$\begin {array} {1 1} (a)\;0.12 \times 10^{-6}N & \quad (b)\;12 \times 10^{-6}N \\ (c)\;12 \times 10^{-8}N & \quad (d)\;1.2 \times 10^{-6}N \end {array}$

The total energy falling on the surface is $ U = (18\: W/cm^2 ) \times ( 20\: cm^2) \times (30 \times 60) = 6.48 \times 10^5J$

Therefore, the total momentum delivered (for complete absorption) is

$ p = \large\frac{U}{c}$$= \large\frac{(6.48 \times 10^5J) }{3 \times 10^8 m/s)}$ $ =m/s) = 2.16 \times 10^{-3}\: kgm/s$

The average force exerted on the surface is

$ F = \large\frac{p}{t}$$ \large\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}$$ = 1.2 \times 10^{-6}N$

Ans : (d)

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