Force due to Liquid in pipe Questions in English

(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dM = (M/L)dx$.
The centripetal force required for this element to rotate in a circle of radius $x$ is $dF = (dM)\omega^2 x$.
Substituting $dM$,we get $dF = (M/L)dx \cdot \omega^2 x = (M\omega^2/L)x dx$.
The total force $F$ exerted by the liquid at the outer end is the integral of these centripetal forces from $x = 0$ to $x = L$:
$F = \int_0^L \frac{M\omega^2}{L} x dx = \frac{M\omega^2}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{M\omega^2}{L} \cdot \frac{L^2}{2} = \frac{1}{2}ML\omega^2$.

(D) The rate of flow of water is given by $\frac{dV}{dt} = 10\,cm^3/s = 10 \times 10^{-6}\,m^3/s$.
Density of water $\rho = 10^3\,kg/m^3$.
The cross-sectional area of the pipe is $A = \pi r^2 = \pi (0.5 \times 10^{-3}\,m)^2 = \pi \times 0.25 \times 10^{-6}\,m^2$.
The velocity of water $v = \frac{dV/dt}{A} = \frac{10 \times 10^{-6}}{\pi \times 0.25 \times 10^{-6}} = \frac{40}{\pi}\,m/s$.
The reactionary force $F$ is equal to the rate of change of momentum: $F = \frac{dm}{dt} v = (\rho \frac{dV}{dt}) v$.
Substituting the values: $F = (10^3) \times (10 \times 10^{-6}) \times \frac{40}{\pi} = 10^{-2} \times \frac{40}{3.14} \approx 0.127\,N$.

(B) The pressure at the base of the vessel due to a liquid column of height $h$ is given by $P = h \rho g$.
Given:
Height $h = 0.4 \, m$
Density $\rho = 900 \, kg/m^3$
Acceleration due to gravity $g = 10 \, m/s^2$
Area of the base $A = 2 \times 10^{-3} \, m^2$
The force $F$ acting on the base is given by $F = P \times A$.
Substituting the values:
$F = (h \rho g) \times A$
$F = 0.4 \times 900 \times 10 \times (2 \times 10^{-3})$
$F = 3600 \times 2 \times 10^{-3}$
$F = 7200 \times 10^{-3} = 7.2 \, N$.
Therefore,the force acting on the base is $7.2 \, N$.

(D) The force exerted by the liquid on the bottom of a vessel is equal to the weight of the liquid column if the walls of the vessel are vertical.
Since the same amount of liquid (same mass $m$) is poured into both vessels,the weight of the liquid $W = mg$ remains constant.
For a vessel with vertical walls,the force on the bottom is equal to the weight of the liquid,$F = mg$.
Therefore,$F_1 = mg$ and $F_2 = mg$.
Thus,$F_1 = F_2$.

(B) The force exerted by the water jet on the wall is equal to the rate of change of momentum of the water.
Since the collision is elastic,the velocity component parallel to the wall remains unchanged,while the component perpendicular to the wall reverses direction.
The change in momentum per unit time is given by $F = \frac{\Delta p}{\Delta t} = \frac{2mv \cos \theta}{\Delta t}$.
Here,$\frac{m}{\Delta t} = \rho A v$,where $\rho = 10^3 \ kg/m^3$ is the density of water,$A = 6 \times 10^{-4} \ m^2$ is the cross-sectional area,and $v = 12 \ m/s$ is the velocity.
Substituting the values: $F = 2(\rho A v) v \cos 60^{\circ} = 2 \rho A v^2 \cos 60^{\circ}$.
$F = 2 \times 10^3 \times (6 \times 10^{-4}) \times (12)^2 \times \cos 60^{\circ}$.
$F = 2 \times 10^3 \times 6 \times 10^{-4} \times 144 \times 0.5$.
$F = 10^3 \times 6 \times 10^{-4} \times 144 = 6 \times 0.144 \times 100 = 86.4 \ N$.

(D) Let the mass of water flowing in time $t$ be $m = \rho L t$.
The initial momentum of the water in the vertical direction is $p_i = m V = \rho L t V$ (upward).
The final momentum of the water in the vertical direction is $p_f = 0$.
Change in momentum in the vertical direction: $\Delta p_y = p_f - p_i = -\rho L V t$.
The initial momentum of the water in the horizontal direction is $p_i = 0$.
The final momentum of the water in the horizontal direction is $p_f = m V = \rho L t V$ (horizontal).
Change in momentum in the horizontal direction: $\Delta p_x = p_f - p_i = \rho L V t$.
The net change in momentum in time $t$ is $\Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} = \sqrt{(\rho L V t)^2 + (-\rho L V t)^2} = \sqrt{2} \rho L V t$.
The force exerted by the water on the corner is equal to the rate of change of momentum:
$F = \frac{\Delta p}{t} = \sqrt{2} \rho L V$.

(A) The mass flow rate $\dot{m}$ is given by $\rho A v$,where $\rho = 1000 \ kg/m^3$,$A = 10 \ cm^2 = 10^{-3} \ m^2$,and $v = 20 \ m/s$.
$\dot{m} = 1000 \times 10^{-3} \times 20 = 20 \ kg/s$.
The change in velocity vector $\Delta \vec{v}$ for a $90^{\circ}$ turn is $\sqrt{v^2 + v^2} = v\sqrt{2}$.
The magnitude of the force exerted by the water on the pipe is $F = \dot{m} \Delta v = \dot{m} v \sqrt{2}$.
$F = 20 \times 20 \times \sqrt{2} = 400 \times 1.4142 = 565.68 \ N \approx 565.7 \ N$.

(B) The force exerted by a fluid jet on a flat plate is equal to the rate of change of linear momentum of the water.
$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \frac{dm}{dt}$
Since the mass flow rate $\frac{dm}{dt} = \rho A v$,where $\rho$ is the density of water $(1000 \ kg/m^3)$,$A$ is the cross-sectional area,and $v$ is the velocity.
$F = (\rho A v) v = \rho A v^2$
Given:
$A = 2 \ cm^2 = 2 \times 10^{-4} \ m^2$
$v = 10 \ m/s$
$\rho = 1000 \ kg/m^3$
Substituting the values:
$F = 1000 \times (2 \times 10^{-4}) \times (10)^2$
$F = 1000 \times 2 \times 10^{-4} \times 100$
$F = 20 \ N$

(A) The force exerted by the liquid on the object is the vertical component of the pressure force acting on its surface.
According to the principle of hydrostatics,the downward force exerted by the liquid on the object is equal to the weight of the liquid column that would have occupied the volume $V$ of the object inside the container.
This force is given by $F = \rho Vg$.
Since the object is fitted at the base,the liquid exerts a downward force on the top surface of the object.
Therefore,the net force due to the liquid on the object is equal to the weight of the displaced liquid,which is $\rho Vg$.

(A) The force exerted by the water on the palm is equal to the rate of change of momentum of the water.
$F = \frac{dp}{dt} = v \frac{dm}{dt}$
Since the mass flow rate $\frac{dm}{dt} = A \rho v$,where $A$ is the cross-sectional area,$\rho$ is the density,and $v$ is the velocity.
Substituting this into the force equation:
$F = v(A \rho v) = A \rho v^2$
Given values: $v = 1.5\, ms^{-1}$,$A = 10^{-2}\, m^2$,$\rho = 10^3\, kgm^{-3}$.
$F = 10^{-2} \times 10^3 \times (1.5)^2$
$F = 10 \times 2.25 = 22.5\, N$.

(A) The mass of liquid striking the wall per unit time is given by $\frac{dm}{dt} = A \cdot v \cdot \rho$,where $A$ is the cross-sectional area,$v$ is the velocity,and $\rho$ is the density.
Since the jet makes an angle $\theta$ with the wall,the component of velocity normal to the wall is $v_n = v \sin \theta$.
The liquid rebounds elastically,meaning the normal component of velocity changes from $v \sin \theta$ to $-v \sin \theta$.
The change in momentum per unit time (force) exerted by the wall on the liquid is $F = \frac{dp}{dt} = \frac{dm}{dt} \cdot \Delta v_n$.
$F = (av\rho) \cdot (v \sin \theta - (-v \sin \theta)) = (av\rho) \cdot (2v \sin \theta) = 2av^2\rho \sin \theta$.
By Newton's third law,the force exerted by the liquid on the wall is equal in magnitude,so the normal force is $2av^2\rho \sin \theta$.

(D) Speed of the water stream,$v = 15\; m s^{-1}$.
Cross-sectional area of the tube,$A = 10^{-2}\; m^2$.
Volume of water coming out from the pipe per second,$V = A \times v = 10^{-2} \times 15 = 0.15\; m^3 s^{-1}$.
Density of water,$\rho = 10^3\; kg m^{-3}$.
Mass of water flowing out through the pipe per second,$\frac{dm}{dt} = \rho \times V = 10^3 \times 0.15 = 150\; kg s^{-1}$.
The water strikes the wall and does not rebound. Therefore,the force exerted by the water on the wall is given by Newton's second law of motion as:
$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \times \frac{dm}{dt}$.
$F = 15\; m s^{-1} \times 150\; kg s^{-1} = 2250\; N$.

(D) The force exerted by a fluid jet hitting a wall and coming to rest is given by the rate of change of momentum.
$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \frac{dm}{dt}$.
Since the mass flow rate $\frac{dm}{dt} = \rho A v$,the force is $F = \rho A v^{2}$.
Given:
Density $\rho = 1000 \, kg/m^{3}$
Area $A = 10 \, cm^{2} = 10 \times 10^{-4} \, m^{2} = 10^{-3} \, m^{2}$
Velocity $v = 20 \, m/s$
Substituting the values:
$F = 1000 \times 10^{-3} \times (20)^{2}$
$F = 1 \times 400 = 400 \, N$.

(D) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dm = (m/L) dx$.
The centrifugal force $dF$ acting on this element is $dF = (dm) \omega^2 x = (m/L) \omega^2 x dx$.
To find the total force $F$ exerted at the outer end,we integrate from $x = 0$ to $x = L$:
$F = \int_{0}^{L} \frac{m}{L} \omega^2 x dx = \frac{m \omega^2}{L} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{m \omega^2 L}{2}$.
Given $m = 250\,g = 0.25\,kg$ and $L = 50\,cm = 0.5\,m$,we have:
$F = \frac{0.25 \times \omega^2 \times 0.5}{2} = \frac{0.125}{2} \omega^2 = 0.0625 \omega^2$.
Thus,$\omega^2 = F / 0.0625 = 16F$,which gives $\omega = 4 \sqrt{F}$.
Comparing this with $\omega = x \sqrt{F}$,we get $x = 4$.

(B) The pressure at a depth $x$ below the surface of the water is given by $P = \rho g x$.
Consider a small horizontal strip of the dam at depth $x$ with thickness $dx$ and width $w$.
The area of this strip is $dA = w \cdot dx$.
The force $dF$ on this strip is $dF = P \cdot dA = (\rho g x)(w \cdot dx)$.
To find the total resultant force $F$ on the dam,we integrate this expression from $x = 0$ to $x = H$:
$F = \int_{0}^{H} \rho g w x \, dx$
$F = \rho g w \int_{0}^{H} x \, dx$
$F = \rho g w \left[ \frac{x^2}{2} \right]_{0}^{H}$
$F = \frac{1}{2} \rho g w H^2$.

(A) The pressure $P$ at a depth $H$ from the free surface of the liquid is given by $P = \rho g H$.
To stop the liquid from gushing out,we must apply a force equal to the force exerted by the liquid on the hole.
The force $F$ exerted by the liquid on the hole is given by $F = P \times A$,where $A$ is the area of the hole.
The area of the hole of radius $r$ is $A = \pi r^2$.
Therefore,the required horizontal force is $F = (\rho g H) \times (\pi r^2) = \rho g H \pi r^2$.

(B) The total force $F$ exerted on the bottom of the tube is the sum of the force due to atmospheric pressure and the force due to the weight of the mercury column.
$F = (P_0 + \rho gh) A$
Here,the area $A = \pi r^2 = \frac{22}{7} \times (2 \times 10^{-2} \,m)^2 = \frac{22}{7} \times 4 \times 10^{-4} \,m^2 \approx 1.257 \times 10^{-3} \,m^2$.
Force due to atmospheric pressure $F_{atm} = P_0 A = 10^5 \times 1.257 \times 10^{-3} = 125.7 \,N$.
Force due to mercury column $F_{Hg} = \rho gh A = (1.36 \times 10^4) \times 10 \times (30 \times 10^{-2}) \times (1.257 \times 10^{-3}) = 13600 \times 10 \times 0.3 \times 1.257 \times 10^{-3} \approx 51.3 \,N$.
Total force $F = 125.7 + 51.3 = 177 \,N$.

(D) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dm = \frac{M_{total}}{L} dx = \frac{2M}{1} dx = 2M dx$.
The centrifugal force $dF$ acting on this element is $dF = (dm) \omega^2 x = (2M dx) \omega^2 x$.
The total force $F$ exerted by the liquid at the outer end is the integral of these forces from $x = 0$ to $x = L = 1 \ m$:
$F = \int_{0}^{L} 2M \omega^2 x dx = 2M \omega^2 \left[ \frac{x^2}{2} \right]_{0}^{1} = 2M \omega^2 \left( \frac{1}{2} \right) = M \omega^2$.
Given that the angular velocity is $\omega = \sqrt{\frac{F}{\alpha M}}$,we have $\omega^2 = \frac{F}{\alpha M}$.
Comparing $F = M \omega^2$ with $\omega^2 = \frac{F}{M}$,we get $\alpha = 1$.

(C) The force exerted by the water on the wall is equal to the rate of change of momentum of the water.
Given:
Area of cross-section $A = 2 \times 10^{-3} \,m^2$
Velocity $v = 12 \,m \,s^{-1}$
Density of water $\rho = 1000 \,kg \,m^{-3}$
The mass of water hitting the wall per unit time is given by $\frac{dm}{dt} = \rho A v$.
Substituting the values:
$\frac{dm}{dt} = 1000 \times (2 \times 10^{-3}) \times 12 = 24 \,kg \,s^{-1}$.
The force $F$ is the rate of change of momentum: $F = \frac{dp}{dt} = \frac{dm}{dt} \times v$.
Since the water does not rebound, the final velocity is $0$.
$F = (24 \,kg \,s^{-1}) \times (12 \,m \,s^{-1}) = 288 \,N$.
Therefore, the force acting on the wall is $288 \,N$.

(B) The pressure at a depth '$y$' from the surface is given by $P(y) = P_{atm} + \rho gy$.
The total force '$F$' acting on the dam of width '$b$' is the integral of pressure over the area:
$F = \int_{0}^{h} (P_{atm} + \rho gy) b \, dy = (P_{atm} h + \frac{1}{2} \rho g h^2) b$.
The point of application '$y_{cp}$' (center of pressure) from the surface is given by the ratio of the moment of force to the total force:
$y_{cp} = \frac{\int_{0}^{h} y (P_{atm} + \rho gy) b \, dy}{F} = \frac{b [P_{atm} \frac{h^2}{2} + \rho g \frac{h^3}{3}]}{(P_{atm} h + \frac{1}{2} \rho g h^2) b}$.
If we consider the force due to water only (ignoring $P_{atm}$ or assuming it acts on both sides),the center of pressure is at a depth of $\frac{2h}{3}$ from the surface,which is $\frac{h}{3}$ from the base '$O$'.
Given the standard context of such problems where the net force due to water pressure distribution (triangular) is considered,the center of pressure is at $\frac{h}{3}$ from the base '$O$'.