Topics

We consider an open container of a liquid that is moving along a straight line with a constant acceleration a as shown in Fig.

Lets consider a small horizontal cylinder of length dx and crossectional area A located y below the free surface of the fluid. This cylinder is accelerating in ground frame with accleration a hence the net horizontal force acting on it should be equal to the product of mass(dm) and acceleration.
\[dm = Adx\rho\]

\[P_{2}\text{ }A - P_{1}\text{ }A = (Adx\rho)a\ P_{2} - P_{1} = Padx\]

If we say that the right face of the cylinder is \(y\) below the free surface of the fluid then the left surface is \(y + dy\) below the surface of liquid. Thus

\[\begin{matrix} & P_{2} - P_{1} = \rho gdy \\ \therefore & \frac{dy}{dx} = \frac{a}{g} \end{matrix}\]

Since the slope of the free surface is coming out to be constant we can say that it must be straight line.

\[tan\theta = \frac{a}{g}\]

If the container have acceleration along \(y\) also than the slope of this line is given by the relationship.

\[\frac{dy}{dx} = - \frac{a_{x}}{g + a_{y}}\]

Along a free surface the pressure is constant, so that for the accelerating mass shown in Fig. () the free surface will be inclined if \(a_{x} \neq 0\). In addition, all lines parallel to the free surface will have same presure. For the special circumstance in which \(a_{x} = 0,a_{y} \neq 0\). which corresponds to the mass of fluid accelerating in the vertical direction, Eq. () indicates that the fluid surface will be horizontal. However, from Eq. () we see that the pressure variation is not \(\rho\) gdy, but is given by the equation.

\[dP = \rho\left( g + a_{y} \right)dy\]

Thus, the pressure on the bottom of a liquid-filled tank which is resting on the floor of an elevator that is accelerating upward will be more than, if the, tank would have been at rest (or moving with a constant velocity). It is to be noted that for a freely falling fluid mass \(\left( a_{y} = - g \right)\), the pressure variation in all three coordinate directions are zero, which means that the pressure throughout will be same. The pressure throughout a "blob" of a liquid floating in an orbiting space shuttle (a form of free fall) is zero. The only force holding the liquid together is surface tension.

Consider a cylinderical vessel, rotating at constant angular velocity about its axis. If it contains fluid then after an initial irregular shape, it will rotate with the tank as a rigid body. The acceleration of fluid particles located at a distance \(r\) from the axis of rotation will be equal to \(\omega^{2}r\), and the direction of the acceleration is toward the axis of rotation as shown in the figure. The fluid particles will be undergoing circular motion.

Lets consider a small horizontal cylinder of length dr and crossectional area A located y below the free surface of the fluid and \(r\) from the axis. This cylinder is accelerating in ground frame with accleration \(\omega^{2}r\) towards the axis hence the net horizontal force acting on it should be equal to the product of mass (dm) and acceleration.
dm \(=\) Adr \(\rho\)

\[P_{2}A - P_{1}A = (Adr\rho)\omega^{2}r\]

If we say that the left face of the cylinder is \(y\) below the free surface of the fluid then the right surface is \(y + dy\) below the surface of liquid. Thus

\[P_{2} - P_{1} = \rho gdy\]

Thus solving we get,

\[\frac{dy}{dr} = \frac{r\omega^{2}}{g}\]

and, therefore, the equation for surfaces of constant pressure is

\[y = \frac{\omega^{2}r^{2}}{2g} + \text{~}\text{constant}\text{~}\]

This equation means that these surfaces of constant pressure are parabolic as shown in Fig

The pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical direction as shown in fig.

When you squeeze one end of a tube to get toothpaste out the other end, you are watching Pascal's principle in action. This principle is also the basis for the Heimlich maneuver, in which a sharp pressure increase properly applied to the abdomen is transmitted to the throat, forcefully ejecting food lodged there. The principle was first stated clearly in 1652 by Blaise Pascal (for whom the unit of pressure is named):

A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.

Demonstrating Pascal's Principle

Consider the case in which the incompressible fluid is a liquid contained in a tall cylinder, as in Fig. The cylinder is fitted with a piston on which a container of lead shot rests. The atmosphere, container, and shot exert pressure \(p_{\text{ext}\text{~}}\) on the piston and thus on the liquid. The pressure p at any point P in the liquid is then

\[P = P_{ext} + \rho_{gh}\]

Let us add a little more lead shot to the container to increase \(P_{\text{ext}\text{~}}\) by an amount \(P_{\text{ext}\text{~}}\). The quantities \(P_{\text{ext}\text{~}}\), \(g\) and \(h\) in Eq. are unchanged, so the pressure change at \(P\) is

\[\Delta p = \Delta p_{ext}\]

This pressure change is independent of h , so it must hold for all points within the liquid, as Pascal's principle states.

Pascal's Principle and the Hydraulic Lever

Figure shows how Pascal's principle can be made the basis of a hydraulic lever. In operation, let an external force of magnitude \(F_{i}\) be directed downwardon the left-hand (or input) piston, whose surface area is \(A_{i}\). An incompressible liquid in the device then produces an upward force of magnitude \(F_{0}\) on the right-hand (or output) piston, whose surface area is \(A_{0}\). To keep the system in equilibrium, there must be a downward force of magnitude \(Fo\) on the output piston from an external load (not shown). The force \({\overrightarrow{F}}_{1}\) applied on the left and the downward force \({\overrightarrow{F}}_{0}\) from the load on the right produce a change \(\Delta p\) in the pressure of the liquid that is given by

\[\begin{matrix} & \Delta p = \frac{F_{i}}{{\text{ }A}_{i}} = \frac{F_{0}}{{\text{ }A}_{0}} \\ & {\text{ }F}_{0} = F_{1}\frac{{\text{ }A}_{0}}{{\text{ }A}_{i}} \end{matrix}\]

Equation shows that the output force \(F_{0}\) on the load must be greater than the input force \(F_{i}\) if \(A_{0} > A_{i}\) as is the case in figure.
If we move the input piston downward a distance \(d_{i}\), the output piston moves upward a distance do, such that the same volume \(V\) of the incompressible liquid is displaced at both pistons.Then

\[V = A_{i}{\text{ }d}_{i} = A_{0}{\text{ }d}_{0}\]

which we can write as

\[d_{0} = d_{i}\frac{{\text{ }A}_{i}}{{\text{ }A}_{0}}\]

This shows that, if \(Ao > Ai\) (as in Figure), the output piston moves a smaller distance than the input piston moves.

From Eqs. we can write the output work as

\[W = F_{0}{\text{ }d}_{0} = \left( F_{i}\frac{{\text{ }A}_{0}}{{\text{ }A}_{i}} \right)\left( d_{i}\frac{{\text{ }A}_{i}}{{\text{ }A}_{0}} \right) = F_{i}{\text{ }d}_{i}\]

which shows that the work W done on the input piston by the applied force is equal to the work W done by the output piston in lifting the load placed on it.
The advantage of a hydraulic lever is this:

With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force applied over a smaller distance.

The product of force and distance remains unchanged so that the same work is done. However, there is often tremendous advantage in being able to exert the larger force. Most of us, for example, cannot lift an automobile directly but can with a hydraulic jack, even though we have to pump the handle farther than the automobile rises in a series of small strokes.

Practice Exercise

Q. 1 The passenger are advised to remove the ink from their pens whils going up in an aeroplane. Explain why?
Q. 2 A hydraulic press has a ram (weight arm) 12.5 cm in diameter and plunger (Force arm) of 1.25 cm diameter what force would be required by plunger to raise a weight of 1 tonn on the ram.
Q. 3 Pressure 3 m below free surface of a liquid is \(15KN/m^{2}\) in excess of atmosphere pressure. Deternine its density and specific gravity. [g \(= 10\text{ }m/\sec^{2}\) ]

Figure shows a student in a swimming pool, manipulating a very thin plastic sack (of negligible mass) that is filled with water. She finds that the sack and its contained water are in static equilibrium, tending neither to rise nor to sink. The downward gravitational force \({\overrightarrow{F}}_{g}\) on the contained water must be balanced by a net upward force from the water surrounding the sack.

(b)

This net upward force is a buoyant force \({\overrightarrow{F}}_{b}\). It exists because the pressure in the surrounding water increases with depth below the surface.Thus, the pressure near the bottom of the sack is greater than the pressure near the top, which means the forces on the sack due to this pressure are greater in magnitude near the bottom of the sack than near the top. Some of the forces are represented in Fig. (a), where the space occupied by the sack has been left empty. Note that the force vectors drawn near the bottom of that space (with upward components) have longer lengths than those drawn near the top of the sack (with downward components). If we vectorially add all the forces on the sack from the water, the horizontal components cancel and the vertical components add to yield the upward buoyant force \({\overrightarrow{F}}_{b}\) on the sack. (Force \({\overrightarrow{F}}_{b}\) is shown to the right of the pool in Fig. (a.) Because the sack of water is in static equilibrium, the magnitude \({\overrightarrow{F}}_{b}\) of is equal to the magnitude \(m_{f}g\) of the gravitational force \({\overrightarrow{F}}_{g}\) on the sack of water: \(F_{b} = m_{f}g\). (Subscript \(f\) refers to fluid, here the water.) In words, the magnitude of the buoyant force is equal to the weight of the water in the sack.

In Fig. (b), we have replaced the sack of water with a stone that exactly fills the hole in Fig. (a). The stone is said to displace the water, meaning that the stone occupies space that would otherwise be occupied by water. We have changed nothing about the shape of the hole, so the forces at the hole's surface must be the same as when the water-filled sack was in place. Thus, the same upward buoyant force that acted on the water-filled sack now acts on the stone; that is, the magnitude \({\overrightarrow{F}}_{b}\) of the buoyant force is equal to \(m_{g}g\), the weight of the water displaced by the stone.
Unlike the water-filled sack, the stone is not in static equilibrium. The downward gravitational force \({\overrightarrow{F}}_{g}\) on the stone is greater in magnitude than the upward buoyant force, as is shown in the free-body diagram in Fig. (b). The stone thus accelerates downward, sinking to the bottom of the pool. Let us next exactly fill the hole in Fig. (a) with a block of lightweight wood, as in Fig. (c). Again, nothing has changed
about the forces at the hole's surface, so the magnitude \({\overrightarrow{F}}_{b}\) of the buoyant force is still equal to \(m_{f}g\), the weight of the displaced water. Like the stone, the block is not in static equilibrium. However, this time the gravitational force \({\overrightarrow{F}}_{g}\) is lesser in magnitude than the buoyant force (as shown to the right of the pool), and so the block accelerates upward, rising to the top surface of the water. Our results with the sack, stone, and block apply to all fluids and are summarized in Archimedes' principle:

When a body is fully or partially submerged in a fluid, a buoyant force \({\overrightarrow{F}}_{b}\), from the surrounding fluid acts on the body. The force is directed upward and has a magnitude equal to the weight \(m_{j}g\) of the fluid that has been displaced by the body. The buoyant force on a body in a fluid has the magnitude

\[F_{b} = m_{f}\text{ }g\text{~}\text{(buoyant force)}\text{~}\]

where \(m_{f}\) is the mass of the fluid that is displaced by the body.

Practice Exercise

Q. 1 A boy is carrying a fish in one hand and a bucket full of water in the other hand. He then place the fish in the bucket and thinks that in accordance with Archimedes's prinicple he is now carrying less weight as weight of fish will reduce due to upthrust. Is he thinking right?
Q. 2 Ice flows in water nine tenth of its volume submerged. What is the frctional volume submerged for an iceberg floating on a fresh water lake of a (hypothetical) planet whose gravity is ten times of earth?
Q. 3 If the body is non-homogeneous, then the body rotates in the fluid why?
Q. 4 A cube of wood supporting a 200 gm mass just floats in water. When the mass is removed the cube rise by 2 cm . Find the size of cube
Q.5 A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enter water. Upto what depht will the ball go? How much time will it take to come again to the water surface ? Neglect air resistance and viscosity effects in water.
Q. 6 A balloon filled with hydrogen has a volume of 1000 liters and its mass of 1 kg . What would be volume of the block of a very light material which it can just lift? One litre of the material has a mass of 91.3 gm . (Density of air \(= 1.3gm/\) litre)
Q. 7 An iceberg of density \(915\text{ }kg/m^{3}\) extends above the surface of sea water of density \(1030\text{ }kg/m^{3}\). What percent-age of the total volume of iceberg is visibel to an obserber.

Answers

Q. 1 it's density is high because of salt Q. 2 Same
Q. 3 Centre of Buoyancy and centre of gravity are diffrent resulting torque.
Q. 410 cm Q. \(519.6\text{ }m,4sec\) Q. 6 3.33 litre. Q. \(711.15\%\)

When we release a block of lightweight wood just above the water in a pool, the block moves into the water because the gravitational force on it pulls it downward.
As the block displaces more and more water, the magnitude \(F_{b}\) of the upward buoyant force acting on it increases. Eventually, \(F_{b}\) is large enough to equal the magnitude \(F_{g}\) of the downward gravitational force on the block, and the block comes to rest. The block is then in static equilibrium and is said to be floating in the water. In general, When a body floats in a fluid, the magnitude \(F_{b}\) of the buoyant force on the body is equal to the magnitude \(F_{g}\) of the gravitational force on the body.
We can write this statement as

\[F_{b} = F_{g}(\text{~}\text{Floating}\text{~})\]

From Eq. we know that \(F_{b} = m_{f}\text{ }g\). Thus,
When a body floats in a fluid, the magnitude \(F_{g}\) of the gravitational force on the body is equal to the weight \(m_{f}\text{ }g\) of the fluid that has been displaced by the body.
We can write this statement as

\[F_{g} = m_{f}\text{ }g\]

In other words, a floating body displaces its own weight of fluid.
The location of the line of action of the buoyant force can be determined by adding torques of the forces due to pressure forces, with respect to some convenient axis. The buoyant force must pass through the center of mass of the desplaced volume, as shown in Fig. (c), as it was in translational and rotational equilibirium. The point through which the buoyant force acts is called the center of buoyancy.

(c)

(D)

These same results apply to floating bodies which are only partially submerged, as shown in Fig.(d), if the density of the fluid above the liquid surface is very small compared with the liquid in which the body floats. Since the fluid above the surface is usually air, for practial purposes this condition is satisfied. In the above discussion, the fluid is assumed to have a constant density. If a body is immersed in a fluid in which density varies with depth, such as having multiple layers of fluid, the magnitude of the buoyant force remains equal to the weight of the displaced fluid and the buoyant force passes through the center of mass of the displaced volume.

Stability

The center of buoyancy and center of gravity do not necessarily coincide so the floating or submerged body may not be in stable equilibirium. A small rotation can cause the buoyant force to produce either a restoring or overturning torque. For example, for the completely submerged body shown in Fig., which has a center of gravity below the center of buoyancy, a rotation from its equilibrium position will create a restoring torque by the buoyant force, \(F_{B}\), which causes the body to rotate back to its original position. Thus, if the center of gravity falls below the center of buoyancy, the body is stable.
However, as shown in Fig., if the center of gravity of the completely submerged body is above the center of buoyancy, the resulting torque formed by the weight and the buoyant force will cause the body to overturn and move to a new equilibrium position. Thus, a completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position.
For floating bodies the stability problem is more complicated, since as the body rotates the location of the center of buoyancy may change.

Ideal Fluids in Motion

The motion of real fluids is very complicated and not yet fully understood. Instead, we shall discuss the motion of an ideal fluid, which is simpler to handle mathematically and yet provides useful results. Here are four assumptions that we make about our ideal fluid; they all are concerned with flow:

1. Steady flow In steady (or laminar) flow, the velocity of the moving fluid at any fixed point does not change with time, either in magnitude or in direction. The gentle flow of water near the center of a quiet stream is steady; the flow in a chain of rapids is not. The speed of the smoke particles increases as they rise and, at a certain critical speed, the flow changes from steady to nonsteady.

2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is incompressible; that is, its density has a constant, uniform value.

3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of how resistive the fluid is to flow. For example, thick honey is more resistive to flow than water, and so honey is said to be more viscous than water. Viscosity is the fluid analog of friction between solids; both are mechanisms by which
   the kinetic energy of moving objects can be transferred to thermal energy. In the absence of friction, a block could glide at constant speed along a horizontal surface. In the same way, an object moving through a nonviscous fluid would experience no viscous drag force-that is, no resistive force due to viscosity; it could move at constant speed through the fluid.

4. Irrotational flow : Although it need not concern us further, we also assume that the flow is irrotational. To test for this property, let a tiny grain of dust move with the fluid. Although this test body may (or may not) move in a circular path, in irrotational flow the test body will not rotate about an axis through its own center of mass. For a loose analogy, the motion of a Ferris wheel is rotational; that of its passengers is irrotational. T8hat the velocity of a particle is always tangent to the path taken by the particle. Here the particle is the fluid element, and its velocity is \(\overrightarrow{v}\) always tangent to a streamline (Figure). For this reason, two streamlines can never intersect; if they did, then an element arriving at their intersection would have two different velocities simultaneously an impossibility.

The Equation of Continuity

You may have noticed that you can increase the speed of the water emerging from a garden hose by partially closing the hose opening with your thumb. Apparently the speed v of the water depends on the cross-sectional area A through which the water flows.

Here we wish to derive an expression that relates v and A for the steady flow of an ideal fluid through a tube with varying cross section, like that in Figure. The flow there is toward the right, and the tube segment shown (part of a longer tube) has length \(L\). The fluid has speeds \(v_{1}\) at the left end of the segment and \(v_{2}\) at the right end. The tube has cross-sectional areas \(A_{1}\) at the left end and \(A_{2}\) at the right end. Suppose that in a time interval \(\Delta t\) a volume \(\Delta V\) of fluid enters the tube segment at its left end (that volume is colored purple in Figure. Then, because the fluid is incompressible, an identical volume \(\Delta V\) must emerge from the right end of the segment (it is colored green in Figure). We can use this common volume \(\Delta V\) to relate the speeds and areas. To do so, we first consider Fig., which shows a side view of a tube of uniform crosssectional area A. In Fig.(a), a fluid element e is about to pass through the dashed line drawn across the tube width. The element's speed is \(v\), so during a time interval \(\Delta t\), the element moves along the tube a distance \(\Delta x = v\Delta t\). The volume \(\Delta v\) of fluid that has passed through the dashed line in that time interval \(\Delta t\) is

(6) Time \(t + \Delta t\)

(b) Time \(t + \Delta t\)

\[\Delta V = A\Delta x = Av\Delta t.\]

Applying Eq. to both the left and right ends of the tube segment in Fig., we have

Or

\[\begin{matrix} & \Delta V = A_{1}v_{1}\Delta t = A_{2}v_{2}\Delta t \\ & A_{1}v_{1} = A_{2}v_{2}\text{~}\text{(equation of continuity)}\text{~} \end{matrix}\]

This relation between speed and cross-sectional area is called the equation of continuity for the flow of an ideal fluid. It tells us that the flow speed increases when we decrease the cross-sectional area through which the fluid flows (as when we partially close off a garden hose with a thumb). Equation applies not only to an actual tube but also to any so-called tube of flow, or imaginary tube whose boundary consists of streamlines. Such a tube acts like a real tube because no fluid element can cross a streamline; thus, all the fluid within a tube of flow must remain within its boundary. Figure shows a tube of flow in which the cross-sectional area increases from area \(A_{1}\) to area \(A_{2}\) along the flow direction. From Eq. we know that, with the increase in area, the speed must decrease, as is indicated by the greater spacing between streamlines at the right in Fig. . Similarly, you can see that in Fig. the speed of the flow is greatest just above and just below the cylinder. We can rewrite Eq. as
\(R_{V} = Av = a\) constant (volume flow rate, equation of continuity), in which \(R_{V}\) is the volume flow rate of the fluid (volume past a given point per unit time). Its SI unit is the cubic meter per second ( \(m^{3}/s\) ). If the density \(\rho\) of the fluid is uniform, we can multiply Eq. by that density to get the mass flow rate \(R_{m}\) (mass per unit time):

\[R_{m} = \rho R_{V} = \rho Av = a\text{~}\text{constant (mass flow rate).}\text{~}\]

The SI unit of mass flow rate is the kilogram per second (kg/s). Equation says that the mass that flows into the tube segment of Fig. each second must be equal to the mass that flows out of that segment each second.

Figure represents a tube through which an ideal fluid is flowing at a steady rate. In a time interval \(\Delta t\), suppose that a volume of fluid \(\Delta V\), colored purple in Fig., enters the tube at the left (or input) end and an identical volume, colored green in Fig., emerges at the right (or output) end. The emerging volume must be the same as the entering volume because the fluid is incompressible, with an assumed constant density \(\rho\).

\[p_{1} + \frac{1}{2}\rho v_{1}^{2} + \rho{gy}_{1} = p_{2} + \frac{1}{2}\rho v_{2}^{2} + \rho{gy}_{2}\]

(B)

Let \(y_{1},v_{1}\), and \(p_{1}\) be the elevation, speed, and pressure of the fluid entering at the left, and \(y_{2},v_{2}\), and \(p_{2}\) be the corresponding quantities for the fluid emerging at the right. By applying the principle of conservation of energy to the fluid, we shall show that these quantities are related by In general, the term \(\frac{1}{2}\rho v^{2}\) is called the fluid's kinetic energy density (kinetic energy per unit volume).We can also write Eq. as

\[p + \frac{1}{2}\rho v^{2} + \rho gy\text{~}\text{a constant (Bernoulli's equation).}\text{~}\]

Equations are equivalent forms of Bernoulli's equation, after Daniel Bernoulli, who studied fluid flow in the 1700s.* Like the equation of continuity Eq., Bernoulli's equation is not a new principle but simply the reformulation of a familiar principle in a form more suitable to fluid mechanics. As a check, let us apply Bernoulli's equation to fluids at rest, by putting \(v_{1} = v_{2} = 0in\) Eq. The result is

\[p_{2} = p_{1} + \rho g\left( y_{1} - y_{2} \right)\]

Which is equation.
A major prediction of Bernoulli's equation emerges if we take \(y\) to be a constant \((y = 0\), say \()\) so that the fluid does not change elevation as it flows. Equation then becomes

\[p_{1} + \frac{1}{2}\rho v_{1}^{2} = p_{2} + \frac{1}{2}\rho v_{2}^{2}\]

Which tells us that :
If the speed of a fluid element increases as the element travels along a horizontal streamline, the pressure of the fluid must decrease, and conversely.
Put another way, where the streamlines are relatively close together (where the velocity is relatively great), the pressure is relatively low, and conversely. The link between a change in speed and a change in pressure makes sense if you consider a fluid element. When the element nears a narrow region, the higher pressure behind it accelerates it so that it then has a greater speed in the narrow region. When it nears a wide region, the higher pressure ahead of it decelerates it so that it then has a lesser speed in the wide region. Bernoulli's equation is strictly valid only to the extent that the fluid is ideal. If viscous forces are present, thermal energy will be involved. We take no account of this in the derivation that follows.

Proof of Bernoulli's Equation

Let us take as our system the entire volume of the (ideal) fluid shown in Fig. We shall apply the principle of conservation of energy to this system as it moves from its initial state to its final state. The fluid lying between the two vertical planes separated by a distance L in Fig. does not change its properties during this process; we need be concerned only with changes that take place at the input and output ends. First, we apply energy conservation in the form of the work-kinetic energy theorem,

\[W = \Delta K,\]

which tells us that the change in the kinetic energy of our system must equal the net work done on the system. The change in kinetic energy results from the change in speed between the ends of the tube and is

\[\Delta K = \frac{1}{2}\Delta{mv}_{2}^{2} - \frac{1}{2}\Delta{mv}_{1}^{2} = \frac{1}{2}\rho\Delta\text{ }V\left( v_{2}^{2} - v_{1}^{2} \right)\]

in which \(\Delta m( = \rho\Delta V)\) is the mass of the fluid that enters at the input end and leaves at the output end during a small time interval \(\Delta t\).
The work done on the system arises from two sources. The work \(W_{g}\) done by the gravitational force \((\Delta m\overrightarrow{g})\) on the fluid of mass \(\Delta m\) during the vertical lift of the mass from the input level to the output level is

\[\begin{matrix} & W_{g} = \Delta mg\left( y_{2} - y_{1} \right) \\ & \ = - \rho g\Delta\text{ }V\left( y_{2} - y_{1} \right) \end{matrix}\]

This work is negative because the upward displacement and the downward gravitational force have opposite directions.

Work must also be done on the system (at the input end) to push the entering fluid into the tube and by the system (at the output end) to push forward the fluid that is located ahead of the emerging fluid. In general, the work done by a force of magnitude F , acting on a fluid sample contained in a tube of area A to move the fluid through a distance \(\Delta x\), is

\[F\Delta x = (pA)(\Delta x) = p(\text{ }A\Delta x) = p\Delta v\]

The work done on the system is then \(p_{1}\Delta\text{ }V\), and the work done by the system is \(- p_{2}\Delta\text{ }V\). Their sum Wp is Wp

\[\begin{matrix} & Wp = p_{1}\Delta\text{ }V - p_{2}\Delta\text{ }V \\ & \ = - \left( p_{2} - p_{1} \right)\Delta V \end{matrix}\]

The work-kinetic energy theorem of Eq. now becomes

\[W = Wg + Wp = \Delta K\]

Substituting from Eqs. yields

\[- \rho g\Delta\text{ }V\left( y_{2} - y_{1} \right) - \Delta V\left( p_{2} - p_{1} \right) = \frac{1}{2}\rho\Delta v\left( v_{2}^{2} - v_{1}^{2} \right)\]

This, after a slight rearrangement, matches Eq., which we set out to prove.

Practice Exercise

Q. 1 During wind storm. light roofs are blown off. Why?
Q. 2 A man standing on the platform just near the raliway line be sucked in by a fast moving train. Explain.
Q. 3 Air is streaming past a horizontal airplane wing such that its speed is \(120{\text{ }ms}^{- 1}\) over the upper surface and \(90{\text{ }ms}^{- 1}\) at the lower surface. If the density of air is \(1.3{kgm}^{- 3}\), find the difference in pressure between the top and bottom of the wing. If the wing is 10 m long and has average width of 2 m . Calculate the gross lift of the wing.
Q. 4 A liquid is kept in cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev per sec. Find the difference in the height of the liquid at the cntre of the vessel and at its sides.
Q. 5 The pressures of water in a water pipe when tap is open and closed respectively \(3 \times 10^{5}\text{ }N/m^{2}\) and \(3.5 \times 10^{5}\text{ }N/m^{2}\). If tap is opened, then find out-
(a) velocity of water flowing (b) rate of volume of water flowing if area of cross-section of tap is \(2{\text{ }cm}^{2}\).
Q. 6 Water flows through a horizontal tube of variable cross-section (figure). The area of cross-section at A and \(B\) are \(4{\text{ }mm}^{2}\) and \(2{\text{ }mm}^{2}\) respectively. If 1 cc of water eneters persecond through \(A\), find (a) the speed of water at \(A\), (b) the speed of water at \(B\) and (c) the pressre differnece \(P_{A} - P_{B}\).

Q. 7 Water flows through the tube shown infigure. The areas of cross-section of the wide and the narrow portion of the tube are \(5{\text{ }cm}^{2}\) and \(2{\text{ }cm}^{2}\) respectively. The rate of flow of water through the tube is 500 \({cm}^{3}/s\). Find the difference of mercury levels in the u-tube.
(density of mercury \(= 13.6gm/{cm}^{3}\) )

Answers

Q. 1 Due to high velocity of wind above roof, pressure decreases resulting upward force.
Q. 2 Due to decreases in air pressure between preson and train.
Q. 3 Due to decrease in pressure in between.
Q. \(4\text{ }h = 2\text{ }cm\)
\[\begin{matrix} \text{~}\text{Q.}\text{~}5 & \text{~}\text{(a)}\text{~}10\text{ }m/s & \text{~}\text{(b)}\text{~}2 \times 10^{- 3}{\text{ }m}^{3}/s \end{matrix} \]Q. \(6\ \) (a) \(25\text{ }cm/s\ \) (b) \(50\text{ }cm/s\ \) (c) \(94\text{ }N/m^{2}\ \) Q. \(7\ 2.13\text{ }cm\)

Surface Tension

Surface Tension is a property of liquid at rest by virtue of which a liquid surface gets contracted to a minimum area and behaves like a stretched membrane.
Surface Tension of a liquid is measured by force per unit length on either side of any imaginary line drawn tangentially over the liquid surface, force being normal to the imaginary line as shown in fig. i.e. Surface tension.

\[(T) = \frac{\text{~}\text{Total force on either of the }\text{imginary}\text{ line}\text{~}(F)}{\text{~}\text{Length of the line}\text{~}(l)}\]

Unit of Surface Tension

In C. G. S. system the unit of surface tension is dyne/cm (dyne \({cm}^{- 1}\) ) and SI system its units is \({Nm}^{- 1}\)

Explanation of some observed phenomena

1. Lead balls are spherical in shape.

2. Rain drops and a globule of mercury placed on glass plate are spherical.

3. Hair of a shaving brush/painting brush, when dipped in water spread out, but as soon as it is taken out. Its hair stick together.

4. A greased needle placed gently on the free surface of water in a beaker does not sink.

5. Similarly, insects can walk on the free surface of water without drowning.

6. Bits of Camphor gum move inregularly when placed on water surface.

Surface energy

The course of reasoning given below is usually followed to prove that the molecules of the surface layer of a liquid have surplus potential energy. A molecule inside the liquid is acted upon by the forces of attraction from the other molecules which compensate each other on the average. If a molecule is singled out on the surface, the resulting force of attraction from the other molecule is directed into the liquid. For this reason the molecule tends to move into the liquid, and definite work should be done to bring it to the surface. Therefore, each molecule of the surface layer has excess potential energy equal to this work. The average force that acts on any molecule from the side of all the others, however, is always equal to zero if the liquid is in equilibrium. This is why the work done to move the liquid from a depth to the surface should also be zero. What is the origin, in this case, of the surface energy?
The forces of attraction acting on a molecule in the surface layer from all the other molecules produce a resultant directed downward. The closest neighbours, however, exert a force of repulsion on the molecule which is therefore in equilibrium.

Owing to the forces of attraction and repulsion, the density of the liquid is smaller in the surface layer than inside. Indeed, molecule 1 (figure) is acted upon by the force of repulsion from molecule 2 and the forces of attraction from all the other molecules \((3,4,\ldots\ldots)\). Molecule 2 is acted upon by the forces of repulsion from 3 and 1 and the forces of attraction from the molecules in the deep layers. As a result, distance 1-2 should be greater than 2-3, etc.
(1)
(2)
(3)
(4)
(5)
(6)

This course of reasoning is quite approximate (thermal motion, etc. is disregarded), but nevertheless it gives a qualitatively correct result.

An increase in the surface of the liquid causes new sections of the rarefied surface layer to appear. Here work should be performed against the forces of attraction between the molecules. It is this work that constitutes the surface energy.

We know that the molecules on the liquid surface experience net downward force. So to bring a molecule from the interior of the liquid to the free surface, some work is required to the done against the intermolecular
force of attraction, which will be stored as potential energy of the molecule on the surface. The potential energy of surface molecules per unit area of the surface is called surface is called surface energy. Unit of surface energy is \(erg{cm}^{- 2}\) in C.G..S. system and \({Jm}^{- 2}\) in SI system. Dimensional formula of surface energy is \(\left\lbrack {ML}^{\circ}T^{- 2} \right\rbrack\) surface energy depends on number of surfaces e.g. a liquid drop is having one liquid air surface while bubble is having two liquid air surface.

Relation between surface tension and surface energy

Consider a rectangular frame PQRS of wire, whose arm RS can slide on the arms PR and QS . If this frame is dipped in a soap solution. then a soap film is produced in the frame PQRS in fig. Due to surface tension(T), the film exerts a force on the frame (towards the interior of the film). Let \(l\) be the length of the arm RS, then the force acting on the arm RS towards the film is \(F = T \times 2l\) [Since soap film has two surfaces, that is way the length is taken twice.]
\(\therefore\ \) work done, \(W = Fx = 2\text{ }T/x\)
Increase in potential energy of the soap film.
\(= EA = 2Elx =\) work done in increasing the area \((\Delta W)\)
where \(E =\) surface energy of the soap film per unit area.
According the law of conservation of energy, the work done must be equal to the increase in the potential energy.
\(\therefore\ 2\text{ }Tlx = 2Elx\) or \(T = E = \frac{\Delta W}{A}\)

Thus, surface tension is numerically equal to surface energy or work done per unit increase surface area.

1. Inside a bubble : Consider a soap bubble of radius r . Let p be the pressure inside the bubble and \(p_{a}\) outside. The excess pressure \(= p - p_{a}\). Imagine the bubble broken into two halves, and consider one half of it as shown in fig. Since there are two surface, inner and outer, so the force due to surface tension is

\(F =\) surface tension \(\times\) length \(= T \times 2\) (circumference of the bubble) \(= T \times 2(2\pi r)\)

The excess pressure \(\left( p - p_{a} \right)\) acts on a cross-sectional area \(\pi r^{2}\), so the force due to excess pressure is \(\Rightarrow \ F = \left( p - p_{a} \right)\pi r^{2}\)

The surface tension force given by equation (1) must balance the force due to excess pressure given by equation (2) to maintain the equilibrium, i.e. \(\left( p - p_{a} \right)\pi r^{2} = T \times 2(2\pi r)\)
or

\[\left( p - p_{a} \right) = \frac{4\text{ }T}{r} = p_{\text{excess}\text{~}}\]

above expression can also be obtained by equation of excess pressure of curve surface by putting \(R_{1} = R_{2}\).
2. Inside the drop : In a drop, there is only one surface and hence excess pressure can be written as

\[\left( p - p_{a} \right) = \frac{2\text{ }T}{r} = p_{\text{excess}\text{~}}\]

3. Inside air bubble in a liquid :

\[\left( p - p_{a} \right) = \frac{2\text{ }T}{r} = p_{\text{excess}\text{~}}\]

4. A charged bubble : If bubble is charged, it's radius increases. Bubble has pressure excess due to charge too. Initially pressure inside the bubble

\[= p_{a} + \frac{4\text{ }T}{r_{1}}\]

for charge bubble, pressure inside \(= p_{a} + \frac{4\text{ }T}{r_{2}} - \frac{\sigma^{2}}{2\epsilon_{0}}\), where \(\sigma\) surface is surface charge density. Taking temperature remains constant, then from Boyle's law

\[\left( p_{a} + \frac{4\text{ }T}{r_{1}} \right)\frac{4}{3}\pi r_{1}^{3} = \left\lbrack p_{a} + \frac{4\text{ }T}{r_{2}} - \frac{\sigma^{2}}{2\epsilon_{0}} \right\rbrack\frac{4}{3}\pi r_{2}^{3}\]

From above expression the radius of charged drop may be calculated. It can conclude that radius of charged bubble increases, i.e. \(r_{2} > r_{1}\).

The force of attraction between the molecules of the same substance is called cohesion.
In case of solids, the force of cohesion is very large and due to this solids have definite shape and size. On the other hand, the force of cohesion in case of liquids is weaker than that of solids. Hence liquids do not have definite shape but have definite volume. The force of cohesion is negligible in case of gases. Because of this fact, gases have neither fixed shape nor volume.

Example

(i) Two drops of a liquid Two drops of a liquid coalesce into one when brought in mutual contact because of the cohesive force.
(ii) It is difficult to separate two sticky plates of glass wetted with water because a large force has to be applied against the cohesive force between the molecules of water.
(iii) It is very difficult to break a drop of mercury into small droplets because of large cohesive force between mercury molecules.

The force of attraction between molecules of different substance is called adhesion.

Example

(i) Adhesive force enables us to write on the black board with a chalk.
(ii) Adhesive force helps us to write on the paper with ink.
(iii) Large force of adhesion between cement and bricks helps us in construction work.
(iv) Due to force of adhesive, water wets the glass plate.
(v) Fevicol and gum are used in gluing two surfaces together because of adhesive force.

The angle which the tangent to the liquid surface at the point of contact makes with the solid surface inside the liquid is called angle of contact. Those liquids which we the wall of the container (say in case of water and glass) have meniscus concave upwards and their value of angle of contact is less than \(90^{\circ}\) (also called acute angle). However, those liquids which don't wet the walls of the container (say in case of mercury and glass) have meniscus convex upwards and their value of angle of contact is greater than \(90^{\circ}\) (also called obtuse angle). The angle of contact of mercury with glass about \(140^{\circ}\), whereas the angle of contact of water with glass is about \(8^{\circ}\). But, for pure water, the angle of contact \(\theta\) with glass is taken as \(0^{\circ}\).

Let us now see why the liquid surface bands near the contact with a solid. A liqued in equilibrium cannot sustain tangential stress. The resultant force onany small part of the surface layer must be perpendicular to the surface there. Consider a small part of the liquid surface near its contact with the solid

(a)

(b)

The ;force acting on this part are
(a) \(F_{s}\), attracting due to the molecules of the solid surface near it,
(b) \(F_{p}\), the force due to the liquid molecules near this part, and
(b) W , the weight of the part considered.

The force between the molecules of the same material is known as cohesive force and the force between the molecules of different kinds of material is called adhesive force. Here \(F_{5}\) is adhesive force and \(F_{l}\) is cohesive force.

As is clear from the figure, the adhesive force \(F_{s}\) is perpendicular to the solid surface and is into the solid. The cohesive force \(F_{l}\) is in the liquid, its direction and magnitude depends on the shape of the liquid surface as this determines the distribution of the molecules attracting the part considered. Of course, \(F_{s}\) and \(F_{l}\) depend on the nature of the substance especialy on their densitites.
The direction of the resultant of \(F_{s},{\text{ }F}_{l}\) and W decides the shape of the surface near the constact. The liquid rests in such a way that the surface is through the solid, the surface is concave upward and the liquid rises along the solid. If the resultant passes through the liquid, the surface is convex upwards and the liquid is demressed near the solid.

Let us consider a small element ABCD (fig.) of a curved liquid surface which is convex on the upper side. \(R_{1}\) and \(R_{2}\) are the maximum and minimum radii of curvature respectively. They are called the 'principal radii of curvature' of the surface. Let p be the excess pressure on the concave side.
then \(p = T\left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)\). If instead of a liquid surface, we have a liquid film, the above expression will be \(p = 2\text{ }T\left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)\), because a film has two surface.

Excess of pressure inside a curved surface

1. Plane surface: If the surface of the liquid is plane [as shown in fig. (a)], the molecule on the liquid surface is attracted equally in all directions. The resultant force due to surface tension is zero. The pressure, therefore on the liquid surface is normal.

2. Concave surface : If the surface is concave upward [as shown in fig. (b)], there will be upward resultant force due to surface tension acting on the molecule. Since the molecule on the surface is in equilibrium, there must be an excess of pressure on the concave side in the downward direction to balance the resultant force of surface tension \(p_{A} - p_{B} = \frac{2T}{r}\).

3. Convex surface : If the surface is convex [as shown fig.(c)], the resultant force due to surface tension acts in the downward direction. Since the molecule on the surface are in equilibrium, there must be an excess of pressure on the concave side of the surface acting in the upward direction to balance the downward resultant force of surface tension. Hence there is always in excess of pressure on concave side of a curved surface over that on the convex side.

\[p_{B} - p_{A} = \frac{2\text{ }T}{r}\]

A glass tube of very fine bore throughout the length of the tube is called capillary tube. If the capillary tube is dipped in water, the water wets the inner side of the tube and rises in it [shown in figure (a)]. If the same capillary tube is dipped in the mercury, then the mercury is depressed [shown in figure (b)]. The phenomenon of rise or fall of liquids in a capillary tube is called capillarity.

(b)

Practical applications of capillarity

1. The oil in a lamp rises in the wick by capillary action.

2. The tip of nib of a pen is split up, to make a narrow capillary so that the ink rises upto the tin or nib continuously.

3. Sap and water rise upto the top of the leaves of the tree by capillary action.

4. If one end of the towel dips into a bucket of water and the Other end hangs over the bucket the towel soon becomes wet throughout due to capillary action.

5. Ink is absorbed by the blotter due to capillary action.

6. Sandy soil is more dry than clay. It is because the capillaries between sand particles are not so fine as to draw the water up by capillaries.

7. The moisture rises in the capillaries of soil to the surface, where it evaporates. To preserve the moisture m the soil, capillaries must be broken up. This is done by ploughing and leveling the fields.

8. Bricks are porous and behave like capillaries.

Capillary rise (height of a liquid in a capillary tube) ascent formula

consider the liquid which wets the wall of the tube, forms a concave meniscus shown in figure. Consider a capillary tube of radius r dipped in a liquid of surface tension \(T\) and density \(\rho\). Let \(h\) be the height through which the liquid rises in the tube. Let \(p\) be the pressure on the concave side of the meniscus and \(p_{a}\) be the pressure on the convex side of the meniscus. The excess pressure
\(\left( p - p_{a} \right)\) is given by \(\left( p - p_{a} \right) = \frac{2\text{ }T}{R}\)

Where \(R\) is the radius of the meniscus. Due to this excess pressure, the liquid will rise in the capillary tube till it becomes equal to the hydrostatic pressure hpg. Thus in equilibrium state.
Excess pressure \(=\) Hydrostatic pressure or \(\frac{2\text{ }T}{R} = hpg\)
Let \(\theta\) be the angle of contact and \(r\) be the radius of the capillary tube shown in the fig.
From \(\bigtriangleup OAC,\frac{OC}{OA} = cos\theta\) or \(R = \frac{r}{cos\theta} \Rightarrow h = \frac{2\text{ }Tcos\theta}{r\rho g}\)
The expression is called Ascent formula.

Discussion.

(i) For liquids which wet the glass tube or capillary tube, angle of contact \(\theta < 90^{\circ}\). Hence \(cos\theta =\) positive \(\Rightarrow h =\) positive. It means that these liquids rise in the capillary tube. Hence, the liquids which wet capillary tube rise in the capillary tube. For example, water milk, kerosene oil, patrol etc.

We know, the height through which a liquid rises in the capillary tube of radius \(r\) is given by

\[\therefore\ h = \frac{2\text{ }T}{R\rho\text{ }g}\text{~}\text{or}\text{~}hR = \frac{2\text{ }T}{\rho\text{ }g} = \text{~}\text{constant}\text{~}\]

When the capillary tube is cut and its length is less then h (i.e. \(h^{'}\) ). then the liquid rises upto the top of the tube and spreads in such a way that the radius \(\left( R^{'} \right)\) of the liquid meniscus increases and it becomes more flat so that \(hR = h^{'}R^{'} =\) Constant. Hence the liquid does not overflow.

\[\begin{matrix} \text{~}\text{If}\text{~} & h^{'} < h\text{~}\text{then}\text{~}R^{'} > R & \text{~}\text{or}\text{~} & \frac{r}{cos\theta^{'}} > \frac{r}{cos\theta} \\ \Rightarrow & cos\theta^{'} < cos\theta & \Rightarrow & \theta^{'} > \theta \end{matrix}\]

(i) The wetting property is made use of in detergents and waterproofing. When the detergent materials are added to liquids, the angle of contact decreases and hence the wettability increases. On the other hand, when water proofing material is added to a fabric, it increases the angle of contact, making the fabric water-repellant.
(ii) The antiseptics have very low value of surface tension. The low value of surface tension prevents the formation of drops that may otherwise block the entrance to skin or a wound. Due to low surface tension the antiseptics spreads properly over the wound. The lubricating oils and paints also have low surface tension. So they can spread properly.
(iii) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area.
(iv) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water.
(v) A rough sea can be calmed by pouring oil on its surface.

Effect of temperature and impurities on surface tension

The surface tension of a liquid decreases with the rise in temperature and vice versa. According to Ferguson, \(T = T_{0}\left( 1 - \frac{\theta}{\theta_{C}} \right)^{n}\) where \(T_{0}\) is surface tension at \(0^{\circ}C,\theta\) is absolute temperature of the liquid, \(\theta_{C}\) is the critical temperature and n is a constant varies slightly from liquid and has means value 1.21. This formula shows that the surface tension becomes zero at the critical temperature, that is why machinery parts get jammed in winter.
The surface tension of a liquid change appreciably with addition of impurities. For example, surface tension of water increases with addition of highly soluble substances like \(NaCl,{ZnSO}_{4}\) etc. On the other hand surface tension of water gets reduced with addition of sparingly soluble substances like phenol, soap etc.

Viscosity

When a layer of a fluid slips or tens to slip on another layer in contact, the two layers exert tangential forces on each other. The directions are such that the relative motion between the layers is opposed. this property of a fluid to oppose relative motion between its layers is called viscosity. The forces between the layers opposing relative motion between them are known as the forces of viscosity. Thus, viscosity may be thought of as the internal friction of a fluid in motion.
If a solid surface is kept in contact with a fluid and is moved, forces of viscosity appear between the solid surface and the fluid layer in contact. the fluid in contact is dragged with the solid. If the viscosity is sufficient, the layer moves with the solid and there is no relative slipping. When a boat moves slowly on
the water of a calm river, the water in contact with the boat is dragged with it, whereas the water in contact with the bed of the river remains at rest. Velocities of different layers are different. Let \(v\) be the velocity of the layer at a distance \(z\) from the bed and \(v + dv\) be the velocity at a distance \(z + dz\) (figure).

Thus, the velocity differs by dv in going through a distance dz perpendicular to it. The quantity \(dv/dz\) is called the velocity gradient.
The force of viscosity between two layers of a fluid is proportional to the velocity gradient in the direction perpendicular to the layers. Also the force is proportional to the area of the layer.
Thus, if F is the force exerted by a layer of area A on a layer in contact,

\[\begin{matrix} & F\alpha\text{ }A\text{~}\text{and}\text{~}F\alpha dv/dz \\ & \text{ }F = - \eta Adv/dz \end{matrix}\]

or,
The negative sign is included as the force is frictional in nature and opposes relative motion. The constant of proportionality \(\eta\) is called the coefficient of viscosity.
The SI unit of viscosity can be easily worked out from equation. It is \(N - s/m^{2}\). However, the corresponding CGS unit dyne-s/cm \(\ ^{2}\) is in common use and is called a poise in honour of the French scientist Poiseuille. We have

\[1\text{~}\text{poise}\text{~} = 0.1\text{ }N - s/m^{2}\]

Terminal velocity

The viscous force on a solid moving through a fluid is proportional to its velocity. When a solid is dropped in a fluid, the forces acting on it are
(a) weight W acting vertically downward,
(b) the viscous force F acting vertically upward and
(c) the buoyancy force B acting vertically upward.

The weight W and the buoyancy B are constant but the force F is proportional to the velocity v , initially, the velocity and hence the viscous force \(F\) is zero and the solid is accelerated due to the force \(W - B\). Because of the acceleration, the velocity increases. Accordingly, the viscous force also increases. At a certain instant the viscous force becomes equal to W-B. the net force then becomes zero and the solid falls with constant velocity. This constant velocity is known as the terminal velocity.
Consider a spherical body falling through a liquid. Suppose the density of the body \(= \rho\), density of the liquid \(= \sigma\), radius of the sphere \(= r\) and the terminal velocity \(= v_{0}\). The viscous force is

\[F = 6\pi\eta{rv}_{0}\]

the weight

\[W = \frac{4}{3}\pi r^{3}\rho\text{ }g\]

and the buoyancy force

\[B = \frac{4}{3}\pi r^{3}\sigma\text{ }g\]

We have
or

\[\begin{matrix} 6\pi\eta{rv}_{0} = W - B = \frac{4}{3}\pi r^{3}\rho\text{ }g - \frac{4}{3}\pi r^{3}\sigma\text{ }g \\ v_{0} = \frac{2r^{2}(\rho - \sigma)g}{9\eta} \end{matrix}\]

Q. 1 A conical glass capillary tube \(A\) of length 0.1 m has diameters \(10^{- 3}\text{ }m\) and \(5 \times 10^{- 4}\text{ }m\) at the ends. When it is just immersed in a liquid at \(0^{\circ}C\) with larger in contact with it, te liquid rises to \(8 \times 10^{- 2}\text{ }m\) in the tube. In another cylindrical capillary tube B, when immersed in the ame liquid at \(0^{\circ}C\), the liquid rises to \(6 \times 10^{-}\ ^{2}\text{ }m\) height. The rise of liquid in tube B is only \(5.5 \times 10^{- 2}\text{ }m\) when the liquid is at \(50^{\circ}C\). Find the rate at which the surface tension chages with temperature considering the chage to be linear. The density of liquid is \((1/14) \times 10^{4}\text{ }kg/m^{3}\) and the angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible.
Sol. The situation is shown if figure.
Let \(r_{1}\) and \(r_{2}\) be radii of upper and lower ends of the conical capillary tube. The radius \(r\) at the meniscus is given by

\[\begin{matrix} & r = r_{1} + \left( r_{2} - r_{1} \right)\left( \frac{l - h}{l} \right) \\ & \ = \left( 2.5 \times 10^{- 4} \right) + \left( 2.5 \times 10^{- 4} \right)\left( \frac{0.1 - 0.08}{0.1} \right) \\ & \ = 3.0 \times 10^{- 4}\text{ }m \end{matrix}\]

The surface tension at \(0^{\circ}C\) is given by

\[\begin{matrix} & T_{0} = \frac{rh\rho\text{ }g}{2} \\ & \ = \frac{\left( 3.0 \times 10^{- 4} \right)\left( 8 \times 10^{- 2} \right)\left( 1/4 \times 10^{4} \right)(9.8)}{2} \\ & \ = 0.084\text{ }N/m \end{matrix}\]

For tube B

\[\frac{T_{0}}{{\text{ }T}_{50}} = \frac{h_{0}}{{\text{ }h}_{50}} = \frac{6 \times 10^{- 3}}{5.5 \times 10^{- 3}} = \frac{12}{11}\]

or \(\ T_{50} = \frac{11}{12} \times T_{0} = \frac{11}{12} \times 0.084 = 0.077\text{ }N/m\)
Considering the change in surface tension as linear, the change ; in surface tension with temperature is given by

\[\begin{matrix} & \alpha = \frac{T_{50} - T_{0}}{{\text{ }T}_{0}{\text{ }T}_{50}} = \frac{0.077 - (0.084)}{0.084 \times 0.077} \\ & \ = \frac{1}{60}\text{~}\text{per}\text{~}K. \end{matrix}\]

Q. 2 A non-viscous liquid of constant density \(1000\text{ }kg/m^{3}\) flows in a streanline motion along a tube of variable cross-section. The tube is kept inclined inthe vertical plane as shown in the figure.the area of crosssectionof the tube at two points \(P\) and \(Q\) at heights of 2 mater and 5 meter are respectively \(4 \times 10^{- 3}{\text{ }m}^{2}\) and \(8 \times 10^{- 3}{\text{ }m}^{2}\). The velocity of the liquid at point P is \(1\text{ }m/s\). Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P and Q .

Sol. As gravitational field is conservative i.e., \(W = - U\)
So, \(\ \left( \frac{dW}{dV} \right)_{g} = - \frac{dU}{dV} = - \frac{mg\left( h_{2} - h_{1} \right)}{V} = - \rho g\left( h_{2} - h_{1} \right)\)
So work done by the force of gravity per unit volume

\[\begin{array}{r} \left( \frac{dW}{dV} \right)_{g} = \rho g\left( {\text{ }h}_{2} - h_{1} \right) = - 10^{3} \times 9.8(5 - 2) = - 2.94 \times 10^{4}\frac{\text{ }J}{{\text{ }m}^{3}}\#(i) \end{array}\]

Now in case of ideal fluid motion by conservation of mass, i.e.

\[\left( \frac{dm}{dt} \right)_{1} = \left( \frac{dm}{dt} \right)_{2}\ \text{~}\text{and}\text{~}\ (\rho Av)_{1} = (\rho Av)_{2}\]

or \(\ (\rho Av)_{1} = (\rho Av)_{2}\ \lbrack as\rho = constant(\) given \()\rbrack\)
so \(\ v_{2} = \frac{A_{1}v_{1}}{A_{2}} = \frac{4 \times 10^{- 3} \times 1}{8 \times 10^{- 3}} = \frac{1}{2}\text{ }m/s\)

Now as work done per unit volume by pressure,

\[\left( \frac{dW}{dV} \right)_{P} = \frac{PdV}{dV} = P = \left( p_{1} - p_{2} \right)\ \lbrack\text{~}\text{as}\text{~}dW = PdV\rbrack\]

But by Bernoulli's theorem,

\[p_{1} + \rho{gh}_{1} + \frac{1}{2}\rho v_{1}^{2} = p_{2} + {rgh}_{2} + \frac{1}{2}\rho v_{2}^{2}\]

so \(\ \left( \frac{dW}{dV} \right)_{P} = \left( p_{1} - p_{2} \right) = \rho g\left( h_{2} - h_{1} \right) + \frac{1}{2}\rho\left( v_{2}^{2} - v_{1}^{2} \right)\)

Using (i), (ii) and (iii) we get

\[\left( \frac{dW}{dV} \right)_{P} = 2.94 \times 10^{4} + \frac{1}{2} - 10^{3}\left\lbrack (0.5)^{2} - 1^{2} \right\rbrack = 29025\text{ }J\]

Q. 3 A cylindrical tank I m in radius rests on a platform 5 m high. Initially the tank is filled with water up to a height of 5 m . A plug whose area is \(10^{- 4}{\text{ }m}^{2}\) is removed from an orifice on the side of the tank at the bottom. Calculate (a) initial speed with which the water flows from the orifice
(b) initial speed with which the water strikes the ground (c) time taken to empty the tank to half its original value (d) Does the time to empty the tank depend upon the height of stand.
Sol. (a) As speed of efflux is given by

\[v_{H} = \sqrt{(2gh)}\text{~}\text{so here}\text{~}u = \sqrt{2 \times 10 \times 5} = 10\text{ }m/s\]

(b) As vertical speed with which water strikes the gournd,

\[v_{v} = \sqrt{v_{H}^{2} + v_{v}^{2}} = 10\sqrt{2} = 14.1\text{ }m/s\]

(c) When the height of water level above the hole is \(y\), velocity of flow will be \(v = \sqrt{2\text{~}\text{gy}\text{~}}\) and so rate flow

\[\begin{matrix} & \frac{dV}{dt} = A_{0}v = A_{0}\sqrt{2gy} \\ & \text{~}\text{or}\text{~} - Ady = (\sqrt{2gu})A_{0}dt\ \lbrack\text{~}\text{as}\text{~}dV = - Ady\rbrack \end{matrix}\]

Which on integration gives

\[t = \frac{A}{{\text{ }A}_{0}}\sqrt{\frac{2}{\text{ }g}}\left\lbrack \sqrt{H} - \sqrt{H^{'}} \right\rbrack\]

So

\[t = \frac{\pi \times 1^{2}}{10^{- 4}}\sqrt{\frac{2}{10}}\lbrack\sqrt{5} - \sqrt{(5/2)}\rbrack = 9.2 \times 10^{3}\text{ }s = 2.5\text{ }h\]

(d) No, as expression of t is independent of height of stand.
Q. 4 Under isothermal condition two soap bubbles of radii \(a\) and \(b\) coalesce to from a single bubble of radius
c. If the external pressure is \(p_{0}\) show that surface tension,

\[T = \frac{p_{0}\left( c^{3} - a^{3} - b^{3} \right)}{4\left( a^{2} + b^{2} - c^{2} \right)}\]

Sol. As excess pressure for a soap bubble is \((4\text{ }T/r)\) and exteranal pressure \(p_{0}\),
so

\[\begin{matrix} & & p_{i} = p_{0} + (4\text{ }T/r) \\ & p_{a} = \left\lbrack p_{0} + \frac{4\text{ }T}{a} \right\rbrack,\ p_{b} = \left\lbrack p_{0} + \frac{4\text{ }T}{\text{ }b} \right\rbrack\ \text{~}\text{and}\text{~}\ p_{c} = \left\lbrack p_{0} + \frac{4\text{ }T}{c} \right\rbrack & \text{(i)} \end{matrix}\]

and

\[\begin{array}{r} V_{a} = \frac{4}{3}\pi a^{3},\ {\text{ }V}_{b} = \frac{4}{3}\pi{\text{ }b}^{3}\ \text{~}\text{and}\text{~}\ V_{c} = \frac{4}{3}\pi c^{3}\#(ii) \end{array}\]

Now as mass is conserved,
i.e., \(\frac{p_{a}V_{a}}{R_{a}} + \frac{p_{b}V_{b}}{R_{b}} = \frac{p_{c}V_{c}}{R_{c}}\ \left\lbrack \right.\ \) as \(PV = \mu RT\), i.e., \(\left. \ \mu = \frac{pV}{RT} \right\rbrack\)
As temp, is constant, i.e., \(T_{a} = T_{b} = T_{c}\), the above expression reduces to

\[p_{a}{\text{ }V}_{a} + p_{b}{\text{ }V}_{b} = p_{c}{\text{ }V}_{c}\]

Which in the light of Eqn. (i) and (ii) becomes

\[\left\lbrack p_{0} + \frac{4\text{ }T}{a} \right\rbrack\left\lbrack \frac{4}{3}\pi a^{3} \right\rbrack + \left\lbrack p_{0} + \frac{4\text{ }T}{\text{ }b} \right\rbrack\left\lbrack \frac{4}{3}\pi{\text{ }b}^{3} \right\rbrack = \left\lbrack p_{0} + \frac{4\text{ }T}{c} \right\rbrack\left\lbrack \frac{4}{3}\pi c^{3} \right\rbrack\]

i.e., \(\ 4\text{ }T\left( a^{2} + b^{2} - c^{2} \right) = p_{0}\left( c^{3} - a^{3} - b^{3} \right)\)
i.e., \(\ T = \frac{p_{0}\left( c^{3} - a^{3} - b^{3} \right)}{4\left( a^{2} + b^{2} - c^{2} \right)}\).
Q. 5 The fresh water behind a reservoir dam is 15 m deep. A horizontal pipe 4.0 cm in diameter passed through the dam 6.0 m below the water surface as shown in figure. A plug secures the pipe opening. (a) Find the friction force between theplug and pipe wall. (b) The plug is remove. What volume of water flows out of the pipe in 3.0 hour?
Sol. (a) As the plug secures the pipe opening, the force of friction between plug and pipe wall.

\[F = A\left( p_{2} - p_{1} \right)\]

But \(p_{1} = p_{0}\) and \(p_{2} = p_{0} + h\rho g\)
so \(\ F = Ah\rho g\)
i.e., \(\ F = \pi \times \left( 2 \times 10^{- 2} \right)^{2} \times 6510^{3} \times 9.8 = 74\text{ }N\)
(b) As the velocity of efflux.

\[v = \sqrt{2gh} = \sqrt{2 \times 10 \times 6} = 11\text{ }m/s\]

so assuming the level of water in the tank to be constant (i.e., area \(= \infty\) ) as it is not given the volume coming out per second will be

\[R = \frac{dV}{dt} = A_{0}v = \pi\left( 2 \times 10^{- 2} \right)^{2} \times 11{\text{ }m}^{3}/s\]

so the volume of the water flowing through the pipe in 3 hours

\[V = R \times t = 44 \times 3.14 \times 10^{- 4} \times 3 \times (60 \times 60) = 150{\text{ }m}^{3}.\]

Q. 6 A cylindrical vessel of base are A has a small hole of cross-section 'a' punched near its base. At time \(t =\) 0 , water is supplied into the vessel at a constant rate ' \(\alpha\) ' \(m^{3}/s\). Find
(a) The maximum water level \(h_{\text{max}\text{~}}\) in the vessel
(b) The time ' t ' when water level becomes \(h\left( < h_{\text{max}\text{~}} \right)\).

Sol. (a) Water level will have maximum height when inflow rate \(=\) outflow rate and there will be no futher change in level.
\(\therefore\ \alpha =\) av
or, \(\alpha = a\sqrt{2{gh}_{\text{max}\text{~}}}\ \lbrack\because v = \sqrt{2gh}\rbrack\)
or, \(\ h_{\max} = a^{2}/2{ga}^{2}\)
(b) Let the water level by y at time t .
\[\therefore\ A\left( \frac{dy}{dt} \right) = \alpha - av = \alpha - \sqrt{2gy}\]

Here \(\frac{dy}{dt}\) is positive as \(y\) increase with time instantaneous effux velocity \(v = \sqrt{2gy}\).
Rearranging the above equation

\[\int_{0}^{h}\mspace{2mu}\frac{dy}{\alpha - \alpha\sqrt{2gy}} = \frac{1}{\text{ }A}\int_{0}^{t}\mspace{2mu} dt\]

Integration under the given limits, we get the required time

\[t = \frac{A}{ag}\left\lbrack \frac{\alpha}{a}In\left( \frac{\alpha - a\sqrt{agh}}{\alpha} \right) - \sqrt{2gh} \right\rbrack\]

This gives the time t as a function of h . For any valume \(h\left( \leq h_{\max} \right)\), the corresponding time t and be evaluated.
Q. 7 A cylindrical vessel of (radius r ) containing a liquid spins continously with constant angular velocity \(\omega\) as shown in the figure. Shown that the pressure at a radial distance \(r\) from the axis is

\[P = P_{0} + \frac{1}{2}\rho\omega^{2}r^{2}\]

where \(P_{0} =\) atmospheric pressure.

Sol. Consider a particel of the fluid at a point \(P(x,y)\) w.r.t. the corrdinate axes as shown in the figure. The force acting on This particle are \(m\omega^{2}x\) (the centrifugal force) and the weight \(mg\).
The net force F acting at P should be perpendicular to the free surface, so that

\[tan\theta = \frac{m\omega^{2}x}{mg} = \frac{\omega^{2}x}{g}\]

or,

\[\frac{dy}{dx} = \frac{m\omega^{2}}{\text{ }g}\ \left\lbrack \because\text{~}\text{slope}\text{~} = tan\theta = \frac{dy}{dx} \right\rbrack\]

or,

\[y = \frac{\omega^{2}}{2g}x^{2}\]

This equation represents a parabola; for which the elevation from origin at \(x = r\) will be \(y = \frac{\omega^{2}}{2\text{ }g}r^{2}\).

\[\therefore\ \text{~}\text{Pressure}\text{~}P(r) = P_{0} + \rho gy = P_{0} + \frac{\rho\omega^{2}r^{2}}{2}\]

Q. 8 Avertical U-tube of uniform cross-section contains mercury in both arms. Aglycerine (relative density=1.3) column of length 10 cm is introduced into one of the arms. Oil of density \(800\text{ }kg{\text{ }m}^{- 3}\) is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is \(13.6 \times 10^{3}{kgm}^{- 3}\).
Sol. Pressure at A and B must be same
Pressure at \(A = p_{0} + 0.1 \times (1.3 \times 1000) \times g\)
where \(p_{0} =\) atmospheric pressure
Pressure at \(B = p_{0} + h \times 800 \times g + (0.1 - h) \times 13.6 \times 1000g\)

\[\begin{matrix} \therefore\ & p_{0} + 0.1 \times 1300 \times g \\ & = p_{0} + 800gh + 1360\text{ }g - 13600 \times g \times h \end{matrix}\]

or \(\ h = 9.6\text{ }cm\)
Q. 9 For the arrangement shown in the figure, what is the density of oil?

Sol. \(p_{\text{surface}\text{~}} = p_{0} + \rho_{w}.gl\)
\[{p_{\text{surface}\text{~}} = p_{0} + \rho_{\text{oil}\text{~}}(l + d)g }{\Rightarrow \rho_{\text{oil}\text{~}} = \frac{\rho_{\omega} \cdot l}{(l + d)} = \frac{1000(135)}{(135 + 12.3)} = 916\text{ }kg/m^{3} }\]Q. 10 A pipe of copper having an internal cavity weighs 264 gm in air and 221 gm in water. Find the volume of the cavity. [Density of copper is \(8.8gm/cc\).]
Sol. The buoyant force on the copper piece, \(F = V\rho g\)
Hence, volume of the copper piece \(V = \frac{F}{\sigma g} = \frac{(264 - 221)g}{1 \times g} = 43cc\)
The volume of the material of the copper piece

\[V_{0} = \frac{\text{~}\text{mass of copper piece}\text{~}}{\text{~}\text{density of material}\text{~}} = \frac{264}{8.8} = 30cc\]

Hence, volume of the cavity \(= V - V_{0} = 43 - 30 = 13cc\)
Q. 11 A piece of brass (alloy of copper and zinc) weighs 12.9 gm in air. When completely immersed in water, it weighs 11.3 gm . What is the mass of copper contained in the alloy?
[Specific gravities of copper and zinc are 8.9 and 7.1, respectively.]
Sol. Let the mass of copper in alloy \(= x\) gm.
\(\therefore\ \) Amount of zinc \(= (12.9 - x)gm\)
Volume of copper, \(V_{Cu} = \frac{X}{\rho_{Cu}} = \frac{X}{8.9}\) and

Volume of zinc \(V_{Zn} = \frac{(12.9 - x)}{7.1}\)
\(\therefore\ \) Total volume of the alloy, \(V = V_{Cu} + V_{Zn}\)
or \(\ V = \frac{(12.9 - x)}{7.1} + \frac{x}{8.9}\)

Buoyant force \(F = V_{\rho}g =\) loss of weight

\[= (12.9 - 11.3)g = 1.6\text{ }g\]

Substituting the value of V in equation (i), we get

\[x = 7.6gm\]

Q. 12 A cubical block of iron of edge 5 cm is floating on mercury in a vessel.
(a) What is the height of the block above mercury level?
(b) Water is poured into the vessel so that it just covers the iron block. What is the height of the water column?
[Relative density of \(Hg = 13.6\) and that of \(Fe = 7.2\) ]

Sol. (a) Let h be the height of the iron block above mercury.
In case of flotation,
Weight of the block = buoyant force
i.e., \(x^{3}\rho g = \lbrack(x - h)\sigma g\rbrack x^{2}\)
or \(h = x\left( 1 - \frac{\rho}{\sigma} \right) = 5\left( 1 - \frac{7.2}{13.6} \right) = 2.35\text{ }cm\)

(b) Let y be the height of the water level.

For equilibrium of the block,
\[{x^{3}\rho g = \left\lbrack \sigma_{w}gy + \sigma_{Hg}g(x - y) \right\rbrack x^{2} }{x\sigma = (x - y)\sigma_{Hg} + y\sigma_{w} }\]

or \(\ y = x\left( \frac{\sigma_{Hg} - \sigma}{\sigma_{Hg} - \sigma_{w}} \right) = 5\left( \frac{13.6 - 7.2}{13.6 - 1} \right) = 2.54\text{ }cm\)
Q. 13 The 'tip of the iceberg' in popular speech has come to mean a small visible fraction of something that is mostly hidden. For real icebergs, what is this fraction? \(\left( \rho_{\text{ice}\text{~}} = 917\text{ }kg/m^{3},\rho_{\text{sea water}\text{~}} = 1024\text{ }kg/m^{3} \right)\)
Sol. \(\ W_{\text{ice}\text{~}} = \rho_{i}V_{i}g,W_{\text{sea water}\text{~}} = \rho_{w}V_{w}g\)
For floatation, \(\rho_{i}V_{i}g = \rho_{w}V_{w}g\)
Fraction of the volume submerged \(= \frac{V_{i} - V_{w}}{V_{i}} = 1 - \frac{917}{1024} = \frac{107}{1024} = 10.45\%\)
Q. 14 A bent tube is lowered into the stream as shown. The velocity of the stream relative to the tube is equal to V . The closed upper end of the tube is located at height \(h_{0}\). To what height h will the water jet spurt?

Sol. Let tube's entrance be a depth ' \(y\) ' below the surface. Take point 1 at entry and point 2 at the maximum height of the fountain. This is a tube of flow. Now let's apply Bemoulli's theorem,

\[P_{1} + \rho gh_{1} + \frac{1}{2}\rho v_{1}^{2} = P_{2} + \rho gh_{2} + \frac{1}{2}\rho v_{2}^{2}\]

Taking, \(h_{1} = 0,\ {\text{ }h}_{2} = \left( y + h_{0} + h \right),V_{1} = V,V_{2} = 0\)

\[P_{1} = P_{0} + \rho gy,P_{2} = P_{0},\]

Substituting, \(P_{0} + \rho gy + \rho g \times 0 + \frac{1}{2}\rho{\text{ }V}^{2} = P_{0} + \rho g\left( y + h_{0} + h \right) + \frac{1}{2}\rho \times 0^{2}\)
\(\Rightarrow \frac{1}{2}\rho{\text{ }V}^{2} = \rho g\left( h_{0} + h \right)\) or \(h = \left( \frac{V^{2}}{2\text{ }g} - h_{0} \right)\)
Q. 15 Water enters a house through a pipe with inlet diameter of 2.0 cm at an absolute pressure of \(4.0 \times 10^{5}\) Pa (about 4 atm ). A 1.0 cm diameter pipe leads to the second floor bathroom 5.0 above. When flow speed at the inlet pipe is \(1.6\text{ }m/s\), find the flow speed, pressure and volume flow rate in the bathroom.
[Sol. Let point 1 and 2 be at the inlet pipe and the bathroom, then from continuity equation

\[a_{1}v_{1} = a_{2}v_{2}\ \Rightarrow v_{2} = 6.0\text{ }m/s\]

Now, applying Bernoulli's equation at the inlet \((y = 0)\) and at the bathroom \(\left( y_{2} = 5.0\text{ }m \right)\).
As \(p + \frac{1}{2}\sigma v^{2} + \sigma gy =\) constant
Hence, \(p_{2} - p_{1} = \frac{1}{2}p\left( v_{2}^{2} - v_{1}^{2} \right) - \rho g\left( y_{2} - y_{1} \right)\)

Which gives \(p_{2} = 3.3 \times 10^{5}\text{ }Pa\)

The volume flow rate \(= A_{2}v_{2} = A_{1}v_{1} = \frac{\pi}{4}(0.1)^{2}6 = 4.7 \times 10^{- 4}{\text{ }m}^{3}/s\). ]
Q. 16 Water coming out the jet having across sectional area a , with a speed v strikes a stationary plate and stops after striking. Find the force exerted by the water jet on the plate.
Sol. The change of momentum of water in time \(dt = 0 - \rho av^{2}dt\widehat{i} = - \rho av^{2}dt\widehat{i}\) where \(\widehat{i}\) is a unit vector in the direction of the velocity of the jet. The rate of change of momentum of water jet \(= - \rho av^{2}\widehat{i}\)
Thus the force exerted on the water jet by the plate \(= - \rho av^{2}\widehat{i}\)
The force exerted on the plate by the water jet \(= \rho{av}^{2}\widehat{i}\).
Q. 17 What is the surface energy of an air bubble inside a soap solution?

Sol. \(\ E = T \times A = 4\pi r^{2}T\), as it has only one surface.
Q. 18 A metal plate \(0.04{\text{ }m}^{2}\) in area is lying on liquid layer of thickness \(10^{- 3}\text{ }m\) and co-efficient of viscosity 140 poise. Calculate the horizontal force needed to move the plate with a speed of \(0.040\text{ }m/s\).
Sol. Area of the plate, \(A = 0.04{\text{ }m}^{2}\)
Thickness, \(\Delta x = 10^{- 3}\text{ }m\)
\(\Delta x\) is the distance of the free surface with respect to the fixed surface.
Velocity gradient, \(\frac{\Delta v}{\Delta x} = \frac{0.04}{10^{- 3}} = 40{\text{ }s}^{- 1}\)
Co-efficient of viscosity, \(\eta = 14\text{ }kg{\text{ }ms}^{- 1}{\text{ }s}^{- 1}\)
Let F be the required force.
Then, \(F = \eta A\frac{\Delta v}{\Delta x} = 22.4\text{ }N\)