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 |

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?

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 2 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 3 |

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 3 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 4 |

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**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 4 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 5 |

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 5 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.

Question 6 |

Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary layer on a flat plate?

Nikuradse stresses | |

Reynolds stresses | |

Boussinesq stresses | |

Prandtl stresses |

Question 6 Explanation:

Reynolds stresses are responsible for eddy viscosity

\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}

where \eta = Eddy viscosity.

\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}

where \eta = Eddy viscosity.

Question 7 |

Consider fully developed, steady state incompressible laminar flow of a viscous fluid between two large parallel horizontal plates. The bottom plate is fixed and the top plate moves with a constant velocity of U = 4 m/s. Separation between the plates is 5 mm. There is no pressure gradient in the direction of flow. The density of fluid is 800 kg/m^3, and the kinematic viscosity is 1.25 \times 10^{-4} \; m^2/s. The average shear stress in the fluid is ________Pa (round off to the nearest integer).

58 | |

64 | |

92 | |

80 |

Question 7 Explanation:

\begin{array}{l} V=4 \mathrm{~m} / \mathrm{s} \\ \rho=800 \mathrm{~kg} / \mathrm{m}^{3} \\ \mathrm{v}=1.25 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\ h=5 \mathrm{~mm} \\ \tau=\mu \cdot \frac{d u}{d y} \\ \tau=\left[800 \times 1.25 \times 10^{-4}\right] \times \frac{4}{5 \times 10^{-3}} \\ \tau=80 \mathrm{~N} / \mathrm{m}^{2} \\ =80 \mathrm{~Pa} \end{array}

Question 8 |

Water (density 1000 kg/m^3) flows through an inclined pipe of uniform diameter. The
velocity, pressure and elevation at section A are V_A = 3.2 m/s, p_A =186 kPa and z_A=24.5 m respectively, and those at section B are V_B = 3.2 m/s, p_B =260 kPa and z_B=9.1 m, respectively. If acceleration due to gravity is 10 m/s^2 then the head lost
due to friction is _________ m (round off to one decimal place).

8 | |

-8 | |

-4 | |

4 |

Question 8 Explanation:

Energy at 'A' head =\frac{P_{A}}{\rho g}+\frac{V_{A}^{2}}{2 g}+Z_{A}=\frac{186 \times 10^{3}}{1000 \times 10}+\frac{3.2^{2}}{2 \times 10}+24.5

=18.6+0.512+24.5=43.612

Energy at 'b' head =\frac{P_{b}}{\rho g}+\frac{V_{b}^{2}}{2 g}+Z_{b}=\frac{286 \times 10^{3}}{1000 \times 10}+\frac{3.2^{2}}{2 \times 10}+9.1

=26+0.512+9.1=35.612

E_{A}>E_{B^{\prime}} so flow from 'A' to 'B'

Heat loss =E_{A}-E_{B}=43.612-35.612=8 \mathrm{m} of water head

Question 9 |

Consider steady, viscous, fully developed flow of a fluid through a circular pipe of internal
diameter D. We know that the velocity profile forms a paraboloid about the pipe centre
line, given by: V=-C\left (r^2-\frac{D^2}{4} \right ) m/s, where C is a constant. The rate of kinetic energy
(in J/s) at the control surface A-B, as shown in the figure, is proportional to D^n. The value
of n is________.

2 | |

4 | |

8 | |

16 |

Question 9 Explanation:

\begin{aligned} v &=-C\left[r^{2}-\frac{D^{2}}{4}\right]=-C\left[r^{2}-R^{2}\right] \\ v &=C\left[R^{2}-r^{2}\right]=C R^{2}\left[1-\frac{r^{2}}{R^{2}}\right] \\ \left[K\dot{E}\right]&=\frac{1}{2} \int_{0}^{R} \rho v^{3} d A=\rho \int_{0}^{R} C^{3} R^{6}\left[1-\frac{r^{2}}{R^{2}}\right]^{3} \cdot 2 \pi r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{2}}{R^{2}}\right]^{3} r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{6}}{R^{6}}-\frac{3 r^{2}}{R^{2}}\left[1-\frac{r^{2}}{R^{2}}\right]\right] r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{6}}{R^{6}}-\frac{3 r^{2}}{R^{2}}+\frac{3 r^{4}}{R^{4}}\right] r \cdot d r \\ &=\rho C^{3} 2 \pi R^{6}\left[\frac{R^{2}}{2}-\frac{R^{2}}{8}-\frac{3 R^{2}}{4}+\frac{R^{2}}{2}\right] \\ &=\rho C^{3} 2 \pi R^{6}\left[R^{2}-\frac{R^{2}}{8}-\frac{3 R^{2}}{4}\right]=\rho C^{3} 2 \pi R^{6}\left[\frac{R^{2}}{8}\right] \\ &=\frac{\rho C^{3} 2 \pi}{8} R^{8}=\frac{\rho C^{3} 2 \pi}{8 \times(2)^{8}} \times D^{8} \\ n &=8 \end{aligned}

Question 10 |

Match the following non-dimensional numbers with the corresponding definitions:

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

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

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

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

There are 10 questions to complete.

IN Q.1. ITS BOUSINESSQ THEORY NOT REYNOLD’S…PLS CONFIRM AND RECTIFY!

I have also the same doubt , but everywhere the answer given is Reynolds forces ,

Yes bro u are ✅️ right