Mechanical Properties Of Fluids
Class 11th Physics NCERT Exemplar Solution
Mcq I
Question 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
Answer:
When a pebble is dropped in a fluid, a variable force called viscous force acts on it. This force increases the velocity of the pebble until it becomes constant. This constant velocity is called the critical velocity. Since, the force is variable, acceleration is variable and so v-t graph is curved and not a straight line to which critical velocity vc is an asymptote.
Question 2.
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
Answer:
When a pebble is dropped in a fluid, a variable force called viscous force acts on it. This force increases the velocity of the pebble until it becomes constant. This constant velocity is called the critical velocity. Since, the force is variable, acceleration is variable and so v-t graph is curved and not a straight line to which critical velocity vc is an asymptote.
Question 3.
Which of the following diagrams (Fig. 10.2) does not represent a streamline flow?
A.
B.
C.
D.
Answer:
The following diagram does not represent a streamline flow.
A streamline can be straight or curved and tangent at which gives the direction of the flow.
As two streamlines cannot cross each other, the given diagram does not represent a streamline.
Question 4.
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:
According to the law of continuity,
Along a streamline,
Av=constant.
This means that at a particular cross-section, the velocity of all fluid particles is constant.
Hence, we can say that the velocity of all fluid particles crossing a given position is constant.
Question 5.
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:
Given:
Diameter of section1 = d1 = 2.5 cm
Diameter of section2 = d2 = 3.75 cm
According to the law of continuity,
Along a streamline,
Av=constant.
Question 6.
The angle of contact at the interface of water-glass is 0°, Ethyl alcohol-glass is 0°, Mercury-glass is 140° and Methyliodideglass 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. ethyl alcohol
C. mercury
D. methyl iodide.
Answer:
It is given that the meniscus is convex. This means the angle is obtuse. This indicates that the liquid in the trough is mercury. This can also, be understood from the fact as: -
The surface is always perpendicular to the resultant force. Since the meniscus is convex, the resultant force is inside the liquid. That means that cohesive force is greater than adhesive force like in the case of mercury.
Mcq Ii
Question 1.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:
Molecules inside a liquid experience a similar force of attraction due to liquid molecules around it from all directions. Molecules on the surface experience force of attraction by the liquid molecules below it and the air molecules above it. The force of attraction of liquid-liquid is greater than that of liquid-gas. Hence, it experiences a net downward force. As a result of this, the potential energy of a surface molecule is more than that of a molecule inside.
Question 2.
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:
During calculation of pressure, only the component of force that is normal to the surface is taken into consideration. It is defined as the ratio of magnitude of force normal to the magnitude of the area of the surface.
Hence, 
Hence, pressure is a scalar quantity as it is a ratio of two scalars.
Question 3.
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.
Answer:
When the coin is on the top of the block, it displaces volume V1 of water whose weight is equal to the weight of the coin. Whereas, in the second case, the coin displaces the volume of water V2 whose volume is equal to the volume of the coin.
According to case 1,
Here, 
So, 
This means in the second case less volume is displaced and hence h decreases.
As the coin falls into the water, total weight of the block-coin system decreases and less water is displaced than before and hence I decreases.
Question 4.
With increase in temperature, the viscosity of
A. gases decrease.
B. liquids increase.
C. gases increase.
D. liquids decrease.
Answer:
With increase in temperature, the rate of diffusion of gases increases and hence, the viscosity of gases increases.
With increase in temperature, the cohesive force between the liquid molecules decreases and hence the viscosity decreases.
Question 5.
Streamline flow is more likely for liquids with
A. high density.
B. high viscosity.
C. low density.
D. low viscosity
Answer:
Critical velocity is the velocity of the liquid up to which liquid can be streamlined and above which it is turbulent.
It is related to coefficient of viscosity and density as,
Hence, for a streamline flow the critical velocity should be high and hence density should be low and coefficient of viscosity should be high. High coefficient of viscosity indicates high viscosity of a liquid.
Vsa
Question 1.Is viscosity a vector?
Answer:
Viscosity is the measure of thickness of a liquid. It is a property of the liquid. It has magnitude but no direction. Hence, viscosity is a scalar and not a vector quantity.
Question 2.
Is surface tension a vector?
Answer:
Surface tension is defined as the force per unit length acting along the imaginary line drawn perpendicular to the tangent line on the surface of the liquid. Its direction is unique and defined. It only depends upon the magnitude force and the length taken into consideration. As it is a ratio of two scalars, surface tension is a scalar and not a vector quantity.
Question 3.
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 volume of iceberg be V1.
We know,
Weight of iceberg = weight of displaced water
Volume of submerged iceberg = volume of displaced water = V2.
Since,
Weight of iceberg = weight of displaced water
Therefore, 0.917 fraction of the total volume of the iceberg is submerged.
Question 4.
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 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:
In the above given case, a vessel filled with water is kept on a weighing pan and the scale is adjusted to zero. Then, a block is suspended by a spring and submerged into the water.
The net force on the block zero as it is in equilibrium and it is given by,
, where x is the compression of the spring.
The reading of the scale as a result of this is the buoyant force on the block due to the water. This force is given by,
Since, the scale is adjusted to zero before the block is suspended into the vessel, the new reading is,
Question 5.
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:
Let ρ be the density of the block, L be its height and A be its area and ρw be the density of water.
Let l be the length of the submerge part, then l = Lx
Here, in case1,
Weight of the block = weight of submerged water
In the case2, in the frame relative to the elevator, the effective coefficient of gravity becomes g’ = g + a
Weight of the block = weight of submerged water
Sa
Question 1.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:
Given:
Radius = r = 2.5×10-5 m
The surface tension of sap = T = 7.28×10-2 Nm-1
angle of contact = 0°
The maximum height h to which sap can rise due to capillary action is given by,
where ρ is density of water.
Since, a lot of trees have heights greater than 0.6m, surface tension does not alone account for the supply of water to the top of all trees.
Question 2.
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 m s-2, what will be the slope of the free surface?
Answer:
Every surface aligns itself such that it is perpendicular to the net force so that the net force along the surface is zero.
When the tanker moves with acceleration a, a pseudo force acts on the oil in the opposite direction.
Here,
So, the slope of the free surface is a/g
Question 3.
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 N m-1.
Answer:
Given:
Radius1 = r1 = 0.1cm = 10-3m
Radius2 = r2 = 0.2cm = 2×10-3m
Here, two droplets form a bigger drop.
So, volume of bigger drop = sum of the volumes of the smaller drops.
Now, energy released is equal to the difference in the surface energies.
Hence,
of energy is absorbed.
Question 4.
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:
Here, N droplets form a bigger drop.
So, volume of bigger drop = sum of the volumes of the smaller drops.
Now, energy released is equal to the difference in the surface energies.
Now, if 
is the temperature drop, m is the mass and c is the specific heat then,
is negative as R>r.
Hence, the drop-in temperature is, 
Question 5.
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:
Given:
Surface Tension = T =7.28×10-2 Nm-1
Vapour pressure = P = 2.33×103 Pa
Radius = r
Now,
Excess pressure = 2T/r
We know, the drop will evaporate when excess pressure is equal to vapour pressure.
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Question 1.(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 N m-2, ρ0 = 1.29 kg m-3 and g = 9.8 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) As we go up in the atmosphere, pressure decreases.
Consider a layer of atmosphere at height h and area A.
Let the pressure at h be p + dp.
Let the pressure at h + dh be p.
This layer is in equilibrium and hence, upward and downward forces are balanced.
Pressure × Area = Force = ma = 
Negative sign indicates that pressure decreases with height.
(b) Here, 
Hence, 
Integrating on both sides,
(c) 
(d) We can assume only under isothermal conditions which is valid near the surface of the earth and not at great heights.
Question 2.
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 ρ w = 103 kg l –1, Avogadro’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 [
]1/3 and find its value.
(c) 1 g of water in the vapor state at 1 atm occupies 1601cm3. 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.
Answer:
(a) Given:
Latent heat of vaporisation of water = Lv = 540 k Cal kg-1 = 540 ×103 ×4.2 J = 2268×103J
Energy required to evaporate 1kmole of water is given by,
E
We know,
The number of molecules in 1kmole of water are NA.
Energy required for evaporation of 1 molecule is given by,
(b) Assume molecules to be at a distance d from each other.
Volume around 1 molecule = d3
Also, we know, volume of NA whose mass is MA is given by,
So, volume of 1 molecule = 
Equating we get,
(c) Volume of 1g of water =1601 cm3 = 1.601 × 10-3m3
There are NA molecules in 18kg of water. So, volume occupied by 1 molecule is given by,
If d’ is the intermolecular distance and d’3 is the volume of one molecule, then
(d) The work done by force in move a molecule from distance d to d’ is equal to the energy required to evaporate one molecule.
(e) Here,
Question 3.
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 -1 K-1, 1 atm. = 1.013×105 Pa; the membrane tension is 5 N m-1.)
Answer:
Given:
Inside temperature = Ti = 60°C = 333K
Outside temperature = To= 20°C = 293K
Outside pressure = Po = 1.013×105Nm-2
Let the pressure inside the balloon be Pi, T be the surface tension and r be the radius of the balloon, then by formula for excess pressure,
Assume air to be ideal gas, then by formula for an ideal gas,
If there are ni moles of air inside the balloon, then
where Mi is the mass of air inside the balloon and MA is the molar mass.
Similarly,
where Mo is the mass of air inside the balloon and MA is the molar mass.
Now, in order to calculate Pi, lets calculate excess pressure.
Now, since this is negligible, we can assume,
Pi = Po = 1.013×105Nm-2
Assume, 21% of O2 and 79% of N2 in the atmosphere.
We can calculate MA as,
We know, the mass M lifted by the balloon can be found as,
So, the balloon can carry a mass of 301.46kg.