A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of 10 N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _________ N.

Solution

Physics sir · 3 steps

Pascal's law: equal pressure on both pistons, so force scales with area, and area goes as diameter squared.

1. Equal pressure. Both water surfaces are at the same level, so by Pascal's law the pressure on them must be equal for equilibrium. That gives F₁/A₁ = F₂/A₂.

   F₂ = F₁ × A₂/A₁ = F₁ × (d₂/d₁)²

2. Area ratio. The diameter ratio is (14 cm)/(1.4 cm) = 10. Area goes as the square of the diameter, so A₂/A₁ = 10² = 100.
3. So. F₂ = 10 N × 100 = 1000 N on the thicker arm.

The common slip: 100. Using the diameter ratio 10 instead of the area ratio 100 gives 100 N, not 1000 N.

The idea to keep. In a hydraulic press, force is in proportion to piston area. When you are given diameters, always square their ratio before multiplying.

Answer from the NTA final answer key. Solution by PrepWiser.