# NEET Questions Solved

NEET - 2015

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c=velocity of light)

(a)E/c

(b)2E/c

(c)2E/c2

(d)E/c2

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The radiation energy is given by

E=hc/λ

Initial momentum of the radiation is

Pi=h/λ=E/c

The reflected momentum is

Pr=-h/λ=-E/c

So, the change in momentum of light is
ΔPlight=Pr-Pi=-2E/c

Thus, the momentum transferred to the surface is
ΔPlight=2E/c

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NEET - 2014

Light with an energy flux of 25 x 104 Wm-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15cm2 the average force exerted on the surface is

(a)1.25x10-6N

(b)2.50x10-6N

(c)1.20x10-6N

(d)3.0x10-6N

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Energy flux=25x104/sm2

Force on unit area= momentum transferred in unit time on area=25x104/C

=250x104/3x108=8.3x10-4N/m2

Force on the 15x10-4m2 area

=8.3x10-4N/m2x15x10-4m2

=124.5x10-8N=1.2x10-6N

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