a
(a) 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,\(v = \sqrt {4gr} \)\( = \sqrt {4 \times 9.8 \times 2.5} \)\( = \sqrt {98} \,m/s\)