A hydraulic lift has a leak so that it is only 6.0 × 101% efficient in raising its load. If the large piston exerts a force of 1.00 × 102 N when the small piston is depressed with a force of 15 N and the radius of the small piston is 18 cm, what is the radius of the large piston?

Answer 1

pressure on each side is the same.

forcesmall*.60/areasmall= forcelarge/arealarge

15/pi*18^2=100/(pi*rl^2*.6)

solve for rlarge.