In a hydraulic lift- the force exerted on the piston with small cross-sectional area F 1 is less than the force exerted on the piston with large cross-sectional area F 2- Then-A- Work done by F 1 is greater than work done by F 2B- Work done by F 2 is greater than work done by F 1C- No work is done by F1 or F2D- Work done by F1 is equal to work done by F2

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Standard IX

Physics

Pascal's Law

In a hydrauli...

Question

In a hydraulic lift, the force exerted on the piston with small cross-sectional area $(F_{1})$ is less than the force exerted on the piston with large cross-sectional area $(F_{2})$ . Then,

Work done by $F_{1}$ is greater than work done by $F_{2}$

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Work done by $F_{2}$ is greater than work done by $F_{1}$

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No work is done by $F_{1}$ or $F_{2}$

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Work done by $F_{1}$ is equal to work done by $F_{2}$

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Solution

The correct option is D
Work done by $F_{1}$ is equal to work done by $F_{2}$

The energy can neither be created nor destroyed. Therefore, work done by the force, $F_{1}$ is equal to the work done by the force, $F_{2}$ .

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In a hydraulic lift, the force exerted on the piston with small cross-sectional area(F1) is less than the force exerted on the piston with large cross-sectional area(F2). Then,

The correct option is D Work done by F1 is equal to work done by F2 The energy can neither be created nor destroyed. Therefore, work done by the force, F1 is equal to the work done by the force, F2.

In a hydraulic lift, the force exerted on the piston with small cross-sectional area $(F_{1})$ is less than the force exerted on the piston with large cross-sectional area $(F_{2})$ . Then,

The correct option is D
Work done by $F_{1}$ is equal to work done by $F_{2}$

The energy can neither be created nor destroyed. Therefore, work done by the force, $F_{1}$ is equal to the work done by the force, $F_{2}$ .