One end of a string of length \(l\) as connected to a particle of mass \(m\) and the other to a small peg on…
- \(T\)
- 0
- \(\frac{T+m v^2}{1}\)
- \(\frac{T-m v^2}{I}\)
Solution

Here, \(N=m g \text { and } T=\frac{m v^2}{l}\) Tension in the string provides necessary centripetal force. Hence, net force on the particle towards centre of circular path is equal to \(T\).
Asked in: AP EAMCET 2020 (17 Sep Shift 2)
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