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Value of temperature gradient is 80^(@) C/m on a rod of 0.5 m length. Temperature of hot end is 30^(@)C, then what is the temperature of cold end ? |
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Answer» Solution :`(T_(1)-T_(2))/(L)=80` `30-T_(2)=80xxL` `=80xx0.5` `30-T_(2)=40` `:.T_(2)=30-40` `:.T_(2)=-10^(@)C` |
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