Use Pascal’s law for a hydraulic press

In a hydraulic press, $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ where:

• $F_1$ = force applied on smaller piston
• $A_1$ = area of smaller piston
• $F_2$ = force exerted by larger piston
• $A_2$ = area of larger piston

Initial condition

A mass $m$ is placed on the smaller piston, and the press can lift $100\,\text{kg}$.

So, $\frac{mg}{A_s} = \frac{100g}{A_L}$

Cancelling $g$, $\frac{m}{A_s} = \frac{100}{A_L}$ Hence, $100 = m\cdot \frac{A_L}{A_s}$

So the lifting mass is proportional to $\frac{A_L}{A_s}$

Effect of change in diameters

Area of a piston is proportional to square of its diameter: $A \propto d^2$

• Larger piston diameter is increased by $4$ times: $d_L' = 4d_L \Rightarrow A_L' = 16A_L$

• Smaller piston diameter is decreased by $4$ times: $d_s' = \frac{d_s}{4} \Rightarrow A_s' = \frac{A_s}{16}$

New area ratio

$\frac{A_L'}{A_s'} = \frac{16A_L}{A_s/16} = 256\cdot \frac{A_L}{A_s}$

Therefore, the lifting capacity becomes $256$ times the original.

New lifted mass

Original lifted mass = $100\,\text{kg}$

New lifted mass: $100 \times 256 = 25600\,\text{kg}$

Final answer

$\boxed{25600}$