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

Options

A. `0^(@)`C
B. `-10^(@)`C
C. `10^(@)`C
D. `40^(@)`C

Answer

B

Explanation

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