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S.No Question Option A Option B Option C Option D Answer Solution Comments Status Action
1 In order to prevent formation of carbon on the injector, the temperature (in ᵒC) of nozzle tip should be less than 100 between 100 – 250 between 250 – 300 between 400 – 500 c Carbon buildup on the tip of injector can restrict fuel delivery causing the engine to run lean. This can cause several issue such as misfire, poor fuel economy and increased risk of detonation and preignition. To prevent carbon buildup injector temperature should be kept between 250℃ to 300℃. Comments Active
2 Installation of supercharger on a four – cycle diesel engine can result in the following percentage incrase in power upto 25% upto 35% upto 50% None of these d A superchanger (Turbocharger) works by compressing an engine’s airflow before it actually enters the engine. By adding more, the engine can also mix in more fuel, which results in more power.
* A supercharger can improve an engine’s performance by 50% but it can also reduce engines efficiency by 20%.		 Comments Active
3 Which of the following is a false statement? Excess quantities of Sulphur in diesel fuel are objectionable because it may cause the following: piston ring and cylinder wear formation of hard coating on piston skirts oil sludge in the engine crank case detonation d Ammount of sulphur in diesel fuel and gasoline must be kept very small amount because –
* Too much sulphur will cause wear of cylinder and piston due to formation of sulphuric acid.
* Sulphur may react with lubrication and form sludge.
However, detonation (or knocking) is generally caused by factors such as:
Low cetane number of fuel,
Incorrect air-fuel mixture,
High compression ratio, etc.		 Comments Active
4 The accumulation of carbon in a cylinder result in increase of clearance volume volumetric efficiency ignition time effective compression ratio d * Carbon deposits on the top of cylinder results in decreased clearance volume.
* Compression ratio defined as ratio of total volume to the clearance volume.
\(r=\frac{V_{c}+V_{s}}{V_{c}}=1+\frac{V_{s}}{V_{c}}\)
Hence compression ratio increases.		 Comments Active
5 Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of the surrounding atmosphere providing forced cooling air injecting excess fuel for raising more load supplying compressed air to remove combustion products fully a Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of surrounding atmosphere so that each intake cycle of engine gets more oxygen, which burns more fuel thus generating more power . Comments Active
6 The diameter of tubes for natural circulation boiler as compared to controlled circulation boiler is more less same could be more or less depending on other factors a In natural circulation boilers, the circulation of water depends on the difference between the density of an ascending mixture of hot water and steam and a descending body of relatively cool and steam free water.
Controlled circulation boilers depend upon pumps rather than upon natural differences in density, for circulationof wtaer within the boiler.
The diameter of tubes for natural circulation boiler as compared to controlled circulation boilers is more.		 Comments Active
7 Cochran boiler is a horizontal fire – tube boiler horizontal water – tube boiler vertical water – tube boiler vertical fire tube boiler d Cochran boiler is a multi-tubular vertical fire tube boiler having numbers of horizontal fire tubes. It is the modification of a simple vertical boiler where the heating surface has been increased by means of numbers of fire tubes. Comments Active
8 At which pressure (in kg/cm2) the properties of water and steam become identical 0.1 1 100 225.6 d At the critical point,the saturated liquid and saturated vapour state are identical or same.
Critical point:221.2 bar and 225.5kg/cm2		 Comments Active
9 One kg of steam sample contains 0.8 kg dry steam; it’s dryness fraction is 0.2 0.8 0.6 0.5 bSolution:given mass of vapour =0.8kg
mass of vapour+mass of liquid=1kg,
Dryness fraction= mass of vapour/mass of vapour+mass of liquid =0.8/1=0.8		 Comments Active
10 Coke is produced by pulverizing coal in inert atmosphere heating wood in a limited supply of air at temperature below 300ᵒC strongly heating coal continuously for about 48 hours in the absence of air in a closed vessel binding the pulverized coal into briqettes c Coke is produced by destructive distillation of coal in coke ovens. Prepared coal is "coked", or heated in an oxygen-free atmosphere until all volatile components in the coal evaporate. The material remaining is called coke. Comments Active
11 The ratio of actual discharge to theoretical discharge through an orifice is \(C_{c}/D_{d}\) \(C_{d}/C_{v}\) \(C_{v}/C_{d}\) \(C_{c}C_{v}\) d Coefficient of discharge (cd) = Coefficient of Contraction (CC) Cofficient of velocity (CV) \(×\)
\(c_{d}=\frac{Actual discharge}{Theoretical discharge}\)		 Comments Active
12 Seperation of flow occurs when pressure gradient tends to approach zero becomes negative reduces to a value when vapor formation starts None of these d Flow separation occures when the boundary layer travels far enough against an adverse pressure gradient that the speed of the boundary layer relative to the object falls almost to zero.
It has been observed that the flow is reversed in the vicinity of the wall under certain conditions.

A favorable pressure gradient is one in which the pressure decreases in the flow direction (i.e., dp/dx < 0)
It tends to overcome the slowing of fluid particles caused by friction in the boundary layer
This pressure gradient arises when the freestream velocity U is increasing with x, for example, in the coverging flow field in a nozzle
On the other hand, an adverse pressure gradient is one is which pressrue increases in the flow direction (i.e., dp/dx > 0)
It will cause fluid particles in the boundary – layer to slow down at a greater rate than that due to bundary – layer friction alone
If the adverse pressure gradient is severe enough, the fluid particles in the boundary layer will actually be brought to rest.
When this occur, the particles will be forced away form the body surface (a phenomenon called flow separation) as they make room for following particles, ultimately leading to a wake in which flow is turbulent		 Comments Active
13 The rate of change of moment of momentum represent the force exerted by fluid torque applied by the fluid work done by the fluid power developed by the fluid b The rate of change of momentum is called force according to Newton’s second law of motion. But the rate of change of moment of momentum is called torque. Comments Active
14 The discharge through a semi-circular weir is proportional H(-1/2) H(1/2) H3/2) None of these c Eg = General relationship for discharge over weir is expressed as \(Q∝H^{n}\)
\(n=3/2 for Rectangular Weir (H)^{3/2}\)
\(n=5/2 for Traingular Weir (H)^{5/2}\)
\(n=1 For proportionla weir or Sutro Weir (H)\)
For Semi Circular Weir:
\(Q=C_{1}H^{3/2}+C_{2}H^{5/2}+C_{3}H^{7/2}\)
As coefficients of \(H^{5/2} and H^{7/2} reduce\)
To zero very quickly: \(Q=C_{1}H^{3/2}\)		 Comments Active
15 When a liquid rotates at constant angular velocity about a vertical axis as a rigid body, the pressure increases linearly as its radial distance varies inversely as the altitude along any vertical line varies as the square of the radial distance decreases as the square of the radial distance c A fluid is rotating at constant angular velocity about the central vertical axis of a cylinderical container. \(ω\)
The Variation of pressrue in the radial direction is given by:
\(\frac{dp}{dr}=rω^{2}ρ\)
It is given that the pressure at the axis of rotation is P(c)
Therefore, the required pressrue at any point r is :
\(p=p_{c}+\frac{1}{2}ρr^{2}ω^{2}\)
Hence we can say that pressure varies as the square of the radial distance.		 Comments Active
16 The discharge over a sharp edged rectangular notch of width h is equal to \(\frac{2}{3}C_{d}w2g h^{5/2}\) \(\frac{2}{3} C_{d}w2gh\) \(\frac{2}{3}C_{d}w2g h^{3/2}\) \(\frac{8}{15}C_{d}w2g h^{3/2}\) c
\(dQ=C_{d}×L×dh×2gh\)
\(Q=C_{d}×L2g0hhdh=\frac{2}{3}C_{d}L2g(H)^{\frac{3}{2}}\)		 Comments Active
17 The value of coefficient of velocity for a sharp edged orifice is of the order of 0.45 0.5 0.62 None of these d * Coefficient of velocity is defined as the ratio between actual velocity of a jet of liquid at vena-contracta and the theoretical velocity of jet.
* It varies form 0.95 to 0.99 for different orifice, depending on the shape, size of the oriface and on the head under which flow takes place.
* The value of coefficient of velocity is 0.98 for sharp edged orifice.		 Comments Active
18 For best hydraulic rectangular cross-section of an open channel, its depth should be equal to width two times the width half of the width tree – eighth of the width c
Wetted Area (A) = By
\(B=\frac{A}{y}\)
Wetted Perimeter (P) =B+2y
\(P=\frac{A}{y}+2y\)
For most economical, Perimeter should be minimum.
\(\frac{dP}{dy}=0=-\frac{A}{y^{2}}+2\)
\(A=2y^{2}, By=2y^{2}, b=2y\)
\(y=\frac{B}{2}\)		 Comments Active
19 Bluff body is the body of such a shape that pressure drag as compared to friction drag is same more less zero b A bluff body is defined as that body whose surface does not coincide with the streamlines when placed in a flow.

Then the flow is separated form the surface of the body much ahead of its trailing edge with the result of a very large wake formation. Then the drag due to pressrue will be very large as compared to the drag due to friction on the body.		 Comments Active
20 Chezy’s equation is used to determine velocity of flow in open channel velcoity of flow in pipe flow over weirs discharge through notch a The Chezy equation can be used to calculate mean flow velcity in open channel.
The loss of head in pipes due to friction considering the wetted perimeter is given by:
(1) \(h_{f}=\frac{f'}{ρg}×\frac{P}{A}×L×V^{2}\)
Where \(h_{f}=loss of head due to friction, f^{'}=frictional resistance per unit wetted area per unit velocity, P=wetted perimeter of pipe, L=Length of pipe\)
\(A=area of cross-section of pipe and V=mean velocity of flow.\)
Hydraulic Mean Depth (m):
It is the ratio of the area of flow to the wetted perimeter of flowing flui(d) It is given by-
\(m=\frac{Area of flow}{Wetted perimeter}=\frac{\frac{Ï€}{4}d^{2}}{Ï€d}=\frac{d}{4}\)
Rearranging the above equation (1):
\(V^{2}=\frac{ρg}{f'}×\frac{A}{P}×\frac{h_{f}}{L}\)
\(V=\frac{ρg}{f'}×m×\frac{h_{f}}{L}\)
\(V=\frac{ρg}{f'}×m×\frac{h_{f}}{L}\)
\(V=Cmi\)
Where \(i=\frac{h_{f}}{L}=Loss of head per unit length of pipe\)
\(C=\frac{ρg}{f'}=Chezy^{'}s constant\)		 Comments Active
21 Which of the following represents steady uniform flow? flow through an expanding tube at an increasing rate flow through an expading tube at constant rate flow through a long pipe at decreasing rate flow through a long pipe at constant rate d Steady Uniform flow
Flow at a constant rate through a duct of uniform cross-section (The region close to the walls of the duct is disregarded)
Steady non-uniform
Flow at a constant rate through a duct of non –uniform cross-section (tapering pipe)
Unsteady Uniform flow
Flow at varying rates through a long straight pipe of uniform cross-section. (Again the region close to the walls in ignored.)
Unsteady non-uniform flow
Flow at varying rates throught a duct of a non-uniform crss-section.		 Comments Active
22 Time of flow one tank in which water level is to another tank having level will be proportional to \(h_{1}\) \(h_{2}\) \(h_{1}-h_{2}\) \(h_{1}-h_{2} \) \(h_{1}-h_{2}\) - \(h13/2\) \(h23/2 \) c
Tank having cross-sectional Area “A”
* Let h is the height of water at any instant in the container an “dh” is change is height in “dt” time.
* Decreasing Rate of water in tank = Rate of discharge from orifice
\(-Adh=cd2gh.a.dt\)
\(-Ah_{1}h_{2}\frac{dh}{2gh}=cd.a0tdt\)
\(-A[\frac{1}{2g}\frac{h^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=cd.a.t\)
\(-\frac{A}{2g} 2(h_{2}-h_{1})=cd.a.t\)
\(t=\frac{2A}{cd2g.a}(h_{1}-h_{2})\)
\(t∝(h_{1}-h_{2})\)		 Comments Active
23 Friction factor of pipes depends on rate of flow fluid density viscosity All options are correct d For laminar flow friction factor depends on renolds number only while for turbulent flow friction factor depends on renold’s number and relative roughness of pipe both.
\(f=f(R,\frac{ε}{D})\)
Darcy friction factor, f depends on the flow’s Reynolds number Re
\((Re=\frac{ρVD}{μ})\)
And on pipe’s relative roughness , where is the pipe’s effective roughness height and D the pipe (inside) diameter \(ε/D\) \(ε\)		 Comments Active
24 In case of a two dimensional flow the components of velocity are given by the streamlines will consist of a series of \(u=ax;v=by,\) circular arcs parabolic arcs hyperbolic arcs elliptical arcs c Equation of streamline is given by –
\(\frac{dx}{u}=\frac{dy}{v}\)
Given u = ax and v = by
\(\frac{dx}{ax}=\frac{dy}{by}\)
Intergating both side –
\(\frac{1}{a}\frac{dx}{x}=\frac{1}{b}\frac{dy}{d}\)
\(\frac{1}{a}l_{n}(x)=\frac{1}{b}l_{n}(y)+l_{n}(c)\)
\(b l_{n}(x)=a l_{n}(y)+al_{n}(c)\)
Let \(a l_{n}(c)=c_{1}\)
\(l_{n}x^{b}-l_{n}y^{a}=c_{1}\)
\(l_{n}(\frac{x_{b}}{y^{a}})=c_{1}\)
\(x^{b}y^{-a}=c_{2}\)		 Comments Active
25 Reynolds number for non – circular cross section is:
(V = mean velocity
= kinematic velocity \(v\)
P = Ratio of cross sectional area to the wetter perimeter)		 \(\frac{V.4P}{v}\) \(\frac{V.P}{v}\) \(\frac{V.2P}{4vn}\) \(\frac{V.P}{4v}\) a The Reynolds number is the ratio of inertial forces to viscous forces witin a fluid.
\(Re=\frac{ρVD_{h}}{μ}=\frac{VD_{h}}{ν}\)
Where is density, U is the velocity , Dh is a hydraulic diameter, is the dynamic viscosity and is the kinematic viscosity of the fluid. \(ρ\) \(μ\) \(ν\)
\(D_{h}=\frac{4×cross section area}{Wetted perimeter}=\frac{4A}{P_{w}}=4\frac{A}{P_{w}}=4P\)
\(R_{e}=\frac{VD_{h}}{ν}=V\frac{4P}{ν}\)		 Comments Active
26 The critical velocity as maximum attainable velocity terminal velocity velocity when hydraulic jump occurs velocity above which the flow ceases to be streamlined d Critical velocity is that velocity of liquid flow up to which its flow is streamlilned and above which its flow becomes turbulent.
Hence the critical velocity of a liquid is that velocity of flow above which the flow ceases to be streamlined.
\(v_{c}=K\frac{η}{ρr}\)
Where is the coefficient of viscosity, is the density of the liquid and r is the radius of the tube. K is constant, known as the Reynolds number. \(η\) \(ρ\)		 Comments Active
27 Borda’s mouthpiece is a short cylindrical tube projecting inward, having length of ½ diameter a convergent tube having length of 2 – 3 diameters most commoly used rarely used a Borda’s mouthpiece is a short cylindrical tube projecting inward. The edge of the tube must be relatively thin and sharp to ensure perfect contraction, and its length, about ½ diameter, such that the jet will not touch the sides of the tube.
 Comments Active
28 The most economical section of circular channel for maximum discharge is obtained when (where, d is the diameter of circular section). depth of water = 0.95d wetter perimeter = 2.6d hydraulic mean depth = 0.29d Any one of these a \(D=Depth of flow\)
d = diameter of pipe
\(R_{m}=Hydraulic mean depth=A/P \)

Perimeter, P =d \(α\)
Condition of maximum discharge for circular section:
\(α=154°=2.65 radians\)
\(D=0.95 d\)
\(R_{m}=0.29 d\)
Wetter perimeter \(=2αr=αd=2.65d\)
Condition for Maximum Velocity for Circular Section:
\(α=128.75°=2.25 radians\)
\(D=0.81 d\)
\(R_{m}=0.3 d\)		 Comments Active
29 The flow of any fluid, real or ideal, must fullfill the following Newton’s law of viscosity Newton’s secodn law of viscosity Velocity at boundary must be zero relative to the boundary the continuity equation d Continuity equation is based on mass conservation “ mass can neither be created nor be destroyed.” This is the basic condition of fluid flow. Ideal fluid does not posses viscocity, thus does not obeys neutons law of viscosity.
\(Av=Q\)		 Comments Active
30 The magnitude of rise of pressure due to water hammer in a rigid and non – elastic pipe carrying water of density and bulk modules k will be equial to \(ρ\) \(\frac{k}{ρ}\) \(kρ \) \(\frac{ρ}{k} \) \(\frac{k}{ρ}\) b The magnitude of pressure rise due to water hammer in a rigid, non-elastic pipe is given by the formula:. And velocity would be \(∆P=kρ\) \(v=\frac{k}{ρ}\) Comments Active
31 In an isothermal atmosphere the pressure decreases linearly with elevation remains constant varies in the same way as the density increases exponentially elevation c \(\frac{PV}{T}=cons→P ∝\frac{1}{V} when T=C\)
\(P ∝\frac{1}{V} ∝ \frac{ρ}{m}→P ∝ ρ \)
So in an isothermal atmosphere, the pressure varies in the same way as the density.		 Comments Active
32 The rise or depression of liquid in a tube due to surface tension with increase in size of tube will increase remain unaffected may increase or decrease depending on the characteristics of liquid decrease d , rise and depression of liquid is inversely depend on dia or size of tube. \( h=\frac{4σcosθ}{ρgd}\) Comments Active
33 The capillary rise at 20 degree celsius in a clean glass tube of 1 mm bore contaning water is approximately 3 mm 5 mm 10 mm 30 mm d \(h=\frac{4σcosθ}{ρgd}=\frac{4×0.0736×cos0}{9.81×10^{3}×10^{-3}}=0.030m=30mm\) Comments Active
34 is the equation to determine kinematic viscosity of liquids by \(V=0.0022t-\frac{1.8}{t}\) Redwood viscometer Engler viscometer saybolt universal viscometer Newton viscometer c The equation 0.0022t - 1.8/t is used to determine the kinematic viscosity of liquids with the Say bolt Universal Viscometer. Here, t represents the time taken for the liquid to flow through a calibrated tube, and the equation helps to convert that time into kinematic viscosity.
Redwood Viscometer and Engler Viscometer are used to measure the viscosity of liquids, but they follow different empirical formulas specific to their calibration.
Newton Viscometer is not typically associated with a time-based formula for kinematic viscosity.		 Comments Active
35 The resultant upward pressure of the fluid on an immersed body is called upthrust buoyancy centre of pressure None of these b Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force is equal to the weight of the fluid displaced by the object, and it is what makes objects float or rise in a fluid. The term "upthrust" is sometimes used colloquially, but buoyancy is the more accurate scientific term. Comments Active
36 For manometer, a better liquid combination is one having higher surface tension lower surface tenison surface tension is no criterion high density and viscosity b Surface tension can introduce errors in pressure readings due to the formation of meniscus at the liquid interface. Especially in narrow tubes, higher surface tension leads to greater capillary rise or depression, causing inaccurate pressure measurements.
Therefore, lower surface tension helps in:
Reducing meniscus curvature
Minimizing capillary effects
Improving accuracy of pressure measurement		 Comments Active
37 A fluid in equilibrium can’t sustain tensile stress compressive stress shear stress bending stress c When a fluid is at rest (static equilibrium), the only stresses it can sustain are normal stresses (either tensile or compressive), not shear stresses.
In static conditions, shear stress = 0.		 Comments Active
38 a perfect gas has constant viscosity has zero viscosity is incompressible None of these d A perfect gas or ideal gas obeys all gas laws under all condition of temperature and pressure. The volume occupied by the molecules is negligible as compared to the total volume occupied by the gas. The force of attraction among the molecules is negligible. Perfect gas obeys ideal ideal gas equation i.e PV=nRT Comments Active
39 Density of water is maximum at 0áµ’C 0áµ’K 4áµ’C 100áµ’C c Volume of water is minimum at 4áµ’C. density is maas per unit volume. Comments Active
40 Fluid is a substance which offers no resistane to change of pressure flow shape volume c Fluid is substance that can flow. It does not posses any definite shape , conforms the shape of containing vessel. Comments Active
41 _____ is obtained by isothermal hardening operation. Cementite Sorbite Acicular troostite Bainite c Acicular troostite is a structural component of steel, a highly dispersed mixture of ferrite and iron carbide. It forms during the isothermal hardening process as a result of decomposition of the austenite in the temperature range of 250 - 450. \(℃\) Comments Active
42 ________ is the hardest known material. Cemented carbide Ceramic Diamond Alloy steel c Diamond is a noble material which is so costly that its application becomes limited. It is the hardest material. It can be used for cutting at a speed 50 times greater than H.S.S. tools. It has a low coefficient of friction and high heat conductivity.
Cemented carbides have a very high hardness (second only to diamond) and high wear resistance to abrasion.		 Comments Active
43 _________ has high tendency to get work hardened. Lead Aluminium Brass Silver c Strain hardening or work hardening is the strengthening of a metal by plastic deformation.
This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.
Due to the dislocation movement of atoms, the density increases and hardens the work material.
Due to strain hardening, yield strength increases, and ductility decreases.		 Comments Active
44 Ball bearings are generally made up of Carbon steel Carbon chrome steel Stainless steel grey cast iron b The most common chrome steels contain from 0.5 to 2% chromium and 0.1 to
1.5% carbon. The chrome steel is used for balls, rollers and races for bearings.
Nickel-Chrome steel containing 3.25% nickel, 1.5% chromium and 0.25% carbon is much used for armour plates.
Chrome nickel steel is extensively used for motor car crank shafts, axles and gears requiring great strength and hardness.		 Comments Active
45 ________ is commonly used for making household utensils. Duralumin Hindalium γ – alloy Magnalium b Duralumin:
It is used in the wrought conditions for forging, stamping, bars, sheets, tubes, bolts, and rivets.
Due to its higher strength and lighter weight, this alloy is widely used in
automobile and aircraft components.
It is also employed in surgical and orthopedic work, non-magnetic work and measuring instrument parts constructing work.
Y -alloy:
It has better strength than duralumin at high temperatures, therefore it is much used in aircraft engines for cylinder heads, pistons, crank cases of internal combustion engines die casting, pump rods et(c)
Magnalium:
Due to its light weight and good mechanical properties, it is mainly used for making aircraft and automobile components.
Hindalium:
Hindalium is mainly used for manufacturing anodized utensils. Utensils manufactured by this alloys are strong and hard, easily cleaned, low cost than stainless steels, having a fine finish, having good scratch resistance, do not absorb much heat etc.		 Comments Active
46 The melting point is the lowest for low carbon steel high carbon steel cast iron wrough iron c Cast iron is a group of iron-carbon alloys with a carbon content greater than 2%. Its usefulness derives from its relatively low melting temperature. The melting point of Wrought iron lies near to 1500°C and of steel lies near 1400°C, whereas cast iron has a melting point near to 1200°C Comments Active
47 Preheating is essential in welding high speed steel cast iron all non – ferrous materials None of these b Preheating involves heating the base metal, either in its entirety or just the region surrounding the joint, to a specific desired temperature.
Preheating is often employed when welding cast iron, high carbon steel, or alloy steel since preheating slows down the cooling rate of that area of parent metal close to the weld itself and the weld itself and thereby prevents the formation of martensite which accounts for hardness across the weld
The change of cracking in the heat-cooling is also minimized by preheating.		 Comments Active
48 _________ structure is obtained by austempering process of heat treatment. Sorbite Bainite Martensite Troostite b Bainite is produced by an austempering process. In this process, the sample is heated and quenched to room temperature below the nose of the T-T-T diagram and this temperature is maintained for a substantial period of time so that cooling curve enters into T-T-T diagram. Comments Active
49 Under microscope ferrite appears as White Light Dark None of these b Type of grain
Appearance in the optical microscope
Ferrite
Lite
Austenite
White
Cementite
White
Pearlite
Dark
Martensite (fresh)
Mostly white
Bainite and tempered matensite
Dark		 Comments Active
50 ___________ does not cantain tin as an alloying element. Babbitt metal White metal Solder All options are incorrect d Babbitt metal contains Sn (Tin) = 88%, Sb (antimony) = 8%, Cu (copper) = 4%. It possesses excellent antifriction properties and sufficient mechanical strength, so most commonly used in bearing metal.
White Metal is an alloy of Antimony, Tin and Lead
Solders are essentially alloys of lead and tin.		 Comments Active
51 Which of the relation represents an irreversible and possible process? \(\frac{dQ}{T}=0\) \(\frac{dQ}{T}>0\) \(\frac{dQ}{T}<0\) None of these c For irreversible process \(\frac{dQ}{T}<0\)
For Reversible process \(\frac{dQ}{T}=0\)		 Comments Active
52 In Clausius theorem the reversible path is substituted by _____. reversible isobars reversible isotherms reversible isochoric None of these b Clausius theorem states that: Any reversible path may be substituted by a reversible isotherm and a reversible adiabatic between the same end states such that the heat transferred during the isothermal process is the same as that transferred during the original process. Comments Active
53 Which gas can attain the highest efficiency for the same compression rise? Any of the gases Diatomic gases Mono atomic gases Tri – atomic gases c Gas Type
Cv
Cp
γ=Cp /Cv ​​
Monoatomic
3/2 R
5/2 R
1.67
Diatomic
5/2 R
7/2 R
1.4
Triatomic
7/2 R
9/2 R
1.28		 Comments Active
54 Which equation defines the enthalpy (h) of a system? \(U+\frac{pv}{J}\) \(U-\frac{pv}{J}\) \(U+\frac{R}{Jpv}\) \(U+Jpv\) a h = \(U+\frac{pv}{J}\)
J≈4186 J/kcal (or 1 kcal = 4186 J)		 Comments Active
55 A reversible engine operates between temperature T1 and T2. The energy rejected by this engine acts as an input for another reversible engine at temperature T2, which rejects to a reservoir at temperature T3. What is the relation between T1 T2 and T3? \(T_{2}=\frac{T_{1}+T_{3}}{2}\) \(T_{2}=T12+T32\) \(T_{2}=T_{1}T_{3} \) \(T_{2}=\frac{T_{1}-T_{3}}{2}\) c \(T_{2}=T_{1}T_{3}\) Comments Active
56 For the same heat added and the same compression ration, ____________. Otto cycle is more efficent than diesel cycle Diesel cycle is more efficient than Otto cycle Both Diesel and Otto cycle are equally efficient Cannot be determined a
\(=1-\frac{Heat Rejected}{Heat Supplied}\)
Since all the cycles rejected their heat at the same specific volume, process line from state 4 to 1, the quantity of heat rejected from each cycle is represented by the appropriate area under the line 4 to 1 on the T – S diagram.
As is evient from the equation
The cycle which has the least heat rejected will have the highest efficiency.
Thus, the Otto cycle is the most efficient and the diesel cycle is the least efficient of the three cycles.
\(η_{otto}>η_{dual}>η_{diesel}\)
For the same compression ratio and the same heat input
\(η_{otto}>η_{dual}>η_{diesel}\)
For constant maximum pressure and heat supplied
\(η_{diesel}>η_{dual}>η_{otto}\)		 Comments Active
57 Clasius’ statement and Kelvin – Planck’s statement are ______. not connected two parallel statements of the second law violation of one does not violates the other false statements b . The Clausius Statement was expressed as “Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time”, and the Kelvin–Planck Statement as “It is impossible to construct a device that operates in a cycle and produces no other effect than the production of work and the transfer of heat from a single body”. Comments Active
58 If the heat rejected from the system is zero, then which of the following statements will hold TRUE? When network is equal to the heat absorbed, work efficiency is 100% Heat is exchanged from one heat reservoir only It violates Kelvin – Plank statement All options are correct d The Kelvin–Planck statement of the second law of thermodynamics is also known as the heat engine statement. According to this statement, it is impossible to construct an engine operating in a cycle that will produce no effect other than extracting heat from a single reservoir and completely converting it into work.Hence, Kelvin Planck’s law deals with heat conservation in work. Comments Active
59 In Mollier diagram, the isotherm in the superheated region at low pressure becomes _____. diverge from one another horizontal parallel vertical b An isotherm, or equal temperature line, has a downward slope until is hits the liquid-vapor dome. Once the line hits the liquid-vapor dome, the isotherm has no slope and is horizontal. Comments Active
60 Which gas will produce the highest efficiency in an ideal Otto cycle for same compression ratio? Air Carbon dioxide Helium Oxygen c The efficiency of the Otto cycle is given as \(η=1-(\frac{1}{r})^{γ-1}\)
Higher the value of γ more will be the efficiency for the same compression ratio.
The value of γ for the different type of gases is given below.		 Comments Active
61 Which relation is the basis of Mollier Diagram? \(\frac{m_{v}}{m_{v}+m_{l}}\) \(\frac{m_{l}}{m_{v}+m_{I}}\) \(\frac{m_{v+m_{l}}}{m_{v}}\) \( \frac{m_{v}+m_{l}}{m_{I}}\) a Dryness fraction(x) is the ratio of mass of steam to total mass (mass of steam +mass of water). Comments Active
62 The combustion in a compression ignition engine is ______. heterogeneous homogeneous laminar turbulent a Homogeneous compression ignition is a form of internal combustion in which air and fuel are well mixed at the point of combustion. On the other hand, in heterogenous combustion engine, the air and fuel are not mixed till the point of combustion. In CI engines or diesel engine, only air is compressed and then the fuel is injected at the last stage. So no proper mixing before combustion takes place. So the CI engines are heterogenous combustion engines. Comments Active
63 In an Otto cycle, air is compressed from 3 litres to 2.4 litres from a starting pressure of 1.5 kg/cm2. The net output per cycle is 400 kJ. What is the mean effective pressure (kPa) of the cycle? 500 567 667 700 c Mean Effective Pressure: It is defined as the ratio of the net – work done to the displacement volume of the piston.
where, mean effective pressure, swept volume \(W_{net}=P_{m}×V_{s}\) \(P_{m}=\) \(V_{s}=\)
Stroke Volume, \(V_{s}=3-2.4=0.6 litres=0.6×10^{-3}m^{3}, W_{net}=400 kJ\)
Therefore \(P_{m}=\frac{W_{net}}{V_{s}}=\frac{400×10^{3}}{0.6×10^{-3}}=667 MPa\)		 Comments Active
64 An ideal gas with heat capacity ratio of 2 is used in an ideal Otto – cycle which operates between minimum and maximum temperature of 200 K and 1800 K. What is the compression ratio of the cycle for maximum work output? 1.5 2 3 4 c Heat capacity ration \(γ=\frac{C_{p}}{C_{v}}=2\)
\(T_{max}=1800K, T_{min}=200K\)
For maximum work output for otto cycle
\(r=(\frac{T_{max}}{T_{min}})^{\frac{1}{2(r-1)}}=(\frac{1800}{200})^{\frac{1}{2(2-1)}}=9^{1/2}\)
\(r=3\)		 Comments Active
65 A heat engine working between the source at 200ᵒC and rejects heat at 25ᵒC receives 5 kW of heat. Work done for this engine is equal to 0 kW. Does this satisfy the inequality of Clausius? Yes No Cannot be determind None of these a \(T_{1}=200℃=473K\)
\(T_{2}=25℃=298K\)
Clausius Inequality
\(\frac{dQ}{T}≤0\)
The equality holds good for a reversible cycle and the inequaliy holds good for an irreversible cycle.
\((\frac{Q_{h}}{T_{h}}-\frac{Q_{c}}{T_{c}})=\frac{5}{473}-\frac{5}{298}=-0.0062≤0\)
Thus, the engine satisfies the inequality of clausius.		 Comments Active
66 While working between temperatures 150 K and 300 K, the entropy change experienced by Carnot engine during heat addition is 1 KJ/k, the work produced (kJ) by the engine is ________. 100 150 300 600 b Given
\(T_{1}=300K, T_{2}=150K\)
\(∆S=1KJ/K\)
\(W=Q_{1}-Q_{2}=T_{1}∆S-T_{2}∆S\)
\(W=(T_{1}-T_{2})∆S\)
\(W=(300-150)1\)
\(W=150 KJ\)		 Comments Active
67 If the COP of Carnot refrigerator is 4, then the thermal efficiency of the Carnot engine would be _____. 0.33 0.25 0.2 0.18 c Given
COP of refrigerator=4
COP of Heat pump=1+COP of refrigerator
COP of heat pump=1+4=5
thermal efficieny of heat engine =1/COP of H.P=1/5=0.2		 Comments Active
68 Which of the following statement related to entropy is TRUE? Minimum entropy is observed when the system is in equilibrium with the surrounding At absolute zero temperature, the solid solutions have non – zero entropy Substance in solid phase has the least entropy Entropy conservation takes place in all irreversible processes. c Entropy is the degree of randomness in a system. The entropies of gases are much larger than those of liquids,which are larger than those of solid. Comments Active
69 The triple point on a P – V diagram is ______. a line a point a triangle not present b Comments Active
70 A rod of dimension 20 mm × 20 mm is carrying an axial tensile load of 10 kN. The tensile stress developed is _______. \(J=\frac{π}{64}d^{4}\) \(J=\frac{π}{32}d^{4}\) \(J=\frac{π}{16}d^{2}\) \(J=\frac{π}{16}d^{4}\) d Polar moment of inertia of the solid shaft is;
\(J=\frac{Ï€}{32}d^{4}\)		 Comments Active
71 Which of the following is the CORRECT bending moment diagram for the cantilever beam carrying uniformly varying load from zero at free and w/unit length at the fixed end? b The cantilever beam carrying uniformly varying load from zero at free and w/unit length at the fixed end is given below:
 Comments Active
72 The correct shear force diagram for the cantilever beam with uniformly distributed load over the whole length of the beam is a Comments Active
73 The property of the material to regain its original shape after deformation when the external forces are removed is ____. plasticity elasticity durabilty None of these b Elasticity is the property due to which the body regains its original shape and size after the removal of applied force. Comments Active
74 What is the volumetric strain in the thin cylinder subjected to internal pressure having hoop stress of 200 MPa, modulus of elasticity, E = 200 GPa and poissons ratio = 0.25? 20/1000 2/1000 0.2/1000 0.02/1000 b volumetric strain is= \( ε_{v}\) \(Pd/4tE[5-4μ\)]
=200MPa, modulus of elasticity, E = 200 GPa=200, \(σ_{h}=\frac{Pd}{2tE}\) \(×10^{3}\) \( μ=0.25\)
\(ε_{v}=\) \(\frac{200}{2×200×10^{3}}[5-4(0.25)\)]
= \(\frac{1}{2×10^{3 }}[5-1\) ]
=2/1000		 Comments Active
75 What is the maximum shear stress on the wall of a thin cylinder, if it has a diameter of d, thickness of t and the gauge pressure in the cylinder is p? \(\frac{pd}{t}\) \(\frac{pd}{4t}\) \(\frac{pd}{2t}\) \(\frac{pd}{8t}\) d Maximum shear stress on the wall is maximum in-plane shear stress and is given by:
\(τ_{max}=(σ_{1}-σ_{2})/2\)
\(σ_{1}=\frac{Pd}{2t};σ_{2}=Pd/4t\)
\(Ï„_{max}=\frac{\frac{Pd}{2t}-\frac{Pd}{4t}}{2}=Pd/8t\)		 Comments Active
76 Which of the following assumption is INCORRECT about the long column? The column behaves elastically The load acts perfectly axial and passes through the centroid of the column section The weight of the column is neglected The material is non – homogeneous and anisotropic d The assumption of long column is:
1)The column is initially perfectly straight, and the load is applied axially.
2)The cross-section of the column is uniform throughout its length.
3) The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke's law.		 Comments Active
77 Rankine theory is applicable to the ______. Short strut/column Long column Both short and long column None of these c Rankine formula (Rankine Gordon formula) for columns is a semi-empirical equation that assumes both crushing and buckling failure of the column and is therefore used for the calculation of failure load for both long as well as short columns. Comments Active
78 The slendeness ratio of the columns is _______. directly proportional to the effective length directly proportional to the least radius of gyration directly proportional to the square of effective length directly propotioalto the square of least radius of gyration a The slenderness ratio is defined as the ratio of length (l) to the least radius of (k), represented as l/k. Comments Active
79 Maximum shear stress theory was postulated by ________. 1 1/2 1/4 2 b Support Conditions
Effective length (Le)
Both ends hinged/pinned
Le = L
One end hinged other end fixed
Le = L/√2
Both ends fixed
Le = L/2
One end fixed and the other end free
Le = 2L		 Comments Active
80 Which term states the S.I. unit of stress? kN/mm N/mm2 N/mm3 m3/sec b S.I. unit of Stress is N/mm2 ( force per unit area) Comments Active
81 Which of the following load does not act on the considerable length of the beam? Uniformly distributed Triangular Point Uniformly varying c Point load is that load which acts over a small distance. Comments Active
82 What term is used for the combined effect of all the forces on a body? Load Stress Strain None of these a The combined effect of all the external forces acting on a body is known as a loa(d) The force of resistance per unit area, offered by a body against deformation is known as stress. Comments Active
83 Which of the following term defines the size of the cam? Base circle Prime circle Pitch curve Pitch curve a Base Circle The smallest circle drawn, tangential to the cam profile, with its center on the axis of the cam shaft. The size of the base circle determines the size of the cam. Comments Active
84 What will be the vertical height (m) of a watt governor, if the speed of rotation is 80 rpm? 1.4 1.14 0.14 0.11 c Given
N = 80 rpm
\(h=\frac{895}{N^{2}}=\frac{895}{80^{2}}\)
\(h=0.14m\)		 Comments Active
85 Which of the following governors does not have central load attached to their sleeves? Porter governor Watt governor Proell governor None of these b Watt governor is the simplest and gravity controlled form of the centrifugal governors. It consists of two fly balls attached to the sleeve of negligible mass. The upper sides of arms are pivoted so that its balls can move upward and downward as they revolve with a vertical spindle. The engine drives the spindle through bevel gears. Comments Active
86 Which of the following is the type of pendulum governor? Hartnell governor Proell governor Porter governor Watt governor d Comments Active
87 Which of the following statement is TRUE about the contact ratio? Varies directly to the length of the arc of contact Inversely proportional to the module Inversely proportional to the circular pitch All options are correct d (i) Arc of contact is the distance traveled by a point on either pitch circle of the two wheels during the period of contact of a pair of teeth.
Arc of contact = \(\frac{path of contact }{cos∅}\)
(ii) Contact ratio: It is defined as the average number of tooth pairs in contact during one rotation.
Contact ratio = \(\frac{Arc of contact }{Circular pitch}\)
For continuous transmission of motion, at least one tooth of one wheel must be in contact with another tooth of the second wheel.		 Comments Active
88 What is the radial distance of a tooth from the pitch circle to the top of the tooth known as? Dedendum Addendum Pitch circle diamete Module b The radial distance from the pitch circle to the top of the tooth is known as the addendum. In gear terminology, the addendum refers to the height of the tooth above the pitch circle. Comments Active
89 When the friction lining is new, the wear varies ______. directly to radius inversely ot radius directly to the square of radius inversely to the square of radius a When the friction lining is new, uniform pressure theory is applicable.Therefore,
P = constant
\(T=μWR\)
\(R_{m}=\frac{2}{3}(\frac{R13-R23}{R12-R22})\)
When the friction line is new, the wear varies directly to the radius.		 Comments Active
90 The pressure distribution in the uniform wear theory is ____________. directly proportional to radius directly proportional to the square of radius inversely proportional to radius inversely proportional to the square of radius c When the friction lining is new, uniform pressure theory is applicable.Therefore,
P = constant
\(T=μWR\)
\(R_{m}=\frac{2}{3}(\frac{R13-R23}{R12-R22})\)
Pressure distribution in the uniform wear theory I sinversely proportional to the radius.
\(T=μWR\)
Pr = constant
\(R_{m}=\frac{R_{1}+R_{2}}{2}\)		 Comments Active
91 The rotary internal combustion engine is the inversion of ______. four bar link chain double slider crank chain single slider crank mechanism Rocker crank mechanism c Single Slider Crank Mechanism
First Inversion
Reciprocating engine or compressor.
Second Inversion
Whitworth quick return mechanism, Rotary engine et(c)
Third Inversion
Slotted crank mechanism, Oscillatory engine.
Fouth Inversion
Hand pump, Pendulum pump or Bull engine.		 Comments Active
92 A pulley is driven by a flat belt and the maximum tension produced in the belt is of 1400 N. The belt has the density of 1000 kg/m3, 100 mm wide and 5mm thick, What is the speed (m/sec) of the belt for the maximum power? 32 31 30.55 3.05 c Given
T = 1000 Kg /m3
e = 1000 kg/m3
b = 100 mm = 0.1 m
t = 5 mm = 0.005 m
Power transmitted by the belt:
P = ( \(T_{1}-T_{2})V\)
For maximum power.
\(T_{1}=\frac{T}{3}\)
Where T1 = Tension on tight side
T = Maximum Tension to which the belt is subjected velocity of the belt for maximum power:
Where m is the mass of belt per unit length. \(V=\frac{T}{3m}\)
m = Area length density \(×\) \(× \)
\(m=0.1×0.005×1×1000=0.5Kg\)
\(V=\frac{T}{3m}=\frac{1400}{3×0.5}=30.55 m/s\)		 Comments Active
93 The mass of flywheel of a steam engine is 3250 kg with the radius of gyration of 1 m. The starting torque of the engine is 4500 N-m. What is the angular acceleration (rad/s2) of the flywheel? 3.4 2 2.48 1.38 d Given
M = 3250 kg
r =1 m
T = 4500 N-m
I = mK2 = 3250 1 = 3250 Kg-m2 \(×\)
Angular acceleration of the flywheel:
\(α=\frac{T}{I}=\frac{4500}{3250}\)
\(=\frac{18}{13}=1.38 rad/s^{2}\)		 Comments Active
94 The graph of turning moment diagram is drawn between ________ crank angle and crank radius crank angle and crank effort crank effort and crank angle crank radius and crank angle c
A turning moment diagram or crank effort diagram shows the relationship between the turning moment or torque and the crank angle at various crank positions. It is drawn in cartesian coordinates, with the turning moment on the vertical (Y-axis) and the crank angle on the horizontal (X-axis).		 Comments Active
95 What is the degree of freedom of the mechanism shown below?
 1 2 3 4 b Given
L = 5
j = 5
h = 0
According to Kutzbach equation
The degree of freedom is given as.
F = 3 (l-1)-2j-h
Therefore, F = 3 (5-1)-25=2 \(×\)		 Comments Active
96 What is the total number of links and joints in the mechanism as shown in figure?
 3 and 3 3 and 2 4 and 3 4 and 4 d There are four links, three binary joints asn one higher pair, i.e. L= 4 j = 3 and h =1
Degree of freedom:
n = 3 (L – 1) – 2 j – h
n = 3 (4 – 1) - 23 – 1 = 2 \(∴\) \(×\)		 Comments Active
97 The number of links (l) which is required to form a kinematic chain can be expressed in term of the number of pairs (p) as ______. l = 2p – 4 l = 2p – 3 l = 2p – 2 l = 2p – 5 a If each link is assumed to form two pairs with two adjacent links, then the relation between the number of pairs (p) forming a kinematic chain and the number of links (L) may be expressed in the form of an equation:
L= 2 p – 4		 Comments Active