Correct Answer - A
If s calorie of heat is received from the sun per `cm^(2)` per minute, the heat collected by the lens of radius `2.5 cm^(2)` in `20` mintues is
`Q_(1) = s xx A xx t = s xx pi xx (2.5)^(2) xx (20) = (392.5) s cal`
Heat required to melt `10 g` of ice is
`Q_(2) = mL = 10 xx 80 = 800 cal`
According to given problem `Q_(1) = Q_(2)`
`:. 392.5 s = 800` or `s= 2.04 cal cm^(-2) min^(-1)`