a
(a)Work output of engine \(= mgh = \) \(100 \times 10 \times 10 = {10^4}J\)

Efficiency (\(\eta\)) = \(\frac{{{\rm{output}}}}{{{\rm{input}}}}\)

Input energy = \(\frac{{{\rm{outupt}}}}{\eta }\)

\( = \frac{{{{10}^4}}}{{60}} \times 100 = \frac{{{{10}^5}}}{6}J\)

 Power = \(\frac{{{\rm{input energy}}}}{{{\rm{time}}}}\)= \(\frac{{{{10}^5}/6}}{5} = \frac{{{{10}^5}}}{{30}} = 3.3\;kW\)