Quiz: Mechanical Engineering 2nd June

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. For the linear programming problem:
Maximize,z=3x_1+2x_2 Subject to
-2x_1+3x_2≤9
x_1-5x_2≥ -20
x_1,x_2≥0
The above problem has
(a) unbounded solution
(b) infeasible solution
(c) alternative optimum solution
(d) degenerated solution

Q2. The demand and forecast for February are 12000 and 10275, respectively. Using single exponential smoothening method (smoothening coefficient = 0.25), forecast for the month of march is
(a) 431
(b) 9587
(c) 10706
(d) 11000

Q3. A GO-No GO plug gauge is to be designed for measuring a hole of nominal diameter 25 mm with a hole tolerance of ± 0.015 mm. considering 10% of work tolerance to be the gauge tolerance and no wear condition, the dimension (in mm) of the GO plug gauge as per the unilateral tolerance system is
(a) 〖24.985〗^■(+0.003@-0.003)
(b) 〖25.015〗^■(+0.000@-0.006)
(c) 〖24.985〗^■(+0.03@-0.03)
(d) 〖24.985〗^■(+0.003@-0.000)

Q4. Two identical cylindrical jobs are turned using
(a) a round nosed tool of nose radius 2 mm and
(b) a sharp corner tool having principal cutting edge angle = 45° and auxiliary Cutting edge angle = 10°. If the operation is carried out of the feed of 0.08 mm/rev, the height at micro irregularities on the machined surfaces (in mm) in the two cases will be
(a) 0.0001, 0.024
(b) 0.0002, 0.012
(c) 0.0003, 0.024
(d) 0.0004, 0.012

Q5. The figure below shows a steel rod of 25 mm² cross sectional area. It is loaded at four points. K, L, M and N. Assume E_steel = 200 GPa. The total change in length of the rod due to loading is

(a) 1 μm
(b) -10 μm
(c) 16 μm
(d) -20 μm

Q6. While cooling, a cubical casting of side 40 mm undergoes 3%, 4% and 5% volume shrinkage during the liquid state, phase transition and solid state, respectively. The volume of metal compensated from the riser is
(a) 2%
(b) 7%
(c) 8%
(d) 9%

Q7. Spherical roller bearings are normally used
(a) for increased radial load
(b) for increased thrust load
(c) when there is less radial space
(d) to compensate for angular misalignment

Q8. During a TIG welding process, the arc current and arc voltage were 50A and 60V, respectively, when the welding speed was 150 mm/min. in another process, the TIG welding is carried out at a welding speed of 120 mm/mm at the same arc voltage and heat input to the material so that weld quality remains the same. The welding current (in A) for the process is
(a) 40.00
(b) 44.72
(c) 55.90
(d) 62.25

Q9. The non-dimensional fluid temperature profile near the surface of a convectively cooled flat plate is given by (T_W-T)/(T_W-T_∞ )=a+b y/L+C(Y/L)^2, where y is measured perpendicular to the plate, L is the plate length, and a, b and c are arbitrary constants, T_W and T_∞ are wall and ambient temperatures, respectively. If the thermal conductivity of the fluid is k and the wall heat flux is q”, the Nusselt number Nu=q”/(T_W-T_∞ ) L/K is equal to
(a) a
(b) b
(c) 2c
(d) (b+2c)

Q10. The part of a gating system which regulates the rate of pouring of molten metal is
(a) pouring basin
(b) runner
(c) choke
(d) ingate

Solutions

S1. Ans.(a)
Sol.

z=3x_1+2x_2
-2x_1+3x_2≤9
x_1-5x,≥-20
x_1,x_2≤=0
∵changing the inequities to equalities
-2x,+3x_2=9
x_1-5x_2= -20
∵ the solution is unbounded

S2. Ans.(c)
Sol. F_t=F_(t-1)+α (D_(t-1) -F_(t-1) )
F_t = Current month forecast
F_(t-1) = last month forecast
D_(t-1) = last month demand
α = smoothening constant
F_t=10275+0.25(12000-10275)
F_t=10706.25
F_t≈10706.

S3. Ans.(d)
Sol. Hole dimension are 24.985 mm and 25.015 mm
Total tolerance = 0.015 + 0.015 = 0.03 m
Work tolerance = 0.03 × 0.1 = 0.003 mm
Dimension of GO plug gauge
= 24.985 – 0.000 mm
= 24.985 + 0.003 mm

S4. Ans.(d)
Sol. R_t1=f^2 \/8r=(0.08)^2/(8×2)=0.0004 mm
R_t2=f/(tan C_s+cot C_e )
=0.08/tan⁡〖45°+cot10〗 =0.012

S5. Ans.(b)
Sol.

Given,
A=25 m^2
E_steel=200 GPa
P_1=100 N (Force in KL)
P_2= -150 N (Force in LM)
P_3=50 N (Force in MN)
∵ Total change in length,
δ_total=δ_1+δ_2+δ_3
=(P_1 L_1)/AE+(P_2 L_2)/AE+(P_3 L_3)/AE
=1/AE [100×0.5-15050.8+50×0.4]
=(-50)/(25×〖10〗^(-6)×200×〖10〗^9 )
= -10×〖10〗^(-6) m or 10 μm
▭(δ_total= -10 μm)

S6. Ans.(b)
Sol. the volume of metal compensated by river does constitute cooling during solid state. Riser can for volume shrinkage only in liquid stage and stage
∵ volume compensated by riser
= 3 + 4 = 7%

S7. Ans.(d)
Sol. Spherical roller bearing are used mainly for adjusting the angular misalignment.

S8. Ans.(a)
Sol. we know that power required for welding (P)=(V×I)/ν
Where
V = Voltage of arc
I = Current of arc.
ν = welding speed in mm/min
So,P=(50×60)/150
= 20 W
As, welding speed has been changed but heat input has remained constant then
20=(60×I)/120
▭(I=40 A)

S9. Ans.(b)
Sol.
q”= -k├ ∂T/∂Y┤|(y=0) =b(T_W-T∞ )
T_W-T=(T_W-T_∞ )[ a+by/L+C(y/L)^2 ]
-∂T/∂Y=(T_W-T_∞ )[b/L+2c/L^2 y]
├ -∂T/∂Y┤|(y=0) =(T_W-T∞ )[b/L]
or,
q”=k(T_W-T_∞ ) b/L
∵Nu=q”/(T_w-T_∞ ) .L/K=b

S10. Ans.(a)
Sol. pouring basing regulates the rate of molten metal, maintains the required rate of liquid metal flow and also reduces turbulence at the sprue entrance.

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