A) \[\sqrt{12}\,m/s\]
B) \[\sqrt{18}\,m/s\]
C) \[\sqrt{24}\,m/s\]
D) \[\sqrt{32}\,m/s\]
Correct Answer: B
Solution :
\[\eta =\frac{{{E}_{o}}}{{{E}_{in}}}\] \[{{\operatorname{E}}_{o}}=\frac{75}{100}{{E}_{in}}=\frac{3}{4}\times 12=9\,\,Joule\] This \[{{\operatorname{E}}_{o}}\]get converted into K.E during fall by energy conbservation \[{{\operatorname{E}}_{o}}=\frac{1}{2}m{{v}^{2}}\] \[9=\frac{1}{2}\times 1{{v}^{2}}\] \[\operatorname{v}=\sqrt{18}\,\,m/s\]
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