WBJEE · Physics · Electrostatics
A particle with charge \(Q\) coulomb, tied at the end of an inextensible string of length \(R\) metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude \(Q\) coulomb. The mass of the moving charge \(M\) is such that \(M g=\frac{Q^{2}}{4 \pi \in_{0} R^{2}}\). If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be
- A 0
- B \(2 \sqrt{g R}\)
- C \(\sqrt{2 g R}\)
- D \(\sqrt{5 g R}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{g R}\)
Step-by-step Solution
Detailed explanation
According to the question, when tension in the string is zero. As \(T=0\) \(M g-\frac{K Q^{2}}{R^{2}}=\frac{m v^{2}}{R} \quad\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)\) \(\frac{m v^{2}}{R}=\) centripetal force required to keep the body in a circular path…
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