Gate 2020: Mechanical Engineering

Memory Based Questions of

GATE 2020 Mechanical Engineering Detailed Solutions

Date of Exam : 1/2/2020 Afternoon Session Scroll down to view www.madeeasy.in Corporate Office: 44-A/1, Kalu Sarai, New Delhi - 110016 | Ph: 011-45124612, 9958995830 Delhi | Hyderabad | Noida | Bhopal | Jaipur | Lucknow | Indore | Pune | Bhubaneswar | Kolkata | Patna

Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.1

In MRP, if inventory cost is very high and setup cost is zero, which of the following lot sizing approach should be used? (a) fixed period quantity for period (b) EOQ (c) base stock level (d) lot for lot

Ans.

(d) End of Solution

Q.2

A → B → C → D → E → Customer Initial price = `120 After every stage price increased by 25%. Calculate the cost when the product in the hand of customer. (a) (b) `366.210 (c) `292.96 (d)

Ans.

(b) Cost = 1.255 × 120 = Rs. 366.210 End of Solution

Q.3

Consider the following welding processes: arrange the following in increasing order of HAZ 1. Arc welding 2. MIG 3. Laser beam welding 4. Submerged arc welding (a) 1 - 2 - 3 - 4 (b) 1 - 4 - 2 - 3 (c) 3 - 2 - 4 - 1 (d) 4 - 3 - 2 - 1

Ans.

(c) End of Solution

Q.4

There are two identical shaping machine S1 and S2. In S2 the width of workpiece is increased by 10 percent and feed is decreased by 10 percent with respect to S1. If all other condition remains same then ratio of total time per pass in S1 and S2 will be:

Ans.

(0.8182) In shaper total time of total time of machining

∴

T=

W (1 + m ) 1000 f

T∝

W f

T1 W f = 1× 2 T2 W2 f1

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Given: W2 = 1.1 W, ∴

f2 = 0.9f1 W1 0.9f1 0.9 T1 × = = 0.8182 = 1.1W1 f1 1.1 T2 End of Solution

Q.5

Keeping all the parameter identical the compression ratio (CR) of an air std. diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle ρ = 2. The difference between the new and old efficiencies is ________ %.

Ans.

(4.79) Keeping all the parameter identical (CR)1 = 21 (CR)2 = 15 ρ=2 γ = 1.3 η = 1−

1

(r )

γ −1

⎡ ρr − 1⎤ 1 ×⎢ ⎥ ⎣ ρ −1 ⎦r

⎛ ρr − 1⎞ 1 ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤⎥ ⎢ 1 1 − − − (ηnew|CR = 21) – (ηold|CR = 15) = ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎝ ρ − 1 ⎠ r ⎢ ⎜⎝ (r ) γ −1⎟⎠ ⎜⎝ (r ) γ −1⎟⎠ ⎥ 1 2⎦ ⎣ ⎛ 21.3 − 1⎞ 1 ⎡⎛ 1 1 ⎞⎤ ⎢ ⎥ − = ⎜⎝ 2 − 1 ⎟⎠ 1.3 ⎢⎜⎜ 1.3 −1 1.3 −1 ⎟ ⎟⎠ ⎥ 15 21 ) ) ( ( ⎝ ⎣ ⎦ ⎛ 21.3 − 1⎞ = ⎜ 1.3 ⎟ × [0.04260 ] = 4.79% ⎝ ⎠ End of Solution

Q.6

The given graph is between 3 variable x, y and z. Which of following is incorrect? xy

z

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session (a) (b) (c) (d) Ans.

when y is constant, x is inversely proportional to z. when x is constant, z is inversely proportional to y. when z is constant, x is inversely proportional to y. None of these

(c) End of Solution

Q.7

An inventory model is as shown Demand = 10,000 unit/year Carrying cost = Rs. 4 /unit/year Order cost = Rs. 300/order Back order cost = Rs. 25/unit/year Assume as soon as order Q is received, the back order is supplied to customer. What is maximum amount of inventory reached? Q

Qs

Ans.

(1137) Maximum inventory level, D = 10000 units/year Holding cost, Ch = `4/unit/year Back order, Cb = `25/unit/year Ordering cost, Co = `300/unit/year EOQ = Q * =

=

2DC o = ⎛ Cb ⎞ Ch × ⎜ ⎝ C h + C b ⎟⎠

2 × 10000 × 300 × (25 + 4 ) 4 × 25

2 × 10000 × 300 25 4× 29

Q* = 1319.0905 units (Q – S)*Ch = S × Cb S* =

Q * Ch ⎡1319.090 × 4 ⎤ =⎢ ⎥⎦ = 181.943 units 29 (Ch + Cb ) ⎣

Maximum inventory level = Q* – S* = 1319.0905 – 181.943 = 1137.14697 units End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.8

Circular roller of radius ‘R’ is pure rolling without slip at P and Q. Calculate angular velocity of disc. v

P

R

2v

Q

Ans.

(a)

v R

(b)

v 2R

(c)

2v R

(d)

3v 2R

(d) v

P

V0 w 2v

Q

For pure rolling condition P and Q velocity should be zero. Assume roller moves with linear velocity with V0 and rotate with ω angular velocity So, for P: V0 + V = Rω V0 – R ω = – v ...(i) For Q, V0 + R ω = 2 V ...(ii) Solve Equation (i) and (ii), ω=

3v 2R End of Solution

Q.9

At both the surfaces, coefficient of static friction is μ. Draw FBD of the body when ball is about to leave the lower surface.

F

W

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session R

F

(a)

(b)

W

(c) Ans.

(d)

(a) End of Solution

Q.10

m1 = 10 kg m2 = 30 kg After collision, both masses get stuck. Find the maximum compression of spring. Vo = 1 m/s m1 Frictionless

Ans.

m2 1m

0.25 m

() ∵ There is no friction, so no energy loss. 1 1 2 m1V02 = k ⋅ xmax 2 2

where compression will be maximum, velocity of combined blocks will be zero. ⇒

2 10 × 12 = k ⋅ xmax

xmax =

10 k

Data is not complete. End of Solution

Q.11

The recent measures to improve the output would ______ the level of production to our satisfaction. (a) speed (b) decrease (c) equalize (d) increase

Ans.

(d) End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.12

A fair coin tossed 20 times. The probability that head will appear exactly 4 times in the 1st ten tosses, and tail will appear exactly 4 times in the next ten tosses is _______.

Ans.

(0.0420) Fair coin tossed 20 times First 10 times probability that head will appear exactly 4 times 4

⎛ 1⎞ ⎛ 1⎞ P(x = 4) = 10C4 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

6

next 10 times probability that tail will appear exactly 4 times 4

⎛ 1⎞ ⎛ 1⎞ P(y =4) = 10C4 ⎜ ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

6

Total probability, P = P(x) × P(y) 4 6 ⎡ ⎛ 1⎞ ⎛ 1⎞ ⎤ = ⎢10 C 4 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎥ ⎢⎣ ⎥⎦

2

2

2

1 ⎤ ⎡10 × 9 × 8 × 7 ⎡ 210 ⎤ × 10 ⎥ = ⎢ 10 ⎥ = 0.0420 = ⎢ ⎣ 4× 3× 2×1 2 ⎦ ⎣2 ⎦ End of Solution

Q.13

A mould cavity of 1200 cm3 volume has to be filled through a sprue of 10 cm length feeding a horizontal runner. Cross-section area at the base of sprue is 2 cm2. Consider acceleration due to gravity as 9.81 m/s2. Neglecting fractional losses due to molten metal flow, the time taken to fill the mould cavity is ______ sec.

Ans.

(4.28) 1 hc 2 ht

Sprue

hs 3

Assuming top gate: A3 = Ag = As Neglecting height of pouring basin (hc) ht = hs + hc = hs

tf =

V 1200 = = 4.28 s Agv g 2 2 × 981 × 10 End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.14

Phase diagram does not represent (a) heat transfer rate (b) temperature at which phase change takes place (c) phase transformation rate (d)

Ans.

(c) End of Solution

Q.15

Initial tool geometry (ASA) 0 - 9 - re - rs - ce - 0 - R After some time 0 - 9 - re - rs - ce - 30 - R Specific energy consumption (e) = V0 (t)0.4 where t = uncut chip thickness ⎛ FC 2 − FC1 ⎞ Initial cutting force is FC1 and final cutting force is FC2 find the value of ⎜ F ⎟⎠ × 100 . ⎝ C2

Ans.

(5.022) We know, specific energy consumption, (e) =

f = f sinλ λ = 90 = Cs ⇒

Now, ∴

∴

FC 1000fd

λ1 = 90 – 30 = 60° λ2 = 90 – 0 = 90° 0.4 e = V0(t)0.4 = V0 (f sin λ) =

FC 1000fd

FC = V0(f sinλ) × 1000 × fd FC ∝ (sinλ)0.4 ⎡⎛ sin λ ⎞ 0.4 ⎤ FC 2 − FC 1 2 − 1⎥ × 100% × 100% = ⎢⎜ ⎟ ⎢⎝ sin λ1 ⎠ ⎥ FC 2 ⎣ ⎦ ⎡⎛ sin90° ⎞ 0.4 ⎤ = ⎢⎢⎜⎝ sin60°⎠⎟ − 1⎥⎥ × 100% = 5.022% ⎣ ⎦ End of Solution

Q.16

Module, m = 1.25, teeth on pinion = 20, teeth on gear = _____, pressure angle = 20°. Maximum allowable stress = σ0 = 400 MPa, Lewis form factor = y = 0.322, tooth thickness = __________. Find maximum power that can be transmitted?

Ans.

() Teeth of gear is not received. So will be calculated after response sheet. End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.17

The spectral distribution of radiation from a block body at T1 = 3000 K has a maximum wavelength λmax. The body cools down to a temp T2. If the wavelength corresponding to the maximum wavelength at T2 is 1.2 times of the original wavelength at Tmax, then the temperature T2 is __________ k. (round off to the nearest integer)

Ans.

(2500) State-1: Spectral distribution T1 = 3000 K (λ1) = λmax State-2: T2 = ? (λ2) = 1.2λmax As we know, from Wien’s displacement law λT = Constant λ1T1 = λ2T2 λmax × 300K = 1.2 λmaxT2 T2 = 2500 K End of Solution

Q.18

Equation of motion of spring-mass damper is given as

d2x

3d x + 9 x = 10sin5t dt dt The damping factor for the system is _________. 2

Ans.

+

(0.5) By comparing equation of motion, 2ξωn = 3 ωn2 = 9 2ξωn = 3 2ξ = 1 ξ = 0.5 End of Solution

Q.19

Total qualitative inversion of Grashof linkage if all the joints are revolute? (a) 4 (b) 3 (c) 2 (d) 1

Ans.

(b) Total qualitative inversion when Grashof’s law satisfied. (i) Crank-crank mechanism when shortest link at the fixed position. (ii) Crank-rocker mechanism when shortest link at adjacent to the fixed position. (iii) Rocker mechanism when shortest link opposite to the fixed link that is coupler position. End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.20

If number of teeth in sun and planet is equal and speed of ring (ωR) and speed of sun is (ωS) and direction of gear P and R is same then arm speed is given by P C = Centre S = Sun P = Planet R = Race A = Arm

(a)

A

3ω R ω S + 4 4

(b)

(c) Ans.

S

C

R

1 3 ω + ω 4 R 4 S

(d)

(a) Let teeth of sun gear Ts = T Ts = Tp = T RR = rs + 2rp TR = T + 2T = 3T As we know, y + x = ωS

y−

On subtraction,

x = ωR 3

4x = ωS – ωR 3

x=

3 3 ωS − ωR 4 4

y + x = ωS Speed of arm,

y = ωS −

3 3 ωS + ωR 4 4

ω ⎞ ⎛3 = ⎜ ωR + S ⎟ ⎝4 4 ⎠

End of Solution

Q.21

In a machine member, stress is varying as σ = 350 sin (8πt) in MPa FOS = 3.5 find maximum endurance limit.

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Ans.

(1225) 1 σm σv = + FOS Syt Se

∵

σm = 0,

⇒

1 350 = 3.5 Se

σv = 350

Se = 350 × 3.5 = 1225 MPa End of Solution

Q.22

dy d 2y − y = 1, which satisfies y t = 0 = 2 dt dt 1 1 (a) s s + 1 (b) s − 1 ( ) ( )

Solution of

1 (c) s s − 1 ( )

Ans.

t =0

= 0 in Laplaces domain is

1 (d) s s + 1 s − 1 ( )( )

(d) Apply Laplace transform on both sides, L [Y ′′] – L[Y ] = L [1]

s2Y(s) – sY(0) – sY ′(0) – Y(s) =

1 s

1 s(s − 1)(s + 1)

Y(s) =

End of Solution

Q.23

A beam of negligible mass hinged at support P and has a roller at Q. Q

1200 N 5m

5m

4m

R

P

Point load at R = 1200 N, then find magnitude of reaction force at Q. Ans.

(1500) 1200 N

RQ

P RHP RVP

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session ΣMP = 0,

RQ × 4 = 1200 × 5 RQ =

1200 × 5 = 1500 N 4 End of Solution

Q.24

One kg of air in closed system is undergoing a reversible process from P1 = 1 bar [abs], T1 = 27°C to P2 = 3 bar [abs] and T2 = 127°C. If R = 287 J/kgK, γ = 1.4, then calculate the change in entropy in J/kgK.

Ans.

(–26.2) At state -1: P1 = 1 bar, T1 = 300 K, At state -1: P2 = 3 bar, T2 = 400 K, R = 287 J/kgK, γ = 1.4 ⎛T ⎞ ⎛P ⎞ Change in specific entropy, Δs = c p ln ⎜ 2 ⎟ − R ln ⎜ 2 ⎟ ⎝ T1 ⎠ ⎝ P1 ⎠ ⎛ 400 ⎞ ⎛ 3⎞ = 1005ln ⎜ − 287ln ⎜ ⎟ ⎝ 300 ⎟⎠ ⎝ 1⎠

= –26.18 ≈ –26.2 J/kgK End of Solution

Q.25

A Carnot cycle working between 27°C and –3°C then ratio of COPref to COPheat pump is __________.

Ans.

(0.9) COP of refrigerator, COPref =

270 =9 30

COP of heat pump, COPheatpump =

COPref /COPheat pump =

300 = 10 30

9 = 0.9 10 End of Solution

Q.26

A cantilever beam, a load is acting at the free end and spring is also connected at free end. What is the transverse deflection under load in terms of E, I and W ? W

K N/m

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

(a)

WL3

(b)

3E I + KL3

(c) Ans.

(d)

(a) W A

B FS FS

K N/m

Let deflection at B will be δ So, FS = Kδ

FS =δ K For cantilever beam and end load (W – FS) δ=

WL3 FSL3 − 3E I 3E I

=

WL3 KδL3 − 3E I 3E I

⎡ KL3 ⎤ WL3 δ ⎢1 + ⎥ = 3E I ⎣ 3E I ⎦ ⎛ WL3 ⎞ ⎛ WL3 ⎞ ⎜ ⎟ E 3 I δ= ⎜ = ⎜ ⎟ ⎟ 3 3 ⎝ 3E I + KL ⎠ ⎜ 3E I + KL ⎟ ⎜⎝ ⎟⎠ 3E I End of Solution

Q.27

Diameter = 2r, both ends are fixed and Gas gauge pressure is p and t is thickness of pipe. Find maximum increase in temperature for which there is no leakage of gas. t

α, p, E, poisson’s ratio Coefficient of linear expansion

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

⎛ (a) ⎜⎝ ν −

(c)

Ans.

1 ⎞ Pr ⎟ 4 ⎠ E αt

⎛ (b) ⎜⎝ ν +

pD ⎡ 1 ⎤ − μ⎥ 2tE α ⎢⎣ 2 ⎦

1 ⎞ Pr ⎟ 2 ⎠ E αt

(d)

(c) Following are the state of stresses developed when pressure (p) and temperature change are acting alone: σh

σh

σL

σL

(+)

σth

σth

=

σL – σth

Fig: due to temperature change

σh Fig: due to (p)

σL – σth

σh Fig: Resultant or Actual state of stress due to pressure and temperature

σ1 = σh; σ2 = σL – σth =

σh − σ th 2

ε2 =

1 [ σ2 − μσ1] = 0 E

ε2 =

⎤ 1 ⎡⎛ σh ⎞ − σth ⎟ − (μσh )⎥ = 0 ⎠ E ⎢⎣⎜⎝ 2 ⎦

σh σth μσh − − =0 2E E E σh ⎡ 1 ⎤ σ − μ ⎥ = th ⎢ E ⎣2 E ⎦

pD 2tE

⎡1 ⎤ αTE ⎢ 2 − μ⎥ = E ⎣ ⎦

T=

pD ⎡ 1 ⎤ − μ⎥ 2tE α ⎢⎣ 2 ⎦ End of Solution

Q.28

A helical spring of stiffness k, if wire diameter spring diameter and number of coils are all doubled, then knew is equal to (a) 16 k (b) 8 k (c) k

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(d)

|

k 2

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Ans.

(c) We know that,

k= knew =

Gd 4 8D 3N

G × (2)4d 4 Gd 4 =k = 8 × (2)3D 3 × 2N 8D 3N End of Solution

Q.29

A vessel containing water is placed in ambient atmospheric conditions at 25°C, 1 bar Oven Valve

Water vapour

Water

25°C, 1 atm

80°C, 1 atm

Then it is transferred to an oven at 80°C, 1 bar, till it attain equilibrium. What will happen immediately after opening the valve inside oven? (a) The water vapour will condense (b) The water vapour will escape out (c) The hot air will enter into vessel (d) Nothing will happen, everything would be in equilibrium Ans.

(b) End of Solution

Q.30

If D1 > D2 then α

α

α

α

D1

D2 x2

x1

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

D1 − D2 (a) sin α = x − x 1 2 (c) sinα =

Ans.

x1 − x2 (b) cos α = 2D − D 1 2

(D1 − D2 ) 2( x1 − x2 ) − (D1 − D2 )

(d)

(c)

A D1

x1

B

2

α O

OA = x1 – AB =

D1 2

D1 2

sinα =

AB as ∠ ABO = 90° OA

sinα =

D1 /2 x1 − D1 /2

...(i)

Similarly,

A1 x2

D2 2

B1 90°

α O1

O1A1 = x2 –

D2 2

A1B1 =

D2 2

sinα =

A1B1 D2 /2 = O1A1 x2 − D2 /2

...(ii)

For eq. (i) and (ii)

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

D1 /2 D2 /2 sinα = x − D /2 = = x 1 1 2 − D2 /2 =

(D1 − D2 ) ( x1 − x2 ) − (D1 − D2 ) /2

(D1 − D2 ) 2 ( x1 − x2 ) − (D1 − D2 ) End of Solution

Q.31

A bolt has to be made at the end of rod of d = 12 mm by localised forging (upsetting). Length of unsupported portion of rod is 40 mm. To avoid buckling of rod, a closed forging operation to be performed at max die diameter of ______mm.

Ans.

() Data is incomplete. Solution will be given after the response sheet. End of Solution

Q.32

In a CNC machine G00 command is given and in travel point P (200, 300) to Q(300, 600) Two separate stepper motor are connected to stage with a lead screw of 0.5 mm pitch, motor rotates with 800 rpm. What is the time of travel in min? 1000 Q(300, 600)

200 100

P(200, 300)

100 200

Ans.

(0.75) y

Q(300, 600)

600 500

Δy = 300 mm

400

P (200, 300) Δx = 100 mm

300 200 100

For

600

500

400

300

200

100

x

Δ x = 100 mm movement

tx =

Δx 100 = min = 0.25 min PN 0.5 × 800

ty =

Δy 300 = = 0.75 min PN 0.5 × 800

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session For this complete movement 0.75 min time needed. x-axis motor and y-axis motor both will work 0.25 min then x-axis motor will stop and y-axis motor will run 0.5 min more. Therefore total time for this operation will be 0.75 min. End of Solution

Q.33

Calculate the tooling cost for per piece of bar of length 250 mm and diameter 25 mm. Feed is 0.2 mm/rev. For maximum production rate if tool change time is 1 min, Taylor equation VT 0.2 = 24 and tool required cost is Rs. 20 (a) (b) (c) (d)

Ans.

() ⎛ 1− n⎞ ⎛ 1 − 0.2 ⎞ = 1⎜ = 4 min T0 = TC ⎜ ⎝ n ⎟⎠ ⎝ 0.2 ⎟⎠

Optimum speed, (V0) = Machining Time (Tm) =

C 24 = = 18.1886 m/min T0n 4 0.2 L fN

Cost of operation = C m × Tm + (C A + TC × C m ) ×

Tm T0

Some more data needed, will be confirmed after response coming. End of Solution

Q.34

Given below is the representation of surface roughness symbol. Match the following (i) – (iii) (iii)

(vi) (v) (iv)

(i) (ii) (iii) (iv) (v) (vi) (a) P Q S R T U (b) P Q R S U T (c) P Q R S T U (d) R S P Q T U Where, P = Maximum waviness height; Q = maximum width R = Roughness width cutoff; S = Maximum roughness width; T = Maximum roughness height; U = Minimum roughness height. Ans.

(b) End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.35

Initially pressure inside cylinder is 105 kPa, the piston moves for 8 cm against atmospheric and another 8 cm against a spring of stiffness k = 2 × 105 N/mm. Find total work done

8 cm

8 cm

P1 = 105 kPa

Patm = 100 kPa, k = 2 × 105 N/mm, Piston area = 300 cm2. Ans.

(544) Area of piston = 300 cm2 k = 12.5 kN/m W1-2 = P1 × A × x = 105 × 300 × 104 × 8 × 10–2 = 0.252 kJ P2 = 105 kPa

3 P2 = 8 cm

kx

2 x = 8 cm 1 P1 = 105 kPa P 3

2

V2

V3

V

P3A = P2A + kx P3 = P2 +

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kx 12.5 × 8 × 10 −2 = 105 + = 138.33 kPa 2 300 × 10 −4

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

W2-3 =

=

1 1 (P + P3 ) × (V3 − V2 ) = 2 (105 + 138.33) A × x 2 2 1 (243.33) × 300 × 10 −4 × 8 × 10 −2 2

W2-3 = 0.2919 kJ Wtot = W1-2 + W2-3 = 0.5439 kJ = 543.91 ≈ 544 J

∴

End of Solution

Q.36

Total student are 5000 and girl are 1500 Art 30%

Art 20%

Management = 15%

Students

Girl

Find ratio of boys registered in art to girl registered in management. Ans.

(2.44) Ratio = =

Total student in art − Girls in arts Girls in management 5000 × 0.2 − 1500 × 0.3 1000 − 450 = = 2.44 1500 × 0.15 225 End of Solution

Q.37

Directional derivative of f(x) = xyz in direction of i – 2y + 2k at (–1, 1, 3) is:

Ans.

(7/3) Directional derivative = ( ∇f ) p × nˆ Here, (∇f)(–1,

∇f = yzi + xzj + xyk 1, 3) = 3i – 3j – k

nˆ i − 2 j + 2k i − 2 j + 2k = nˆ =  = 3 n 1+ 4 + 4 ∴ Directional derivative =

1 7 [3 + 6 − 2 ] = 3 3 End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.38

Given that function is analytical f(x) = ux i + v(x y) Where, ux = x3y2 – xy3. Find v(x, y) = ? (a) (b) (c) (d)

Ans.

() u(x, y) is not properly given. So, we cannot guess the answer. End of Solution

Q.39

Let I be a 100 multidimensional matrix, and E is set of its distinct real eigen value, then number of elements in E ______? (a) (b) (c) (d)

Ans.

() Data is incomplete. Solution will be given after the response sheet. End of Solution

1 x2

Q.40

Let I =

∫ ∫

xy 2dyd x, then I may also be expressed as,

x =0 y =0 1

(a)

1

∫ ∫

1

xy 2d xdy

(b)

y =1 x = y

1

(c)

1

xy 2d xdy

(d)

y =0 x =0

Ans.

y x 2d xdy

y =1 x = y

y

∫ ∫

1

∫ ∫

y

∫ ∫

xy 2d xdy

y =0 x =0

() 1 y=x

0

x=1

Unit is given wrong in problem phase change. End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session

π/2

Q.41

∫ (8 + 4cos(8t ))dt

What is the absolute error in % if integral is evaluated using

0

trapezoidal rule only by considering Ist and last term. Ans.

(5.187) π /2

y=

∫ (8 + 4cos(8t ))dt 0

By trapezoidal method 0

π/2

12

8

yo

y1

y1 =

h ⎡y + y 1⎦⎤ 2⎣ o

y1 =

π [12 + 8] = 15.707 2×2

By integration: π /2

y2 =

π /2

0

= 8 Absolute error =

⎛ π⎞

∫ (8 + 4cos x)d x = 8 ⎜⎝ 2 ⎟⎠ + 4[sin x]0 π + 4 = 16.566 2

16.566 − 15.707 = 0.05187 = 5.187% 16.566 End of Solution

Q.42

The sum of normally distributed random variable x and y if (a) normally distributed, only if x and y have same mean. (b) normally distributed, only if x and y have same standard deviation. (c) normally distributed, only if x and y are independent. (d) always normally distributed.

Ans.

(c) End of Solution

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Q.43

A matrix P is decomposed into its symmetric part ‘S ’ and skew symmetric part ‘V ’ if 2 ⎞ ⎛ −4 4 −2 3 ⎞ ⎛0 , ⎜ ⎟ ⎜ S= 4 3 7 / 2 , symmetric matrix V = 2 0 7 / 2⎟ ,skew symmetric matrix . ⎜ ⎟ ⎜ ⎟ ⎝ 2 7/2 2 ⎠ ⎝ −3 −7 / 2 0 ⎠

then matrix P is

Ans.

−1 ⎞ ⎛ −2 9/ 2 ⎜ −1 81/ 4 11 ⎟ (a) ⎜ ⎟ ⎝ −2 45/ 2 73/ 4⎠

⎛ −4 2 5⎞ ⎜ 6 3 7⎟ (b) ⎜ ⎟ ⎝ −1 0 0⎠

⎛ −4 2 5⎞ ⎜ ⎟ (c) ⎜ 6 3 7⎟ ⎝ −1 0 2⎠

⎛ 4 −6 1 ⎞ (d) ⎜ −2 −3 0 ⎟ ⎜ ⎟ ⎝ −5 −7 −2⎠

(c) For matrix P its symmetric part is

S=

P + PT 2

For matrix P its skew symmetric part is

V=

P − PT 2

⎛ −4 2 5⎞ P = S + V = ⎜ 6 3 7⎟ ⎜ ⎟ ⎝ −1 0 2⎠ End of Solution

Q.44

Water (density 1000 kg/m3) flows through on inclined pipe. The velocity, pressure and elevation at section are VA = 3.2 m/s, PA = 186 kPA and ZA = 24.5 m respectively, and those at section B are VB = 3.2 m/s, PB = 260 kPa and ZB = 9.1 m respectively. If acceleration due to gravity is 10 m/s2 then the head lost due to friction is_______ m. (round off to one decimal place)

Ans.

(8) A

VB = 3.2 m/s PB = 260 kPa ZB= 9.1 m B

VA = 3.2 m/s PA = 186 kPa ZA = 24.5 m

2

g = 10 m/s

3

ρ = 1000 kg/m Datum

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Memory Based Questions of GATE 2020 Mechanical Engg. | Afternoon Session Energy head at A :

EA =

PA V A2 186 × 10 3 3.2 2 + + ZA = + + 24.5 = 43.612 m ρg 2g 10 3 × 10 2 × 10

EB =

PB VB2 260 × 10 3 3.2 2 + + ZB = + + 9.1 = 35.61m ρg 2g 10 3 × 10 2 × 10

Energy head at B:

EA > EB, so flow from A to B Friction head, hf = EA – EB = 8 m



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