Q2GATE 2024MCQ

The following figure shows a plot between shear stress and velocity gradient for materials/fluids P, Q, R, S, and T. Which one of the following options is CORRECT?

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📖 Explanation

Various types of newtonian & non-newtonian fluids are shown in the figure. Fluids which obeys Newton's law of viscosity $\left(\tau=\mu \frac{\mathrm{du}}{\mathrm{dy}}\right)$ are called Newtonian Fluids and those fluids which do not obey this rule are called Non-Newtonian Fluids. General relationship between shear stress and velocity gradient is given by $\tau=\mathrm{A}\left(\frac{\mathrm{du}}{\mathrm{dy}}\right)^{\mathrm{n}}+\mathrm{B}$ In the figures shown above, slope of the curve is called apparent viscosity. Fluid for which apparent viscosity increases with du/dy are called Dilatant. Dilatant fluids are also called shear thickening fluids. Examples of dilatant fluids are solution with suspended starch or sand, sugar in water. Fluids for which apparent viscosity decreases with du/dy are called Pseudo Plastic. Pseudo plastic fluid are also called shear thinning fluid. Examples are paints, polymer solutions, blood, paper pulp, syrup, molasses, milk, gelatine. Bingham Plastic (ideal plastic) fluids require a certain minimum shear stress $t_{y}$ (yield stress) before they start flowing. Examples: tooth paste, sewage sludge, drilling mud have time dependent Newtonian Behaviour.

Q3GATE 2023NAT

The pressure in a pipe at $X$ is to be measured by an open manometer as shown in figure. Fluid $A$ is oil with a specific gravity of 0.8 and Fluid $B$ is mercury with a specific gravity of 13.6. The absolute pressure at $X$ is $\mathrm{kN} / \mathrm{m}^{2}$ (round off to one decimal place). [Assume density of water as $1000 \mathrm{~kg} / \mathrm{m}^{3}$ and acceleration due to gravity as $9.81 \mathrm{~m} / \mathrm{s}^{2}$ and atmospheric pressure as $\left.101.3 \mathrm{kN} / \mathrm{m}^{2}\right]$

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📖 Explanation

Equating pressure at $\mathrm{A}-\mathrm{A}^{\prime}$

\begin{aligned} & P_{a t m}+\rho_{B} \gamma_{w} \times \frac{25}{100}=P_{x}-\rho_{A} \gamma_{w} \times \frac{75}{100} \\ & \therefore \quad P_{x}=P_{a t m}+\rho_{B} \gamma_{w} \times 0.25+\rho_{A} \gamma_{w} 0.75 \end{aligned}

$\begin{gathered} =101.3+13.6 \times 9.81 \times 0.25+0.8 \times 9.81 \times 0.75 \\ =140.54 \mathrm{kN} / \mathrm{m}^{2} \end{gathered}$

Q4GATE 2018NAT

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 ______

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📖 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}

Q5GATE 2018MCQ

Bernoulli's equation is applicable for

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Q6GATE 2018MCQ

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

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📖 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}

Q7GATE 2017NAT

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_____

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📖 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 $=20+0.74=20.74 \mathrm{cm}$

Q8GATE 2016MCQ

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

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Q9GATE 2014MCQ

The dimension for kinematic viscosity is

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📖 Explanation

The Sl unit of kinematic viscosity is $\mathrm{m}^{2} / \mathrm{s}$ Dimension of kinematic viscosity $[\mathrm{v}]=\mathrm{L}^{2} / \mathrm{T}$

Q10GATE 2014MCQ

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

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📖 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, $F_{1}=F_{2}=F_{3}$

Q11GATE 2004MCQ

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

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📖 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}

Q12GATE 2003MCQ

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/ $s^{2}$ )

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📖 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}

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