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A hydraulic automobile lift is designed to lift vehicles of mass 5000 \mathrm{~kg}. The area of cross section of the cylinder carrying the load is 250 \mathrm{~cm}^{2}. The maximum pressure the smaller piston would have to bear is [Assume \left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]:

Option: 1

2 \times 10^{+5} \mathrm{~Pa}


Option: 2

20 \times 10^{+6} \mathrm{~Pa}


Option: 3

200 \times 10^{+6} \mathrm{~Pa}


Option: 4

2 \times 10^{+6} \mathrm{~Pa}


Answers (1)

best_answer


From pascal law same \Delta \mathrm{P} transmitted through out liquid

\Delta \mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{5000 \times 10}{250 \times 10^{-4}}

=2 \times 10^{6} \mathrm{~Pa}

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