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Q.
An electric motor drill, rated 350 W has an efficiency of 35%. The torque produced, if it is working at 3000 rpm.
System of Particles and Rotational Motion
Solution:
Angular velocity of the motor drill,
$\omega=\frac{2\pi\upsilon}{60}=\frac{2\pi\times3000}{60}$
Let $τ$ be the torque produced by the motor.
Power produced $= τ\omega=\tau\times\frac{2\pi\times3000}{60}$
Now, $\tau\times\frac{2\pi\times3000}{60}=35\%$ of $350\, W$
or $\tau\times2\pi\times50=\frac{35}{100}\times350$
or $\tau=\frac{35\times350}{100\times2\pi\times50}=0.39\,N\,m$