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Standard IX

Physics

Buoyancy

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 are hung 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 condition is required if $θ_{1} = θ_{2} ?$

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

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

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m_{1} = m_{2}

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l_{1} = l_{2}

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Solution

The correct option is C $m_{1} = m_{2}$
Since both the small spheres are at the same horizontal level, the electrostatic forces on both spheres are in horizontal direction. The $F B D$ of the left sphere is shown in the figure.
Diagram

The sphere is in equilibrium, so
$T cos θ = m g \dots (i)$
And $T sin θ = F_{E} \dots (i i)$
From $(i)$ and $(i i)$ , $tan θ = \frac{F_{E}}{m g}$
The magnitude of electrostatic force on each sphere is the same irrespective of its charge.
For $θ_{1} = θ_{2}$ the necessary condition is
$m_{1} = m_{2}$
Hence, option (a) is correct.

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Standard IX Physics

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