alleryA hydraulic lift uses a small piston (area 1 cm²) and a large piston (area 100 cm²). To lift a 1000 kg load, the minimum force needed on the small piston is:
A98 N
B980 N
C9.8 N
D98,000 N
Answer & Solution
Correct answer: A. 98 N
Load weight = $1000 \times 9.8 = 9800$ N. Pascal: $F_1/A_1 = F_2/A_2$
⇒ $F_1 = F_2 A_1/A_2 = 9800 \times (1/100) = 98$ N. Mechanical advantage = $A_2/A_1 = 100$.
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