Light with an energy flux of 18 text{ W/cm}^2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 text{ cm}^2, th…

Total energy received $U = \text{flux} \times \text{Area} \times \text{time} = 18 \times 20 \times (30 \times 60) = 6.48 \times 10^5 \text{ J}$. For a completely absorbing surface, the momentum delivered is $p = \frac{U}{c} = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3} \text{ kg m/s}$? Wait, let's recalculate: $18 \text{ W/cm}^2 \times 20 \text{ cm}^2 = 360 \text{ W}$. Time $t = 1800 \text{ s}$. $U = 360 \times 1800 = 6.48 \times 10^5 \text{ J}$. $p = U/c = 6.48 \times 10^5 / 3 \times 10^8 = 2.16 \times 10^{-3} \text{ kg m/s}$. Option D is correct.