A) 225
B) 450
C) 900
D) 1800
Correct Answer: D
Solution :
\[P=\sigma {{T}^{4}}(4\pi {{r}^{2}})\propto {{T}^{4}}{{r}^{4}}\] \[\therefore \]\[\frac{{{P}_{1}}}{{{P}_{2}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}{{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{r}}={{(2)}^{2}}\times {{\left( \frac{1}{2} \right)}^{2}}=4\] or \[{{P}_{2}}=4{{P}_{1}}=4\times 50=1800\,\text{watt}\] Hence, the correction option is [d].
You need to login to perform this action.
You will be redirected in
3 sec
