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Mechanical Properties of Fluids

143223 Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as $v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$ . If coefficient of viscosity for water is $η$ , what will be shear stress between layers of water at $y = a$ ?

1

\frac{η k}{a}

2

\frac{η}{ka}

3

\frac{η a}{k}

4 None of these

Explanation:

A Given,
$v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$
Differentiating with respect to $y$ ,
$\frac{d v}{d y} = k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
From Newton's law of viscosity,
$F = η A \frac{dv}{dy}$
$σ = η (\frac{d v}{d y})$
$[∴ σ = \frac{F}{A}]$
$σ = η k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
At $y = a$ ,
$σ = η k (\frac{4}{a} - \frac{3}{a})$
$σ = (\frac{η k}{a}) N / m^{2}$

BITSAT-2015

Mechanical Properties of Fluids

143224 The rate of flow of liquid through a capillary tube under a constant pressure difference is $Q$ . On doubling the length of tube and reducing the diameter of tube to half, the rate of flow will become

1

\frac{Q}{4}

2

\frac{Q}{32}

3

\frac{Q}{8}

4

16 Q

Explanation:

B Given,
$P = constant$
$l_{2} = 2 l$
$r_{2} = \frac{r}{2}$
We know,
Rate of flow of liquid $(Q) = \frac{π \Pr^{4}}{8 η L}$
When a diameter of tube reduced to half and length is doubled,
$Q_{1} = \frac{π P {(\frac{r}{2})}^{4}}{8 η \times (2 l)}$
$Q_{1} = \frac{π \Pr^{4}}{8 η ℓ} \times \frac{1}{32}$
$Q_{1} = \frac{Q}{32}$

CG PET- 2011

Mechanical Properties of Fluids

143226 A cylindrical vessel is filled with water upto the height $1 m$ from the base. A small orifice is opened at some height in the cylinder and the water level is reduced to height of orifice in 20 s. If the base area of the cylinder is $100$ times the area of orifice, then the height of orifice from the base is (take, $g = 10 m / s^{2}$ )

1

80 cm

2

60 cm

3

40 cm

4

20 cm

Explanation:

D Given,
Height of vessel filled with water $(h) = 1 m$
Time taken to reduce the water level $(t) = 20 \sec$
Area of base $(A) = 100 \times$ area of orifice $(A_{0})$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know,
Time taken to reduce the water level up to orifice,
$t = \frac{A}{A_{0}} \sqrt{\frac{2 x}{g}}$
$20 = 100 \sqrt{\frac{2 x}{10}}$
$\frac{1}{5} = \sqrt{\frac{x}{5}}$
$x = 0.2 m = 20 cm$
Height of orifice from base $(h) = 100 - 20$
$h = 80 cm$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143227 A tank of height $15 m$ and cross-section area $10 m^{2}$ is filled with water. There is a small hole of cross-section area ' $a$ ' which is much smaller than the container, located at a height of $12 m$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 m$ ? (Take, density of water $= 1000 kg m^{- 3}$ )

1

233 kN

2

200 kN

3

320 kN

4

400 kN

Explanation:

A
Given,
Range of water jet, $R = 16 m ., A = 10 m^{2}$
Let the velocity of efflux be $v$ .
Then, $v t = R$
$v \times \sqrt{\frac{2 H}{g}} = 16$
Where $H = 12 m$
We also know,
$v = \sqrt{2 g h + \frac{2 (p - p_{a})}{ρ}}$
Where $ρ = 1000 {kgm}^{- 3}$
and $h = 3 m$
Putting the value of $v$ in equation (i),
$\sqrt{2 gh + \frac{2 (p - p_{a})}{ρ}} \times \sqrt{\frac{2 H}{g}} = 16$
Squaring both side,
$(2 gh + \frac{2 (p - p_{a})}{ρ}) (\frac{2 H}{g}) = 256$
Putting the value in the equation,
$[2 \times 10 \times 3 + \frac{2 Δ p}{1000}] \times [\frac{2 \times 12}{10}] = 256$
$Δ p = 23.3 k N / m^{2}$
So,
Force applied $(F) = Δ p \times A$
$= 23.3 \times 10$
$F = 233 kN$

TS- EAMCET-04.05.2018

Mechanical Properties of Fluids

143223 Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as $v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$ . If coefficient of viscosity for water is $η$ , what will be shear stress between layers of water at $y = a$ ?

1

\frac{η k}{a}

2

\frac{η}{ka}

3

\frac{η a}{k}

4 None of these

Explanation:

A Given,
$v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$
Differentiating with respect to $y$ ,
$\frac{d v}{d y} = k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
From Newton's law of viscosity,
$F = η A \frac{dv}{dy}$
$σ = η (\frac{d v}{d y})$
$[∴ σ = \frac{F}{A}]$
$σ = η k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
At $y = a$ ,
$σ = η k (\frac{4}{a} - \frac{3}{a})$
$σ = (\frac{η k}{a}) N / m^{2}$

BITSAT-2015

Mechanical Properties of Fluids

143224 The rate of flow of liquid through a capillary tube under a constant pressure difference is $Q$ . On doubling the length of tube and reducing the diameter of tube to half, the rate of flow will become

1

\frac{Q}{4}

2

\frac{Q}{32}

3

\frac{Q}{8}

4

16 Q

Explanation:

B Given,
$P = constant$
$l_{2} = 2 l$
$r_{2} = \frac{r}{2}$
We know,
Rate of flow of liquid $(Q) = \frac{π \Pr^{4}}{8 η L}$
When a diameter of tube reduced to half and length is doubled,
$Q_{1} = \frac{π P {(\frac{r}{2})}^{4}}{8 η \times (2 l)}$
$Q_{1} = \frac{π \Pr^{4}}{8 η ℓ} \times \frac{1}{32}$
$Q_{1} = \frac{Q}{32}$

CG PET- 2011

Mechanical Properties of Fluids

143226 A cylindrical vessel is filled with water upto the height $1 m$ from the base. A small orifice is opened at some height in the cylinder and the water level is reduced to height of orifice in 20 s. If the base area of the cylinder is $100$ times the area of orifice, then the height of orifice from the base is (take, $g = 10 m / s^{2}$ )

1

80 cm

2

60 cm

3

40 cm

4

20 cm

Explanation:

D Given,
Height of vessel filled with water $(h) = 1 m$
Time taken to reduce the water level $(t) = 20 \sec$
Area of base $(A) = 100 \times$ area of orifice $(A_{0})$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know,
Time taken to reduce the water level up to orifice,
$t = \frac{A}{A_{0}} \sqrt{\frac{2 x}{g}}$
$20 = 100 \sqrt{\frac{2 x}{10}}$
$\frac{1}{5} = \sqrt{\frac{x}{5}}$
$x = 0.2 m = 20 cm$
Height of orifice from base $(h) = 100 - 20$
$h = 80 cm$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143227 A tank of height $15 m$ and cross-section area $10 m^{2}$ is filled with water. There is a small hole of cross-section area ' $a$ ' which is much smaller than the container, located at a height of $12 m$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 m$ ? (Take, density of water $= 1000 kg m^{- 3}$ )

1

233 kN

2

200 kN

3

320 kN

4

400 kN

Explanation:

A
Given,
Range of water jet, $R = 16 m ., A = 10 m^{2}$
Let the velocity of efflux be $v$ .
Then, $v t = R$
$v \times \sqrt{\frac{2 H}{g}} = 16$
Where $H = 12 m$
We also know,
$v = \sqrt{2 g h + \frac{2 (p - p_{a})}{ρ}}$
Where $ρ = 1000 {kgm}^{- 3}$
and $h = 3 m$
Putting the value of $v$ in equation (i),
$\sqrt{2 gh + \frac{2 (p - p_{a})}{ρ}} \times \sqrt{\frac{2 H}{g}} = 16$
Squaring both side,
$(2 gh + \frac{2 (p - p_{a})}{ρ}) (\frac{2 H}{g}) = 256$
Putting the value in the equation,
$[2 \times 10 \times 3 + \frac{2 Δ p}{1000}] \times [\frac{2 \times 12}{10}] = 256$
$Δ p = 23.3 k N / m^{2}$
So,
Force applied $(F) = Δ p \times A$
$= 23.3 \times 10$
$F = 233 kN$

TS- EAMCET-04.05.2018

Mechanical Properties of Fluids

143223 Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as $v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$ . If coefficient of viscosity for water is $η$ , what will be shear stress between layers of water at $y = a$ ?

1

\frac{η k}{a}

2

\frac{η}{ka}

3

\frac{η a}{k}

4 None of these

Explanation:

A Given,
$v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$
Differentiating with respect to $y$ ,
$\frac{d v}{d y} = k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
From Newton's law of viscosity,
$F = η A \frac{dv}{dy}$
$σ = η (\frac{d v}{d y})$
$[∴ σ = \frac{F}{A}]$
$σ = η k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
At $y = a$ ,
$σ = η k (\frac{4}{a} - \frac{3}{a})$
$σ = (\frac{η k}{a}) N / m^{2}$

BITSAT-2015

Mechanical Properties of Fluids

143224 The rate of flow of liquid through a capillary tube under a constant pressure difference is $Q$ . On doubling the length of tube and reducing the diameter of tube to half, the rate of flow will become

1

\frac{Q}{4}

2

\frac{Q}{32}

3

\frac{Q}{8}

4

16 Q

Explanation:

B Given,
$P = constant$
$l_{2} = 2 l$
$r_{2} = \frac{r}{2}$
We know,
Rate of flow of liquid $(Q) = \frac{π \Pr^{4}}{8 η L}$
When a diameter of tube reduced to half and length is doubled,
$Q_{1} = \frac{π P {(\frac{r}{2})}^{4}}{8 η \times (2 l)}$
$Q_{1} = \frac{π \Pr^{4}}{8 η ℓ} \times \frac{1}{32}$
$Q_{1} = \frac{Q}{32}$

CG PET- 2011

Mechanical Properties of Fluids

143226 A cylindrical vessel is filled with water upto the height $1 m$ from the base. A small orifice is opened at some height in the cylinder and the water level is reduced to height of orifice in 20 s. If the base area of the cylinder is $100$ times the area of orifice, then the height of orifice from the base is (take, $g = 10 m / s^{2}$ )

1

80 cm

2

60 cm

3

40 cm

4

20 cm

Explanation:

D Given,
Height of vessel filled with water $(h) = 1 m$
Time taken to reduce the water level $(t) = 20 \sec$
Area of base $(A) = 100 \times$ area of orifice $(A_{0})$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know,
Time taken to reduce the water level up to orifice,
$t = \frac{A}{A_{0}} \sqrt{\frac{2 x}{g}}$
$20 = 100 \sqrt{\frac{2 x}{10}}$
$\frac{1}{5} = \sqrt{\frac{x}{5}}$
$x = 0.2 m = 20 cm$
Height of orifice from base $(h) = 100 - 20$
$h = 80 cm$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143227 A tank of height $15 m$ and cross-section area $10 m^{2}$ is filled with water. There is a small hole of cross-section area ' $a$ ' which is much smaller than the container, located at a height of $12 m$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 m$ ? (Take, density of water $= 1000 kg m^{- 3}$ )

1

233 kN

2

200 kN

3

320 kN

4

400 kN

Explanation:

A
Given,
Range of water jet, $R = 16 m ., A = 10 m^{2}$
Let the velocity of efflux be $v$ .
Then, $v t = R$
$v \times \sqrt{\frac{2 H}{g}} = 16$
Where $H = 12 m$
We also know,
$v = \sqrt{2 g h + \frac{2 (p - p_{a})}{ρ}}$
Where $ρ = 1000 {kgm}^{- 3}$
and $h = 3 m$
Putting the value of $v$ in equation (i),
$\sqrt{2 gh + \frac{2 (p - p_{a})}{ρ}} \times \sqrt{\frac{2 H}{g}} = 16$
Squaring both side,
$(2 gh + \frac{2 (p - p_{a})}{ρ}) (\frac{2 H}{g}) = 256$
Putting the value in the equation,
$[2 \times 10 \times 3 + \frac{2 Δ p}{1000}] \times [\frac{2 \times 12}{10}] = 256$
$Δ p = 23.3 k N / m^{2}$
So,
Force applied $(F) = Δ p \times A$
$= 23.3 \times 10$
$F = 233 kN$

TS- EAMCET-04.05.2018

Mechanical Properties of Fluids

143223 Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as $v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$ . If coefficient of viscosity for water is $η$ , what will be shear stress between layers of water at $y = a$ ?

1

\frac{η k}{a}

2

\frac{η}{ka}

3

\frac{η a}{k}

4 None of these

Explanation:

A Given,
$v = k (\frac{2 y^{2}}{a^{2}} - \frac{y^{3}}{a^{3}})$
Differentiating with respect to $y$ ,
$\frac{d v}{d y} = k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
From Newton's law of viscosity,
$F = η A \frac{dv}{dy}$
$σ = η (\frac{d v}{d y})$
$[∴ σ = \frac{F}{A}]$
$σ = η k (\frac{4 y}{a^{2}} - \frac{3 y^{2}}{a^{3}})$
At $y = a$ ,
$σ = η k (\frac{4}{a} - \frac{3}{a})$
$σ = (\frac{η k}{a}) N / m^{2}$

BITSAT-2015

Mechanical Properties of Fluids

143224 The rate of flow of liquid through a capillary tube under a constant pressure difference is $Q$ . On doubling the length of tube and reducing the diameter of tube to half, the rate of flow will become

1

\frac{Q}{4}

2

\frac{Q}{32}

3

\frac{Q}{8}

4

16 Q

Explanation:

B Given,
$P = constant$
$l_{2} = 2 l$
$r_{2} = \frac{r}{2}$
We know,
Rate of flow of liquid $(Q) = \frac{π \Pr^{4}}{8 η L}$
When a diameter of tube reduced to half and length is doubled,
$Q_{1} = \frac{π P {(\frac{r}{2})}^{4}}{8 η \times (2 l)}$
$Q_{1} = \frac{π \Pr^{4}}{8 η ℓ} \times \frac{1}{32}$
$Q_{1} = \frac{Q}{32}$

CG PET- 2011

Mechanical Properties of Fluids

143226 A cylindrical vessel is filled with water upto the height $1 m$ from the base. A small orifice is opened at some height in the cylinder and the water level is reduced to height of orifice in 20 s. If the base area of the cylinder is $100$ times the area of orifice, then the height of orifice from the base is (take, $g = 10 m / s^{2}$ )

1

80 cm

2

60 cm

3

40 cm

4

20 cm

Explanation:

D Given,
Height of vessel filled with water $(h) = 1 m$
Time taken to reduce the water level $(t) = 20 \sec$
Area of base $(A) = 100 \times$ area of orifice $(A_{0})$
Acceleration due to gravity $(g) = 10 m / \sec^{2}$
We know,
Time taken to reduce the water level up to orifice,
$t = \frac{A}{A_{0}} \sqrt{\frac{2 x}{g}}$
$20 = 100 \sqrt{\frac{2 x}{10}}$
$\frac{1}{5} = \sqrt{\frac{x}{5}}$
$x = 0.2 m = 20 cm$
Height of orifice from base $(h) = 100 - 20$
$h = 80 cm$

TS- EAMCET-07.05.2018

Mechanical Properties of Fluids

143227 A tank of height $15 m$ and cross-section area $10 m^{2}$ is filled with water. There is a small hole of cross-section area ' $a$ ' which is much smaller than the container, located at a height of $12 m$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 m$ ? (Take, density of water $= 1000 kg m^{- 3}$ )

1

233 kN

2

200 kN

3

320 kN

4

400 kN

Explanation:

A
Given,
Range of water jet, $R = 16 m ., A = 10 m^{2}$
Let the velocity of efflux be $v$ .
Then, $v t = R$
$v \times \sqrt{\frac{2 H}{g}} = 16$
Where $H = 12 m$
We also know,
$v = \sqrt{2 g h + \frac{2 (p - p_{a})}{ρ}}$
Where $ρ = 1000 {kgm}^{- 3}$
and $h = 3 m$
Putting the value of $v$ in equation (i),
$\sqrt{2 gh + \frac{2 (p - p_{a})}{ρ}} \times \sqrt{\frac{2 H}{g}} = 16$
Squaring both side,
$(2 gh + \frac{2 (p - p_{a})}{ρ}) (\frac{2 H}{g}) = 256$
Putting the value in the equation,
$[2 \times 10 \times 3 + \frac{2 Δ p}{1000}] \times [\frac{2 \times 12}{10}] = 256$
$Δ p = 23.3 k N / m^{2}$
So,
Force applied $(F) = Δ p \times A$
$= 23.3 \times 10$
$F = 233 kN$

TS- EAMCET-04.05.2018

Question Bank Series

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