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A sphere at temperature 600 K is placed ...

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

A

`(3//16)H`

B

`(16//3)H`

C

`(9//27)H`

D

`(1//16)H`

Text Solution

Verified by Experts

The correct Answer is:
A
Rate of cooling `prop (T^4 - T_0^4)`
`implies (H)/(H.)=((T_1^4 -T_0^4)/(T_2^4 -T_0^4))=(((600)^4-(200)^4)/((400)^4-(200)^4)) implies H. =(3)/(16) H`
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