Chapter 10

MECHANICAL PROPERTIES OF FLUIDS

MCQ I

10.1 A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in Fig. 10.1, indicate the one that represents the velocity $(v)$ of the pebble as a function of time $(t)$.

(a)

(b)

(c)

(d)

Fig. 10.1

10.2 Which of the following diagrams (Fig. 10.2) does not represent a streamline flow?

Fig. 10.2

10.3 Along a streamline

(a) the velocity of a fluid particle remains constant.

(b) the velocity of all fluid particles crossing a given position is constant.

(c) the velocity of all fluid particles at a given instant is constant.

(d) the speed of a fluid particle remains constant.

10.4 An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters $2.5 \mathrm{~cm}$ and $3.75 \mathrm{~cm}$. The ratio of the velocities in the two pipes is

(a) $9: 4$

(b) $3: 2$

(c) $\sqrt{3}: \sqrt{2}$

(d) $\sqrt{2}: \sqrt{3}$

10.5 The angle of contact at the interface of water-glass is $0^{\circ}$, Ethylalcohol-glass is $0^{\circ}$, Mercury-glass is $140^{\circ}$ and Methyliodideglass is $30^{\circ}$. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is

(a) water

(b) ethylalcohol

(c) mercury

(d) methyliodide.

MCQ II

10.6 For a surface molecule

(a) the net force on it is zero.

(b) there is a net downward force.

(c) the potential energy is less than that of a molecule inside.

(d) the potential energy is more than that of a molecule inside.

10.7 Pressure is a scalar quantity because

(a) it is the ratio of force to area and both force and area are vectors.

(b) it is the ratio of the magnitude of the force to area.

(c) it is the ratio of the component of the force normal to the area.

(d) it does not depend on the size of the area chosen.

10.8 A wooden block with a coin placed on its top, floats in water as

shown in Fig.10.3.

The distance $l$ and $h$ are shown in the figure. After some time the coin falls into the water. Then

(a) $l$ decreases.

(b) $h$ decreases.

(c) $l$ increases.

(d) $h$ increase.

Fig. 10.3

10.9 With increase in temperature, the viscosity of

(a) gases decreases.

(b) liquids increases.

(c) gases increases.

(d) liquids decreases.

10.10 Streamline flow is more likely for liquids with

(a) high density.

(b) high viscosity.

(c) low density.

(d) low viscosity.

VSA

10.11 Is viscosity a vector?

10.12 Is surface tension a vector?

10.13 Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is $\rho_{i}=$ $0.917 \mathrm{~g} \mathrm{~cm}^{-3}$ ?

10.14 A vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass $M$ and density $\rho$ is suspended by a massless spring of spring constant $k$. This block is submerged inside into the water in the vessel. What is the reading of the scale?

10.15 A cubical block of density $\rho$ is floating on the surface of water. Out of its height $L$, fraction $x$ is submerged in water. The vessel is in an elevator accelerating upward with acceleration $a$. What is the fraction immersed?

SA

10.16 The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r=2.5 \times 10^{-5} \mathrm{~m}$. The surface tension of sap is $T=7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?

10.17 The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta$. If the acceleration is $a \mathrm{~m} \mathrm{~s}^{-2}$, what will be the slope of the free surface?

10.18 Two mercury droplets of radii $0.1 \mathrm{~cm}$. and $0.2 \mathrm{~cm}$. collapse into one single drop. What amount of energy is released? The surface tension of mercury $T=435.5 \times 10^{-3} \mathrm{~N} \mathrm{~m}^{-1}$.

10.19 If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.

10.20 The sufrace tension and vapour pressure of water at $20^{\circ} \mathrm{C}$ is $7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and $2.33 \times 10^{3} \mathrm{~Pa}$, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{\circ} \mathrm{C}$ ?

LA

10.21 (a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure $\mathrm{d} p$ over a differential height $\mathrm{d} h$ ?

(b) Considering the pressure $p$ to be proportional to the density, find the pressure $p$ at a height $h$ if the pressure on the surface of the earth is $p_{0}$.

(c) If $p_{0}=1.03 \times 10^{5} \mathrm{~N} \mathrm{~m}^{-2}, \rho_{0}=1.29 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$, at what height will the pressure drop to $(1 / 10)$ the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

10.22 Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water $L_{v}=540 \mathrm{k} \mathrm{cal} \mathrm{kg}^{-1}$, the mechanical equivalent of heat $\mathrm{J}=4.2 \mathrm{~J} \mathrm{cal}^{-1}$, density of water $\rho_{\mathrm{w}}=10^{3} \mathrm{~kg} t^{-1}$, Avagadro’s No $N_{\mathrm{A}}=6.0 \times 10^{26} \mathrm{k} \mathrm{mole}^{-1}$ and the molecular weight of water $M_{\mathrm{A}}=18 \mathrm{~kg}$ for $1 \mathrm{kmole}$.

(a) estimate the energy required for one molecule of water to evaporate.

(b) show that the inter-molecular distance for water is $d=\left[\frac{M_{\mathrm{A}}}{N_{\mathrm{A}}} \times \frac{1}{\rho_{w}}\right]^{1 / 3}$ and find its value.

(c) $1 \mathrm{~g}$ of water in the vapor state at $1 \mathrm{~atm}$ occupies $1601 \mathrm{~cm}^{3}$. Estimate the intermolecular distance at boiling point, in the vapour state.

(d) During vaporisation a molecule overcomes a force $F$, assumed constant, to go from an inter-molecular distance $d$ to $d^{\prime}$. Estimate the value of $F$.

(e) Calculate $F / d$, which is a measure of the surface tension.

10.23 A hot air balloon is a sphere of radius $8 \mathrm{~m}$. The air inside is at a temperature of $60^{\circ} \mathrm{C}$. How large a mass can the balloon lift when the outside temperature is $20^{\circ} \mathrm{C}$ ? (Assume air is an ideal gas, $R=8.314 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}, 1 \mathrm{~atm} .=1.013 \times 10^{5} \mathrm{~Pa}$; the membrane tension is $5 \mathrm{~N} \mathrm{~m}^{-1}$.)

Solutions 10

10.1 (c)

10.2 (d)

10.3 (b)

10.4 (a)

10.5 (c)

10.6 (a), (d)

10.7 (c), (d)

10.8 (a), (b)

10.9 (c), (d)

10.10 (b), (c)

10.11 No.

10.12 No.

10.13 Let the volume of the iceberg be $V$. The weight of the iceberg is $\rho_{i} V g$. If $x$ is the fraction submerged, then the volume of water displaced is $x V$. The buoyant force is $\rho_{\mathrm{w}} x V g$ where $\rho_{\mathrm{w}}$ is the density of water.

$\rho_{i} V g=\rho_{\mathrm{w}} x V g$

$\therefore x=\frac{\rho_{i}}{\rho_{\mathrm{w}}}=0.917$

10.14 Let $x$ be the compression on the spring. As the block is in equilibrium $M g-\left(k x+\rho_{w} V g\right)=0$ where $\rho_{\mathrm{w}}$ is the density of water and $V$ is the volume of the block. The reading in the pan is the force applied by the water on the pan i.e.,

$m_{\text {vessel }}+m_{\text {water }}+\rho_{\mathrm{w}} V g$.

Since the scale has been adjusted to zero without the block, the new reading is $\rho_{\mathrm{w}} V g$.

10.15 Let the density of water be $\rho_{w}$.

Then $\rho a L^{3}+\rho L^{3} g=\rho_{\mathrm{w}} x L^{3}(g+a)$

$\therefore x=\frac{\rho}{\rho_{\mathrm{w}}}$

Thus, the fraction of the block submerged is independent of any acceleration, whether gravity or elevator.

10.16 The height to which the sap will rise is

$h=\frac{2 T \cos 0^{\circ}}{\rho g r}=\frac{2\left(7.2 \times 10^{-2}\right)}{10^{3} \times 9.8 \times 2.5 \times 10^{-5}} \quad 0.6 \mathrm{~m}$

This is the maximum height to which the sap can rise due to surface tension. Since many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.

10.17 If the tanker acclerates in the positive $x$ direction, then the water will bulge at the back of the tanker. The free surface will be such that the tangential force on any fluid parcel is zero.

Consider a parcel at the surface, of unit volume. The forces on the fluid are

$-\rho g \hat{\mathbf{y}}$ and $\quad-\rho a \hat{\mathbf{x}}$

The component of the weight along the surface is $\rho g \sin \theta$

The component of the acceleration force along the surface is

$\rho a \cos \theta$

$\therefore \rho g \sin \theta=\rho a \cos \theta$

Hence, $\tan \theta=a / g$

10.18 Let $v_{1}$ and $v_{2}$ be the volume of the droplets and $v$ of the resulting drop.

Then $v=v_{1}+v_{2}$

$\Rightarrow r^{3}=r_{1}^{3}+r_{2}^{3}=(0.001+0.008) \mathrm{cm}^{3}=0.009 \mathrm{~cm}^{3}$

$\therefore r \quad 0.21 \mathrm{~cm}$

$\therefore \Delta U=4 \pi T\left(r^{2}-\left(r_{1}^{2}+r_{2}^{2}\right)\right)$

$=4 \pi \times 435.5 \times 10^{-3}\left(0.21^{2}-0.05\right) \times 10^{-4} \mathrm{~J}$

$$ -32 \times 10^{-7} \mathrm{~J} $$

10.19 $R^{3}=N r^{3}$

$\Rightarrow r=\frac{R}{N^{1 / 3}}$

$\Delta U=4 \pi T\left(R^{2}-N r^{2}\right)$

Suppose all this energy is released at the cost of lowering the temperature. If $s$ is the specific heat then the change in temperature would be,

$$ \begin{aligned} & \Delta \theta=\frac{\Delta U}{m s}=\frac{4 \pi T\left(R^{2}-N r^{2}\right)}{\frac{4}{3} \pi R^{3} \rho s}, \text { where } \rho \text { is the density. } \ & \therefore \Delta \theta=\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^{2}}{R^{3}} N\right) \ & =\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^{2} R^{3}}{R^{3} r^{3}}\right)=\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{1}{r}\right) \end{aligned} $$

10.20 The drop will evaporate if the water pressure is more than the vapour pressure. The membrane pressure (water)

$$ \begin{aligned} & p=\frac{2 T}{r}=2.33 \times 10^{3} \mathrm{~Pa} \ & \therefore r=\frac{2 T}{p}=\frac{2\left(7.28 \times 10^{-2}\right)}{2.33 \times 10^{3}}=6.25 \times 10^{-5} \mathrm{~m} \end{aligned} $$

10.21 (a) Consider a horizontal parcel of air with cross section $A$ and height $d h$. Let the pressure on the top surface and bottom surface be $p$ and $p+d p$. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.

i.e. $(p+\mathrm{d} p) A-p A=-P g A d h$

$\Rightarrow \mathrm{d} p=-\rho g \mathrm{~d} h$

(b) Let the density of air on the earth’s surface be $\rho_{0}$, then

$$ \begin{aligned} & \frac{p}{p_{o}}=\frac{\rho}{\rho_{o}} \ & \Rightarrow \rho=\frac{\rho_{o}}{p_{o}} p \end{aligned} $$

$\therefore d p=-\frac{\rho_{o} g}{p_{o}} p d h$

$\Rightarrow \frac{d p}{p}=-\frac{\rho_{o} g}{p_{o}} d h$

$\Rightarrow \int_{p_{o}}^{p} \frac{d p}{p}=-\frac{\rho_{o} g}{p_{o}} \int_{o}^{h} d h$

$\Rightarrow \ln \frac{p}{p_{o}}=-\frac{\rho_{o} g}{p_{o}} h$

$\Rightarrow p=p_{o} \exp \left(-\frac{\rho_{o} g}{p_{o}} h\right)$

(c) $\ln \frac{1}{10}=-\frac{\rho_{o} g}{p_{o}} h_{o}$

$\therefore h_{o}=-\frac{p_{o}}{\rho_{o} g} \ln \frac{1}{10}$

$=\frac{p_{o}}{\rho_{o} g} \times 2.303$

$=\frac{1.013 \times 10^{5}}{1.29 \times 9.8} \times 2.303=0.16 \times 10^{5} \mathrm{~m}=16 \times 10^{3} \mathrm{~m}$

(d) The assumption $p \propto \rho$ is valid only for the isothermal case which is only valid for small distances.

10.22 (a) $1 \mathrm{~kg}$ of water requires $\mathrm{L}_{v} \mathrm{k}$ cal

$\therefore M_{\mathrm{A}} \mathrm{kg}$ of water requires $M_{\mathrm{A}} \mathrm{L}_{v} \mathrm{k}$ cal

Since there are $N_{\mathrm{A}}$ molecules in $M_{\mathrm{A}} \mathrm{kg}$ of water the energy required for 1 molecule to evaporate is

$$ \begin{aligned} u= & \frac{M_{A} L_{v}}{N_{A}} \mathrm{~J} \ = & \frac{18 \times \not 940 \times 4.2 \times 10^{3}}{\not 6 \times 10^{26}} \mathrm{~J} \ = & 90 \times 18 \times 4.2 \times 10^{-23} \mathrm{~J} \ & 6.8 \times 10^{-20} \mathrm{~J} \end{aligned} $$

(b) Consider the water molecules to be points at a distance $d$ from each other.

$N_A$ molecules occupy $\frac{M_A}{\rho_w} l$

Thus, the volume around one molecule is $\frac{M_{A}}{N_{A} \rho_{w}} l$

The volume around one molecule is $d^{3}=\left(M_A / N_A \rho_w\right)$

$$ \begin{aligned} \therefore d & =\left(\frac{M_{A}}{N_{A} \rho_{w}}\right)^{1 / 3}=\left(\frac{18}{6 \times 10^{26} \times 10^{3}}\right)^{1 / 3} \ & =\left(30 \times 10^{-30}\right)^{1 / 3} \mathrm{~m} \quad 3.1 \times 10^{-10} \mathrm{~m} \end{aligned} $$

(c) $1 \mathrm{~kg}$ of vapour occupies $1601 \times 10^{-3} \mathrm{~m}^{3}$.

$\therefore 18 \mathrm{~kg}$ of vapour occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^{3}$

$\Rightarrow 6 \times 10^{26}$ molecules occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^{3}$

$\therefore 1$ molecule occupies $\frac{18 \times 1601 \times 10^{-3}}{6 \times 10^{26}} \mathrm{~m}^{3}$

If $d^{\prime}$ is the inter molecular distance, then

$$ \begin{aligned} & d^{3}=\left(3 \times 1601 \times 10^{-29}\right) \mathrm{m}^{3} \ & \therefore \quad d^{\prime}=(30 \times 1601)^{1 / 3} \times 10^{-10} \mathrm{~m} \ & =36.3 \times 10^{-10} \mathrm{~m} \end{aligned} $$

(d) $F\left(d^{\prime}-d\right)=\mathrm{u} \Rightarrow F=\frac{u}{d^{\prime}-d}=\frac{6.8 \times 10^{-20}}{(36.3-3.1) \times 10^{-10}}=0.2048 \times 10^{-10} \mathrm{~N}$

(e) $\quad F / d=\frac{0.2048 \times 10^{-10}}{3.1 \times 10^{-10}}=0.066 \mathrm{~N} \mathrm{~m}^{-1}=6.6 \times 10^{-2} \mathrm{~N} \mathrm{~m}^{-1}$

10.23 Let the pressure inside the balloon be $P_{i}$ and the outside pressure be $P_{\mathrm{o}}$ $P_{i}-P_{o}=\frac{2 \gamma}{r}$

Considering the air to be an ideal gas

$P_{i} V=n_{i} R T_{i}$ where $V$ is the volume of the air inside the balloon, $n_{\mathrm{i}}$ is the number of moles inside and $T_{i}$ is the temperature inside, and $P_{o} V=n_{o} R T_{o}$ where $\mathrm{V}$ is the volume of the air displaced and $n_{\mathrm{o}}$ is the number of moles displaced and $T_{\mathrm{o}}$ is the temperature outside.

$n_{\mathrm{i}}=\frac{P_i V}{R T_{i}}=\frac{M_i}{M_A}$ where $M_i$ is the mass of air inside and $M_A$ is the molar mass of air and $n_{\mathrm{o}}=\frac{P_{O} V}{R T_{O}}=\frac{M_{O}}{M_{A}}$ where $M_{\mathrm{o}}$ is the mass of air outside that has been displaced. If $W$ is the load it can raise, then $W+M_{i} g=M_{o} g$ $\Rightarrow W=M_{o} g-M_{i} g$

Air is $21 \% O_2$ and $79 \% N_2$

$\therefore$ Molar mass of air $M_{\mathrm{A}}=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}$.

$\Rightarrow W=\frac{M_{A} V}{R}\left(\frac{P_{O}}{T_{o}}-\frac{P_{i}}{T_{i}}\right) g$

$=\frac{0.02884 \times \frac{4}{3} \pi \times 8^{3} \times 9.8}{8.314}\left(\frac{1.013 \times 10^{5}}{293}-\frac{1.013 \times 10^{5}}{333}-\frac{2 \times 5}{8 \times 313}\right) \mathrm{N}$ $\frac{0.02884 \times \frac{4}{3} \pi \times 8^{3}}{8.314} \times 1.013 \times 10^{5}\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8 \mathrm{~N}$

$=3044.2 \mathrm{~N}$.