Q4. How tall would a barometer be if it used water instead of mercury? Take atmospheric pressure as 1.013 × 10 Pa and g = 9.8 m s.
h = Pa/(ρg) = 1.013 × 10/(1000 × 9.8), about 10.3 m, which is why mercury, 13.6 times denser, is used.
Q5. A dam holds water to a depth of 10 m behind a vertical wall 20 m wide. The gauge pressure rises steadily from zero at the surface to ρgH at the bottom. Find the net force on the wall, with atmospheric air on its other side. Take g = 9.8 m s and the density of water as 1000 kg m.
The average gauge pressure is ρgH/2, so, with H = 10 m and W = 20 m, F = (ρgH/2) × HW = 1000 × 9.8 × 10 × 20/2, about 9.8 × 10 N. Dams are made thicker at the bottom, where the pressure is greatest.
Q6. In the lift of Q2, the small piston is pushed down 10 cm. How far does the load rise?
Equal volumes: L2 = L1A1/A2 = 10/50 = 0.2 cm. The force is multiplied by 50, the distance divided by 50.