1.

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 isA. `0^(@)`CB. `-10^(@)`CC. `10^(@)`CD. `40^(@)`C

Answer» Correct Answer - B
Temperature gradient = `(theta_(1)-theta_(2))/l`
Here, `theta_(1) = 30^(@)`C rArr l =0.5 cm`
`therefore 80=(30-theta_(2))/0.5`
`theta_(2) = 30- 80 xx 0.5 rArr theta_92) = 30-40 = 10^(@)`C


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