A hydraulic lift has a large cross-section and a small cross-section. The large cross-sectional area is 20 times the small cross-sectional area. If on the small cross-section, an input force of 25 N is applied, then the output force is

200 N

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400 N

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300 N

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500 N

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Solution

The correct option is D
500 N

Let the small cross-sectional area, $A_{1} = A$

Large cross-sectional area, $A_{2} = 20 A$

Input force, $F_{1} = 25 N$

Let the output force be $F_{2}$ .

According to Pascal's law,

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

$\frac{25}{A} = \frac{F_{2}}{20 A}$

On soving, we get, $F_{2} = 500 N$

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A hydraulic lift has a large cross-section and a small cross-section. The large cross-sectional area is 20 times the small cross-sectional area. If on the small cross-section, an input force of 25 N is applied, then the output force is

The correct option is D 500 N Let the small cross-sectional area, A1=A Large cross-sectional area, A2=20A Input force, F1=25 N Let the output force be F2. According to Pascal's law, F1A1=F2A2 25A=F220A On soving, we get, F2=500 N