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Temperature gradient in a rod 0.5 m long...

Temperature gradient in a rod `0.5` m long is `80^(@)Cm^(-1)` The temperature of the hotter end is `60^(@)C`. The temperature of cooler end will be :

A

`40^(@)C`

B

`10^(@)C`

C

`20^(@)C`

D

`12^(@)C`.

Text Solution

Verified by Experts

The correct Answer is:
D
Here `(T_(1)-T_(2))/(x)=80`
`rArr(60-T_(2))/(0*5)=80" "rArrT_(2)=+20^(@)C`.
Thus correct choice is (d).
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