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A collimated beam of light of flux density `3kWm^(-2)` is incident normally on `100mm^(2)` completely absorbing screen. If P is the pressure exerted on the screen and `trianglep` is the momentum tranferred to the screen during a 1000 s interval, thenA. `P=10^(-3)Nm^(-2)`B. `P=10^(-4)Nm^(-2)`C. `trianglep=10^(-4)kgms^(-1)`D. `trianglep=10^(-5)kgms^(-1)` |
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Answer» Correct Answer - B::D Since `P=(I)/(c )=10^(4)Nm^(-2)` `P=(E)/(A)=(1)/(A)(trianglep)/(trianglet)` `trianglep=Patrianglet=10^(-5)kgms^(-1)` |
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