QBManager

Mechanical Properties of Fluids

143445 Radius of one arm of hydraulic lift is four times of radius of other arm. What force should be applied on narrow arm to lift $100 kg$ .

1

26.5 N

2

62.5 N

3

6.25 N

4

8.3 N

Explanation:

B Given that,
Mass (m) $= 100 kg$
By using the Pascal's law
$\frac{F}{A} = \frac{f}{a}$
$f = \frac{F a}{A}$
$f = \frac{100 g \times π r^{2}}{π (4 r)^{2}}$
$f = \frac{100 \times 10 \times π r^{2}}{π \times 16 r^{2}}$
$f = \frac{1000}{16}$
$f = 62.5 N$

DCE-2007

Mechanical Properties of Fluids

143446 Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$ . If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

A Area of hole $A = A_{A}$
Area of hole $B = A_{B}$
According to questions-
$A_{B} = 2 A_{A}$
Velocity of rejection for both $= \sqrt{2 gh}$
Mass flux $= ρ A V$
Some mass flux for both hole
$ρ_{A} A_{A} V_{A} = ρ_{B} A_{B} V_{B}$
$V_{A} = V_{B}$
$\frac{ρ_{A}}{ρ_{B}} = \frac{A_{B}}{A_{A}} = \frac{2 A_{A}}{A_{A}} = 2$

EAMCET-2002

Mechanical Properties of Fluids

143447 There are two holes one each along the opposite sides of a wide rectangular tank. The crosssection of each hole is $0.01 m^{2}$ and the vertical distance between the holes is $1 m$ . The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 kg m^{- 3}$ )

1 100

2 200

3 300

4 400

Explanation:

B Net force on the tank $= ρ a v^{2}$
Where $ρ =$ density, $a =$ area, $v =$ velocity
$v = \sqrt{2 gh}$
$v^{2} = 2 gh [\begin{aligned} where g = 10 m / \sec^{2} \\ h = 1 m \end{aligned}$
$v^{2} = 2 \times 10 \times 1 [\begin{array}{l} ρ = 1000 {kgm}^{- 3} \\ a = 0.01 m^{2} \end{array}$
$v^{2} = 20$
Net force on the tank $= ρ v^{2}$
$= 1000 \times 0.01 \times 20$
$= 200 N$

EAMCET-2004

Mechanical Properties of Fluids

143448 A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1 ${ms}^{- 1}$ when pressure is $50 kPa$ at a point. If the velocity of flow has to be $2 {ms}^{- 1}$ at some other point, the pressure at that point should be

1

50 kPa

2

100 kPa

3

48.5 kPa

4

24.25 kPa

Explanation:

C Given that,
$v_{1} = 1 m / \sec$
$P_{1} = 50 k Pa$
$v_{2} = 2 m / \sec$
$P_{2} = ?$
From Bernoulli's equation -
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + Z_{2}$
For horizontal pipe $(Z_{1} = Z_{2})$
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{v^{2}}{2}$
$\frac{50}{1000} + \frac{1^{2}}{2} = \frac{P_{2}}{1000} + \frac{2^{2}}{2}$
$\frac{P_{2}}{1000} = \frac{50 \times 1000}{1000} + \frac{1}{2} - 2$
$\frac{P_{2}}{1000} = 48.5 Pa$
$P_{2} = 48.5 \times 1000 Pa$
$P_{2} = 48.5 k Pa$

EAMCET-2007

Mechanical Properties of Fluids

143450 An aircraft has a mass $4 \times 10^{5} kg$ with total wing area $500 m^{2}$ flying at a speed of $720 km / h$ . The density of air at its height is $1.2 kg / m^{3}$ . Estimate the fractional increase in air speed on the upper surface of its wing relative to lower surface.

1 0.04

2 0.08

3 0.17

4 0.32

Explanation:

C Given data,
$Mass (m) = 4 \times 10^{5} kg$
$Area (A) = 500 m^{2}$
$Speed (v) = 720 km / h$
$Density (ρ) = 1.2 kg / m^{3}$
Acceleration due to gravity $(g) = 10 m / s^{2}$
The weight of the aircraft is balanced by the upward force due to the pressure difference
$Δ P = \frac{w}{A}$
$Δ P = \frac{mg}{A} [w = mg]$
$Δ P = \frac{4}{5} \times 10^{4} {Nm}^{- 2} = 8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure,
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Δ P = ρ \frac{(v_{2} + v_{1})}{2} (v_{2} - v_{1}) = ρ v_{avg} (v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{Δ P}{ρ v_{avg}}$
Here, $v_{avg} = \frac{v_{1} + v_{2}}{2} = 720 km / h$
$= 720 \times \frac{5}{18} = 200 m / \sec$
$∴ \frac{v_{2} - v_{1}}{v_{avg}} = \frac{Δ P}{ρ v_{avg}^{2}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times (200)^{2}}$
$= \frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}}$
$= 0.17$

AMU-2012

Mechanical Properties of Fluids

143445 Radius of one arm of hydraulic lift is four times of radius of other arm. What force should be applied on narrow arm to lift $100 kg$ .

1

26.5 N

2

62.5 N

3

6.25 N

4

8.3 N

Explanation:

B Given that,
Mass (m) $= 100 kg$
By using the Pascal's law
$\frac{F}{A} = \frac{f}{a}$
$f = \frac{F a}{A}$
$f = \frac{100 g \times π r^{2}}{π (4 r)^{2}}$
$f = \frac{100 \times 10 \times π r^{2}}{π \times 16 r^{2}}$
$f = \frac{1000}{16}$
$f = 62.5 N$

DCE-2007

Mechanical Properties of Fluids

143446 Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$ . If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

A Area of hole $A = A_{A}$
Area of hole $B = A_{B}$
According to questions-
$A_{B} = 2 A_{A}$
Velocity of rejection for both $= \sqrt{2 gh}$
Mass flux $= ρ A V$
Some mass flux for both hole
$ρ_{A} A_{A} V_{A} = ρ_{B} A_{B} V_{B}$
$V_{A} = V_{B}$
$\frac{ρ_{A}}{ρ_{B}} = \frac{A_{B}}{A_{A}} = \frac{2 A_{A}}{A_{A}} = 2$

EAMCET-2002

Mechanical Properties of Fluids

143447 There are two holes one each along the opposite sides of a wide rectangular tank. The crosssection of each hole is $0.01 m^{2}$ and the vertical distance between the holes is $1 m$ . The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 kg m^{- 3}$ )

1 100

2 200

3 300

4 400

Explanation:

B Net force on the tank $= ρ a v^{2}$
Where $ρ =$ density, $a =$ area, $v =$ velocity
$v = \sqrt{2 gh}$
$v^{2} = 2 gh [\begin{aligned} where g = 10 m / \sec^{2} \\ h = 1 m \end{aligned}$
$v^{2} = 2 \times 10 \times 1 [\begin{array}{l} ρ = 1000 {kgm}^{- 3} \\ a = 0.01 m^{2} \end{array}$
$v^{2} = 20$
Net force on the tank $= ρ v^{2}$
$= 1000 \times 0.01 \times 20$
$= 200 N$

EAMCET-2004

Mechanical Properties of Fluids

143448 A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1 ${ms}^{- 1}$ when pressure is $50 kPa$ at a point. If the velocity of flow has to be $2 {ms}^{- 1}$ at some other point, the pressure at that point should be

1

50 kPa

2

100 kPa

3

48.5 kPa

4

24.25 kPa

Explanation:

C Given that,
$v_{1} = 1 m / \sec$
$P_{1} = 50 k Pa$
$v_{2} = 2 m / \sec$
$P_{2} = ?$
From Bernoulli's equation -
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + Z_{2}$
For horizontal pipe $(Z_{1} = Z_{2})$
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{v^{2}}{2}$
$\frac{50}{1000} + \frac{1^{2}}{2} = \frac{P_{2}}{1000} + \frac{2^{2}}{2}$
$\frac{P_{2}}{1000} = \frac{50 \times 1000}{1000} + \frac{1}{2} - 2$
$\frac{P_{2}}{1000} = 48.5 Pa$
$P_{2} = 48.5 \times 1000 Pa$
$P_{2} = 48.5 k Pa$

EAMCET-2007

Mechanical Properties of Fluids

143450 An aircraft has a mass $4 \times 10^{5} kg$ with total wing area $500 m^{2}$ flying at a speed of $720 km / h$ . The density of air at its height is $1.2 kg / m^{3}$ . Estimate the fractional increase in air speed on the upper surface of its wing relative to lower surface.

1 0.04

2 0.08

3 0.17

4 0.32

Explanation:

C Given data,
$Mass (m) = 4 \times 10^{5} kg$
$Area (A) = 500 m^{2}$
$Speed (v) = 720 km / h$
$Density (ρ) = 1.2 kg / m^{3}$
Acceleration due to gravity $(g) = 10 m / s^{2}$
The weight of the aircraft is balanced by the upward force due to the pressure difference
$Δ P = \frac{w}{A}$
$Δ P = \frac{mg}{A} [w = mg]$
$Δ P = \frac{4}{5} \times 10^{4} {Nm}^{- 2} = 8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure,
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Δ P = ρ \frac{(v_{2} + v_{1})}{2} (v_{2} - v_{1}) = ρ v_{avg} (v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{Δ P}{ρ v_{avg}}$
Here, $v_{avg} = \frac{v_{1} + v_{2}}{2} = 720 km / h$
$= 720 \times \frac{5}{18} = 200 m / \sec$
$∴ \frac{v_{2} - v_{1}}{v_{avg}} = \frac{Δ P}{ρ v_{avg}^{2}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times (200)^{2}}$
$= \frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}}$
$= 0.17$

AMU-2012

Mechanical Properties of Fluids

143445 Radius of one arm of hydraulic lift is four times of radius of other arm. What force should be applied on narrow arm to lift $100 kg$ .

1

26.5 N

2

62.5 N

3

6.25 N

4

8.3 N

Explanation:

B Given that,
Mass (m) $= 100 kg$
By using the Pascal's law
$\frac{F}{A} = \frac{f}{a}$
$f = \frac{F a}{A}$
$f = \frac{100 g \times π r^{2}}{π (4 r)^{2}}$
$f = \frac{100 \times 10 \times π r^{2}}{π \times 16 r^{2}}$
$f = \frac{1000}{16}$
$f = 62.5 N$

DCE-2007

Mechanical Properties of Fluids

143446 Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$ . If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

A Area of hole $A = A_{A}$
Area of hole $B = A_{B}$
According to questions-
$A_{B} = 2 A_{A}$
Velocity of rejection for both $= \sqrt{2 gh}$
Mass flux $= ρ A V$
Some mass flux for both hole
$ρ_{A} A_{A} V_{A} = ρ_{B} A_{B} V_{B}$
$V_{A} = V_{B}$
$\frac{ρ_{A}}{ρ_{B}} = \frac{A_{B}}{A_{A}} = \frac{2 A_{A}}{A_{A}} = 2$

EAMCET-2002

Mechanical Properties of Fluids

143447 There are two holes one each along the opposite sides of a wide rectangular tank. The crosssection of each hole is $0.01 m^{2}$ and the vertical distance between the holes is $1 m$ . The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 kg m^{- 3}$ )

1 100

2 200

3 300

4 400

Explanation:

B Net force on the tank $= ρ a v^{2}$
Where $ρ =$ density, $a =$ area, $v =$ velocity
$v = \sqrt{2 gh}$
$v^{2} = 2 gh [\begin{aligned} where g = 10 m / \sec^{2} \\ h = 1 m \end{aligned}$
$v^{2} = 2 \times 10 \times 1 [\begin{array}{l} ρ = 1000 {kgm}^{- 3} \\ a = 0.01 m^{2} \end{array}$
$v^{2} = 20$
Net force on the tank $= ρ v^{2}$
$= 1000 \times 0.01 \times 20$
$= 200 N$

EAMCET-2004

Mechanical Properties of Fluids

143448 A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1 ${ms}^{- 1}$ when pressure is $50 kPa$ at a point. If the velocity of flow has to be $2 {ms}^{- 1}$ at some other point, the pressure at that point should be

1

50 kPa

2

100 kPa

3

48.5 kPa

4

24.25 kPa

Explanation:

C Given that,
$v_{1} = 1 m / \sec$
$P_{1} = 50 k Pa$
$v_{2} = 2 m / \sec$
$P_{2} = ?$
From Bernoulli's equation -
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + Z_{2}$
For horizontal pipe $(Z_{1} = Z_{2})$
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{v^{2}}{2}$
$\frac{50}{1000} + \frac{1^{2}}{2} = \frac{P_{2}}{1000} + \frac{2^{2}}{2}$
$\frac{P_{2}}{1000} = \frac{50 \times 1000}{1000} + \frac{1}{2} - 2$
$\frac{P_{2}}{1000} = 48.5 Pa$
$P_{2} = 48.5 \times 1000 Pa$
$P_{2} = 48.5 k Pa$

EAMCET-2007

Mechanical Properties of Fluids

143450 An aircraft has a mass $4 \times 10^{5} kg$ with total wing area $500 m^{2}$ flying at a speed of $720 km / h$ . The density of air at its height is $1.2 kg / m^{3}$ . Estimate the fractional increase in air speed on the upper surface of its wing relative to lower surface.

1 0.04

2 0.08

3 0.17

4 0.32

Explanation:

C Given data,
$Mass (m) = 4 \times 10^{5} kg$
$Area (A) = 500 m^{2}$
$Speed (v) = 720 km / h$
$Density (ρ) = 1.2 kg / m^{3}$
Acceleration due to gravity $(g) = 10 m / s^{2}$
The weight of the aircraft is balanced by the upward force due to the pressure difference
$Δ P = \frac{w}{A}$
$Δ P = \frac{mg}{A} [w = mg]$
$Δ P = \frac{4}{5} \times 10^{4} {Nm}^{- 2} = 8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure,
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Δ P = ρ \frac{(v_{2} + v_{1})}{2} (v_{2} - v_{1}) = ρ v_{avg} (v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{Δ P}{ρ v_{avg}}$
Here, $v_{avg} = \frac{v_{1} + v_{2}}{2} = 720 km / h$
$= 720 \times \frac{5}{18} = 200 m / \sec$
$∴ \frac{v_{2} - v_{1}}{v_{avg}} = \frac{Δ P}{ρ v_{avg}^{2}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times (200)^{2}}$
$= \frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}}$
$= 0.17$

AMU-2012

Mechanical Properties of Fluids

143445 Radius of one arm of hydraulic lift is four times of radius of other arm. What force should be applied on narrow arm to lift $100 kg$ .

1

26.5 N

2

62.5 N

3

6.25 N

4

8.3 N

Explanation:

B Given that,
Mass (m) $= 100 kg$
By using the Pascal's law
$\frac{F}{A} = \frac{f}{a}$
$f = \frac{F a}{A}$
$f = \frac{100 g \times π r^{2}}{π (4 r)^{2}}$
$f = \frac{100 \times 10 \times π r^{2}}{π \times 16 r^{2}}$
$f = \frac{1000}{16}$
$f = 62.5 N$

DCE-2007

Mechanical Properties of Fluids

143446 Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$ . If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

A Area of hole $A = A_{A}$
Area of hole $B = A_{B}$
According to questions-
$A_{B} = 2 A_{A}$
Velocity of rejection for both $= \sqrt{2 gh}$
Mass flux $= ρ A V$
Some mass flux for both hole
$ρ_{A} A_{A} V_{A} = ρ_{B} A_{B} V_{B}$
$V_{A} = V_{B}$
$\frac{ρ_{A}}{ρ_{B}} = \frac{A_{B}}{A_{A}} = \frac{2 A_{A}}{A_{A}} = 2$

EAMCET-2002

Mechanical Properties of Fluids

143447 There are two holes one each along the opposite sides of a wide rectangular tank. The crosssection of each hole is $0.01 m^{2}$ and the vertical distance between the holes is $1 m$ . The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 kg m^{- 3}$ )

1 100

2 200

3 300

4 400

Explanation:

B Net force on the tank $= ρ a v^{2}$
Where $ρ =$ density, $a =$ area, $v =$ velocity
$v = \sqrt{2 gh}$
$v^{2} = 2 gh [\begin{aligned} where g = 10 m / \sec^{2} \\ h = 1 m \end{aligned}$
$v^{2} = 2 \times 10 \times 1 [\begin{array}{l} ρ = 1000 {kgm}^{- 3} \\ a = 0.01 m^{2} \end{array}$
$v^{2} = 20$
Net force on the tank $= ρ v^{2}$
$= 1000 \times 0.01 \times 20$
$= 200 N$

EAMCET-2004

Mechanical Properties of Fluids

143448 A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1 ${ms}^{- 1}$ when pressure is $50 kPa$ at a point. If the velocity of flow has to be $2 {ms}^{- 1}$ at some other point, the pressure at that point should be

1

50 kPa

2

100 kPa

3

48.5 kPa

4

24.25 kPa

Explanation:

C Given that,
$v_{1} = 1 m / \sec$
$P_{1} = 50 k Pa$
$v_{2} = 2 m / \sec$
$P_{2} = ?$
From Bernoulli's equation -
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + Z_{2}$
For horizontal pipe $(Z_{1} = Z_{2})$
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{v^{2}}{2}$
$\frac{50}{1000} + \frac{1^{2}}{2} = \frac{P_{2}}{1000} + \frac{2^{2}}{2}$
$\frac{P_{2}}{1000} = \frac{50 \times 1000}{1000} + \frac{1}{2} - 2$
$\frac{P_{2}}{1000} = 48.5 Pa$
$P_{2} = 48.5 \times 1000 Pa$
$P_{2} = 48.5 k Pa$

EAMCET-2007

Mechanical Properties of Fluids

143450 An aircraft has a mass $4 \times 10^{5} kg$ with total wing area $500 m^{2}$ flying at a speed of $720 km / h$ . The density of air at its height is $1.2 kg / m^{3}$ . Estimate the fractional increase in air speed on the upper surface of its wing relative to lower surface.

1 0.04

2 0.08

3 0.17

4 0.32

Explanation:

C Given data,
$Mass (m) = 4 \times 10^{5} kg$
$Area (A) = 500 m^{2}$
$Speed (v) = 720 km / h$
$Density (ρ) = 1.2 kg / m^{3}$
Acceleration due to gravity $(g) = 10 m / s^{2}$
The weight of the aircraft is balanced by the upward force due to the pressure difference
$Δ P = \frac{w}{A}$
$Δ P = \frac{mg}{A} [w = mg]$
$Δ P = \frac{4}{5} \times 10^{4} {Nm}^{- 2} = 8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure,
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Δ P = ρ \frac{(v_{2} + v_{1})}{2} (v_{2} - v_{1}) = ρ v_{avg} (v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{Δ P}{ρ v_{avg}}$
Here, $v_{avg} = \frac{v_{1} + v_{2}}{2} = 720 km / h$
$= 720 \times \frac{5}{18} = 200 m / \sec$
$∴ \frac{v_{2} - v_{1}}{v_{avg}} = \frac{Δ P}{ρ v_{avg}^{2}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times (200)^{2}}$
$= \frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}}$
$= 0.17$

AMU-2012

Mechanical Properties of Fluids

143445 Radius of one arm of hydraulic lift is four times of radius of other arm. What force should be applied on narrow arm to lift $100 kg$ .

1

26.5 N

2

62.5 N

3

6.25 N

4

8.3 N

Explanation:

B Given that,
Mass (m) $= 100 kg$
By using the Pascal's law
$\frac{F}{A} = \frac{f}{a}$
$f = \frac{F a}{A}$
$f = \frac{100 g \times π r^{2}}{π (4 r)^{2}}$
$f = \frac{100 \times 10 \times π r^{2}}{π \times 16 r^{2}}$
$f = \frac{1000}{16}$
$f = 62.5 N$

DCE-2007

Mechanical Properties of Fluids

143446 Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$ . If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

1 2

2

\frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

A Area of hole $A = A_{A}$
Area of hole $B = A_{B}$
According to questions-
$A_{B} = 2 A_{A}$
Velocity of rejection for both $= \sqrt{2 gh}$
Mass flux $= ρ A V$
Some mass flux for both hole
$ρ_{A} A_{A} V_{A} = ρ_{B} A_{B} V_{B}$
$V_{A} = V_{B}$
$\frac{ρ_{A}}{ρ_{B}} = \frac{A_{B}}{A_{A}} = \frac{2 A_{A}}{A_{A}} = 2$

EAMCET-2002

Mechanical Properties of Fluids

143447 There are two holes one each along the opposite sides of a wide rectangular tank. The crosssection of each hole is $0.01 m^{2}$ and the vertical distance between the holes is $1 m$ . The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 kg m^{- 3}$ )

1 100

2 200

3 300

4 400

Explanation:

B Net force on the tank $= ρ a v^{2}$
Where $ρ =$ density, $a =$ area, $v =$ velocity
$v = \sqrt{2 gh}$
$v^{2} = 2 gh [\begin{aligned} where g = 10 m / \sec^{2} \\ h = 1 m \end{aligned}$
$v^{2} = 2 \times 10 \times 1 [\begin{array}{l} ρ = 1000 {kgm}^{- 3} \\ a = 0.01 m^{2} \end{array}$
$v^{2} = 20$
Net force on the tank $= ρ v^{2}$
$= 1000 \times 0.01 \times 20$
$= 200 N$

EAMCET-2004

Mechanical Properties of Fluids

143448 A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1 ${ms}^{- 1}$ when pressure is $50 kPa$ at a point. If the velocity of flow has to be $2 {ms}^{- 1}$ at some other point, the pressure at that point should be

1

50 kPa

2

100 kPa

3

48.5 kPa

4

24.25 kPa

Explanation:

C Given that,
$v_{1} = 1 m / \sec$
$P_{1} = 50 k Pa$
$v_{2} = 2 m / \sec$
$P_{2} = ?$
From Bernoulli's equation -
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + Z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + Z_{2}$
For horizontal pipe $(Z_{1} = Z_{2})$
$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$
$\frac{P_{1}}{ρ} + \frac{v_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{v^{2}}{2}$
$\frac{50}{1000} + \frac{1^{2}}{2} = \frac{P_{2}}{1000} + \frac{2^{2}}{2}$
$\frac{P_{2}}{1000} = \frac{50 \times 1000}{1000} + \frac{1}{2} - 2$
$\frac{P_{2}}{1000} = 48.5 Pa$
$P_{2} = 48.5 \times 1000 Pa$
$P_{2} = 48.5 k Pa$

EAMCET-2007

Mechanical Properties of Fluids

143450 An aircraft has a mass $4 \times 10^{5} kg$ with total wing area $500 m^{2}$ flying at a speed of $720 km / h$ . The density of air at its height is $1.2 kg / m^{3}$ . Estimate the fractional increase in air speed on the upper surface of its wing relative to lower surface.

1 0.04

2 0.08

3 0.17

4 0.32

Explanation:

C Given data,
$Mass (m) = 4 \times 10^{5} kg$
$Area (A) = 500 m^{2}$
$Speed (v) = 720 km / h$
$Density (ρ) = 1.2 kg / m^{3}$
Acceleration due to gravity $(g) = 10 m / s^{2}$
The weight of the aircraft is balanced by the upward force due to the pressure difference
$Δ P = \frac{w}{A}$
$Δ P = \frac{mg}{A} [w = mg]$
$Δ P = \frac{4}{5} \times 10^{4} {Nm}^{- 2} = 8 \times 10^{3} {Nm}^{- 2}$
Let $v_{1}, v_{2}$ are the speed of air on the lower and upper surface of the wings of the aircraft and $P_{1}, P_{2}$ are the pressure,
Using Bernoulli's theorem, we get
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
$Δ P = ρ \frac{(v_{2} + v_{1})}{2} (v_{2} - v_{1}) = ρ v_{avg} (v_{2} - v_{1})$
$v_{2} - v_{1} = \frac{Δ P}{ρ v_{avg}}$
Here, $v_{avg} = \frac{v_{1} + v_{2}}{2} = 720 km / h$
$= 720 \times \frac{5}{18} = 200 m / \sec$
$∴ \frac{v_{2} - v_{1}}{v_{avg}} = \frac{Δ P}{ρ v_{avg}^{2}} = \frac{\frac{4}{5} \times 10^{4}}{1.2 \times (200)^{2}}$
$= \frac{4 \times 10^{4}}{5 \times 1.2 \times 4 \times 10^{4}}$
$= 0.17$

AMU-2012