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A body cools at the rate of 0.75^(@)C//s...

A body cools at the rate of `0.75^(@)C//s` when it is `50^(@)C` above the surrounding. Its rate of cooling when it is `30^(@)C` above the same surrounding is

A

`0.32^(@)C//s`

B

`0.36^(@)C//s`

C

`0.40^(@)C//s`

D

`0.45^(@)C//s`

Text Solution

Verified by Experts

The correct Answer is:
D
`R_(1)=K(theta_(1)-theta_(0))`
`R_(2)=K(theta_(2)-theta_(0))`
`(R_(2)))/(R_(1))=((theta_(2)-theta_(0)))/((theta_(1)-theta_(0)))`
`R_(2)=(0.75xx30)/(50)=0.45^(@)C//s`.
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