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If a motor running at a rate of 1200 rev...

If a motor running at a rate of 1200 revolutions per minute can supply torque of 80 Nm, then the power required will be

A

`10 pi` kwatt

B

`192 pi` kwatt

C

`3.2 pi` kwatt

D

`40 pi ` kwatt

Text Solution

Verified by Experts

The correct Answer is:
C
`P = tau omega = 80xx 2pi xx n`
`= (80xx pi xx 1200)/(60)=3.2 pi K w`
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