Question

what force is applied on a piston of area of cross section 2cm² to obtain a force 150N on the piston of area of cross section 12cm² in hydraulic machine?​

Answer 1

Answer:

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Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.・. P1 = P2

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