Q.A fully loaded Boeing aircraft has a mass of $3.3 \times 10^{5}\ \text{kg}$. Its total wing area is $500\ \text{m}^{2}$. It is in level flight with a speed of $960\ \text{km/h}$.

Concept and Intuition

By Torricelli's law, water escapes a small hole at the bottom of a tank with speed $v=\sqrt{2gh}$ when the level is at height $h$. Using continuity (area of tank $A$, area of hole $a$), the rate of fall of the level is $\dfrac{dh}{dt} = -\dfrac{a}{A}\sqrt{2gh} = -k\sqrt{h}$ for a constant $k = \dfrac{a}{A}\sqrt{2g}$. As $h$ decreases, the level falls more slowly (since $\sqrt h$ decreases), so the tank takes longer to drain the lower portion than an equal height of the upper portion — the ratio should come out less than 1, consistent with $\sqrt2-1\approx0.414$.

Step-by-Step Solution

1. From $\dfrac{dh}{dt}=-k\sqrt h$, separate variables: $\dfrac{dh}{\sqrt h} = -k\,dt$.
2. Integrating between height $h_1$ (start) and $h_2$ (end) gives the time: $t = \dfrac{2}{k}(\sqrt{h_1}-\sqrt{h_2})$.
3. Time to fall from $h$ to $h/2$: $t_1 = \dfrac{2}{k}\left(\sqrt h - \sqrt{h/2}\right) = \dfrac{2\sqrt h}{k}\left(1-\dfrac{1}{\sqrt2}\right)$. …

Concept and Intuition

When a ball spins while moving through air (or any fluid), it drags a thin layer of fluid around with it due to viscosity. On the side where the ball's surface motion adds to the oncoming flow, the relative fluid speed is higher; on the opposite side it's lower. By Bernoulli's principle, higher local flow speed corresponds to lower local pressure, so a net pressure difference — and hence a net force perpendicular to the direction of travel — arises. While Bernoulli's principle is the underlying pressure-speed relation used to explain why it happens, the named physical phenomenon of spin-induced deflection is specifically the Magnus effect (seen in curving cricket/football/baseball trajectories).

Step-by-Step Solution

1. A spinning ball moving through a fluid creates asymmetric relative flow speeds on its two sides due to the combination of translational and spin motion.
2. Faster relative flow on one side → lower pressure there (Bernoulli); slower relative flow on the other side → higher pressure. …

Concept and Intuition

An aeroplane's wings are shaped so that air flows faster over the top surface than the bottom, creating a pressure difference (Bernoulli's principle) that produces an upward lift force. In steady, constant-height (level) flight, this lift force exactly balances the plane's weight. The lift force is just this pressure difference multiplied by the total wing area, so dividing the required lift (weight) by the area recovers the pressure difference.

Step-by-Step Solution

1. In level flight, lift = weight: $F_{\text{lift}} = mg = 4.5\times10^4\times10 = 4.5\times10^5\ \mathrm{N}$. …

Concept and Intuition

With the valve closed, water is static and the gauge reads the full static pressure. Opening the valve lets water flow; part of the pressure energy converts into kinetic energy, so the measured pressure at the same point drops. Bernoulli's equation (same height, same point) relates this pressure drop directly to flow speed.

Step-by-Step Solution

1. Bernoulli (same elevation): $P_1 = P_2 + \frac12\rho v^2$, so $\Delta P = P_1-P_2 = \frac12\rho v^2$. …

Concept and Intuition

For streamline flow of an incompressible fluid, the equation of continuity ($A v = \text{const}$) fixes speed changes as the pipe narrows, and Bernoulli's equation (constant total energy per unit volume along a streamline) then trades that speed change for a pressure change. In a horizontal pipe there's no elevation term, so pressure and kinetic energy per unit volume trade off directly.

Step-by-Step Solution

1. Continuity equation: $A_1v_1 = A_2v_2$. Here $A_1=10\ \mathrm{cm^2}$, $v_1=1\ \mathrm{ms^{-1}}$, $A_2=5\ \mathrm{cm^2}$.

2. Bernoulli's equation for a horizontal pipe (same height, so gravitational PE terms cancel): $P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2$ …

We have a steady flow of an ideal liquid (incompressible, non‑viscous). The pipe changes both cross‑section and height, so both kinetic and gravitational potential energy terms matter. Bernoulli’s principle tells us that along a streamline, the sum of pressure energy, kinetic energy per unit volume, and gravitational potential energy per unit volume is constant. That’s the natural tool here.

Let’s work through it step by step.

1. Write Bernoulli’s equation between point 1 (upper, wider section) and point 2 (lower, narrower section).

Rearranging to isolate the pressure difference:

We are told the pressure difference $P_1 - P_2 = 100\,\mathrm{N\,m^{-2}}$ (positive, so pressure at 1 is higher). The height difference is given as $10\,\mathrm{cm} = 0.1\,\mathrm{m}$, with point 1 higher than point 2, so $h_1 - h_2 = 0.1\,\mathrm{m}$. Hence $h_2 - h_1 = -0.1\,\mathrm{m}$.

2. Use continuity to find $v_2$. For steady flow of an incompressible fluid, $A_1 v_1 = A_2 v_2$. Here $A_1 = 2A$, $A_2 = A$, and $v_1 = \sqrt{2}\,\mathrm{m/s}$.

3. Substitute into Bernoulli’s equation.

Compute the squares:

$(2\sqrt{2})^2 = 4 \times 2 = 8$, $(\sqrt{2})^2 = 2$. So $v_2^2 - v_1^2 = 8 - 2 = 6$.

Then: …

Concept and Intuition

Bernoulli's equation says that along a streamline, $P+\frac12\rho v^2+\rho gh$ is constant. Fast-moving air just above the roof has high $v$, so its pressure $P_{out}$ is lower than the pressure of the still air $P_{in}$ trapped inside the house (approximately atmospheric, since it isn't moving). This pressure imbalance pushes the roof upward from inside — the same physics as lift on an aircraft wing.

Step-by-Step Solution

1. Apply Bernoulli between still air inside (speed $\approx0$) and moving air just outside (speed $40\ m.s^{-1}$), same height: $P_{in} - P_{out} = \frac12\rho v^2 = \frac12(1.2)(40)^2 = \frac12(1.2)(1600)=960\ Pa$.
2. This is the net outward (upward, since the roof is horizontal and wind blows over its top) pressure difference pushing the roof up. …

Concept and Intuition

Bernoulli's principle says that along a streamline, $P + \tfrac12\rho v^2 + \rho g h$ stays constant for an ideal fluid. With the tap closed, water is static ($v=0$) and its pressure is entirely "static" pressure. When the tap opens, some of that pressure converts into kinetic energy of the moving water (at the same height, so the $\rho g h$ term is unchanged). The pressure reading drops because part of it now shows up as motion.

Step-by-Step Solution

1. Bernoulli's equation (same height, comparing closed vs open tap): $P_{closed} = P_{open} + \tfrac12 \rho v^2$.
2. So $\Delta P = P_{closed} - P_{open} = \tfrac12 \rho v^2$. …

Concept and Intuition

An aeroplane wing generates lift because air moves faster over the curved upper surface than along the flatter lower surface. By Bernoulli's equation (applied along a streamline at roughly the same height), faster-moving air has lower pressure, so the lower surface (slower air) has higher pressure than the upper surface (faster air). This pressure difference, multiplied by the wing area, produces the net upward lift force. For level flight at constant speed, this lift exactly balances the weight of the plane.

Step-by-Step Solution

1. Convert speeds to SI units: $v_{upper} = 270\ kmph = \dfrac{270\times1000}{3600} = 75\ m/s$; $v_{lower} = 234\ kmph = \dfrac{234\times1000}{3600} = 65\ m/s$.
2. Apply Bernoulli's equation (same height, so no gravitational PE term): $P_{lower} - P_{upper} = \dfrac12\rho\left(v_{upper}^2 - v_{lower}^2\right)$.
3. Compute: $v_{upper}^2 - v_{lower}^2 = 75^2 - 65^2 = 5625 - 4225 = 1400$.
4. $\Delta P = \dfrac12 \times 1 \times 1400 = 700\ Pa$.
5. Total wing area (both wings, each $25\ m^2$): $A_{total} = 2\times25 = 50\ m^2$. …

Concept and Intuition

An aerofoil (wing cross-section) is shaped with a more curved (convex) upper surface and a flatter (or concave) lower surface. Air flowing over the top must travel a longer path in the same time as air flowing under the flatter bottom, so by continuity the air moves FASTER over the top. By Bernoulli's principle, faster-moving fluid exerts LOWER pressure. So the pressure above the wing is lower than the pressure below, and this pressure difference produces the net upward force — lift.

Step-by-Step Solution

1. Assertion: "upper surface convex, lower surface concave" — this correctly describes the classic aerofoil cross-section. True.
2. Reason as stated: "air currents at the top have smaller velocity and thus less pressure at the bottom than at the top."
3. Check velocity claim: actual physics says top velocity is GREATER, not smaller — so the reason's premise is wrong. …