## Tili Kannada Text Book Class 8 Vyakarana Dvirukti – Jodi Nudi Nudigattugalu

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## Karnataka 1st PUC Physics Question Bank Chapter 10 Mechanical Properties Of Fluids

### 1st PUC Physics Mechanical Properties Of Fluids Textbook Questions and Answers

Question 1.
Explain why
(a) The blood pressure in humans is greater at the feet than at the brain.
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half 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 the pressure is force divided by area, and force is a vector.
(a) P = h pg, where h is the height of the liquid column. Since the height of blood tolumn at the feet is greater than at the brain, therefore, the blood pressure in humans is greater at the feet than at the brain.
(b) The density of air in the atmosphere does not decrease linearly with height. It decreases exponentially with height. For this reason, atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level.
(c) Hydrostatic pressure is transmitted equally in all directions. Hence no definite direction is associated with pressure. Therefore, the pressure is a scalar quantity.

Question 2.
Explain why
1. The angle of contact of mercury with glass Is obtuse, while that of water with glass is acute.
When a small quantity of liquid is poured on solid, three interfaces, namely liquid – air, solid – air, and solid-liquid are formed. The surface tension corresponding to these three layers are SLA, SSA, and SSL respectively. Let θ be the angle of contact between the liquid and solid.

The molecules in the region, where the three interfaces meet is in equilibrium. It means that net force acting on them is zero. For the molecule at K to be in equilibrium, we have
SSL + SLA cos θ = SSA or cos θ = $$\frac{\mathrm{S}_{\mathrm{SA}}-\mathrm{S}_{\mathrm{SL}}}{\mathrm{S}_{\mathrm{LA}}}$$ In case of mercury-glass, SSA < SSL, therefore cos θ is negative or θ > 90° i.e. (obtuse). In case of water-glass SSA > SSL, therefore cos θ is positive or θ < 90° (acute).

2. Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops.
In the case of mercury the mercury glass, angle of contact is obtuse. To obtain this obtuse value of angle of contact, mercury tends to form a drop. But in the case of water, the water-glass angle of contact is acute. Thus the water tends to spread out to achieve this acute angle of contact.

3. The surface tension of a liquid is independent of the area of the surface.
The surface tension of a liquid is the force acting per unit length on a line drawn tangentially to the liquid surface which is at rest. As this force is independent of the area of the liquid surface, the surface tension is also independent of the area of the liquid surface.

4. Water with detergent dissolved in ‘ it should have small angles of contact.
The cloth has narrow spaces in the form of capillares. The rise of liquid in a capillary tube is directly proportional to cos θ. If θ is small cos θ will be large. Hence capillary rise will be more so that the detergent will penetrate more in cloth and hence gets dissolved.

5. A drop of liquid under no external forces is always spherical in shape.
In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. For a given volume, the surface area of the sphere is the least and hence the liquid drop takes the spherical shapes.

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
1. Surface tension of liquids generally …….with temperatures (increases /decreases)
decreases

2. Viscosity of gases ……with temperature, whereas Viscosity of liquids …….with temperature (increases/decreases).
increases, decreases

3. 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).
Shear strain, rate of shear strain.

4. For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle).
Both conservation of mass and Bernoulli’s principle.

5. For the model of a plane in a wind tunnel, turbulence occurs at a ….. speed for turbulence for an actual plane (greater / smaller).
Greater

Question 4.
Explain why.
1. To keep a piece of paper horizontal, you should blow over, not under, it.
If we blow over a piece, velocity of air above the paper becomes more than that below the paper. This reduces the pressure of the air above the paper in accordance with Bernoulli’s theorem. But the pressure of air below the paper is still atmospheric and is higher than the pressure above. Hence the paper remains horizontal and does not fall.

2. When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
When we try to close the water top with our fingers, the area of outlet of water eject gets reduced. Therefore in accordance with the principle of Continuity (AV = constant), the velocity of the water will increase creating fast jets of water.

3. The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
The size of the needle controls the velocity of flow and the thumb pressure controls the pressure. According to Bernoulli’s equation:
P + ρgh+ $$\frac{1}{2} \rho$$ v2 = constant
we can note that the equation varies linearly with pressure, P but varies as the square of the velocity. Hence the contribution of the velocity of flow is greater, due to which the needle has better control over the flow.

4. A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
When a fluid flows out of a small hole in a vessel, it acquires a large velocity and hence has a large momentum. Since no external force is acting on the system, a backward velocity must be acquired by the vessel (By-law of conservation of momentum). As a result backward thrust is experienced by the vessel.

5. A spinning cricket ball In air does not follow a parabolic trajectory.
A spinning ball displaces air. The ball moves formed and relative to it, the air moves backward. Hence, the velocity of air above the ball relative to it is larger and below it is smaller. This difference in the velocities of air results in a pressure difference between the lower and upper faces leading to a net upward force on the ball. Therefore the ball does not follow a parabolic trajectory.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel Is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Force exerted by the heel due to the weight of the girl, F = mg = 50 × 10 = 500N
Diameter of the heel’s circular are, d = 1 cm = 10-2m
∴ Area of the heel $$\frac{\pi \mathrm{d}^{2}}{4}$$
= $$\frac{\pi \times\left(10^{-2}\right)^{2}}{4}=7.85 \times 10^{-5} \mathrm{m}^{2}$$
∴ Pressure exerted by the heel = $$\frac{F}{A}$$
$$=\frac{500}{7.85 \times 10^{-5}}$$
= 6.366 × 106 Pa.

Question 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. (Assume g = 9.8 m/s2)
Normal atmospheric pressure = 1 atm
⇒ gauge pressure of wine = 1 atm
ρgh = 1.013 × 105 Pa (Since 1 atm = 1.013 × 105pa)
⇒ h = $$\frac{1.013 \times 10^{5}}{984 \times 9.8}$$
⇒ h = 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand maximum stress of 109 Pa. 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.
The depth of ocean, h = 3 km = 3000 m.
Density of water a 1000 kg m 3 (assumed) ; g = 9.8 ms’2.
∴ Pressure, P = ρgh = 1000 x 9.8 x 3000 = 2.94 x 107 Pa.
This value of pressure is less than the stress of 109 Pa which the structure can withstand. Therefore, the structure is suitable.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of cross¬section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear.
Let Fmax be the maximum force experienced by the Bigger piston.
Fmax = mg = 3000 × 9.8 = 29400 N
Area of the bigger piston = 425 cm2
= 425 x 10-4m2
Maximum pressure on the bigger piston
pmax = $$\frac{F_{\max }}{A}$$
$$P=\frac{29400}{425 \times 10^{-4}}$$ = 6.92 × 105 Pa
Since pressure is transmitted uniformly throughout the liquid the smaller piston will also bear a pressure of 6.92 × 105Pa.

Question 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?

$$\rho_{1}$$ → density of water
$$\rho_{2}$$ → density of spirit
h1 → height of water column
h2 → height of spirit column
Let Po be the atmospheric pressure
⇒ PA = Po + $$\rho_{1}$$ gh1 ….(1)
PB = Po + $$\rho_{2}$$ gh2 …..(2)
Since the mercury levels are same at point A and B we can say that,
PA = PB
⇒ Po + $$\rho_{1}$$ gh1 = Po + $$\rho_{2}$$ gh2
from (1) and (2)
⇒ $$\rho_{1}$$ gh1 = $$\rho_{2}$$ gh2
⇒ $$\frac{\rho_{2}}{\rho_{1}}=\frac{h_{1}}{h_{2}}=\frac{10}{12.5}=0.8$$
∴ specific gravity of spirit
$$=\frac{\text { density of spirit }}{\text { density of water }}=\frac{\rho_{2}}{\rho_{1}}=0.8$$

Question 10.
In the previous problem, 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)
Since we are putting equal amount of water and spirit and specific gravity of spirit is < 1 we can assume that the mercury level will rise on the spirit side,
$$\rho_{1}$$ → density of water,
$$\rho_{2}$$ → density of spirit,
$$\rho_{3}$$ → density of mecury,

Since A and B are at the same height,
PA = PB
we know that, PA = $$\rho_{1}$$gh1
and PB = $$\rho_{2}$$ gh2 + $$\rho_{3}$$ gx
⇒ $$\rho_{1}$$ gh1 = $$\rho_{2}$$ gh2 + $$\rho_{3}$$ gx
⇒ x = $$\frac{\rho_{1} h_{1}-\rho_{2} h_{2}}{\rho_{3}}$$
x = $$\frac{\mathrm{h}_{1}-\frac{\rho_{2}}{\rho_{1}} \mathrm{h}_{2}}{\frac{\rho_{3}}{\rho_{1}}}$$
where $$\frac{\rho_{2}}{\rho_{1}}$$ ⇒ specific gravity of spirit and $$\frac{\rho_{3}}{\rho_{1}}$$ ⇒ specific gravity of mercury.
Thus x = $$\frac{25-0.8 \times 27.5}{13.6}$$ = 0.221 cm
Therefore the difference in the mercury level in the arms is 0.221 cm.

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Bernoulli’s theorem is applicable only for the ideal fluids in streamlined motion. Since the flow of water in a river in a rapid way (i.e., turbulent) can not be treated as streamlined motion and hence the theorem can not be used.

Question 12.
Does it matter if one uses gauge Instead of absolute pressures in applying Bernoulli’s equation? Explain.
No, unless the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1cm. If the amount of glycerine collected per second at one end is 4.0 x 10-3 kg s-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 103 kg m-3 and viscosity of glycerine = 0.83 Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
The volume of liquid floawing out per second is given by, $$Q=\frac{\pi R^{4}\left(p_{1}-p_{2}\right)}{8 h l}$$
⇒ pressure difference at tube ends, $$\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{8 \mathrm{Q} \eta}{\pi \mathrm{r}^{4}}$$
where Q is the volume of liquid flowing per second = $$\frac{\text { mass of liquid flowing per second }}{\text { density of liquid }}$$
$$=\frac{4 \times 10^{-3} \mathrm{kg} / \mathrm{s}}{1.3 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}}$$
= 3.077 x 10-6 m3/s
Therefore,
P1 – P2 = $$=\frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{3.14 \times(1 \times 10)^{3}}$$
= 9.76 x 102 N/m2
A pressure difference of 9.76x 102 N/m2 is obtained.
Reynolds Number,
R = $$\frac{4 \rho V}{\pi d n}$$ = $$\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{-6}}{\pi \times(0.02) \times 0.83}$$ = 0.3
Since the Reynolds number is about 0.3, the flaw is laminar.

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s-1 and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m-3.
P1 = pressure at the lower surface of wings
P2 = pressure at the lower surface of wings
ρ = Density of air
V1 = Speed of wind at the lower surface = 63 m/s.
V2 = speed of wind at the upper surface = 70 m/s
According to Bernoulli’s theorem,
P1 + $$\frac{1}{2} \rho V_{1}^{2}$$ = P2 + $$\frac{1}{2} \rho V_{2}^{2}$$
or P1 – P2 = $$\frac{1}{2} \rho\left(V_{1}^{2}-V_{2}^{2}\right)$$
P1 – p2 = $$\frac{1}{2}$$ × 1.3 × (702 – 632) = 605.15 pa
Force on wings = (P1 – P2) × area
= 605.15 × 2.5 = 1512.87 N
∴ The lift on the wings is about 1.512 × 103N

Question 15.
Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Figure (a) is incorrect. From the equation of continuity AV = constant. Thus, when the area of constriction is narrow, velocity of streamline is high. If the velocity is high, we can say that the pressure head is low at that point (from Bernoulli’s equation). Correspondingly the height of the liquid column should be lesser at the narrow construction than that of the wide construction. Hence figure (a) is incorrect.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the. tube is 1.5 m min-1, what is the speed of ejection of the liquid through the holes?
Number of holes = 40
Diameter of each hole, D = 10-3m
Area of cross section of each hole
$$=\frac{\pi D^{2}}{4}$$
$$=\frac{\pi \times 10^{-6}}{4}$$
Total area of cross section of 40 holes,
a2 = $$\frac{40 \times \pi \times 10^{-6}}{4}$$ = π × 10-5m2
speed of liquid inside the tube,
V1 = 1.5 m/min = $$\frac{1.5}{60}$$ m/s
Area of cross section of tube,
a1 = 8.0cm2 = 8 × 10-4m2
If V2 is the velocity of ejection of liquid through the holes then from equation of
continuity,
a1 v1 = a2 v2
⇒ v2 = $$\frac{a_{1} v_{1}}{a_{2}}=\frac{8 \times 10^{-4} \times 1.5}{60 \times \pi \times 10.5}$$
⇒ v2 = 0.637 m/s
∴ the speed of ejection is 0.637 m/s.

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10-2N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
The soap film has two free surfaces.
∴ Total length of the film to he supported,
l = 2 × 30 = 60 cm = 0.6
Total force on the slider due to surface tension will be, F = S × 2l = S × 0.6N
At equilibrium , the force F on slider due to surface tension should balance the weight mg
∴ F = mg = 1.5 × 10-2 N
⇒ S × 0.6 = 1.5 × 10-2
⇒ S = 2.5 × 10-2 N/m.

Question 18.
Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.

In all three figures, the liquid is at the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and (c) is the same as in figure.
1. Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20 °C) is 4.65 × 10-1 N m-1. The atmospheric pressure is 1.01 × 105 Pa. Also, give the excess pressure inside the drop.
r = 3mm = 3 × 10-3m Surface tension of mercury, S = 4.65 × 10-1 Nm-1. Atmospheric pressure, Po = 1.01 × 105 Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= $$\frac{2 \mathrm{S}}{\mathrm{r}}$$ + Po
= $$\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}$$ + 1.01 × 105
Total Pressure = 1.013 × 105 Pa
Excess pressure = $$\frac{2 \mathrm{S}}{\mathrm{r}}$$
$$=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}$$
= 310 Pa.

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10-2 N/m? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Radius of the soap bubble, r = 5mm
= 5 × 103m
Surface tension of the soap solution,
S = 2.5 × 10-2 N/m
Since the soap Bubble has two surfaces,
Excess pressure, P = $$\frac{4 \mathrm{S}}{\mathrm{r}}$$
$$=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$$
P = 20 Pa
Excess pressure inside the soap bubble is 20 Pa
Radius of air bubble, r = 5 mm = 5 × 10-3
Depth, h = 40 c = 0.4 m
Density of soap solution, ρ = 1.2 × 10-3 kg/m2
Atmospheric pressure, Po = 1.01 × 105 Pa
g = 9.8 m/s2
Excess pressure inside air bubble,
P1 = $$\frac{25}{r}$$
⇒ P1 = $$\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$$
P1 = 10Pa
At a depth of 0.4m, the total pressure inside the air bubble,
PTotal = Po + ρgh + P1
PTotal = 1.01 × 105 + 1.2 × 103 × 9.8 × 0.4 + 10
⇒ PTotal = 1.057 × 105Pa
∴ The pressure inside the air bubble is 1.06 × 105Pa.

Question 21.
A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4 m. compute the force necessary to keep the door closed.
Assume g = 9.8 m/s2
Density of water, $$\rho_{1}$$ = 103 kg/m3
height of water column, h1 = 4m
∴ Pressure due to water, P1 = $$\rho_{1}$$ gh1
= 103 × 9.8 × 4
= 3.92 × 104 Pa
Density of acid
= Relative, densityx density of water = 1.7 × 103 kg/m3
height of acid column, h2 = 4m
∴ Pressure due to acid, P2 = $$\rho_{2}$$ gh2
= 1.7 × 103 × 9.8 × 4
⇒ P2 = 6.664 × 104Pa
Pressure difference between the water and acid columns :
∆ P = P2 – P1
∆ P = 6.664 × 104 – 3.92 × 104
⇒ ∆ P = 2.744 × 104 Pa
Area of the door, a = 20 cm2
= 20 × 10-4m2
∴ Force exerted on the door, F = ∆ P × a
⇒ F = 2.744 × 104 × 20 × 10-4
F = 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury,

1. Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), In units of cm of mercury.
2. How would the levels change in case of (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

1. Difference between the mercury levels in the two limbs gives gauge pressure
∴ Gauge pressure, Pa = 20 cm of mercury
Atmospheric pressure, Po = 76 cm of mercury
∴ Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm + 20 cm
= 96 cm of mercury

2. Difference between the levels of mercury in the two limbs = – 18 cm
∴ Gauge pressure = – 18cm of mercury
∴ Absolute pressure = Atmospheric pressure + Gauge pressure
= 76cm – 18 cm
= 58 cm of mercury.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Two vessels having water-filled up to the same height, exert equal pressure on their bases. Also since the base area is the same, they exert an equal amount of force on the base. (F = PA). Force exerted on the sides of the vessels has non-zero vertical components. When their vertical components are added, the total force experienced by one vessel will be different than the other vessel depending on the shape of the vessels. That is why the two vessels filled with water to the same vertical height show different readings on a weighing machine.

Question 24.
During a blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
Gauge pressure, P = 2000 Pa
Density of blood, ρ = 1.06 × 103 kg/m3
g = 9.8 m/s2
Let the height of the blood container be h
∴ pressure of the blood container, P = ρgh
⇒  ρgh = 2000
h = $$\frac{2000}{1.06 \times 10^{3} \times 9.8}$$
⇒ h = 0.1925 m
The blood may just enter the vein if the blood container is kept at a height slightly greater than 0.1925 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.

1. What is the largest average velocity of blood flow in an artery of diameter 2 × 10-3 m if the flow must remain laminar?
2. How does the pressure change as the fluid moves along the tube if dissipative forces are present?
3. Do the dissipative forces become more important as the fluid velocity Increases? Discuss qualitatively.

1. Diameter of artery = 2 × 10-3m
Viscosity of blood, = 2.084 × 10-3Pa s
Density of blood, ρ = 1.06 × 103kg/m3
Reynolds number for laminar flow,
NR = 2000
The largest average velocity of blood is
$$\mathrm{V}_{\mathrm{avg}}=\frac{\mathrm{N}_{\mathrm{R}} \eta}{\rho \mathrm{d}}$$
$$=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}$$
= 1.966 m/S
The largest average velocity of blood is 1.966 m/s.

2. If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces due to which the pressure drop becomes large.

3. The dissipative forces become more important with increasing flow velocity. This is because of the rise of turbulence. Turbulent flow causes dissipative losses in fluid.

Question 26.

1. What is the largest average velocity of blood flow in an artery of radius 2 × 10-3m if the flow must remain laminar?
2. What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10-3 Pas).

1. Diameter of artery, d = 2 × 2 × 10-3 m
= 4 × 10-3 m
Viscosity of blood, h = 2.084 × 10-3 m Pas
Density of blood, ρ = 1.06x 10-3 kg/m3
Reynold’s number for laminar flow,
NR = 2000
The largest average velocity of blood is,
$$\mathrm{V}_{\mathrm{avg}}=\frac{\mathrm{N}_{\mathrm{R}} \eta}{\rho \mathrm{d}}$$
$$V_{\text {avg }}=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}$$
V = 0.983 m/s
Hence, the largest average velocity of blood is 0.983 m/s.

2. Flow rate is given by
R= πr2Vavg
R = 3.14 × (2 × 10-3)2 × 0.983
⇒ R = 1.235 × 10-5m3/s
Therefore, the corresponding flow rate is 1.235 × 10-5m3/s

Question 27.
A plane is in level flight at constant speed and each of Its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m-3).
Total area of wings, A = 2 × 25
= 50m2
speed of air lover the longer wing,
V1 = 180 km/h = 50 m/s
speed of air over the upper wings,
V2 = 243 km/h = 65 m/s
Density of air, ρ =1kg/m3
Let P1 and P2be the pressure of air over the lower wing and upper wing respectively.
From Bimoull’s equation,
P1 + $$\frac{1}{2} \rho V_{1}^{2}$$ = P2 + $$\frac{1}{2} \rho V_{2}^{2}$$
P1 – P2 = $$\frac{1}{2} \rho\left(V_{2}^{2}-V_{1}^{2}\right)$$
∴ pressure difference, ∆ P
= $$\frac{1}{2} \rho\left(V_{2}^{2}-V_{1}^{2}\right)$$
= $$\frac{1}{2}$$ × 1 × (652 – 502)
= 862.5 Pa
The upward force on the plane, F = ( ∆ P) A
= 862.5 × 50
=43125 N
We know that the upward force balances the weight of the plane
∴ F = mg
43125 = m × 9.8
⇒ m = 4400.51 kg.

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kg m-3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.(Take g = 9.8 ms2)
Radius of the given unchanged drop, r = 2 × 10-5 m
Density of the uncharged drop,
ρ = 1.2 × 103 kg/m-3
Viscosity of air, $$\eta$$ = 1.8 × 10-5 Pa S
The density of air ($$\rho_{0}$$) can be taken as zero in order to neglect the buoyancy of air.
Terminal velocity (v) is given by the relation:

V = 5.807 × 10-2 m/s
V = 5.8 cm/s
The terminal speed of the drop is 5.8 cm/s
Viscous force, F = π 6 hrv
∴ F = 6 × 3.14 × 1.8 × 10-5 × 2 × 105 × 5.8 × 102
⇒ F = 3.9 × 10-10N
The viscous force on the drop is
= 3.9 × 10-10 N

Question 29.
Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? The surface tension of mercury at the temperature of the experiment is 0.465 N m. Density of mercury = 13.6 × 103 kg m-3.
Angle of contact 1 θ = 140°
Radius of the tube, r = 1 mm = 10-3m
Surface tepsion of mercury S = 0.465 N/m
Density of mercury, ρ = 13.6 × 103 kg/m3
g = 9.8 m/s2
Let the Dip in height of mercury be h Surface tension is given by,
S= $$\rho_{o} e^{-y / y_{o}}$$
⇒ h = $$\frac{2 \mathrm{S} \cos \theta}{\rho \mathrm{gr}}$$
⇒ h = $$\frac{2 \times 0.465 \times \cos 140^{\circ}}{13.6 \times 10^{3} \times 9.8 \times 10^{-3}}$$
h = – 0.00534 m = -5.34 mm

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are Joined together to form a U-tube open at both ends. If the U-tube contains water, what Is the difference In its levels in the two limbs of the tube? The surface tension of water at the temperature of the experiment is 7.3 × 10-2 N/m. Take the angle of contact to be zero and the density of water to be 1.0 × 103 kg m-3 (g= 9.8 m s-2).
Radius of the first bore, r1 = $$\frac{3 \mathrm{mm}}{2}$$ = 1.5 × 10-3 m
Radius of the second bore, r2 = $$\frac{6 \mathrm{mm}}{2}$$ = 3 × 10-3 m
Surface tension of water, S = 7.3 × 10-2 N/m
Angle of contact, θ = 0°
(give in the question) Density of water, ρ = 1 × 103 kg/m3
Let h1 and h2 he the heights to which the water rises in the 3 mm and 6 mm diameter bores respectively.We have,
h1 = $$\frac{2 \mathrm{S} \cos \theta}{\mathrm{r}_{1} \rho \mathrm{g}}$$
h2 = $$\frac{2 \mathrm{S} \cos \dot{\theta}}{\mathrm{r}_{2} \rho \mathrm{g}}$$
Difference in water level = h1 – h2

= 4.966 × 10-3m
= 4.97 mm
The difference in the water levels in the cores is 4.97 mm.

Question 31.
1. It is known that density P of air decreases with height y as$$\rho_{\mathrm{o}} \mathrm{e}^{-y / \mathrm{y}_{0}}$$ where $$\rho_{0}$$ = 1.25 kg/m-3 is the density at sea level, and yo is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains constant (isothermal conditions). Also, assume that the value of g remains constant.
2. A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains a constant radius as It rises. How high does it rise?
[Take y0 = 8000m and $$\rho_{\mathrm{He}}$$ = 0.18kg/m3].
1. Volume of the balloon, V = 1425m3
Payload mass, m = 400 kg
g = 9.8 m/s2
Give that, yo = 8000m
$$\rho_{\mathrm{He}}$$ =0.18 kg/m3
$$\rho_{\mathrm{o}}$$ = 1.25 kg/m3
Density of the balloon = ρ
Let the height to which the balloon rises be y Density (ρ) of air decreases with height (y) as,

we can infer from (1), that rate of decrease of density with height is directly proportional to ρ,i.e.,

$$\frac{\mathrm{d} \rho}{\rho}=-\mathrm{kdy}$$
where, K is the constant of proportionality Height changes from o to y, while density changes from $$\rho_{\mathrm{o}}$$
to ρ by integrating.
we get:

from (1) and (2),
yo = 1/K
⇒ K = $$\frac{1}{y_{0}}$$
From (2) and (3) we get,

2. Density,

⇒ ρ = 0.46 kg/m3
we have proved that,

⇒ y = – 8000 × loge $$\frac{0.46}{1.25}$$
⇒ y = 8000m = 8km .
Hence balloon will rise to a height of 8 km.

### 1st PUC Physics Mechanical Properties Of Fluids One Mark Questions and Answers

Question 1.
Pascal’s law is valid for dynamic fluids. True/False.
False. Valid only for static fluid.

Question 2.
What is the relation between absolute pressure and gauge pressure?
Absolute pressure = atmospheric pressure ± gauge pressure.

Question 3.
Angle of contact for a glass tube dipped in mercury is obtuse. True / False.
True.

Question 4.
Water is an ideal fluid. True/False.
False. It is real fluid.

Question 5.
Define thrust.
The total normal force exerted by the fluid at rest on a surface is called thrust.

Question 6.
What is the dimensional formula for pressure?
ML-1T-2

Question 7.
What happens to the viscosity of a gas when the temperature is increased?
Viscosity of gases increases as the temperature is increased.

Question 8.
How does the coefficient of viscosity of liquid vary with temperature?
The coefficient of viscosity of liquids decreases as temperature is increased.

### 1st PUC Physics Mechanical Properties Of Fluids Two Marks Questions and Answers

Question 1.
Name two applications of Pascal’s law.
Hydraulic lifts, hydraulic breaks.

Question 2.
What is streamline? What are its properties?
The path taken by a fluid particle under a steady flow is a streamline.
Properties:

1. The tangent at any point in streamline gives direction of fluid velocity at that point.
2. No two streamlines can cross.

Question 3.
Define the specific gravity of a fluid. What is its unit?
Specific gravity is defined as a ratio of a density of a fluid to the density of water at 4°C. It has no unit.

Question 4.
Which are the factors on which viscous force depends?
Viscous force acting between two layers of liquids depends on,

1. area of the layers
2. relative velocity of the two layers and
3. distance between two layers.

Question 5.
A force 6f 1 kN is applied on a road surface by the tyre of a car. If the contact area of the tyre is 10 cm2, find the pressure on the contact area.
Pressure = $$\frac{\text { Force }}{\text { area }}$$
$$\frac{1000 N}{10 \times 10^{-4} M^{2}}$$ = 106 Pa

Question 6.
State whether true or false :

1. An ideal liquid has non zero viscosity but zero compressibility.
2. With increase in temperature, the viscosity of liquid decreases but the viscosity of gases increases.

1. False
2. True.

Question 7.
A bigger raindrop falls faster than a smaller one (True/False) Explain.
True.
The raindrop moves with terminal velocity due to viscous drag of the air. The terminal velocity of drop varies as the square of its radius. Hence, a bigger drop has a higher terminal velocity than a smaller one.

Question 8.
Machine parts get jammed during winter. Why?
In winter, due to low temperature, the viscosity of oil between the machine parts increases considerably resulting in jamming of machine parts.

Question 9.
Discuss the effect of temperature on the viscosity of gases and liquids.
Viscosity of liquid decreases with increase in temperature and viscosity of gases increases with increase in temperature.

Question 10.
What is an ideal liquid?

1. An ideal liquid is incompressible meaning density of liquid remains irrespective of the pressure.
2. An ideal liquid is non-viscous. No tangential forces between layers of liquid in relative motion.
3. Ideal liquid cannot withstand any shearing stress.

Question 11.
Explain the effect of density and pressure on the viscosity of liquids/gases.
Liquids:
1. Viscosity increases with increase in density of the liquid. Viscosity increases with increase in pressure (except water).

2. Gases:
Viscosity decreases with increase in density. Viscosity decreases with increase in pressure.

Question 12.
The excess pressure inside a soap bubble is four times the excess pressure inside a moving soap bubble. What is the ratio of the first and second bubble?
Let r1 and r2 be radius of soap bubbles. Excess pressure in them are
$$\frac{4 \sigma}{r_{1}}$$ and $$\frac{4 \sigma}{r_{2}}$$ where σ is surface tension.
Given P1 = 4P2 $$\frac{4 \sigma}{r_{1}}$$ = 4 $$\frac{4 \sigma}{r_{2}}$$
ratio of volume of I to II
$$=\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{r_{1}^{3}}{64 r_{1}^{3}}=\frac{1}{64}$$

Question 13.
Explain why water with detergent dissolved in it should have small angles of contact.
Cloth has narrow spaces inform of capillaries. The rise of liquid in a capillary tube is directly proportional to cos e.
θ ∝ $$\frac{1}{\cos \theta}$$. Hence e should be less in order that detergent penetrates more in clothes.

Question 14.
What is a venturi meter? Name two applications.
Venturi meter is a device to measure the flow speed of incompressible fluid. The principle is used in filter pumps, Bunsen burner, atomiser.

Question 15.
Water is filled as shown In following containers. In which case highest force is exerted on the base of the container? Why?

Water exerts same pressure at the bottom of the container, regardless of its shape. Since height, his constant, equal pressure is exerted.
P = ρgh (h = constant)

Question 16.
Is it better to wash clothes in hot soap solution? Why?
Yes. Surface tension decreases with temperature. Hence, in hot soap solution, spreading of solution over the surface of clothes happens easily. Hot soap solution can penetrate and clean clothes better.

Question 17.
Explain why a drop of liquid under no external force is always spherical in shape.
Due to surface tension, a drop of liquid tries to acquire a shape with minimum surface area. For a given volume, surface area of sphere is least. Hence, liquid drop takes a spherical shape.

Question 18.
Point out any two properties required for a fluid to be used in a barometer.

1. Density of fluid should be high so as to avoid use of long barometer,
2. Temperature variation of fluid should be minirial.

Question 19.
Why does water flow faster that honey?
The coefficient of viscosity of water is less compared to that of honey. This makes water to move faster.

Question 20.
What are the factors affecting viscosity?

1. Increase in temperature decreases viscosity.
2. Increases in pressure increases viscosity in liquids. In water, it decreases whereas, in gases, it remains the same.

Question 21.
What do you mean by angle of contact of a liquid with a solid surface. What are the factors that effect it?
The angle made by the tangent drawn to the meniscus from the point of contact with the walls of the container measured from within the liquid is called angle of contact. It depends on the atmosphere pressure, adhesive and cohesive forces of the liquid in the tube.

Question 22.
Bernoulli’s equation can be used to describe the flow of a river rapidly. True/False. Explain why?
False. Bernoulli’s equation can be applied only to streamline flow.

Question 23.
Archimedes’s Principle holds good in a vessel at free fall. True/False. Give reason.
False
The principle does not hold good in this particular case as the vessel in free fall is in a condition of weightlessness, where the buoyant force accounting for the Archimedes’s principle does not exist.

Question 24.
Why are cars and aeroplanes streamlined?
The shape of the aeroplane is streamlined because when it flies in air. The velocity on the top surface is more than the bottom surface. Hence the pressure on the top surface decreases. This causes an upward thrust on the wings of the place which gives uplift to the aeroplane.

Question 25
Why do some liquids rise and some liquids dip in a capillary tube? Explain
The rise or fall of liquid in a capillary tube is due to surface tension. If the liquid wets the glass then there is rise in the liquid inside the capillary tube.

Question 26.
Define cohesive force and adhesive, force of molecules.
Force of attraction between molecules of the same substance is called cohesive force. Force of attraction between molecules of the different substances is called adhesive force.

Question 27.
Diameter of a ball A is thrice that of B. What will be the ratio of their terminal velocities in water?
Terminal velocity × (rad. of ball)2 Required ratio is $$\frac{3^{2}}{1}$$ = 9

Question 28.
A hole of area 2 mm opens near bottom of a large water storage tank and stream of water shoots from it. If top of the water in the tank is to be kept 30 m above leaking point, how much water in litres Is should be added to reservoir tank to keep this level? Take g = 10 m/s2
Velocity of the outflowing water
v = $$\sqrt{2 g h}=\sqrt{2 \times 10 \times 30}$$ = 24.49 m/s
Quantity of water flowing out per sec.
= av = 10-5 × 4 × 24.49
= 97.96 × 10-6
= 9.796 × 10-5 × 103 litres/s
= 97.96 ml/s.

Question 29.
A small ball of mass m and density ρ is dropped in a viscous liquid having density $$\rho_{0}$$. After some time, ball falls with a constant velocity. Calculate viscous force acting on the ball.
Volume of the ball, v = m/ρ; mass of liquid displaced by the ball.
$$\mathrm{m}^{\prime}=\left(\frac{\mathrm{m}}{\rho}\right) \rho_{0}$$
ρ = density of ball
When the ball falls with a constant velocity
⇒ viscous force = effective weight of the ball or F = mg = m’g = (m – m’)g
$$=\left(\mathrm{m}-\frac{\mathrm{m} \rho_{0}}{\rho}\right) \mathrm{g}=\mathrm{mg}\left(1-\frac{\rho_{0}}{\rho}\right)$$

Question 30.
What is Reynold’s number? What is its significance?
Turbulent flow is less likely for viscous fluid flowing at low rates. A dimensionless number, whose value gives one an approximate idea whether the flow would be turbulent. This number is called the Reynold’s Re.
$$R_{e}=\rho v d / \eta$$
ρ → density of fluid
speed v,d stands for the dimension of the pipe.
$$\eta$$ → viscosity of the fluid. Significance is that,
flow is lamina’s/streamline for Re <1000 flow is turbulent for Re > 2000.

Question 31.
A wooden log 100 mm × 100 mm × 5 m hangs vertically from a vertical string so that 3 m length of log Is submerged in water. Find the tension in string. Take specific gravity of wood = 0.65.

FB + T = W
(0.1 × 0.1 × 3) × 9810 + T
= (0.1 × 0.1 × 5) × 0.65 × 9810
T = 24.52 N
(0.1 × 0.1 × 3) × 9810 + T
= (0.1 × 0.1 × 5) × 0.65 × 9810
∴ T = 24.52 N

Question 32.
A piston of small cross-section A1 is used to exert a force F1 on liquid to transfer it from the existing cylinder to a larger cylinder attached with a larger piston of area A2. What is the weight, say if a truck that can be placed on a platform supported by the larger piston due to the force exerted on the liquid in existing cylinder?
F2 = $$\frac{\mathrm{F}_{1} \mathrm{A}_{2}}{\mathrm{A}_{1}}=\frac{100 \mathrm{kN} \times 1 \mathrm{m}^{2}}{10 \times 10^{-4} \mathrm{m}^{2}}$$
= 105 kN
∴ Weight of truck = 105 kN.

Question 33.
The pressure inside a droplet of water is 10.3 N/m2 in excess of atmospheric pressure. Its diameter is 50 mm. Find surface tension of the water film.
P = $$\frac{2 \lambda}{r}$$
λ = $$\frac{P r}{2}=\frac{10.3 \times 25 \times 10^{3}}{2}$$
= 128.75 N/m

### 1st PUC Physics Mechanical Properties Of Fluids Four/Five Marks Questions and Answers

Question 1.
Define viscosity. Discuss the cause the viscosity.
Viscosity is the property of a fluid, (liquid or gas) by virtue of which an internal frictional force comes into play when the fluid is in motion and opposes the relative motion of its different layers. Viscosity is due to the intermolecular forces which are effective when the different layers of the liquid are moving with different velocities.

These forces are of Vander waal type. Due to these forces, every fast-moving liquid layers tends to accelerate the adjoining slow-moving layers and every slow-moving layer tends to retard the adjoining fast-moving layer of liquid. As a result a backward dragging force called viscous drag comes into play which accounts for viscosity of liquid.

Question 2.
Derive the dimensional formula for co-efficient of viscosity and hence its unit.

Units:
We have $$\eta=\frac{F}{A d v / d n}$$
in cgs system, the unit of $$\eta$$ is called poise.
1 poise $$=\frac{1 \mathrm{dyne}}{1 \mathrm{cm}^{2} \times(1(\mathrm{cm} / \mathrm{s}) / \mathrm{cm})}$$
= dyne cm-2 sec.
The S.I.unit of $$\eta$$ is Pa.s (Pascal second) or deca poise.
1 dec poise = $$\frac{1 \mathrm{N}}{1 \mathrm{m}^{2}\left(1 \mathrm{ms}^{-1} / \mathrm{m}\right)}$$ = 1Nsm-2

Question 3.
State Stoke’s law? What are the factors on which viscous drag depends?
Stoke’s law states the backward dragging force (F) acting on a small spherical body of radius r, moving through a medium having coefficient of viscosity $$\eta$$ and velocity v is given by
F = 6π$$\eta$$rv
viscous drag (F) depends on

1. coefficient of viscosity of the medium $$\eta$$
2. velocity of the body (v)
3. radius of the spherical body (r)

Question 4.
How is the rise of liquid affected if top of the capillary is closed?
There will be a small rise in the capillary tube if top of capillary tube is closed. Because the rise of liquid in capillary tube due to surface tension will be opposed by the downward force exerted by the compressed air above the liquid in tube. This downward force increases with increase in height of liquid column. Therefore only a small rise of liquid column is possible in a capillary tube with a closed top.

Question 5.
What is terminal velocity? What are the factors on which terminal velocity depends?
The terminal velocity of an object is the maximum constant velocity acquired by the object while falling freely in a viscous medium.
Terminal velocity (v) of an object of radius (r) density $$(\rho)$$ moving through a viscous medius of viscosity $$\eta$$ and density $$\rho_{0}$$ is given by
$$\mathrm{v}=\frac{2}{9} \frac{\mathrm{r}^{2}}{\eta}\left(\rho-\rho_{0}\right) \mathrm{g}$$
Hence, terminal velocity depends on

2. coefficient of viscosity of the medium
3. density of the object.
4. density of the medium

Question 6.
Derive expression for capillary rise in a tube in terms of surface tension, radius of tube angle of contact and density of fluid.
Consider a tube of radius ‘a’ dipped in a container filled with fluid of density of ρ as shown.

h = capillary rise
s = surface tension
θ = angle of contact
The weight of column of liquid of height h in the tube is balanced by vertical component of the surface tensile force of fluid. Vertical component of the surface tensile force
= (S cos θ) × (length over which it acts)
= (S cos θ) × circumference of tube
= S cos θ × 2πa ……..(1)
Weight of column of fluid balanced by surface tension = volume of fluid × density × g
= (πa2h) × ρg ……(2)
Equating (1)and(2),
S cosθ × 2πa = πa2h ρg
h $$=\frac{S \cos \theta \times 2}{a \rho g}$$
$$=\frac{2 \mathrm{S} \cos \theta}{\mathrm{a} \rho \mathrm{g}}$$

Question 7.
State and prove the Archimedes principle?
Archimedes principle states that when a body is wholly partially immersed in a liquid at rest. The loss of weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body.
Proof:

Let A be the cross-sectional area of the top or bottom face of the object.
Volume of liquid displaced, v = volume of the object = Ah
∴ mass of liquid displaced, m = vρ = Ahρ
ρ is the density of the liquid.
Liquid pressure on the top of the body
P1 = xρg
Vertical downward thrust (Force f1) on top face of the object.
F1 = P1A = xgAρ
Liquid pressure on the bottom face of body P2 = (x + h) ρg
Vertical upward thrust (Force F2) on bottom force
F2 = P2 A(X + h) ρgA.
Since F2 >F1, the net upward force acting is F = F2 – F = (x + h) ρ gA – x ρgA
= hρgA = hA(ρg)
= mg
weight of liquid displaced.
True weight of the object = Mg
Upward thrust (Force F) on the body F = mg
Apparent weight of the body in liquid.
= W-F = Mg – mg
This means that observed weight at the body (object) immersed in a liquid is less than its true weight by an amount equal to weight of the liquid displaced by body.

Question 8.
State and prove Bernoulli’s theorem.
Bernoulli theorem states that for streamlined flow of an ideal liquid, the total energy (Pressure energy + potential energy + kinetic energy) per unit mass remains constant at every cross-section throughout the flow.
Proof:
Consider a tube AB of varying cross-section through which an ideal liquid is in streamlined flow.

Let P1 be the pressure applied on the liquid at A and P2 be the pressure against which liquid has to move out at B
a1, a2 be the cross-sectional area at tube at A and B
h1 and h2 are the mean height at section A and B from the reference level.
v1 and v2 are the velocity of liquid flow at A and B
ρ is the density of the ideal liquid flowing through the tube.
The liquid flows from A to B hence,
P1 > P2
The mass m of the liquid crossing per second through any section at the tube with respect to equation of continuity is,
a1v1 ρ = a2v2 = ρm
a1v1 = a2v2 = V (say)
Force on liquid at A = P1a1
Force on liquid at B = P2a2
Work done per second by liquid at A = P1a1 × v1 = P1V
Work done per second by liquid at B = P2a2 × v2 = P2V
Net work done per second on liquid by pressure energy moving from A to B is
= P1V- P2V = v (P1-P2)
Increase in potential energy per second as the liquid flows from A to B
= mgh2 – mgh,
(since v2 is greater than v1)
Increase in kinetic energy per second of liquid from A to B = 1/2 m$$v_{2}^{2}$$ – 1/2 m$$v_{1}^{2}$$
According to work – energy principle, work done per second by pressure energy
= increase in P.E per second + increase in K.E per second
or P1v – P2v = (mgh2 – mgh1) + $$\left(\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}\right)$$
i.e., P1v + mgh1 + $$\frac{1}{2} m v_{1}^{2}$$
= P2v + mgh2 + $$\frac{1}{2} m v_{2}^{2}$$
Dividing by m1

Hence, $$\frac{\mathrm{P}}{\rho}+\mathrm{gh}+\frac{1}{2} \mathrm{v}^{2}$$ = constant.
But, $$\frac{P}{\rho}$$ is the pressure energy per unit mass, gh is the potential energy per unit mass and $$\frac{1}{2} v^{2}$$ is the kinetic energy
per unit mass Hence,
Pressure Energy + Potential Energy + Kinetic Energy remains constant.
Hence proved.

Question 9.
Derive the equation at continuity.
Consider a non-viscous liquid in streamlined flow through a tube AB at varying cross-section.

Let a1, a2 be the area of cross-section of the tube at A and B.
v1, v2 = velocity of flow of liquid at A and B.
Volume of the liuid entering per second at A = a1v1
Mass of liquid entering per second at A = a1v1$$\rho_{1}$$
Similarly mass of liquid leaving per second at B = a2v2 $$\rho_{2}$$
Assuming, no loss of liquid in the tube and
steady flow, then mass of liquid entering at A/sec = mass of liquid leaving at B/sec.
⇒ a1v1$$\rho_{1}$$ = a2v2$$\rho_{2}$$
Assuming liquid is incompressible,
$$\rho_{1}$$ = $$\rho_{2}$$
⇒ a1v1 = a2v2
av = constant
This is the equation of continuity.

Question 10.
Explain the limitations of Bernoulli theorem.

1. While deriving the Bernoulli theorem, it is assumed that the velocity of every particle of liquid across any crosssection is uniform. Practically this is not possible.
2. Viscous drag of the liquid which comes into play when the liquid is in motion has not been considered.
3. While deriving the equation, it is assumed that there is no loss of energy when liquid is in motion. In fact, some KE is converted to heat energy.
4. If a liquid is flowing along a curved path, energy due to centrifugal force should also be considered.

Question 11.
Explain why the angle of contact of water with glass is acute while that mercury with glass is obtuse?

SSL + SLA cos θ = SSA
or cos θ = $$\frac{S_{S A}-S_{S L}}{S_{L A}}$$
In case of mercury – glass, SSA < SSL
∴ cos θ is negative or θ>90° i.e., obtuse.
In case of water – glass, SSA > SSL
cos θ is positive or θ<90° i.e., acute.
Note:
SSL:
Surface tension corresponding to solid-liquid interface.
SLA:
Surface tension corresponding to liquid – air interface.
SSA:
Surface tension corresponding to solid-air interface.

Question 12.
Derive an expression for velocity of fluid at wide neck in a venturi meter.
Speed of fluid flowing through tube at broad neck = V1
Speed of fluid flowing through tube at narrow neck = V2
By equation of continuity,
A1V1 = A2V2
V2 = $$\frac{A_{1} V_{1}}{A_{2}}=\frac{A V_{1}}{a}$$
From Bernoulli’s equation,

Question 13.
What is the Magnus effect? Explain.
A ball Which is spinning drags air along with it. If the surface is rough, more air will be dragged.

Ball is moving forward and relative to it air is moving backwards. This results in a larger velocity of air relative to the ball above it and a smaller velocity below it. This difference in velocities of air results in pressure difference between lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called the Magnus effect.

Question 14.
Define terminal velocity. Give reasons why a sphere attains this velocity. Derive a relation for the terminal velocity.
The constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by
v = $$\frac{2}{9} \frac{r^{2} g}{\eta}(\rho-\sigma)$$, where ρ and σ are densities of the body and liquid respectively, $$\eta$$ is the coefficient of viscosity of liquid and r is the radius of the spherical body.
The net force on the sphere becomes zero as the viscous force equals the apparent weight.
Consider a long column of dense liquid-like glycerine. As the ball is dropped in it, the forces experienced are;

1. weight = mg = $$\frac{4}{3} \pi r^{3} \rho g$$, where ρ is the density of the ball.
2. upthrust = U = $$\frac{4}{3} \pi r^{3} \rho_{1} g$$, where$$\rho_{1}$$ is the density of the liquid.
3. viscous force Fv = $$6 \pi \eta \rho v$$, where v is the terminal velocity.

Net force and the acceleration should be 0.
∴ mg – U – Fv = 0

Question 15.
Write the differences between laminas/streamline flow and turbulent flow.
1. Laminar Flow

• Layered flow with one layer of fluid sliding smoothly over the other.
• Reynold’s No. < 1000
• Observed usually in flow through porous materials.

2. Turbulent

• Haphazard flow with rapid and continuous mixing of fluid resulting in momentum transfer.
• Reynold’s No. > 2000
• Pipeline flow, river flow, etc.

Question 16.
A razor blade can be made to float on water. What forces act on this blade? Is Archimedes’s principle applicable?
Three forces are acting on the blade when it is made to float on water:

1. Weight of blade acting vertically downwards.
2. Reaction in blade exerted by the liquid surface acting vertically upwards.
3. Force of surface tension on circumference, of the blade acting tangentially to the liquid surface.
In this case, as no portion of razor blade is immersed in water, hence Archimedes’s principle is not applicable.

1st PUC Physics Mechanical Properties Of Fluids Numerical Problems Questions and Answers

Question 1.
A capillary tube of diameter 1.5 mm is dipped in

1. mercury.
2. water.

Find capillary rise for each case. Surface tension for water and mercury can be taken as 0.07 N/m and 0.52 N/m respectively. The contact angle for water and mercury can be taken as 0° and 130°.
1. h = $$\frac{4 \lambda \cos \theta}{\mathrm{dr}}$$
$$=\frac{4 \times 0.52 \times \cos (130)}{1.5 \times 10^{-3} \times 13.65 \times 9810}$$
= 6.66 mm (fall)

2. h = $$\frac{4 \lambda \cos \theta}{\mathrm{dr}}$$
$$=\frac{4 \times 0.07 \times 1}{1.5 \times 10^{-3} \times 9810}$$
= 19 mm (rise).

Question 2.
Calculate the energy evolved when 8 droplets of water (surface tension 0.72 N/m) of radius 0.5 mm combine into one.
Here, s = 0.072 N/m
r = 0.5 mm = 0.5 × 10-3 m.
Let R be the radius of big drop formed.
Volume of the big drop = volume of 8 small drops.
$$\frac{4}{3}$$ π R3 = 8 × $$\frac{4}{3}$$ πr3
or R = 2 r = 2 × 0.5 × 10-3= 10-3 m
Surface area of big drop :
= 4π R2 = 4π × (10-3)2
= 4π × 10-6m2
Surface area of 8 small drops:
= 8 × 4π × r2
= 8 × 4π × (0.5 × 10-3)2
= 8π × 10-6m2
Energy evolved = S.T × decrease in area.
= 0.072 × 4π × 10-6 = 9.05 × 107J.

Question 3.
A plate 0.025 mm distant from a fixed plate moves at 60 cm/s and requires a force of 2 N per meter square to maintain its speed. Determine fluid viscosity between the plates.
dy = 0.025 mm
= 0.025 × 10-3m
Velocity of upper plate .
= 60 cm/s = 0.6 m/s
Force on upper plate
= 2$$\frac{\mathrm{N}}{\mathrm{m}^{2}}$$ = Shear stress , τ
du = change of velocity = u – 0 = 0.6 m/s
dy = change of distance = 0.025 × 10-3m
Viscosity,
µ = $$\frac{\tau}{\mathrm{du} / \mathrm{dy}}=\frac{2}{\left(0.6 / 0.25 \times 10^{-3}\right)}$$
8.33 × 10-5 $$\frac{\mathrm{Ns}}{\mathrm{m}^{2}}$$ = 8.33 × 10-5 Pas

Question 4.

In the above figure, if area of plate is A, viscosity of fluid is µ, distance of plate from top and bottom plane surfaces are y1 and y2 respectively, find the expression for force F required to drag the plate at a velocity v.

Required force F = F1 + F2
$$=\frac{\mu_{1} A_{1} v_{1}}{y_{1}}+\frac{\mu_{2} A_{2} v_{2}}{y_{2}}$$
$$=\mu \mathrm{Av}\left(\frac{1}{\mathrm{y}_{1}}+\frac{1}{\mathrm{y}_{2}}\right)$$

Question 5.
An open tank contains water up to a depth of 3m and above it an oil of sp.gr 0.9 for a depth of 1m. Find pressure intensity

1. at the interface of two liquids
2. at the bottom of the tank.

Height of water, z1 = 3 m
Height of oil, z2 = 1 m
Density of oil
= Sp.gr of oil × Density of water
= 0.9 × 1000 kg/m3 = 9000 kg/m3
= $$\rho_{2}$$
Density of water = $$\rho_{1}$$ = 1000 kg/m3

1. At the interface, i.e., at A,
P= $$\rho_{2}$$ g × z2 = 900 × 9.81 × 1
= 8829 N/m2

2. At the bottom i.e., at B,
p =$$\rho_{2}$$gz2 + $$\rho_{1}$$gz1
= 900 × 9.81 × 1 + 1000 × 9.81 × 3
= 8829 + 29430 = 38259 N/m2

Question 6.
Determine pressure difference between two points A and B for the setup as shown in figure.

G = specific gravity,
PA = pressure at A,
PB = Pressure at B
Pressure at any point is the same horizontal line x – x should be same.
Hence, pressure at C in left column = pressure at D in right column.
Pressure at C
= (2 × 1000 × g × 2) + (1 × 1000 × g × 4) + PA …..(1)
pressure at D
= (4 × 1000 × g × 2) + PB ……(2)
Equating (1) and (2),
8000 g + PA = 8000 g + PB
PA – PB = 0

Question 7.
A closed tank contains 1m of mercury, 2m of water and 3 m of oil of sp.gravity 0.6. There is an unknown fluid in space above oil. If gauge pressure at the bottom of the tank is 200 kPa. What is the pressure on the top surface of oil?

Let pressure on top of oil be PA
Total pressure = PA + Poil +Pwater + Pmercury
P= ρgh
200 × 103 =PA + 0.6 × 1000 × g × 3 + 1000 × g × 2 + 13.6 × 1000 × g × 1
PA = 200 × 103 – 170694
= 29.306 kPa

Question 8.
The flow rate of water from a tap of diameter 1.2 cm is 0.48 L/min. The coefficient of viscosity of H20 is 10-3 Pa s. After some time the flow rate is increased to 4 L/min. Characterise the flow for both the flow rates.
Let the speed of the flow be v and the area of tap be d = 1.2 cm. The volume of the water flowing out per second is
Q = v × πd2 /4
v = 4Q/d2
Reynold’s number is
Re = 4P2 / πd$$\eta$$
= 4 × 103kg/m3× Q/(3.14 × 1.2 × 10-2m × 10-1Pa s)
= 1.061 × 108 m-3s   Q = 1.061 × 108Q
Initially,
Q = 0.48 L/min = 8.0 cm3/s
= 8.00 × 10-6m3/s
we obtain
R =484.8
Since this is below 1000, the flow is steady.
After sometime when Q = 4L/min,
= 66.67 cm3/s
= 6.67 × 10-5 m3s-1
we obtain
R = 1.061 × 108 × 6.67 × 10-5
= 7076.9
The flow now is turbulent.

Question 9.
Calculate the total energy possessed by one kg of water at a point where pressure is 30 gm wt/sq.mm. Velocity of 0.1 ms-1 and height is 60 cm above the ground.
Given, P = 30 g wt/sq.mm
= $$\frac{30}{1000}$$ × 9.8 × 106N/m2
v = 0.1ms-1 h = 0.6 m
Total energy per unit mass
$$\frac{P}{\rho}$$ + gh + 1/2 v2
= $$\frac{30 \times 9.8 \times 10^{3}}{10^{3}}$$ + 9.8 × 0.6 + 1/2 × (0.1)2
= 294 + 5.88 + 0.005
= 299.885 J.

Question 10.
Water flows at the rate of 4 litres per second through an orifice at the bottom of the tank which contains water 720 cm deep. Find the rate of escape of water If additional pressure of 20 kg/cm2 is applied to the surface of water.
Given, h = 720 cm v = $$\sqrt{2 g h}$$
v = $$\sqrt{2 \times 980 \times 720}$$ cm/s ;
= 1187.93 cm/s v
v = 4 litres
= 20000 g/cm2
$$\frac{20000 \times 980}{980}$$ Cm 0f Water
= 20,000 cm of water column.
Now, pressure head (h1) = 20,000 + 720
= 20,720 cm
New velocity v1 = $$\sqrt{2 \mathrm{gh}_{1}}$$
$$\sqrt{2 \times 980 \times 20,720}$$ cm/s
= 6372.69 cm/s
As v = av and V1 = av1
v1 = V$$\frac{v_{1}}{v}$$ = 4 × $$\sqrt{\frac{20720}{720}}$$
= 21.45 litres/s

Question 11.
A venturi meter is connected to two points in the mains where its radii are 20 cm and 10 cm. and the levels of water column in the tubes differ by 10 cms. How much water flows through the pipe per minute?
Volume of water flowing per second
v = a1a2$$\sqrt{\frac{2 g h}{a_{1}^{2}-a_{2}^{2}}}$$
where a1 = $$\pi r_{1}^{2}$$ = 22/2 × 2012cm2
a2 = $$\pi r_{2}^{2}$$ 22/7 × 1012cm2
g = 980 cm/s2;
h = 10 cm .
Volume of water flowing per minute

= 2726.58 litres/minute.

Question 12.
When a drop of mercury of radius R is split into n similar drops, What is the change in surface energy? Assume σ as surface tension of mercury.
Volume of initial mercury drop = $$\frac{4}{3}$$ πR3
r = radius of smaller drops,
volume conservation
⇒ $$\frac{4}{3}$$ πR3 = n $$\frac{4}{3}$$ πr3
r = Rn-1/3
Initial Surface Energy = σ × surface area
= σ × 4πR2
Final Surface Energy = n × σ × 4πr2
= nσ 4π[Rn-1/3]2
= nσ 4πR2n-2/3
= n1/34σ πR2
Change in surface energy
= 4σ πR2[1 – n1/3]

Question 13.

1. The diameter of pipe at section (1) and (2) are 10 and 15 cm respectively. Find discharge through pipe if the velocity of water flowing through pipe at section (1) is 5 m/s. Find also velocity at section (2)
2. Find the work done in blowing a soup bubble of surface tension 0.06 N/m from a 2 cm radius to a 5 cm radius.

1.

Q = A1V1 = π × $$\frac{0.1^{2}}{4}$$ × 5
= 0.0393 m3/s
V2 = $$\frac{A_{1} V_{1}}{A_{2}}$$ (continuity equation)
= $$\frac{0.0393}{\frac{\pi}{4} \times 0.15^{2}}$$ = 2.22 m/s

2. Here, s = 0.06 N/m
r1 = 2 cm = 0.02 m ;
r2 = 5 cm = 0.05 m
Since bubble has 2 surfaces, initial surface area of the bubble.
= 2 × 4πr12= 2 × 4π(0.02)2
= 32π × 10-4m2
Final surface area of the bubble
= 2 × 4πr22 = 2 × 4π(0.05)2
= 200π × 10-4m2
Increase in surface area
= 200π × 10-4 – 32π × 10-4
= 168π × 10-4m2
∴ work done = S × increase in surface area
= 0.06 × 168π × 10-4 = 0.003168 J

Question 14.
A tank is filled with water to a height H. At a depth ‘h’ from the free surface a hole is made so that the water comes out of it. What is the velocity of efflux? Also, what is the maximum range and the position of hole for the same? Find the time taken by a water molecule to reach the ground. Determine the horizontal length covered by the molecule.

Take two points at the same height (H- h) from ground one inside and one outside the hole. Applying Bernoulli’s theorem for the points, we have
$$\frac{P_{0}+h p g}{\rho g}$$ + 0 + (H – h) = (H – h) + $$\frac{v^{2}}{2 g}+\frac{P_{0}}{\rho g}$$
On Solving, we get, v = $$\sqrt{2 \mathrm{gh}}$$
The velocity of efflux thus depends on the depth of which the hole is made from the surface of the liquid. Time taken to reach
ground = t = $$\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}}$$,
Since v = $$\sqrt{2 \mathrm{gh}}$$ is horizontal and ax = 0;
R = vt = $$\sqrt{2 \mathrm{gh}}$$ $$\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}}$$
R = $$2 \sqrt{\mathrm{h}(\mathrm{H}-\mathrm{h})}$$
To maximise Range, $$\frac{\mathrm{d} \mathrm{R}}{\mathrm{dh}}$$ = 0
i.e., h = H/2
Maximum Range,
$$R_{\max }=2 \sqrt{\frac{H}{2}(H-H / 2)}=H$$.

Question 15.
Water flows through a horizontal pipe of varying cross-section. If the pressure is 1 cm of mercury when the velocity is 0.35 m/s. Find the pressure at a point where velocity is 0.65 m/s.
At 1st point, P1 = 1 cm of mercury
= 0.01m of Hg.
= 0.01 × (13.6 × 103) × 9.8 P a;
v1 = 0.35 m/s
At the second nd point,
p2 = ?
v2 = 0.65 m/s and
ρ = 103 hg/m3.
According to Bernoulli’s theorem,
p1 + 1/2 $$\rho \mathrm{V}_{1}^{2}$$ = p2 + 1/2 $$\rho \mathrm{V}_{2}^{2}$$
or P2 = P1 + 1/2 $$\rho\left(v_{2}^{2}-v_{1}^{2}\right)$$
= 0.01 × 13.6 × 103 × 9.8 – 1/2 × 103 × [(0.65)2 – (0.35)2]
= 1182.8 Pa ⇒ $$\frac{1182.8}{9.8\left(13.6 \times 10^{3}\right)}$$
= 0.00887 m of Hg.

Question 16.
If excess pressure inside a soap bubble of diameter 2 cm is balanced by that due to a column of oil of sp.gravity 0.8, 2mm high, find the surface tension of the soap bubble.
R = 1 cm = 10-2 m
ρ of oil = 0.8 × 103kg/m3
h = 2 mm = 2 × 10-3m
Pressure due to 2 mm column of oil,
P = hρg – (2 × 10-3)(0.8 × 10-3) × 9.8
= 2 × 0.8 × 9.8 Pa
In case of soap bubble, P = $$\frac{4 \mathrm{s}}{r}$$
or s = $$\frac{P r}{4}=\frac{2 \times 0.8 \times 9.8 \times 10^{-2}}{4}$$
= 3.92 × 10-2N/m

Question 17.
A spray pump having a cylindrical tube of cross-section 8 cm2 has 50 fine holes of radius 0.5 mm. Consider the flow of liquid inside the tube to be 1.5 m/min. What is the speed of ejection of the liquid through the whole?
Area of cross-section of the tube:
a1 = 8 cm2 = 8 × 10-4m2
Number of holes = 50
Diameter of each hole = 0.5 × 2 = 1 mm
= 10-3
= 0.5 mm = 5 × 10-4m
Area of cross-section of each hole
= π r2 = π (5 × 10-4)2 m2
Total area of cross-section of 50 holes,
a2= 50 × π(5 × 10-4) m2
Speed of the liquid inside the tube,
v1 = 1.5 m/min = $$\frac{1.5}{60}$$ m/s
If v2 is the velocity of ejection of the liquid through the holes,
then, a1v1 = a2v2 or v2 = a1v1/a2
∴ v2 = $$\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 50 \times \pi \times\left(5 \times 10^{-4}\right)^{2}}$$
= 0.5096 m/s.

Question 18.
Calculate the total energy/unit mass possessed by water at a point, where the pressures is 10gm f/sq mm, velocity is 0.1 m/s and height of water level from the ground is 0.2 m (g = 9.8 m/s2).
Here, P = 10 gm f/sq mm
= $$\frac{10}{1000}$$ × 9.8 × (103)2
= 98 × 103Nm-2
v = 0.1 m/s ;
ρ = 103 kg/m3;
h = 0.20 m
1. Pressure energy / unit mass
= $$\frac{P}{\rho}=\frac{9.8 \times 10^{3}}{10^{3}}$$ = 98 J/kg

2. Gravitational potential energy / unit mass = gh = 9.8 × 0.2 = 1.96 J/kg

3. Kinetic energy / unit mass
= 1/2 v2= $$\frac{1}{2}$$(0.1)2 = 0.0053/kg
∴ Total energy / unit mass
= $$\frac{P}{\rho}$$ + gh + 1/2 v2 = 98 + 1.96 + 0.005
= 99.965 J/kg

Question 19.
A mass of 15 kg is placed on the wider tube of a U tube as shown in the figure. Given that the area of the wider tube is 5 m2, find the difference in the water levels in the two tubes, (density of water = 1 kg/m3).

Area of the wider tube, A = 500 cm2
= 5 × 10-2 m2
The pressure applied by the mass at A is transmitted through the liquid and leads to a difference in the water levels.
Since the water level is same a A and B,
PA = PB ……(1)
But PA = $$\frac{F}{A}=\frac{m g}{A}=\frac{15 \times 9.8}{5}$$ = 29.4 Pa
Also we know that PB = ρgh = 1 × 9.8 × h
From (1) PB = PA
9.8 × h = 29.4
h = 3m
∴ The difference in water level obtained is 3 m.

Question 20.
If two liquids of mass m1 and m2 and density $$\rho_{1}$$, and $$\rho_{2}$$ are mixed together, then what is the density of the resulting mixture?
Volume occupied by first liquid,
V1 = $$\frac{m_{1}}{\rho_{1}}$$
Volume occupied by second liquid,
V2 = $$\frac{m_{2}}{\rho_{2}}$$
∴ density of the mixture,

Question 21.
The container shown in the figure below is filled with a liquid of density ρ. Note that the container has a height h1 and area of cross-section A1 for the upper half height h2 and area of cross-section A2 for the lower half. Find:

1. The pressure at the base of the container.
2. Force exerted by the liquid on the base of the container.

1. The pressure at the base of the container is due to the liquids both at the upper and lower half.
∴ PTotal = P1 + P2
PTotal = ρgh1 + ρgh2
PTotal = ρg(h1 + h2)

2. Force exerted on the bottom of the container,
P = PTotal × A2
F = ρg(h1 + h2) × A2.

## Karnataka 2nd PUC Physics Question Bank Chapter 6 Electromagnetic Induction

### 2nd PUC Physics Electromagnetic Induction NCERT Text Book Questions and Answers

Question 1.
Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

(a) According to Lenz’s law south pole forms at q and north pole a to p. Therefore, the induced current is from p to q.
(b) South pole develops at end q and at end x of the two coils as per Lenz’s law. Therefore, the induced current is from q to p in coil PQ and x to y in the coil.
(c) The induced current should flow clockwise in the right loop (when seen from the left) to oppose the current in the left loop. Hence, induced current flows along yzx.
(d) When the rheostat is adjusted to increase the current, magnetic flux through the neighbouring coil increases. As per Lenz’s law, the induced current in the neighbouring coil should produce magnetic flux in opposite direction to oppose the original magnetic flux. Hence, the induced current flows along
(e) When the circuit breaks, magnetic flux decreases. The induced current should flow along xzy to increase the magnetic flux.
(f) The magnetic flux threading the coil in the perpendicular direction is zero. Any change in current will not change this magnetic flux. Hence, no induced current is set up in the coil.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.

(a) The flux linked with the loop increases in this case so flux produced by the induced current should decrease it i.e. the direction of induced current should produce a magnetic field in a direction outwards the plane of the paper. The direction of current as per the right-hand thumb rule comes out to be along adcba. (i.e. anti­clockwise)

(b) In this case, the flux linked with the loop is decreasing so the direction of the induced current is anticlockwise so as to oppose the decrease in flux. Thus, the direction of the induced current is along a’d’c’b’.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
The magnetic field produced inside the solenoid,
B = µ0 nl
If A is the area of the loop placed inside the solenoid, then magnetic flux linked with the loop,
Φ = BA = µ0 n IA
If e is the induced e.m.f. produced due to change in current through the solenoid, then

Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving, out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Here, B=0.3 T,V=1 cm S-1 10-2 ms-1
In the above expression, 1 is the length of the arm, which is perpendicular to the direction of motion.
(a) B = 0.3 T, l = 8 x 10-2 m, v = 1 x 10-2 ms-1
$$\mathscr{E}$$ = Blv = 0.3 x 10-2 x 8 x 10-2 = 0.24 mV
v = $$\frac { l }{ t }$$
∴ t = $$\frac { l }{ v }$$
= $$\frac { 2 cm }{ 1 cm/s }$$
= 8 sec

(b) The emf developed when the velocity of loop is in the direction normal to the shorter side
$$\mathscr{E}$$ = 0.3 x 10-2 x 2 x 10-2 = 0.06 mV
t = $$\frac { l }{ v }$$
= $$\frac { 8 cm }{ 1 cm/s }$$
= 8 sec

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
The induced e.m.f. between the center and the circumference of the ring is given by
e = $$\frac { 1 }{ 2 }$$ Bl2 ω[in magnitude]
Here, B = 0.5 T; 1 = tn; ω=400 rad s-1
∴e= $$\frac { 1 }{ 2 }$$ x 0.5 x l2 x 400 = 100V

Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s_1 in a uniform horizontal magnetic field of magnitude
3.0 x 10-2 Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 q, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

The external agent, which keeps the coil rotating, is the source of this power.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 104 Wbm2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
(a) $$\mathscr{E}$$ = Blv = 0.3 x 10-4 x 10 x 5 = 1.5 mv
(b) West to east
(c) Eastern end

Question 8.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self­inductance of the circuit.

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Using dΦ = Mdl, we get
dΦ= 1.5 X (20 – 0) = 30 Wb

Question 10.
A jet plane is traveling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 30°?
$$\mathscr{E}$$ = Bv lv = (B sin 30)lv
= 5 x 10-4 x $$\frac { 1 }{ 2 }$$ x 25 x $$\frac { 180×5 }{ 18 }$$
= 3.125 V

### 2nd PUC Physics Electromagnetic Induction Additional Exercises

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 Ts-1. If the cut is joined and the loop has a resistance of 1.6 q, how much power is dissipated by the loop as heat? What is the source of this power?
Here, area of the loop,

Question 12
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s-1 in the positive x-direction in an environment containing a magnetic field in the positive Z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3 T cm-1 along the negative x-direction (that is it increases by 103 T cm-1 as one move in the negative x-direction), and it is decreasing in time at the rate of 10 3 Ts-1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Here, area of the loop,

The induced e.m.f produced due to spatial variation of the magnetic field.

According to Lenz’s law, the induced current flows through the loop in a direction so as to cause an increase in the magnetic flux along the positive Z-direction. To an observer, the current will appear to be flowing in an anti-clock-wise direction, if the loop is moving towards the right.

Question 13.
It is desired to measure the magnitude of the field between the poles of a powerful loudspeaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flew in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Q. Estimate the field strength of the magnet.
Here, number of turns, N = 25
area of coil, A= 2.0 cm2 =2.0 x 10-4 m2
charge flown in the coil, q = 7.5 m
C =7.5 × 10-3 C; resistance of coil R= 0.50 mΩ.
Let B be the strength of the magnetic field, then, magnetic flux linked with the search coil initially.
φ= BAN
When the search coil is brought out of the magnetic field, the magnetic flux linked with the coil finally,

Question 14.
Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed-loop containing the rod = 9.0 Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s_1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s-1) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed-circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
(a) Using e = Bυl, we get
e = 0.5 x 12 x 10.2 x 15 x 102
= 9 x 10-3 V
The electrons in the rod will experience force along PQ, so end P becomes positive and Q becomes negative.

(b) On closing the key K, the number of electrons becomes more at end Q. Therefore, excess charge is maintained by the continuous current.

(c) Magnetic force gets cancelled by electric force due to excess charges of opposite sign at the ends of the rod.

(e) When the switch K is closed, the power required by the external agent against the retarding force.
P = Fv = 7.5×10-2×12×10-2 = 9×10-3W

(f) Power dissipated as heat

The source of this power is the power of external agent against the retarding force,
P = Fv = 7.5×10-2 ×12×10-2= 9×10-3 W

(g) The motion of rod does not cut field lines, hence no induced e.m.f. is produced.

Question 15.
An air-cored solenoid with a length of 30 cm, area of cross-section 25 cm2, and a number of turns 500, carries a current of 2.5 A, The current is suddenly switched off in a brief time of 10-3. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in the magnetic field near the ends of the solenoid.
Here, area of cross-section of the solenoid,

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side an as shown in Fig. 6.21.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

(a) Consider as an elementary portion of the area of the loop in the form of a strip parallel to the straight wire having a width dx and at a distance x from the wire as shown in the figure.

If dA is are of this strip, then dA = a dx
The magnetic field at the strip due to current-carrying wire,

The magnetic flux linked with the whole loop can be found by integrating the above between the limits x = r to x = r + b
i.e.

Question 17.
A line charge X per unit length is lodged uniformly onto the rim of a wheel of mass M and radius The wheel has light non­conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by, B = – B0 k (r < a; a < R)= 0 (otherwise).
What is the angular velocity of the wheel after the field is suddenly switched off?

Question 1.
Two identical loops, one of copper and another of aluminium are rotated at the same speed in the same magnetic field. In which case, the
(a) induced emf and
(b) induced current will be more and why? (CBSE 2010)
The induced emf will be the same in both but the induced current will be more in the copper loop as its resistance will be lesser as compared to that of aluminium loop.

Question 2.
A plot of magnetic flux (φ) versus current (I) is shown in the figure, for two conductors A and B

Which of the two has a larger value of self-inductance?
From the relation,
φ = LI
it follows that the self-inductance of a conductor is equal to the slope of its graph between φ and I. Therefore, the self-inductance of conductor A has a larger value.

Question 3.
A metal conductor of length lm rotates vertically about one of its ends at angular velocity 5 rad s-1. If the horizontal component of the earth’s magnetic field is 0.2 x 10-4T, then emf developed between the two ends of the conductor is
(A) 5μV
(B) 50μV
(C) 5mV
(D) 50 mV (AIEEE 2004)
The induced emf produced across the two ends of the conductor
e= $$\frac { 1 }{ 2 }$$ Bl2 W=$$\frac { 1 }{ 2 }$$×0.2×10-4×l2 ×5
= 5 ×10-5V = 50 µV

Question 4.
The self-inductance of the motor of a fan is 10H. In order to impart maximum power at 50Hz, it should be connected to a capacitance of
(A) 4µF
(B) 8µF
(C) 1µF
(D) 2µF (AIEEE 2005)

Question 5.
Two coils are placed closed to each other. The mutual inductance of the pair of coil depends upon:
(A) The rates at which currents are changing in the two coils
(B) Relative position and orientation of the two coils
(C) The material of wires of the coils
(D) the currents in the two coils
(B) Relative position and orientation of the two coils

Question 6.
A 100 mH coil carries a current of 1A. Energy stored in the form of magnetic field
(A) 0.5J
(B) 1J
(C) 0.05J
(D) 0.1J
L = 100 mH = 0.1. H 1=1 A
The energy stored is given by
U = $$\frac { 1 }{ 2 }$$LI2 = $$\frac { 1 }{ 2 }$$ x 0.1 x l2 = 0.05 J 2

Question 7.
A straight line conductor of length 0.4 m is moved with a speed of 7ms1 perpendicular to the magnetic field of intensity 0.9 Wb nr2. The induced emf across the conductor is.
(A) 1.26 V
(B) 2.52 V
(C) 5.24 V
(D) 25.2 V
B = 0.9 Wb m2
v = 7 ms-1
l = 0.4m
The induced emf produced
φ = BVl = 0.9x 7×0.4
= 2.52 V

Question 8.
What is the power factor of the LCR circuit at resonance?
At resonance, the impedance of the circuit is equal to the resistance in the circuit,
i.e. Z = R
∴ power factor of the circuit,
cos φ= $$\frac { R }{ Z } =\frac { R }{ R } =1$$

Question 9.
In a resonant circuit, at which angular frequency, potential difference leads the current?
In a resonant circuit,

Question 10.
In Lenz’s law, there is the conservation of
(A) Charge
(B) Momentum
(C) Current
(D) Energy.
(D) Energy.                                                                            ‘

## Karnataka 2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers

### 2nd PUC Chemistry Alcohols, Phenols and Ethers NCERT Textbook Questions

Question 1.
Write IUPAC names of the following compounds:

(i) 2, 2, 4 – Trimethyl pentan – 3 – ol
(ii) 5 – Ethyl heptane – 2, 4 – diol
(iii) Butane – 2, 3 – diol
(iv) Propane – 1, 2, 3 – triol
(v) 2 – Methyl phenol
(vi) 4 – Methyl phenol
(vii) 2,5 – Dimethyl phenol
(viii) 2,6 – Dimethyl phenol
(ix) 1 – Methoxy – 2 – Methyl propane
(x) Ethoxy benzene
(xi) 1 – Phenoxyheptane
(xii) 2 – Ethoxy butane

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methyl butan-2-ol
(ii) phenyl propan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-trioi
(iv) 2,3 – Diethyl phenol
(v) 1 – Ethoxy propane
(vi) 2-Ethoxy-3-methyipentane
(vii) Cyclohexylmethanol
(viii) 3-CycIohexyipentan-3-oI
(ix) Cyclopent-3-en-l-ol
(x) 3-Chloromethylpentan-l-ol.

Question 3.
1. Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
2. Classify the. isomers of alcohols in question 3 (i) as primary, secondary and tertiary alcohols.
1. The structures of all isomeric alcohols of molecular formula, C5H12O are shown below
(a)

2. Primary alcohol:
Pentan -1 -ol; 2-Methylbutan-1- ol;
3-Methylbutan- 1-ol; 2,2—Dimethylproparv4-ol;

Secondary alcohol:
Pentan -2 -ol; 3 -Methylbutan- 2 – ol; Pentan-3-ol

Tertiary alcohol:
2 – Methylbutan -2 -ol

Question 4.
Explain why propanol has a higher boiling point than of hydrocarbon, butane?
Propane undergoes intermolecular H – bonding because of the presence of the -OH group. On the other hand, butane does not.

Therefore, extra energy is required to break hydrogen bonds. For this reason, propand has a higher boiling point than hydrocarbon butane.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Alcohols form H-bonds with water due to the presence of -OH group. However, hydrocarbons cannot form H-bonds with water.

As a result, alcohols are comparatively more soluble in water than hydrocarbons, of comparable molecular masses.

Question 6.
What is meant by hydroboration- oxidation reaction? Illustrate it with an example.
The addition of diborane to the alkene to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohol is called hydroboration-oxidation. For example:

The alcohols obtained by this process appear to have been formed by the direct addition of water to the alkene against the MarkownikofFs rule.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Orthonitrophenol is steam volatile while para-nitrophenol is not. This is on ac¬count of chelation (intramolecular H—bonding) in the molecule of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol, which is not steam volatile and its molecules are linked by intermolecular H—bonding.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
To prepare phenol, cumene is first oxidized in the presence of air of cumene hydroperoxide.

Then cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone as by-products.

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.

Chlorobenzene is fused with NaOH to produce sodium phenoxide, which gives phenol on acidification.

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
The mechanism of hydration of ethane to form ethanol involves three steps.
Step 1:
Protonation of ethane to form carbocation by the electrophilic attack of H3O+
H2O + H+ → H3O+

Step 2:
Nucleophilic attack of water on carbocation:

Step 3:
Deprotonation to form ethanol

Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.

Question 13.
Show how well you synthesise:
(i) 1-phenyl ethanol from a suitable alkene.
(ii) cyclohexyl methanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
The acidic nature of phenol can be represented by the following two reactions
1. Phenol reacts with sodium to give sodium phenoxide, liberating H2

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Question 15.
Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?

The nitro-group is the art electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. Also, the O-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, O-nitrophenol is a stronger acid.

On the other hand, the methoxy group is an electron releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For this reason, O-nitrophenol is more acidic than O-methoxy phenol.

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution?
The -OH group exerts a +R effect on the benzene ring under the effect of attacking electrophile. As a result, there is an increase in the electron density in the ring particularly at ortho and para positions, therefore electrophilic substitution occurs mainly at o-and p-positions. (For resonance hybrid structures of phenol refer to NCERT textbook.)

Question 17.
Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol,
(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Question 18.
Explain the following with an example.
1. Kolbe’s reaction.
2. Reimer-Tiemann reaction
3. Williamson ether synthesis.
4. unsymmetrical ether.
1. Kolbe’s reaction:
When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon-di-oxide, followed by acidification, undergoes electrophilic substitution to give o-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.

2. Reimer-Tiemann reaction: When phenol is treated with chloroform (CHCl3) in the presence of sodium hydroxide, a – CHo group is introduced at the ortho position of the benzene ring.

This reaction is known as the Reimer – Tiemann reaction.
The intermediate is hydrolyzed in the presence of alkalis to produce salicylic dehyde.

3. Williamson ether synthesis:
Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.
R-X + R- ONa → R-O-R + NaX
This reaction involves SN2 attack of the alkoxides ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

Ifthealkyl halide is 2° or 3°, then elimination competes over substitution.

4. Unsymmetrical ether:
An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For e.g: ethyl methyl ether

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
The mechanisms of acid dehydration of ethanol to yield ethene involves the following three steps.
Steps 1:
Protonation of ethanol to form ethyl oxonium ion

Steps 2:
Formation of carbocation (rate-determining step)

Steps 3:
Elimination of proton to form ethene

The acid consumed in step 1 is released in step 3. After the formation of ethene, it is removed to shift the equilibrium in the forwarding direction.

Question 20.
How are the following conversions carried out?
1. Propene → Propan-2-ol.
2. Benzyl chloride → Benzyl alcohol.
3. Ethyl magnesium chloride → Propan-1-ol.
4. Methyl magnesium bromide → 2-Methylpropan-2-ol.
1. If propene is allowed to react with water in the presence of an acid as a catalyst, then propan – 2-ol is obtained.

2. If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

3. When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propan-l-ol on hydrolysis.

4. When methyl magnesium bromide is treated with propane, an adduct is the product which give 2- methyl propane – 2-ol hydrolysis.

Question 21.
Name the reagents used in the following reactions:

1. Oxidation of primary alcohol to a carboxylic acid.
2. Oxidation of primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-tribromophenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.

1. Acidified KMnO4
2. Pyridinium chlorochromate (PCC)
3. Bromine water
4. Acidified KMnO4
5. 85% phosphoric acid
6. NaBH4 or Li AiH4

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxy -methane.
The boiling point of ethanol is higher than methoxymethane because of the presence of strong intermolecular hydrogen bonding between ethanol molecules. As a result, ethanol exists as associates molecules. However, no such H-bonding is present in methoxy methane.

Question 23.
Give IUPAC names of the following ethers:
1.

2. CH3OCH2CH2CL
3. O2N-C6H4-OCH3(p)
4. CH3CH2CH2OCH3
5.

6.

1. 1- Ethoxy – 2 – methyl propane
2. 2 – chloro -1 – methoxy ethane
3. 4 – Nitroanisole
4. 1 – Methoxypropane
5. 1 – Ethoxy – 4,4 – dimethyl cyclohexane
6. Ethoxy benzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxy propane
(ii) Ethoxy benzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide

But if secondary or tertiary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence they react with alkyl halides, which results in an elimination reaction.

Question 26.
How is 1-propoxypropane synthesised from propan-l-ol? Write the mechanism of this reaction.
1-Propoxy propane can be synthesised from propan-l-ol by dehydration.

The mechanism of this rxn involves the following three steps.
Steps 1: Protonation

Step 2: Nucleophilic attack

Step 3: Deprotonation

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohol, the alkyl group is hindered. As a result, elimination dominates over substitution. Hence, in place of ethers, alkenes are formed.

Question 28.
Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane
(ii) methoxy benzene and
(iii) benzyl ethyl ether.

Question 29.
Explain the fact that in aryl alkyl ethers
1. the alkoxy group activates the benzene ring towards electrophilic substitution and
2. it directs the incoming substituents to ortho and para positions in the benzene ring.
1.

In aryl alkyl ethers, due to +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

2. It can be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 30.
Write the mechanism of the reaction of HI with methoxymethane.
The mechanism of the reaction of HI with methoxymethane involves the following steps.
Step 1: Protonation of methoxymethane

Step2: Nucleophilic attack of I

Step 3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide.

Question 31.
Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.

Question 32.
Show how would you synthesise the following alcohols from appropriate alkenes?

Acid hydration of Pent-2-ene produces Penton – 3-ol along with penton-2-ol. Hence first rxn using Pent-l-ene is preferred.

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from the 3rd carbon atom.
The mechanism of the given reaction involves the following steps:
Step 1: Protonation

Step 2: Formation of 2° carbocation by the elimination of water molecule

Step 3: Rearrangement by the hydride-ion shift

Step 4: Nucleophilic attack