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  4. The cross - section of two pistons in a hydraulic lift are 100 cm 2 and 800 cm 2 respectively. What is the minimum force required to lift 5000 kg -wt on large piston surface? (in kN)

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Q. The cross - section of two pistons in a hydraulic lift are $100 cm ^{2}$ and $800 cm ^{2}$ respectively. What is the minimum force required to lift $5000 kg$ -wt on large piston surface? (in kN)

Mechanical Properties of Fluids

Solution:

As we know $F _{1} A _{1}= F _{2} A_{2}$
$F _{1}= F _{2} \times \frac{ A _{2}}{ A _{1}}=5000 \times 9.8 \times \frac{800}{100}$
$F_{1}=5000 \times 9.8 \times 8$
$F _{1}=392000 N$