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This mock test of GATE Mock Test Mechanical Engineering (ME) - 9 for GATE helps you for every GATE entrance exam.
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QUESTION: 1

Direction: Study the following pie-chart and tables carefully and answer the questions given below.

Total Number of email received by the organization = 90000

Ratio of Read emails to Unread emails received by the organization

What is the ratio of the number of emails read in January to those unread in the month of April in the organization?

Solution:

Number of read emails in the month of January = (90000*17/100*8/15)

Number of unread emails in the month of April = (90000* 8/100 * 5/12)

Ratio = 68:25

QUESTION: 2

Direction: Two sentences with one blank in each, followed by five alternatives, are given. Choose that option as the answer which can fill both the blanks of both the sentences.

I. Most of our employees get _____ abroad at some stage

II. The aircraft and its crew were ______ missing.

Solution:

Post** **means to send somebody to a place for a period of time as part of their job and to announce something publicly or officially.

Hence, option D is correct.

QUESTION: 3

In the following question, out of the given four alternatives, select the one which is opposite in the meaning of the given word.

__Scurrilous__

Solution:

The meanings of the given words are as follows:

Scurrilous— abusive

Coarse— rough or harsh in texture

Sophisticated— developed to a high degree of complexity.

Insolent— showing a rude and arrogant lack of respect

Complimentary— expressing a compliment; praising or approving

Clearly, option D is the antonym of the given word.

QUESTION: 4

**Statements:**

All pillows are beds.

No fruit is pillow.

Some foods are fruits.

**Conclusions:**

I. At least some foods are pillows.

II. Some beds are definitely fruits.

Solution:

Consider the following least possible Venn diagram,

**Conclusions:**

I. At least some foods are pillows → it’s possible but not definite, hence false.

II. Some beds are definitely fruits → it’s possible but not definite, hence false.

QUESTION: 5

**Direction: **Study the following information carefully and answer the questions given below:

P, Q, R, S, T, U, V and F are sitting around a circle facing the centre. U is third to the right of Q, who is third to the right of F. P is third to the left of F. R is fourth to the left of P. T is third to the right of S. S is not a neighbour of P.

Four of the following five are similar in a certain way based on their positions in the seating arrangement and so form a-group. Which of the following does not belong to that group?

Solution:

Option D does not belong the group.

QUESTION: 6

Following bar graphs shows time taken by different pipes to fill that particular percentage of tank which is described in tabular data. For Ex- Pipe 1 fills the 10 % of tank in 6 mins.

Now, read the following data and answer the question carefully:

A Tank has three pipes; two of them are used to fil the tank while another one used to empty the tank namely: pipe 3, pipe 5 to fill the tank and a third pipe for making the tank empty. When all three pipes are open, 7/18^{th} part of the tank is filled in 1 hours. How much time will the third pipe take to empty the completely filled tank 10 times the bigger than the previous one.

Solution:

part of work in one hour=

(P3+P5+Third pipE. X 1 = 7/18^{th} part of work

⇒ 7/18^{th} part of work = 1 hr

⇒ Total work done = 18/7 hr by all three pipes

Pipe 3 does 50% of work in 90 mins

100 % of work = 90/50 * 100 = 180 mins = 3 hr

Pipe 5 does 50% of work in 60 mins

100 % of work = 60/50 * 100 = 120 mins = 2 hr

Time Tw Efficiency

Pipe -3 = 3 hr Total work/3 = 6

Pipe-5 = 2 hr Total work/2 = 9

All three = 18/7 hr Total work/(18/7)= 7

__For Total work = LCM of (3,2,18/7) = 18__

Efficiency of pipe 3 + Efficiency of pipe 5 + Efficiency of third pipe = total efficiency

6+9+X= 7

X = 7-15 = -8 (Negative sign shows the negative work herE.

Efficiency X Time = Total work

8 X Time = 18 * 10 ( as tank is 10 times bigger than the previous onE.

Time = 180/8 = 22.5 hrs

QUESTION: 7

**Directions:** Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph: then answer the questions given below them.

A). Indian Government remain concerned about potential spillover effects from the unconventional monetary policies of the advanced economies, which could cause disruptive volatility of exchange rates, asset prices and capital flows.

B). The challenges are related to high public debt and unemployment, poverty and inequality, lower investment and trade, negative real interest rates along with signs of prolonged low inflation in advanced economies.

C). In this context, emerging markets and developing countries (EMDCs) continue to be major drivers of global growth.

D). The global recovery continues, although growth remains fragile, with considerable divergences across countries and regions.

E). It is important to strengthen the framework of international financial cooperation, including through instruments such as swap-lines, to mitigate the negative impacts of monetary policy divergence in reserve currency issuing countries.

F). Structural reforms, domestic adjustment and promotion of innovation are important for sustainable growth and provide a strong and sustainable contribution to the world economy.

Which of the following will be the Second sentence?

Solution:

Refer to the last question of the series.

QUESTION: 8

In a school, 12th class consists of 30% male students of which 30% male students failed in the class. Total 82% students passed in 12th examination out of 900 students. Calculate the total number of female passed students?

Solution:

**Short Trick:**

Total students = 900

Boys : Girls = 30 % : 70% = 3 : 7

So, exactly Boys : Girls = 270 : 630

Total failed students = 18% of 900 = 162

Failed boys = 30% of 270 = 81

Failed girls = 162 - 81 = 81

Passed female students = 630 - 81 =**549
Basic Method:**

⇒ Let us consider total number of students be 100x

⇒ ∴ Males students = 30x , female students = 70x

⇒ According to condition given in the problem, of this 30x male students, 70 % male students pass the exam.

QUESTION: 9

**Direction:** In the following number series, only one number is wrong. Find out the wrong number.

17 25 34 98 121 339

Solution:

The pattern followed here is,

⇒ 17 + 2^{3} = 17 + 8 = 25

⇒ 25 + 3^{2} = 25 + 9 = 34

⇒ 34 + 4^{3} = 34 + 64 = 98

⇒ 98 + 5^{2} = 98 + 25 = 123

⇒ 123 + 6^{3} = 123 + 216 = 339

QUESTION: 10

There are 3 red balls and 5 green balls in a bag. If two balls are picked at random from the bag then find the probability of getting a red ball?

Solution:

Required probability

=^{3}C_{1}*^{5}C_{1}/^{8}C_{2}

= 3*5/28

= 15/28

*Answer can only contain numeric values

QUESTION: 11

In the parallel-plate configuration shown, steady-flow of an incompressible Newtonian fluid is established by moving the top plate with a constant speed, U_{0} = 1m/s. If the force required on the top plate to support this motion is 0.75 N per unit area (in m2 ) of the plate then the viscosity of the fluid between the plates is __________ N-s/m2

Solution:

*Answer can only contain numeric values

QUESTION: 12

The figure shows reducing area conduit carrying water. The pressure p and velocity V are uniform across sections 1 and The density of water is 1000 kg/m^{3} . If the total loss of head due to friction is just equal to the loss of potential head between the inlet and the outlet, then V2 in m/s will be _____

Solution:

QUESTION: 13

The stress-strain curve of an ideal elastic strain hardening material will be as

Solution:

The figure d is the stress-strain curve of an ideal elastic strain hardening material

QUESTION: 14

The strain hardening exponent ‘n’ of stainless steel relates to

N →column-1)

P) Zero

Q) between zero and one

R) One

Column-2

1) Plastic deformation

2) elastic solid

3) Perfect moldable solid

Solution:

For n = 0 → Perfect moldable solid

O < n< 1 → Plastic deformation

n = 1 → elastic solid

QUESTION: 15

Mohr’s circle for the state of stress defined by MPa. is a circle with radius of

Solution:

*Answer can only contain numeric values

QUESTION: 16

Water flows through a tube of diameter 25mm at an average velocity 1 m/sec. The properties of water ρ = 1000 kg/m^{3}, μ = 6x 10^{-4} N-s/m^{2} , K= 0.625 W/mK, P_{r}= 4, Given Dittus – Boelter equation, N_{u} =0.023R_{e}^{0.8} P_{r}^{n} and constant heat flux N_{u} = 3.66. What is Nusset number when fluid is cooling?

Solution:

Given, P_{r} = 4, fluid is cooling so, n= 0.3

So the flow is turbulent, we have to use Dittus- Boelter equation i.e

*Answer can only contain numeric values

QUESTION: 17

A machine tool company decides to make four sub-assemblies through three contractors differing in efficiency. Given below are the estimates of the time in hours) matrix. The minimum time in seconds is?

Solution:

Here, we have three contractors to make four sub-assemblies. The problem is unbalanced, we have to add dummy column to make it balanced.

Subtracting by the smallest element of each column.

S_{1 }→ M_{1}

S_{2 }→ M_{3}

S_{3 }→ M_{2
}

The minimum time is = 7+4+18 = 29 hour = 1740 seconds

*Answer can only contain numeric values

QUESTION: 18

The average service rate at a Service Centre is 8 customer/hr, which varies exponentially. Customers arrive at random order with an average of 6 customer/hr. The probability of at least 3 customers in the system is____

Solution:

At least 3 in system means 3 or more customers in system,

QUESTION: 19

In an Electrochemical machining process an iron( atomic weight=56 gm and valency=2) is used having cross section of 0.5x0.5 cm2. A current of 1000 A is passed through the electrolytic solution and the density of iron is given as 7.86gm/cm3. Find the feed rate of the electrode in cm/sec.

Solution:

QUESTION: 20

A mould has downsprue whose length is 15cm and cross-section area at the base of downsprue is 1.5 cm^{2}. The downsprue feeds a horizontal runner leading into mould cavity of 1200 cm^{2}. The time required to fill the mould cavity will be

Solution:

Length of sprue h= 15 cm = 0.15 m

QUESTION: 21

Match the following:

a) Drop forging i-material is only upset

b) Machine forging ii-repeated hammer blows

c) Press forging iii-carried out manually

d) Upset forging iv-squeezing action

e) Smith forging v-metal gripped & pressure applied at heated end

Solution:

In Drop forging repeated hammer blows, for machine forging material is only upset, in press forging squeezing action takes place.

*Answer can only contain numeric values

QUESTION: 22

The maximum reduction (upto 2 decimal points) in cross-sectional area per pass (R) of a cold wire drawing process is R = 1 - e ^{-(n+1) }

where n represents the strain hardening coefficient. For the case of a perfectly plastic material, R is

Solution:

Perfectly plastic material implies that strain hardening coefficient n is zero.

Hence

*Answer can only contain numeric values

QUESTION: 23

In spot welding of two stainless steel sheets, each of 5mm thickness, welding current 15000A is applied for 0.4sec, the heat dissipated to the base metal is 22000J . Assuming that the heat required for melting 2mm^{3} volume steel is 30J and contact resistance between sheets is 0.0004Ω. The volume (in mm^{3}) of weld nugget is__________.

Solution:

Total heat generated = I^{2}RT

= (15000)^{2 }x 0.0004 x 0.4

=36000

Heat required to melt the nugget = 36000 -22000

=14000J

Volume of nugget = 14000/15 = 933.33 mm^{3}

QUESTION: 24

Which of the following is false about comparator :

Solution:

Comparators are more reliable as compare to other measuring method.

QUESTION: 25

Which types of stress strain relationship best describes the behaviour of brittle materials as ceramics and thermosetting plastics?

Solution:

Ceramics and thermosetting plastics does not undergo plastic strain since they are brittle materials. The stress strain curve reached to the elastic point and then material fails.

QUESTION: 26

Gyroscopic acceleration is due to

Solution:

Gyroscopic acceleration is due to the change in the direction of angular velocity.

So, the correct option is (b).

*Answer can only contain numeric values

QUESTION: 27

A paddle wheel is rotated inside a sealed container containing 20kg of air. If the applied torque on paddle wheel is 80Nm and is rotated at 120rpm for 3 minutes. The temperature rise of air is ________ ℃

Solution:

Using energy balance,

QUESTION: 28

According to second law of thermodynamics, the entropy change of the system for any process can be

Solution:

We know that

is always a positive value and 0 for reversible process.

can be +ve or –ve depending upon heat addition and heat release respectively. So, the correct option is (d).

QUESTION: 29

The Mollier diagram of a pure substance is a

Solution:

The Mollier diagram of a pure substance is a h-s diagram. So, the correct option is (a).

QUESTION: 30

Match List-1 (Refrigerant) with List-2 (Uses) and select the correct answer using the code given below the List

Solution:

So, the correct option is (c).

*Answer can only contain numeric values

QUESTION: 31

In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 10^{0}C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 25^{0}C in the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .

Solution:

Bell Coleman cycle is also known as Reversed Brayton cycle

QUESTION: 32

Find the minimum number of stages required when air enters at 1 bar and 27^{0}C and delivers at 260 bar. The maximum temperature at the outlet of compressor is limited to 327^{0}C . Take the perfect intercooling in between stages and γ = 1.3.

Solution:

In multistage compression, if there are N stages then pressure ratio in each stage will be equal for perfect intercooling.

From this we can write ,

Now, for perfect intercooling the inlet temperature for each compressor will be same. And as we know that pressure ratio will be same for each stage. So for perfect intercooling case we can write:

Solving , we get N = 1.85 ~ 2

So, the correct option is (a).

QUESTION: 33

The free body diagram of member AB is :-

Solution:

Free body diagram of member DC, BC and AB is shown below :Therefore, correct option is (a).

QUESTION: 34

Consider the function f = |5-4x| and refer to the following options.

Solution:

Since the function is polynomial in both the cases, it is continuous.

Left Hand Derivative=-4

Right hand Derivative= +4

Since LHD≠RHD, the function is not differentiable at x=5/4.

QUESTION: 35

The value of integral

Solution:

*Answer can only contain numeric values

QUESTION: 36

A small boat is floating in a small pool as shown in the figure, initially the valve is closed, find the time taken (in seconds) by water to spill out of the pool as soon as the valve is opened ?

Solution:

Apply continuity equation

*Answer can only contain numeric values

QUESTION: 37

A tank contains water up to a height of 1m and oil of specific gravity 0.9 is filled on the top of water up to 1 m height as shown in figure). Calculate the net force on the gate. (in kN)

Solution:

Pressure at the top end of the gate.

= 0.9 × 9.81×1 =8.829 kPa.

Pressure at the lower end of the gate.

= 0.9 × 9.81×1 + 1 × 9.81 × 1 = 18.639 kPa.

QUESTION: 38

A lawn sprinkler has 0.6 cm diameter nozzle at the end of the rotating arm and discharges water at the rate of 10m/sec velocity. Determine the torque required to hold the rotating arm stationary

Solution:

Rate of change for Angular momentum for arm ‘A’ =

Rate of change forAngular momentum for arm ‘B’ =

Net Rate of change of angular momentum = net Torque.

QUESTION: 39

Calculate the mean bucket speed(m/s) of a Pelton wheel working under a head of 52m and producing 320kW. The diameter of nozzle is 178mm and coefficient of velocity is 0.98. The jet is deflected by 165 °.

Solution:

QUESTION: 40

A rigid bar ABC is supported by three rods of same material and cross section, outer rods are of copper and inner rod is of steel →as shown in figure). Calculate the forces in the rod due to an applied load 400kN, If the ABC remain horizontal after the load has been applied →neglect wt. of rods)

Solution:

2P_{cu} + P_{s} =400 kN …→1)

Since rods are fixed to rigid bar and bar remain horizontal after the load has been applied, so deformation in rod will be same.

⇒ from →1) and →2) P_{s} = 200kN P_{cu} = 100 kN

*Answer can only contain numeric values

QUESTION: 41

The circular shaft of constant cross- section carries four pulleys and belts which are vertical. Calculate and Un Known tension on the belt as shown in figure.

Solution:

Shaft is in equilibrium, Hence ∑T = 0

T_{1} + T_{2} +T_{3} + T_{4} = 0

T_{1} = →8-6) 0.1 = 0.4 kNm

T_{2} = →2P-3) 0.05 = →0.1P – 0.15) kNm

T_{3} = →2-P) 0.05 = →0.1 – 0.05P) kNm

T_{4} = →4-8) 0.1 = -0.4 kN-m

T_{2} = →2P-3) 0.05 = →0.1P – 0.15) kNm

⇒ 0.4 + 0.1P – 0.15 + 0.1 –0.05P – 0.4 =0 ⇒ P = 1 kN

QUESTION: 42

Find the vertical reaction at the point A as shown in figure.

Solution:

Given Beam

R_{A} + R_{C} = wL 1

Now from superposition we will break this beam into two i.e.

QUESTION: 43

The inlet temperature of hot water is 800K and outlet temperature is 400 K and the inlet temperature of cold water is 200 K, both are flowing at some mass flow rate. Taking counter flow heat exchanger the NTU and LMTD respectively are

Solution:

Here, this is a case of balanced heat exchanger

M_{h}C_{h} =m_{c}C_{c} = C [Both the fluids are water]

NTU = 2

*Answer can only contain numeric values

QUESTION: 44

Two products A & B are manufactured by two machines. Capacity of machines are limited to 18 hours and 40 hours per week. The requirement for product A & B on each machinery from the table

The profit from each unit of A & B is 30 Rs and 20 Rs per unit respectively. The optimum profit in Rs is?

Solution:

Z = 30A + 20B Maximize

Constraints

⇒3A + 2B ≤ 18 →1)

⇒4A + 8B ≤ 40 →2)

Z_{p}= 30x0+ 20x0 = 0 Rs

Z_{Q} = 30x0 + 20x5 = 100 Rs

Z_{R} = 30x4 + 20x3 = 180 Rs

Z_{S} = 30x6 + 20x0 = 180 Rs

Optimality occurs at R & S

Cost = 180 Rs

QUESTION: 45

Annual demand of a product A is 4000 units. Cost of order is Rupees 900 per order. Handling charge per year is 10% of the unit price. Item can be purchased in lot as shown below.

The feasible EOQ should be

Solution:

Staring from the lowest cost available,

Since it is not feasible, moving to next

Again not feasible, moving to next

This lies in feasible range. Total cost at this EOQ,

It can be seen that minimum cost is obtained at Order Quantity of 1000 units hence our feasible EOQ is 1000 units.

*Answer can only contain numeric values

QUESTION: 46

Job A, B, C, D are done on two machines. The percentage utilization of machine 2 is?

Solution:

Optimum Sequence: - A-D-B-C

Make span time = 23 hours

Idle time = 2+1+6 = 9

= 60.8%

*Answer can only contain numeric values

QUESTION: 47

There is a rotating bar made up of steel (S_{ut} = 650 MPA. which is subjected to a completely reversed bending stress. The corrected endurance limit of the bar is 300 MPa. The fatigue strength of the bar for a life of 1,00,000 cycles will be _____ MPa

Solution:

Given:

S_{ut} = 650 MPa

Corrected endurance limit , S_{e} = 300 MPa

N = 1,00,000 cycles

We will draw the S-N diagram for it:

log_{10}(0.9S_{ut}) = log_{10}(585) = 2.7672

log_{10}(S_{e}) = log_{10}(300) = 2.4771

log_{10}(N) = log_{10}(100000) = 5

Fatigue strength for 1,00,000 cycles:

Equation of line AB:

Therefore the fatigue strength of the bar will be 468.274 MPa.

QUESTION: 48

Consider a single plate clutch, consisting of only one pair of connecting surfaces. It develops a maximum torque of 150 N-m. The permissible intensity of pressure is 400 kPa and the coefficient of friction is 0.3. Assume uniform wear theory and the ratio of inner to outer diameters of the friction lining is 0.577. The inner and outer radius of the friction lining will be:

Solution:

Let the inner diameter be ‘d’ and outer diameter be ‘D’.

Given :

M_{t} = 150 N-m = 150000 N-mm

Permissible intensity of pressure i.e., Maximum

pressure, P = 400 kPa = 0.4 N/mm^{2}

µ = 0.3

In uniform wear theory case Pressure is inversely proportional to radius, therefore maximum pressure will occur at inner radius.

For Uniform Wear Theory,

Putting all values in equation (i) we will get

D = 202.35 mm, so outer radius = 101.175 mm

d = 116.75 mm, so inner radius = 58.375 mm

Therefore, correct option is (c).

*Answer can only contain numeric values

QUESTION: 49

There is a riveted joint which consist of four identical rivets. It is subjected to an eccentric force of 50 kN as shown in figure. If the permissible shear stress of rivet is 60 MPa, then the minimum diameter of rivets will be _____ mm.

Solution:

Firstly we will locate the centre of gravity of the group of rivets.

Here ‘O’ is centre of gravity

Now, we will take 2 equal and opposite forces passing through centre of gravity (P_{1} and P_{2}).

Here P_{1} = P_{2} = P = 50 kN

P_{1} will act equally on all rivets in the same direction of applied load and P_{2} and P causes a twisting couple in clockwise direction. Due to P_{1} there will be a primary shear force acting on each rivet and due to twisting couple there will be a secondary shear force acting on each rivet.

Primary shear force acting on each rivet

P_{2} and P causes a twisting couple of magnitude, T = P × e = 50 × 200 = 10000 kN-mm

And this twisting couple will lead to a secondary shear force on rivet 1, 2 , 3 and 4 respectively as shown in figure:

Now,

And we know that

Now, Rivet 3 and Rivet 4 are critical rivets because when we will calculate the resultant shear force on each rivet then we will find it maximum for rivet 3 and 4. The angle between primary and secondary shear force is 45^{0}for rivet 3 and 4. The angle between rimary and secondary shear force is 135^{0} for rivet 1 and 2.

Putting A= 12.5 kN

B = 70.72 kN

And θ = 45^{0}

We will get R_{max} = 80.05 kN

Now , for safe condition:

Therefore minimum diameter of rivet is 41.22 mm.

*Answer can only contain numeric values

QUESTION: 50

The length L key required to transmit 500 Nm torque for 60mm diameter shaft. The key has a cross section of 20x10mm (Syt=Syc=240 MPa, FOS=4) is?

Solution:

Key Length → Shear design

So length of key should be 56mm

QUESTION: 51

In orthogonal turning of low carbon steel pipe, with shear strength of 300MPa, Following condition are taken.

Axial feed = 0.25 mm/rev

Radial depth of cut = 4mm

Chip thickness = 0.5mm

Cutting velocity = 150 m/min

Orthogonal rake angle 8^{0}

The shear force and MRR respectively are

Solution:

*Answer can only contain numeric values

QUESTION: 52

There is a vertical engine, having piston at the top and crank at the bottom, running at 2400 rpm. The length of crank and connectivity rod is 50 mm and 300 mm respectively. The diameter of the pistons is 100 mm and the mass of the reciprocating parts is 1.5 kg. The crank has turned 30^{0} from the top dead centre during the expansion stoke and the gas pressure is 600 kN/m^{2}. The net force on the piston is _____ N.

Solution:

Given:

Length of crank, r = 50 mm = 0.05 m

Length of connecting rod, l = 300 mm = 0.3 m

N = 2400 rpm

Gas pressure, P = 600 kN/m^{2}

Mass of reciprocating parts, m = 1.5 kg

Diameter of piston, d = 0.1 m

θ = 30^{0}

Now,

n = l/r = 0.3/0.05 = 6

Force due to gas pressure, F_{p} = Area x Pressure

= 4712.39 N

Net force on the piston, F

F = F_{p} – F_{b} + mg = 4712.39 – 4497.59 + 1.5 x 9.81 = 229.52 N

QUESTION: 53

For single degree of freedom, condition for getting peak (maxima) in "magnification factor" Vs "ω / ωn" curveThe damping factor should be less than

Solution:

Frequency response curves

As we know

Now, for having maxima

So, we will yet the peak only it our damping is less than 0.707

QUESTION: 54

Air enters an insulated diffuser operating at steady state with a pressure of 0.65 bar, temperature of 67^{0}C and a velocity of 300 m/s. The inlet area is 0.5 m^{2}. At the exit, the pressure is 1 bar and the area is 0.6 m^{2}. Potential energy changes can be neglected. Assume ideal gas behavior for air. The mass flow rate and temperature of the air at exit is

Solution:

P_{1} = 0.65 bar = 65 kPa

T_{1} = 67^{0}C = 340 K

V_{1} = 300 m/s

A_{1} = 0.5 m^{2}

R = 0.287 kJ/kgK

Cp = 1.008 kJ/kgK

P_{2} = 1 bar = 100 kPa

A_{2} = 0.6 m^{2}

Given Steady state

Now applying steady flow energy equation between 1 and 2:

Putting the value of V_{2} in equation (i), we will get

*Answer can only contain numeric values

QUESTION: 55

There are two blocks connected by a light string placed on the inclined parts of a triangular structure as shown in the figure. The coefficients of static friction is 0.3 at each of the surfaces. The minimum value of ‘M’ for which the system remains at rest is ___ kg.

Solution:

The tendency of the mass ‘M’ will be to slide in the upward direction for its minimum value.

We will draw free body diagram for both object.

QUESTION: 56

Which one of the following is an eigen vector of the matrix:-

Solution:

Firstly we will find its eigen values by solving its characteristics equation

|A - λI| = 0

Where A is given matrix and I is identity matrix[

now we will find the eigen vector corresponding to λ = -5

So, the correct option is (c).

QUESTION: 57

is a Partial differential equation whose solution is given by

The value of K is

Solution:

The objective here is to find the Particular integral as Complementary function is already given in the solution.

QUESTION: 58

The Laurent series for f(z) = that is valid in the region |z| > 10 is

Solution:

QUESTION: 59

In a batch of three students marks obtained in a certain test are 50, 60 and 70. The standard deviation of the marks obtained in test is

Solution:

Standard Deviation,

where n= 3 in this case.

QUESTION: 60

The surface integral of N ds over the sphere given by x^{2} + y^{2} + z^{2} = 4

Solution:

From Gauss – Divergence theorem, we can write:

= 33.51 (2a + 3b + 4c)

= 67a + 100.53b + 134.04c

So, the correct option is (a).

QUESTION: 61

Water flows over a flat plate at a free stream velocity of 0.15m/s. There is no pressure gradient and laminar boundary layer is 6mm thick. Assume velocity profile as for the flow conditions stated above, calculate the skin friction coefficient.

Solution:

*Answer can only contain numeric values

QUESTION: 62

A light bulb sphere shaped of diameter 80 mm is hanged inside a brick room of cube shaped of side as 0.4 m. The light bulb is at a temperature of 327 ^{0}C and the brick is at 27^{0}C. Take emissivity (Ԑ) of light bulb and cube brick as 1 and 0.8 respectively, Determine the rate of heat loss (in kW) by radiation _____________ (upto two decimal).

Solution:

Given, d = 0.07 m , r = 0.035 m

T_{1} = 327 + 273 = 600 K, T_{2} = 300K

*Answer can only contain numeric values

QUESTION: 63

In a single-cylinder reciprocating engine, the mass of reciprocating parts is 55 kg and the mass of revolving parts is 30 kg. The speed of the engine is 200 rpm and the stroke length is 300 mm. If 65 % of the reciprocating parts and all the revolving parts are to be balanced then the required balance mass at a radius of 270 mm is ___ kg.

Solution:

Given,

Mass of reciprocating parts, m_{1} = 55 kg

Mass of revolving parts, m_{2} = 30 kg

Speed of engine, N = 200 rpm

Stoke length, l = 300 mm

% of reciprocating parts to be balanced, k = 65 %

Angular speed, ω = (2πN) / 60

= (2π × 200) / 60

= 20.94 rad/s

Radius of crank, r = l / 2 = 300/2 = 150 mm

Total mass to be balanced, m = k × m_{1} + m_{2}

= 0.65 × 55 + 30 = 65.75 kg

Let the required mass at radius (r_{p}) 270 mm be m_{p.}

So from balancing we can write,

m_{p} × r_{p} = m × r

m_{p} = (m × r) / r_{p} = (65.75 × 150) / 270 = 36.53 kg.

QUESTION: 64

Find the amount of work supplied by the engine (in kW) if fuel of Calorific Value 45 MJ burns at a rate of 3kg/min in an Air Standard Otto cycle of compression ratio 7.

Solution:

*Answer can only contain numeric values

QUESTION: 65

A figure is shown below. The view factor F_{12} is

Solution:

- GATE Mock Test Mechanical Engineering (ME) - 9
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