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An automobile sack is lifted by a hydraulic jack that consists of two pistons. The larger piston is 1m in diameter and small piston is 10cm in diameter. If

W be the weight of the car, the minimum force is needed on small piston to lift the car is


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To lift the car, the pressure must be atleast same in both pistons.

Diameter of larger piston = 1m = 100cm
Radius of larger piston, R = 100/2 cm = 50cm
Area(A₁) = πR² = π×50² = 2500π cm²

Diameter of smaller piston = 10cm
Radius of smaller piston, r = 10/2 cm = 5cm
Area(A₂) = πR² = π×5² = 25π cm²

Car of weight W is on larger piston,
Let the force on smaller piston required to lift it = F
Since pressure on both pistons is same,

 \frac{F}{A_2} = \frac{W}{A_1} \\ \\ \Rightarrow F=W \times  \frac{A_2}{A_1} \\ \\ \Rightarrow F=W \times  \frac{25 \pi }{2500 \pi } \\ \\ \Rightarrow F= \frac{W}{100}

Minimum force required is W/100.
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