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Question

Physics Question on System of Particles & Rotational Motion

An automobile engine develops 100kW100 \,kW power when rotating at a speed of 1800rpm1800\, rpm. The torque delivered by the engine is

A

1026πNm\frac{10^{2}}{6\pi} \,N \,m

B

1046πNm\frac{10^{4}}{6\pi} \,N \,m

C

1066πNm\frac{10^{6}}{6\pi} \,N \,m

D

1086πNm\frac{10^{8}}{6\pi} \,N \,m

Answer

1046πNm\frac{10^{4}}{6\pi} \,N \,m

Explanation

Solution

Here, P=100kWP = 100 \,kW =100×103W=105W= 100\times10^{3} W = 10^{5}W υ=1800rpm=180060rps=30rps\upsilon = 1800 \,rpm = \frac{1800}{60}\, rps = 30 \,rps ω=2πυ=2π(30)=60πrads1 \therefore\omega = 2 \,\pi \,\upsilon =2\pi \left(30\right) = 60 \pi \,rad \,s^{-1} As P=τωP= \tau\omega τ=pω=105W60πrads1\therefore\tau = \frac{p}{\omega} = \frac{10^{5} W}{60 \pi \,rad \,s^{-1}} =1046πNm = \frac{10^{4}}{6\pi} \,N \,m