An electric motor drill, rated 350 W has an efficiency of 35%. The torque produced, if it is working at 3000 rpm.

Options

A. 0.25 N m
B. 0.35 N m
C. 0.39 N m
D. 0.30 N m

Answer

C

Explanation

Angular velocity of the motor drill ,
`omega = (2pi epsilon)/(60) = (2 pi xx 3000)/(60)`
Let `tau` be the torque produced by the motor.
Power produced = `tau omega = tau xx (2pi xx 3000)/(60)`
Now `tau xx (2 pi xx 3000)/(60) = 35%` of 350 W
or `tau xx 2pi xx 50 = (35)/(100) xx 350`
or `tau = (35 xx 350)/(100 xx 2pi xx 50) = 0.39 Nm`