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# Light with an energy flux of $18\: W/cm^2$ surface has an area of $20\: cm^2$ falls on a non-reflecting surface at normal incidence. If the , the average force exerted on the surface during a $30$ minute time span is

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

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)