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Light with an energy flux of 20 "W/cm"^(...

Light with an energy flux of `20 "W/cm"^(2)` falls on a non-reflecting surface at normal incidence. If the surface has an area of `"30 cm"^(2)`, the total momentum delivered (for complete absorption) during 30 minutes is

A

`36 xx 10^(-5) kg-m//s`

B

`36 xx 10^(-4) kg-m//s`

C

`108 xx 10^(4) kg-m//s`

D

`1.08 xx 10^(7) kg-m//s`

Text Solution

Verified by Experts

The correct Answer is:
B

Given energy flux `phi =20W//cm^(2)`
Area `A = 30cm^(2)`
Time , `t= 30min =30 xx 60s`
Now , total energy falling on the surface in time is `u=phiAt =20 xx 30 xx (30 xx 60)J`
Momentum of the incident light `=U/C`
`=(20 xx 30 xx (30 xx 60))/(3xx 10^(8)) rArr = 36 xx 10^(-4) kg-ms^(-1)`
Momentun of the reflected light =0
`:.` Momentum delivered to the suface
`=36 xx 10^(-4) -0 =36 xx 10^(-4) kg-ms^(-1)`
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