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Question Number 36548 by rahul 19 last updated on 03/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${Q}=\frac{{kA}\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right){t}}{{l}}=\frac{{i}^{\mathrm{2}} {Rt}}{{J}}=\frac{\left({V}\right)^{\mathrm{2}} {t}}{{R}\:{J}} \\ $$$${A}=\left(\mathrm{40}−\mathrm{2}×\mathrm{0}.\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1584}.\mathrm{04}{cm}^{\mathrm{2}} \\ $$$$\theta_{\mathrm{2}} −\theta_{\mathrm{1}} =\mathrm{100} \\ $$$${l}=\mathrm{0}.\mathrm{1}{cm} \\ $$$${V}=\mathrm{100}{Volt} \\ $$$${J}= \\ $$$${R}=\frac{\left({V}\right)^{\mathrm{2}} {l}}{{J}\:{kA}\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right)} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${heat}\:{transfer}\:...{attaching}\:{theorem}... \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
