A hydraulic press can lift 100 kg when a mass 'm' is placed on the smaller piston. It can lift ___________ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass 'm' on the smaller piston.

Solution

Physics sir · 4 steps

Pistons work through their areas, and area goes as diameter squared, so each 4 times change counts as 16.

1. Pascal's law. The pressure under both pistons is the same, so mg/a = Mg/A. This gives M = mA/a, where a and A are the areas of the small and large pistons.
2. Areas change. Area is (π d²)/4, so it scales as d². Making the large diameter 4 times bigger makes A 16 times bigger. Making the small diameter 4 times smaller makes a 16 times smaller.

   A'/a' = 16A/(a/16) = 256 A/a

3. Same m. With the same mass m on the small piston, the liftable mass grows by the same factor: M' = 256 × 100 kg.
4. So. The press can now lift M' = 25600 kg, so the answer is 25600.

The common slip: 1600. Using the diameter ratio in place of the area ratio gives a factor of 4 × 4 = 16 and an answer of 1600, not 25600.

The idea to keep. In a hydraulic press the force ratio equals the area ratio, which is the square of the diameter ratio. Square every diameter change before you multiply.

Answer from the NTA final answer key. Solution by PrepWiser.