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S.No Question Option A Option B Option C Option D Answer Solution Comments Status Action
1 A boat goes down stream 60 km in 2 hours and goes upstream 18 km in 3 hours. The speed of the boat in still water is 12 km/hr. 30 km/hr. 6 km/hr. 18 km/hr. d Let the speed of boat be B and the speed of stream be S.
⇒ Speed of the boat in downstream = 60/2 = 30km/hour
⇒ Speed of the boat in upstream = 18/3 = 6km/hour
⇒ Speed of the boat = (30 +6)/2 = 18km/hour		 Comments Active
2 The equation of the line joining the points (3,-1) and (-4, 5) is ? 6x + 7y + 11 = 0 6x - 7y - 11 = 0 -6x + 7y + 11 = 0 6x + 7y - 11 = 0 d Y - Y1 = (x - ) \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) \(x_{1}\)
Y + 1 = (x -3) \(\frac{5-1}{-4-3}\)
-7y - 7 = 6(x-3)
6x + 7y - 11 = 0		 Comments Active
3 An aeroplane starts from a place and flies 10M in a straight line, at 45 to the horizontal. Find the horizontal distance covered? \(2\) \(°\) 20 10 10 \(2\) 10/ \(2\) b Let horizontal distance be 'b'Cos45°=b/101/√2=b/10B=10 \(2\) \(2\) Comments Active
4 The sixth term in the expansion of (x + 1/x)10 is ? 250 252 10C6 10C7 b We know that Tr+1 = Cr an-rbr.
In the given binomial expression ( x + )10, n = 10, a = x and b = . \(\frac{1}{x}\) \(\frac{1}{x}\)
Tr+1 = 10Cr x10 - r = 10Cr (1)r \(∴\) \((\frac{1}{x})^{r}\) \((x)^{10-2r}\)
For the term to be independent of x, we must have 10 - 2r = 0
R =5.
The required term is:
10C5(1)5 = 10C5 = 252		 Comments Active
5 If x% of 450 is 67.5, the value of x is 15 25 85 50 a We already have our first value 450 and the second value 67.5. Let's assume the unknown value is Y which answer we will find out.
As we have all the required values we need, Now we can put them in a simple mathematical formula as below:
Y = 67.5/450
By multiplying both numerator and denominator by 100 we will get:
Y = 67.5/450 × 100/100 = 15 /100
Y = 15
Finally, we have found the value of Y which is 15
Or
67.5 ÷ 450 × 100 = 15		 Comments Active
6 Two coins are tossed simultaneously. Find the probability of getting at least one tail? 1/4 3/4 4/3 1/2 b Elementary events associated to random tossing of two coins is HH, HT, TH, TT = 4
At least one tail means 1 or more than 1 tail I.e. TH, HT, TT.
So, no. Of favourable events = 3
Required probability = \(\frac{3}{4}\)		 Comments Active
7 Pascal’s law states that the pressure at a point is equal in all directions for: Confined compressible fluid Confined Incompressible fluid Fluid having laminar /Viscous flow All fluids b Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. Comments Active
8 The pressure P of an ideal gas and its mean kinetic energy E per unit volume are related as P = E/2 P = E P = 3E/2 P = 2E/3 d Comments Active
9 A vibrating system is said to be critically damped when the damping factor is Zero Unity 1/2 Infinity b Damping ratio (ζ): The ratio of the actual damping coefficient (C) to the critical damping coefficient Cc) is known as damping factor or damping ratio.

Over damped System: ζ > 1
Underdamped: ζ < 1
Critical Damping: ζ = 1 
The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion.		 Comments Active
10 Which mechanism produces mathematically exact straight line motion Watt’s mechanism Peaucellier mechanism Ackermann mechanism Grasshopper mechanism b Exact straight-line motion mechanisms :
Mechanisms in which coincidence of the coupler curve arcs with a straight line is theoretically exact Example: Peaucellier mechanism, Hart’s mechanism, Scott Russell’s Mechanism
Approximate straight-line motion mechanisms :
Mechanisms in which coincidence of the coupler curve arcs with a straight line is only approximate Example: Watt’s mechanism, Modified Scott-Russell mechanism		 Comments Active
11 In a cantilever beam with point load at its free end, the maximum bending moment occur at Centre of the beam Free end of the beam Fixed end of the beam At the point of application of the load c Comments Active
12 If an unconstrained steel bar is heated uniformly, there develops Thermal stress Shear stress Tensile stress No stress d In other words, if the material is constrained (i.e. Body is not allowed to expand or contract freely), change in length due to rise or fall of temperature is prevented, stresses are developed in the body which is kNown as thermal stress. Comments Active
13 The CLA value is used for the measurement of Surface flatness Hardness Surface roughness Internal voids c Comments Active
14 The stress due to unit strain is Modulus of elasticity Modulus of rigidity Principal stress Normal stress a Comments Active
15 The state of stress at a point under plane stress condition is xx = 40MPa, yy = 100MPa and xy = 40MPa. The radius of the Mohr’s circle representing the given state of stress in MPa is \(σ\) \(σ\) \(τ\) 40 50 60 100 b Maximum shear stress is given by
max = = \(Τ\) \(\frac{Σ_{Max}-Σ_{Min}}{2}\) \((\frac{Σ_{Xx}-Σ_{Yy}}{2})^{2}+ ΤXy2\)
max = R = \(Τ\) \((\frac{60-120}{2})^{2}+ 40^{2}\)
-> R = 50 MPa		 Comments Active
16 The potential energy of a particle executing SHM varies sinusoidally. If the frequency of oscillation of particle is n. That of potential energy is n/2 n/ \(2\) n 2n d
In a single cycle that is from point 12341 kinetic energy reaches its maximum value two times. So we can say that if the frequency of the SHM is f then its frequency will be 2 times that is 2f when its kinetic energy changes to potential energy.		 Comments Active
17 Friction at the tool-chip interface can be reduced by Decreasing the rake angle Increasing the depth of cut Decreasing the cutting speed Increasing the cutting speed d High cutting speed results in lesser chip tool contact time and better heat dissipation. These two effects make it infavourable condition to form a bond between tool and chip and the coefficient of friction is reduced. Comments Active
18 What type of lubricant is graphite Solid Liquid Semi solid \(σ\) None of the above a Comments Active
19 Which of the following welding uses non consumable electrodes SAW MIG GTAW Thermit c Comments Active
20 If a closed system is undergoing an irreversible process, the entropy of the system must increase always remains constant must decrease can increase , decrease or remain constant d If a closed system is undergoing an irreversible process, the entropy of the system can increase, decrease or remain constant. This is because the change in entropy of a system undergoing an irreversible process depends on the nature of the process and the initial and final states of the system. Comments Active
21 Dew point temperature is the temperature at which condensation starts when air is cooled at constant: Volume Pressure Enthalpy None of the above b Comments Active
22 A concentrated mass m is attached at the centre of a rod of length 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the undamped natural frequency of the system is \(\frac{K}{M}\) \(\frac{2k}{M}\) \(\frac{K}{2m}\) \(\frac{4k}{M}\) d Displacing rod by small distance x

Taking moment about ‘O’
Kx 2L = mg L \(×\) \(×\)
X = \(\frac{Mg}{2K}\)
From similar triangle property
\(\frac{X}{2L} = \frac{Δ}{L}\)
= \(Δ\) \(\frac{X}{2} = \frac{Mg}{4k}\)
Using static deflection of the mass ‘m’
n = \(Ω\) \(\frac{G}{Δ}=\) \(\frac{4K}{M}\)		 Comments Active
23 The height to which a liquid will rise in an open capillary tube is inversely proportional to Temperature of liquid Density of liquid Air pressure Surface tension b Comments Active
24 A condenser of refrigeration system rejects heat at rate of 120kW, while its compressor consumes power of 30kW, The COP of the system is 4 1/4 3 1/3 c COP = \(\frac{Desired effect}{Work input} =\) \(\frac{90}{30} = 3\) Comments Active
25 In order to have interference fit, it is essential that the lower limit of the shaft should be Greater than the upper limit of the hole Lesser than the upper limit of the hole Greater than the lower limit of hole Lesser than the lower limit of the hole a The diameter of the shaft is always larger than the hole diameter.
 Comments Active
26 If two metallic plates of equal thickness and thermal conductivities K1 and K2 are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be \(\frac{K_{1}K_{2}}{K_{1}+K_{2}}\) \(\frac{2K_{1}K_{2}}{K_{1}+K_{2}}\) \(\frac{(K12+K22)^{3/2}}{K_{1}K_{2}}\) \(\frac{(K12+K22)^{3/2}}{2K_{1}K_{2}}\) b Using Kirchhoff’s lawEquivalent resistance = R1 + R2
R1 = , R2 = \(\frac{1}{K_{1}}\frac{L_{1}}{A_{1}}\) \(\frac{1}{K_{2}}\frac{L_{1}}{A_{2}}\)
From question, l1 = l2 and A1 = A2
Req = \(\frac{2l}{K_{Eq}A} = \frac{L}{K_{1}A} + \frac{L}{K_{2}A}\)
\(K_{Eq} = \frac{2K_{1}K_{2}}{K_{1}+K_{2}}\)		 Comments Active
27 Two metal strips that constitute a thermostat must necessarily differ in their Mass Length Resistivity Coefficient of linear expansion d Comments Active
28 Surface tension has the units of: Force Force per unit length Force per unit Area Force per unit volume b Comments Active
29 In stroboscopic measurement, a flywheel having a mark on the rim was seen as two marks at 100Hz signal, its rotational speed is 100 rpm 6000 rpm 3000 rpm 1.67 rpm c Comments Active
30 Two 1mm thick steel sheets are to be spot welded at a current of 5000A. Assuming effective resistance to be 200 micro-ohms and current flow time of 0.2 second, heat generated during the process is 0.2 Joule 1 Joule 5 Joule 1000 Joule d I = 5000AmpR = 200 micro ohm = 200 \(×10^{-6}Ω\)
T = 0.2 sec, H = ?
Since, H = I2RT
H = 0.2 \((5000)^{2}× 200 ×10^{-6}×\)
H = 1000J		 Comments Active
31 Babbit is an alloy of Sn and Cu Cu and Zn Sn, Cu, Sb and Pb Sn, Cu and Pb c Comments Active
32 Within the HAZ in fusion welding, the material undergoes: Microstructure changes however, material will not melt. Neither melting not microstructure changes since the temperature at HAZ is less than the upper critical temperature. Both melting and microstructure changes Material melts however, after solidification, original microstructure is retained a The heat-affected zone (HAZ) is the area of the base material, either a metal or a thermoplastic, which is not melted and has had its microstructure and properties altered by welding or heat intensive cutting operations. The extent and magnitude of property change depend primarily on the base material, the weld filler metal, and the amount and concentration of heat input by the welding process. Comments Active
33 Maximum power transmission in a belt is met when the total tension in the belt equals centrifugal tension 2 times centrifugal tension \(Ï€\) Thrice centrifugal tension Half the centrifugal tension c Comments Active
34 To find fits on hole basis system Shaft is kept constant and hole size is changed Hole is kept constant and shaft size is changed Upper deviation of shaft is zero Lower limit of hole is not same as basic size b Size of the hole is kept constant and the size of the shaft is varied to get the different class of fits. In this system, a lower deviation of the hole is zero, i.e. The low limit of the hole is same as basic size. The high limit of the hole and the two limits of size for the shaft are then varied to give the desired type of fit. Comments Active
35 Double hemispherical buckets are used in Kaplan turbine Francis turbine Propeller turbine Pelton turbine d Comments Active
36 Two bodies m1 and m2 (m1>m2) have the same kinetic energy. Then their momentum P1 and P2 satisfy P1 = P2 P1 > P2 P1 < P2 P1 = -P2 b m1 > m2 and,
\(\frac{1}{2}m_{1}v12=\frac{1}{2}m_{2}v22\)
We can rewrite it as,
\(\frac{1}{2}×\frac{(m_{1}×v_{1})^{2}}{m_{1}}=\frac{1}{2}×\) \(\frac{(m_{2}×v_{2}) ^{3}}{m_{2}}\)
\(\frac{p12}{m_{1}}=\frac{p22}{m_{2}} \)
……(3) \(\frac{p_{1}}{p_{2}}= \frac{m_{1}}{m_{2}}\)
Since m1 > m2 RHS becomes greater that 1, which implies
\(\frac{p_{1}}{p_{2}}>1\)		 Comments Active
37 For a Newtonian fluid Shear stress is proportional to shear strain Rate of shear stress is proportional to shear strain Shear stress is proportional to rate of shear strain Rate of shear stress is proportional to rate of shear strain c Comments Active
38 Presence of Hydrogen in steel results in: Corrosion resistance Improved weldability No effect on steel due to small size of hydrogen atom Embrittlement d Hydrogen is a common impurity in steel, and its presence can have significant effects on the material properties. Here are the effects of hydrogen in steel:
1. Embrittlement
2. Reduced Ductility
3.DelayedCracking4. Corrosion		 Comments Active
39 A satellite of mass 1000kg orbits around the earth at an altitude of 100km, will experience a weight of 1000 kgf Zero 750 kgf 9800 kgf b As acceleration due to gravity is zero in space
Therefore weight of a planet which orbits around the earth = m*g = m*0 = 0		 Comments Active
40 In 18-4-1 HSS, what is the percentage of chromium? 1% 4% 18% None of the above b Comments Active
41 A Gantt chart provides information about Material handling Production schedule Proper utilization of man power All of the above b Comments Active
42 The most appropriate failure theory for the ductile materials is Maximum principle theory Maximum shear stress theory Maximum principal strain theory Maximum shear strain energy theory b Comments Active
43 A hydraulic turbine develops 1000kW power for a head of 40m. If the head is reduced to 20m, the power developed (in kW) is 177 354 500 707 b Since \(\frac{P^{2}}{H^{3}D^{4}} = Constant\)
\(P22 = (P_{1})^{2}×(\frac{H_{2}}{H_{1}})^{3}\)
= 354kW \(P22 = (1000)^{2}×(\frac{20}{40})^{3}\)		 Comments Active
44 Bending moment at a hinge is: 0 Maximum Minimum Depends on the loading a Comments Active
45 A bullet is fired to get maximum range. The angle at which it is to be fired (neglecting drag in air) with respect to the horizontal is: = 90 \(θ\) \(°\) = 30 \(θ\) \(°\) = 60 \(θ\) \(°\) = 45 \(θ\) \(°\) d Rmax = \(\frac{u^{2}}{g}\) Comments Active
46 In a close-coiled helical spring subjected to an axial load other quantities remaining the same, if the wire diameter is doubled, then the stiffness of the spring when compared to the original one will become: Twice Four times Eight times Sixteen times d Stiffness of the spring is given by:
 Comments Active
47 The pressure within the cylinder of hydraulic press is 9MPa. The inside diameter of the cylinder is 500mm. If the permissible tensile stress is 10N/mm2, the minimum wall thickness of the cylinder is 12.5 mm 16 mm 9 mm 18 mm a \(Σ_{H} = \frac{Pd}{2t}\)
T = = 12.5mm \(\frac{9}{2}×\frac{500}{180}\)		 Comments Active
48 The condition for self-locking in screw is ( = coefficient of friction, = helix angle) \(μ\) \(α\) Tan \(μ≥\) \(α\) Tan \(μ\) \(≤\) \(α\) = \(μ\) \(α\) \(μ\) \(≥\) \(α\) a Comments Active
49 In an automobile car where is the Hook’s joint used? Between clutch and gear box Between differential gear and wheels Flywheel and clutch Between gear box and propeller shaft d Comments Active
50 Which of the following does not change during throttling process Internal energy Pressure Entropy Enthalpy d Comments Active
51 A body falls freely from rest from the top of a tall cliff. The distance it travelled in the first second is 9.8 m 10 m 4.9 m 1 m c Applying second equation of motion
S = 4.9m		 Comments Active
52 For a stability of a floating body, under the influence of gravity alone, which of the following is TRUE? Metacentre should be below centre of gravity. Metacentre should be above centre of gravity. Metacentre and centre of gravity must lie on the same horizontal line. Metacentre should be at canter of gravity. b Comments Active
53 A column has a rectangular cross-section of 10*20mm and a length of 1m. The slenderness ratio of the column is close to 200 346 477 1000 b Slenderness ratio = \(\frac{Effective length L_{E}}{Minimum radius of gyration k}\)

Ixx = = = 6666.67mm4 \(\frac{Bd^{3}}{12}\) \(\frac{10×(20)^{3}}{121}\)
IYY = = = 1666.67mm4 \(\frac{Db^{3}}{12}\) \(\frac{20×(10)^{3}}{12}\)
Since, IYY < Ixx
I = A k2 \(∴ \)
K = \(\frac{I}{A}\)
Kmin = = \(\frac{I_{Y-y}}{A}\) \(\frac{1666.67}{200}\)
Kmin = 2.888mm
Therefore Slenderness ratio = = 346.26 \(\frac{1000}{2.888}\)		 Comments Active
54 The generalized Hook’s law is applicable up to Yield point Braking point Linear elastic point Plastic Limit c Comments Active
55 In a state of stress, principal planes are planes of Zero normal stress Zero shear stress Equal normal and shear stresses Maximum shear stress b Comments Active
56 The locus of a point on a string unwound from a circular disc is A circle A cycloid An involute A parabola c Comments Active
57 A cantilever beam of length L is subjected to a moment M at the free end. The moment of inertia of the beam cross section about the neutral axis is 1 and the Young’s modulus is E. The magnitude of the maximum deflection is \(\frac{ML^{2}}{2EI}\) \(\frac{ML^{2}}{EI}\) \(\frac{2ML^{2}}{2EI}\) \(\frac{4ML^{2}}{2EI}\) a Comments Active
58 A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter? Young’s modulus Shear modulus Poisson’s ratio Both a and b d Lateral deformation (ϵd) = -μ × Longitudinal deformation (ϵl)
So, to calculate the change in diameter (ϵd) we need both μ and longitudinal deformation.
From the relation E = 2G(1+μ) we can calculate Poisson's ratio if we know the value of both Shear Modulus (G) and Young's Modulus (E).		 Comments Active
59 Ring rolling is used For producing seamless tubes To increase thickness of a ring To decrease thickness of a ring For producing large cylinder c Comments Active
60 What is the percentage increase in tool life when the cutting speed is halved? 300% 400% 5% 200% a Comments Active
61 Torsion in a shaft produces Tensile stress Compressive stress Shear stress Bending c Comments Active
62 A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of T. The Young’s modulus is E and the coefficient of linear expansion is . The thermal stress in the rod is \(∆\) \(α\) 0 \(α\) ET \(α∆\) ETL \(α∆\) c Comments Active
63 This material is used for producing laser Graphite Ruby Diamond Emerald b Comments Active
64 Which of the following metal has the lowest melting point? Antimony Tin Silver Zinc b Comments Active
65 A spherical black body of radius r radiates power P, and its rate of cooling is R, then P r2 \(α\) P r \(α\) R r2 \(α\) R 1/r \(α\) a We know that rate of radiation heat transfer per unit area is proportional to T4 .
P ∝ AT4 ( P is radiation power )
Also surface area, A ∝ r 2
⇒ P ∝ r 2 . . . . . (i)		 Comments Active
66 A liquid flows in the tube from left to right as shown in figure. A1 and A2 are the cross-sections of the portions of the tube as shown. Then the ratio of speeds v1/v2 will be \(\frac{A_{1}}{A_{2}}\) \(\frac{A_{2}}{A_{1}}\) \(\frac{A_{2}}{A_{1}}\) \(\frac{A_{1}}{A_{2}}\) b Comments Active
67 In the case of perfectly elastic collision, the coefficient of restitution is 1.0 0.5 Less than 1.0 Zero a For a collision between two objects, the coefficient of restitution is the ratio of the relative speed after to the relative speed before the collision.
The coefficient of restitution is a number between 0 (perfectly inelastic collision) and 1 (elastic collision) inclusive.
Coefficient of restitution (e)
= \(\frac{Relative velocity after collosion }{Relative velocity before collosion}\) \(\frac{v_{b}-v_{a}}{v_{a}-v_{b}}\)
In the case of a ball bouncing off a flat, stationary surface, the coefficient of restitution turns out to be: v/u		 Comments Active
68 Bernoulli’s equation is the statement about fluid flow for Conservation of mass Conservation of momentum Force balance Conservation of energy d The Bernoulli Equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids. It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline. Comments Active
69 Emissivity of a black body is: 1 0 infinite Cannot be determined a Emissivity = e/E
Where e is emissive power and E is the emissive power of the black body
The value of emissivity lies between 0 and 1, where the emissivity of the perfect black body is 1.		 Comments Active
70 Which of the following theories is related to the theory of the thermocouple? Piezoelectric effect Seeback effect Skin effect Faraday’s Law b The See beck effect is a phenomenon in which a temperature difference between two dissimilar electrical conductors or semiconductors produces a voltage difference between the two substances. Comments Active
71 The thermal offered by a rectangular wall of thermal conductivity k, thickness t and cross sectional area A (assume no heat generation) is given by: t/(k/A) (kA)/t (kt)/A (At)/k a Comments Active
72 Reynolds’s number is the ratio of the inertia force to the surface tension force viscous force gravity force pressure force b Comments Active
73 The most efficient section of a flow channel is Circular Rectangular Square Trapezoidal d Comments Active
74 Charpy test is conducted to measure Hardness Malleability Brittleness Machinability c Charpy tests show whether a metal can be classified as being either brittle or ductile. This is particularly useful for ferritic steels that show a ductile to brittle transition with decreasing temperature Comments Active
75 Ball bearings are generally made of Cast Iron Aluminium Chrome steel None of the above c Comments Active
76 A tank is filled with water up to a height H. Water is allowed to come of a hole P in one of the walls at a depth ‘h’ below the surface of water. Express the horizontal distance x in terms of H and h
 x = (H-h) \(H\) x = \((\frac{H(H-h)}{2})\) x = (H-h) \(2H\) x = 4(H-h) \(H\) c Let’s assume (H - h) = y
Since, Cv = \(\frac{X^{2}}{4yh}\)
Squaring both side
= \(CV2\) \(\frac{X^{2}}{4yh}\)
\(CV2×\) \(4yh=x^{2}\)
Taking square root both side
Cv = x \(×\) \(4yh\)
Considering Cv = 1
X = (H-h) \(2H\)		 Comments Active
77 Two bodies with moment of inertia I1 and I2 (I1>I2) have equal angular momentum. If Kinetic Energy of rotation are E1 and E2 then E1 E1>E2 E1 = E2 E1 E2 \(≥\) a
I1>I2
Angular momentum (J) = Iω
I1ω1 = I2ω2 …………(1)
Since kinetic energy E1 = …………(2) \(\frac{1}{2}I_{1}Ω12\)
E2 = …………(3) \(\frac{1}{2}I_{2}Ω22\)
Dividing equation 2 by 3
= ……….(4) \(\frac{E_{1}}{E_{2}}\) \(\frac{1/2I_{1}Ω12}{1/2I_{2}Ω22}\)
Putting value from 1 to 4
\(\frac{E_{1}}{E_{2}} = \frac{I_{2}}{I_{1}}\)
Therefore, E 1/I \(∝\)
Since I1>I2
Therefore E1 < E2		 Comments Active
78 A vertical column has two moments of inertia Ixx and Iyy, The column will tend to buckle in the direction of the axis of load perpendicular to the axis of load maximum moment of inertia minimum moment of inertia d Comments Active
79 Positive displacement flow meters are ………… flow meters. Variable area flow Differential pressure flow Quantity flow None of these c Comments Active
80 If the resultant of two forces P and Q acting at an angle makes an angle with the force P, then \(θ\) \(α\) tan = \(α\) \(\frac{P sinθ}{P + Q cosθ}\) tan = \(α\) \(\frac{P cosθ}{P + Q cosθ}\) tan = \(α\) \(\frac{Q sinθ}{P + Q cosθ}\) tan = \(α\) \(\frac{Q cosθ}{P + Q sinθ}\) c Comments Active