In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of $20 : 1$ . What weight placed on the small piston will be sufficient to lift a car of mass $1500 k g$ ?

3.75 k g

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37.5 k g

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7.5 k g

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75 k g

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Solution

The correct option is A $3.75 k g$
According to Pascal's law
Pressure is constant.
hence $P_{1} = P_{2}$ given : $F_{1}$ = 1500 kg
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ $\frac{A_{1}}{A_{2}} = {(\frac{20}{1})}^{2}$
So $\frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{1500}{F_{2}} = {(\frac{20}{1})}^{2} \Rightarrow F_{2} = 3.75 K g$

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