Correct Answer - C
At exterme position, `T_("min") - mg cos theta_(0) = 0`
`T_("min") = mg cos theta_(0)`
At lowest position, `T_("max") = mg = (mv^(2))/(l)` … (1)
`mg l (1 - cos theta_(0)) = (1)/(2) mv^(2) - 0` ….(2)
from (1) & (2)
`T_("max") = 3 mg - 2 mg cos theta_(0)`
`T_("max") = 1.2 T_("min")`
Solving, `cos theta_(0) = (15)/(16)`