A) 1000 watt
B) 1200 watt
C) 1500 watt
D) 1600 watt
Correct Answer: B
Solution :
\[{{P}_{t}}={{P}_{c}}\left( 1+\frac{m_{a}^{2}}{2} \right)\,;\] Here ma = 1 Þ \[1800={{P}_{c}}\left( 1+\frac{{{(1)}^{2}}}{2} \right)\] Þ \[{{P}_{c}}=1200W\]You need to login to perform this action.
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