Practice question
Question
A hydraulic lift raises a 1900kg load using a small piston of radius 4.5cm and a large piston of radius 18cm. What force is applied on the small piston? (Take g\=9.8m/s2)
Explanation
F1 = A1A2F2, F2 = 1900×9.8 = 18620N. A1 = π(0.045)2, A2 = π(0.18)2, A1A2 = 0.0020260.0324 = 116. F1 = 1862016 = 1163.75N≈1164N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1164 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Discussion
Comments
Share your thoughts. New comments appear after admin approval.
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.