# A plate of mass 10gm is in equilibrium in air due to force exerted by light beam on plate . Assuming plate is perfectly absorbing the power of beam is :

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

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
$F_{photon}=\large\frac{P}{C}\;$
$\large\frac{P}{C}=10^{-1}$
$P=3\times10^{8}\times10^{-1}$
$P=3 \times10^{7}W\;.$