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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360796 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 c m$ of $H g$ where the velocity is $35 c m s^{- 1}$ . At a point where the velocity is $65 c m s^{- 1}$ , the pressure will be

1

0.89 c m

of

H g

2

8.9 c m

of

H g

3

0.5 c m

of

H g

4

1 c m

of

H g

Explanation:

Using Bernoulli's theorem,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2} (1)$
Here, $p_{1} = ρ_{m} g h_{1} = 13600 \times 9.8 \times 10^{- 2}$
$p_{2} = 13600 \times 9.8 \times h$
$ρ = 1000 k g m^{- 3}$
$v_{1} = 35 \times 10^{- 2} m s^{- 1}, v_{2} = 65 \times 10^{- 2} m s^{- 1}$
$∴$ From Eq.(1), we get
$\Rightarrow 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} \times 1000 \times (0.35)^{2}$
$= 13600 \times 9.8 \times h + \frac{1}{2} \times 1000 \times (0.65)^{2}$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} 10^{3} [(0.35)^{2} - (0.65)^{2}]$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} - 500 (0.3)$
After solving, $h = 0.89 c m$ of $H g$ .

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360797 At what speed will the velocity of a stream of water be equal to $20 c m$ of mercury column ? (Take, $g = 10 m s^{- 2}$ )

1

6.4 m s^{- 1}

2

7.3756 m s^{- 1}

3

6.4756 m s^{- 1}

4 None of these

Explanation:

Here, velocity head $= 20 c m$ of $H g$
$= 20 \times 13.6 cm$ of water
As, velocity head $= v^{2} / 2 g$
$∴ 20 \times 13.6 = \frac{v^{2}}{2 \times 1000}$
$v = \sqrt{20 \times 13.6 \times 2 \times 1000}$
$= 737.56 c m s^{- 1} = 7.3756 m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360798 The rate of flow of glycerine of density $1.25 \times 10^{3} k g / m^{3}$ through the conical section of a pipe, if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length is $10 N / m^{2}$ is

1

5.28 \times 10^{- 4} m^{3} / s

2

6.28 \times 10^{- 4} m^{3} / s

3

7.28 \times 10^{- 4} m^{3} / s

4

8.28 \times 10^{- 4} m^{3} / s

Explanation:

: According to equation of continuity
$\begin{matrix} (1) & \frac{v_{2}}{v_{1}} = \frac{A_{1}}{A_{2}} = \frac{π \times (0.1)^{2}}{π \times (0.04)^{2}} = \frac{25}{4} \end{matrix}$
According to Bernoulli's equation for horizontal tube,
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(p_{1} - p_{2})}{ρ}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(2 \times 10)}{(1.25 \times 10^{3})} = 16 \times 10^{- 3}$ $(2)$
Substituting the value of $v_{2}$ from equation (1) in
(2), we get
${(6.25 v_{1})}^{2} - v_{1}^{2} = 16 \times 10^{- 3}$ or $v_{1} = 0.02 m / s$
So rate of flow through the tube
$= A_{1} v_{1} = A_{2} v_{2} = π \times (0.1)^{2} \times 0.02$
$= 6.28 \times 10^{- 4} m^{3} / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360799 Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will
supporting img

1 remain same

2 increase

3 may increase or decrease, depending on speed of air

4 decrease

MHTCET - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360800 At what speed, the velocity head of the water is twice the pressure head of $75 c m$ of $H g$ ?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

Pressure head $= \frac{p}{ρ g};$ velocity head $= \frac{V^{2}}{2 g}$
$\frac{V^{2}}{2 g} = \frac{2 p}{ρ g} \Rightarrow V^{2} = \frac{4 p^{2}}{ρ}$
$V^{2} = \frac{4 \times 75 \times 10^{- 2} \times 13.6 \times 10^{3} \times 9.8}{10^{3}}$
$= 39984 \times 10^{- 2} = 399$
$V \approx 20 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360796 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 c m$ of $H g$ where the velocity is $35 c m s^{- 1}$ . At a point where the velocity is $65 c m s^{- 1}$ , the pressure will be

1

0.89 c m

of

H g

2

8.9 c m

of

H g

3

0.5 c m

of

H g

4

1 c m

of

H g

Explanation:

Using Bernoulli's theorem,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2} (1)$
Here, $p_{1} = ρ_{m} g h_{1} = 13600 \times 9.8 \times 10^{- 2}$
$p_{2} = 13600 \times 9.8 \times h$
$ρ = 1000 k g m^{- 3}$
$v_{1} = 35 \times 10^{- 2} m s^{- 1}, v_{2} = 65 \times 10^{- 2} m s^{- 1}$
$∴$ From Eq.(1), we get
$\Rightarrow 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} \times 1000 \times (0.35)^{2}$
$= 13600 \times 9.8 \times h + \frac{1}{2} \times 1000 \times (0.65)^{2}$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} 10^{3} [(0.35)^{2} - (0.65)^{2}]$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} - 500 (0.3)$
After solving, $h = 0.89 c m$ of $H g$ .

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360797 At what speed will the velocity of a stream of water be equal to $20 c m$ of mercury column ? (Take, $g = 10 m s^{- 2}$ )

1

6.4 m s^{- 1}

2

7.3756 m s^{- 1}

3

6.4756 m s^{- 1}

4 None of these

Explanation:

Here, velocity head $= 20 c m$ of $H g$
$= 20 \times 13.6 cm$ of water
As, velocity head $= v^{2} / 2 g$
$∴ 20 \times 13.6 = \frac{v^{2}}{2 \times 1000}$
$v = \sqrt{20 \times 13.6 \times 2 \times 1000}$
$= 737.56 c m s^{- 1} = 7.3756 m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360798 The rate of flow of glycerine of density $1.25 \times 10^{3} k g / m^{3}$ through the conical section of a pipe, if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length is $10 N / m^{2}$ is

1

5.28 \times 10^{- 4} m^{3} / s

2

6.28 \times 10^{- 4} m^{3} / s

3

7.28 \times 10^{- 4} m^{3} / s

4

8.28 \times 10^{- 4} m^{3} / s

Explanation:

: According to equation of continuity
$\begin{matrix} (1) & \frac{v_{2}}{v_{1}} = \frac{A_{1}}{A_{2}} = \frac{π \times (0.1)^{2}}{π \times (0.04)^{2}} = \frac{25}{4} \end{matrix}$
According to Bernoulli's equation for horizontal tube,
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(p_{1} - p_{2})}{ρ}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(2 \times 10)}{(1.25 \times 10^{3})} = 16 \times 10^{- 3}$ $(2)$
Substituting the value of $v_{2}$ from equation (1) in
(2), we get
${(6.25 v_{1})}^{2} - v_{1}^{2} = 16 \times 10^{- 3}$ or $v_{1} = 0.02 m / s$
So rate of flow through the tube
$= A_{1} v_{1} = A_{2} v_{2} = π \times (0.1)^{2} \times 0.02$
$= 6.28 \times 10^{- 4} m^{3} / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360799 Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will
supporting img

1 remain same

2 increase

3 may increase or decrease, depending on speed of air

4 decrease

MHTCET - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360800 At what speed, the velocity head of the water is twice the pressure head of $75 c m$ of $H g$ ?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

Pressure head $= \frac{p}{ρ g};$ velocity head $= \frac{V^{2}}{2 g}$
$\frac{V^{2}}{2 g} = \frac{2 p}{ρ g} \Rightarrow V^{2} = \frac{4 p^{2}}{ρ}$
$V^{2} = \frac{4 \times 75 \times 10^{- 2} \times 13.6 \times 10^{3} \times 9.8}{10^{3}}$
$= 39984 \times 10^{- 2} = 399$
$V \approx 20 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360796 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 c m$ of $H g$ where the velocity is $35 c m s^{- 1}$ . At a point where the velocity is $65 c m s^{- 1}$ , the pressure will be

1

0.89 c m

of

H g

2

8.9 c m

of

H g

3

0.5 c m

of

H g

4

1 c m

of

H g

Explanation:

Using Bernoulli's theorem,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2} (1)$
Here, $p_{1} = ρ_{m} g h_{1} = 13600 \times 9.8 \times 10^{- 2}$
$p_{2} = 13600 \times 9.8 \times h$
$ρ = 1000 k g m^{- 3}$
$v_{1} = 35 \times 10^{- 2} m s^{- 1}, v_{2} = 65 \times 10^{- 2} m s^{- 1}$
$∴$ From Eq.(1), we get
$\Rightarrow 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} \times 1000 \times (0.35)^{2}$
$= 13600 \times 9.8 \times h + \frac{1}{2} \times 1000 \times (0.65)^{2}$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} 10^{3} [(0.35)^{2} - (0.65)^{2}]$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} - 500 (0.3)$
After solving, $h = 0.89 c m$ of $H g$ .

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360797 At what speed will the velocity of a stream of water be equal to $20 c m$ of mercury column ? (Take, $g = 10 m s^{- 2}$ )

1

6.4 m s^{- 1}

2

7.3756 m s^{- 1}

3

6.4756 m s^{- 1}

4 None of these

Explanation:

Here, velocity head $= 20 c m$ of $H g$
$= 20 \times 13.6 cm$ of water
As, velocity head $= v^{2} / 2 g$
$∴ 20 \times 13.6 = \frac{v^{2}}{2 \times 1000}$
$v = \sqrt{20 \times 13.6 \times 2 \times 1000}$
$= 737.56 c m s^{- 1} = 7.3756 m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360798 The rate of flow of glycerine of density $1.25 \times 10^{3} k g / m^{3}$ through the conical section of a pipe, if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length is $10 N / m^{2}$ is

1

5.28 \times 10^{- 4} m^{3} / s

2

6.28 \times 10^{- 4} m^{3} / s

3

7.28 \times 10^{- 4} m^{3} / s

4

8.28 \times 10^{- 4} m^{3} / s

Explanation:

: According to equation of continuity
$\begin{matrix} (1) & \frac{v_{2}}{v_{1}} = \frac{A_{1}}{A_{2}} = \frac{π \times (0.1)^{2}}{π \times (0.04)^{2}} = \frac{25}{4} \end{matrix}$
According to Bernoulli's equation for horizontal tube,
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(p_{1} - p_{2})}{ρ}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(2 \times 10)}{(1.25 \times 10^{3})} = 16 \times 10^{- 3}$ $(2)$
Substituting the value of $v_{2}$ from equation (1) in
(2), we get
${(6.25 v_{1})}^{2} - v_{1}^{2} = 16 \times 10^{- 3}$ or $v_{1} = 0.02 m / s$
So rate of flow through the tube
$= A_{1} v_{1} = A_{2} v_{2} = π \times (0.1)^{2} \times 0.02$
$= 6.28 \times 10^{- 4} m^{3} / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360799 Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will
supporting img

1 remain same

2 increase

3 may increase or decrease, depending on speed of air

4 decrease

MHTCET - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360800 At what speed, the velocity head of the water is twice the pressure head of $75 c m$ of $H g$ ?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

Pressure head $= \frac{p}{ρ g};$ velocity head $= \frac{V^{2}}{2 g}$
$\frac{V^{2}}{2 g} = \frac{2 p}{ρ g} \Rightarrow V^{2} = \frac{4 p^{2}}{ρ}$
$V^{2} = \frac{4 \times 75 \times 10^{- 2} \times 13.6 \times 10^{3} \times 9.8}{10^{3}}$
$= 39984 \times 10^{- 2} = 399$
$V \approx 20 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360796 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 c m$ of $H g$ where the velocity is $35 c m s^{- 1}$ . At a point where the velocity is $65 c m s^{- 1}$ , the pressure will be

1

0.89 c m

of

H g

2

8.9 c m

of

H g

3

0.5 c m

of

H g

4

1 c m

of

H g

Explanation:

Using Bernoulli's theorem,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2} (1)$
Here, $p_{1} = ρ_{m} g h_{1} = 13600 \times 9.8 \times 10^{- 2}$
$p_{2} = 13600 \times 9.8 \times h$
$ρ = 1000 k g m^{- 3}$
$v_{1} = 35 \times 10^{- 2} m s^{- 1}, v_{2} = 65 \times 10^{- 2} m s^{- 1}$
$∴$ From Eq.(1), we get
$\Rightarrow 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} \times 1000 \times (0.35)^{2}$
$= 13600 \times 9.8 \times h + \frac{1}{2} \times 1000 \times (0.65)^{2}$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} 10^{3} [(0.35)^{2} - (0.65)^{2}]$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} - 500 (0.3)$
After solving, $h = 0.89 c m$ of $H g$ .

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360797 At what speed will the velocity of a stream of water be equal to $20 c m$ of mercury column ? (Take, $g = 10 m s^{- 2}$ )

1

6.4 m s^{- 1}

2

7.3756 m s^{- 1}

3

6.4756 m s^{- 1}

4 None of these

Explanation:

Here, velocity head $= 20 c m$ of $H g$
$= 20 \times 13.6 cm$ of water
As, velocity head $= v^{2} / 2 g$
$∴ 20 \times 13.6 = \frac{v^{2}}{2 \times 1000}$
$v = \sqrt{20 \times 13.6 \times 2 \times 1000}$
$= 737.56 c m s^{- 1} = 7.3756 m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360798 The rate of flow of glycerine of density $1.25 \times 10^{3} k g / m^{3}$ through the conical section of a pipe, if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length is $10 N / m^{2}$ is

1

5.28 \times 10^{- 4} m^{3} / s

2

6.28 \times 10^{- 4} m^{3} / s

3

7.28 \times 10^{- 4} m^{3} / s

4

8.28 \times 10^{- 4} m^{3} / s

Explanation:

: According to equation of continuity
$\begin{matrix} (1) & \frac{v_{2}}{v_{1}} = \frac{A_{1}}{A_{2}} = \frac{π \times (0.1)^{2}}{π \times (0.04)^{2}} = \frac{25}{4} \end{matrix}$
According to Bernoulli's equation for horizontal tube,
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(p_{1} - p_{2})}{ρ}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(2 \times 10)}{(1.25 \times 10^{3})} = 16 \times 10^{- 3}$ $(2)$
Substituting the value of $v_{2}$ from equation (1) in
(2), we get
${(6.25 v_{1})}^{2} - v_{1}^{2} = 16 \times 10^{- 3}$ or $v_{1} = 0.02 m / s$
So rate of flow through the tube
$= A_{1} v_{1} = A_{2} v_{2} = π \times (0.1)^{2} \times 0.02$
$= 6.28 \times 10^{- 4} m^{3} / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360799 Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will
supporting img

1 remain same

2 increase

3 may increase or decrease, depending on speed of air

4 decrease

MHTCET - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360800 At what speed, the velocity head of the water is twice the pressure head of $75 c m$ of $H g$ ?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

Pressure head $= \frac{p}{ρ g};$ velocity head $= \frac{V^{2}}{2 g}$
$\frac{V^{2}}{2 g} = \frac{2 p}{ρ g} \Rightarrow V^{2} = \frac{4 p^{2}}{ρ}$
$V^{2} = \frac{4 \times 75 \times 10^{- 2} \times 13.6 \times 10^{3} \times 9.8}{10^{3}}$
$= 39984 \times 10^{- 2} = 399$
$V \approx 20 m / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360796 Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 c m$ of $H g$ where the velocity is $35 c m s^{- 1}$ . At a point where the velocity is $65 c m s^{- 1}$ , the pressure will be

1

0.89 c m

of

H g

2

8.9 c m

of

H g

3

0.5 c m

of

H g

4

1 c m

of

H g

Explanation:

Using Bernoulli's theorem,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2} (1)$
Here, $p_{1} = ρ_{m} g h_{1} = 13600 \times 9.8 \times 10^{- 2}$
$p_{2} = 13600 \times 9.8 \times h$
$ρ = 1000 k g m^{- 3}$
$v_{1} = 35 \times 10^{- 2} m s^{- 1}, v_{2} = 65 \times 10^{- 2} m s^{- 1}$
$∴$ From Eq.(1), we get
$\Rightarrow 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} \times 1000 \times (0.35)^{2}$
$= 13600 \times 9.8 \times h + \frac{1}{2} \times 1000 \times (0.65)^{2}$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} + \frac{1}{2} 10^{3} [(0.35)^{2} - (0.65)^{2}]$
$\Rightarrow 13600 \times 9.8 \times h$
$= 13600 \times 9.8 \times 10^{- 2} - 500 (0.3)$
After solving, $h = 0.89 c m$ of $H g$ .

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360797 At what speed will the velocity of a stream of water be equal to $20 c m$ of mercury column ? (Take, $g = 10 m s^{- 2}$ )

1

6.4 m s^{- 1}

2

7.3756 m s^{- 1}

3

6.4756 m s^{- 1}

4 None of these

Explanation:

Here, velocity head $= 20 c m$ of $H g$
$= 20 \times 13.6 cm$ of water
As, velocity head $= v^{2} / 2 g$
$∴ 20 \times 13.6 = \frac{v^{2}}{2 \times 1000}$
$v = \sqrt{20 \times 13.6 \times 2 \times 1000}$
$= 737.56 c m s^{- 1} = 7.3756 m s^{- 1}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360798 The rate of flow of glycerine of density $1.25 \times 10^{3} k g / m^{3}$ through the conical section of a pipe, if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length is $10 N / m^{2}$ is

1

5.28 \times 10^{- 4} m^{3} / s

2

6.28 \times 10^{- 4} m^{3} / s

3

7.28 \times 10^{- 4} m^{3} / s

4

8.28 \times 10^{- 4} m^{3} / s

Explanation:

: According to equation of continuity
$\begin{matrix} (1) & \frac{v_{2}}{v_{1}} = \frac{A_{1}}{A_{2}} = \frac{π \times (0.1)^{2}}{π \times (0.04)^{2}} = \frac{25}{4} \end{matrix}$
According to Bernoulli's equation for horizontal tube,
$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(p_{1} - p_{2})}{ρ}$
or $v_{2}^{2} - v_{1}^{2} = \frac{(2 \times 10)}{(1.25 \times 10^{3})} = 16 \times 10^{- 3}$ $(2)$
Substituting the value of $v_{2}$ from equation (1) in
(2), we get
${(6.25 v_{1})}^{2} - v_{1}^{2} = 16 \times 10^{- 3}$ or $v_{1} = 0.02 m / s$
So rate of flow through the tube
$= A_{1} v_{1} = A_{2} v_{2} = π \times (0.1)^{2} \times 0.02$
$= 6.28 \times 10^{- 4} m^{3} / s$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360799 Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will
supporting img

1 remain same

2 increase

3 may increase or decrease, depending on speed of air

4 decrease

MHTCET - 2019

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

360800 At what speed, the velocity head of the water is twice the pressure head of $75 c m$ of $H g$ ?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

Pressure head $= \frac{p}{ρ g};$ velocity head $= \frac{V^{2}}{2 g}$
$\frac{V^{2}}{2 g} = \frac{2 p}{ρ g} \Rightarrow V^{2} = \frac{4 p^{2}}{ρ}$
$V^{2} = \frac{4 \times 75 \times 10^{- 2} \times 13.6 \times 10^{3} \times 9.8}{10^{3}}$
$= 39984 \times 10^{- 2} = 399$
$V \approx 20 m / s$