Explanation

(a) $Minimum\ tension\ T_{1} = \frac{{mv}^{2}}{r} - mg$
$Maximum\ tension\ T_{2} = \frac{{mv}^{2}}{r} + mg$
$Let\ \ \ \frac{{mv}^{2}}{r} = x$
$So,\ \ \ T_{1} = x - mg$ …(i)
$T_{2} = x + mg$ …(ii)
Diving Eq. (i) by Eq. (ii)
$\frac{T_{1}}{T_{2}} = \frac{x - mg}{x + mg}\ \ \ \ \ \left( \because Given\frac{T_{1}}{T_{2}} = \frac{3}{5} \right)$
$\therefore\frac{3}{5} = \frac{x - mg}{x + mg}$
$or\ \ 3x + 3mg = 5x - 5mg$
$or\ \ x = 4\ mg$
$ie,\ \frac{{mv}^{2}}{r} = 4mg$
$\therefore\ \ \ v^{2} = 4\ rg$
$or\ \ \ \ v = \sqrt{4rg}$
$or\ \ v = \sqrt{4 \times 2.5 \times 9.8}$
$v = \sqrt{98}\ {ms}^{- 1}$