KCET · Physics · Electrostatics
Two small spheres of masses \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) are suspended by weightless insulating threads of lengths \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\). The spheres carry charges \(Q_{1}\) and \(Q_{2}\) respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of \(\theta_{1}\) and \(\theta_{2}\) as shown. Which one of the following conditions is essential, if \(\theta_{1}=\theta_{2}\) ?
- A \(\mathrm{M}_{1} \neq \mathrm{M}_{2}\), but \(\mathrm{Q}_{1}=\mathrm{Q}_{2}\)
- B \(\mathrm{M}_{1}=\mathrm{M}_{2}\)
- C \(Q_{1}=Q_{2}\)
- D \(\mathrm{L}_{1}=\mathrm{L}_{2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{M}_{1}=\mathrm{M}_{2}\)
Step-by-step Solution
Detailed explanation
For sphere 1, in equilibrium
\(\text {and } \mathrm{T}_{1} \cos \theta_{1}=\mathrm{M}_{1} \mathrm{~g}\)
\(\mathrm{T}_{1} \sin \theta_{1}=\mathrm{F}_{1}\)
\(\therefore \tan \theta_{1}=\frac{\mathrm{F}_{1}}{\mathrm{M}_{1} g}\)
Similarly for sphere \(2, \tan \theta_{2}=\frac{F_{2}}{M_{2} g}\)
F is same on both the charges, \(\theta\) will be same only if their masses \(\mathrm{M}\) are equal.
\(\text {and } \mathrm{T}_{1} \cos \theta_{1}=\mathrm{M}_{1} \mathrm{~g}\)
\(\mathrm{T}_{1} \sin \theta_{1}=\mathrm{F}_{1}\)
\(\therefore \tan \theta_{1}=\frac{\mathrm{F}_{1}}{\mathrm{M}_{1} g}\)
Similarly for sphere \(2, \tan \theta_{2}=\frac{F_{2}}{M_{2} g}\)
F is same on both the charges, \(\theta\) will be same only if their masses \(\mathrm{M}\) are equal.
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