ1=m1g So, tanθ1=Fm1g−−−−−(1) Similarly, tanθ2=Fm2g−−−−−(2) For θ1=θ2, from (1) and (2), m1=m2