# Fluid Properties

 Question 1
A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as shown in the figure below. A thin oil film of thickness h = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is 0.04 m2. The velocity distribution within the oil film gap is linear as shown in the figure. Take dynamic viscosity of oil as $7\times 10^{-3}$ Pa-s and acceleration due to gravity as 10 $m/s^{2}$. Neglect weight of the oil. The terminal velocity V (in m/s) of the block is _________ (correct to one decimal place).
 A 8.2 B 8.8 C 10.8 D 11.2
GATE ME 2018 SET-1   Fluid Mechanics
Question 1 Explanation:
Terminal velocity is a constant velocity i.e. the net acceleration is zero.
\begin{aligned} \text{So,} \qquad \Sigma \mathrm{F}_{\mathrm{net}}&=m a \\ \mathrm{mg}-\tau \mathrm{A} &=0 \\ \tau \mathrm{A} &=\mathrm{mg} \\ \mu \frac{V}{h} A &=\mathrm{mg} \\ \mathrm{T} \times 10^{-3} \times \frac{V}{0.15 \times 10^{-3}} \times 0.04&=2 \times 10 =10.714 \mathrm{m} / \mathrm{s} \end{aligned}
 Question 2
In a Lagrangian system, the position of a fluid particle in a flow is described as $x=x_{o}e^{-kt}$ and $y=y_{o}e^{kt}$ where t is the time while $x_{o}$ ,$x_{o}$, and k are constants. The flow is
GATE ME 2018 SET-1   Fluid Mechanics
Question 2 Explanation:
x direction scalar of velocity field,
$\begin{array}{l} u=\frac{d x}{d t} \\ u=-k x \cdot e^{-k t} \end{array}$
y direction scalar of velocity field
\begin{aligned} v&=\frac{d y}{d t}\\ v &=k y_{0} e^{k t} \\ \vec{V} &=u \hat{i}+v \hat{j} \\ \vec{v} &=-k x_{0} e^{-k t} \hat{i}+k y_{0} e^{k t} \hat{j} \end{aligned}
u & v are non zero scalar $t \geq 0$ so it is 2 D flow.
2D possible flow field
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \frac{\partial}{\partial x}\left(-k x_{o} e^{-k t}\right)+\frac{\partial}{\partial y}\left(k y_{0} e^{k t}\right)&=0\\ 0+0&=0 \\ \text{ continuity satisfied.} & \\ \frac{\partial u}{\partial t}&=+k^{2} x_{0} e^{-k t} \\ \frac{\partial v}{\partial t}&=k^{2} y_{0} e^{k t} \\ \frac{\partial u}{\partial t} &\neq 0 \\ \frac{\partial v}{\partial t} &\neq 0 \end{aligned}
 Question 3
Assuming constant temperature condition and air to be an ideal gas, the variation in atmospheric pressure with height calculated from fluid statics is
 A linear B exponential C quadratic D cubic
GATE ME 2016 SET-2   Fluid Mechanics
Question 3 Explanation:
Air pressure falls with altitude exponentially

 Question 4
Which combination of the following statements about steady incompressible forced vortex flow is correct?
P: Shear stress is zero at all points in the flow.
Q: Vorticity is zero at all points in the flow.
R: Velocity is directly proportional to the radius from the centre of the vortex.
S: Total mechanical energy per unit mass is constant in the entire flow field.
 A P and Q B R and S C P and R D P and S
GATE ME 2007   Fluid Mechanics
Question 4 Explanation:
For forced Vortex flow the relation is given by,
$V=r\omega$
From above equation, it is shown easily that velocity is directly proportional to the radius from the centre of the vortex (Radius of fluid particle from the axis of rotation)
And also for forced vortex flow,
\begin{aligned} \frac{1}{2} \rho \omega ^2(r_2^2-r_1^2)-\rho g(z_2-z_1)&=0 \\ \Delta K.E. -\Delta P.E.&=0 \\ \Rightarrow \;\; \Delta K.E.&= \Delta P.E. \end{aligned}
Now total mechanical energy per unit mass is constant in the entire flow field.
 Question 5
For a Newtonian fluid
 A shear stress is proportional to shear strain B rate of shear stress is proportional to shear strain C shear stress is proportional to rate of shear strain D rate of shear stress is proportional to rate of shear strain
GATE ME 2006   Fluid Mechanics
Question 5 Explanation:
For a Newtonian fluid
$\tau \propto \frac{d u}{d y}$
$\tau=\mu \frac{d u}{d y}$
where $\quad \tau=$ shear stress
$\frac{d u}{d y}=$ rate of shear strain
There are 5 questions to complete.