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physics ncert solutions class 11th

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.29 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? Surface tension of mercury at the temperature of the experiment is 0.465 N m^–1. Density of mercury = 13.6 * 10³kg m^–3.

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Vishal Baghel

Contributor-Level 10

Radius of the narrow tube, r = 1 mm= 1 $* 10^{- 3}$ m

Surface tension of mercury at the given temperature, s= 0.465 N/m

Density of mercury $ρ = 13.6 * 10^{3}$ kg/ $m^{3}$

Dipping depth = h

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Surface tension related with angle of contact and dipping depth is given by:

s = $\frac{h ρ g r}{2 \cos ? θ}$ or h = $\frac{2 scos ? θ}{ρ g r}$ = $\frac{2 * 0.465 * \cos ? 140 °}{13.6 * 10^{3} * 9.8 * 1 * 10^{- 3}}$ = -5.345 $* 10^{- 3}$ m= -5.345 mm

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.28 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.

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Vishal Baghel

Contributor-Level 10

Radius of the uncharged drop, r = 2.0 $* 10^{- 5}$ m

Density of the uncharged drop, $ρ$ = 1.2 $* 10^{3}$ kg/ $m^{3}$

Viscosity of air, $η$ = 1.8 $* 10^{- 5}$ Pas

Density of air $ρ_{o}$ , can be taken as zero in order to neglect the buoyancy of air

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Terminal velocity, v can be written as

v = $\frac{2 * r^{2} * (ρ - ρ_{o}) g}{9 * η}$ = $\frac{2 * ({2.0 * 10^{- 5})}^{2} * (1.2 * 10^{3} - 0) * 9.8}{9 * 1.8 * 10^{- 5}}$ = 0.05807 m/s = 5.8 cm/s

The viscous force on the drop is given by

F = 6 $π η r v$ = 6 $* 3.1416 *$ 1.8 $* 10^{- 5} *$ 2.0 $* 10^{- 5} *$ 0.05807

= 3.9 $* 10^{- 10}$ N

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.27 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).

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Vishal Baghel

Contributor-Level 10

The area of the wing of the plane, A = 2 $* 25 m^{2}$ = 50 $m^{2}$

Speed of air over the lower wing, $V_{1}$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_{2}$ = 234 km/h = 65 m/s

Density of air, $ρ$ = 1 kg/ $m^{3}$

Let us assume, pressure over the lower wing = $P_{1}$ and pressure over the upper wing = $P_{2}$

The upward force on the plane can be obtained using Bernoulli's equation:

On lower wing= $P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ , On upper wing = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

From equilibrium of momentum

$P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

( $P_{1}$ - $P_{2}) = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ )

The net upward force = ( $P_{1}$ - $P_{2}) A$ = $\frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ ) $* A$

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The area of the wing of the plane, A = 2 $* 25 m^{2}$ = 50 $m^{2}$

Speed of air over the lower wing, $V_{1}$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_{2}$ = 234 km/h = 65 m/s

Density of air, $ρ$ = 1 kg/ $m^{3}$

Let us assume, pressure over the lower wing = $P_{1}$ and pressure over the upper wing = $P_{2}$

The upward force on the plane can be obtained using Bernoulli's equation:

On lower wing= $P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ , On upper wing = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

From equilibrium of momentum

$P_{1}$ + $\frac{1}{2} ρ V_{1}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ V_{2}^{2}$

( $P_{1}$ - $P_{2}) = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ )

The net upward force = ( $P_{1}$ - $P_{2}) A$ = $\frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}$ ) $* A$

= $\frac{1}{2} 1 (65^{2} - 50^{2}$ ) $* 50$ = 43125 N

Using the relation F = mg, we get m = F/g

= (43125/9.8) Kg = 4400.51 kg

Hence the mass of the plane is about 4400 kg.

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.26 (a) What is the largest average velocity of blood flow in an artery of radius 2*10^–3m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 * 10^–3 Pa s).

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Vishal Baghel

Contributor-Level 10

Radius of the artery, r = 2 $* 10^{- 3}$ m

Diameter of the artery, d = 4 $* 10^{- 3}$ m

Viscosity of blood, $η$ = 2.084 $* 10^{- 3}$ Pa-s

Density of the blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Reynolds's number for laminar flow, $R_{e}$ = 2000

The largest velocity is given by the relation: $V_{a v g}$ = $\frac{R_{e} η}{ρ d}$

= $\frac{2000 * 2.084 * 10^{- 3}}{1.06 * 10^{3} * 4 * 10^{- 3}}$ = 0.983 m/s

Flow rate is given by the relation:

R = $π r^{2} V_{a v g}$ = 3.1416 $* ({2 * 10^{- 3})}^{2} * 0.983$ = 1.235 $* 10^{- 5}$ $m^{3}$ /s

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.25 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. (a) What is the largest average velocity of blood flow in an artery of diameter 2 * 10^–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

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Vishal Baghel

Contributor-Level 10

Diameter of the artery, d = 2 $* 10^{- 3}$ m

Viscosity of the blood, $η$ = 2.084 $* 10^{- 3}$ Pa-s

Density of the blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Reynolds's number for laminar flow, $R_{e}$ = 2000

$V_{a v g}$ = $\frac{R_{e} η}{ρ d}$ = $\frac{2000 * 2.084 * 10^{- 3}}{1.06 * 10^{3} * 2 * 10^{- 3}}$ = 1.966 m/s

As the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.24 During 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? [Use the density of whole blood from Table 10.1].

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Vishal Baghel

Contributor-Level 10

Gauge pressure, P = 2000 Pa

Density of whole blood, $ρ$ = 1.06 $* 10^{3}$ kg/ $m^{3}$

Acceleration due to gravity, g = 9.8 m/s

Let the height of the blood container = h

Pressure on the blood container P = $ρ g h$

By equating, we get h = (2000)/ ( 1.06 $* 10^{3} * 9.8)$ = 0.1925m

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.23 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?

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Vishal Baghel

Contributor-Level 10

Two vessels, having the same base area, will have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components.

When these vertical components are added, the total force on one vessel comes out to be more than on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on the weighing scale.

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.22 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.

(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury

(b) How would the levels change in case (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)

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Vishal Baghel

Contributor-Level 10

Atmospheric pressure, $P_{0}$ = 76 cm of Hg

For figure (a)

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = 20 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg

For figure (b),

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = - 18 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg

When 13.6 cm of water is poured into the right limb of figure (b)

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm of Mercury.

Let h be

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Atmospheric pressure, $P_{0}$ = 76 cm of Hg

For figure (a)

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = 20 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg

For figure (b),

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = - 18 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg

When 13.6 cm of water is poured into the right limb of figure (b)

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm of Mercury.

Let h be the difference between the levels of mercury in the two limbs

The pressure in the right limb is given as:

$P_{g}$ = Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg

The mercury column will rise in the left limb. Hence, pressure in the left limb

$P_{l}$ = 58 + h

Since $P_{g}$ = $P_{l}$ , we get h = 77 – 58 = 19 cm

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physics ncert solutions class 11th Thermal Properties of Matter

11.22 Answer the following questions based on the P – T phase diagram of CO₂:

(a) CO₂ at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?

(b) What happens when CO₂at 4 atm pressure is cooled from room temperature at constant pressure ?

(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.

(d) CO₂ is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe?

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Payal Gupta

Contributor-Level 10

11.22

The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves. It departs from ideal gas behavior as pressure increases.

(a) At 1 atm pressure and at – 60 °C, it lies left of -56.6 $?$ (triple point O). Hence, it lies in the region of vapour and solid phases. Thus, CO₂ condenses into solid state directly, without going through liquid phases.

(b) At 4 atm pressure, CO₂ lies below 5.11 atm (triple point O). Hence it lies in the region of vaporous and solid phases. Thus, CO₂condenses into solid state directly, without going through liquid state.

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11.22

The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves. It departs from ideal gas behavior as pressure increases.

(c) When the temperature of a mass of solid CO₂ ( at 10 atm pressure and at -65 $?$ is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves.

(d) If CO2 is heated to 70 $?)$ compressed isothermally, then it will not exhibit any transition to the liquid state. This because 70 $? a n d$ is higher than the critical temperature of CO₂. It will remain in the vapour state, but will depart from its ideal behavior as pressure increases.

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physics ncert solutions class 11th Mechanical Properties of Fluids

10.21 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 4.0 m. compute the force necessary to keep the door close.

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Vishal Baghel

Contributor-Level 10

Base area of the given tank, A = 1.0 $m^{2}$

Area of the hinged door, a = 20 ${c m}^{2}$ = 20 $* 10^{- 4}$ $m^{2}$

Density of water, $ρ_{1}$ = $10^{3}$ kg/ $m^{3}$

Density of acid, $ρ_{2}$ = 1.7 $* 10^{3}$ kg/ $m^{3}$

Height of the water column, $h_{1}$ = 4 m

Height of the acid column, $h_{2}$ = 4 m

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

The pressure due to Water column, $P_{1}$ = $ρ_{1} g h_{1}$ = $10^{3} * 9.8 * 4$ = 3.92 $* 10^{4}$ Pa

The pressure due to Acid column, $P_{2}$ = $ρ_{2} g h_{2}$ = 1.7 $* 10^{3} * 9.8 * 4$ = 6.64 $* 10^{4}$ Pa

Pressure difference between two columns ΔP = $P_{2}$ - $P_{1}$ = 2.774 $* 10^{4}$ Pa

Force exerted on the small hi

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Base area of the given tank, A = 1.0 $m^{2}$

Area of the hinged door, a = 20 ${c m}^{2}$ = 20 $* 10^{- 4}$ $m^{2}$

Density of water, $ρ_{1}$ = $10^{3}$ kg/ $m^{3}$

Density of acid, $ρ_{2}$ = 1.7 $* 10^{3}$ kg/ $m^{3}$

Height of the water column, $h_{1}$ = 4 m

Height of the acid column, $h_{2}$ = 4 m

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

The pressure due to Water column, $P_{1}$ = $ρ_{1} g h_{1}$ = $10^{3} * 9.8 * 4$ = 3.92 $* 10^{4}$ Pa

The pressure due to Acid column, $P_{2}$ = $ρ_{2} g h_{2}$ = 1.7 $* 10^{3} * 9.8 * 4$ = 6.64 $* 10^{4}$ Pa

Pressure difference between two columns ΔP = $P_{2}$ - $P_{1}$ = 2.774 $* 10^{4}$ Pa

Force exerted on the small hinged door = ΔP $* a$ = 2.774 $* 10^{4} *$ 20 $* 10^{- 4}$ N = 55.48 N

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