1.

Radiated energy per second gain by an electric bulb is 25 Joule/second and velocity of electromagnetic wave is c then force gain by the surface per second is ….

Answer»

`8.33xx10^(-8)J`
`8.33xx10^(-8)J`
`75xx10^(8)N`
`75xx10^(8)N`

Solution :`F=(U)/(C )=(25)/(3XX10^(8))=8.33xx10(-8)N`


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