Solving the Mystery of Wrench Work Ratios

[Saitama]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

If a force a F acts on the farther end of the wrench, the torque due to it Fl where l is the length of wrench. The work done by this torque for one full turn is Fl*(2$\pi$)
For wrench A, l=10 cm, and for wrench B, l=20cm.
Therefore the ratio of work done is 0.5. But this is wrong.
I don't understand why they have given the radius of the nut. I haven't used this information and I have no idea about where to use this.

Any help is appreciated. Thanks!

 

[Doc Al]

If a force F is required to tighten the nut using the 10 cm wrench, what force is required with the 20 cm one?

 

[Saitama]

F/2?

 

[Doc Al]

F/2?

Right.

 

[Saitama]

Right.

But what next? Does that mean the ratio is 1?

 

[Doc Al]

But what next? Does that mean the ratio is 1?

That's what I would say.

 

[Saitama]

That's what I would say.

Thanks a lot Doc Al!