$(a)\;3\times10^{9} W\qquad(b)\;3\times10^{6} W\qquad(c)\;3\times10^{8} W\qquad(d)\;3\times10^{7} W$

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Answer : (d) $\;3\times10^{7} W$

Explanation :

Since plate is in air , so gravitational force will act on this

$F_{gravitational }=mg\;(downward)$

$=10\times10^{-3}\times10$

$=10^{-1} N$

For equilibrium force exerted by light beam should be equal to $\;F_{gravitational}\;.$

$F_{photon}=F_{gravitational}$

Let power of light beam be P

$F_{photon}=\large\frac{P}{C}\;$

$\large\frac{P}{C}=10^{-1}$

$P=3\times10^{8}\times10^{-1}$

$P=3 \times10^{7}W\;.$

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