Question 1 |
A uniform wooden rod (specific gravity = 0.6,
diameter = 4 cm and length = 8 m) is immersed
in the water and is hinged without friction at point
A on the waterline as shown in the figure. A solid
spherical ball made of lead (specific gravity = 11.4)
is attached to the free end of the rod to keep the
assembly in static equilibrium inside the water.
For simplicity, assume that the radius of the ball is
much smaller than the length of the rod.
Assume density of water = 10^3 kg/m^3 and \pi = 3.14.
Radius of the ball is _______ cm (round off to 2 decimal places).
Assume density of water = 10^3 kg/m^3 and \pi = 3.14.
Radius of the ball is _______ cm (round off to 2 decimal places).
2.42 | |
3.62 | |
8.26 | |
3.59 |
Question 1 Explanation:
As the system is in equilibrium,
\Sigma M_o=0
(F_{B1}-W_1) \times \frac{L}{2} \cos \theta +F_{B2}-W_2) \times L \times \cos \theta =0
(\rho gV_1-\rho _1gV_1) \times \frac{1}{2}+(\rho gV_2-\rho _2gV_2) =0
(\rho -\rho _1) \times g \times \frac{\pi d^2L}{4} \times \frac{1}{2}+(\rho -\rho _2) \times \frac{4}{3} \times \\pi R^3 \times g=0
(100-600) \times \frac{0.04^2 \times 8}{8} +(1000-11400) \times \frac{4}{3} \times R^3=0
\Rightarrow R=0.03587m=3.59cm
Question 2 |
The steady velocity field in an inviscid fluid of
density 1.5 is given to be \vec{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}
Neglecting body forces, the pressure gradient at
(x = 1, y = 1) is
10\hat{j} | |
20\hat{i} | |
-6\hat{i}-6\hat{j} | |
-4\hat{i}-4\hat{j} |
Question 2 Explanation:
By Euler's equation of motion,
\begin{aligned} x:-\frac{\partial p}{\partial x}+\rho g_x&=\rho \frac{Du}{Dt}\\ y:-\frac{\partial p}{\partial y}+\rho g_y&=\rho \frac{Dv}{Dt}\\ \end{aligned}
Neglecting body forces (i.e. g_x=g_y=0 )
\begin{aligned} \frac{\partial p}{\partial x}&=-\rho \frac{Du}{Dt}=-\rho \left [ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(-2x)+(2xy)(2y)]\\ &=-\rho[-2xy^2+2x^3+4xy^2]\\ &=-\rho(2xy^2+2x^3)\\ &=-1.5 \times (2 \times 1 \times 1^2 +2\times1^3 )\\ &=-6 Pa/m \end{aligned}
Similarly,
\begin{aligned} \frac{\partial p}{\partial y}&=-\rho \frac{Dv}{Dt}\\ &=-\rho \left [ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(2y)+(2xy)(2x)]\\ &=-\rho[2y^3-2x^2y+4x^2y]\\ &=-\rho(2y^3+2x^2y)\\ &=-1.5 \times (2 \times 1^3 +2\times1^2 \times 1 )\\ &=-6 Pa/m \end{aligned}
The pressure gradient vector is given by
\triangledown p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}=-6\hat{i}-6\hat{j}
\begin{aligned} x:-\frac{\partial p}{\partial x}+\rho g_x&=\rho \frac{Du}{Dt}\\ y:-\frac{\partial p}{\partial y}+\rho g_y&=\rho \frac{Dv}{Dt}\\ \end{aligned}
Neglecting body forces (i.e. g_x=g_y=0 )
\begin{aligned} \frac{\partial p}{\partial x}&=-\rho \frac{Du}{Dt}=-\rho \left [ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(-2x)+(2xy)(2y)]\\ &=-\rho[-2xy^2+2x^3+4xy^2]\\ &=-\rho(2xy^2+2x^3)\\ &=-1.5 \times (2 \times 1 \times 1^2 +2\times1^3 )\\ &=-6 Pa/m \end{aligned}
Similarly,
\begin{aligned} \frac{\partial p}{\partial y}&=-\rho \frac{Dv}{Dt}\\ &=-\rho \left [ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(2y)+(2xy)(2x)]\\ &=-\rho[2y^3-2x^2y+4x^2y]\\ &=-\rho(2y^3+2x^2y)\\ &=-1.5 \times (2 \times 1^3 +2\times1^2 \times 1 )\\ &=-6 Pa/m \end{aligned}
The pressure gradient vector is given by
\triangledown p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}=-6\hat{i}-6\hat{j}
Question 3 |
A tube of uniform diameter D is immersed in a
steady flowing inviscid liquid stream of velocity V,
as shown in the figure. Gravitational acceleration is
represented by g. The volume flow rate through the
tube is ______.
\frac{\pi}{4}D^2V | |
\frac{\pi}{4}D^2\sqrt{2gh_2} | |
\frac{\pi}{4}D^2\sqrt{2g(h_1+h_2)} | |
\frac{\pi}{4}D^2\sqrt{V^2-2gh_2} |
Question 3 Explanation:
By applying Bernoulli's equation between (1) and (2)
\begin{aligned} \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1&=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \frac{P_{atm}+\rho gh_1}{\rho g}+\frac{V_1^2}{2g}+0&=\frac{P_{atm}}{\rho g}+\frac{V_2^2}{2g}+h_1+h_2 \\ &[\because \;\; V_1=V]\\ \frac{V_1^2}{2g}&=\frac{V}{2g}+h_2-(h_1+h_2)\\ \therefore \; V_2&=\sqrt{V^2-2gh_1}\\ \therefore \; Q&=A_2V_2=\frac{\pi d^2}{4} \times \sqrt{V^2-2gh_1} \end{aligned}
Question 4 |
The velocity field in a fluid is given to be
\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}
Which of the following statement(s) is/are correct?
MSQ
\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}
Which of the following statement(s) is/are correct?
MSQ
The velocity field is one-dimensional. | |
The flow is incompressible | |
The flow is irrotational | |
The acceleration experienced by a fluid particle is zero at (x = 0, y = 0). |
Question 4 Explanation:
For given flow,
u=4xy, v = 2(x^2- y^2)
As velocity field is function of two space variables, flow is two dimensional.
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=4y-4y
Therefore, flow is incompressible.
\omega _z=\frac{1}{2}\left ( \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y}\right )=\frac{1}{2}(4x-4x)=0
Therefore, flow is irrotational.
\begin{aligned} a_x &=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}\\ &= 4xy(4y)+2(x^2-y^2)(4x)\\ &= 16xy^2+8x^3-8xy^2 \\ &=16 \times 0 \times 0^2+8 \times 0^3-8 \times 0 \times 0^2 \\ &= 0 \end{aligned}
\begin{aligned} a_y &=u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\\ &= 4xy(4x)+2(x^2-y^2)(-4y)\\ &= 16x^2y-8x^2y+8y^3 \\ &=16 \times 0^2 \times 0-8 \times 0^2 \times 0+8 \times 0^3 \\ &= 0\\ |\vec{a}|&=\sqrt{a_x^2+a_y^2}=0 \end{aligned}
u=4xy, v = 2(x^2- y^2)
As velocity field is function of two space variables, flow is two dimensional.
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=4y-4y
Therefore, flow is incompressible.
\omega _z=\frac{1}{2}\left ( \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y}\right )=\frac{1}{2}(4x-4x)=0
Therefore, flow is irrotational.
\begin{aligned} a_x &=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}\\ &= 4xy(4y)+2(x^2-y^2)(4x)\\ &= 16xy^2+8x^3-8xy^2 \\ &=16 \times 0 \times 0^2+8 \times 0^3-8 \times 0 \times 0^2 \\ &= 0 \end{aligned}
\begin{aligned} a_y &=u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\\ &= 4xy(4x)+2(x^2-y^2)(-4y)\\ &= 16x^2y-8x^2y+8y^3 \\ &=16 \times 0^2 \times 0-8 \times 0^2 \times 0+8 \times 0^3 \\ &= 0\\ |\vec{a}|&=\sqrt{a_x^2+a_y^2}=0 \end{aligned}
Question 5 |
Consider a steady flow through a horizontal
divergent channel, as shown in the figure, with
supersonic flow at the inlet. The direction of flow is
from left to right.
Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?
MSQ
Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?
MSQ
Since volume flow rate is constant, velocity at
B is lower than velocity at A | |
Normal shock | |
Viscous effect | |
Boundary layer separation |
Question 5 Explanation:
If the supersonic flow enters to the diverging duct,
it will act as nozzle the pressure of the flow is
decreases in the nozzle. But it is given the pressure
of the flow at B is more than that A
It will happen only with the development of normal shoot in the diverging flow (the normal shock wave is the characteristic of only supersonic flow) Because of normal shock wave in the diverging nozzle pressure increases and velocity decreases. But mass flow rate remains unchanged.
It will happen only with the development of normal shoot in the diverging flow (the normal shock wave is the characteristic of only supersonic flow) Because of normal shock wave in the diverging nozzle pressure increases and velocity decreases. But mass flow rate remains unchanged.
Question 6 |
A steady two-dimensional flow field is specified by
the stream function
\psi =kx^3y
where x and y are in meter and the constant k = 1 \; m^{-2}s^{-1}. The magnitude of acceleration at a point (x,y) = (1 m, 1 m) is ________ m/s^2 (round off to 2 decimal places).
\psi =kx^3y
where x and y are in meter and the constant k = 1 \; m^{-2}s^{-1}. The magnitude of acceleration at a point (x,y) = (1 m, 1 m) is ________ m/s^2 (round off to 2 decimal places).
2.42 | |
1.25 | |
3.62 | |
4.24 |
Question 6 Explanation:
Given,
Stream function,
\begin{aligned} \psi &=kx^3y; \; k=1m^{-2}s^{-1}\\ u&=-\frac{\partial \psi }{\partial y}\Rightarrow u=-x^3\\ v&=-\frac{\partial \psi }{\partial x}\Rightarrow v=3x^2y\\ \vec{V}&=-x^3\hat{i}+3x^2y\hat{j}\\ a_x&=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\\ a_x&=-x^3(-3x^2)\Rightarrow a_x=3x^5\\ At\; &(1,1), \; a_x=3m/sec^2\\ a_y&=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\\ a_y&=-x^3(6xy)+3(x^2y)(3x^2)\\ &=-6x^4y+9x^4y=3x^4y\\ \vec{a}&=3x^5\hat{i}+3x^4y\hat{j}\\ At& \;\;(1,1)\\ \vec{a}&=3\hat{i}+3\hat{j}\\ \Rightarrow |a|=\sqrt{3^2+3^2}&=4.24m/s^2 \end{aligned}
Stream function,
\begin{aligned} \psi &=kx^3y; \; k=1m^{-2}s^{-1}\\ u&=-\frac{\partial \psi }{\partial y}\Rightarrow u=-x^3\\ v&=-\frac{\partial \psi }{\partial x}\Rightarrow v=3x^2y\\ \vec{V}&=-x^3\hat{i}+3x^2y\hat{j}\\ a_x&=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\\ a_x&=-x^3(-3x^2)\Rightarrow a_x=3x^5\\ At\; &(1,1), \; a_x=3m/sec^2\\ a_y&=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\\ a_y&=-x^3(6xy)+3(x^2y)(3x^2)\\ &=-6x^4y+9x^4y=3x^4y\\ \vec{a}&=3x^5\hat{i}+3x^4y\hat{j}\\ At& \;\;(1,1)\\ \vec{a}&=3\hat{i}+3\hat{j}\\ \Rightarrow |a|=\sqrt{3^2+3^2}&=4.24m/s^2 \end{aligned}
Question 7 |
Consider steady, one-dimensional compressible
flow of a gas in a pipe of diameter 1 m. At one
location in the pipe, the density and velocity are 1 kg/m^3
and 100 m/s, respectively. At a downstream
location in the pipe, the velocity is 170 m/s. If the
pressure drop between these two locations is 10 kPa,
the force exerted by the gas on the pipe between
these two locations is ____________ N.
350 \pi ^2 | |
750 \pi | |
1000 \pi | |
3000 |
Question 7 Explanation:
Applying momentum equation to the pipe flow,
\Sigma \vec{F}=(\dot{m}\vec{V})_{out}-(\dot{m}\vec{V})_{in}+\frac{\partial }{\partial t}(m\vec{V})_{e,v}
P_1A-P_2A+F=\dot{m}(V_2)-\dot{m}(V_1)
\begin{aligned} \therefore \; F&=\dot{m}(V_2-V_1)-(P_1-P_2)A\\ &=\rho _1AV_1(V_2-V_1)-(P_1-P_2)A\\ &=\frac{ \pi d^2}{4}\left [ \rho (V_1V_2-V_1^2)-(P_1-P_2) \right ]\\ &=\frac{\pi \times 1^2}{4}\left [ 1 \times (100 \times 170 -100^2)-(10 \times 10^3) \right ]\\ &=-750 \pi N \end{aligned}
Negative sign shows that assumed direction of force is opposite of actual direction.
Question 8 |
A solid spherical bead of lead (uniform density =11000\; kg/m^3 ) of diameter d = 0.1 mm sinks with
a constant velocity V in a large stagnant pool of a
liquid (dynamic viscosity =1.1 \times 10^{-3} kg.m^{-1}.s^{-1} ).
The coefficient of drag is given by C_D=\frac{24}{Re}, where
the Reynolds number (Re) is defined on the basis of
the diameter of the bead. The drag force acting on
the bead is expressed as D=(C_D)(0.5 \rho V^2)\left ( \frac{\pi d^2}{4} \right ), where \rho is the density of the liquid. Neglect the
buoyancy force. Using g = 10 m/s^2, the velocity V
is __________ m/s.
\frac{1}{24} | |
\frac{1}{6} | |
\frac{1}{18} | |
\frac{1}{12} |
Question 8 Explanation:
As buoyancy force is neglected we can neglect
density of fluid ( ) in the following equation.
\begin{aligned} V&=\frac{1}{18\mu _f}(\rho _b-\rho _f)gd^2\\ &=\frac{11000 \times 10 \times (10^{-4})^2}{18 \times 1.1 \times 10^{-3}}\\ &=\frac{1}{18}m/s \end{aligned}
\begin{aligned} V&=\frac{1}{18\mu _f}(\rho _b-\rho _f)gd^2\\ &=\frac{11000 \times 10 \times (10^{-4})^2}{18 \times 1.1 \times 10^{-3}}\\ &=\frac{1}{18}m/s \end{aligned}
Question 9 |
The figure shows a purely convergent nozzle with
a steady, inviscid compressible flow of an ideal gas
with constant thermophysical properties operating
under choked condition. The exit plane shown in
the figure is located within the nozzle. If the inlet
pressure (P_0) is increased while keeping the back
pressure (P_{back}) unchanged, which of the following
statements is/are true?
MSQ
MSQ
Mass flow rate through the nozzle will remain unchanged. | |
Mach number at the exit plane of the nozzle will remain unchanged at unity | |
Mass flow rate through the nozzle will increase. | |
Mach number at the exit plane of the nozzle will become more than unity. |
Question 9 Explanation:
\begin{aligned}
\frac{\dot{m}}{A}&=\sqrt{\frac{\gamma }{R}}\frac{P_0}{\sqrt{T_0}}\frac{1}{\left ( \frac{\gamma +1}{2} \right )^{\frac{\gamma +1}{2(\gamma -1)}}}\\
\frac{\dot{m}}{A}&=0.685 \times \sqrt{P_0\rho _0} \;\;\;...\text{ for Air}\\
\frac{T_c}{T_0}&=\frac{T^*}{T_0}=\frac{2}{\gamma +1}=0.8333\\
&(\text{critical temp ratio})\\
\frac{P_c}{P_0}&=\frac{P^*}{P_0}=\left ( \frac{2}{\gamma +1} \right )^{\frac{\gamma }{\gamma -1}}=0.5282\\
&(\text{critical pressure ratio})
\end{aligned}
where \gamma= Adiabatic index = 1.4
If P_0 increases , mass flow rate increases. Mach number at the exit plane of the nozzle will remain unchanged at unity.
Since the maximum expected velocity from the converging nozzle is sound velocity.
where \gamma= Adiabatic index = 1.4
If P_0 increases , mass flow rate increases. Mach number at the exit plane of the nozzle will remain unchanged at unity.
Since the maximum expected velocity from the converging nozzle is sound velocity.
Question 10 |
In the following two-dimensional momentum
equation for natural convection over a surface
immersed in a quiescent fluid at temperature T_\infty ( g is the gravitational acceleration, \beta is the volumetric
thermal expansion coefficient, v is the kinematic
viscosity, u and v are the velocities in x and y
directions, respectively, and T is the temperature)
u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=g\beta (T-T_\infty )+v\frac{\partial u^2}{\partial y^2}
the term g\beta (T-T_\infty ) represents
u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=g\beta (T-T_\infty )+v\frac{\partial u^2}{\partial y^2}
the term g\beta (T-T_\infty ) represents
Ratio of inertial force to viscous force. | |
Ratio of buoyancy force to viscous force. | |
Viscous force per unit mass. | |
Buoyancy force per unit mass. |
Question 10 Explanation:
u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=g\beta (T-T_\infty )+v\frac{\partial^2 u}{\partial y^2}
This is the equation that governs the fluid motion in the boundary layer due to effect of Buoyancy
g\beta (T-T_\infty )=\frac{m}{s^2} \times \frac{1}{K} \times K=m/s^2
Unit of g\beta (T-T_\infty )= m/s^2
Buoyancy force is due to density difference and gravitational effect
g\beta (T-T_\infty ) represents Buoyancy force per unit mass.
This is the equation that governs the fluid motion in the boundary layer due to effect of Buoyancy
g\beta (T-T_\infty )=\frac{m}{s^2} \times \frac{1}{K} \times K=m/s^2
Unit of g\beta (T-T_\infty )= m/s^2
Buoyancy force is due to density difference and gravitational effect
g\beta (T-T_\infty ) represents Buoyancy force per unit mass.
There are 10 questions to complete.
q142 diagram missing
There is no diagram in the official question paper.
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