Quiz: Mechanical Engineering

Exam: GATE

Topic: Miscellaneous

Each question carries 2 mark

Negative marking: 1/3 mark

Time: 20 Minutes

Q1. Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributed exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is

(a) 0.3

(b) 0.5

(c) 0.7

(d) 0.9

Q2. The bolts in a rigid flanged coupling connecting two shafts transmitting power are subjected to

(a) shear force and bending moment

(b) axial force

(c) torsion

(d) torsion and bending moment

Q3. In orthogonal turning of low carbon steel pipe with principal cutting edge angle of 90°, the main cutting force is 1000N and the feed force is 800 N. the shear angle is 25° and orthogonal rake angle is zero. Employing Merchant’s theory, the ratio of friction force to normal force acting on the cutting tool is

(a) 1.56

(b) 1.25

(c) 0.80

(d) 0.64

Q4. A resistance – capacitance relaxation circuit is used in an electrical discharge machining process. The discharge voltage is 100 V. At a spark cycle time of 25 μs, the average power input required is 1 kV. The capacitance (in μF) in the circuit is

(a) 2.5

(b) 5.0

(c) 7.5

(d) 10.0

Q5. In sand casting of hollow part of lead, a cylindrical core of diameter 120 mm and height 180 mm is placed inside the mould cavity. The densities of core material and lead are 1600 kg/m³ and 11300 kg/m³ respectively. The net force (in N) that tends to lift he core during pouring of molten metal will be

(a) 19.7

(b) 64.5

(c) 193.7

(d) 257.6

Q6. A small bore is designated 25H7. The lower (minimum) and upper (maximum) limits of the bore are 25.000 mm and 25.021 respectively. When the bore is designated as 25H8, then the upper limit is 25.033 mm. when the bore is designated as 25H6, then the upper limit of the bore in mm is

(a) 25.001

(b) 25.005

(c) 25.009

(d) 25.013

Q7. A diesel engine is usually more efficient than a spark ignition engine because

(a) diesel being a heavier hydrocarbon, releases more heat per kg than gasoline

(b) the air standard efficiency of diesel cycle is higher than the Otto cycle, at a fixed compression ratio

(c) the compression ratio of a diesel engine is higher than that of an SI engine

(d) self-ignition temperature of diesel is higher than that of gasoline

Q8. Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is

(a) 242.0

(b) 166.5

(c) 121.0

(d) 74.0

Q9. A Carnot cycle is having an efficiency of 0.75. if the temperature of the high temperature reservoir is 727°C. what is the temperature of low temperature reservoir?

(a) 23°C

(b) -23°C

(c) 0°C

(d) 250°C

Q10. In a DC are welding operation, the voltage-arc length characteristic was obtained as V_arc=20+5l where the arc length l was varied between 5 mm and 7mm. here V_arc denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are

(a) 45 V, 450 A

(b) 75 V, 750 A

(c) 95 V, 950 A

(d) 150 V, 1500 A

Solutions

S1. Ans.(c)

Sol. λ = 0.1 per min, μ = 0.33 per min

ρ=0.1/0.33=0.3

ρ_0=1-ρ=0.7

S2. Ans.(a)

Sol. Bolts in coupling are subjected to both shear and bending loads.

S3. Ans.(c)

Sol. Given, cutting force: f_C=1000 N,Feed force:F_T=800 N

Rake angle α = 0°, Shear angle: Φ = 25°

We know that, μ=F/N=(F_c sinα+F_T cosα)/(F_c cosα-F_T sinα)

=(1000 Sin0°+800 Cos 0°)/(1000 Cos0°-800 Sin〖0°〗 )

=800/1000=0.8

S4. Ans.(b)

Sol. Power = work/time

P=1/2 (CV^2)/t

C=(P×2t)/V^2 =(1000× 2×25×10^(-6))/(100)^2 =5μf

S5. Ans.(c)

Sol. net force acting on core = (ρ-σ)gv

Where,

V=π/4×120^2×180

=2035752 mm^2

=2.035×10^(-3) m^3

∵ Net force = (11300-1600) × 9.81 × 2.035 × 10^(-3)

= 193.71 N

S6. Ans.(d)

Sol. 25H7 → Grade of tolerance

= IT7 = 16i = 21μm

⇒ i =21/16=1.3125

For same diameter step, i will be same

For 25 P – 16 → IT6 = 10i

= 10×1.3125≈13μm

Upper limit for 25H6 = 250.13 mm

S7. Ans.(c)

Sol. for same compression ratio and the same heat supplied otto cycle is most efficient and diesel cycle is least efficient in practice, however, the compression ratio of the diesel engine ranges between 14 and 25 where as that of the otto engine between 6 and 12. Because of its higher efficiency than the otto engine.

S8. Ans.(b)

Sol. tooth geometry factor, y = 0.32

Combined effect of dynamic load and allied factory intensifying the stress is f_s=1.5

F_t=(σ_b by_m)/f_s

3552N=(σ_b (25mm)(0.32)(4mm))/1.5

⇒σ_b=166.5 MPa

S9. Ans.(b)

Sol. Given data; η_cannot=0.75

T_1=727°C=(727+2730)K=100 K=T_L= ?

η_cannot=1-T_2/T_1

0.75=1-T_2/T_1

0.75=1-T_2/1000

or,T_2/1000=0.25

T_2=0.25×1000=250 K

=(250-273)°C= -23°C

S10. Ans.(c)

Sol. V_arc=20+5l

l=5mm,V_arc=45 V

l=7 mm,V_arc=55 V

V=V_0-V_0/I_0 I

Where V_O = open circuit voltage

I_s = Short circuit current

45=V_0=V_0/I_s ×500

55=V_0-V_0/I_s =400

⇒V_0=95 voltas; I_S=950 amp.