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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 :

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

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Answer : (d) $\;3\times10^{7} W$
Explanation :
Since plate is in air , so gravitational force will act on this
$F_{gravitational }=mg\;(downward)$
$=10^{-1} N$
For equilibrium force exerted by light beam should be equal to $\;F_{gravitational}\;.$
Let power of light beam be P
$P=3 \times10^{7}W\;.$
answered Feb 22, 2014 by yamini.v

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