Physics, asked by amar7226, 10 months ago

A force of 50 kgf is applied to the smaller piston of a hydraulic
machine. Neglecting friction, find the force exerted on the large
piston, if the area of the piston are 5 cm^2 and 25 cm^2.

Answers

Answered by Thakshayini
0

Answer:

hi there!!!

Explanation:

happy helping ❤️

refer to the picture... Here pressure is constant.

Attachments:
Answered by Anonymous
1

Ratio of diameter of smaller piston to bigger piston = 5:25

.•. Ratio of area of smaller piston to bigger piston = 25:625

Force applied on smaller piston, F1 = 50kgf

Let F2 be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = Pressure on wider piston

 \frac{F1}{A1}  =  \frac{F2}{A2}  \\  \\  \implies \:   \frac{F1}{F2}  =  \frac{A1}{A2}  \\  \\  \implies \:  \frac{50}{F2}  =  \frac{25}{625}  \\  \\  \implies \: F2 \:  =  \:  \cancel{50}  \:  \: ²\times  \frac{625}{ \cancel{25} } \\  \\  \ = 1250 \: kgf \:  \:  \:  \:  \: (ans)

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