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An effort applied on the small piston of a hydraulic jack lifts the load through a distance of 50 cm and the effort is displaced through 2 m. If the diameter of the larger piston is 28 cm, calcualte the area of the cross section of the smaller piston. |
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Answer» SOLUTION :`d_(1)=50 CM` `THEREFORE M.A. (L)/(E )=(d_(E ))/(d_(L))=(2)/(0.5)=4` `4=(pi_(2)^(2))/(pi r_(1)^(2))` `4=((22)/(7)xx0.14xx0.14)/(A_(1))` `A_(1)=0.0154` |
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