Question

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.

Hint:

The force multiplication in a hydraulic press is proportional to the square of the ratio of the diameters: $F_2 = F_1 (\frac{D_2}{D_1})^2$.

Answer 1

Correct Answer: 25600

Step 1: Pascal's Law: $\frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow M = m \frac{A_2}{A_1} = m (\frac{D}{d})^2$.
Step 2: Initially, $100 = m (\frac{D}{d})^2$.
Step 3: New diameter of larger piston $D' = 4D$. New diameter of smaller piston $d' = d/4$.
Step 4: New ratio of diameters $\frac{D'}{d'} = \frac{4D}{d/4} = 16 \frac{D}{d}$.
Step 5: New mass $M' = m (16 \frac{D}{d})^2 = 256 \times m (\frac{D}{d})^2$.
Step 6: $M' = 256 \times 100 = 25600$ kg.