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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. `40^(@)C`B. `- 10^(@)C`C. `10^(@)C`D. `0^(@)C` |
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Answer» Correct Answer - 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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