NCERT Exemplar Solutions Class 11 Physics Mechanical Properties of Fluids — Very Short, Short and Long Answer Questions Q11–23

NCERT Exemplar Problems, Class 11 Physics — Mechanical Properties of Fluids, Very Short Answer Questions 11 to 15, Short Answer Questions 16 to 20 and Long Answer Questions 21 to 23. All 13 questions solved. The book numbers them 10.11 to 10.23.

Three ideas carry this page. Buoyancy is the weight of the displaced fluid, and by Newton’s third law the fluid feels an equal push back, which is why a scale under the vessel registers it. Surface tension stores energy in area: making new surface costs $T$ per unit area, merging drops releases it, and a curved surface has an excess pressure $2T/r$ inside. Pressure in a fluid at rest changes only because of weight, $dp = -\rho g\,dh$, and that one line gives both the hydrostatic and the atmospheric results.

Key insight. In Questions 18 and 19 the same energy balance runs in opposite directions: merging drops lose area and release energy, splitting a drop creates area and must take that energy from the liquid’s own heat.

Very Short Answer Questions

Question 11

Is viscosity a vector?

Solution. No. The coefficient of viscosity $\eta$ is defined by $F = \eta A\,\dfrac{dv}{dx}$: it is the constant of proportionality between the tangential stress and the velocity gradient. It is a property of the fluid, like density, with a magnitude and a unit ($\text{Pa s}$) but no direction. The viscous force has a direction, but the coefficient does not.

Question 12

Is surface tension a vector?

Solution. No. Surface tension is the force per unit length acting on a line drawn in the surface, or equally the surface energy per unit area. The force on a particular line acts along the surface and perpendicular to that line, but its direction is fixed by the line chosen, not by the surface tension itself; at a point it is the same for a line in any direction. So surface tension has only a magnitude and is a scalar.

Question 13

An iceberg floats in water with part of it submerged. What is the fraction of the volume of the iceberg submerged if the density of ice is $\rho_i = 0.917\ \text{g cm}^{-3}$?

Solution. Let the iceberg’s volume be $V$ and the submerged fraction $x$. A floating body displaces its own weight of water, so

$$\rho_i V g = \rho_w (xV) g \quad\Rightarrow\quad x = \frac{\rho_i}{\rho_w} = \frac{0.917}{1.00} = 0.917.$$

About 92% of an iceberg is below the surface, which is why only its “tip” is seen. (In sea water, density about $1.03\ \text{g cm}^{-3}$, the fraction is slightly smaller, about $0.89$.)

Question 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?

Solution. The block is held by the spring from above, so it does not rest on the vessel. Its volume is $V = M/\rho$, and the water pushes up on it with the buoyant force $\rho_w V g$. The spring carries the rest: $kx = Mg – \rho_w Vg$, where $x$ is the stretch of the spring.

By Newton’s third law the block pushes down on the water with the same force $\rho_w Vg$, and the water passes that push on to the bottom of the vessel and the pan. The scale was set to zero with the vessel and water alone, so the new reading is just this extra force:

$$\text{reading} = \rho_w V g = \frac{\rho_w}{\rho}\,Mg.$$

The spring constant $k$ does not enter; it only decides how far the spring stretches.

Question 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?

Solution. In the elevator the block accelerates upward with $a$, so the net upward force on it is $\rho L^3 a$. The pressure in the water is also larger, because the water itself must be accelerated: in the elevator’s frame everything behaves as if gravity were $g + a$. Let the new immersed fraction be $x’$. Then

$$\rho_w (x’L^3)(g+a) – \rho L^3 g = \rho L^3 a \quad\Rightarrow\quad \rho_w x'(g+a) = \rho(g+a) \quad\Rightarrow\quad x’ = \frac{\rho}{\rho_w}.$$

This is the same as at rest, so the fraction immersed is unchanged at $x$. The weight and the buoyant force both scale with the effective gravity, so it cancels.

Short Answer Questions

Question 16

The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r = 2.5\times10^{-5}$ m. The surface tension of sap is $T = 7.28\times10^{-2}\ \text{N m}^{-1}$ and the angle of contact is $0^\circ$. Does surface tension alone account for the supply of water to the top of all trees?

Solution. The capillary rise is

$$h = \frac{2T\cos\theta}{\rho g r} = \frac{2\times7.28\times10^{-2}\times1}{10^3\times9.8\times2.5\times10^{-5}} = 0.594\ \text{m} \approx 0.6\ \text{m}.$$

This is the greatest height to which surface tension alone can lift the sap. Many trees are tens of metres tall, so capillary action cannot by itself account for the rise of water to the top of all trees; other mechanisms (the pull from evaporation at the leaves, and root pressure) must do most of the work.

Question 17

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

Solution. Work in the frame of the tanker, which accelerates with $a$ along $+x$. In this frame each parcel of oil feels its weight $\rho g$ per unit volume downwards and a pseudo-force $\rho a$ per unit volume backwards ($-x$). A free liquid surface at rest settles perpendicular to the net effective force, because any component along the surface would push the liquid along it.

Take the surface inclined at $\theta$ to the horizontal, rising towards the back. Along the surface, the component of the weight is $\rho g\sin\theta$ (down the slope, forwards) and the component of the pseudo-force is $\rho a\cos\theta$ (up the slope, backwards). Setting the tangential force to zero,

$$\rho g\sin\theta = \rho a\cos\theta \quad\Rightarrow\quad \tan\theta = \frac{a}{g}.$$

The oil piles up at the back of the tanker.

Question 18

Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury $T = 435.5\times10^{-3}\ \text{N m}^{-1}$.

Solution. The volume is conserved, so the radius $R$ of the new drop satisfies

$$R^3 = r_1^3 + r_2^3 = (0.001 + 0.008)\ \text{cm}^3 = 0.009\ \text{cm}^3 \quad\Rightarrow\quad R = 0.2080\ \text{cm} = 2.080\times10^{-3}\ \text{m}.$$

The surface area falls, and the surface energy released is $T$ times the loss of area:

$$\Delta E = 4\pi T\left(r_1^2 + r_2^2 – R^2\right) = 4\pi\times0.4355\times\left(1 + 4 – 4.327\right)\times10^{-6}\ \text{J} = 3.68\times10^{-6}\ \text{J}.$$

Note on the printed key. The key rounds $R$ to $0.21$ cm and gets $3.2\times10^{-6}$ J. Because the answer is a small difference of nearly equal areas ($0.05$ against $0.0433\ \text{cm}^2$), that rounding changes it by about 13%; the unrounded value is $3.7\times10^{-6}$ J.

Question 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.

Solution. The volume is conserved: $R^3 = Nr^3$. Breaking up the drop increases the surface area from $4\pi R^2$ to $4\pi Nr^2$, and the extra surface energy

$$\Delta U = 4\pi T\left(Nr^2 – R^2\right)$$

must come from the liquid’s own internal energy, since no heat is supplied. If the drop has density $\rho$ and specific heat $s$, its mass is $m = \tfrac{4}{3}\pi R^3\rho$, and the fall in temperature is

$$\Delta\theta = \frac{\Delta U}{ms} = \frac{4\pi T\left(Nr^2 – R^2\right)}{\tfrac{4}{3}\pi R^3\rho s} = \frac{3T}{\rho s}\left(\frac{Nr^2}{R^3} – \frac{1}{R}\right).$$

Using $N/R^3 = 1/r^3$, so that $Nr^2/R^3 = 1/r$,

$$\Delta\theta = \frac{3T}{\rho s}\left(\frac{1}{r} – \frac{1}{R}\right).$$

This is positive because $r < R$: the droplets are cooler. (If $s$ is given in calories, divide by $J$ to keep the units consistent.)

Question 20

The surface tension and vapour pressure of water at $20^\circ$C is $7.28\times10^{-2}\ \text{N m}^{-1}$ and $2.33\times10^3$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^\circ$C?

Solution. A drop of radius $r$ has an excess pressure $2T/r$ inside it because of its curved surface. The smaller the drop, the larger this pressure. In this simple model the drop evaporates if that pressure exceeds the vapour pressure of the water, so the smallest stable drop has

$$\frac{2T}{r} = p_v \quad\Rightarrow\quad r = \frac{2T}{p_v} = \frac{2\times7.28\times10^{-2}}{2.33\times10^3} = 6.25\times10^{-5}\ \text{m}.$$

Droplets smaller than about $0.06$ mm evaporate at $20^\circ$C.

Long Answer Questions

Question 21

(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure $dp$ over a differential height $dh$?

(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 $p_0 = 1.03\times10^5\ \text{N m}^{-2}$, $\rho_0 = 1.29\ \text{kg m}^{-3}$ and $g = 9.8\ \text{m s}^{-2}$, 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.

Solution. (a) Take a thin horizontal slab of air of area $A$ and thickness $dh$, with pressure $p$ on its lower face and $p + dp$ on its upper face. It is in equilibrium, so the net upward pressure force balances its weight:

$$pA – (p+dp)A = \rho A\,dh\,g \quad\Rightarrow\quad dp = -\rho g\,dh.$$

(b) With $p \propto \rho$, $\rho = \rho_0\,p/p_0$, where $\rho_0$ is the density at the surface. Then

$$\frac{dp}{p} = -\frac{\rho_0 g}{p_0}\,dh \quad\Rightarrow\quad \int_{p_0}^{p}\frac{dp}{p} = -\frac{\rho_0 g}{p_0}\int_0^h dh \quad\Rightarrow\quad p = p_0\exp\!\left(-\frac{\rho_0 g h}{p_0}\right).$$

(c) Put $p = p_0/10$:

$$h = \frac{p_0}{\rho_0 g}\ln 10 = \frac{1.03\times10^5}{1.29\times9.8}\times2.303 = 8147\times2.303\ \text{m} \approx 1.88\times10^4\ \text{m} \approx 19\ \text{km}.$$

(d) Taking $p \propto \rho$ assumes the temperature of the air is the same at all heights (an isothermal atmosphere), since for an ideal gas $p = \rho RT/M$. In reality the temperature falls by several degrees per kilometre in the lower atmosphere, so the model holds only over heights small enough for the temperature to be nearly constant. (It also treats $g$ as constant, which is a much smaller error at these heights.)

Note on the printed key. The key evaluates $\dfrac{1.013\times10^5}{1.29\times9.8}\times2.303$ as $16\times10^3$ m. That product is $1.85\times10^4$ m; with the question’s own $p_0 = 1.03\times10^5$ Pa it is $1.88\times10^4$ m. The method is right, the arithmetic is not.

Question 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 $L_v = 540\ \text{k cal kg}^{-1}$, the mechanical equivalent of heat $J = 4.2\ \text{J cal}^{-1}$, density of water $\rho_w = 10^3\ \text{kg m}^{-3}$, Avogadro’s number $N_A = 6.0\times10^{26}\ \text{k mole}^{-1}$ and the molecular weight of water $M_A = 18$ kg for 1 k mole.

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

(b) Show that the inter-molecular distance for water is $d = \left[\dfrac{M_A}{N_A}\times\dfrac{1}{\rho_w}\right]^{1/3}$ and find its value.

(c) 1 g of water in the vapour state at 1 atm occupies $1601\ \text{cm}^3$. 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 $d’$. Estimate the value of $F$.

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

Solution. (a) One kilomole (18 kg) of water needs $M_A L_v$ of heat to evaporate, and it contains $N_A$ molecules, so

$$u = \frac{M_A L_v J}{N_A} = \frac{18\times540\times10^3\times4.2}{6.0\times10^{26}}\ \text{J} = 6.8\times10^{-20}\ \text{J}.$$

(b) Treat each molecule as occupying a small cube of side $d$. One kilomole has mass $M_A$, so it occupies volume $M_A/\rho_w$, shared among $N_A$ molecules. The volume per molecule is therefore

$$d^3 = \frac{M_A}{N_A\rho_w} \quad\Rightarrow\quad d = \left[\frac{M_A}{N_A}\times\frac{1}{\rho_w}\right]^{1/3} = \left(\frac{18}{6.0\times10^{26}\times10^3}\right)^{1/3}\ \text{m} = (3.0\times10^{-29})^{1/3}\ \text{m} = 3.1\times10^{-10}\ \text{m}.$$

(c) In the vapour, 1 g occupies $1601\ \text{cm}^3$, so 18 kg occupies $18\times1601\times10^{-3}\ \text{m}^3$, shared among $6.0\times10^{26}$ molecules. The volume per molecule is

$$d’^3 = \frac{18\times1601\times10^{-3}}{6.0\times10^{26}}\ \text{m}^3 = 4.80\times10^{-26}\ \text{m}^3 \quad\Rightarrow\quad d’ = 3.63\times10^{-9}\ \text{m} = 36.3\times10^{-10}\ \text{m},$$

about twelve times the spacing in the liquid.

(d) The work done against the constant force $F$ over the distance $d’ – d$ equals the energy $u$ found in (a):

$$F = \frac{u}{d’ – d} = \frac{6.8\times10^{-20}}{(36.3 – 3.1)\times10^{-10}}\ \text{N} = 2.05\times10^{-11}\ \text{N}.$$

(e) Force per unit length:

$$\frac{F}{d} = \frac{2.05\times10^{-11}}{3.1\times10^{-10}}\ \text{N m}^{-1} = 0.066\ \text{N m}^{-1} = 6.6\times10^{-2}\ \text{N m}^{-1}.$$

This crude estimate is close to the measured surface tension of water, about $7\times10^{-2}\ \text{N m}^{-1}$, which shows that surface tension really is the molecular attraction that has to be overcome in evaporation.

Question 23

A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of $60^\circ$C. How large a mass can the balloon lift when the outside temperature is $20^\circ$C? (Assume air is an ideal gas, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$, 1 atm $= 1.013\times10^5$ Pa; the membrane tension is $5\ \text{N m}^{-1}$.)

Solution. Pressures. The membrane is a single curved surface under tension $\gamma = 5\ \text{N m}^{-1}$, so the pressure inside exceeds the outside pressure by

$$P_i – P_o = \frac{2\gamma}{r} = \frac{2\times5}{8} = 1.25\ \text{Pa},$$

with $P_o = 1.013\times10^5$ Pa. This is tiny compared with the atmosphere, so it hardly affects the answer.

Masses of air. The volume is $V = \tfrac{4}{3}\pi(8)^3 = 2144.7\ \text{m}^3$. For an ideal gas the mass in volume $V$ is $PVM_A/(RT)$, where $M_A$ is the molar mass of air. The question does not give $M_A$; for air that is 21% oxygen and 79% nitrogen, $M_A = 0.21\times32 + 0.79\times28 = 28.84\ \text{g mol}^{-1}$.

Lift. The balloon is held up by the weight of the outside air it displaces (mass $M_o$, at $T_o = 293$ K) and pulled down by the weight of the hot air inside (mass $M_i$, at $T_i = 333$ K). The load it can just lift has weight

$$W = (M_o – M_i)g = \frac{M_A V g}{R}\left(\frac{P_o}{T_o} – \frac{P_i}{T_i}\right) = \frac{0.02884\times2144.7\times9.8}{8.314}\left(\frac{1.013\times10^5}{293} – \frac{1.013\times10^5 + 1.25}{333}\right)\ \text{N}.$$

The bracket is $345.73 – 304.21 = 41.52\ \text{Pa K}^{-1}$, so

$$W \approx 72.91\times41.52\ \text{N} \approx 3.03\times10^3\ \text{N},$$

which is a mass of $W/g \approx 309$ kg (this includes the balloon’s own fabric and basket).

Note on the printed key. The key sets the problem up the same way but prints $3044.2$ N; its own expression evaluates to $3027.5$ N. It also writes $8\times313$ for $8\times333$ in the membrane term, which makes no visible difference. The difference is under 1%, so the answer about $3\times10^3$ N, or about 300 kg, stands.

Common mistakes

• Forgetting Newton’s third law in Question 14. The block is held by the spring, but the water still feels the buoyant reaction, and the scale reads $\rho_w Vg$.
• Expecting the fraction to change in the lift in Question 15. Weight and buoyancy both scale with the effective gravity $g + a$.
• Rounding too early in Question 18. The energy released is a small difference of two nearly equal areas; rounding $R$ to $0.21$ cm loses 13% of it.
• Getting the sign of the temperature change in Question 19. Splitting a drop creates surface, so energy is taken from the liquid and it cools.
• Using $4T/r$ for the balloon in Question 23. A soap bubble has two surfaces; the balloon’s membrane is a single surface with tension $5\ \text{N m}^{-1}$, so the excess pressure is $2\gamma/r$.

Practise next

• A glass capillary of radius $0.1$ mm is dipped in water ($T = 0.072\ \text{N m}^{-1}$, angle of contact $0^\circ$, $g = 9.8\ \text{m s}^{-2}$). How high does the water rise?

Show answer

About $14.7$ cm. $h = 2T\cos\theta/(\rho g r) = 2\times0.072/(10^3\times9.8\times10^{-4}) = 0.147$ m.

• One thousand water droplets, each of radius $1$ mm, merge into a single drop. Taking $T = 0.072\ \text{N m}^{-1}$, how much surface energy is released?

Show answer

About $8.1\times10^{-4}$ J. The new radius is $R = 1000^{1/3}\times1$ mm $= 10$ mm, so the area falls by $4\pi(1000\times10^{-6} – 10^{-4}) = 1.131\times10^{-2}\ \text{m}^2$, and $T\Delta A = 8.14\times10^{-4}$ J.

• In the isothermal atmosphere of Question 21, with $p_0 = 1.013\times10^5$ Pa, $\rho_0 = 1.29\ \text{kg m}^{-3}$ and $g = 9.8\ \text{m s}^{-2}$, at what height does the pressure fall to half its surface value?

Show answer

About $5.55$ km. $h = (p_0/\rho_0 g)\ln 2 = 8013\times0.693 = 5554$ m.

• Exemplar Mechanical Properties of Fluids: Multiple Choice Questions — the same ideas in objective form: terminal velocity, streamlines, continuity and floating bodies.
• NCERT textbook Chapter 9 — Mechanical Properties of Fluids for the theory of pressure, viscosity, surface tension and capillary rise.
• Exemplar Thermal Properties of Matter — the next unit, where the latent heat used in Question 22 is treated in full.