A sphere at temperature `600 K` is placed in an enviroment to temperature is `200 K`. Its cooling rate is `H`. If its temperature reduced to `400 K` then cooling rate in same enviorment will become

Options

A. `(3)/(16)R`
B. `(16)/(3)R`
C. `(9)/(27)R`
D. None of these

Answer

A

Explanation

From Stefan.s law, net rate of heat energy lost per second
`R= epsi sigma A (T^(4)- T_(0)^(4))`
Where T is the temperature of the body and `T_(0)` is the temperature of the surroundings.
Here `R= epsi sigma A (600^(4) - 200^(4))`
`R. = epsi sigma A (400^(4) - 200^(4)) rArr (R.)/(R ) = [(400^(4) - 200^(4))/(600^(4) - 200^(4))]`
`R. = ((4^(4) -2^(4))/(6^(4) -2^(4)))R`
Hence, `R. = (3)/(16) R rArr`. a is correct