A) \[400\,In\frac{1.5}{1.3}J\]
B) 300 J
C) \[200\,In\frac{2}{3}J\]
D) \[400\,In\frac{5}{6}J\]
Correct Answer: D
Solution :
Bonus \[\frac{{{(200)}^{2}}}{{{R}_{0}}(1+\alpha (T-{{T}_{0}})}\] T \[\to \]?temperature at 't' T0 \[\to \]?temperature at t = 300 K T\[T-{{T}_{0}}=\frac{500-300}{30}(t)\] \[T-{{T}_{0}}=\frac{200}{30}t\] \[T-{{T}_{0}}=\frac{20t}{3}\] \[\int\limits_{0}^{30}{\frac{{{(200)}^{2}}}{100(1+\alpha \frac{20t}{3})}dt=\frac{200\times 200}{100}\,\,\,\int\limits_{0}^{30}{\frac{dt}{1+\frac{20\alpha }{3}}t}}\] \[=\frac{400\times 3}{20\alpha }\ell n\left( \frac{\frac{1+20\alpha }{3}\times 30}{1} \right)\] \[120=100\,(1+\alpha (200))\] \[1+(200)\alpha =\frac{6}{5}\] \[(200\alpha )=\frac{1}{5}\]\[\alpha =\frac{1}{1000}\] \[=60,000\,\ell n\left( \frac{6}{5} \right)\]You need to login to perform this action.
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