Q.

An automobile engine develops 100 kW power when rotating at a speed of 1800rpm. The torque delivered by the engine is

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a

1026πNm

b

1046πNm

c

1086πNm

d

1066πNm

answer is B.

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Detailed Solution

Here,P=100 kW=100×103 W=105 W

v=1800rpm=180060rps=30rps

ω=2πv=2π(30)=60πrads-1

As P=τω

τ=Pω=105 W60πradss-1=1046πNm

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