← MES Jr. Grade
Past Questions
Which process is used for fine hole drilling and contour machining?
A)
EBM
B)
LBM
C)
ECM
D)
USM
Ultrasonic Machining (USM) is used for fine hole drilling and contour machining, especially in hard, brittle materials. In USM, abrasive slurry and ultrasonic vibrations remove material accurately with low thermal damage. EBM is mainly for micro-drilling using electron beam, LBM is for laser drilling/cutting, and ECM is used for contouring and shaping conductive materials but not specifically fine hole drilling in brittle materials.
Which of the following is always a cold-working operation?
A)
rolling
B)
wire drawing
C)
forging
D)
extrusion
Wire drawing is always a cold-working process because it is carried out by pulling metal through a die at room temperature to reduce its cross-section. In contrast, rolling, forging, and extrusion can be done either hot or cold depending on the material and process requirements. Therefore, the operation that is always cold-working is wire drawing.
The order in which different jobs are being taken up in a machine or process is called
A)
scheduling
B)
sequencing
C)
routing
D)
aggregate planning
The order in which different jobs are taken up on a machine or process is called sequencing. It decides the job priority and the sequence of operations. Scheduling is broader and deals with assigning time slots and dates; routing is the path/process flow a job follows; aggregate planning is a medium-term production planning activity.
Permeability of a moulding sand is increased with
A)
increasing the moisture content
B)
coarse grain size
C)
increasing the compact factor
D)
fine grain size
Permeability is the ability of moulding sand to allow gases to escape through its voids. It increases when the sand has larger grains, because coarse grains create larger inter-granular spaces for air and gases to pass through. In contrast, increasing moisture content or compacting the sand reduces pore spaces and hence lowers permeability. Fine grain size also decreases permeability because it packs more tightly and leaves smaller voids.
Which of the following processes are heat treatment processes?
A)
normalizing
B)
annealing
C)
quenching
D)
all of these
Normalizing, annealing, and quenching are all heat treatment processes used to alter the microstructure and mechanical properties of metals. Annealing involves heating and slow cooling, normalizing involves heating and air cooling, and quenching involves rapid cooling after heating. Since all three are heat treatment processes, the correct answer is 'all of these'.
In a network analysis technique under project management, CPM is based on
A)
single time
B)
two times
C)
three times
D)
none of these
CPM (Critical Path Method) is a deterministic network analysis technique. It assumes a single, fixed time estimate for each activity. In contrast, PERT uses three time estimates (optimistic, most likely, pessimistic). Therefore, CPM is based on single time.
For grinding steels, the preferred abrasive is
A)
silicon carbide
B)
aluminium oxide
C)
diamond
D)
cubic boron nitride
For grinding steels, aluminium oxide is the preferred abrasive because it is tougher and more suitable for ferrous metals. Silicon carbide is generally preferred for non-ferrous materials like cast iron, brass, and aluminum. Diamond is not suitable for steel because it reacts with iron at grinding temperatures, and cubic boron nitride is excellent for hardened steels but the standard preferred abrasive for general steel grinding is aluminium oxide.
The stepping stone method in transportation problem is used to find
A)
a basic solution
B)
a feasible solution
C)
the initial basic feasible solution
D)
the optimal solution
In a transportation problem, the stepping stone method is used after obtaining an initial basic feasible solution. It evaluates unused routes by checking the change in total cost and helps improve the current solution. This process is continued until no further cost reduction is possible, which gives the optimal solution. Hence, it is not used to find the initial or basic feasible solution.
German Silver is an alloy of
A)
silver and some impurities
B)
nickel, copper and zinc
C)
refined silver
D)
silver and gold
German silver, also called nickel silver, is an alloy made mainly of nickel, copper, and zinc. It contains no actual silver despite its name. The other options are incorrect because (a) refers to impure silver, (c) is pure/refined silver, and (d) is an alloy of precious metals not German silver.
The simplex method is the basic method for
A)
queuing theory
B)
network analysis
C)
value analysis
D)
linear programming
The simplex method is the standard algorithm used to solve linear programming problems, especially when there are multiple constraints and an objective function to maximize or minimize. It works by moving from one feasible corner point to another until the optimal solution is reached. The other options are incorrect because queuing theory deals with waiting lines, network analysis with project scheduling/PERT-CPM, and value analysis with cost reduction and functional improvement.
In four high rolling mill the bigger rollers are called
A)
guide rolls
B)
back up rolls
C)
main rolls
D)
support rolls
In a four-high rolling mill, the two small rolls are the working or main rolls that actually contact the metal, while the larger rolls are placed behind them to provide support and reduce deflection. These larger rolls are called back up rolls. Hence, option (b) is correct. Guide rolls are used for guiding the strip, and support rolls is a more general term, but the standard term in four-high mills is back up rolls.
A firm manufactures a product which is sold for Rs.10.50 per unit and the fixed cost of the assets is Rs.50000/- with a variable cost of Rs.6.50 per unit. How many units must be produced to earn a profit of Rs.1000/-? What would be the profit for sales volume of 20,000 units?
Manganese in steel increases its
A)
ductility
B)
hardness
C)
fluidity
D)
tensile strength
Manganese is added to steel mainly to increase its strength and toughness. It improves the tensile strength of steel and also helps remove sulfur by forming manganese sulfide, reducing brittleness. Among the given options, tensile strength is the best answer. Ductility is not primarily increased, hardness is not the main effect here, and fluidity is more associated with cast iron or molten metal behavior.
The most economical furnace to melt cast iron in large volume is
A)
reverberatory furnace
B)
cupola furnace
C)
induction furnace
D)
electric furnace
A cupola furnace is the most economical furnace for melting cast iron in large quantities. It is widely used in foundries because it has a high melting rate, low operating cost, and is suitable for continuous production. Reverberatory furnaces are less efficient for cast iron, while induction and electric furnaces are generally costlier for this specific large-volume melting application.
Eye bolts are used for
A)
locking devices
B)
lifting and transportation of machines and cubicles
C)
absorbing shock and vibration
D)
transmission of power
Eye bolts have a looped head designed to attach hooks, ropes, or chains, so they are commonly used for lifting, hoisting, and transporting heavy equipment such as machines and cubicles. Option (a) is incorrect because locking devices are different fastening elements. Option (c) is incorrect because shock and vibration are absorbed by springs, rubber pads, or similar devices. Option (d) is incorrect because power transmission is done by shafts, belts, gears, etc., not eye bolts.
The strain energy stored in a solid circular shaft of torsional rigidity GJ and length L, subjected to a twisting moment T is given by
A)
TL/GJ
B)
T2L/GJ
C)
T2L/2GJ
D)
GJ/TL
For a shaft in torsion, the angle of twist is theta = TL/GJ. The strain energy stored is U = (1/2) × T × theta. Substituting theta, we get U = (1/2) × T × (TL/GJ) = T^2L/2GJ. Hence option (c) is correct. Option (a) is only the angle of twist, option (b) misses the factor 1/2, and option (d) is dimensionally incorrect for strain energy.
The bearing characteristic number of a journal bearing depends upon
A)
length, width and load
B)
length, width and speed
C)
viscosity, speed and load
D)
viscosity, speed and bearing pressure
The bearing characteristic number for a journal bearing is the Sommerfeld number, commonly written as S ∝ (ηN/P) × (r/c)^2, where η = viscosity of lubricant, N = speed, and P = bearing pressure (load per unit projected area). Thus, it depends directly on viscosity and speed, and inversely on bearing pressure.
Why others are incorrect:
(a) Length, width and load: these are not the main variables in the characteristic number itself.
(b) Length, width and speed: viscosity and pressure are essential, so this is incorrect.
(c) Viscosity, speed and load: the correct term is bearing pressure, not just total load.
Hence, the correct option is (d).
Maximum shear stress in a Mohr’s circle is
A)
equal to diameter of the Mohr’s circle
B)
equal to radius of the Mohr’s circle
C)
less than the radius of the Mohr’s circle
D)
more than the radius of the Mohr’s circle
In Mohr’s circle for plane stress, the maximum shear stress is represented by the maximum vertical distance from the center of the circle to its outermost point. This value is exactly the radius of the circle. Hence, maximum shear stress = radius of Mohr’s circle.
Why others are incorrect:
(a) Diameter is twice the radius, so it is too large.
(c) It is not less than the radius; it is exactly equal.
(d) It is not more than the radius; it is exactly equal.
100 kW of power is to be transmitted by each of two separate shafts, made of the same material. Shaft A is turning at 250 rpm and shaft B at 300 rpm. Which shaft must have greater diameter?
A)
A
B)
B
C)
Both will have the same diameter
D)
None of these
For a shaft transmitting power, torque is given by P = 2πNT/60, so T = 60P/(2πN). For the same power, torque is inversely proportional to speed. Therefore, the shaft running at lower speed must transmit higher torque. Shaft A runs at 250 rpm, which is lower than shaft B at 300 rpm, so shaft A carries greater torque. Since shaft diameter depends on the torque it must withstand, shaft A must have the greater diameter. Hence, option (a) A is correct.
When the relative motion between two elements is completely or successfully constrained, then these two elements form a
A)
mechanism
B)
machine
C)
kinematic pair
D)
none of these
When the relative motion between two elements is completely or successfully constrained, the two elements together are called a kinematic pair. A mechanism is a combination of links arranged to transmit motion, and a machine is a mechanism or collection of mechanisms that transmits and modifies energy. Hence, the correct answer is kinematic pair.
Critical damping is a function of
A)
mass and stiffness
B)
mass and damping coefficient
C)
stiffness and natural frequency
D)
stiffness and damping coefficient
For a single degree of freedom system, the critical damping coefficient is c_c = 2mω_n. Since ω_n = √(k/m), substituting gives c_c = 2√(km). Thus, critical damping depends on mass (m) and stiffness (k). इसलिए correct option is (a).
Why others are incorrect:
(b) Critical damping itself is not a function of the existing damping coefficient; rather, it is the reference value used to compare actual damping.
(c) Though c_c can be written using natural frequency, ω_n itself depends on mass and stiffness; the standard direct dependence is on m and k.
(d) It does not depend on the actual damping coefficient.
The ratio of the torsional strength of a hollow shaft to that of a solid shaft of equal length and weight and made of same material is
A)
more than one
B)
less than one
C)
equal to one
D)
none of these
For shafts of equal length, equal weight, and same material, the cross-sectional areas are equal. A hollow shaft places more material away from the center, which increases the polar section modulus and therefore the torque-carrying capacity.
Torsional strength is proportional to polar section modulus Zp = J/R. For the same area, a hollow shaft has larger Zp than a solid shaft. Hence, the hollow shaft can transmit greater torque before reaching the same allowable shear stress.
Therefore, the ratio of torsional strength of hollow shaft to solid shaft is greater than 1.
Why others are incorrect:
- (b) less than one: incorrect, because hollow shafts are stronger in torsion for the same weight.
- (c) equal to one: incorrect, strength is not the same.
- (d) none of these: incorrect, since option (a) is correct.
In a solid circular shaft subjected to pure torsional moment, the ratio of maximum shear stress to maximum normal stress at any point is:
A)
2
B)
1
C)
2/3
D)
1/2
In a solid circular shaft under pure torsion, the stress at a point is pure shear stress only. So the maximum shear stress is τmax, and the principal normal stresses developed are +τmax and -τmax. Hence the maximum normal stress in magnitude is also τmax. Therefore, the ratio of maximum shear stress to maximum normal stress is τmax/τmax = 1. So option (b) is correct. Other options are incorrect because they do not match the stress transformation result for pure shear.
Columns which fail by direct stress only are called
A)
short column
B)
medium column
C)
weak column
D)
long column
A short column fails primarily by crushing due to direct compressive stress, with negligible buckling effect. Medium columns fail by a combination of direct stress and buckling, while long columns fail mainly by buckling. Therefore, the correct answer is short column.
If a rectangular beam is loaded transversely maximum stress develops at
A)
neutral layer
B)
bottom layer
C)
middle layer
D)
top layer
In a rectangular beam subjected to transverse loading, bending stress varies linearly from zero at the neutral axis to maximum at the extreme outer fibres. Thus, the maximum stress develops at the top and bottom layers, not at the middle or neutral layer. Since only one option is to be chosen and the bottom layer is one of the extreme fibres, the correct option is (b).
Why others are incorrect:
(a) Neutral layer: stress is zero here.
(c) Middle layer: same as near neutral axis, not maximum.
(d) Top layer: also an extreme fibre, but in usual single-answer convention for a sagging beam, bottom layer has maximum tensile stress.