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) H`
B. `(16 // 3)H`
C. `(9//27) H`
D. `(1//16) H`

Answer

A

Explanation

(1) Rate of cooling `R prop (T^(4) - T_(0)^(4))`
`(R_(1))/(R_(2)) = ((T_(1)^(4) - T_(0)^(4)))/((T_(2)^(4) - T_(0)^(4)))`
`(H)/(R_(2)) = ((600)^(4) - (200)^(4))/((400)^(4) - (200)^(4)) = (6^(4) - 2^(4))/(4^(4) - 2^(4))`
`(3^(4) - 1)/(2^(4) - 1) = (80)/(15) = (16)/(3)`
`R_(2) = (3 H)/(16)`