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The temperature gradient in a rod of 0.5...

The temperature gradient in a rod of `0.5 m` length is `80^(@)C//m`. It the temperature of hotter end of the rod is `30^(@)C`, then the temperature of the cooler end is

A

`40^(@)C`

B

`-10^(@)C`

C

`10^(@)C`

D

`0^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
B
Temperature gradient
`(dT)/(dx)=80^(@)C//m`
`implies int_(30)^(T)dT=-80int_(0)^(0.5)dx`
`implies T-30=-80(0.5)`
`implies T-30=-40`
`T=-10^(@)C`
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