Bernoulli's Theorem and Applications: notes and previous year questions

Answer: Option D

Continuity first, in consistent units: $1.0\ \mathrm{cm^{2}} = 100\ \mathrm{mm^{2}}$, five times B's $20\ \mathrm{mm^{2}}$, so the speed at B is five times A's: $v_{B} = 5 \times 0.1 = 0.5\ \mathrm{m/s}$. Then Bernoulli, with A and B at the same height:

$P_{A} - P_{B} = \dfrac{1}{2} \times 600 \times \left(0.5^{2} - 0.1^{2}\right) = 300 \times 0.24 = 72\ \mathrm{Pa}$.

The trap is the units: $10\ \mathrm{cm/s}$ must become $0.1\ \mathrm{m/s}$, and the two areas must be in the same unit before they are compared. $144\ \mathrm{Pa}$ forgets the $\dfrac{1}{2}$ in $\dfrac{1}{2}\rho v^{2}$; $36$ and $18\ \mathrm{Pa}$ halve once and twice too often. Check the sign: B is narrower, faster, so its pressure is the lower one.

Answer: 3

Tank 1: $t_{1} = \dfrac{A}{a}\sqrt{\dfrac{2h}{g}}$.

Tank 2: when the water stands $y$ above the hole, it leaves the pipe at ground level, a drop $H + y$ below the surface, at $\sqrt{2g(H + y)}$. So $A(-dy) = a\sqrt{2g(H + y)}\,dt$, and as $y$ falls from $h$ to $0$,

$t_{2} = \dfrac{A}{a\sqrt{2g}} \times 2\left(\sqrt{H + h} - \sqrt{H}\right)$.

With $H = \dfrac{16}{9}h$: $\sqrt{H + h} = \dfrac{5}{3}\sqrt{h}$ and $\sqrt{H} = \dfrac{4}{3}\sqrt{h}$, so the bracket is $\dfrac{1}{3}\sqrt{h}$ and $t_{2} = \dfrac{1}{3} \times \dfrac{A}{a}\sqrt{\dfrac{2h}{g}} = \dfrac{t_{1}}{3}$.

So $\dfrac{t_{1}}{t_{2}} = 3$: the pipe's extra drop makes the water leave much faster. This is the last question's rule with $c = H$: tank 2 empties in the time a bare tank's level would take to fall from $H + h$ to $H$. The trap is thinking the pipe changes nothing because the hole is the same size — the speed is set where the water leaves into the air, at the bottom of the pipe.

Answer: Option D

With the tank full, the hole near the bottom is $2\ \mathrm{m}$ below the surface, so the water leaves at $v = \sqrt{2 \times 10 \times 2} = \sqrt{40} \approx 6.32\ \mathrm{m/s}$. The area is $2\ \mathrm{mm^{2}} = 2 \times 10^{-6}\ \mathrm{m^{2}}$ (a square millimetre is $10^{-6}\ \mathrm{m^{2}}$), so

$Q = 2 \times 10^{-6} \times 6.32 \approx 12.6 \times 10^{-6}\ \mathrm{m^{3}/s}$.

The trap is $8.9 \times 10^{-6}$, which uses $\sqrt{gh} = \sqrt{20}$ and loses the $2$ in $2gh$. $6.4 \times 10^{-6}$ is close to the speed alone, forgetting that the area is $2$ square millimetres, not $1$.

Answer: Option D

Halving the diameter quarters the area, so the speed goes up four times: $1.5 \times 4 = 6\ \mathrm{m/s}$.

The trap is $3\ \mathrm{m/s}$: letting the speed follow the diameter rather than the area. $0.75$ and $0.375\ \mathrm{m/s}$ have the water slowing down in the narrow part, the ratio turned upside down. Check: $d^{2}v$ is $1 \times 1.5 = 1.5$ before and $\left(\dfrac{1}{2}\right)^{2} \times 6 = 1.5$ after, in any unit of diameter.

Answer: Option C

Q's area is twice P's, and its depth four times P's gives a speed $\sqrt{4} = 2$ times P's. Together $2 \times 2 = 4$ times the flow.

The trap is $8$: multiplying by the depth instead of its square root — the speed grows only as $\sqrt{h}$. $2$ counts the larger area but forgets the faster jet, and $1$ imagines every hole in a tank letting out the same flow.

Answer: Option B

The pressure is atmospheric at both ends, so only speed and height change. Per cubic metre: $\dfrac{1}{2}\rho v^{2} = \dfrac{1}{2} \times 1000 \times 100 = 5 \times 10^{4}\ \mathrm{J}$, and $\rho g h = 1000 \times 10 \times 5 = 5 \times 10^{4}\ \mathrm{J}$, together $1.0 \times 10^{5}\ \mathrm{J/m^{3}}$. Times $2 \times 10^{-3}\ \mathrm{m^{3}/s}$: $200\ \mathrm{W}$.

$100\ \mathrm{W}$ counts only one of the two energies. $1.0 \times 10^{5}\ \mathrm{W}$ stops at the energy per cubic metre, which is in $\mathrm{J/m^{3}}$, not watts. $400\ \mathrm{W}$ forgets the $\dfrac{1}{2}$ in $\dfrac{1}{2}\rho v^{2}$.