Mechanical Properties of Fluids

Class 11 NCERT Physics

NCERT

1   Explain why$\\$ (a)The blood pressure in human s is greater at the feet than at the brain$\\$ (b)Atmospheric pressure at a height of about $6 km$ decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than $100 km .$$\\$ (c)Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Solution :

(a)The pressure of a liquid is given by the relation:$\\$ $P = h \rho g$$\\$ Where,$\\$ $P =$ Pressure$\\$ $h = $Height of the liquid column$\\$ $\rho=$ Density of the liquid$\\$ $g =$ Acceleration due to the gravity$\\$ It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, the blood pressure at the feet is more than it is at the brain.$\\$ (b)Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about $6 km,$ density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of $6 km$ from the surface, it decreases to nearly half of its value at the sea level.$\\$ (c)When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

2   Explain why$\\$ (a)The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.$\\$ (b)Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)$\\$ (c)Surface tension of a liquid is independent of the area of the surface$\\$ (d)Water with detergent dissolved in it should have small angles of contact.$\\$ (e)A drop of liquid under no external forces is always spherical in shape

Solution :

The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact $(\theta)$, as shown in the given figure.$\\$ $S_{la} , S_{sa} ,$ and $S_{sl}$ are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i.e., $\\$ $\cos \theta =\dfrac{S_{sa}-S_{sl}}{S_{la}}$$\\$ The angle of contact $\theta$ , is obtuse if $S_{sa} < S_{la}$ (as in the case of mercury on glass). This angle is acute if $S_{sl} < S_{la}$ (as in the case of water on glass).$\\$ (b)Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.$\\$ On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.$\\$ (c)Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.$\\$ (d)Water with detergent dissolved in it has small angles of contact $(\theta)$ . . This is because for a small $\theta$ , there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact $(\theta)$ . If $ \theta $ is small, then $\cos \theta $ will be large and the rise of the detergent water in the cloth will be fast.$\\$ (e)A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

3   Fill in the blanks using the word(s) from the list appended with each statement:$\\$ (a)Surface tension of liquids generally . . . with temperatures (increases / decreases)$\\$ (b)Viscosity of gases. with temperature, whereas viscosity of liquids . . . with temperature (increases / decreases)$\\$ (c)For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to . .. (shear strain / rate of shear strain)$\\$ (d)For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)$\\$ (e)For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller)

Solution :

(a)decreases$\\$ The surface tension of a liquid is inversely proportional to temperature.$\\$ (b)increases; decreases$\\$ Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.$\\$ (c)Shear strain; Rate of shear strain$\\$ With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.$\\$ (d)Conservation of mass/Bernoulli’s principle$\\$ For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bernoulli’s principle.$\\$ (e)Greater$\\$ For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds’ numbers are associated with the motions of the two planes.

4   Explain why$\\$ (a)To keep a piece of paper horizontal, you should blow over, not under, it$\\$ (b)When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers$\\$ (c)The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while ad ministering an injection$\\$ (d)A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel$\\$ (e)A spinning cricket ball in air does not follow a parabolic trajectory

Solution :

(a)When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.$\\$ (b) According to the equation of continuity:$\\$ Area × Velocity = Constant$\\$ For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.$\\$ (c)The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.$\\$ (d)When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity:$\\$ Area × Velocity = Constant$\\$ According to the law of conservation of momentum, the vessel attains a back ward velocity because there are no external forces acting on the system.$\\$ (e) A spinning cricket ball has two simultaneous motions - rotatory and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.$\\$

5   A $50 kg$ girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter $1.0 cm.$ What is the pressure exerted by the heel on the horizontal floor?

Solution :

Mass of the girl,$m=50 kg$$\\$ Diameter of the heel,$d=1 cm =0.01 m$$\\$ Radius of the heel,$r=\dfrac{d}{2}=0.005 m$$\\$ Area of the heel$=\pi r^2$$\\$ $=\pi (0.005)^2\\ =7.85 * 10^{-5}m^2$$\\$ Force exerted by the heel on the floor:$\\$ $F = mg$$\\$ $=50*9.8\\ 490 N$$\\$ Pressure exerted by the heel on the floor:$\\$ $P=\dfrac{\text{Force}}{\text{Area}}\\ =\dfrac{490}{7.85*10^{-5}}\\ =6.24 * 10^6N m^{-2}$$\\$ Therefore, the pressure exerted by the heel on the horizontal floor is$6.24*10^6N m^{-2}.$

6   Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984 kg m^{-3}$ . Determine the height of the wine column for normal atmospheric pressure.

Solution :

Density of mercury, $\rho_1 = 13. 6 * 10^ 3 kg / m^3$$\\$ Height of the mercury column,$ h_ 1 = 0.76 m$$\\$ Density of French wine, $\rho_ 2 = 984 kg / m^3$$\\$ Height of the French wine column $= h_ 2$$\\$ Acceleration due to gravity, $g = 9.8 m / s^2$$\\$ The pressure in both the columns is equal, i.e.,$\\$ Pressure in the mercury column = Pressure in the French wine column$\\$ $\rho_1h_1g=\rho_2h_2g$$\\$ $h_2=\dfrac{\rho_1h_1}{\rho_2}$$\\$ $=\dfrac{13.6*10^3*0.76}{984}$$\\$ $=10.5$$\\$ Hence, the height of the French wine column for normal atmospheric pressure is $10.5 m.$

7   A vertical off-shore structure is built to withstand a maximum stress of $10 ^9 P a$. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly $3 km,$ and ignore ocean currents.

Solution :

Answer: Yes$\\$ The maximum allowable stress for the structure, $P =10^ 9 Pa$$\\$ Depth of the ocean, $d = 3 km = 3 * 10^3 m$$\\$ Density of water,$\rho=10^3 kg / m^3$$\\$ Acceleration due to gravity, $g = 9.8 m / s ^2$$\\$ The pressure exerted because of the sea water at depth, $d =\rho dg$$\\$ $= 3 * 10^ 3 * 10^3 * 9.8$$\\$ $= 2.94*10^ 7 Pa$$\\$ The maximum allowable stress for the structure ($10^9 Pa$) is greater than the pressure of the sea water $(2.94 *10^7 Pa )$ . The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

8   A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000 kg$. The area of cross-section of the piston carrying the load is $425 cm ^2$ . What maximum pressure would the smaller piston have to bear?

Solution :

The maximum mass of a car that can be lifted, $m = 3000 kg$$\\$ Area of cross-section of the load-carrying piston, $A = 425 cm^2 = 425 * 10 ^{-4} m ^2$$\\$ The maximum force exerted by the load, $F = mg$$\\$ $= 3000 × 9.8$$\\$ $= 29400 N$$\\$ The maximum pressure exerted on the load-carrying piston,$P=\dfrac{F}{A}$$\\$ $=\dfrac{29400}{425*10^{-4}}\\ =6.917*10^5 Pa$$\\$ Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is $6.917 * 10^5 Pa$

9   A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with $10.0 cm$ of water in one arm and $12.5 cm$ of spirit in the other. What is the specific gravity of spirit?

Solution :

The given system of water, mercury, and methylated spirit is shown as follows:

Height of the spirit column, $h _1 =12.5 cm = 0.125 m$$\\$ Height of the water column, $h_ 2 =10 cm = 0.1 m$$\\$ $P _0 =$ Atmospheric pressure$\\$ $\rho _1 =$ Density of spirit $\\$ $\rho_ 2=$ Density of water $\\$ Pressure at point $B = P_0 + h _1 \rho_ 1 g$$\\$ Pressure at point $D = P_ 0 + h_ 2 \rho_ 2 g$$\\$ Pressure at points B and D is the same.$\\$ $P _0 + h _1 \rho_ 1 g = h_ 2 \rho_ 2 g$$\\$ $\dfrac{\rho_1}{\rho_2}=\dfrac{h_2}{h_1}$$\\$ $=\dfrac{10}{12.5}=0.8$$\\$ Therefore, the specific gravity of spirit is $0.8.$

10   In problem $10.9$, if $15.0 cm$ of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury$ = 13.6$)

Solution :

Height of the water column, $h _1 = 10 + 15 = 25 cm$$\\$ Height of the spirit column, $h_ 2 =12.5 + 15 = 27.5 cm$$\\$ Density of water, $\rho_ 1 = 1 g cm^{-3}$$\\$ Density of spirit, $\rho_2 = 0.8 g cm^{-3}$$\\$ Density of mercury $=13.6 g cm^{-3}$$\\$ Let h be the difference between the levels of mercury in the two arms. Pressure exerted by height h, of the mercury column:$\\$ $= h \rho g\\ = h * 13.6 g .........(i )$$\\$ Difference between the pressures exerted by water and spirit:$\\$ $h _1 \rho _1 g - h _2 \rho_ 2 g\\ =g(25 × 1 - 27.5 × 0.8)\\ =3g ... (ii)$$\\$ Equating equations (i) and (ii), we get:$\\$ $13.6 hg = 3g\\ h = 0.220588 \approx 0.221 cm$$\\$ Hence, the difference between the levels of mercury in the two arms is $0.221 cm.$