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Assuming a particle to have the form of a sphere and to absorb all incident light, the radius (in mm) of a particle for which its gravitational attraction to the Sun is
Counterbalanced by the force that light exerts on it is------. The power of light radiated by the sun equals `P=4xx10^(26)W` and the density of the particle is `r=1.0 g//cm^(3)`. Use `G=(20)/(3)xx10^(-11)Nm^(2)//kg^(2)`, `pi=(25)/(8)` and mass of the sun `=2xx10^(30)kg`

A

`0.8`

B

`0.6`

C

`0.1`

D

`0.4`

Text Solution

Verified by Experts

The correct Answer is:
B

energy incident/sec on the particle
`=(P)/(4pir^(2))xxpiR^(2)=(PR^(2))/(4r^(2))`
`vecF` due to striking of photon `vecF_(1)=(vecdp)/(dt)=` change in lin. Mom. Of `1photon` in collisionxxNo. Of photons striking per sec `=(hv)/(C )xx(PR^(2))/(4r^(2)xxhv)=(PR^(2))/(4r^(2)C)`
Gravitational force on the particle
`vecF_(2)=(GM_(s)m)/(r^(2))=(GM_(s))/(r^(2))xx(4)/(3)piR^(3)rho`, `|vecF_(1)|=|vecF_(2)|`
`(PR^(2))/(4r^(2)C)=(GM_(s)xx4piR^(3)rho)/(3r^(2))impliesR=(3P)/(4GM_(s)pirhoC)=0.6`
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