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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
(2) Temperature gradient
`(Delta theta)/(Delta x) = (theta_(1) - theta_(2))/(l)`
`80 = (30 - theta_(2))/(0.5)`
`theta_(2) = 10^(@)C`
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