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11 months ago

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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11 months ago

Class 11th Mechanical Properties of Fluids

10.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 * 10^–2 N m^–1 ? If an air bubble of the same dimension were formed at 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).

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Vishal Baghel

Contributor-Level 10

Soap bubble radius, r = 5.0 mm = 5 $* 10^{- 3}$ m

Surface tension of the soap bubble, S = 2.50 $* 10^{- 2}$ N/m

Relative density of soap solution = 1.20, hence density of soap solution, $?$ = 1.2 $* 10^{3}$ Kg/ $m^{3}$

Air bubble formed at a depth, h = 40 cm = 0.4 m

1 atmospheric pressure = 1.01 $* 10^{5}$ Pa

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

We know, the excess pressure inside the soap bubble is given by the relation:

P = $\frac{4 S}{r}$ = $\frac{4 * 2.50 * 10^{- 2}}{5 * 10^{- 3}}$ Pa = 20 Pa

Hence, the excess pressure inside the air bubble is given by the relation, P' = $\frac{2 S}{r}$ = 10 Pa

At a depth of h, the total pressure inside the air bubble = Atmospheric press

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Soap bubble radius, r = 5.0 mm = 5 $* 10^{- 3}$ m

Surface tension of the soap bubble, S = 2.50 $* 10^{- 2}$ N/m

Relative density of soap solution = 1.20, hence density of soap solution, $?$ = 1.2 $* 10^{3}$ Kg/ $m^{3}$

Air bubble formed at a depth, h = 40 cm = 0.4 m

1 atmospheric pressure = 1.01 $* 10^{5}$ Pa

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

We know, the excess pressure inside the soap bubble is given by the relation:

P = $\frac{4 S}{r}$ = $\frac{4 * 2.50 * 10^{- 2}}{5 * 10^{- 3}}$ Pa = 20 Pa

Hence, the excess pressure inside the air bubble is given by the relation, P' = $\frac{2 S}{r}$ = 10 Pa

At a depth of h, the total pressure inside the air bubble = Atmospheric pressure + h $?$ g +P'

= 1.01 $* 10^{5}$ + (0.4 $* 1.2 * 10^{3} * 9.8)$ + 10 Pa = 105714 Pa = 1.06 $* 10^{5}$ Pa

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11 months ago

Class 11th Thermal Properties of Matter

11.21 Answer the following questions based on the P-T phase diagram of carbon dioxide:

(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium ?

(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂ ?

(c) What are the critical temperature and pressure for CO₂ ? What is their significance?

(d) Is CO₂ solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm 15 °C under 56 atm ?

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Payal Gupta

Contributor-Level 10

11.21

(a) The P-T phase diagram for CO₂is shown here. O is the triple point of the CO₂phase diagram. This means that at the temperature and pressure corresponding to this point, the solid, liquid and vaporous phases of CO₂exists in equilibrium.

(b) The fusion and the boiling points of CO₂decreases with a decrease in pressure.

(c) The critical temperature and critical pressure of CO₂ are 31 $?$ 72.9 atm respectively. Even if it is compressed to a pressure greater than 72.9 atm, CO₂ will not liquefy above the critical temperature.

(d) It can be concluded from the P-T phase diagram of CO2 that:

CO₂ is gaseous at -70 $?$ , under 1 atm pr

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11.21

(b) The fusion and the boiling points of CO₂decreases with a decrease in pressure.

(d) It can be concluded from the P-T phase diagram of CO2 that:

CO₂ is gaseous at -70 $?$ , under 1 atm pressure.

CO₂ is solid at -60 $?$ , under 10 atm pressure.

CO2 is liquid at 15 $?$ , under 56 atm pressure.

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11 months ago

Class 11th Mechanical Properties of Fluids

10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? 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.

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Vishal Baghel

Contributor-Level 10

Radius of the mercury droplet, r = 3.00 mm = 3 $* 10^{- 3}$ m

Surface tension, S = 4.65 $* 10^{- 1}$ N/m

Atmospheric pressure, $P_{o}$ = 1.01 $* 10^{5}$ Pa

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

= $\frac{2 S}{r}$ + $P_{o}$ = $\frac{2 * 4.65 * 10^{- 1}}{3 * 10^{- 3}}$ + 1.01 $* 10^{5}$ = 1.01310 $* 10^{5}$ Pa

Excess pressure inside mercury = $\frac{2 S}{r}$ = $\frac{2 * 4.65 * 10^{- 1}}{3 * 10^{- 3}}$ = 310 Pa

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11 months ago

Class 11th Mechanical Properties of Fluids

10.18 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.

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Vishal Baghel

Contributor-Level 10

Let us take the case (a):

The length of the liquid film supported by the weight, l = 40 cm = 0.4 m

The weight supported by the film, W = 4.5 $* 10^{- 2}$ N

Since a liquid film has two free surfaces,

Surface tension = $\frac{W}{2 l}$ = $\frac{4.5 * 10^{- 2}}{2 * 0.4}$ = 5.625 $* 10^{- 2}$ N/m

In all 3 figures, the liquid is the same, temperature is also the same. Hence the surface tension in (b) and (c) are also going to be the same, with the value of 5.625 $* 10^{- 2}$ N/m and weight supported in each case is also going to be the same, since length of the film is same.

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11 months ago

Class 11th Mechanical Properties of Fluids

10.17 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 * 10^–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

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Vishal Baghel

Contributor-Level 10

The weight that the soap film supports, W = 1.5 $* 10^{- 2}$ N

Length of the slider, l = 30 cm = 0.3 m

A soap film has two free surfaces, hence total length = 2l = 0.6 m

Surface tension, S= $\frac{F o r c e s / W e i g h t}{2 l}$ = $\frac{1.5 * 10^{- 2}}{2 * 0.3}$ = 2.5 $* 10^{- 2}$ N/m

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11 months ago

Class 11th Thermal Properties of Matter

11.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

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Payal Gupta

Contributor-Level 10

11.20 According to Newtonâ€™s law of cooling, we have $- \frac{d T}{d t}$
= K(T - $T_{o}$ ) or $\frac{d T}{K (T - T_{o})}$ = -Kdtâ€¦â€¦(i)Where, temperature of the body = T

Temperature of the surroundings = $T_{o}$ = 20 °C

K is a constant

The temperature of the body falls from 80 $?$ to 50 $? i n t i m e t = 5 \min ? = 300 s$

Integrating equation (i), we get

$5080 \frac{d T}{K (T - T_{o})}$ = - $\int_{0}^{300} K d t∫$

$? {[\log_{e} ? (T - T_{o})]}_{30}^{80}$ = -K ${[t]}_{0}^{300}$

$? \frac{2.3026}{K} \log_{10} ? \frac{80 - 20}{50 - 20}$ = -300

$? \frac{2.3026}{300} \log_{10} ? 2$ = K â€¦.(ii)

The temperature of the body falls from 60 $?$ to 30 $? i n t i m e = t'$

Hence, we get:

$\frac{2.3026}{K} \log_{10} ? \frac{60 - 20}{30 - 20}$ = -t'

$? \frac{- 2.3026}{t'} \log_{10} ? 4$ = K.(iii)

Equating equations (ii) and (iii), we get,

$\frac{2.3026}{300} \log_{10} ? 2$ = $\frac{- 2.3026}{t'} \log_{10} ? 4$

$? t^{'} = 300 * 2 = 600 s$ = 10 min

Hence, the time taken to cool the body from 60 $?$ to 30 $?$ is 10 minutes.

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Class 11th Mechanical Properties of Fluids

10.16 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?

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Vishal Baghel

Contributor-Level 10

Area of cross-section of the spray pump, $A_{1}$ = 8 ${c m}^{2}$ = 8 $* 10^{- 4}$ $m^{2}$

Number of holes, n = 40

Diameter of each hole, d = 1 mm = 1 $* 10^{- 3}$ m

Radius of each hole, r = d/2 = 0.5 $* 10^{- 3}$ m

Area of cross-section of each hole, a = $π r^{2}$ = $π ({0.5 * 10^{- 3})}^{2}$

Total area of 40 holes, $A_{2}$ = 40 $π * ({0.5 * 10^{- 3})}^{2}$ = 3.14 $* 10^{- 5}$ $m^{2}$

Speed of liquid inside the tube, $V_{1}$ = 1.5 m/min = 0.025 m/s

Speed of ejection of liquid = $V_{2}$

According to law of continuity, we have $A_{1} V_{1}$ = $A_{2} V_{2}$ or $V_{2} = \frac{A_{1} V_{1}}{A_{2}}$ = $\frac{8 * 10^{- 4} * 0.025}{3.14 * 10^{- 5}}$

= 0.637 m/s

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11 months ago

Class 11th Mechanical Properties of Fluids

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

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Vishal Baghel

Contributor-Level 10

Take the case given in figure (b)

$A_{1}$ = Area of pipe 1, $A_{2}$ = Area of pipe 2, $V_{1}$ = Speed of fluid in pipe 1, $V_{2}$ = Speed of fluid in pipe 2

From the law of continuity, we have

$A_{1} V_{1}$ = $A_{2} V_{2}$

When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more. According to Bernoulli's principle, if speed is more, the pressure is less. Pressure is directly proportional to height, hence the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

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11 months ago

Class 11th Mechanical Properties of Fluids

10.14 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^–1and 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.

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Vishal Baghel

Contributor-Level 10

Speed of the wind on the upper surface of the wing, $V_{1}$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_{2}$ = 63 m/s

Area of the wing, A = 2.5 $m^{2}$

Density of air, $ρ$ = 1.3 kg/ $m^{3}$

According to Bernoulli's theorem, we have the relation:

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

( $P_{2}$ - $P_{1}$ )= $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ , where $P_{1}$ = pressure on the upper surface of the wing and $P_{2}$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_{2}$ - $P_{1}$ )A = $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ A = $\frac{1}{2} * 1.3 * (70^{2}$ - $63^{2}) * 2.5$ N =

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Speed of the wind on the upper surface of the wing, $V_{1}$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_{2}$ = 63 m/s

Area of the wing, A = 2.5 $m^{2}$

Density of air, $ρ$ = 1.3 kg/ $m^{3}$

According to Bernoulli's theorem, we have the relation:

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

( $P_{2}$ - $P_{1}$ )= $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ , where $P_{1}$ = pressure on the upper surface of the wing and $P_{2}$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_{2}$ - $P_{1}$ )A = $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ A = $\frac{1}{2} * 1.3 * (70^{2}$ - $63^{2}) * 2.5$ N = 1512.8 N = 1.5 $* 10^{3}$ N

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