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Page No 72:

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

Answer:

In fluids, when the pebble is dropped from the top of a tall cylinder filled with viscous oil, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant, that constant velocity is called terminal velocity.

When the pebble is falling through the viscous oil, the viscous force is F= ηrv, where r is the radius of the pebble, v is instantaneous speed, η is coefficient of viscosity. As the force is variable, hence acceleration is also variable so v-t graph will not be a straight line. First velocity increases and then becomes constant known as terminal velocity.
Hence the correct answer is option  (c).



Page No 73:

Question 10.2:

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

Answer:

Streamline flow: Streamline flow of a liquid is that flow in which each element of the liquid passing through
a point travels along the same path and with the same velocity as the preceding element passes through
that point.
A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the
direction of the flow of liquid at that point.
The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the
greater is the velocity of liquid particles at that place. If we consider a cross-sectional area, then a point on
the area cannot have different velocities at the same time.
In figure (d) lines cross each other which is not possible for streamline flow.
Hence, the correct answer is option (d).

Page No 73:

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:

For a streamline flow of a liquid velocity of each particle at a particular cross-section is constant.
So we can say that along a streamline, the velocity of every fluid particle while crossing a given position is the same.
Hence, the correct answer is option (b).

Page No 73:

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) 3 : 2
(d) 2 : 3

Answer:

Consider a pipe having cross-sections at two positions 1 and 2 are as shown in figure,

Here,  d1 = diameter at 1 = 3.75
d2 = diameter at 2 = 2.5
a1 = area of pipe at 1
a2area of pipe at 2

According to equation of continuity
a1v1=a2v2

v1v2=a2a1=πd224πd124=d2d12
v1v2=3.752.52=94

Hence, the correct answer is option (a)
 

Page No 73:

Question 10.5:

The angle of contact at the interface of water-glass is 0°, Ethylalcohol-glass is 0°, Mercury-glass is 140° and Methyliodide-glass is 30°. 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:

When a liquid has concave meniscus, the angle of contact is acute. 
​The meniscus of liquid in a capillary tube will be convex upwards if the angle of contact is obtuse. It is so when one end of a glass capillary tube is immersed in a trough of mercury.
​Hence, the correct answer is option (c)

Page No 73:

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:


It can be observed in the diagram that the molecule on the surface is pulled by the molecules present in the sphere of influence and the resulting net force will be in downward direction.
The Surface Tension acts on the molecules on the surface of liquid but not on the molecules present in the bulk of liquid, so potential energy of a molecules on the surface of a liquid is larger than compared to the one inside the liquid.
Hence the correct answers are options (b) and (d). 
 



Page No 74:

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

Answer:


The pressure is then defined in a limiting sense as
P = limA0FA.
Pressure is a scalar quantity. Here F is the component of the force normal to the area under consideration and not the (vector) force that appears in the numerator.
Hence the correct answers is option (c).

Page No 74:

Question 10.8:


A wooden block with a coin placed on its top, floats in water as shown in the 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 increase.

Answer:

Mass of block  = m
Volume of coin  = Vc
Density of coin = ρc
Density of water = ρw
Assuming, ρ> ρw
When the coin is on the block the volume of water displaced is:
  V1 = m + ρc Vc ρw = mρw + ρc Vcρw

When the coin falls in water the volume of water displaced is:
  V2  = mρw +Vc
since    ρc ρw > 1
V1 > V2
Therefore when coin falls both h and l decrease.
Hence the correct options are (a) and (b).

Page No 74:

Question 10.9:

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

Answer:

With increase in temperature, the coefficient of viscosity of liquids decreases but that of gases increases. The reason is that as temperature rises, the atoms of the liquid become more mobile, whereas in case of a gas, the collision frequency of atoms increases as their motion becomes more random.
Hence the correct options are (c) and (d).

Page No 74:

Question 10.10:

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

Answer:

Reynold's number is used to help predict flow patterns in different fluid flow situations. At low Reynolds numbers, flows tend to be dominated by laminar (sheet-like) flow, while at high Reynolds numbers turbulence results from differences in the fluid's speed and direction.
Re = ρ.ν.rη
ρ = density of fluid
v = speed of fluid
r = radius of pipe
η = viscosity of fluid
for high value of viscosity and low value of density flow tends to be streamline/laminar
Hence, the correct options are (b) and (c).

 

Page No 74:

Question 10.11:

Is viscosity a vector?

Answer:

No, viscosity is not a vector quantity. Viscosity is the quantity that describes a fluid's resistance to flow. Fluids resist the relative motion of immersed objects through them as well as to the motion of layers with differing velocities within them. It is a scalar quantity as it does not have any direction.

Page No 74:

Question 10.12:

Is surface tension a vector?

Answer:

No, surface tension is a scalar quantity because it has no specific direction.
 

Page No 74:

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 ρi = 0.917 g cm–3?

Answer:

Let, ρw = density of water,
ρi= density of ice,
V = total volume of ice,
Vs= volume of ice submerged.
When ice is floating ,
Weight of ice  = Buoyant force
ρiVg = ρwVsg
Fraction of iceberg submerged = VsV=ρiρw                                              =0.9171= 0.91




 

Page No 74:

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 ρ is suspended by a mass-less spring of spring constant k. This block is submerged inside into the water in the vessel. What is the reading of the scale?

Answer:

Let, ρw = density of water,
ρ = density of block,
Vs = volume of block submerged.
When the block is submerged in water it will experience a buoyant force FB , in turn its reaction pair will act on the water which results in increase in reading of scale. Since before submerging the block the reading of scale was set to zero, after submerging the reading of scale will be equal to FB.
FB = VsρwgVs = MρFB=Mρρwg
 
 



Page No 75:

Question 10.15:

A cubical block of density ρ 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?

Answer:

ρ = density of block
ρ= density of water
L = side of cube
x = fraction of side submerged
m = mass of cube L3
a = acceleration of elevator

when the vessel is at rest and the block floats with fraction x submerge

buoyant force  = weight of block
ρwL2(xL)g = ρL3g
x=ρρw

when the vessel is in elevator which is accelerating in upward direction with acceleration a, and we look from the frame of reference of vessel then their will be pseudo force acting on the block in the opposite direction of acceleration which has a magnitude of m.a.
Note that when the vessel accelerates in upward direction the pressure at depth h in fluid will become ρh(g + a), where (g + a) is the effective acceleration due to gravity. Thus the buoyant force on volume submerged will be ρV(g + a

In equilibrium with respect to frame of vessel the equation of motion will become
F'B = mg + maρwL2(x'L)(g+a) =ρL3 g + ρL3aon solving we will get x' = ρρw=x 
It means, the fraction of the block submerged is independent of any acceleration, whether gravity or elevator.


 

Page No 75:

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 × 10–5 m. The surface tension of sap is T = 7.28 × 10–2 Nm–1 and the angle of contact is 0°. Does surface tension alone account for the supply of water to the top of all trees?

Answer:

Radius, = 2.5⨯10-5 m
Surface Tension, = 7.28⨯10-2 N/m
density, ρ = 103
Angle of contact (θ) = 0°
Maximum height to which sap can rise in trees through capillarity action,
h =2Scosθrρg   =2×7.28×10-2×cos 02.5×10-5×1×10-3×9.8   = 0.6 m

from above we can observe that maximum height to which the sap can rise due to surface tension is 0.6 m. Many trees have heights much greater than 0.6 m, so only this action is not sufficient for supply of water to the top of such long trees.

Page No 75:

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 θ. If the acceleration is a ms–2, what will be the slope of the free surface?.

Answer:


Let us consider consider a small element of water contained in a small cross-sectional area A having length L as shown in diagram:


Net force acting on it,
F1-F2=ma=ALρaPatm+ρgh1A-Patm+ρgh2A=ALρah1-h2g=Lah1-h2L=ag
As shown in the diagram, 
tanθ=h1-h2L =ag       θ= tan-1 ag
Hence, the required slope of the free surface is tan-1 ag.





 

Page No 75:

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 × 10–3 Nm–1.

Answer:

When two drops collapse to form a bigger drop, volume remains conserved.
Radius of smaller drop, r1 = 0.1 cm
Radius of bigger drop, r2 = 0.2 cm 
Surface tension, T = 0.4355 Nm-1
Let V1 and V2 be volume of smaller and bigger drop respectively.
Then volume of bigger drop (V) of radius (R) will be given by 
VV1+V2
43πR3 = 43πr13+ 43πr23R3 = r13+ r23     = (0.1)3 + (0.2)3 = 0.009R = 0.21 cm

Then, change in surface area:
A = 4πR2 -(4πr12  + 4πr22) 
Hence, the energy released in the process,
E = T×A     =T×4πR2 -(4πr12  + 4πr22)    =T×4πR2 -(r12  + r22)    =0.4355×4×3.14(0.21×10-3)2 -(0.1×10-3)2+(0.2×10-3)2    =-32.23×10-7 J







 

Page No 75:

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:

when the big drop breaks into N droplets its volume remains conserved,
Volume of big drop = N⨯Volume of one droplet
43πR3=N×43πr3R = N13 r

Energy released,
E =T×A=T[4πR2-N(4πr2)]       =4πT(R2 - Nr2)
Due to this release of energy the temperature is lowered
If c is the specific heat of liquid, ρ is the density and its temperature is lowered by T , then
mcT=ET=Emc      =4πT(R2 - Nr2)43πR3ρc      =3Tρc1R - Nr2R3      =3Tρc1R - 1r
since,
R > r ⇒ 1R<1r ⇒ 1R-1r<0
T will be negative. Hence, temperature will decrease. 

 

Page No 75:

Question 10.20:

The surface tension and vapour pressure of water at 20°C is 7.28 × 10–2 Nm–1 and 2.33 × 103 Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?

Answer:

Surface Tension, (T) = 7.28⨯10-2 N/m
Vapour Pressure,  (P) = 2.33⨯103 Pa
Radius of drop = r
The excess pressure in the drop 2Tr should be less than the vapour pressure, then the drop can form without evaporating
for smallest spherical droplet to form,
Vapour pressure = Excess pressure in drop
2Tr=Pr=2TP=2×7.28×10-22.33×103=6.25×10-5 m
 

Page No 75:

Question 10.21:

(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, 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 p0.
(c) If p0 = 1.03 × 105 Nm–2, ρ0 = 1.29 kg m–3 and g = 9.8 ms–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) Pressure will decrease as we are go up the atmosphere because the thickness of the gases above us decreases.
                                                     
      Let us consider a horizontal layer of air with cross-section A and height dh.
      Let , the pressure at the top of layer is p and pressure at the bottom of layer is  p+dp.
      Where dp is the change in pressure of top most layer and bottom layer.
      The layer is in equilibrium . So, the net upward force must be balanced by the weight.
      ∴  net upward force = net downward force
      (p + dp)A − PA = − ρgAdh
      dp= −ρgAdh
      Negative sign shows that pressure decreases with height.

(b) Let ρo be the density of air on the surface of the earth.
      According to the problem, pressure is proportional to density,
        p∝ ρ
      Pressure at some height (p)Pressure at the surface of Earth (po) =ρρoρ=ρopop       [ dp =-ρgdh]dp=ρopogdhpopdpp=-ρogpo0hdh     ln ppop=-ρogpoh0hlnppo = -ρogpohp = poeρoghpo

c) As  p= poeρoghpo
and it is given that po = 1.013⨯105 Nm-2
ρo = 1.29 kgm-3, g =9.8 ms-2, p = po10, h = ?
By substituting the values in the above relation we get,
lnppo=-ρogpohln110ppo=-ρogpohln110=-1.29×9.81.013×105×hon solving we geth=0.1843×105 m   =18.43 km

d) The assumption p ∝ ρ is valid only for the isothermal (constant temperature) case which is only valid for small distances.


 

Page No 75:

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 vaporisation for water Lv = 540 k cal kg–1, the mechanical equivalent of heat J = 4.2 J cal–1, density of water ρw = 103 kg l–1, Avagadro’s No NA = 6.0 × 1026 k mole–1 and the
molecular weight of water MA = 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=MANA×1ρw13 and find its value.
(c) 1 g of water in the vapor state at 1 atm occupies 1601cm3. Estimate the inter-molecular 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.

Answer:

(a) According to the problem, Lv= 540 kcal =540⨯103⨯4.2 J = 2268⨯103 J
Therefore, energy required to evaporate 1 k mole (18 kg) of water
= 2268⨯103⨯18
=4.0824⨯107 J

Since there are NA molecules in MA kg of water. the energy required for 1 molecule to evaporate is
U=MALvNA Jwhere NA = 6×1026U=4.0824×1076×1026J   =6.8×10-20 J

(b) Let water molecules to be point masses and are placed at distance d from each other,
Volume of NA molecule of water =MAρw
Thus the volume around one molecule is 
=volume of 1 kmolnumber of molecules/kmol=MANAρw
also volume around 1 molecule  = d3  

Thus by equating these, we get:
d3=MANAρwd=MANAρw13=186×1026×103   =3.1×10-10 m

(c) Volume occupied by 1 k mole (18 kg) of water molecules
=1601×10-6×18×103=28818×10-3 m3
Since 6⨯1026 molecules occupies 18⨯1601⨯10-3 m3 
Volume occupied by 1 molecule,
==28818×10-36×1026=48030×10-30 m3if d' is the intermolecular distance, thend'3=48030×10-30 m3so, d'=36.3×10-10 m

(d) Work done to change the distance from d to d' is U=F(d'-d),
 This work done is equal to energy required to evaporate 1 molecule,
F (d' − d) = 6.8×10-20 m3
F=6.8×10-20d'-d   =6.8×10-2036.3×10-10-3.1×10-10   =2.05×10-11 N

(e) Surface Tension = Fd=2.05×10-113.1×10-10=6.6×10-2 N/m



Page No 76:

Question 10.23:

A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is 20°C? (Assume air is an ideal gas, R = 8.314 J mole–1K-1, 1 atm. = 1.013 × 105 Pa; the membrane tension is 5 Nm–1.)

Answer:

Inside temperature of the balloon = Ti = 60°C = (60+273) K = 333 K
Outside temperature =To = 20°C =(20+273) K = 293K
Outside pressure = Po = 1.013×105 Nm-2
Tension in the membrane, S = 5 Nm–1
If , Pi  be the pressure inside, then
Pi-Po=2Sr  ( where, r = radius of the balloon)Also, for ideal gas,PiV=niRTini=PiVRTi where,ni=no of moles of air inside the balloonMass of ni mole of air,Mi=niM=PiVRTiM     (where, M= molar mass) Simillarly, Mo=noM=PoVRToM 
As, air contains  21% O2 and 79% N2 ,
So, molar mass of the air,
M=0.21×32+0.79×28=28.84 gWeight of the mass lifted by the balloon,W=no-niMg=MVRPoTo-PiTig=M43πr3RPoTo-PiTigBy putting the required value in the above expression in SI unit we get, W=3044.2 N
Hence, required mass that can be lifted,
=Wg=3044.210=304.4 kg



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