Light with an energy flux of 20W/cm2 falls on a non – reflecting surface at normal incidence. If the surface has an area of 30cm2, the total momentum delivered (for complete absorption) during 30 minutes is:

P=IAtCP=20×10+4×30×10−4×30×603×108=36×10−4kg−m/sec

Light with an energy flux of 20W/cm2 falls on a non – reflecting surface at normal incidence. If the surface has an area of 30cm2, the total momentum delivered (for complete absorption) during 30 minutes is:

P=IAtCP=20×10+4×30×10−4×30×603×108=36×10−4kg−m/sec

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Light with an energy flux of 20W/cm² falls on a non – reflecting surface at normal incidence. If the surface has an area of 30cm², the total momentum delivered (for complete absorption) during 30 minutes is:

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$36 \times 10^{- 5} k g m / s$

$108 \times 10^{4} k g m / s$

$36 \times 10^{- 4} k g m / s$

$1.08 \times 10^{7} k g m / s$

answer is B.

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Detailed Solution

$P = \frac{I A t}{C}$

$P = \frac{20 \times 10^{+ 4} \times 30 \times 10^{- 4} \times 30 \times 60}{3 \times 10^{8}}$

$= 36 \times 10^{- 4} k g - m / \sec$

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Light with an energy flux of 20W/cm2 falls on a non – reflecting surface at normal incidence. If the surface has an area of 30cm2, the total momentum delivered (for complete absorption) during 30 minutes is:

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Light with an energy flux of 20W/cm² falls on a non – reflecting surface at normal incidence. If the surface has an area of 30cm², the total momentum delivered (for complete absorption) during 30 minutes is: