Official Paper

GATE Mechanical Engineering (ME) Official Paper (Held On: 03 Feb, 2024) (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

The real variable x, y, z, and the real constants p, q, r satisfy

$\frac{x}{p q-r^{2}}=\frac{y}{q r-p^{2}}=\frac{z}{r p-q^{2}}$

Given the denominator are non-zero, the value of px + qy + rz.

((a))
pqr
((b))
1
((c))
p² + q² + r²
((d))
0

Show Answer

Answer: ((d))

Explanation:

Let,

$\frac{x}{p q-r^{2}}=\frac{y}{q r-p^{2}}=\frac{z}{r p-q^{2}}=K$

$\frac{p x}{p^{2} q-p r^{2}}=K$ ..... (1)

$\frac{q y}{q^{2} r-q p^{2}}=K$ ...... (2)

From 1 & 2:

px + qy + rz = kp²q - kpr²+ kqr² - kqp² + kr²p - krq²

= 0

Take two long dice (rectangular parallelepiped), each having four rectangular faces labelled as 2, 3, 5, and 7. If thrown, the long dice cannot land on the square faces and has 1/4 probability of landing on any of the four rectangular faces. The lebel on the top face of the dice is the score of the throw.

If thrown together, what is the probability of getting the sum of the two long dice scores greater than 11?

((a))
$\frac{3}{16}$
((b))
$\frac{1}{16}$
((c))
$\frac{1}{8}$
((d))
$\frac{3}{8}$

Show Answer

Answer: ((a))

$\frac{3}{16}$

Explanation:

Total outcomes : (2, 2), (2, 3), (2, 5), (2, 7)

(3, 2), (3, 3), (3, 5), (3, 7)

(5, 2), (5, 3), (5, 5), (5, 7)

(7, 2), (7, 3), (7, 5), (7, 7)

Favorable outcomes (sum > 11)

: (5, 7), (7, 5), (7, 7)

$\text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}=\frac{3}{16}$

In the following series, identify the number that needs to be changed to form the Fibonacci series.

1, 1, 2, 3, 6, 8, 13, 21, ....

((a))
13
((b))
8
((c))
21
((d))
6

Show Answer

Answer: ((d))

Explanation:

In Fibonacci series, every number is equal to the sum of previous two numbers of the series. i.e. 0, 1, 1, 2, 3, 5, 8, 13, 21, ....

0 + 1 = 1

1 + 1 = 2

2 + 1 = 3

3 + 2 = 5

5 + 3 = 8

8 + 5 = 13

13 + 8 = 21

Find the odd one out in the set: (19, 37, 21, 17, 23, 29, 31, 11)

((a))
23
((b))
21
((c))
29
((d))
37

Show Answer

Answer: ((b))

Explanation:

Expect 21 all are prime number.

Prime Number:

Prime numbers are numbers greater than 1.
They only have two factors, 1 and the number itself.
This means these numbers cannot be divided by any number other than 1 and the number itself without leaving a remainder.

If '→' denotes increasing order of intensity, then the meaning of the words [smile → giggle → laugh] is analogous to [disapprove → _______ → chide].

Which one of the given options is appropriate to fill the blank?

((a))
reprove
((b))
grieve
((c))
reprise
((d))
praise

Show Answer

Answer: ((a))

reprove

Explanation:

[smile giggle laugh] is similarly related as [disapprove → reprove → chide]

Reprove : To criticize somebody or not approve of something that somebody has done.

Disapprove : To refuse approval

Chide : criticize somebody

Four equilateral triangles are used to form a regular closed three-dimensional object by joining along the edges. The angle between any two faces is

((a))
45°
((b))
30°
((c))
60°
((d))
90°

Show Answer

Answer: ((c))

60°

Explanation:

Tetrahedron Angles:

In a regular tetrahedron, all the faces are equilateral triangles. Therefore, all the interior angles of a tetrahedron are 60° each.

In the given text, the blanks are numbered (i)-(iv). Select the best match for all the blanks.

Prof. P (i) merely a man who narrated funny stories (ii) in his blackest moments he was capable of self-depreciating humor.

Prof. Q (iii) a man who hardly narrated funny stories (iv)_ in his blackest moments was he able to find humor.

((a))
(i) wasn't (ii) Even, (iii) was, (iv) Only
((b))
(i) wasn't (ii) Only, (iii) was, (iv) Even
((c))
(i) was (ii) Only, (iii) wasn't, (iv) Even
((d))
(i) was (ii) Even, (iii) wasn't, (iv) Only

Show Answer

Answer: ((a))

(i) wasn't (ii) Even, (iii) was, (iv) Only

Explanation:

Prof. P wasn't merely a man who narrated funny stories even in his blackest moments he was capable of self-depreciating humor.

Prof. Q was a man who hardly narrated funny stories only in his blackest moments was he able to find humor.

How many combinations of non-null sets A, B, C are possible from the subsets of (2, 3, 5) satisfying the condition: (i) A is a subset of B, and (ii) B is a subset of C?

((a))
18
((b))
27
((c))
28
((d))
37

Show Answer

Answer: ((d))

Explanation:

The question is about non-null sets A, B, C $\Rightarrow 4^3$ – (Any set is Empty)

Consider $A = \phi$

Universal set $= {2,3,5}$ contains 3 elements $\Rightarrow~ B~ has~ 2^3$ possible choices, and for each possible B set, we need to calculate possible sets of C. As $B \subseteq C$ , the elements present in the B, should be present in C. Remaining elements of Universal set has two choices; present in C or not present in C.

$if ~B = \phi$ (number of elements in B = 0) $\Rightarrow$ number of possible sets for $C = 2^{n-0} = 2^3$

$\textbf{if B = {1}}$ (number of elements in B = 1) $\Rightarrow$ number of possible sets for $C = 2^{n-1} = 2^2$

$\textbf{if B = {2}}$ (number of elements in B = 1) $\Rightarrow$ number of possible sets for $C = 2^{n-1} = 2^2$

$\textbf{if B = {3}}$ (number of elements in B = 1) $\Rightarrow$ number of possible sets for $C = 2^{n-1} = 2^2$

$\textbf{if B = {1, 2}}$ (number of elements in B = 2) $\Rightarrow$ number of possible sets for $C = 2^{n-2} = 2^1$

$\textbf{if B = {1, 3}}$ (number of elements in B = 2) $\Rightarrow$ number of possible sets for $C = 2^{n-2} = 2^1$

$\textbf{if } B = {2, 3}$ (number of elements in B = 2) $\Rightarrow$ number of possible sets for $C = 2^{n-2} = 2^1$

$\textbf{if } B = {1, 2, 3}$ (number of elements in B = 3)} $\Rightarrow$ number of possible sets for $C = 2^{n-3} = 2^0$

∴ $\binom{n}{0} \cdot 2^n + \binom{n}{1} \cdot 2^{n-1} + \cdots + \binom{n}{r} \cdot 2^{n-r} \cdots + \binom{n}{n} \cdot 2^0 = (1 + 2)^n = 3^n = 3^3 = 27$

$∴ \text{when } A = \phi, \text{ possible sets of B and C are 27}$

Final answer = $4^3 - 3^3 = 37$

The bar chart gives the batting averages of VK and RS for 11 calender years from 2012 to 2022. Considering that 2015 and 2019 are world cup years, which one of the following options is true?

((a))
VK has a higher yearly batting average than that of RS in every world cup year.
((b))
VK's yearly batting average is consistently higher than that of RS between the two world cup years.
((c))
RS's yearly batting average is consistently higher than that of VK in the last three years.
((d))
RS has a higher yearly batting average than that of VK in every world cup year.

Show Answer

Answer: ((b))

VK's yearly batting average is consistently higher than that of RS between the two world cup years.

Explanation:

From graph:

In between 2015 and 2019, VK's batting average = 35 + 90 +75 +130 + 60 = $\frac{390}{5}$ = 78

In between 2015 and 2019, RS's batting average = 50 + 65 + 70 + 70 + 55 = $\frac{320}{5}$ = 64

So, VK yearly batting average is consistently higher then RS's between two World Cup years (2015 and 2019).

A planar rectangular paper has two V-shaped pieces attached as shown below:

This piece of paper is folded to make the following closed three-dimensions object.

The number of folds required to form the above object is

((a))
8
((b))
7
((c))
9
((d))
11

Show Answer

Answer: ((c))

Explanation:

Hence, the total number of folds required to form the above object is 9.

Mechanical Engineering (55 questions)

The most suitable electrode materials used for joining low alloy steels using Gas Metal Arc Welding (GMAW) process is

((a))
Copper
((b))
Cadmium
((c))
Low alloy steel
((d))
tungsten

Show Answer

Answer: ((c))

Low alloy steel

Explanation:

Gas metal arc welding (GMAW) or Metal inert gas arc welding (MIG)

In this process, the arc is formed between a continuous, automatically fed, metallic consumable electrode and welding job in an atmosphere of inert gas, and hence this is called metal inert gas arc welding (MIG) process.
The shielding gases for MIG welding are mixtures of argon, oxygen, and CO2, and special gas mixtures may contain helium.
In GMAW, consumable electrodes are used which has to be same composition as base metal. Thus, low alloy steels electrodes are the most suitable electrode material.

A ram in the from of a rectangular body of size, l = 9 cm and b = 2 m is suspended by two parallel ropes of lengths 7 m. Assume the center-of-mass of the body is at its geometric center and g = 9.81 m/s². For striking the object P with horizontal velocity of 5 m/s, what is the angle with the vertical from which the ram should be released from rest?

((a))
67.1°
((b))
35.1°
((c))
40.2°
((d))
79.5°

Show Answer

Answer: ((b))

35.1°

Explanation:

Given, AB = 7 m, u = 0, v = 5 m/s

According to question, the rope will become vertical on striking the object, hence AB = 7 m.

From geometry,

AD = 7 cosθ

∴ h = AB - AD = 7 - 7 cosθ

= 7(1 - cosθ)

From energy conservation,

Initial potential energy = Final kinetic energy

⇒ $m g h=\frac{1}{2} m v^{2}$

$9.81 \times 7(1-\cos \theta)=\frac{1}{2} \times 5^{2}$

$\cos \theta=\frac{9.81 \times 7-12.5}{9.81 \times 7}$

⇒ θ = 35.1°

A furnace can supply heat steadily at 1200 K at a rate of 24000 kJ/min. The maximum amount of power (in kW) that can be produced by using the heat supplied by this furnace in an environment at 300 K is

((a))
0
((b))
150
((c))
300
((d))
18000

Show Answer

Answer: ((c))

300

Explanation:

Q_in = 24000 kJ/min = 400 kW

T₁ = 1200 K, T₂ = 300 K

Maximum power can be obtained when the heat supplied

in used in Carnot engine,

∴ W_max = Q_in × η_carnot

= $400\left(1-\frac{300}{1200}\right)$

W_max = 300 kW

The value of the surface integral

∯_s zdxdy

where S is the external surface of the sphere x² + y² + z² = R² is

((a))
4πR3
((b))
0
((c))
$\rm \frac{4πR^3}{3}$
((d))
πR3

Show Answer

Answer: ((c))

$\rm \frac{4πR^3}{3}$

Explanation:

To evaluate, ∯_s z dx dy

where S(sphere) = x² + y² + z² = R²

= ∯z dxdy

$\int_S \vec{F} \cdot \hat{n} , ds = \iint_R \left(F_1 , dy , dz + F_2 , dx , dz + F_3 , dx , dy\right)$

$F_1 = 0, F_2 = 0, F_3 = z$

$\vec{F} = Z \hat{K}$

$\nabla \cdot \vec{F} = 1$

Using Gauss Divergence Theorem:

$\int_S \vec{F} \cdot \hat{n} , ds = \int_V \nabla \cdot \vec{F} , dV = \int_V dV$

= Volume of sphere = $\frac{4 \pi}{3} R^{3}$

Let f(.) be a twice differentiable function from ℝ² → ℝ. If p, x₀ ∈ ℝ² where ||p|| is sufficiently small (here, II.II is the Euclidean norm or distance function), then f(x₀ + p)

= $f\left(x_{o}\right)+\nabla f\left(x_{o}\right)^{T} p+\frac{1}{2} p^{T} \nabla^{2} f(\psi) p$ where we ℝ2 is a point on the line segment joining

x₀and x₀ + p. If x₀ is a strict local minimum of f(x), then which one of the following statements is TRUE?

((a))
∇f(x₀)^T p = 0 and p^T∇²f(ψ)p > 0
((b))
∇f(x0)T p = 0 and pT∇2f(ψ)p = 0
((c))
∇f(x0)T p > 0 and pT∇2f(ψ)p = 0
((d))
∇f(x0)T p = 0 and pT∇2f(ψ)p < 0

Show Answer

Answer: ((a))

∇f(x₀)^T p = 0 and p^T∇²f(ψ)p > 0

Explanation:

Let the function be parabolic nature, i.e. a form of: Ax² + Bx + C

Now, the conditions for strictly local minimum

to exist are:

The coefficient of x² (i.e. A) should be positive.

First derivative of the function should be zero.

Second derivative of the function should be positive.

Now, After the observation of all four options,

above these conditions will be satisfied by option (a)

For a ball bearing the fatigue life in millions of revolutions is given by L = $\left(\frac{C}{P}\right)^{n}$ , where P is the constant applied load and C is the basic dynamic load rating. Which one of the following statements is TRUE?

((a))
n = 3, assuming that the inner racing is fixed and outer racing is revolving.
((b))
n = 1/3, assuming that the inner racing is fixed and outer racing is revolving.
((c))
n = 3, assuming that the outer racing is fixed and inner racing is revolving.
((d))
n = 1/3, assuming that the outer racing is fixed and inner racing is revolving.

Show Answer

Answer: ((c))

n = 3, assuming that the outer racing is fixed and inner racing is revolving.

Explanation:

Dynamic load carrying Capacity:

It is defined as the radial load in radial bearing (Thrust load in thrust bearing) that can be carried for a minimum life of one million revolutionss.
It is based on the assumption that the inner race is rotating and the outer race is stationary.

Equivalent / Actual bearing Load:

The equivalent dynamic load is defined as the contact radial load in radial bearings (Thrust load in thrust bearing), which if applied to the bearing would give the same life as that which the bearing would attain in actual condition.

Bearing life under variable load:

Life, L = $(\frac{C}{P_e})^n$ million revolution

n = 3 for ball bearing, n = 10/3 for roller bearing, C = basic dynamic load, Pe = Dynamic load

Which one of the following failure theories is the most conservative design approach against fatigue failure?

((a))
Modified Goodman line
((b))
Yield line
((c))
Gerber line
((d))
Soderberge line

Show Answer

Answer: ((d))

Soderberge line

Explanation:

Soderberg line:

Soderberg line for ductile materials gives upper limit for any combination of mean and alternating stress. Soderberg line is the most conservative fatigue failure criterion. Following diagram depicts the same.

Considering following notations

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

N = Factor of safety

Soderberg line:

The line joining Syt (yield strength of the material) on the mean stress axis and Se (endurance limit of the component) on stress amplitude axis is called as Soderberg line. This line is used when yielding defines failure (Ductile materials).

The equation for the Soderberg line:

$\frac{{{{\rm{\sigma }}{\rm{m}}}}}{{{{\rm{S}}{{\rm{yt}}}}}} + \frac{{{{\rm{\sigma }}{\rm{a}}}}}{{{{\rm{S}}{\rm{e}}}}} = \frac{1}{N}$

Additional Information

Goodman line:

Line joining Se on stress amplitude axis and Sut on mean stress axis is known as Goodman line. The triangular region below this line is considered a safe region.

The equation for Goodman line:

$\frac{{{\sigma m}}}{{{S{ut}}}} + \frac{{{\sigma _a}}}{{{S_e}}} = \frac{1}{N}$

Gerber Line:

Line joining Se on stress amplitude axis and Sut on mean stress axis is joined by a parabolic curve.

The equation for Gerber line:

Consider the system of linear equations

x + 2y + z = 5

2x + ay + 4z = 12

2x + 4y + 6z = b

The values of a and b such that there exists a non-trivial null space and the system admits infinite solutions are

((a))
a = 4, b = 12
((b))
a = 8, b = 12
((c))
a = 4, b = 14
((d))
a = 8, b = 14

Show Answer

Answer: ((c))

a = 4, b = 14

Explanation:

x + 2y + z = 5

2x + ay + 4z = 12

2x + 4y + 6z = b

Performing row operations on augmented matrix,

$\left[\begin{array}{ccc:c} 1 & 2 & 1 & 5 \ 2 & a & 4 & 12 \ 2 & 4 & 6 & b \end{array}\right] \xrightarrow[R_{3}=R_{2}-2 R_{1}]{\substack{R_{1}=R_{2}-2 R_{1}}}\left[\begin{array}{ccc:c} 1 & 2 & 1 & 5 \ 0 & a-4 & 2 & 2 \ 0 & 0 & 4 & b-10 \end{array}\right]$

For infinitely many solutions,

ρ(A) = ρ(AB) = 2

a - 4 = 0

a = 4

and, for this

ρ(AB) = 2

R₃ = 2R₂

b - 10 = 4

b = 14

Let f(z) be an analytic function, where z = x + iy. If the real part of f(z) is coshx cosy, and the imaginary part of f(z) is zero for y = 0, then f(z) is

((a))
cosh x exp (-iy)
((b))
cosh z exp z
((c))
cosh z cos y
((d))
cosh z

Show Answer

Answer: ((a))

cosh x exp (-iy)

Explanation:

Let, f(z) = u + iv

u = cosh z cos y

= $\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y$

$u=\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y$

by Mine Thomson method

$u(x, y)=\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y$

Partial differential w.r.t. x, y

$u_{x}=\left(\frac{e^{x}-e^{-x}}{2}\right) \cos y ; \quad u_{y}=\left(\frac{e^{x}+e^{-x}}{2}\right)(-\sin y)$

$u_{x}=\left(\frac{e^{x}-e^{-x}}{2}\right) \cos y \quad u_{y}=-\frac{\left(e^{x}+e^{-x}\right)}{2} \sin y$

$u_{x}(z, 0)=\frac{e^{z}-e^{-z}}{2}=\sinh z \quad u_{y}(z, 0)=-0$

f(z) = ∫(u_x - iu_y)dz + c

= ∫(sinhz - a)dz + c

= ∫sinh z + c

= cos h z + c

A set of jobs U, V, W, X, Y, Z arrive at time t = 0 to a production line consisting of two workstations in series. Each job must be processed by both workstations in sequence (i.e., the first followed by the second). The process times (in minutes) for each job on each workstation in the production line are given below:

Job	U	V	W	X	Y	Z
Workstation 1	5	7	3	4	6	8
Workstation 2	4	6	6	8	5	7

The sequence in which the jobs must be processed by the production line if the total makespan of production is to be minimized is

((a))
U-Y-V-Z-X-W
((b))
W-X-Z-V-Y-U
((c))
W-X-V-Z-Y-U
((d))
W-U-Z-V-Y-X

Show Answer

Answer: ((b))

W-X-Z-V-Y-U

Explanation:

By Johnson’s rule of sequencing mark minimum time-consuming operation for each process.

$\begin{array}{|c|c|c|} \hline \textbf{Job} & \textbf{Work Stations 1} & \textbf{Work Stations 2} \ \hline U & 5 & 4 \ \hline V & 7 \leftarrow & 6 \ \hline W & {3} & 6 \ \hline X & 4 & 8 \ \hline Y & 6 & 5 \ \hline Z & 8 & 7 \ \hline \end{array}$

According to the question, work on work station (1) is followed by work on workstation (2).

Perform that job in mark station which is minimum time consumption. It is W-X during this Z on work station-2 will be performed. So, W – X – Z

After W-X-Z, V will be performed on workstation (1). So, W–X–Z–V, in between Y-work on the workstation will perform.

As the minimum time in the workstation is U, so will be performed at last. So, W – X – Z – V – Y – U

A plane, solid slab of thickness L, shown in the figure, has thermal conductivity k that varies with the spatial coordinate x as k = A + Bx, where A and B are positive constant (A > 0, B > 0). The slab walls are maintained at fixed temperature of 7(x = 0) = 0 and T(x = L) = T₀ > 0. The slab has no internal heat sources. Considering one-dimensional heat transfer, which one of the following plots qualitatively depicts the steady-state temperature distribution within the slab?

((a))
((b))
((c))
((d))

Show Answer

Answer: ((b))

Explanation:

k = A + Bx

where A and B are positive real number

So with an increase in x, k increases

Energy equation;

$\frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) + \dot{q} = 0$

$\frac{\partial}{\partial x}\left((A+B x) \cdot \frac{\partial T}{\partial x}\right)=0$

On integration

$(A+B x) \cdot \frac{\partial T}{\partial x}=C_{1}$

$\frac{\partial T}{\partial x}=\frac{C_{1}}{A+B x}$

On integration,

T = logarithmic function of x

Hence, variation of T with x is given as;

The "Earing" phenomenon in metal forming is associated with

((a))
deep drawing
((b))
extrusion
((c))
rolling
((d))
forging

Show Answer

Answer: ((a))

deep drawing

Explanation:

Anisotropy plays an important role in the performance of deep drawing processes. The anisotropy is of two types. In normal anisotropy the properties differ in the thickness direction. In planar anisotropy, the properties vary with the orientation in the plane of the sheet. Whereas deep drawability of sheets increases with normal anisotropy, anisotropy leads to the formation of ears in cup drawing. Ears cause the wavy edge of a drawn cup.

Earing:

In deep drawing, the edges of cups may become wavy, called earing. Ears are objectionable on deep-drawn cups because they have to be trimmed off, as they serve no useful purpose, and interfere with further processing of the cup, resulting in scarp. Earing is caused by the planar anisotropy of the sheet metal, and the number of ears produced may be two, four or eight, depending on the processing history and microstructure of the material. If the sheet is stronger in the rolling direction than transverse to the rolling direction, and the strength varies uniformly with respect to orientation, then two ears will form. If the sheet has high strength at different orientations, then more ears will form.

A rigid massless tetrahedron is placed such that vertex O is at the origin and the other three vertices A, B, C lie on the coordinate axes as shown in the figure. The body is acted on by three point loads, of which one is acting at A along x-axis and another at point B along y-axis. For the body to be in equilibrium, the third point load acting at point O must be

((a))
In y-z plane but not along y or z axis
((b))
along z-axis
((c))
in z-x plane but not along z or x axis
((d))
in x-y plane but not along x or y axis

Show Answer

Answer: ((d))

in x-y plane but not along x or y axis

Explanation:

For the body to be in equilibrium, the three forces can be coplanar, parallel or concurrent.

Let the resultant of F₁ and F₂ is R.

For equilibrium, third force shall be in opposite direction of R in same plane (x–y plane)

Consider a hydrodynamically fully developed laminar flow through a circular pipe with the flow along the axis (i.e. z direction). In the following statements, T is the temperature of the fluid, Tw is the wall temperature and Tm is the bulk mean temperature of the fluid. Which one of the following statements is TRUE?

((a))
Nusselt number varies linearly along the z-direction for a thermally fully developed flow.
((b))
For a thermally fully developed flow, $\frac{\partial T}{\partial z}$ = 0, always.
((c))
For constant wall temperature of the duct, $\frac{d T_{m}}{d z}$ = constant.
((d))
For constant wall temperature (Tw > Tm) of the duct, $\frac{d T_{m}}{d z}$ increase exponentially with distance along z-direction.

Show Answer

Answer: ((d))

For constant wall temperature (Tw > Tm) of the duct, $\frac{d T_{m}}{d z}$ increase exponentially with distance along z-direction.

Explanation:

In a thermally fully developed flow, the temperature profile will no longer change along the length of the flow, similar to the velocity profile. It means there is no further growth of the boundary layer. This is the case of a steady state.

$\frac{dT}{dZ} = 0$

$N_u$ = Constant for thermally fully developed flow

$(T_w - T_m) \text{ decays exponential}$

$\frac{dT_m}{dz} \text{ increases exponentially}$

The rate of increase in bulk mean temperature along the flow $\left(\frac{d T_{m}}{d z}\right)$ increase exponentially along the flow direction (z).

A linear spring-mass-dashpot system with a mass of 2 kg is set in motion with viscous damping. If the natural frequency is 15 Hz, and the amplitudes of two successive cycles measured are 7.75 mm and 7.20 mm, the coefficient of viscous damping (in N.s/m) is

((a))
4.41
((b))
6.11
((c))
2.52
((d))
7.51

Show Answer

Answer: ((a))

4.41

Explanation:

m = 2 kg

f_n = 15 Hz

ω_n = 2π(f_n) = 2π × 15 = 30π rad/s

x_n = 7.75 mm, x_n+1 = 7.20 mm

Decrement ratio, $\frac{x_{n}}{x_{n+1}}=\frac{7.75}{7.20}=1.07638$

Logarithmic decrement,

$δ=\log {e}\left(\frac{x{n}}{x_{n+1}}\right)=\log _{e}(1.07638)$

δ = 0.0736

$\frac{2 \pi ξ}{\sqrt{1-ξ^{2}}}=0.0736$

⇒ ξ = 0.0117

But, $2 \xi \omega_{n}=\frac{C}{m}$

⇒ $2 \times 0.0117 \times 30 \pi=\frac{C}{2}$

⇒ C = 4.41 N-s/m

The grinding wheel used to provide the best surface finish is

((a))
A60L5V
((b))
A80L5V
((c))
A36L5V
((d))
A54L5V

Show Answer

Answer: ((b))

A80L5V

Explanation :

In grinding operation, surface finish is decided by the grain size. Fine and very fine grain size will result in best surface finish.

Additional InformationDesignation of Grinding Wheel:

Prefix / Suffix: These are the secret codes used by the manufacturers to represent the wheel by its size and shapes respectively.

Type of Abrasives / Grain type:

It indicates materials used for the manufacturing of abrasive particles.
Out of the abrasives B4C is giving the poor performance during machining and diamond is very costly, therefore Al2O3 or SiC is the most commonly bused abrasives in the grinding wheel.
Al2O3 soft and tougher than the SiC whereas SiC will be hard and brittle than Al2O3
The type of abrasive is selected based on the mechanical properties of workpiece material I.e. for machining of soft and ductile workpieces, Al2O3, and machining of hard and brittle workpiece SiC will be used.
A- Al2O3, B – B4C, C – SiC, D - Diamond

Grain size or Grit size:

It indicates the size of abrasive particles.
i.e. Side if abrasives = 1/ Grain Size Number (GSN)
when the GSN > 600, the size of the abrasive particles becomes very very small and it cannot act like a cutting tool, therefore MRR is less.
When GSN < 600, the actual size of abrasive is increasing, the chip size is increasing and MRR is increasing.
As the GSN is reducing or the size of abrasive is increasing, the MRR is increasing first and then reducing.
The grain size is selected based on the surface finish required on the workpiece i.e. for a rough grinding, course or medium grain size is selected and for finished grinding fine or very fine grain size will be selected.
10-24= Coarse, 30-60 = Medium, 80 -180 = Fine, 220 – 600 = Very fine

Grades of Grinding Wheel:

It indicates the hardness of the grinding wheel.
The grade of the grinding wheel is selected based on the mechanical properties of the workpiece material.
Soft wheels are used for grinding of hard workpiece because the rubbing forces induced by the blunt abrasive particle i.e. the self-sharpening is taking place and no dressing is required.
Hard wheels are used for grinding of the soft workpiece, the abrasive particle will be effectively utilized so that at the end of effective utilization the dressing will be carried for resharpening of grinding wheel.
A –H = Soft, I – P = Medium, Q – Z = Hard

Structure:

The structure is indicating the average gap between the two consecutive abrasive particles.
As the average gap is large, the number of abrasive particle presents per unit area will be small hence it is called the open structure.
The structure of a grinding wheel can be varied by varying the % of abrasive particles and bonding material in the manufacturing of a grinding wheel. i.e. when higher % of abrasives and lower % of bonding material is used in manufacturing it produces the dense structure and vice-versa.
0 – 7 = Dense, 8 – 16 = Open

Bonds:

Bond indicates the bonding material used for the manufacturing of the grinding wheel.
Out of the different bonding materials, vitrified is the most commonly used bonding material because it gives higher bonding strength, high temperature withstanding capability, and high thermal conductivity.
For the manufacturing of flexible grinding wheels also called buffing wheels, shellac or rubber can be used as the bonding material.
V – Vitrified, B – bakelite, S – Silicate, E – Shellac, R - Rubber

A queueing system has one single server workstation that admits an infinitely long queue. The rate of arrival of jobs, to the queueing system follows the Poisson distribution with a mean of 5 jobs/hour. The service time of the server is exponentially distributed with a mean of 6 minutes. In steady state operation of the queueing system, the probability that the server is not busy at any point in time is

((a))
0.83
((b))
0.50
((c))
0.17
((d))
0.20

Show Answer

Answer: ((b))

0.50

Explanation:

Given: λ = 5/hr, μ = 10/hr

∴ $\rho=\frac{\lambda}{\mu}=\frac{5}{10}=\frac{1}{2}$

Now, probability for the system to be idle,

P₀ = 1 - ρ = 1 - 0.5 = 0.5

In order to numerically solve the ordinary differential equation $\frac{d y}{d t}=-y$ for t > 0, with an initial condition y(0) = 1, the following scheme is employed

$\frac{y_{n+1}-y_{n}}{Δ t}=\frac{1}{2}\left(y_{n+1}+y_{n}\right)$

Here, Δt is the time step and y_n = y(nΔt) for n = 0, 1, 2, ..... This numerical scheme will yield a solution with non-physical oscillations for Δt > h. The value of h is

((a))
1
((b))
$\frac{3}{2}$
((c))
$\frac{1}{2}$
((d))
2

Show Answer

Answer: ((d))

Explanation:

Given that,

$\frac{y_{n+1}-y_{n}}{\Delta t}=\frac{1}{2}\left(y_{n+1}+y_{n}\right)$

$y_{n+1}-y_{n}=\frac{\Delta t}{2}\left(y_{n+1}+y_{n}\right)$

$y_{n+1}-\frac{Δ t}{2} y_{n+1}=y_{n}+\frac{Δ t}{2} y_{n}$

$y_{n+1}=\frac{2+Δ t}{2-Δ t} y_{n}$

For stability:

$\left|\frac{2+\Delta t}{2-\Delta t}\right| \leq 1$

⇒ $-1 \leq \frac{2+\Delta t}{2-\Delta t} \leq 1$

⇒ $-1 \leq-1+\frac{4}{2-\Delta t} \leq 1$

⇒ $0 \leq \frac{4}{2-\Delta t} \leq 2$

⇒ $\frac{1}{2} \leq \frac{2-\Delta t}{4} \leq 0$

⇒ 2 ≤ 2 - Δt ≤ 0

⇒ 0 ≤ - Δt ≤ -2

⇒ Δt ≥ 2

The velocity field of a two - dimentional, incompressible flow is given by

$\vec{V}=2 \sinh x ̂{i}+v(x, y) ̂{j}$

where î and ĵ denote the unit vector in x and y direction, respectively. If v(x,0) = cosh x, then (0, -1) is

((a))
1
((b))
2
((c))
4
((d))
3

Show Answer

Answer: ((d))

Explanation:

The velocity field of a two-dimensional, incompressible flow is given by,

$\vec{V}=2 \sinh x \hat{i}+V(x, y) \hat{j}$

and V(x, 0) = coshx,

For an incompressible flow, ∇ . $\vec{V}$ = 0

⇒ $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

= 2coshx + $\frac{\partial v}{\partial y}$ = 0

⇒ $\frac{\partial v}{\partial y}$ = -2coshx

⇒ $\int \partial v=\int-2 \cosh x \cdot \partial y$

V = -2y . coshx + f(x)

For, V(x, 0) = coshx

⇒ -2y × coshx + f(x) = coshx

⇒ f(x) = coshx (at x, 0)

⇒ V = -2 . y × coshx + coshx

V = (1 - 2y) × coshx

V(0, - 1) = [1 - {2 × (-1)}] × cosh(0) = 3

The preparatory functions in Computer Numerical Controlled (CNC) machine programme are denoted by the alphabet

((a))
P
((b))
O
((c))
M
((d))
G

Show Answer

Answer: ((d))

Explanation:

Preparatory function of CNC programming is denoted by G-codes. Preparatory functions are the G-codes that identify the type of activities the machine will execute. The preparatory functions describe the way in which the machine axes have to move, the method of interpolation, the dimension system, the time delay of program execution and the activation of specific operational modes in the control.

Additional Information

G-codes: These are the general-purpose codes. Some of the G codes with their purpose are:

G-Code	Function
G-00	Rapid transverse
G-01	Linear interpolation
G-02	Circular interpolation (CW)
G-03	Circular interpolation (CCW)
G-04	Dwell
G-05	Hold/Delay
G-06	Parabolic Interpolation
G-08	Acceleration of feed rate
G-09	Deceleration of feed rate
G-17	XY Plane designation
G-18	ZX Plane designation
G-19	YZ Plane designation
G-33	Thread cutting, constant lead
G-34	Thread cutting, linearly increasing lead
G-35	Thread cutting, linearly decreasing lead
G-40	Cutter compensation cancels to zero
G-41	Cutter radius compensation - offset left
G-42	Cutter radius compensation - offset right
G-43	Cutter compensation positive
G-44	Cutter compensation negative
G-63	Tapping cycle
G-64	Change in feed rate or speed
G-70	Dimensioning in inch units
G-71	Dimensioning in metric units
G-80	Canned cycle cancelled
G-81	Canned drilling cycle
G-89	Canned boring cycle
G-90	Absolute input dimensions
G-91	Incremental input dimensions
G-97	Spindle speed in revolution per minute

Consider incompressible laminar flow over a flat plate with freestream velocity of u∞. The Nusselt number corresponding to this flow velocity is Nu1. If the freestream velocity is doubled, the Nusselt number changes to Nu2. Choose the correct option for $\frac{N u_{2}}{N u_{1}}$ .

((a))
1
((b))
√2
((c))
2
((d))
1.26

Show Answer

Answer: ((b))

√2

Explanation:

For laminar flow over a flat plate,

Nu ∝ Re^0.5 Pr^1/3, where Re is Reynolds number and Pr is the Prandtl number.

∵ $\mathrm{Re}=\frac{U_{\infty} x}{\mathrm{v}}$

∴ $\mathrm{Nu} \propto \cup_{\infty}^{0.5}$

Given that: $\frac{U_{\infty, 2}}{U_{\infty, 1}}~=~2$

or, $\frac{N u_{2}}{N u_{1}}=\sqrt{\frac{U_{\infty, 2}}{U_{\infty, 1}}}=\sqrt{2}$

Which one of the following statements regarding a Rankine cycle is FALSE?

((a))
Cycle efficiency increases as boiler pressure decreases.
((b))
Superheating the steam in the boiler increases the cycle efficiency.
((c))
The pressure at the turbine outlet depends on the condenser temperature.
((d))
Cycle efficiency increases as condenser pressure decreases.

Show Answer

Answer: ((a))

Cycle efficiency increases as boiler pressure decreases.

Explanation:

• Superheating in Rankine cycle increases the cycle efficiency because of increase in mean temperature of heat addition.

• With increase in pressure of boiler, the cycle efficiency increases. So, the given statement is wrong.

• With decrease in condenser pressure, the cycle efficiency increases because of decrease in mean temperature of heat rejection.

• The pressure of turbine outlet is governed by the condenser temperature. Decreasing the cooling water temperature, creates more vacuum in condenser which results in pressure drop and vice-versa.

Additional InformationRankine cycle

It is the ideal cycle for a vapour power plant.
It comprises four reversible processes:

Process	Process type
1-2	Isentropic compression process(pump work)
2-3	Constant pressure heat addition process
3-4	Isentropic expansion process
4-1	Constant pressure heat rejection process

The efficiency of the Rankine cycle

We know that efficiency η of the Rankine cycle is given as,

$\eta =1-\frac{T_L}{T_{avg}}$

That means to increase the efficiency we should increase the average temperature at which heat is transferred to the working fluid in the boiler.
Another way would be to decrease the average temperature at which heat is rejected from the working fluid in the condenser.

Decreasing the condenser pressure

Lowering the condenser pressure will increase the area enclosed by the cycle on a T - S diagram which indicates that the net-work will increase.
Thus, the thermal efficiency of the cycle will be increased.

Superheating the steam to high temperature

This will increase the net-work output and the efficiency of the cycle.
It also decreases the moisture contents of the steam at the turbine exit.
The temperature to which steam can be superheated is limited by metallurgical considerations (620°C)

Increasing the boiler pressure

Increasing the operating pressure of the boiler leads to an increase in the temperature at which heat is transferred to the steam and thus raises the efficiency of the cycle.

The change in kinetic energy ΔE of an engine is 300 J, and minimum and maximum shaft speeds are ω_min= 220 rad/s and ω_max= 280 rad/s, respectively. Assume that the torque-time function is purely-harmonic. To achieve a coefficient of fluctuation of 0.05, the moment of inertia (in kg.m²) of the flywheel to be mounted on the engine shaft is

((a))
0.053
((b))
0.113
((c))
0.096
((d))
0.071

Show Answer

Answer: ((c))

0.096

Explanation:

Given: ΔE = 300 J, ω_min = 220 rad/s, ω_max = 280rad/s, C_s = 0.05, I = ?

Energy fluctuation, ΔE = lω²C_s

⇒ $\Delta E=I \times\left(\frac{\omega_{\max }+\omega_{\min }}{2}\right)^{2} \times C_{s}$

⇒ $300=I \times\left(\frac{280+220}{2}\right)^{2} \times 0.05$

⇒ I = 0.096 kg-m²

The allowance provided to a pattern for easy withdrawal from a sand mold is

((a))
finishing allowance
((b))
shake allowance
((c))
shrinkage allowance
((d))
distortion allowance

Show Answer

Answer: ((b))

shake allowance

Concept:

A pattern is the replica of the casting to be formed with some modifications.

These modifications are in the form of allowances. The allowances provided to the pattern are discussed below

Shake Allowance: Before the withdrawal from the sand mold, the pattern is rapped all around the vertical faces to enlarge the mold cavity slightly, which facilitates its removal. Since it enlarges the final casting made, it is desirable that the original pattern dimension should be reduced to account for this increase. It is a negative allowance and is to be applied only to those dimensions that are parallel to the parting plane.

Draft or Taper Allowance: By draft is meant the taper provided by the pattern maker on all vertical surfaces of the pattern so that it can be removed from the sand without tearing away the sides of the sand mold and without excessive rapping by the molder.

Machining or Finish Allowance: The finish and accuracy achieved in sand casting are generally poor and therefore when the casting is functionally required to be of good surface finish or dimensionally accurate, it is generally achieved by subsequent machining. Machining or finish allowances are therefore added in the pattern dimension.

Shrinkage or Contraction Allowance: All most all cast metals shrink or contract volumetrically on cooling. The metal shrinkage is of two types:

Liquid Shrinkage: it refers to the reduction in volume when the metal changes from liquid state to solid state at the solidus temperature. To account for this shrinkage; riser, which feed the liquid metal to the casting, are provided in the mold.
Solid Shrinkage: it refers to the reduction in volume caused when metal loses temperature in solid state. To account for this, shrinkage allowance is provided on the patterns.

The phases present in pearlite are

((a))
cementite and austenite
((b))
austenite and ferrite
((c))
ferrite and cementite
((d))
martensite and ferrite

Show Answer

Answer: ((c))

ferrite and cementite

Explanation:

Pearlite is the eutectoid mixture of cementite and ferrite.

Iron carbon equilibrium diagram:

The phase diagram of Fe-Fe3C is not a true equilibrium because the iron carbide is an unstable phase that after prolonged heat treatment decomposes into iron and carbon( the graphite form).
The study is simplified by within assumption that sufficient time has been allowed at each new temperature for any necessary adjustment in phase compositions and relative amounts.

Phase components:

Carbon is an interstitial impurity in iron and forms a solid solution in alpha and delta phases. Different phase components are as follows:

Pearlite:

Microstructure for the eutectoid steel slowly cooled through the eutectoid temperature consist of alternating lamellae of the two phases alpha and Fe3C.
Carbon atoms diffuse away from the 0.22% ferrite, extending from the grain boundaries into unreacted austenite grain.
The resulting structure is called pearlite because it has the appearance of the mother of pearl.

Additional InformationFerrite:

Pure iron contains 0.008% weight of carbon.
Upon heating, pure iron experiences a change in crystal structure before it melts.
The stable form at room temperature is called ferrite.
Alpha iron has a BCC crystal structure.
This phase is relatively soft can be made magnetic at a temperature below 768° C and has the maximum possible density of 7880 kg/m3.

Austenite:

Ferrite experiences a polymorphic transformation to FCC austenite (gamma iron) at 912°C.
The austinite persists up to 1394°C at which it becomes BCC-phase known as delta-iron.
Upon further heating, it melts at 1538°C.
The maximum solubility of carbon in gamma iron is 2.14% approximately 100 times the maximum solubility in the BCC ferrite phase because FCC interstitial positions are larger.
Austenite is a non-magnetic phase.

Cementite:

A very hard and brittle phase is formed when carbon exceeds 6.67%.
It has two phases eutectic and eutectoid.

Which of the following beam(s) is/are statically indeterminate?

((a))
((b))
((c))
((d))

Show Answer

Answer: ((b))

Explanation:

Statically indeterminate structures are those structures that cannot be analyzed using statics or equations of equilibrium. In such cases, the number unknowns exceeds the number of equilibrium equations available.

Checking each option:

For option (1),

Number of unknown = 3

Number of equilibrium equation = 3

For option (2),

Number of unknown = 3

Number of equilibrium equation = 2

For option (3),

Number of unknown = 2

Number of equilibrium equation = 2

For option (4),

Number of unknown = 2

Number of equilibrium equation = 2

So, (2) is statically indeterminate structures.

A heat pump (HP) is driven by the work output of a heat engine (HE) as shown in the figure. The heat engine extracts 150 kJ of heat from the source at 1000 K. The heat pump absorbs heat from the ambient at 280 K and delivers heat to the room which is maintained at 300 K. Considering the combined system to be ideal, the total amount of heat delivered to the room together by the heat engine and heat pump is ________ kJ. [Answer in integer]

A flat surface of a C60 steel having dimensions of 100 mm (length) × 200 mm (width) is produced by a HSS slab mill cutter. The 8-toothed cutter has 100 mm diameter and 200 mm width. The feed per tooth is 0.1 mm, cutting velocity is 20 m/min and depth of cut is 2 mm. The machining time required to remove the entire stock is _________ minutes. [Rounded off to 2 decimal places]

Let X be a continuous random variable defined on [0, 1] such that its probability density function f(x) = 1 for 0 ≤ x ≤ 1 and 0 otherwise. Let Y = log_e(X + 1). Then the expected value of Y is ______. [Rounded off to 2 decimal places]

The matrix $\left[\begin{array}{ll} 1 & a \ 8 & 3 \end{array}\right]$ (where a > 0) has a negative eigen value if a is greater than

((a))
$\frac{1}{8}$
((b))
$\frac{1}{4}$
((c))
$\frac{3}{8}$
((d))
$\frac{1}{5}$

Show Answer

Answer: ((c))

$\frac{3}{8}$

Explanation:

For a negative eigen value, the product of eigen values < 0

$\left[\begin{array}{ll} 1 & a \ 8 & 3 \end{array}\right]$

⇒ Determinant < 0

⇒ 3 - 8a < 0

⇒ 8a > 3

⇒ a > $\frac{3}{8}$

A company orders gears in conditions identical to those considered in the economic order quantity (EOQ) model in inventory control. The annual demand is 8000 gears, the cost per order is 300 rupees, and the holding cost is 12 rupees per month per gear. The company uses an order size that is 25% more than the optimal order quantity determined by the EOQ model. The percentage change in the total cost of ordering and holding inventory from that associated with the optimal order quantity is

((a))
0
((b))
5
((c))
12.5
((d))
2.5

Show Answer

Answer: ((d))

2.5

Explanation:

Given: D = 8000 unitsIyr, C₀ = Rs 300Iorder, C_h = Rs 12 × 12 = Rs. 144 per year

Now, $\mathrm{EOQ}=\sqrt{\frac{2 C_{D} D}{C_{h}}}=\sqrt{\frac{2 × 300 × 8000}{144}}$

∴ Q* = 182.5 units

Now, Actual quantity, Q = 1.25Q* = 1.25 × 182.5

= 228.125 units

Total cost at EOQ, TIC₁ = $\sqrt{2 C_{o} C_{h} D}=\sqrt{2 × 300 × 144 × 8000}$ = Rs 26290.68

Now, Total cost at Q = 228.125 units

$\mathrm{TIC}{2}=\frac{Q}{2} × C{h}+\frac{D}{Q} × C_{o}$

= $(\frac{228.125}{2} × 144)+(\frac{8000}{228.125} × 300)$

= Rs. 26945.54

∴ % increase = $\frac{26945.54-26290.68}{26290.68} × 100=2.5 \%$

Alternate Method

We know, if Q = kQ*

then, $\frac{T I C(Q)}{T I C\left(Q^{*}\right)}=\frac{1}{2}\left[k+\frac{1}{k}\right]$

∴ Percentage increase,

$\frac{T I C(Q)-T I C\left(Q^{\star}\right)}{T I C\left(Q^{}\right)} × 100=\left(\frac{T I C(Q)}{T I C\left(Q^{}\right)}-1\right) × 100$

= $\left(\frac{1}{2}\left[k+\frac{1}{k}\right]-1\right) × 100=\left(\frac{1}{2}\left[1.25+\frac{1}{1.25}\right]-1\right) × 100$

= 0.025 × 100

= 2.5 %

The Levai type-A train illustrated in the figure has gears with module m = 8 mm/tooth. Gears 2 and 3 have 19 and 24 teeth respectively. Gear 2 is fixed and internal gear 4 rotates at 20 rev/min counter-clockwise. The magnitude of angular velocity of the arm is ________ rev/min. (rounded off to 2 decimal places)

If x(t) satisfies the differential equation

$t \frac{d x}{d t}+(t-x)=0$

subject to the condition x(1) = 0, then the value of x(2) is ________. (rounded off to 2 decimal places)

A piston-cylinder arrangement shown in the figure has a stop located 2 m above the base. The cylinder initially contains air at 140 kPa and 350°C and the piston is resting in equilibrium at a position which is 1 m above the stops. The system is now cooled to the ambient temperature of 25°C. Consider air to be an ideal gas with a value of gas constant R = 0.287 kJ/kgK.

The absolute value of specific work done during the process is ________ kJ/kg. (rounded off to 1 decimal place)

Aluminium is casted in a cube-shaped mold having dimensions as 20 mm × 20 mm × 20 mm. Another mold of the same mold material is used to cast a sphere of aluminium having a diameter of 20 mm. The pouring temperature for both cases is the same. The ratio of the solidification times of the cube-shaped mold to the spherical mold is _______. (answer in integer)

At the current basic feasible solution (bfs) vo (vo ∈ ℝ5), the simplex method yields the following form of a linear programming problem in standard form.

Minimize,

st.

z = -x1 - 2x2

x3 = 2 + 2x1 = x2

x4 = 7 + x1 = 2x2

x5 = 3 - x1

x1, x2, x3, x4, x5 ≥ 0

Here the objective function is written as a function of the non-basic variables. if the simplex method moves to the adjacent bfs v1(v1 ∈ ℝ5) that best improves the objective function, which of the following represents the objective function at v1, assuming that the objective function is written in the same manner as above?

((a))
z = -4 - 5x1 + 2x4
((b))
z = -4 - 5x1 + 2x3
((c))
z = -6 - 5x1 + 2x3
((d))
z = -3 - x5 - 2x2

Show Answer

Answer: ((a))

z = -4 - 5x1 + 2x4

Explanation:

z = -x1 - 2x2

x3 = 2 + 2x1 = x2 ...(i)

x4 = 7 + x1 = 2x2 ...(ii)

x5 = 3 - x1 ...(iii)

x1, x2, x3, x4, x5 ≥ 0 ...(iv)

From equation (i);

x2 = 2 + 2x1 - x3

On substituting this value in the objective function,

z = -x1 - 2(2 + 2x1 - x3)

⇒ z = -4 - 5x1 + 2x3

Hence option (2) is correct.

Also from equation (iii);

x1 = 3 - x5

On substituting this value in the objective function,

z = -(3 - x5) - 2x2

z = −3 + x5 - 2x2

Hence option (4) is correct.

As per the given options, (2) and (4) both are correct.

A three-hinge arch ABC in the form of semi-circle is shown in the figure. The arch is in static equilibrium under vertical loads of P = 100 KN and Q = 50 kN. Neglect friction at all the hinges. The magnitude of the horizontal reaction at B is _______ KN. (rounded off to 1 decimal place)

A vibratory system consists of mass m, a vertical spring of stiffness 2k and a horizontal spring of stiffness k. The end A of the horizontal spring is given a horizontal motion x_A = a sin ωt. The other end of the spring is connected to an inextensible rope that passes over two massless pulleys as shown. Assume m = 10 kg, k = 1.5 kN/m, and neglect friction. The magnitude of critical driving frequency for which the oscillations of mass m tend to become excessively large is ________ rad/s. (answer in integer)

The figure shows a thin cylindrical pressure vessel constructed by welding plates together along a line that makes an angle α = 60° with the horizontal. The closed vessel has a wall thickness of 10 mm and diameter of 2 m. When subjected to an internal pressure of 200 kPa, the magnitude of the normal stress acting on the weld is. MPa. (rounded off to 1 decimal place)

A liquid fills a horizontal capillary tube whose one end is dipped in a large pool of the liquid. Experiments show that the distance L travelled by the liquid meniscus inside the capillary in time t is given by

$L=k \gamma^{\alpha} R^{b} \mu^{c} \sqrt{t}$

where γ is the surface tension, R is the inner radius of the capillary, and μ is the dynamic viscosity of the liquid. If k is a dimensionless constant, then the exponent a is _______ (rounded off to 1 decimal place)

A solid massless cylindrical member of 50 mm diameter is rigidly attached at one end, and is subjected to an axial force P = 100 KN and a torque T = 600 Nm at the other end as shown. Assume that the axis of the cylinder is normal to the support. Considering distortion energy theory with allowable yield stress as 300 MPa, the factor of safety in the design is ________ (rounded off to 1 decimal place)

In an arc welding process, the voltage and current as 30 V and 200 A, respectively. The cross-sectional area of the joint is 20 mm² and the welding speed is 5 mm/s. The heat required to melt the material is 20 J/s. The percentage of heat lost to the surrounding during the welding process is ________ (rounded off to 2 decimal places)

Consider an air-standard Brayton cycle with adiabatic compressor and turbine, and a regenerator, as shown in the figure. Air enters the compressor at 100 kPa and 300 K and exits the compressor at 600 kPa and 550 K. The air exits the combustion chamber at 1250 K and exits the adiabatic turbine at 100 kPa and 800 K. The exhaust air exits the regenerator (state 6) at 600 K. There is no pressure drop across the regenerator and the combustion chamber. Also, there is no heat loss from the regenerator to the surroundings. The ratio of specific heats at constant pressure and volume is c_p/c_v = 1.4. The thermal efficiency of the cycle is ________ %. (answer in integer)

In a supplier-retailer supply chain, the demand of each retailer, the capacity of each supplier, and the unit cost in rupees of material supply from each supplier to each retailer are tabulated below. The supply chain manager wishes to minimize the total cost of transportation across the supply chain.

	Retailer I	Retailer II	Retailer III	Retailer IV	Capacity
Supplier A	11	16	19	13	300
Supplier B	5	10	7	8	300
Supplier C	12	14	17	11	300
Supplier D	8	15	11	9	300
Demand	300	300	300	300

The optimal cost of satisfying the total demand from all retailers is ________ rupees. (answer in integer)

Consider a slab of 20 mm thickness. There is a uniform heat generation of q = 100 MW/m³ inside the slab. The left right faces of the slab are maintained at 150°C and 110°C, respectively. The plate has a constant thermal conductivity of 200 W/(mk). Considering a 1-D steady state heat conduction, the location of the maximum temperature from the left face will be at _______ mm. (answer in integer)

Steady, compressible flow of air takes place through an adiabatic converging-diverging nozzle, as shown in the figure. For a particular value of pressure difference across the nozzle, a stationary normal shock wave forms in the diverging section of the nozzle. If E and F denote the flow conditions just upstream and downstream of the normal shock, respectively, which of the following statement(s) is/are TRUE?

((a))
Mach number at E is lower than the Mach number at F
((b))
Density at E is lower than the density at F
((c))
Specific entropy at E is lower than the specific entropy at F
((d))
Static pressure at E is lower than the static pressure at F

Show Answer

Answer: ((a))

Mach number at E is lower than the Mach number at F

Explanation:

After normal shock:

• Stagnation temperature remains the same.

• Density after shock will increase, i.e., ρE < ρF

• Mach number will decrease, i.e., ME > MF.

• Entropy across a shockwave always increases as it is a highly irreversible process.

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A blanking operation is performed on C20 steel sheet to obtain a circular disc having a diameter of 20 mm and a thickness of 2 mm. An allowance of 0.04 is provided. The punch size used for the operation is _________ mm (rounded off to 2 decimal places)

A band brake shown in the figure has a coefficient of friction of 0.3. The band can take a maximum force of 1.5 kN. The maximum braking force (F) that can be safely applied is _______ N (rounded off to the nearest integer)

A horizontal beam of length 1200 mm is pinned at the left end and is resting on a roller at the other end as shown in the figure. A linearly varying distribution load is applied on the beam. The magnitude of maximum bending moment acting on the beam is Nm. (round off to 1 decimal place)

At the instant when OP is vertical and AP is horizontal, the link OD is rotating counter clockwise at a constant rate ω = 7 rad/s. Pin P on link OD slides in the slot BC of link ABC which is hinged at A, and causes a clockwise rotation of the link ABC. The magnitude of angular velocity of link ABC for this instant is _______ rad/s. (rounded off to 2 decimal places)

A condenser is used as a heat exchanger in a large steam power plate in which steam is condensed to liquid water. The condenser is a shell and tube heat exchanger which consists of 1 shell and 20000 tubes. Water flows through each of the tubes at a rate of 1 kg/s with an inlet temperature of 30°C. The steam in the condenser shell condenses at the rate of 430 kg/s at a temperature of 50°C. If the heat of vaporization is 2.326 MJ/kg and specific heat of water is 4 kJ/(kg.K), the effectiveness of the heat exchanger is _________ (rounded off to 3 decimal places)

A cutting tool provides a tool life of 60 minutes while machining with the cutting speed of 60 m/min. When the same tool is used for machining the same material, it provides a tool life of 10 minutes for a cutting speed of 100 m/min. If the cutting speed is changed to 80 m/min for the same tool and work material combination, the tool life computed using Taylor's tool life model is ________ minutes. (rounded off to 2 decimal places)

If the value of the double integral

$\int_{x=3}^{4} \int_{y=1}^{2} \frac{d y d x}{(x+y)^{2}}$

is $log_e(\frac{a}{24})$ , the a is ________ (answer in integer)

In the pipe network shown in the figure, all pipes have the same cross-section and can be assumed to have the same friction factor. The pipes, connecting points W, N, and S with point J have an equal length L. The pipe connecting points J and E has a length 10L. The pressure at the ends N, E, and S are equal. The flow rate in the pipe connecting W and J is Q. Assume that the fluid flow is steady, incompressible, and the pressure losses at the pipe entrance and junction are negligible. Consider the following statements:

I. The flow rate in pipe connecting J and E is Q/21.

II. The pressure difference between J and N is equal to the pressure difference between J and E.

Which one of the following option is CORRECT?

((a))
I is False and II is True
((b))
I is True and II is False
((c))
Both I and II are False
((d))
Both I and II are True

Show Answer

Answer: ((d))

Both I and II are True

Explanation:

P_N- P_J = P_E - P_J

⇒ $\frac{P_{N}-P_{J}}{\rho g}=\frac{P_{E}-P_{J}}{\rho g}$

⇒ $\left(h_{f}\right){N J}=\left(h{f}\right)_{E J}$

⇒ $\frac{32 \mu v_{1} L}{\rho g d}=\frac{32 \mu v_{2}(10 L)}{\rho g d}$

⇒ v₁ = 10v₂

Using continuity equation,

Q = 2Q₁ + Q₂

⇒ $\frac{\pi}{4} d^{2} v=\frac{\pi}{4} d^{2}\left[2 v_{1}+v_{2}\right]$

⇒ v = 2v₁+ v₂

Using equation (i) and (ii)

$v_{2}=\frac{1}{21} v \text { and } v_{1}=10 v_{2}=\frac{10}{21} v$

Discharge, $Q_{2}=\frac{\pi}{4} d^{2} v_{2}=\frac{\pi}{4} d^{2}\left(\frac{1}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right) \frac{1}{21}=\frac{Q}{21}$

Similarly, $Q_{1}=\frac{\pi}{4} d^{2} v_{1}=\frac{\pi}{4} d^{2}\times\left(\frac{10}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right)\times \frac{10}{21}=\frac{10 Q}{21}$

Now, clearly $Q_{2}=\frac{Q}{21}$

Hence, statement I is correct.

and (Pressure)_N = (Pressure)_E

⇒ P_N - P_E

⇒ P_J - P_N = P_J - P_E

Hence, the pressure difference between J and N is equal to the pressure difference between J and E.

So, statement II is true.

Consider a hemispherical furnace of diameter D = 6 m with a flat base. The dome of the furnace has an emissivity of 0.7 and the flat base is a blackbody. The base and the dome are maintained at uniform temperature of 300 K and 1200 K, respectively. Under steady state conditions, the rate of radiation heat transfer from the dome to the base is kW. (rounded off to the nearest integer).

Use Stefan-Boltzmann constant = 5.67 × 10^-8W(m²K⁴)

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