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Q.

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 kg?

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a

7.5 kg

b

37.5 kg

c

75 kg

d

3.75 kg

answer is A.

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Detailed Solution

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

$∴ F_{1} = \frac{A_{1}}{A_{2}} \cdot F_{2} = {(\frac{1}{20})}^{2} (1500) = 3.75 kg$

$(A = π r^{2})$

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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 kg?