NCERT Exemplar Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids

NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQ

Question: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 figure, indicate the one that represents the velocity (v) of the pebble as a function of time (t).

Answer:

The answer is the option (c)
When a cylinder is filled with viscous oil and a pebble is dropped from the top, it falls under the effect of gravity and acceleration due to this force. When it enters the oil, a dragging force $f=6\pi rnv$ acts opposite, and hence the acceleration of the pebble decreases till it becomes zero and the velocity becomes constant. Hence option c is correct.

Question:10.2

Which of the following diagrams does not represent a streamline flow?

Answer:

The answer is the option (d)
The velocity remains constant for a liquid in case of a streamlined flow. So, the layers do not cross each other. And hence, d is the correct choice.

Question: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

Answer:

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

The speed at any point or any cross-section for a liquid is constant in case of a streamlined flow.

Question:10.4

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is
a) 9:4
b) 3:2
c) $\sqrt{3}:\sqrt{2}$
d) $\sqrt{2}:\sqrt{3}$

Answer:

The answer is the option (a) 9:4
a1 v1 = a2 v2
$\frac{v1}{v2}=\frac{a2}{a1}$
$\frac{v1}{v2}=\frac{d{2}^2}{d{1}^2}$
putting values,
$\frac{v1}{v2}=\frac{(3.75\times 3.75)}{2.5\times 2.5}=\frac{9}{4}$
so, v1: v2 = 9:4. Hence option a is correct.

Question: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 methyl iodide-glass 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

Answer:

The answer is the option (c) Mercury.
In the case of mercury glass, the meniscus of liquid is convex, which makes the angle of contact obtuse. Hence option c is the correct choice.

NCERT Exemplar Class 11 Physics Solutions Chapter 10: MCQII

Question: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

Answer:

The answer is the option (b) and (d)
The force exerted on a molecule by all other neighbouring molecules is equal and opposite in all directions. Hence the horizontal component is zero.
A downward force acts vertically, which is not cancelled and hence b is the correct option.
The potential energy of the molecules on the surface is negative as compared to the molecules in the subsequent layers. But the magnitude of this PE is higher than the below molecules. Hence option d is also correct.

Question: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 component of the force normal to the area
d) it does not depend on the size of the area chosen

Answer:

The answer is the option (c)
The pressure is nothing but a component of some force acting along a particular area. The directions here are fixed for both these quantities which make pressure a scalar quantity. Hence option c is the right choice.

Question:10.8

A wooden block with a coin placed on its top, floats in water as shown in figure:

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 increases

Answer:

The answer is the option (a) and (b)
The amount of displaced fluid is equal to the weight of the floating body; this is as per the flotation principle. When the coin is dropped, the net force acting on the block decreases as the coin displaces water in an equivalent amount to its weight. This is the reason that the block rises up. As the block rises, l decreases. The height h will also decrease as the amount of water getting displaced now will be lesser. Hence option a and b are correct.

Question:10.9

With increase in temperature, the viscosity of
a) gases decreases
b) liquids increases
c) gases increases
d) liquids decreases

Answer:

The answer is the option (c) and (d)
If n is the coefficient of viscosity of a liquid, $n\propto \frac{1}{\sqrt{T}}$
In case of the viscosity of gases, $n\propto \sqrt{T}$
As the temperature increases, kinetic energy in the gas molecules also increases and hence due to the collisions between the molecules, the viscosity increases.
Whereas in case of a liquid, the kinetic energy increases with temperature, but the relative speed of the layers of liquid remains fairly the same as before. Hence option c and d are correct.

Question:10.1

Streamline flow is more likely for liquids with
a) high density
b) high viscosity
c) low density
d) low viscosity

Answer:

The answer is the option (b) and (c)
Liquids which have a lesser density are more subjected to a streamline flow. The velocity gradient of liquid increases with its viscosity. Hence the line of flows can be differentiated accordingly.
Reynold number, $R=\frac{pvD}{n}$
As we increase n and decrease p, R decreases. This decrease in R makes the flow more streamlined. Hence options b and c are correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Very Short Answer

Question:10.11

Is viscosity a vector?

Answer:

Viscosity: the dragging force magnitude between two layers of a liquid per unit area, who have their velocity gradient as unity is called viscosity. This quantity just has a magnitude and no direction, and hence it is not a vector quantity.

Question:10.12

Is surface tension a vector?

Answer:

Surface tension = work done/ change in the surface area
These are scalar quantities, and hence surface tension is a scalar and not a vector quantity.

Question: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$\rho i=0.917\frac{g}{c{m}^3}$?

Answer:

The weight of the iceberg is equal to the weight of the displaced liquid, as the iceberg is floating on the surface of the sea.
$VPig=VPwg$
The volume of an iceberg submerged in water/ vol of iceberg = $\frac{V^{{}^{\prime }}}{V}=\frac{Pi}{Pw}=\frac{0.917}{1}$
Hence the portion of the iceberg which is submerged in water is 0.917

Question: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?

Answer:

The given figure has a beaker filled with water. The beaker has been placed on a weighing scale which is adjusted to zero.

As we submerge the block into the water, a force equal to the weight of the block acts upward in the form of a buoyant force. So, the downward force applied by the block equals to V Pw g
V = the water volume which is displaced by the block
Pw = density of water
Mass of block = M = V
$V=\frac{M}{p}$
So, the reading on the weighing scale reads:
$\frac{M}{p}.p_{w}g=\frac{p_w}{p}Mg$

Question: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 upwards with acceleration a. What is the fraction immersed?

Answer:

Through the principle of flotation when the block gets submerged in water
$Vpg=V^{{}^{\prime }}Pwg$
Where V’ = volume of water displaced by block
V’ = base area of block x height of block inside the water
$V^{{}^{\prime }}=L^{2}x$
Let V be the volume of the block and Pb by the density of the block
Hence,
$V$$=L^2\times Pw$
$\frac{Pb}{Pw}=\frac{x}{L}$ ---------(1)
$x=\frac{Pb}{Pw}L$
as the block lifts upward, the acceleration = (g + a)
weight of L³ Pb block = $m(g+a)=V\times Pb(g+a)=L^{3}Pb(g+a)$
let x1 be the part of block which is submerged.
We know that the buoyant force will be equal and opposite to the weight of the block.
$L^{3}Pb(g+a)=x1L^{2}Pw(g+a)$
So,
$\frac{Pb}{Pw}=\frac{x1}{L}$
So, $x1=\frac{Pb}{Pw}L$ ------------(2)
From (1) and (2) we can infer that$\frac{Pb}{Pw}L$ is the part of the block which is submerged in water. This is independent of the acceleration of the block (upwards, downwards or at rest)

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Short Answer

Question:10.16

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

Answer:

r = radius of the capillary
$r=2.5\times {10}^{-5}m$
$S=T=7.28\times {10}^{-2}N/m$
$g=9.8\frac{m}{s^2}$
$\theta =0$, p = 1000 kg/m³
now, $h=\frac{2s\cos \theta }{rpg}=2\times 7.28\times {10}^{-2}\frac{\cos 0}{2.5\times {10}^{-5}\times 1000\times 9.8}$

$h=\frac{104}{175}\times \frac{1000}{1000}=\frac{104}{175}=0.594\approx 0.6m$

Question: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 m/s², what will be the slope of the free surface?

Answer:

Let us assume the tanker is pulled by a force F and it produces an acceleration in the forward direction. Let $\delta$m be the mass of a small element. As the tanker is pulled forward and accelerates, the small element also experiences a force in the same direction. The inertia of rest makes the element to remain at rest and hence as the tank moves forward, the element moves backwards as a result of inertia.
Force acting on $\delta$m, F1 = $\delta$ma in the backward direction
F2 = $\delta$mg which is in a vertically downward direction.
The normal reaction gets balanced by various components: $\delta$$ma\sin \theta$ from F1
The situation when there is a maximum angle inclination by the surface,
$\delta ma\cos \theta =\delta mg\sin \theta$
So required slope, $\frac{\sin \theta }{\cos \theta }=\frac{a}{g}$
So, $\tan \theta =\frac{a}{g}$

Question:10.18

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 $T=435.5\times {10}^{-3}\frac{N}{m}$.

Answer:

Surface tension energy, $E=\sigma \mathrm{\Delta }A$
As we can infer from the law of conservation of mass, the volume of a drop V = v1 + v2
$r_1=0.1cm=0.001m$
$r_2=0.2cm=0.002m$
$\mathrm{\Delta }A=4\pi r_1^2+4\pi r_2^2-4\pi R^2$
$\mathrm{\Delta }A=4\pi [r_1^2+r_2^2-R^2]$
The new bigger drop has a radius of R,
$\frac{4}{3}\pi R^3=\frac{4}{3}\pi [r_1^3+r_2^3]$
$R^3=r_1^3+r_2^3$
$R^3=9\times {10}^{-9}$
$R^3=2.1\times {10}^{-3}m$
Now, $E=\sigma \mathrm{\Delta }A$
$E=4\times 3.14[({10}^{-3}{)}^2+(2\times {10}^{-3}{)}^2-(2.1\times {10}^{-3}{)}^2]\times 435.5\times {10}^{-3}$
$E=1742\times 3.14\times {10}^{-9}[0.59]=5469.88\times 0.59\times {10}^{-9}$
$E=32.27\times {10}^{-7}J$
Finally, the area will be smaller, when energy is released as the bigger drop is formed from smaller drops.

Question: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.

Answer:

now, $\mathrm{\Delta }E=\sigma$
(final area - initial area) of surface
$\mathrm{\Delta }E=ms\mathrm{\Delta }t$
As per the law of conservation of mass,
Final volume =initial volume of one drop which has a radius R which is then split into N drops r
So, $\frac{4}{3}\pi R^3=N.\frac{4}{3}\pi r^3$
$N.\frac{4}{3}\pi r^{3}ps\mathrm{\Delta }t=4\pi \sigma [N{r}^2-R^2]$
$\mathrm{\Delta }t=\frac{4\pi \sigma \times 3}{N.4\pi r^{3}ps}[N{r}^2-R^2]$
$\mathrm{\Delta }t=\frac{3\sigma }{Nps}[\frac{N{r}^2}{r^3}-\frac{R^2}{r^3}]$
$\mathrm{\Delta }t=\frac{3\sigma N}{Nps}[\frac{1}{r}-\frac{1}{R}]$
$\mathrm{\Delta }t=\frac{3\sigma }{ps}[\frac{1}{r}-\frac{1}{R}]$ as R>r
Hence, we can say that $\mathrm{\Delta }t$ will be positive and therefore, $\frac{1}{R}<\frac{1}{r}$
So, the temperature increases as the formation smaller drops take place, and this change takes place through absorption of energy through the surroundings.

Question:10.20

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

Answer:

In case when the water pressure in the liquid is greater than the surface above the liquid, the drop will evaporate. We assume R to be the radius of the droplet before evaporation.
We know that pressure of vapour = pressure which becomes excess inside the drop
$P=\frac{2\sigma }{R}$
$R=2\times 7.28\times {10}^{-2}$$/P$ $=\frac{2\times 7.28\times {10}^{-2}}{2.33\times 1000}=\frac{14560}{233}\times 100000$
Hence, $R=6.25\times {10}^{-5}m$

NCERT Exemplar Class 11 Physics Solutions Chapter 10: Long Answer

Question:10.21

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 ${\rho }_0$.
c) If $p_0=1.03\times {10}^{5}N/m^2$, ${\rho }_0=1.29\frac{kg}{m^3}$ and g is $9.8\frac{m}{s^2}$ 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.

Answer:

a) Let us assume that there is a packet which has a thickness of dh. In a liquid, at any point, the pressure is working in all the directions in an equal manner. The force as a result of the pressure is opposed by the buoyant force of the liquid.
$(P+dP)A-P.A=-Vpg$
$dPA=-Adhpg$
$dP=-pgdh$ -----------(1) {since pressure decreases as height increases, the negative sign is present here}

b) Let us assume Po to be the density of air on Earth.
At any point p, the pressure is directly proportional to the density
Hence,
$p=\frac{P}{P_0}{p}_0$
now, $dP=pgdh$
$dP=\frac{P}{P_0}{p}_{0}gdh$
$\frac{dP}{P}=-p_0\frac{g}{P_0}dh$
on integrating both sides we get,
$\log \frac{P}{P_0}=\frac{-p_0}{P_0}gh$ -----------(3)
$\frac{P}{P_0}=e^{\frac{-p_0}{P_0}gh}$
c)
we know that, $\log \frac{P}{P_0}=\frac{-p_0}{P_0}gh$
but, $P=\frac{P_0}{10}$ (given)
$\log \frac{\frac{P_0}{10}}{P_0}=\frac{-p_{0}gh}{P_0}$
$\log \frac{1}{10}=\frac{-p_{0}gh}{P_0}$
$h=\frac{P_0}{p_{0}g}\log 10$
$h=\frac{1.013\times {10}^5\times 2.303}{1.29\times 9.8}=18.4km$
d)
the temperature does not remain constant at greater height. It only remains constant near the surface of the Earth.

Question: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 vaporization for water $Lv=540kcalk{g}^{-1}$, the mechanical equivalent of heat $J=4.2Jca{l}^{-1}$, density of water $\rho w=103kg{l}^{-1}$, Avagadro’s number $N_A=6.0\times {10}^{26}kmol{e}^{-1}$ and the molecular weight of water $M_A=18kg$ for 1 k mole.
(a) Estimate the energy required for one molecule of water to evaporate.
(b) show that the intermolecular distance for water is
$d={[\frac{M_A}{N_A}\times \frac{1}{{\rho }_w}]}^{\frac{1}{3}}$
(c) 1 g of water in the vapor state at 1 atm occupies 1601 cm³. 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 where d=3.1x10^-10m
(e) Calculate $\frac{F}{d}$, which is a measure of the surface tension.

Answer:

a) $Lv=540kcalk{g}^{-1}$ $=540\times 4.2\times 1000J/Kg$
For evaporation of 1 kg water, energy required = Lv K Cal
For evaporation of Ma kg of water, La Ma K Cal
Number of molecules in Ma kg of water = Na
Hence, for the evaporation of 1 molecule, the energy required (U)= (Lv Ma/Na) K. Cal
$U=\frac{MaLv\times 1000\times 4.2}{Na}=\frac{18\times 540\times 1000\times 4.2}{6\times {10}^{26}}=6.8\times {10}^{-20}J$
b)
water molecules are assumed to be of a point size and have distance d between each other
total volume of Na molecules of water = mass/density
$=\frac{M_a}{p}$
volume occupied by 1 molecule =
$d^3=\frac{M_a}{N_a}\times p$
$d={[\frac{M_a}{N_a}\times p]}^{\frac{1}{3}}$
Hence, the volume occupied by 1kg vapour $=1601c{m}^3=1601\times {10}^{-6}{m}^3$
c)
the volume occupied by 18kg water vapour $=18\times 1601\times {10}^{-6}c{m}^3$
number of molecules in 18 kg water =$6\times {10}^{26}$
so, the volume occupied by 1 molecule = $\frac{18\times 1601\times {10}^{-6}c{m}^3}{6\times {10}^{26}}$
hence $d^{{}^{\prime }3}=\frac{18\times 1601\times {10}^{-6}c{m}^3}{6\times {10}^{26}}$
$d^{{}^{\prime }}=36.3\times {10}^{-10}m$
d) work done to change the distance to d’ from d = F (d’-d)
$6.8\times {10}^{-20}=F(36.3\times {10}^{-10}-3.1\times {10}^{-10})=F\times 33.2\times {10}^{-10}$
Hence, $F=\frac{6.8\times {10}^{-20}}{33.2\times {10}^{-10}}$
$F=2.05\times {10}^{-11}N$

e)
from the formula of surface tension, we know that, surface tension = F/d
Surface tension = $\frac{2.05\times {10}^{-11}N}{3.1\times {10}^{-10}}$
Surface tension =$6.6\times {10}^{-2}N/m$

Question:10.23

A hot air balloon is a sphere of radius 8 m. The surface inside is at a temperature of ${60}^{\circ }C$. How large a mass can the balloon lift when the outside temperature is ${20}^{\circ }C$?

Answer:

The amount of pressure outside a balloon is lesser than the amount of pressure inside the balloon.

$P_i-P_0=\frac{2\sigma }{R}$ {$\sigma$: surface tension, R: balloon’s radius}

$PiV=niRTi$

Where, V is the volume of balloon and ni is number of moles of gas in the balloon

R is the gas constant, Ti is the temperature

$ni=\frac{PiV}{RTi}$= Mass of balloon (Mi)/Molecular mass (Ma)

$Mi=\frac{PiVMa}{RTi}$-------------(1)

In the same way, $n_0=\frac{P_{0}V}{R{T}_0}$

According to principal of floatation, $W+Mig=M_{0}g$, here W is the weight which the balloon lifts

$W=(M_0-M_i)g$

If the mass displaced by balloon is M0 and Ma is the mass inside or outside of the balloon

$n_0=\frac{M_0}{M_a}=\frac{P_{0}V}{R{T}_0}$

$M_0=\frac{P_{0}V{M}_a}{R{T}_0}$---------------(2)

Hence from (1) and (2) we can write,

$W=[\frac{P_{0}VMa}{R{T}_0}-\frac{P_{i}V{M}_a}{R{T}_i}]g$

$W=\frac{V{M}_a}{R}[\frac{P_0}{T_0}-\frac{P_i}{T_i}]g$

Ma = 79% N₂ + 21%O₂

$Ma=0.79\times 28+0.21\times 32=28.84g=0.2884kg$

$P_i-P_0=\frac{2\sigma }{R}$

$P_i=P_o$ + Pressure due to ST of membrane

$Pi=[1.1013\times {10}^5+2\times \frac{5}{8}]=\sim 1.1013\times {10}^5$

$W=\frac{4}{3}\times \frac{3.14\times 8\times 8\times 8\times 0.02884}{3\times 8.314}[\frac{1.013\times {10}^5}{293}-\frac{1.013\times {10}^5}{333}]g$

$W=\frac{4}{3}\times \frac{3.14\times 8\times 8\times 8\times 0.02884\times 1.013\times {10}^5}{3\times 8.314}[\frac{1}{293}-\frac{1}{333}]g$

$W=3044.2N$