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Mechanical Properties of Fluids

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Question : 8 of 31
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A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2^{2}. What maximum pressure would the smaller piston have to bear?
Solution:  
The maximum force, which the bigger piston can bear,
F=3000 kg, f=3000×9.8 NF = 3000\,\mathrm{kg},\ f = 3000 \times 9.8\,\mathrm{N}
∴ Maximum pressure on the bigger piston,
P=FA=3000×9.8425×10−4P= \frac{F}{A}= \frac{3000 \times 9.8}{425 \times 10^{-4}}
=6.92×105 Pa=6.92 \times 10^{5}\,\mathrm{Pa}
Since the liquid transmits pressure equally, therefore the maximumpressure the smaller piston can bear is 6.92×1056.92 \times 10^{5}Pa.
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