A) \[\sqrt{98}\,\,m/s\]
B) \[7\,\,m/s\]
C) \[\sqrt{490}\,\,m/s\]
D) \[\sqrt{4.9}\]
Correct Answer: A
Solution :
In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant. Tension at lowest point \[{{T}_{\max }}=\frac{m{{v}^{2}}}{r}+mg\] Tension at highest point \[{{T}_{\min }}=\frac{m{{v}^{2}}}{r}-mg\] \[\frac{{{T}_{\max }}}{{{T}_{\min }}}=\frac{\frac{m{{v}^{2}}}{r}+mg}{\frac{m{{v}^{2}}}{r}-mg}=\frac{5}{3}\] by solving we get,\[v=\sqrt{4gr}\]\[=\sqrt{4\times 9.8\times 2.5}\] \[=\sqrt{98}\,m/s\]
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