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प्रश्न
Multiple Choice Question.
A hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg. The area of cross-section of piston carrying the load is 2.25 × 10-2 m2. What is the maximum pressure the smaller piston would have to bear?
पर्याय
0.8711 × 106 N/m2
0.5862 × 107 N/m2
0.4869 × 105 N/m2
0.3271 × 104 N/m2
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उत्तर
0.8711 × 106 N/m2
Explanation:
Apply Newton’s second law of motion to find the force:
F = mg
Substitute the value mass as m = 2000 kg and acceleration due to gravity as g = 9.81 m/s2 to find the force exerted by the piston.
F = mg
F = 2000 kg × 9.81 m/s2
F = 19620 N
Use the law of pressure
p = `F/A`
Substitute the value of force as F = 19620 N, and area as A = 2.25 × 10–2m2 to calculate the maximum pressure the hydraulic lift can bear.
Pmax = `(19620N)/(2.25 xx 10^-2m^2)`
Pmax = 0.8711 × 106Pa
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