A) 1.025 MW
B) 2.05 MW
C) 5 MW
D) 6 MW
Correct Answer: D
Solution :
[d] \[\operatorname{Power}=\frac{Work done}{Time}=\frac{\frac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)}{t}\] \[P=\frac{1}{2}\times \frac{2.05\times {{10}^{6}}\times \left[ {{\left( 25 \right)}^{2}}-\left( {{5}^{2}} \right) \right]}{5\times 60}\] \[\operatorname{P} = 2.05\times 1{{0}^{6}}W= 2.05 MW\]
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