Here are the solutions to the questions: 17. Draw a labeled diagram showing the working of a hydraulic lift and explain the transmission of pressure in it. Diagram Description: Imagine two vertical cylinders of different cross-sectional areas, $A_1$ (small) and $A_2$ (large), connected at their bases by a horizontal tube. Each cylinder contains a piston. The entire system is filled with an incompressible fluid, typically oil. A small force $F_1$ is applied to the piston in the smaller cylinder, and a larger load $F_2$ is lifted by the piston in the larger cylinder. ` (Diagram would show two connected cylinders, one narrow with a small piston, one wide with a large piston. Both cylinders are filled with fluid. Labels: Small Piston (Area A1) Large Piston (Area A2) Input Force F1 Output Force F2 (lifting a load) Incompressible Fluid Connecting Tube ) ` Explanation of Pressure Transmission: A hydraulic lift works based on Pascal's Principle, which states that pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. 1. When a small input force $F_1$ is applied to the small piston (with area $A_1$), it creates a pressure $P_1 = \frac{F_1}{A_1}$ in the fluid. 2. According to Pascal's Principle, this pressure $P_1$ is transmitted equally throughout the fluid to the larger piston (with area $A_2$). So, the pressure exerted on the larger piston is $P_2 = P_1$. 3. Since $P_2 = \frac{F_2}{A_2}$, we have $\frac{F_1}{A_1} = \frac{F_2}{A_2}$. 4. Because $A_2$ is much larger than $A_1$, the output force $F_2 = F_1 \left(\frac{A_2}{A_1}\right)$ will be significantly greater than the input force $F_1$, allowing a small force to lift a heavy load. 18. Explain how a drinking straw works using the concept of atmospheric pressure. A drinking straw works by creating a pressure difference. When you suck on a straw, you remove some of the air inside it, which reduces the air pressure within the straw. The atmospheric pressure acting on the surface of the liquid outside the straw is now greater than the pressure inside the straw. This higher external atmospheric pressure pushes the liquid down on its surface and forces it up the straw and into your mouth. 19. a) Describe the mechanism of water pumping using atmospheric pressure. The mechanism of water pumping using atmospheric pressure, as seen in a simple hand pump (suction pump), involves creating a partial vacuum. When the pump handle is lifted, a piston inside the pump barrel moves upwards, increasing the volume and reducing the air pressure above the water in the well pipe. This lower pressure inside the pipe allows the higher atmospheric pressure acting on the surface of the water in the well to push the water up the pipe. As the piston moves down, a valve closes, trapping the water, and another valve opens to allow the water to be expelled. b) Give one example of its application in daily life. Hand pumps* used to draw water from wells. 20. Explain how pressure varies with depth in a liquid and relate this to fluid density. Pressure in a liquid increases proportionally with depth. This is because as you go deeper into a liquid, the weight of the column of liquid above that point increases. The deeper the point, the more liquid there is above it, and thus the greater the force exerted by its weight per unit area. Pressure also varies directly with the fluid's density. For a given depth, a denser liquid will exert greater pressure because it has more mass per unit volume, meaning the column of liquid above that point will have a greater weight. This relationship is summarized by the formula $P = \rho gh$, where $P$ is pressure, $\rho$ is density, $g$ is acceleration due to gravity, and $h$ is depth. 21. A liquid column has a height of 2 m and a density of 1200 kg/m³. Calculate the pressure at the base of the column. (Take $g = 10 \text{ m/s}^2$) Step 1: Identify the given values. Height (depth), $h = 2 \text{ m}$ Density, $\rho = 1200 \text{ kg/m}^3$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$ Step 2: Write down the formula for pressure in fluids. $$P = \rho gh$$ Step 3: Substitute the values into the formula. $$P = (1200 \text{ kg/m}^3) \times (10 \text{ m/s}^2) \times (2 \text{ m})$$ Step 4: Calculate the pressure. $$P = 1200 \times 10 \times 2 \text{ Pa}$$ $$P = 12000 \times 2 \text{ Pa}$$ $$P = 24000 \text{ Pa}$$ The pressure at the base of the column is $\boxed{\text{24000 Pa}}$.