The Workbook

1. Gauge pressure \(= h\rho g = 10 \times 1000 \times 10 = 10^{5}\) Pa, one atmosphere.

   Absolute pressure \(= P_0 + h\rho g = 10^{5} + 10^{5} = 2 \times 10^{5}\) Pa.

2. \(P_0 + h\rho g = 3P_0\) gives \(h\rho g = 2 \times 10^{5}\) Pa, so \(h = \dfrac{2 \times 10^{5}}{10^{4}} = 20\) m. Every \(10\) m of water adds one atmosphere.

1. Gauge pressure at the base \(= h\rho g = 0.5 \times 1000 \times 10 = 5000\) Pa.

   Force \(= P \times A = 5000 \times 2 = 10^{4}\) N, the weight of the \(1000\) kg of water.

2. On the wall the pressure grows from \(0\) at the surface to \(5000\) Pa at the base. The average is \(2500\) Pa.

   Force \(= 2500 \times (2 \times 0.5) = 2500\) N.

1. \(P = P_0 + h\rho g\) depends only on the depth. The three bases are at the same depth \(h\), so the pressures are equal.

2. Force \(= P \times\) base area. Equal pressures on equal areas give equal forces, even though the amounts of water differ.

3. The sloping walls of \(B\) press down on the water, and the water presses up on them. The walls carry the extra weight, and the base carries only the weight of the column directly above it.

1. \(P_0 = h\rho g = 0.76 \times 13600 \times 10 = 103360 \approx 1.03 \times 10^{5}\) Pa.

2. Water is \(13.6\) times less dense, so the column must be \(13.6\) times taller: \(0.76 \times 13.6 = 10.3\) m.

   This is why mercury, not water, is used in barometers.

1. The gas pushes the mercury down on its side and up on the open side, so the gas pressure is above atmospheric.

   At the level of the lower mercury surface, both arms are at the same pressure: \(P_{\text{gas}} = P_0 + h\rho g = 10^{5} + 0.2 \times 13600 \times 10 = 1.272 \times 10^{5}\) Pa.

2. \(P_{\text{gas}} = P_0 - h\rho g = 10^{5} - 27200 = 0.728 \times 10^{5}\) Pa, below atmospheric.

1. Pascal's law: a pressure applied to an enclosed liquid is passed on undiminished to every part of the liquid.

   Pressure \(= \dfrac{200}{10 \times 10^{-4}} = 2 \times 10^{5}\) Pa. Force on the large piston \(= 2 \times 10^{5} \times 500 \times 10^{-4} = 10^{4}\) N, which lifts \(1000\) kg.

2. The liquid volume moved is the same: \(10 \times 50 = 500 \times d\), so \(d = 1\) cm.

3. Work on the small piston \(= 200 \times 0.5 = 100\) J. Work by the large piston \(= 10^{4} \times 0.01 = 100\) J. The lift multiplies force, not energy.

A gauge reads the pressure above atmospheric. Absolute pressure \(= 2 \times 10^{5} + 10^{5} = 3 \times 10^{5}\) Pa.

A flat tyre holds air at atmospheric pressure, so the gauge reads zero.

At the level of the oil-water boundary the pressure is the same in both arms. In the oil arm it is \(P_0 + 0.20\rho_{\text{oil}}g\); in the water arm it is \(P_0 + 0.16 \times 1000 \times g\).

\(0.20\rho_{\text{oil}} = 0.16 \times 1000\), so \(\rho_{\text{oil}} = 800\) kg/m\(^3\). The lighter liquid needs a taller column for the same pressure.

Add the two columns: \(P = h_1\rho_1 g + h_2\rho_2 g = 0.10 \times 800 \times 10 + 0.20 \times 1000 \times 10 = 800 + 2000 = 2800\) Pa.

Consider a thin horizontal cylinder of liquid joining the two points. The only horizontal forces on it are the pressures on its two ends. At rest these must balance, so the two pressures are equal.

Every point of the free surface is at atmospheric pressure. Points at the same pressure lie at the same height, so the whole surface is at one height: it is horizontal.

1. Consider a small drop on the surface. The forces on it are its weight \(mg\) downward and the push of the liquid, perpendicular to the surface. Their resultant must be \(ma\) horizontally.

   The push is perpendicular to the surface and makes angle \(\theta\) with the vertical, so \(\tan\theta = \dfrac{ma}{mg} = \dfrac{a}{g} = 0.5\), giving \(\theta \approx 27^\circ\).

2. The rear end. The liquid needs a net forward force, which comes from a higher pressure at the rear. Higher pressure at the same level means a taller column, so the level rises at the rear.

The surface tilts about its middle. Over half the length, \(1\) m, the level changes by \(1 \times \tan\theta = 0.5\) m.

Rear depth \(= 1 + 0.5 = 1.5\) m. Front depth \(= 1 - 0.5 = 0.5\) m.

Pressure at the base of the rear end \(= 1.5\rho g\); at the front end \(= 0.5\rho g\). Difference \(= \rho g \times 1 = 1000 \times 10 \times 1 = 10^{4}\) Pa.

Also \(\rho a L = 1000 \times 5 \times 2 = 10^{4}\) Pa. The two agree because \(\Delta h = L\tan\theta = La/g\), so \(\rho g\,\Delta h = \rho a L\).

The cylinder has mass \(\rho A\,\Delta x\). The pressure \(P_1\) on its rear face pushes it forward, and \(P_2\) on its front face pushes it back.

Newton's second law: \((P_1 - P_2)A = \rho A\,\Delta x\,a\), so \(P_2 - P_1 = -\rho a\,\Delta x\), that is, \(\dfrac{\Delta P}{\Delta x} = -\rho a\). The pressure falls along the acceleration.

In the lift the liquid behaves as if gravity were \(g_{\text{eff}} = g + a\) (accelerating up) or \(g - a\) (accelerating down). \(P = h\rho g_{\text{eff}}\).

1. \(g_{\text{eff}} = 12\) m/s\(^2\): \(P = 1 \times 1000 \times 12 = 12000\) Pa.

2. \(g_{\text{eff}} = 8\) m/s\(^2\): \(P = 8000\) Pa.

3. \(g_{\text{eff}} = 0\): \(P = 0\). In free fall the water presses on nothing.

1. \(y = \dfrac{\omega^2 R^2}{2g} = \dfrac{100 \times 0.01}{20} = 0.05\) m \(= 5\) cm.

2. Each ring of water moves in a circle and needs an inward force. That force comes from a pressure that grows outward. Higher pressure at the same level needs a taller column, so the liquid stands higher at the rim.

A thin element at radius \(r\), thickness \(dr\), area \(A\) facing the axis, needs an inward force \(dm\,\omega^2 r = \rho A\,dr\,\omega^2 r\). It comes from the pressure difference across it: \(dP\,A = \rho A\omega^2 r\,dr\), so \(\dfrac{dP}{dr} = \rho\omega^2 r\).

Integrate from the axis: \(P(r) - P(0) = \dfrac{1}{2}\rho\omega^2 r^2\) at one horizontal level. A pressure excess \(\dfrac{1}{2}\rho\omega^2 r^2\) needs an extra height \(y\) with \(\rho g y = \dfrac{1}{2}\rho\omega^2 r^2\).

So \(y = \dfrac{\omega^2 r^2}{2g}\), a parabola; turned about the axis it is a paraboloid.

On a smooth incline the box accelerates at \(g\sin 30^\circ\) down the slope. Inside the box the effective gravity is \(\vec g - \vec a\), which points perpendicular to the incline with size \(g\cos 30^\circ\).

The free surface is always perpendicular to the effective gravity, so it lies parallel to the incline.

The pressure difference between the two arms at the base is \(\rho a L\) (Question 13). It is supported by a height difference \(\Delta h\) with \(\rho g\,\Delta h = \rho a L\).

\(\Delta h = \dfrac{aL}{g} = \dfrac{2 \times 0.30}{10} = 0.06\) m \(= 6\) cm. The rear arm is higher.

1. The rim rises \(5\) cm above the centre (Question 16). For a paraboloid the average height is halfway between centre and rim, and the average must stay at \(20\) cm.

   So the centre falls \(2.5\) cm to \(17.5\) cm and the rim rises \(2.5\) cm to \(22.5\) cm.

2. The centre depth is \(20 - \dfrac{\omega^2 R^2}{4g}\) cm. It is zero when \(\omega^2 = \dfrac{4gh}{R^2} = \dfrac{4 \times 10 \times 0.20}{0.01} = 800\), so \(\omega = \sqrt{800} = 20\sqrt{2} \approx 28\) rad/s.