A) 1000 W
B) 1200 W
C) 1500 W
D) 1600 W
Correct Answer: B
Solution :
Total power \[{{P}_{t}}={{P}_{c}}\left[ 1+\frac{m{{a}^{2}}}{2} \right]\] \[\because \] \[m{{a}^{2}}=1\] \[\therefore \] \[1800={{P}_{c}}\left[ 1+\frac{1}{2} \right]\] \[{{P}_{c}}=1200W\]
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