Question

What force is applied on a piston of area of cross section 2 cm square to obtain a force of 150 Newton on the piston of area of cross section 12 cm square in a hydraulic machine

Answer 1

178 n i think so..........

Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.・. P1 = P2

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies\frac{F1}{2 \times {10}^{ - 4} } = \frac{150}{12 \times {10}^{ - 4} } \\ \\ \implies \: F1 \: = \frac{150}{12 \times {10}^{ - 4} } \times 2 \times {10}^{ - 4} \\ \\ \implies \: F1 \: = \frac{ \cancel{150} \: \: ²⁵ }{ \cancel{12 }\times \cancel{ {10}^{ - 4} }} \times \cancel{2 }\times \cancel{{10}^{ - 4} } \\ \\ \ = 25 \: N....... \: \: \: (ans)$