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PhysicsHeat & ThermoRadiationMedium2 minPYQ_2022
PhysicsMediumnumerical

Nearly10%of the power of a110 Wlight bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of1 mfrom the bulb to a distance of5 misa×10-2 W m-2. The value of 'a' will be

Answer:
84.00
Solution:

Power of visible radiation is P'=10100×110 W =11 W

Change in average intensity of visible radiation is  Iradiation=Iradiation1-Iradiation2

I1-I2=P'4πr12-P'4πr22

=114π11-125  =114π×2425

=264π×10-2=84×10-2 W m-2

Hence, the value of a=84.

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation
2mℹ️ Source: PYQ_2022

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