Question

work done in rotating a handle 1 meter long to 360 degree when a force through of 10 newton is applied at the end off the handle and always perpandicular to it.

Answer 1

Torque$=\vec r\times\vec F=1\times 10\times sin90\degree=10 Nm$

Work done$=\tau\times \delta\theta=10\times 2\pi=20\times 3.14=62.8 J$