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A diesel engine transmits 25 KW at the end of the crankshaft, moving at 1650 rev/min, find the torque exerted by the piston.

a) 148.90 Newton-metre
b) 144.75 Newton-metre
c) 150.75 Newton-metre
d) 120.10 Newton-metre

Answer: b
Explanation: Power(W) = \(\frac{2πnT}{60}\)
25*1000=\(\frac{2π*1650*T}{60}\)
T=144.75 Newton-metre.

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