Quiz: Mechanical Engineering 7th July

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. Customers arrive at a ticket counter at a rate of 50 per hr and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1 min. Assuming that customer arrivals form a poison process and service times are exponentially distributed, the average waiting time in queue in min is
(a) 3
(b) 4
(c) 5
(d) 6

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q2. Match the following
Group-I
P. G09
Q. G41
R. G01
S. G03
Group – II

1. Linear interpolation
2. Retardation
3. Circular interpolation
4. Cutter radius compensation
   P Q R S
   (a) 3 1 4 2
   (b) 4 1 3 2
   (c) 3 4 1 2
   (d) 2 4 1 3

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q3. A sine bar has a length of 250 mm, each roller has a diameter of 20 mm, during taper angle measurement of a component, the height from the surface plate to the center off a roller is 100 mm. the calculated taper angle (in degrees) is
(a) 21.1
(b) 22.8
(c) 23.6
(d) 68.9

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q4. The true stress true curve is given by σ=1400 ε^0.33, Where the stress σ is in MPa. The true stress maximum load (in MPa) is
(a) 971
(b) 750
(c) 698
(d) 350

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q5. For a typical sample of ambient air (at 30°C, 75% relative humidity and standard atmospheric pressure), the amount of moisture in kg per kg of dry air will be approximately.
(a) 0.002
(b) 0.027
(c) 0.25
(d) 0.75

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q6. In a spark ignition engine working on the ideal Otto cycle, the compression ratio is 5.5. The work output per cycle (i.e., area of the P-v diagram) is equal to 23.625 × 10^5 × V_c ,J Where V_c is the clearance volume in m³. the indicated mean effective pressure is
(a) 4.295 bar
(b) 5.250 bar
(c) 86.870 bar
(d) 106.300 bar

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q7. For a given set of operating pressure limits of a Rankine cycle the highest efficiency occurs for
(a) Saturated cycle
(b) superheated cycle
(c) Reheat cycle
(d) Regenerative cycle

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q8. Which one of the following is not correct?
(a) intermediate principal stress is ignored when applying the maximum principal stress theory.
(b) the maximum shear stress theory gives the most accurate results amongst all the failure theories.
(c) as per the maximum strain energy theory, failure occurs when the strain energy per unit volume exceeds a critical value.
(d) As per the maximum distortion energy theory, failure occurs when the distortion energy per unit volume exceeds a critical value.

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q9. A compacting machine shown in the figure below is used to create a desired thrust force by using a rack and pinion arrangement. The input gear is mounted on the motor shaft. The gears have involute teeth of 2 mm module.

If the drive efficiency is 80%, the torque required on the input shaft to create 1000 N output thrust is
(a) 20 Nm
(b) 25 Nm
(c) 32 Nm
(d) 50 Nm

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q10. A compacting machine shown in the figure below is used to create a desired thrust force by using a rack and pinion arrangement. The input gear is mounted on the motor shaft. The gears have involute teeth of 2 mm module.

If the pressure angle of the rack is 20°, the force acting along the line of action between the rack and the gear teeth is
(a) 250 N
(b) 342 N
(c) 532 N
(d) 600 N

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Solutions

S1. Ans.(c)
Sol. λ = 50 Customer/hr = 50/60 customer /min = 0.833 customer/min
μ = 1
∵ ρ =λ/μ=0.833
As we know
L_q=λ.ω_q
ρ^2/(1-ρ)=λω_q⇒(0.833)^2/(1-0.833)=0.833 ω_q
⇒ω_q=5 min.

S2. Ans.(d)
Sol. P -2, Q-4, R-1, S-3

S3. Ans.(a)
Sol. Consider ∆ABC, we have
BC = 100 – 20/2 = 90 mm

Sin θ=BC/AB=90/250
θ=21.1

S4. Ans.(a)
Sol. σ_f=Kε^n
K=1400
n=0.33=ε
At, UTS = n= ε
σ_f=1400(0.33)^0.83=971 MPa.

S5. Ans.(b)
Sol. Given data: Dry bulb temperature: T_abt=35°C
Relative humidity, ϕ = 75% = 0.75
AT 35°C, P_s=0.05628 bar
Relative humidity:
ϕ=P_V/P_S
=0.75 =P_V/0.05628
P_V=0.75×0.05628
=0.04221 bar
specific humidity,
ω=(0.6222 P_V)/(P-P_v )
=(0.622×0.04221)/(1.01325-0.04221)
=6.0270 kg\/kg of dry air.

S6. Ans.(b)
Sol.
mep = W/V_s =W/(V_1-V_c )
=(23.625×10^5×V_c)/(5.5V_C-V_c )
=(23.625×10^5 V_C)/(4.5 V_C )
=5.25×10^5 P_a
S7. Ans.(d)
Sol. the thermal efficiency of an ideal regenerative cycle is equal to carnot efficiency.
i.e., η_regenerative=η_carnot
Hence, it is maximum,

S8. Ans.(b)
Sol. the maximum shear stress theory predicts that shear yield value τ_y is 0.5 times the tensile yield value. This is about 15% less than the values predicted by the distortion energy (or the octahedral shear) theory. The maximum shear stress theory gives values for design on the safe side. Then because of its simplicity, this theory is widely used in machine design dealing with ductile materials.

S9. Ans.(b)
Sol. 2F = 1000N
Or F = 500 N

Let T_1 be the torque applied by motor
T_2 be the torque applied by hear
∵ power transmission = 80%
Now, T_1 ω_1=(2T_2 ω_2)/0.8
T_1=(2×f×(D/2))/0.8×(w_2/w_1 )
=2×500×0.16/2×1/0.8×1/4
=25 Nm

S10. Ans.(c)
Sol. we known that
P cos θ = F
∵ P=500/(cos20°)=532N

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