Fluid Pressure and Pascal's Law: notes and previous year questions

Answer: Option A

Absolute pressure is $P_{0} + \rho g h$, so the $100\ \mathrm{atm}$ the hull can stand includes the $1\ \mathrm{atm}$ of air on the sea's surface: $100 \times 10^{5} = 10^{5} + 1000 \times 10 \times h$, which gives $10^{4}h = 99 \times 10^{5}$ and $h = 990\ \mathrm{m}$.

The one step that needs care is that only $99\ \mathrm{atm}$ of the $100$ is water; setting all of it equal to $\rho g h$ gives $1000\ \mathrm{m}$. $9900\ \mathrm{m}$ and $99\ \mathrm{m}$ are slips of a factor of ten in $10^{4}h$. Check with the rule of thumb from the last question: every $10\ \mathrm{m}$ of water adds about $1\ \mathrm{atm}$, so $990\ \mathrm{m}$ is $99\ \mathrm{atm}$ of water, plus $1$ of air.

Answer: Option D

For a block of uniform cross-section, the fraction of its volume under is the fraction of its height under: $\dfrac{h}{H} = \dfrac{\rho_{b}}{\rho_{l}} = \dfrac{600}{900} = \dfrac{2}{3}$, so $h = \dfrac{2}{3} \times 8.0 \approx 5.3\ \mathrm{cm}$.

The trap is to flip the ratio to $\dfrac{900}{600}$, which would put more than the whole block under. $2.7\ \mathrm{cm}$, the part left above, is not offered. Check: the block is two-thirds as dense as the liquid, so two-thirds of it is under.

Answer: Option B

The extra pressure $2.5\ \mathrm{km}$ down is $\rho g h = 10^{3} \times 10 \times 2500 = 2.5 \times 10^{7}\ \mathrm{Pa}$, so $\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B} = \dfrac{2.5 \times 10^{7}}{2 \times 10^{9}} = 1.25 \times 10^{-2} = 1.25\%$.

Volume shrinks by the same small fraction that density rises, so this is the last question asked about volume. The trap is leaving the depth in kilometres, which throws the answer out by a factor of a thousand; $1.0$ and $1.5$ are what depths of $2$ and $3\ \mathrm{km}$ would give. Check against the last question: $1.1\ \mathrm{km}$ with a stiffer $B$ gave $0.5\%$; more than twice as deep, with water a little easier to squeeze, $1.25\%$ is the right size.

Answer: Option D

The water adds $\rho g h = 1000 \times 10 \times 20 = 2.0 \times 10^{5}\ \mathrm{Pa} = 2\ \mathrm{atm}$, and the air presses on the surface with $1\ \mathrm{atm}$ more, so $P = 3\ \mathrm{atm}$. Every $10\ \mathrm{m}$ of water adds about one atmosphere — a rule of thumb worth keeping.

The trap is $2\ \mathrm{atm}$, the gauge pressure: it leaves out $P_{0}$, and the question asks for the absolute pressure. $1.2\ \mathrm{atm}$ comes from leaving $g$ out of $\rho g h$, and $21\ \mathrm{atm}$ from adding an atmosphere for every metre.

Answer: Option B

The fraction under is $\dfrac{\rho_{b}}{\rho_{l}} = \dfrac{0.6}{1} = 0.6$, so the fraction above is $1 - 0.6 = 0.4$.

The trap is A: $0.6$ is the part under water, and the question asks for the part above. C, a half, is the guess that anything floating sits half in. Check: $0.6$ of the log's volume under water displaces $0.6$ of a log-sized volume of water, which weighs exactly what the log weighs, since the log is $0.6$ times as dense.

Answer: Options C, D

Along the slope, pointing down it, real gravity has a component $g\sin 30^{\circ} = 5\ \mathrm{m/s^{2}}$, and taking away the box's acceleration $a$ leaves $5 - a$. That is the part of $\vec{g}_{\mathrm{eff}}$ along the line from Y down to X.

Smooth incline: $a = g\sin 30^{\circ} = 5\ \mathrm{m/s^{2}}$, so nothing is left along the slope and $\vec{g}_{\mathrm{eff}}$ is just $g\cos 30^{\circ}$, at right angles to the incline. The pressure does not change along any line parallel to the slope, so C is true; and the surface, at right angles to $\vec{g}_{\mathrm{eff}}$, lies parallel to the incline, so A is false. The trap in A is to think a liquid's surface is always horizontal: it lies at right angles to the gravity the liquid feels, which is straight down only when the container is not accelerating.

With friction: $a = 2\ \mathrm{m/s^{2}}$ leaves $5 - 2 = 3\ \mathrm{m/s^{2}}$ down the slope, so $P_{X} - P_{Y} = 1000 \times 3 \times 0.4 = 1200\ \mathrm{Pa}$, and D is true.

B is the trap of leaving out the pseudo force: $1000 \times 5 \times 0.4 = 2000\ \mathrm{Pa}$ is the difference in a box at rest, where X is $0.4 \sin 30^{\circ} = 0.2\ \mathrm{m}$ lower than Y and $\rho g \times 0.2 = 2000\ \mathrm{Pa}$.