Question

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

Answer 1

Force applied will be - 25 N

Pascal's law is states that "a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere."

Hydraulic machine works over principle of Pascal's law.

Thus, pressure will be equal.

Pressure is defined as the force acted on a body, perpendicular to the surface area, per unit area.

Pressure is given by the formula-

P = f/a

P denotes pressure, f denotes force and a denotes area.

Now, according to given question,

F1/A1 = F2/A2

F1 and A1 representes given values of force and area. F2 and A2 representes given value of A2 and value to be calculated of F2.

150/0.0012 = F2/0.0002

F2 = 25 N

Thus, force applied to get required result is 25N.

Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.•. P1 = P2

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