In the given figure of a hydraulic lift, the narrow piston area is 6 $c m^{2}$ and the wide piston area is 60 $c m^{2}$ . Find the force required to lift the block if the weight of the block, w is 1200 N.

1200 N

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600 N

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180 N

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120 N

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Solution

The correct option is D
120 N

Let F be the force required on the hydraulic lift.
Given, for narrow piston:
$F_{1} = F, A_{1} = 6 c m^{2}$
For wide piston:
$F_{2} = 1200 N, A_{2} = 60 c m^{2}$
By Pascal’s law,
Pressure on narrow piston = Pressure on wide piston
Or $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$
Or $\frac{F}{6} = \frac{1200}{60}$
Thus, F = $\frac{1200}{60} \times 6$ = 120 N
Force required to lift the block = 120 N

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In the given figure of a hydraulic lift, the narrow piston area is 6 cm2 and the wide piston area is 60 cm2. Find the force required to lift the block if the weight of the block, w is 1200 N.

In the given figure of a hydraulic lift, the narrow piston area is 6 $c m^{2}$ and the wide piston area is 60 $c m^{2}$ . Find the force required to lift the block if the weight of the block, w is 1200 N.