Question

If a shaft rotates at 100 r.p.m and is subjected to a torque of 3000 N-m, the power transmitted in kW would be

• A. $30\pi$
• B. $15\pi$
• C. $20\pi$
• D. $10\pi$

Answer 1

The correct option is D $10\pi$

$P=\frac{2\pi NT}{60,000}kW$

$=\frac{2\pi \times 100\times 3000}{60,000}=10\pi$