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

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Q2. Match the following

Group-I

P. G09

Q. G41

R. G01

S. G03

Group – II

- Linear interpolation
- Retardation
- Circular interpolation
- 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

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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

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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

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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

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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

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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

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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.

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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

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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

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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