Question 1 |

A three-fluid system (immiscible) is connected to a vacuum pump. The specific gravity values of the fluids (S1, S2) are given in the figure.

The gauge pressure value (in kN/m^{2}, up to two decimal places) of p_{1} is ______

The gauge pressure value (in kN/m^{2}, up to two decimal places) of p_{1} is ______

-8.73 | |

-4.78 | |

-2.54 | |

0 |

Question 1 Explanation:

Taking P_{1} is in gauge pressure.

\begin{aligned} P_{A}=& P_{1}+\left(0.88 \times 10^{3}\right) \cdot(9.81)(0.5) \\ &+\left(0.95 \times 10^{3}\right)(9.81)(1) \\ \left(10^{3}\right)(9.81)(0.5)=& P_{1}+\left(0.88 \times 10^{3}\right) \cdot(9.81)(0.5) \\ &+\left(0.95 \times 10^{3}\right)(9.81)(1) \\ P_{1}=&-8.73 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Question 2 |

A closed tank contains 0.5 m thick layer of mercury (specific gravity = 13.6) at the bottom. A 2.0 m thick layer of water lies above the mercury layer. A 3.0 m thick layer of oil (specific gravity = 0.6) lies above the water layer. The space above the oil layer contains air under pressure. The gauge pressure at the bottom of the tank is 196.2 kN/m^{2}. The density of water is 1000 kg/m^{3} and the acceleration due to gravity is 9.81 m/s^{2}. The value of pressure in the air space is

92.214 kN/m^{2} | |

95.644 kN/m^{2} | |

98.922 kN/m^{2} | |

99.321 kN/m^{2} |

Question 2 Explanation:

\begin{aligned} &P_{a i t} \text { is in gauge pressure. }\\ &\begin{array}{l} P_{a i r}+\left(0.6 \times 10^{3}\right)(9.81)(3)+\left(10^{3}\right)(9.81)(2) \\ \quad+\left(13.6 \times 10^{3}\right)(9.81)(0.5)=196.2 \times 10^{3} \\ P_{a i r}=92.214 \mathrm{kN} / \mathrm{m}^{2} \end{array} \end{aligned}

Question 3 |

Bernoulli's equation is applicable for

viscous and compressible fluid flow | |

inviscid and compressible fluid flow | |

inviscid and incompressible fluid flow | |

viscous and incompressible fluid flow |

Question 4 |

The figure shows a U-tube having a 5 mmx5 mm square cross-section filled with mercury (specific gravity = 13.6) up to a height of 20 cm in each limb (open to the atmosphere).

If 5 cm^{3} of water is added to the right limb, the new height (in cm, up to two decimal places) of mercury in the LEFT limb will be_____

If 5 cm^{3} of water is added to the right limb, the new height (in cm, up to two decimal places) of mercury in the LEFT limb will be_____

1.73 | |

19.26 | |

20.74 | |

2.73 |

Question 4 Explanation:

\begin{aligned} P_{A} &=P_{B} \\ \left(13.6 \times 10^{3}\right) g \times(2 x) &=\left(10^{3}\right) g(20) \\ x &=\frac{10}{3.6}=0.735 \mathrm{cm} \end{aligned}

So, the new height (in cm to two decimal place of

mercury in the left limb will be

=20+0.74=20.74 \mathrm{cm}

Question 5 |

Group I contains the types of fluids while Group II contains the shear stress - rate of shear relationship of different types of fluids, as shown in the figure.

The Correct match between Group I and Group II is

The Correct match between Group I and Group II is

P-2, Q-4, R-1, S-5 | |

P-2, Q-5, R-4, S-1 | |

P-2, Q-4, R-5, S-3 | |

P-2, Q-1, R-3, S-4 |

Question 6 |

Three rigid buckets, shown as in the figures (1), (2) and (3), are of identical heights and base areas. Further, assume that each of these buckets have negligible mass and are full of water. The weights of water in these buckets are denoted as W_{1},W_{2}, \; and \; W_{3} respectively. Also, let the force of water on the base of the bucket be denoted as F_{1},F_{2}, \;and \; F_{3} respectively. The option giving an accurate description of the system physics is

W_{2}=W_{1}=W_{3}\; and\; F_{2} \gt F_{1} \gt F_{3} | |

W_{2} \gt W_{1} \gt W_{3}\; and\; F_{2} \gt F_{1} \gt F_{3} | |

W_{2}=W_{1}=W_{3}\; and\; F_{1}=F_{2}=F_{3} | |

W_{2} \gt W_{1} \gt W_{3}\; and \; F_{1}=F_{2}=F_{3} |

Question 6 Explanation:

Bucket \rightarrow identical height

\rightarrow identical base area

W_{2}>W_{1}>W_{3}

Force on the base in each case will be equal

to=\gamma_{w} h A

Hence, F_{1}=F_{2}=F_{3}

\rightarrow identical base area

W_{2}>W_{1}>W_{3}

Force on the base in each case will be equal

to=\gamma_{w} h A

Hence, F_{1}=F_{2}=F_{3}

Question 7 |

The dimension for kinematic viscosity is

\frac{L}{MT} | |

\frac{L}{T^{2}} | |

\frac{L^{2}}{T} | |

\frac{ML}{T} |

Question 7 Explanation:

The Sl unit of kinematic viscosity is \mathrm{m}^{2} / \mathrm{s}

Dimension of kinematic viscosity [\mathrm{v}]=\mathrm{L}^{2} / \mathrm{T}

Dimension of kinematic viscosity [\mathrm{v}]=\mathrm{L}^{2} / \mathrm{T}

Question 8 |

In the inclined manometer shown in the figure below, the reservoir is large. Its
surface may be assumed to remain at a fixed elevation. A is connected to a gas
pipeline and the deflection noted on the inclined glass tube is 100 mm. Assuming
\theta =30^{\circ} and the manometric fluid as oil with specific gravity of 0.86, the pressure at A is

43 mm water (vaccum) | |

43 mm water | |

86 mm water | |

100 mm water |

Question 8 Explanation:

Total pressure at A will be equal to the pressure at the surface of manometric fluid because above the surface, the gas pipeline is connected.

\therefore Pressure at A=100 \sin 30^{\circ} \mathrm{mm} of oil

=50 \mathrm{mm} \text { of oil }

We know that

\begin{aligned} S_{1} h_{1} &=S_{2} h_{2} \\ \Rightarrow \quad 0.86 \times 50 &=1 \times h_{2} \\ \Rightarrow &=43 \mathrm{mm} \text { of water } \end{aligned}

\therefore Pressure at A=100 \sin 30^{\circ} \mathrm{mm} of oil

=50 \mathrm{mm} \text { of oil }

We know that

\begin{aligned} S_{1} h_{1} &=S_{2} h_{2} \\ \Rightarrow \quad 0.86 \times 50 &=1 \times h_{2} \\ \Rightarrow &=43 \mathrm{mm} \text { of water } \end{aligned}

Question 9 |

Two pipelines, one carrying oil (mass density 900 kg/m^{3}) and the other water, are connected to a manometer as shown in figure. By what amount the pressure
in the water pipe should be increased so that the mercury levels in both the limbs
of the manometer become equal ? (mass density of mercury= 13550 kg/m^{3} and g = 9.81 m/s^{2})

24.7 kPa | |

26.5 kPa | |

26.7 kPa | |

28.9 kPa |

Question 9 Explanation:

Initially the pressure difference between water and oil can be given as

\begin{aligned} p_{w}+(1000 &\times 9.81 \times 1.5)+(13550 \times 9.81 \times 20 \\ \left.\times 10^{-2}\right)-(900 &\times 9.81 \times 3)=p_{0}\\ \Rightarrow \; p_{w t}+14715+26585.1-26487=p_{0} \\ \Rightarrow \quad p_{0}-p_{w}&=14813.1 \\ \Rightarrow \quad P_{0}-P_{w}&=14.8 \mathrm{kPa} &\ldots(i) \end{aligned}

To make the same level of mercury in both the limbs, the level of mercury should be decreased by 10 cm in right limb and increased by 10cm in lett limb

\begin{aligned} P_{w t}+(1000 \times 9.81 &\times 1.6)-(900 \times 9.81 \times 2.9)=p_{0}\\ \Rightarrow \quad P_{0}-p_{w t}&=-9.9 \mathrm{kPa} &\ldots(ii)\\ \end{aligned}

Subtracting (ii) from (i) we get

\begin{array}{l} P_{W}-P_{N}=14.8+99 \\ P_{\mathrm{w} t}-P_{\mathrm{w}}=24.7 \mathrm{kPa} \text { (increase) } \end{array}

\begin{aligned} p_{w}+(1000 &\times 9.81 \times 1.5)+(13550 \times 9.81 \times 20 \\ \left.\times 10^{-2}\right)-(900 &\times 9.81 \times 3)=p_{0}\\ \Rightarrow \; p_{w t}+14715+26585.1-26487=p_{0} \\ \Rightarrow \quad p_{0}-p_{w}&=14813.1 \\ \Rightarrow \quad P_{0}-P_{w}&=14.8 \mathrm{kPa} &\ldots(i) \end{aligned}

To make the same level of mercury in both the limbs, the level of mercury should be decreased by 10 cm in right limb and increased by 10cm in lett limb

\begin{aligned} P_{w t}+(1000 \times 9.81 &\times 1.6)-(900 \times 9.81 \times 2.9)=p_{0}\\ \Rightarrow \quad P_{0}-p_{w t}&=-9.9 \mathrm{kPa} &\ldots(ii)\\ \end{aligned}

Subtracting (ii) from (i) we get

\begin{array}{l} P_{W}-P_{N}=14.8+99 \\ P_{\mathrm{w} t}-P_{\mathrm{w}}=24.7 \mathrm{kPa} \text { (increase) } \end{array}

There are 9 questions to complete.