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Mechanical Properties of Fluids

143172 The air pressure at sea level is $101325 Pa$ . At the centre of a rarefaction of a sound wave in air, the pressure is $91000 Pa$ . Which is the most likely pressure at the centre of a compression of the same wave?

1

91000 Pa

2

101000 Pa

3

111600 Pa

4

121000 Pa

Explanation:

C Given,
\section*{}
Air Pressure $(P_{o}) = 101325 Pa$
Pressure at centre of rarefaction $(P_{1}) = 91000 Pa$

From the above figure-
We have,
$P_{1} = P_{o} - d P$
$91000 = 101325 - d P$
$d P = 101325 - 91000$
$d P = 10325 Pa$
Then,
Pressure at the centre of compression $(P_{2}) = P_{o} + dP$
$P_{2} = 101325 Pa + 10325 Pa$
$P_{2} = 111650 Pa$
Hence, most appropriate answer is $111600 Pa$ .

AMU-2018

Mechanical Properties of Fluids

143173 The volumes of containers $A$ and $B$ , connected by a tube and a closed valve, are $V$ and $4 V$ respectively. Both the containers $A$ and $B$ have the same ideal gas at pressures (temperatures) $5.0 \times 10^{5} Pa (300 K)$ and $1.0 \times 10^{5} Pa (400 K)$ , respectively. The valve is opened to allow the pressure to equalize, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers

1

2.5 \times 10^{5} Pa

2

2.0 \times 10^{5} Pa

3

3.0 \times 10^{5} Pa

4

1.5 \times 10^{5} Pa

Explanation:

B

Given,
For container $A$ ,
$P_{1} = 5 \times 10^{5} Pa, T_{1} = 300 K, V_{1} = V$
For container $B$ , Then,
$P_{2} = 1 \times 10^{5} Pa, T_{2} = 400 K, V_{2} = 4 V$
$P_{1} > P_{2}$
To equalise pressure $P$ , there will be movement of gas molecule from container A to container B.
Then,
$P_{A} = P_{1} - P$
$\frac{P_{A} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(P_{1} - P) \times V}{T_{1}} = \frac{P_{2} \times 4 V}{T_{2}}$
$\frac{(5 \times 10^{5} - P) \times V}{300} = \frac{1 \times 10^{5} \times 4 V}{400}$
$\frac{(5 \times 10^{5} - P)}{3} = \frac{1 \times 10^{5} \times 4}{4}$
$P = 5 \times 10^{5} - 3 \times 10^{5}$
$P = 2 \times 10^{5} Pa$

AMU-2014

Mechanical Properties of Fluids

143174 The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is

1

\frac{v R^{2}}{n^{2} r^{2}}

2

\frac{{vR}^{2}}{{nr}^{2}}

3

\frac{vR}{n^{3} r^{2}}

4

\frac{v^{2} R}{nr}

Explanation:

B Given,
$R =$ Radius of spray pump
$r =$ Radius of each holes
$n =$ No. of holes
From the Equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
Where $A_{1} =$ area of spray pump
$v_{1} =$ speed of the liquid in the tube
$A_{2} =$ area of holes
$v_{2} =$ ejection of liquid through holes
Then,
$π R^{2} \times v = n π r^{2} \times v_{2}$
$v_{2} = \frac{v R^{2}}{n r^{2}}$

AIPMT-2015

Mechanical Properties of Fluids

143175 The pressure on the top surface of an aeroplane wing is $0.8 \times 10^{5} Pa$ and the pressure on the bottom surface is $0.75 \times 10^{5} Pa$ . If the area of each surface is $50 m^{2}$ , the dynamic lift on the wing is

1

25 \times 10^{4} N

2

0.5 \times 10^{4} N

3

5 \times 10^{4} N

4

0.25 \times 10^{5} N

Explanation:

A Given,
Pressure on the top surface of a aeroplane wing-
$P_{1} = 0.8 \times 10^{5} Pa$
And pressure on the bottom surface $(P_{2}) = 0.75 \times 10^{5} Pa$ . Area of each surface $= 50 m^{2}$

We know that,
$Pressure = \frac{Force}{Area}$
$Force = Pressure \times Area$
Now,
$Net Pressure (P_{net}) = P_{1} - P_{2}$
$= 0.8 \times 10^{5} - 0.75 \times 10^{5}$
$= 0.05 \times 10^{5} Pa$
Then,
$Force = P_{net} \times Area$
$= 0.05 \times 10^{5} \times 50$
$Force = 25 \times 10^{4} N$

AP EAMCET(Medical)-2007

Mechanical Properties of Fluids

143176 An aeroplane of mass $3 \times 10^{4} kg$ and total wing area of $120 m^{2}$ is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in $kPa$ is $(g = 10$ $m / s^{2}$ )

1 2.5

2 5.0

3 10.0

4 12.5

Explanation:

A Given,
$Mass of plane = 3 \times 10^{4} kg$
$Area of wing = 120 m^{2}$
$Gravity (g) = 10 m / \sec^{2}$
We know that,
$Force (F) = mg$
$F = 3 \times 10^{4} \times 10$
$= 3 \times 10^{5} N$
Now,
Force $=$ Pressure difference $(P_{2} - P_{1}) \times$ Area
$3 \times 10^{5} = (P_{2} - P_{1}) \times 120$
$P_{2} - P_{1} = \frac{3 \times 10^{5}}{120}$
$P_{2} - P_{1} = 2500 Pa$
$P_{2} - P_{1} = 2.5 kPa .$

AP EAMCET(Medical)-2002

Mechanical Properties of Fluids

143172 The air pressure at sea level is $101325 Pa$ . At the centre of a rarefaction of a sound wave in air, the pressure is $91000 Pa$ . Which is the most likely pressure at the centre of a compression of the same wave?

1

91000 Pa

2

101000 Pa

3

111600 Pa

4

121000 Pa

Explanation:

C Given,
\section*{}
Air Pressure $(P_{o}) = 101325 Pa$
Pressure at centre of rarefaction $(P_{1}) = 91000 Pa$

From the above figure-
We have,
$P_{1} = P_{o} - d P$
$91000 = 101325 - d P$
$d P = 101325 - 91000$
$d P = 10325 Pa$
Then,
Pressure at the centre of compression $(P_{2}) = P_{o} + dP$
$P_{2} = 101325 Pa + 10325 Pa$
$P_{2} = 111650 Pa$
Hence, most appropriate answer is $111600 Pa$ .

AMU-2018

Mechanical Properties of Fluids

143173 The volumes of containers $A$ and $B$ , connected by a tube and a closed valve, are $V$ and $4 V$ respectively. Both the containers $A$ and $B$ have the same ideal gas at pressures (temperatures) $5.0 \times 10^{5} Pa (300 K)$ and $1.0 \times 10^{5} Pa (400 K)$ , respectively. The valve is opened to allow the pressure to equalize, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers

1

2.5 \times 10^{5} Pa

2

2.0 \times 10^{5} Pa

3

3.0 \times 10^{5} Pa

4

1.5 \times 10^{5} Pa

Explanation:

B

Given,
For container $A$ ,
$P_{1} = 5 \times 10^{5} Pa, T_{1} = 300 K, V_{1} = V$
For container $B$ , Then,
$P_{2} = 1 \times 10^{5} Pa, T_{2} = 400 K, V_{2} = 4 V$
$P_{1} > P_{2}$
To equalise pressure $P$ , there will be movement of gas molecule from container A to container B.
Then,
$P_{A} = P_{1} - P$
$\frac{P_{A} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(P_{1} - P) \times V}{T_{1}} = \frac{P_{2} \times 4 V}{T_{2}}$
$\frac{(5 \times 10^{5} - P) \times V}{300} = \frac{1 \times 10^{5} \times 4 V}{400}$
$\frac{(5 \times 10^{5} - P)}{3} = \frac{1 \times 10^{5} \times 4}{4}$
$P = 5 \times 10^{5} - 3 \times 10^{5}$
$P = 2 \times 10^{5} Pa$

AMU-2014

Mechanical Properties of Fluids

143174 The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is

1

\frac{v R^{2}}{n^{2} r^{2}}

2

\frac{{vR}^{2}}{{nr}^{2}}

3

\frac{vR}{n^{3} r^{2}}

4

\frac{v^{2} R}{nr}

Explanation:

B Given,
$R =$ Radius of spray pump
$r =$ Radius of each holes
$n =$ No. of holes
From the Equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
Where $A_{1} =$ area of spray pump
$v_{1} =$ speed of the liquid in the tube
$A_{2} =$ area of holes
$v_{2} =$ ejection of liquid through holes
Then,
$π R^{2} \times v = n π r^{2} \times v_{2}$
$v_{2} = \frac{v R^{2}}{n r^{2}}$

AIPMT-2015

Mechanical Properties of Fluids

143175 The pressure on the top surface of an aeroplane wing is $0.8 \times 10^{5} Pa$ and the pressure on the bottom surface is $0.75 \times 10^{5} Pa$ . If the area of each surface is $50 m^{2}$ , the dynamic lift on the wing is

1

25 \times 10^{4} N

2

0.5 \times 10^{4} N

3

5 \times 10^{4} N

4

0.25 \times 10^{5} N

Explanation:

A Given,
Pressure on the top surface of a aeroplane wing-
$P_{1} = 0.8 \times 10^{5} Pa$
And pressure on the bottom surface $(P_{2}) = 0.75 \times 10^{5} Pa$ . Area of each surface $= 50 m^{2}$

We know that,
$Pressure = \frac{Force}{Area}$
$Force = Pressure \times Area$
Now,
$Net Pressure (P_{net}) = P_{1} - P_{2}$
$= 0.8 \times 10^{5} - 0.75 \times 10^{5}$
$= 0.05 \times 10^{5} Pa$
Then,
$Force = P_{net} \times Area$
$= 0.05 \times 10^{5} \times 50$
$Force = 25 \times 10^{4} N$

AP EAMCET(Medical)-2007

Mechanical Properties of Fluids

143176 An aeroplane of mass $3 \times 10^{4} kg$ and total wing area of $120 m^{2}$ is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in $kPa$ is $(g = 10$ $m / s^{2}$ )

1 2.5

2 5.0

3 10.0

4 12.5

Explanation:

A Given,
$Mass of plane = 3 \times 10^{4} kg$
$Area of wing = 120 m^{2}$
$Gravity (g) = 10 m / \sec^{2}$
We know that,
$Force (F) = mg$
$F = 3 \times 10^{4} \times 10$
$= 3 \times 10^{5} N$
Now,
Force $=$ Pressure difference $(P_{2} - P_{1}) \times$ Area
$3 \times 10^{5} = (P_{2} - P_{1}) \times 120$
$P_{2} - P_{1} = \frac{3 \times 10^{5}}{120}$
$P_{2} - P_{1} = 2500 Pa$
$P_{2} - P_{1} = 2.5 kPa .$

AP EAMCET(Medical)-2002

Mechanical Properties of Fluids

143172 The air pressure at sea level is $101325 Pa$ . At the centre of a rarefaction of a sound wave in air, the pressure is $91000 Pa$ . Which is the most likely pressure at the centre of a compression of the same wave?

1

91000 Pa

2

101000 Pa

3

111600 Pa

4

121000 Pa

Explanation:

C Given,
\section*{}
Air Pressure $(P_{o}) = 101325 Pa$
Pressure at centre of rarefaction $(P_{1}) = 91000 Pa$

From the above figure-
We have,
$P_{1} = P_{o} - d P$
$91000 = 101325 - d P$
$d P = 101325 - 91000$
$d P = 10325 Pa$
Then,
Pressure at the centre of compression $(P_{2}) = P_{o} + dP$
$P_{2} = 101325 Pa + 10325 Pa$
$P_{2} = 111650 Pa$
Hence, most appropriate answer is $111600 Pa$ .

AMU-2018

Mechanical Properties of Fluids

143173 The volumes of containers $A$ and $B$ , connected by a tube and a closed valve, are $V$ and $4 V$ respectively. Both the containers $A$ and $B$ have the same ideal gas at pressures (temperatures) $5.0 \times 10^{5} Pa (300 K)$ and $1.0 \times 10^{5} Pa (400 K)$ , respectively. The valve is opened to allow the pressure to equalize, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers

1

2.5 \times 10^{5} Pa

2

2.0 \times 10^{5} Pa

3

3.0 \times 10^{5} Pa

4

1.5 \times 10^{5} Pa

Explanation:

B

Given,
For container $A$ ,
$P_{1} = 5 \times 10^{5} Pa, T_{1} = 300 K, V_{1} = V$
For container $B$ , Then,
$P_{2} = 1 \times 10^{5} Pa, T_{2} = 400 K, V_{2} = 4 V$
$P_{1} > P_{2}$
To equalise pressure $P$ , there will be movement of gas molecule from container A to container B.
Then,
$P_{A} = P_{1} - P$
$\frac{P_{A} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(P_{1} - P) \times V}{T_{1}} = \frac{P_{2} \times 4 V}{T_{2}}$
$\frac{(5 \times 10^{5} - P) \times V}{300} = \frac{1 \times 10^{5} \times 4 V}{400}$
$\frac{(5 \times 10^{5} - P)}{3} = \frac{1 \times 10^{5} \times 4}{4}$
$P = 5 \times 10^{5} - 3 \times 10^{5}$
$P = 2 \times 10^{5} Pa$

AMU-2014

Mechanical Properties of Fluids

143174 The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is

1

\frac{v R^{2}}{n^{2} r^{2}}

2

\frac{{vR}^{2}}{{nr}^{2}}

3

\frac{vR}{n^{3} r^{2}}

4

\frac{v^{2} R}{nr}

Explanation:

B Given,
$R =$ Radius of spray pump
$r =$ Radius of each holes
$n =$ No. of holes
From the Equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
Where $A_{1} =$ area of spray pump
$v_{1} =$ speed of the liquid in the tube
$A_{2} =$ area of holes
$v_{2} =$ ejection of liquid through holes
Then,
$π R^{2} \times v = n π r^{2} \times v_{2}$
$v_{2} = \frac{v R^{2}}{n r^{2}}$

AIPMT-2015

Mechanical Properties of Fluids

143175 The pressure on the top surface of an aeroplane wing is $0.8 \times 10^{5} Pa$ and the pressure on the bottom surface is $0.75 \times 10^{5} Pa$ . If the area of each surface is $50 m^{2}$ , the dynamic lift on the wing is

1

25 \times 10^{4} N

2

0.5 \times 10^{4} N

3

5 \times 10^{4} N

4

0.25 \times 10^{5} N

Explanation:

A Given,
Pressure on the top surface of a aeroplane wing-
$P_{1} = 0.8 \times 10^{5} Pa$
And pressure on the bottom surface $(P_{2}) = 0.75 \times 10^{5} Pa$ . Area of each surface $= 50 m^{2}$

We know that,
$Pressure = \frac{Force}{Area}$
$Force = Pressure \times Area$
Now,
$Net Pressure (P_{net}) = P_{1} - P_{2}$
$= 0.8 \times 10^{5} - 0.75 \times 10^{5}$
$= 0.05 \times 10^{5} Pa$
Then,
$Force = P_{net} \times Area$
$= 0.05 \times 10^{5} \times 50$
$Force = 25 \times 10^{4} N$

AP EAMCET(Medical)-2007

Mechanical Properties of Fluids

143176 An aeroplane of mass $3 \times 10^{4} kg$ and total wing area of $120 m^{2}$ is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in $kPa$ is $(g = 10$ $m / s^{2}$ )

1 2.5

2 5.0

3 10.0

4 12.5

Explanation:

A Given,
$Mass of plane = 3 \times 10^{4} kg$
$Area of wing = 120 m^{2}$
$Gravity (g) = 10 m / \sec^{2}$
We know that,
$Force (F) = mg$
$F = 3 \times 10^{4} \times 10$
$= 3 \times 10^{5} N$
Now,
Force $=$ Pressure difference $(P_{2} - P_{1}) \times$ Area
$3 \times 10^{5} = (P_{2} - P_{1}) \times 120$
$P_{2} - P_{1} = \frac{3 \times 10^{5}}{120}$
$P_{2} - P_{1} = 2500 Pa$
$P_{2} - P_{1} = 2.5 kPa .$

AP EAMCET(Medical)-2002

Mechanical Properties of Fluids

143172 The air pressure at sea level is $101325 Pa$ . At the centre of a rarefaction of a sound wave in air, the pressure is $91000 Pa$ . Which is the most likely pressure at the centre of a compression of the same wave?

1

91000 Pa

2

101000 Pa

3

111600 Pa

4

121000 Pa

Explanation:

C Given,
\section*{}
Air Pressure $(P_{o}) = 101325 Pa$
Pressure at centre of rarefaction $(P_{1}) = 91000 Pa$

From the above figure-
We have,
$P_{1} = P_{o} - d P$
$91000 = 101325 - d P$
$d P = 101325 - 91000$
$d P = 10325 Pa$
Then,
Pressure at the centre of compression $(P_{2}) = P_{o} + dP$
$P_{2} = 101325 Pa + 10325 Pa$
$P_{2} = 111650 Pa$
Hence, most appropriate answer is $111600 Pa$ .

AMU-2018

Mechanical Properties of Fluids

143173 The volumes of containers $A$ and $B$ , connected by a tube and a closed valve, are $V$ and $4 V$ respectively. Both the containers $A$ and $B$ have the same ideal gas at pressures (temperatures) $5.0 \times 10^{5} Pa (300 K)$ and $1.0 \times 10^{5} Pa (400 K)$ , respectively. The valve is opened to allow the pressure to equalize, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers

1

2.5 \times 10^{5} Pa

2

2.0 \times 10^{5} Pa

3

3.0 \times 10^{5} Pa

4

1.5 \times 10^{5} Pa

Explanation:

B

Given,
For container $A$ ,
$P_{1} = 5 \times 10^{5} Pa, T_{1} = 300 K, V_{1} = V$
For container $B$ , Then,
$P_{2} = 1 \times 10^{5} Pa, T_{2} = 400 K, V_{2} = 4 V$
$P_{1} > P_{2}$
To equalise pressure $P$ , there will be movement of gas molecule from container A to container B.
Then,
$P_{A} = P_{1} - P$
$\frac{P_{A} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(P_{1} - P) \times V}{T_{1}} = \frac{P_{2} \times 4 V}{T_{2}}$
$\frac{(5 \times 10^{5} - P) \times V}{300} = \frac{1 \times 10^{5} \times 4 V}{400}$
$\frac{(5 \times 10^{5} - P)}{3} = \frac{1 \times 10^{5} \times 4}{4}$
$P = 5 \times 10^{5} - 3 \times 10^{5}$
$P = 2 \times 10^{5} Pa$

AMU-2014

Mechanical Properties of Fluids

143174 The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is

1

\frac{v R^{2}}{n^{2} r^{2}}

2

\frac{{vR}^{2}}{{nr}^{2}}

3

\frac{vR}{n^{3} r^{2}}

4

\frac{v^{2} R}{nr}

Explanation:

B Given,
$R =$ Radius of spray pump
$r =$ Radius of each holes
$n =$ No. of holes
From the Equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
Where $A_{1} =$ area of spray pump
$v_{1} =$ speed of the liquid in the tube
$A_{2} =$ area of holes
$v_{2} =$ ejection of liquid through holes
Then,
$π R^{2} \times v = n π r^{2} \times v_{2}$
$v_{2} = \frac{v R^{2}}{n r^{2}}$

AIPMT-2015

Mechanical Properties of Fluids

143175 The pressure on the top surface of an aeroplane wing is $0.8 \times 10^{5} Pa$ and the pressure on the bottom surface is $0.75 \times 10^{5} Pa$ . If the area of each surface is $50 m^{2}$ , the dynamic lift on the wing is

1

25 \times 10^{4} N

2

0.5 \times 10^{4} N

3

5 \times 10^{4} N

4

0.25 \times 10^{5} N

Explanation:

A Given,
Pressure on the top surface of a aeroplane wing-
$P_{1} = 0.8 \times 10^{5} Pa$
And pressure on the bottom surface $(P_{2}) = 0.75 \times 10^{5} Pa$ . Area of each surface $= 50 m^{2}$

We know that,
$Pressure = \frac{Force}{Area}$
$Force = Pressure \times Area$
Now,
$Net Pressure (P_{net}) = P_{1} - P_{2}$
$= 0.8 \times 10^{5} - 0.75 \times 10^{5}$
$= 0.05 \times 10^{5} Pa$
Then,
$Force = P_{net} \times Area$
$= 0.05 \times 10^{5} \times 50$
$Force = 25 \times 10^{4} N$

AP EAMCET(Medical)-2007

Mechanical Properties of Fluids

143176 An aeroplane of mass $3 \times 10^{4} kg$ and total wing area of $120 m^{2}$ is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in $kPa$ is $(g = 10$ $m / s^{2}$ )

1 2.5

2 5.0

3 10.0

4 12.5

Explanation:

A Given,
$Mass of plane = 3 \times 10^{4} kg$
$Area of wing = 120 m^{2}$
$Gravity (g) = 10 m / \sec^{2}$
We know that,
$Force (F) = mg$
$F = 3 \times 10^{4} \times 10$
$= 3 \times 10^{5} N$
Now,
Force $=$ Pressure difference $(P_{2} - P_{1}) \times$ Area
$3 \times 10^{5} = (P_{2} - P_{1}) \times 120$
$P_{2} - P_{1} = \frac{3 \times 10^{5}}{120}$
$P_{2} - P_{1} = 2500 Pa$
$P_{2} - P_{1} = 2.5 kPa .$

AP EAMCET(Medical)-2002

Mechanical Properties of Fluids

143172 The air pressure at sea level is $101325 Pa$ . At the centre of a rarefaction of a sound wave in air, the pressure is $91000 Pa$ . Which is the most likely pressure at the centre of a compression of the same wave?

1

91000 Pa

2

101000 Pa

3

111600 Pa

4

121000 Pa

Explanation:

C Given,
\section*{}
Air Pressure $(P_{o}) = 101325 Pa$
Pressure at centre of rarefaction $(P_{1}) = 91000 Pa$

From the above figure-
We have,
$P_{1} = P_{o} - d P$
$91000 = 101325 - d P$
$d P = 101325 - 91000$
$d P = 10325 Pa$
Then,
Pressure at the centre of compression $(P_{2}) = P_{o} + dP$
$P_{2} = 101325 Pa + 10325 Pa$
$P_{2} = 111650 Pa$
Hence, most appropriate answer is $111600 Pa$ .

AMU-2018

Mechanical Properties of Fluids

143173 The volumes of containers $A$ and $B$ , connected by a tube and a closed valve, are $V$ and $4 V$ respectively. Both the containers $A$ and $B$ have the same ideal gas at pressures (temperatures) $5.0 \times 10^{5} Pa (300 K)$ and $1.0 \times 10^{5} Pa (400 K)$ , respectively. The valve is opened to allow the pressure to equalize, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers

1

2.5 \times 10^{5} Pa

2

2.0 \times 10^{5} Pa

3

3.0 \times 10^{5} Pa

4

1.5 \times 10^{5} Pa

Explanation:

B

Given,
For container $A$ ,
$P_{1} = 5 \times 10^{5} Pa, T_{1} = 300 K, V_{1} = V$
For container $B$ , Then,
$P_{2} = 1 \times 10^{5} Pa, T_{2} = 400 K, V_{2} = 4 V$
$P_{1} > P_{2}$
To equalise pressure $P$ , there will be movement of gas molecule from container A to container B.
Then,
$P_{A} = P_{1} - P$
$\frac{P_{A} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{(P_{1} - P) \times V}{T_{1}} = \frac{P_{2} \times 4 V}{T_{2}}$
$\frac{(5 \times 10^{5} - P) \times V}{300} = \frac{1 \times 10^{5} \times 4 V}{400}$
$\frac{(5 \times 10^{5} - P)}{3} = \frac{1 \times 10^{5} \times 4}{4}$
$P = 5 \times 10^{5} - 3 \times 10^{5}$
$P = 2 \times 10^{5} Pa$

AMU-2014

Mechanical Properties of Fluids

143174 The cylindrical tube of a spray pump has radius $R$ , one end of which has $n$ fine holes, each of radius $r$ . If the speed of the liquid in the tube is $v$ , the speed of the ejection of the liquid through the holes is

1

\frac{v R^{2}}{n^{2} r^{2}}

2

\frac{{vR}^{2}}{{nr}^{2}}

3

\frac{vR}{n^{3} r^{2}}

4

\frac{v^{2} R}{nr}

Explanation:

B Given,
$R =$ Radius of spray pump
$r =$ Radius of each holes
$n =$ No. of holes
From the Equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
Where $A_{1} =$ area of spray pump
$v_{1} =$ speed of the liquid in the tube
$A_{2} =$ area of holes
$v_{2} =$ ejection of liquid through holes
Then,
$π R^{2} \times v = n π r^{2} \times v_{2}$
$v_{2} = \frac{v R^{2}}{n r^{2}}$

AIPMT-2015

Mechanical Properties of Fluids

143175 The pressure on the top surface of an aeroplane wing is $0.8 \times 10^{5} Pa$ and the pressure on the bottom surface is $0.75 \times 10^{5} Pa$ . If the area of each surface is $50 m^{2}$ , the dynamic lift on the wing is

1

25 \times 10^{4} N

2

0.5 \times 10^{4} N

3

5 \times 10^{4} N

4

0.25 \times 10^{5} N

Explanation:

A Given,
Pressure on the top surface of a aeroplane wing-
$P_{1} = 0.8 \times 10^{5} Pa$
And pressure on the bottom surface $(P_{2}) = 0.75 \times 10^{5} Pa$ . Area of each surface $= 50 m^{2}$

We know that,
$Pressure = \frac{Force}{Area}$
$Force = Pressure \times Area$
Now,
$Net Pressure (P_{net}) = P_{1} - P_{2}$
$= 0.8 \times 10^{5} - 0.75 \times 10^{5}$
$= 0.05 \times 10^{5} Pa$
Then,
$Force = P_{net} \times Area$
$= 0.05 \times 10^{5} \times 50$
$Force = 25 \times 10^{4} N$

AP EAMCET(Medical)-2007

Mechanical Properties of Fluids

143176 An aeroplane of mass $3 \times 10^{4} kg$ and total wing area of $120 m^{2}$ is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in $kPa$ is $(g = 10$ $m / s^{2}$ )

1 2.5

2 5.0

3 10.0

4 12.5

Explanation:

A Given,
$Mass of plane = 3 \times 10^{4} kg$
$Area of wing = 120 m^{2}$
$Gravity (g) = 10 m / \sec^{2}$
We know that,
$Force (F) = mg$
$F = 3 \times 10^{4} \times 10$
$= 3 \times 10^{5} N$
Now,
Force $=$ Pressure difference $(P_{2} - P_{1}) \times$ Area
$3 \times 10^{5} = (P_{2} - P_{1}) \times 120$
$P_{2} - P_{1} = \frac{3 \times 10^{5}}{120}$
$P_{2} - P_{1} = 2500 Pa$
$P_{2} - P_{1} = 2.5 kPa .$

AP EAMCET(Medical)-2002