1. **Problem Statement:** We have a hydraulic lift with a small disc of area $2.5\ \text{cm}^2$ and a large disc of area $200\ \text{cm}^2$. A force of 30 N is applied to the small disc. We need to find: i) The pressure exerted on the liquid by the small disc. ii) The force exerted by this pressure on the large disc. b) The force needed on the small disc to support a 7000 N load on the large disc. 2. **Relevant Formula:** Pressure $P$ is defined as force $F$ divided by area $A$: $$P = \frac{F}{A}$$ In a hydraulic system, pressure is transmitted equally throughout the liquid, so: $$P_{small} = P_{large}$$ Force on large disc can be found by: $$F_{large} = P \times A_{large}$$ 3. **Unit Conversion:** Convert areas from $\text{cm}^2$ to $\text{m}^2$ because SI units for pressure are in Pascals (N/m²): $$2.5\ \text{cm}^2 = 2.5 \times 10^{-4}\ \text{m}^2$$ $$200\ \text{cm}^2 = 200 \times 10^{-4} = 0.02\ \text{m}^2$$ 4. **Calculations:** i) Calculate pressure on the small disc: $$P = \frac{F}{A} = \frac{30}{2.5 \times 10^{-4}} = 120000\ \text{Pa}$$ ii) Calculate force on the large disc: $$F_{large} = P \times A_{large} = 120000 \times 0.02 = 2400\ \text{N}$$ b) Calculate force needed on small disc to support 7000 N on large disc: Since pressure is equal: $$P = \frac{F_{small}}{A_{small}} = \frac{F_{large}}{A_{large}}$$ Rearranged to find $F_{small}$: $$F_{small} = F_{large} \times \frac{A_{small}}{A_{large}} = 7000 \times \frac{2.5 \times 10^{-4}}{0.02} = 87.5\ \text{N}$$ **Final answers:** i) Pressure on liquid: $120000\ \text{Pa}$ ii) Force on large disc: $2400\ \text{N}$ b) Force needed on small disc: $87.5\ \text{N}$