Bernoulli's Theorem and Applications of Bernoulli's Theory Questions in English

(A) Bernoulli's equation is derived from the work-energy theorem applied to a flowing fluid. It states that for an incompressible,non-viscous,and steady flow,the sum of pressure energy,kinetic energy per unit volume,and potential energy per unit volume remains constant along a streamline. Therefore,it represents the conservation of energy.

(B) Applying Bernoulli's principle between the free surface (point $1$) and the hole (point $2$):
$p_o + \rho g h + \frac{1}{2} \rho v_1^2 = p_a + \rho g(0) + \frac{1}{2} \rho v^2$
Since the hole is very small $(r \ll \sqrt{A/\pi})$,the velocity of the free surface $v_1 \approx 0$.
Thus,$p_o + \rho g h = p_a + \frac{1}{2} \rho v^2$
Rearranging for $v$:
$\frac{1}{2} \rho v^2 = (p_o - p_a) + \rho g h$
$v^2 = \frac{2(p_o - p_a)}{\rho} + 2gh$
$v = \sqrt{2gh + \frac{2(p_o - p_a)}{\rho}}$
The rate of flow (volume flow rate) $Q = \text{Area of hole} \times v = \pi r^2 \sqrt{2gh + \frac{2(p_o - p_a)}{\rho}}$.

(C) Using the equation of continuity,$A_1 V_1 = A_2 V_2$.
Given $A_1 = 10 \text{ cm}^2$,$V_1 = 1 \text{ ms}^{-1}$,and $A_2 = 5 \text{ cm}^2$.
$V_2 = (A_1 V_1) / A_2 = (10 \times 1) / 5 = 2 \text{ ms}^{-1}$.
Using Bernoulli's equation for a horizontal pipe,$P_1 + 0.5 \rho V_1^2 = P_2 + 0.5 \rho V_2^2$.
$P_2 = P_1 + 0.5 \rho (V_1^2 - V_2^2)$.
$P_2 = 2000 + 0.5 \times 1000 \times (1^2 - 2^2)$.
$P_2 = 2000 + 500 \times (1 - 4) = 2000 - 1500 = 500 \text{ Pa}$.

(D) According to the equation of continuity, $A_{1} v_{1} = A_{2} v_{2}$.
Given $A_{1} = 10 \,cm^{2}$, $v_{1} = 1 \,ms^{-1}$, $A_{2} = 5 \,cm^{2}$.
Substituting the values: $10 \times 1 = 5 \times v_{2} \implies v_{2} = 2 \,ms^{-1}$.
Using Bernoulli's equation for a horizontal pipe: $P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2} \rho v_{2}^{2}$.
Here, $\rho = 1000 \,kg/m^{3}$ (density of water).
$2000 + \frac{1}{2} \times 1000 \times (1)^{2} = P_{2} + \frac{1}{2} \times 1000 \times (2)^{2}$.
$2000 + 500 = P_{2} + 2000$.
$2500 = P_{2} + 2000 \implies P_{2} = 500 \,Pa$.

(B) According to the equation of continuity,the product of cross-sectional area and velocity remains constant for an incompressible fluid: $A_1 v_1 = A_2 v_2$.
Given $A_1 = A$,$v_1 = 4 \,m/s$,and $A_2 = \frac{2}{3} A$.
Substituting these values: $A \times 4 = \frac{2}{3} A \times v_2$.
Solving for $v_2$: $v_2 = 4 \times \frac{3}{2} = 6 \,m/s$.
Now,using the principle of conservation of energy (Bernoulli's equation) for the falling stream: $P + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P + \rho g h_2 + \frac{1}{2} \rho v_2^2$.
Since the pressure $P$ is constant (atmospheric pressure) and $h_1 - h_2 = h$,the equation simplifies to: $g h = \frac{1}{2} (v_2^2 - v_1^2)$.
Substituting the known values: $10 \times h = \frac{1}{2} (6^2 - 4^2)$.
$10 h = \frac{1}{2} (36 - 16) = \frac{1}{2} (20) = 10$.
Therefore,$h = 1 \,m$.

(D) Using the equation of continuity:
$A_1 V_1 = A_2 V_2$
$10 \times 1 = 5 \times V_2 \Rightarrow V_2 = 2 \,m/s$
Applying Bernoulli's theorem for a horizontal pipe $(h_1 = h_2)$:
$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$
Given $\rho = 1000 \,kg/m^3$ (density of water):
$2000 + \frac{1}{2} \times 1000 \times (1)^2 = P_2 + \frac{1}{2} \times 1000 \times (2)^2$
$2000 + 500 = P_2 + 2000$
$P_2 = 2500 - 2000 = 500 \,Pa$

(C) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.
Given $A_1 = 2A$,$v_1 = \sqrt{2} \ m \ s^{-1}$,and $A_2 = A$.
Substituting these values: $(2A)(\sqrt{2}) = (A)v_2 \Rightarrow v_2 = 2\sqrt{2} \ m \ s^{-1}$.
For a steady flow,applying Bernoulli's equation at points $1$ and $2$:
$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$.
Rearranging the terms: $P_1 - P_2 = \rho g (h_2 - h_1) + \frac{1}{2} \rho (v_2^2 - v_1^2)$.
Given $P_1 - P_2 = 100 \ N \ m^{-2}$ (since pressure is higher at the wider section $1$ where speed is lower),$h_1 - h_2 = 10 \ cm = 0.1 \ m$,so $h_2 - h_1 = -0.1 \ m$.
$100 = \rho [10(-0.1) + \frac{1}{2}((2\sqrt{2})^2 - (\sqrt{2})^2)]$.
$100 = \rho [-1 + \frac{1}{2}(8 - 2)]$.
$100 = \rho [-1 + 3] = 2\rho$.
$\rho = 50 \ kg \ m^{-3}$.

(A) According to Bernoulli's theorem for an ideal,steady,and incompressible fluid flow,the sum of pressure energy,kinetic energy,and potential energy per unit mass remains constant along a streamline.
For a horizontal tube,the height $h$ is constant,so the equation simplifies to:
$P + \frac{1}{2} \rho v^2 = \text{constant}$
Here,$P$ is the pressure,$\rho$ is the density,and $v$ is the velocity of the fluid.
From the equation of continuity,$A_1 v_1 = A_2 v_2$,which implies that where the cross-sectional area $A$ is smallest,the velocity $v$ is highest.
Since $P + \frac{1}{2} \rho v^2 = \text{constant}$,if the velocity $v$ increases,the pressure $P$ must decrease to keep the sum constant.
Therefore,the pressure is lowest where the velocity is highest.

(C) The shape of an aeroplane wing is designed such that the upper surface is convex and the lower surface is relatively flat or concave.
According to Bernoulli's principle, the air flowing over the curved upper surface travels at a higher velocity compared to the air flowing beneath the wing.
As velocity increases, pressure decreases $(P + \frac{1}{2} \rho v^2 = \text{constant})$.
Therefore, the pressure at the top is lower than the pressure at the bottom, creating an upward force known as lift.
The assertion $(A)$ is true, but the reason $(R)$ is false because the air at the top has higher velocity, not smaller, and the pressure at the top is lower, not higher.

(C) According to Bernoulli's theorem,$p + \frac{1}{2} \rho V^2 + \rho g h = \text{constant}$.
Here,$p$ is the pressure energy per unit volume,$\frac{1}{2} \rho V^2$ is the kinetic energy per unit volume,and $\rho g h$ is the potential energy per unit volume.
For an ideal fluid in streamline flow,the sum of these energies remains constant.
Therefore,Bernoulli's theorem is based on the law of conservation of energy.

(B) Bernoulli’s theorem is derived from the work-energy theorem for a fluid element. It states that for an incompressible,non-viscous,and steady flow of a fluid,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant along a streamline. Therefore,it is a direct consequence of the law of conservation of energy.

(C) The mass of the Boeing aircraft is $M = 3.3 \times 10^5 \text{ kg}$.
The total wing area is $A = 500 \text{ m}^2$.
In level flight,the upward lift force generated by the pressure difference between the lower and upper surfaces of the wings must balance the weight of the aircraft.
Therefore,the upward force $F = \Delta p \times A$,where $\Delta p$ is the pressure difference.
Equating the lift force to the weight: $\Delta p \times A = M \times g$.
Using $g = 9.8 \text{ m/s}^2$,we get:
$\Delta p = \frac{M \times g}{A} = \frac{3.3 \times 10^5 \times 9.8}{500}$.
$\Delta p = \frac{32.34 \times 10^5}{500} = 0.06468 \times 10^5 \text{ N/m}^2$.
$\Delta p = 6.468 \times 10^3 \text{ N/m}^2 \approx 6.5 \times 10^3 \text{ N/m}^2$.

(A) According to the equation of continuity,for an incompressible fluid in streamline flow,the product of the cross-sectional area $A$ and velocity $v$ is constant $(A_1v_1 = A_2v_2)$.
At the narrowest part of the pipe,the cross-sectional area $A$ is minimum,which implies that the velocity $v$ must be maximum.
According to Bernoulli's theorem for horizontal flow,the sum of pressure energy and kinetic energy per unit volume remains constant:
$p + \frac{1}{2} \rho v^2 = \text{constant}$
Since the velocity $v$ is maximum at the narrowest part,the pressure $p$ must be minimum to keep the sum constant.
Therefore,at the narrowest part,velocity is maximum and pressure is minimum.

(D) The lift force $F$ must balance the weight of the aircraft: $F = mg = (3 \times 10^5 \ kg) \times (10 \ ms^{-2}) = 3 \times 10^6 \ N$.
Using Bernoulli's principle,the pressure difference $\Delta P$ between the lower and upper surfaces is $\Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2)$,where $v_1$ is the speed on the lower surface and $v_2$ is the speed on the upper surface.
Since $F = \Delta P \times A$,we have $\Delta P = \frac{F}{A} = \frac{3 \times 10^6 \ N}{400 \ m^2} = 7500 \ Pa$.
Given $v_1 = 540 \ km \ h^{-1} = 540 \times \frac{5}{18} \ ms^{-1} = 150 \ ms^{-1}$.
Thus,$7500 = \frac{1}{2} \times 1.2 \times (v_2^2 - 150^2) \implies 12500 = v_2^2 - 22500 \implies v_2^2 = 35000 \implies v_2 \approx 187.08 \ ms^{-1}$.
The fractional increase is $\frac{v_2 - v_1}{v_1} = \frac{187.08 - 150}{150} = \frac{37.08}{150} \approx 0.247$.
Re-evaluating the calculation: $\Delta P = \frac{1}{2} \rho (v_2 - v_1)(v_2 + v_1) \approx \frac{1}{2} \rho (\Delta v)(2v_1) = \rho v_1 \Delta v$.
$\Delta v = \frac{\Delta P}{\rho v_1} = \frac{7500}{1.2 \times 150} = \frac{7500}{180} = 41.67 \ ms^{-1}$.
Fractional increase $= \frac{\Delta v}{v_1} = \frac{41.67}{150} \approx 0.277$.

(D) For an aeroplane to travel at a constant height,the upward lift force must balance the downward gravitational force (weight) of the aeroplane.
Let $m$ be the mass of the aeroplane,$A$ be the total wing area,$g$ be the acceleration due to gravity,and $\Delta P$ be the pressure difference between the lower and upper surfaces of the wings.
The lift force $F_L$ is given by $F_L = \Delta P \times A$.
The weight of the aeroplane is $W = m \times g$.
Equating the two forces for constant height: $\Delta P \times A = m \times g$.
Substituting the given values: $\Delta P \times 600 = (4.5 \times 10^4) \times 10$.
$\Delta P \times 600 = 4.5 \times 10^5$.
$\Delta P = \frac{4.5 \times 10^5}{600} = \frac{450000}{600} = 750 \,N \,m^{-2}$.
Thus,the pressure difference is $750 \,N \,m^{-2}$.

(A) Let the pressure applied by the piston be $P$ and the atmospheric pressure be $P_a$. The excess pressure is $\Delta P = P - P_a$. The height of the water column above the hole is $h = 15 \ m - 12 \ m = 3 \ m$. The height of the hole from the ground is $H = 12 \ m$.
Using Bernoulli's equation at the top surface and the hole,the velocity of efflux $v$ is given by $v = \sqrt{2gh + \frac{2\Delta P}{\rho}}$.
The time taken for the water to reach the ground is $t = \sqrt{\frac{2H}{g}}$.
The horizontal range $R$ is given by $R = v \times t = \sqrt{2gh + \frac{2\Delta P}{\rho}} \times \sqrt{\frac{2H}{g}}$.
Given $R = 16 \ m$,$H = 12 \ m$,$h = 3 \ m$,$\rho = 1000 \ kg/m^3$,and $g = 10 \ m/s^2$:
$16 = \sqrt{2(10)(3) + \frac{2\Delta P}{1000}} \times \sqrt{\frac{2(12)}{10}}$
$16 = \sqrt{60 + \frac{\Delta P}{500}} \times \sqrt{2.4}$
$256 = (60 + \frac{\Delta P}{500}) \times 2.4$
$106.67 = 60 + \frac{\Delta P}{500}$
$46.67 = \frac{\Delta P}{500} \Rightarrow \Delta P = 23335 \ Pa \approx 23.3 \ kPa$.
Wait,re-calculating: $256 / 2.4 = 106.666...$,so $106.666 - 60 = 46.666$. $\Delta P = 46.666 \times 500 = 23333 \ Pa$.
Force $F = \Delta P \times A = 23333 \times 10 = 233330 \ N = 233.3 \ kN \approx 233 \ kN$.

(B) Given:
Diameter of pipe $D_1 = 4 \,cm$,radius $r_1 = 2 \,cm$.
Diameter of throat $D_2 = 2 \,cm$,radius $r_2 = 1 \,cm$.
Velocity in pipe $V_1 = 10 \,m/s$.
Density of water $\rho = 1000 \,kg/m^3$.
Step $1$: Using the equation of continuity $A_1 V_1 = A_2 V_2$:
$\pi r_1^2 V_1 = \pi r_2^2 V_2$
$(2)^2 \times 10 = (1)^2 \times V_2$
$V_2 = 40 \,m/s$.
Step $2$: Using Bernoulli's theorem for horizontal flow:
$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$
$P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2)$
$P_1 - P_2 = \frac{1}{2} \times 1000 \times (40^2 - 10^2)$
$P_1 - P_2 = 500 \times (1600 - 100)$
$P_1 - P_2 = 500 \times 1500 = 7.5 \times 10^5 \,Pa$.

(C) Given that,the pressure difference between points $A$ and $B$ is $p_A - p_B = 1500 \text{ N m}^{-2}$.
By using Bernoulli's equation for a horizontal tube $(h_A = h_B)$:
$p_A + \frac{1}{2} \rho v_A^2 = p_B + \frac{1}{2} \rho v_B^2$
$p_A - p_B = \frac{1}{2} \rho (v_B^2 - v_A^2) \quad \dots (i)$
Density of water,$\rho = 10^3 \text{ kg m}^{-3}$.
Cross-sectional areas at points $A$ and $B$ are:
$a_A = 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2$
$a_B = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$
By the equation of continuity,the rate of flow of water is constant:
$a_A v_A = a_B v_B \Rightarrow v_B = v_A \left( \frac{a_A}{a_B} \right) = v_A \left( \frac{40}{20} \right) = 2v_A \quad \dots (ii)$
Substituting Eq. $(ii)$ into Eq. $(i)$:
$1500 = \frac{1}{2} \times 10^3 \times ((2v_A)^2 - v_A^2)$
$1500 = 500 \times (4v_A^2 - v_A^2)$
$3 = 3v_A^2 \Rightarrow v_A^2 = 1 \Rightarrow v_A = 1 \text{ m s}^{-1}$
Therefore,the rate of flow of water is:
$Q = a_A v_A = 40 \times 10^{-4} \text{ m}^2 \times 1 \text{ m s}^{-1} = 40 \times 10^{-4} \text{ m}^3 \text{ s}^{-1} = 4000 \text{ cm}^3 \text{ s}^{-1}$.

(C) According to Bernoulli's equation for a horizontal pipe (where height $h_1 = h_2$):
$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$
Given:
$P_1 = 50 \ kPa = 50 \times 10^3 \ Pa$
$v_1 = 1 \ m/s$
$v_2 = 2 \ m/s$
Density of water $\rho = 1000 \ kg/m^3 = 10^3 \ kg/m^3$
Rearranging the equation to solve for $P_2$:
$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)$
$P_2 = 50 \times 10^3 + \frac{1}{2} \times 10^3 \times (1^2 - 2^2)$
$P_2 = 50 \times 10^3 + 0.5 \times 10^3 \times (1 - 4)$
$P_2 = 50 \times 10^3 + 0.5 \times 10^3 \times (-3)$
$P_2 = 50 \times 10^3 - 1.5 \times 10^3$
$P_2 = 48.5 \times 10^3 \ Pa = 48.5 \ kPa$

(B) The power $P$ required to pump water is the sum of the power required to lift the water to height $h$ and the power required to provide kinetic energy to the water.
However, in this context, we calculate the power required to lift the water against gravity at a steady rate.
Power $P = \frac{\text{Work}}{\text{Time}} = \frac{mgh}{t}$.
From the equation of continuity, the mass flow rate is $\frac{m}{t} = A v \rho$.
Substituting this into the power equation: $P = (A v \rho) g h$.
Given values: $r = 1 \ cm = 10^{-2} \ m$, $v = 10 \ m \ s^{-1}$, $\rho = 1000 \ kg \ m^{-3}$, $g = 10 \ m \ s^{-2}$, $h = 3 \ m$.
Area $A = \pi r^2 = \pi (10^{-2})^2 = \pi \times 10^{-4} \ m^2$.
Calculating power: $P = (\pi \times 10^{-4}) \times 10 \times 1000 \times 10 \times 3$.
$P = \pi \times 10^{-4} \times 10^5 \times 3 = 30\pi \ W$.

(C) According to the equation of continuity, the volume flow rate is constant:
$AV = av \implies V = \frac{a}{A}v$
where $V$ is the velocity of the piston and $v$ is the velocity of the liquid emerging from the orifice.
Applying Bernoulli's principle between the inside of the cylinder (near the piston) and the orifice:
$P_{in} + \frac{1}{2} \rho V^2 = P_{out} + \frac{1}{2} \rho v^2$
The pressure inside the cylinder is $P_{in} = P_0 + \frac{F}{A}$, where $P_0$ is the atmospheric pressure. The pressure at the orifice is $P_{out} = P_0$.
Substituting these into the Bernoulli equation:
$(P_0 + \frac{F}{A}) + \frac{1}{2} \rho V^2 = P_0 + \frac{1}{2} \rho v^2$
$\frac{F}{A} = \frac{1}{2} \rho (v^2 - V^2)$
Substituting $V = \frac{a}{A}v$:
$\frac{F}{A} = \frac{1}{2} \rho (v^2 - (\frac{a}{A}v)^2) = \frac{1}{2} \rho v^2 (1 - \frac{a^2}{A^2})$
Since $A \gg a$, we have $\frac{a^2}{A^2} \approx 0$, so:
$\frac{F}{A} = \frac{1}{2} \rho v^2 \implies v = \sqrt{\frac{2F}{\rho A}}$

(A) According to Bernoulli's principle for streamline flow in a horizontal tube, the sum of static pressure and dynamic pressure remains constant at all points along a streamline.
$p + \frac{1}{2} \rho v^{2} = p_{1} + \frac{1}{2} \rho v_{1}^{2}$
Given that at the second point, $p_{1} = p/2$.
Substituting this into the equation:
$p + \frac{1}{2} \rho v^{2} = \frac{p}{2} + \frac{1}{2} \rho v_{1}^{2}$
Rearranging the terms to solve for $v_{1}^{2}$:
$\frac{1}{2} \rho v_{1}^{2} = p - \frac{p}{2} + \frac{1}{2} \rho v^{2}$
$\frac{1}{2} \rho v_{1}^{2} = \frac{p}{2} + \frac{1}{2} \rho v^{2}$
Multiplying both sides by $2/\rho$:
$v_{1}^{2} = \frac{p}{\rho} + v^{2}$
$v_{1} = \sqrt{v^{2} + \frac{p}{\rho}}$

(D) Given: Density $\rho = 600 \text{ kg/m}^3$,Area $A_A = 1.0 \text{ cm}^2 = 100 \text{ mm}^2$,Area $A_B = 20 \text{ mm}^2$,Velocity $v_A = 10 \text{ cm/s} = 0.1 \text{ m/s}$.
Using the equation of continuity,$A_A v_A = A_B v_B$.
$100 \times 0.1 = 20 \times v_B \Rightarrow v_B = 0.5 \text{ m/s}$.
Applying Bernoulli's equation for a horizontal pipe $(h_A = h_B)$:
$P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2$.
The pressure difference is $P_A - P_B = \frac{1}{2} \rho (v_B^2 - v_A^2)$.
Substituting the values: $P_A - P_B = \frac{1}{2} \times 600 \times (0.5^2 - 0.1^2) = 300 \times (0.25 - 0.01) = 300 \times 0.24 = 72 \text{ Pa}$.