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 :
| [d] Bonus |
| \[\frac{{{(200)}^{2}}}{{{R}_{0}}(1+\alpha (T-{{T}_{0}})}\] |
| \[T\to \]temperature at 't' |
| \[{{T}_{0}}\to \]temperature at t = 300 K |
| \[T-{{T}_{0}}=\frac{500-300}{30}(t)\] |
| \[T-{{T}_{0}}=\frac{200}{30}t\] |
| \[T-{{T}_{0}}=\frac{20t}{3}\] |
| \[\int\limits_{0}^{30}{\frac{{{(200)}^{2}}}{100(1+\alpha \frac{20t}{3})}dt=\frac{200\times 200}{100}\,\,\,\int\limits_{0}^{30}{\frac{dt}{1+\frac{20\alpha }{3}}t}}\] |
| \[=\frac{400\times 3}{20\alpha }\ell n\left( \frac{\frac{1+20\alpha }{3}\times 30}{1} \right)\] |
| \[120=100\,(1+\alpha (200))\] |
| \[1+(200)\alpha =\frac{6}{5}\] |
| \[(200\alpha )=\frac{1}{5}\]\[\alpha =\frac{1}{1000}\] |
| \[=60,000\,\ell n\left( \frac{6}{5} \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
