Q2GATE 2025MCQ

For a flowing fluid, a dimensionless combination of velocity ( $V$ ), length scale ( $l$ ), and acceleration due to gravity ( $g$ ) would be

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📖 Explanation

Let us write their dimensional formulas: $[V] = LT^{-1}$ $[l] = L$ $[g] = LT^{-2}$ Now, check the dimensional consistency of each option: Option (A): $\frac{V^2}{gl}$ $= \frac{(LT^{-1})^2}{(LT^{-2})(L)} = \frac{L^2T^{-2}}{L^2T^{-2}} = 1$ This expression is dimensionless. Option (B): $\frac{Vg}{l}$ $= \frac{(LT^{-1})(LT^{-2})}{L} = \frac{L^2T^{-3}}{L} = LT^{-3}$ This is not dimensionless. Option (C): $\frac{gl^2}{V}$ $= \frac{(LT^{-2})(L^2)}{LT^{-1}} = \frac{L^3T^{-2}}{LT^{-1}} = L^2T^{-1}$ This is not dimensionless. Option (D): $\frac{l}{V^2g}$ $= \frac{L}{(LT^{-1})^2 \cdot LT^{-2}} = \frac{L}{L^3T^{-4}} = L^{-2}T^4$ This is not dimensionless. Final Answer: $\frac{V^2}{gl}$ is the correct dimensionless combination.

Q3GATE 2022MCQ

The dimension of dynamic viscosity is:

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📖 Explanation

Unit of dynamic viscosity $=\frac{kg}{m.s} \; or \; \frac{Ns}{m^2}=ML^{-1}T^{-1}$

Q4GATE 2021MCQ

Kinematic viscosity' is dimensionally represented as

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📖 Explanation

Kinematic viscosity $v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s$ $[v]=\frac{m^2}{s }=\frac{L^2}{T}$

Q5GATE 2018NAT

A 1:50 model of a spillway is to be tested in the laboratory. The discharge in the prototype spillway is 1000 $m^{3}/s$ . The corresponding discharge (in $m^{3}/s$ , up to two decimal places) to be maintained in the model, neglecting variation in acceleration due to gravity, is ______

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📖 Explanation

Froude law is valid

\begin{aligned} Q_{r} &=L_{r}^{25} \\ \frac{Q_{m}}{Q_{p}} &=\left(\frac{1}{50}\right)^{25} \\ \frac{Q_{m}}{1000} &=\left(\frac{1}{50}\right)^{25} \\ Q_{m} &=0.0566 \mathrm{m}^{3 / \mathrm{s}} \\ \text{So},\quad Q_{m} & \simeq 0.06 \mathrm{m}^{3 / \mathrm{s}} \end{aligned}

Q6GATE 2018NAT

In a laboratory, a flow experiment is performed over a hydraulic structure. The measured values of discharge and velocity are 0.05 $m^{3}$ /s and 0.25 m/s, respectively. If the full scale structure (30 times bigger) is subjected to a discharge of 270 $m^{3}$ /s, then the time scale (model to full scale) value (up to two decimal places) is ______

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📖 Explanation

\begin{aligned} \text { Froude Law }(F r)_{m}&=(F r)_{p} \\ \left(\frac{V}{\sqrt{L \mathrm{g}}}\right)_{m} &=\left(\frac{v}{\sqrt{L g}}\right)_{p} \quad\left(g_{m}=g_{p}\right) \\ v_{r} &=\sqrt{L_{r}} \\ \text{or}\quad \frac{L_{r}}{T_{r}} &=\sqrt{L} \\ T_{r} &=\sqrt{L_{r}} \\ T_{r} &=\sqrt{\frac{1}{30}}=0.1826 \end{aligned}

Q7GATE 2015MCQ

The drag force $F_{D}$ , on a sphere kept in a uniform flow field depends on the diameter of the sphere, D; flow velocity, V; fluid density, $\rho$ ; and dynamic viscosity, $\mu$ . Which of the following options represents the non-dimensional parameters which could be used to analyze this problem?

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📖 Explanation

\begin{aligned} \text{For sphere,}\quad A&=\frac{\pi}{4} D^{2} \\ F_{D} &=C_{D} \times \frac{1}{2} \times p \times A \times V^{2} \\ F_{D} &=C_{D} \times \frac{1}{2} \times \rho \times \left(\frac{\pi}{4} D^{2}\right) \times V^{2} \\ \frac{F_{D}}{\rho D^{2} V^{2}} &=\frac{1}{2} \times \left(\frac{\pi}{4}\right) \times C_{D} \end{aligned}

Hence, right hand side is dimensionless, so $\frac{F_{0}}{\rho D^{2} V^{2}}$ should also be a dimensionless parameter. Option (C) $\frac{F_{0}}{\rho V^{2} D^{2}}, \frac{\rho V D}{\mu}$

Q8GATE 2015MCQ

The relationship between the length scale ratio $(L_{r})$ and the velocity scale ratio $(V_{r})$ in hydraulic models, in which Froude dynamic similarity is maintained, is:

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📖 Explanation

As per Froude Law

\begin{aligned} \left(\frac{V}{\sqrt{L g}}\right)_{m}&=\left(\frac{V}{\sqrt{L g}}\right)_{p} \\ \Rightarrow \quad V_{r}&=\sqrt{L_{r}} \end{aligned}

Q9GATE 2013MCQ

Group-I contains dimensionless parameters and Group- II contains the ratios. The correct match of dimensionless parameters in Group- I with ratios in Group-II is:

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Q10GATE 2008MCQ

A river reach of 2.0 km long with maximum flood discharge of 10000 $m^{3}/s$ is to be physically modeled in the laboratory where maximum available discharge is 0.20 $m^{3}/s$ . For a geometrically similar model based on equally of Froude number, the length of the river reach (m) in the model is

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📖 Explanation

According to Froude model law

\begin{aligned} \therefore \quad \frac{V_{t}}{\sqrt{L_{r} g_{1}}}&=1 \quad\left[\because F_{r}=1\right]\\ \text{Now, we have}\\ \frac{Q_{m}}{Q_{p}}&=L_{1}^{2} \times V_{r}\\ \Rightarrow \quad \frac{Q_{m}}{Q_{p}}&=L_{r}^{2} \times \sqrt{L_{r}} \\ \Rightarrow \quad \frac{Q_{m}}{Q_{0}}&=\left(L_{r}\right)^{5 / 2} \\ \Rightarrow \quad \frac{0.20}{10000}&=\left(\frac{L_{m}}{2 \times 1000}\right)^{5 / 2} \\ \Rightarrow \quad L_{m}&=26.4 \mathrm{m} \end{aligned}

Q11GATE 2007MCQ

A 1:50 scale model of a spillway is to be tested in the laboratory. The discharge in the prototype is 1000 $m^{3}/s$ . The discharge to be maintained in the model test is

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📖 Explanation

Froude model law will be applicable in this case

\begin{aligned} \frac{Q_{m}}{Q_{p}} &=\left(\frac{L_{m}}{L_{p}}\right)^{5 / 2} \\ \Rightarrow \quad Q_{m} &=1000 \times \left(\frac{1}{50}\right)^{25} \\ &=0.057 \mathrm{m}^{3} / \mathrm{sec} \end{aligned}

Q12GATE 2006MCQ

The flow of glycerin (kinematic viscosity $v = 5 \times 10^{-4}m^{2}/s$ ) in an open channel is to be modeled in a laboratory flume using water ( $v = 10^{-6}m^{2}/s$ ) as the flowing fluid. If both gravity and viscosity are important, what should be the length scale (i.e. ratio of prototype to model dimensions) for maintaining dynamic similarity ?

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📖 Explanation

Equating Reynolds number and Froude number, we get

\begin{aligned} \frac{V_{t} L}{v_{r}} &=\frac{V_{r}}{\sqrt{L}} \\ \therefore\quad L_{r} &=\left(v_{r}\right)^{2 / 3} \\ v_{r} &=\frac{v_{m}}{v_{p}}=\frac{10^{-6}}{5 \times 10^{-4}}=2 \times 10^{-3} \\ \Rightarrow\quad L_{r} &=\left(2 \times 10^{-3}\right)^{2 / 3} \\ \Rightarrow\quad L_{r} &=0.0159 \quad \text { But } \cdot L_{r}=\frac{L_{m}}{L_{p}} \\ \Rightarrow\quad \frac{L_{p}}{L_{m}} &=\frac{1}{L_{r}}=\frac{1}{0.0159}=62.99 \approx 63 \end{aligned}

Q13GATE 2004MCQ

The height of a hydraulic jump in the stilling pool of 1:25 scale model was observed to be 10 cm. The corresponding prototype height of the jump is

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📖 Explanation

\begin{aligned} L_{r} &=\frac{L_{m}}{L_{p}}=\frac{1}{25} \\ \Rightarrow \quad L_{p} &=L_{m} \times 25 \\ \Rightarrow \quad L_{p} &=10 \times 25 \\ &=250 \mathrm{cm} \\ &=2.5 \mathrm{m} \end{aligned}

Q14GATE 2003MCQ

A laboratory model of a river is built to a geometric scale of 1:100. The fluid used in the model is oil of mass density 900 kg/ $m^{3}$ . The highest flood in the river is 10,000 $m^{3}$ /s. The corresponding discharge in the model shall be

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📖 Explanation

Froude model law will be applicable in this case.

\begin{aligned} \frac{Q_{m}}{Q_{p}}&=\frac{\left(L_{m}\right)^{3} / T_{m}}{\left(L_{p}\right)^{3} / T_{D}} \qquad\left[\because \frac{L_{m}}{T_{m}}=V_{m}\right]\\ \Rightarrow \quad \frac{Q_{m}}{Q_{p}}&=\frac{V_{m}\left(L_{m}\right)^{2}}{V_{p}\left(L_{p}\right)^{2}} \quad\left[\because v_{1}=\sqrt{L}\right] \\ \Rightarrow \frac{Q_{m}}{Q_{p}}&=\left(\frac{L_{m}}{L_{D}}\right)^{5 / 2} \\ \Rightarrow \quad Q_{m}&=Q_{p} \times \left(\frac{L_{m}}{L_{p}}\right)^{5 / 2} \\ \Rightarrow \quad Q_{m}&=10.000 \times \left(\frac{1}{100}\right)^{5 / 2} \\ &=0.1 \mathrm{m}^{3}/s \end{aligned}

Q15GATE 2001MCQ

In a 1/50 model of a spillway, the discharge was measured to be 0.3 $m^{3}/sec$ . The corresponding prototype discharge in $m^{3}/sec$ is

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📖 Explanation

\begin{aligned} Q_{r} &=L_{r}^{25} \\ \frac{Q_{\mathrm{p}}}{Q_{m}} &=\left(\frac{L_{p}}{L_{m}}\right)^{25}(50)^{25} \\ \therefore \quad Q_{p} &=0.3 \times (50)^{25}=5303.3 \mathrm{m}^{3 / \mathrm{s}} \end{aligned}

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