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S.No Question Option A Option B Option C Option D Answer Solution Comments Status Action
1 Ignition quality of petrol is expressed by octane number cetane number calorific number self – ignition temperature a Quality of Petrol:
The octane number is used to measure gasoline’s ignition quality and the cetane number is used to measure diesels.
The ignition of gas is typically measured by the octane rating or octane number (gasoline or petrol).
The higher this value, the less likely it is that the gas would “knock” – an explosion brought on by its early burning in the combustion chamber.
When burned in a typical internal combustion engine with spark ignition.		 Comments Active
2 Which of the following is the lightest and most volatile air fuel ratio?
1. 6:1
2. 9:1
3. 12:1
4. 15:1
Options		 Only 1 Only 2 Only 3 Only 4 d Gasoline (petrol) is the lightest and most volatile common liquid fuel.
ï‚· the stoichiometric air-fuel ratio for petrol is about 15:1.		 Comments Active
3 The fuel air ratio in a petrol engine fitted with suction carburetor, operating with dirty air filter as compared to clean filter will be Higher Lower Remain unaffected Unpredictable a The fuel air ratio in a petrol engine fitted with suction carburettor, operating with dirty air filter as compared to clean filter will be higher because dirty air filter leads to lesser inducted air. For proper combustion of fuel, engines require oxygen which is supplied to engine in form of atmospheric air. Comments Active
4 Diesel fuel, compared to petrol is Less difficult to ignite Just about the same difficult to ignite More difficult to ignite Highly ignitable c Fire point of diesel fuel is higher than petrol that’s why diesel fuel is more difficult to ignite. Comments Active
5 An engine indicator is used to determine the following Speed Temperature Volume of cylinder None of these d An engine indicator is used to determine mean effective pressure and indicated horse power. An engine indicator is an instrument for graphically recording the pressure versus piston displacement through an engine stroke cycle. Engineers use the resulting diagram to check the design and performance of the engine.
The IHP can be calculated by integrating the area under the indicator diagram, which represents the work done by the engine during each cycle. - The Mean Indicative Pressure is obtained by dividing the integrated work by the stroke volume of the engine		 Comments Active
6 Which of the following boilers is best suited to meet fluctuating demands? babock and Wilcox locomotive Lancashire Cochran b The Locomotive Boiler is horizontal, multi-tubular, natural circulation, internally fired, fire tube boiler. The maximum pressure range 21bar and streaming rate is as high as 55 to 70 Kg per square meter of heating surface per hour.
Locomotive boilers are best suited for fluctuating demand because of their fast steam generation and compact design.		 Comments Active
7 The diameter (in m) of Cornish boiler shell is of the order of 1 to 2 1.5 to 2.5 2 to 3 2.5 to 3.5 a The diameter of Cornish boiler shell is generally 1 m to 2 m and its length varies from 5 m to 7.5m. The diameter of the tube may be about 0.6 times of shell. The capacity and working pressure of a Cornish boiler is low as compared to Lancashire boiler. Comments Active
8 Adiabatic process is Essentially as isentropic process Non – heat transfer process Reversible process Constant temperature process b An adiabatic process is a thermodynamic process which involves the transfer of energy without transfer of heat or mass to the surrounding. Comments Active
9 The specific heat of superheated steam in kcal/kg is generally of the order of 0.1 0.3 0.5 0.8 c At atmospheric pressure, the specific heat of superheated steam is approximately 0.5 kcal/kg. At higher pressures, the specific heat of superheated steam decreases. Therefore, the specific heat of superheated steam in kcal/kg is generally of the order of 0.5 kcal/kg. Comments Active
10 Heating of dry steam above saturation temperature is known as Enthalpy Superheating Super saturation Latent heat b Steam not containing water droplets, but which is at saturation temperature, is called dry saturated steam. Steam above saturation temperature is known as superheated steam. Comments Active
11 The value of coefficient of discharge in comparison to coefficient of velocity is more less same unpredictable b We know,
\(C_{d}=C_{v}×C_{c}\)
Since the question doesn’t ask for any measuring device like orifice meter or venturimeter.
Also, In practice \(C_{c}<1 and C_{v}<1\)
Product of two quantities having values less than 1 is always a lesser value.
For example,
\(Take C_{c}=0.90 and C_{v}=0.90\)
\(∴\) \(C_{d}=0.90×0.90=0.81Hence, is generally less than \(C_{d}\) \(C_{v}\)		 Comments Active
12 Orifice is an opening with closed perimeter and of regular form through which water flows with prolonged sides having length of 2 to 3 diameters of opening in thick wall with partially full flow in hydraulic structure with regulation provision a • The orifice is a small opening of any cross-section on the side or at the bottom of a tank, through which a fluid is flowing.
• It is used for measuring the rate of flow of fluid by measuring the pressure decrease across the opening.		 Comments Active
13 The energy loss in flow through nozzle as compared to venturimeter is same more less unpredictable b The flow nozzle is essentially a venturimeter with the divergent part omitted. Therefore the basic equations for calculation of flow rate are the same as those for a venturimeter. The dissipation of energy downstream of the throat due to flow separation is greater than that for a venturimeter. But this disadvantage is often offset by the lower cost of the nozzle. Flow nozzle (top) and venturimeter(bottom) diagram is given below.
 Comments Active
14 If flow is an open channel is gradually varied, then the flow will be Steady uniform flow Unsteady uniform flow Steady non-uniform flow Unsteady non-uniform flow c Gradually varied flow is steady non-uniform because the velocity of water remains constant at a specified point, but it changes from one point to another point.
The terms steady and uniform are used frequently in engineering, and thus it is important to have a clear understanding of their meanings.
• The term steady implies no change with time. The opposite of steady is unsteady. Or transient.
• The term uniform, however, implies no change with location over a specified region.
Any devices such as turbines, compressors, boilers, condensers, and heat exchangers operate for long periods of time under the same conditions, and they are classified as steady-flow devices. During steady flow, the fluid properties can change from point to point within a device, but at any fixed point, they remain constant.
 Comments Active
15 Total pressure on the top of a closed cylindrical vessel completely filled with liquid, is directly proportional to radius square of radius cube of radius None of these b When the cylindrical vessel containing liquid is revolved, the surface of the liquid takes the shape of a parabola.
• The rise of liquid along the walls of a revolving cylinder about the initial level is the same as the depression of the liquid at the axis of rotation.
• The total pressure (P) on the top of a closed cylindrical vessel of the radius (r) completely filled up with a liquid of specific weight (w) and rotating about its vertical axis is given by:
\(P=\frac{πwω^{2}r^{2}}{4g}\)
\(w=ρg\)
Where h is the height of the vessel
Putting the value of w in value of P.
\(P=\frac{π(ρg)ω^{2}r^{2}}{4g}=\frac{π^{2}ρω^{2}r^{2}}{4}\)
\(P∝r^{2}\)		 Comments Active
16 The discharge through an orifice fitted in a tank can be increased by Fitting a short length of pipe to the outside Sharpening the edges of orifice Fitting a long length of pipe to the outside Fitting a long length of pipe to the inside a • An orifice is a small aperture through which the fluid passes. The liquid form a tank is usually discharged through a small orifice at its side.
• A drowned or submerged orifice is one which does not discharge into an open atmosphere but discharge into the liquid of the same kind.
• The discharge through an orifice is increased by fitting a short length of pipe to the outside known as an external mouthpiece.
• The discharge rate is increased due to a decrease in the pressure at vena contracta within the mouthpiece resulting in an increase in the effective head causing the flow.		 Comments Active
17 The value of coefficient of velocity depends upon Slope of orifice Size of orifice Head of liquid above orifice Friction at the orifice surface d Coefficient of velocity represents loss in velocity due to friction. Comments Active
18 A triangular section in open channel flow will be most economical when the vertex angle at the triangle base point is 30 degree 50 degree 20 degree None of these d Most economical triangle section:

For a triangular channel, the most economical condition is:
θ=90∘
Where θ is the vertex angle at the base point (the angle between the two sloping sides of the triangle).		 Comments Active
19 For maximum discharge through a circular open channel, the ratio of depth of flow to diameter of channel should be 0.7 0.5 0.65 None of these d \(D=depth of flow\)
\(d=diameter of pipe\)
\(R_{m}=Hydraulic Mean depth=A/P\)

Perimeter, P = \(αd\)
Condition for maximum Discharge for Circular Section:
\(α=154°=2.65 radians\)
\(D=0.95d\)
\(R_{m}=0.29 d\)
Wetter perimeter \(=2αr=αd=2.65d\)
Condition for Maximum Velocity for Circular Section:
\(α=128.75°=2.25 radians\)
\(D=0.81 d\)
\(R_{m}=0.3d\)		 Comments Active
20 Hydraulic grade line for any flow system as compared to energy line is Above Below At same level May be below or above depending upon velocity of flow b Bernolli’s equation along streamlines:
\(H=\frac{P}{ρg}+\frac{v^{2}}{2g}+z\)
Total head (H) = Pressure head (P/) + Kinetic head + Potential head (z) \(ρg\) \((v^{2}/2g)\)
Total head (H) = Hydraulic gradient line \(((P/ρg)+z)+Kinetic head (v^{2}/2g)\)
The line representing the sum of pressure head, datum head, and velocity head with respect to some reference line is known as the total energy line (T.G.L).
The line representing the sum of the pressure head and potential head or datum head with respect to some reference line is the hydraulic gradient line (H.G.L).
The centerline of the pipe or channel through which the fluid is flowing.		 Comments Active
21 In the case of turbulent flow it occurs in open channel losses are proportional to square of velocity velocity at boundary is zero shear stresses are more compared to laminar flow b In turbulent flow, the energy loss due to friction is proportional to the square of the velocity of flow, and frictional resistance is higher than that in laminar flow." Comments Active
22 Maximum efficiency of transmission of power through a pipe is 0.25 0.3333 0.5 0.6667 d The efficiency of power transmission is given by
\(η=\frac{H-H_{L}}{H}\)
Here, H = head available at the inlet, hf = frictional head loss
For maximum efficiency: \(H_{L}=\frac{H}{3}\)
\(η_{max}=\frac{H-\frac{H}{3}}{H}\)
\(η_{max}=66.66%\)		 Comments Active
23 The head loss in case of hot water flow through a pipe compared to cold water flow will be same more less more or less depending on range of temperatures c For hot water viscosity will be less, so less loss in head.
ï‚· Hot water has lower viscosity than cold water.
 Lower viscosity → higher Reynolds number → lower friction factor in turbulent flow.
 lower friction factor → less head loss.		 Comments Active
24 In case of a two dimensional flow the components of velocity are given by u = ax; v = by, the point where no motion occurs, is known as critical point neutral point stagnation point None of these c Point where velocity becomes zero known as stagnation point. Comments Active
25 The most economical section of a rectangular channel for maximum discharge is obtained when its depth is equal to half the breadth twice the breadth same as the breadth three – fourth of the breadth a Most economical Hydraulic section:
The discharge from a channel section increases with increases in hydraulic radius or with decrease in the wetted perimeter.
From hydraulics viewpoint, therefore, the channel section having the least wetted perimeter for a given area has the maximum discharge; such a section is known as the best hydraulic section or most economical hydraulic section.
So for most economical hydraulic section;
Wetted perimeter (P) = Minimum Q= maximum \(→\)		 Comments Active
26 The rise of liquid along the walls of a revolving cylinder as compared to depression at the center with respect to initial level is same more less more or less depending on speed a When the cylindrical vessel containing liquid is revolved, the surface of the liquid take the shapes of parabola.
The rise of liquid along the walls of a revolving cylinder about the initial level is the same as the depression of the liquid at the axis of rotation.
The total pressure (P) on the top of a closed cylindrical vessel of the radius (r) completely filled up with a liquid of specific weight (w) and rotating about its vertical axis is given by:
\(P=\frac{πwω^{2}r^{2}}{4g}\)		 Comments Active
27 A hydraulic ram acts like a centrifugal pump a rotary pump a reciprocating pump an impulse pump d A hydraulic ram is a type of pump which raises water without using any external power.
In a hydraulic ram large quantity of water available at a small height is utilized for lifting a small quantity of water to a greater height.
The working of the hydraulic ram is based on the small water hammer effect developed in the supply pipeline when the exit valve is suddenly closed.
The working of a hydraulic ram is based on the principle of water hammer or inertia pressure developed in the supply pipe.		 Comments Active
28 The flow in venture flume takes place at Atmospheric pressure At pressure greater than atmospheric pressure Vacuum High pressure a Venturi flume: The venture flume is a section of an open channel with a gradually decreasing width followed by a gradually increasing width.
Venturiflume is also known as throat flume and is used for measurement of flow in streams, small channels etc.
As the flow takes place in an open channel so the pressure of the flow is atmospheric pressure.
Venturiflume is especially suited at the location where a large loss of head cannot be permitted, as in irrigation canal.
It can also be used where water is muddy.		 Comments Active
29 Cipolletti notch is designed as trapezoid with its sides sloping at 1 horizontal and 1 vertical 2 vertical 3 vertical 4 vertical d The Cippoletti weir is a trapezoidal weir, having side slopes 1 horizontal to 4 vertical.
The purpose of the slope, on the sides, is to obtain an increased discharge through the triangular portions of the weir, which otherwise would have been decreased due to end contractions in the case of rectangular weirs.
A type of contracted weir that is related to the rectangular sharp-crested weir is Cippoletti weir, which has a trapezoidal cross-section with side slopes 1:4 (H: V).
 Comments Active
30 The losses due to sudden contraction are expressed by
1. \(\frac{V12-V22}{2g}\)
2. \(\frac{V22-V12}{2g}\)
3. \(\frac{(V_{1}-V_{2})^{2}}{2g}\)
Options:		 1 only 2 only 3 only None of these d Losses due to the sudden expansion
\(h_{L}=\frac{(V_{1}-V_{2})^{2}}{2g}=\frac{V_{1}}{2g}(1-\frac{A_{1}}{A_{2}})^{2}\)
Losses due to sudden contraction
\(h_{l}=\frac{(V_{c}-V_{2})^{2}}{2g}=\frac{V22}{2g}(\frac{1}{C_{c}}-1)^{2}\)
If CC is not given then,
\(h_{l}=\frac{0.5V22}{2g}\)
Losses at the exit of the pipe
\(h_{l}=\frac{V^{2}}{2g}\)
Losses at the entrance to the pipe
\(h_{l}=\frac{0.5V^{2}}{2g}\)
Losses due to bends
\(h_{l}=\frac{KV^{2}}{2g}\)
K = 1.2 for 90 \(°\)
K=0.4 for 45 \(°\)		 Comments Active
31 Mercury is often used in barometer because?
1. It is the best liquid
2. The height of barometer will be less
3. Its vapor pressure is so low that it may be neglected
Options		 Only 1 Only 2 Both 1 and 2 Neither 1 not 2 c Mercury is used in the barometer because it is a high-density fluid which gives less height of column for high pressures. A barometer using water, for instance, would need to be 13.6 times taller than a mercury barometer to obtain the same pressure difference. This is because mercury is 13.6 times denser than water. And also vapor pressure of mercury is very low as compared to other fluids. Comments Active
32 Liquids transmit pressure equally in all the directions. This is according to Boyle’s law Archimedes law Pascal’s law Newton’s formula c The external static pressure applied on a confined liquid is distributed or transmitted evenly throughout the liquid in all directions Comments Active
33 If the surface of the liquid is convex, then Cohesion pressure is negligible Cohesion pressure is decreased Cohesion pressure is increased None of these c
* \(θ<90°\)
* Adhesion > Cohesion.
* Liquid will wet the surface.
* Top surface of liquid is concave.

* \(θ<90°\)
* Cohesion > Adhesion
* Liquid does not wet the surface.
* Top surface of liquid is convex.		 Comments Active
34 A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury? The metal piece will simply float over the mercury The metal piece will be immersed in mercury by half Whole of the metal piece will be immersed with its top surface just at mercury level Metal piece will sink to the bottom c It is used to explain the law of flotation or upward thrust experienced when immersed in a fluid.
The principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”
The flotation and sinking of an object are dependent upon the relative density of each other.
If the density of the object is more than the density of the liquid, the object will sink.
On the other side, if the density of an object is less than the liquid, then it will float over it.
If the density of the object and liquid is equal to each other, they are in equilibrium and float and sink both at the same time i.e. the whole of the object will be immersed with its top surface at liquid level.
A piece of the metal having a specific gravity of 13.6 is placed in mercury of specific gravity 13.6, then the whole of the metal piece will be immersed with its top surface just at mercury level.		 Comments Active
35 The conditions of the stable equilibrium of a floating body are The meta – centre should lie above the centre of gravity The centre of buoyancy and the centre of gravity must lie on the same vertical line A righting couple should be formed All options are correct d Condition of stable equilibrium for a floating body in terms of metacentric height (GM) as follows:
Stable equilibrium : GM > 0 (M is above G)
Neutral equilibrium : GM = 0 (M coinciding with G)
Unstable equilibrium: GM < 0 (M is below G)		 Comments Active
36 Which of the following statements is not true? Fluids are capable of flowing Fluids conform to the shape of the containing vessels When in equilibrium, fluids cannot sustain tangential forces When in equilibrium, fluids can sustain shear forces d A solid can resist shear stress by a static deformation; a fluid cannot
Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid
The fluid moves and deforms continuously as long as the shear stress is applied
So, a fluid at rest must be in a state of zero shear stress		 Comments Active
37 The bulk modulus of elasticity has the dimensions of 1/ pressure Increases with pressure is large when fluid is more compressible is independent of pressure and viscosity b Compressibility and Elasticity
Fluids also possess elastic characteristics like elastic solid.
Compressibility of fluid is quantitatively expressed as the inverse of the bulk modulus of elasticity, K of the fluid
Bulk modulus K is defined as,
\(K=\frac{stress}{strain}=\frac{Change in pressure}{(\frac{Change in volume}{Original volume})}=-\frac{dp}{(\frac{dV}{V})}\)
\(K_{water}=2.06×10^{9}N/m^{2}, K_{air}=1.03×10^{5}N/m^{2}\)
The bulk modulus (K) of a fluid is not constant but it increases with an increase in pressure.
For liquids, K decreases with an increase in temperature.
For gases, K increases with an increase in temperature.		 Comments Active
38 For very great pressures, viscosity of most gases and liquids remains same Increases Decreases Shows erratic behavior b Viscosity of gases increases with increase in temperature while viscosity of liquids decreases with increase in temperature.
Viscosity of gases increases with increase in pressure while viscosity of liquids mostly remains constant with respect to change in pressure.		 Comments Active
39 Property of a fluid by which its own molecules are attracted is called Adhesion Cohesion Viscosity Compressibility b Adhesion is the mutual attraction between unlike molecules that cause them to cling to one another
Cohesion is the mutual attraction between like molecules that causes them to stick together.

Capillary action and meniscus (the curved surface which is formed by any liquid in a cylinder) are the effects of adhesion.
Surface tension, Capillary action, and meniscus are the effect of cohesion.		 Comments Active
40 Practical fluids Are viscous Possess surface tension Are compressible Possess all the above properties d Real Fluid
All the fluids are real as all the fluids possess viscosity.
Real fluid is Compressible and it has some value of shear force.
Its surface tension value in non-zero.
Ideal fluid:
Ideal fluid is Incompressible and it has some value of shear force.
The ideal fluid does not actually exist in nature, but sometimes used for fluid flow problems.
An ideal fluid is a fluid that has several properties including the fact that it is:
Incompressible – the density is constant
Irrotational – the flow is smooth, no turbulence
Nonviscous – (Inviscid) fluid has no internal friction		 Comments Active
41 ______ is the most important element which controls the physical properties of steel. Carbon Chromium Vanadium Tungsten a Carbon plays important role in iron, it influence various properties of iron like toughness, ductility, hardness and strength etc. Comments Active
42 ________ is used for bearing liner. Brass Bronze Gun metal Babbitt metal d White metal or Babbit metal:
• The white metal is used for making bearings that are subjected to heavy load.
• The white metal is an alloy of Lead and Lithium metal.
• These alloys usually consist of relatively hard crystals embedded in a softer matrix, a structure important for machine bearings.
• They are composed primarily of tin, copper, and antimony, with traces of other metals added in some cases and lead substituted for tin in others.
• The alkali metals are quite useful.		 Comments Active
43 ____________ structure can be studied by naked eye. Atomic Grain Micro Macro d Macrostructure refers to the overall structure of a material that can be seen with the naked eye without the need for magnification. It includes features such as surface defects, grain boundaries, and general shape, which can be observed directly.
Grain structure, on the other hand, is the arrangement of crystals or grains within the material. While grain structure can often be observed with low magnification (e.g., using a magnifying glass or a low-powered microscope), it typically requires at least some form of magnification to study the detailed arrangement of the grains at higher magnification levels.
Microstructure requires a microscope to be studied as it deals with the structure of materials at the microscopic scale (e.g., the arrangement of atoms or small crystalline structures).
Atomic structure refers to the arrangement of atoms in a material, which requires advanced equipment like scanning electron microscopes (SEM) or transmission electron microscopes (TEM) to observe.		 Comments Active
44 _________ is the essential gradient of any hardened steel. Carbon Pearlite Austenite Martensite d • The essential ingredient of any hardened steel is martensite.
• Martensite is obtained by the rapid quenching of carbon steels and is the transitional substance formed by the rapid decomposition of austenite.
• It is a supersaturated solution of carbon in alpha iron.
• Under the microscope, it appears as a needle-like constituent. The hardness of martensite depends on the amount of carbon present and varies from Rockwell C45 to C67.
• It cannot be machined, is brittle, and is strongly magnetic.
 Comments Active
45 _______ has least coefficient of expansion. Manganin Invar Constantan Duralumin b Co-efficient of expansion:
The coefficient of expansion of a material is numerically equal to the ratio of increase in length, area or volume to its original length, area or volume when the material is heated by 1. \(℃\)
Unit - \(℃^{-1} or K^{-1}\)
Material
Co-efficient of
Invar
1.5 () \(≈0\)
Stainless steel
10 – 17
Silver
19 – 20
Selenium
37		 Comments Active
46 ___________ structure has maximum hardness. Troostile Pearlite Martensite Sorbite c Pearlite: It is formed by the decomposition of austenite at 723°C. The decomposition of austenite leads to the formation of a eutectoid mixture of 87% ferrite and 13% cementite called pearlite. The pearlite constituent consists of alternate lamellae of ferrite and cementite and contains 0.8% carbon.
Martensite: It is a metastable phase of steel formed by the transformation of austenite below 320°C. Martensite is an interstitial supersaturated solid solution of carbon in a - Iron and has body-centred tetragonal lattice. It has a carbon content of up to 2% and is extremely hard and brittle. It is a product of rapid cooling (quenching) and possesses an acicular or needle-like structure.
Austenite: If steel is heated, a change in its structure commences from 723°C. The new structure formed is called Austenite.
Hence Hardness in increasing order is:
Sphrodite < coarse pearlite < Fine pearlite < martensite
The hardness of various structures is given below
ROCKWELL
HARDNESS STRUCTURE
RC 15
Coarse Pearlite
RC 25
Fine Pearlite
RC 65
Maternsite		 Comments Active
47 ________ is not the neutral refractory material Graphite Karnelite Chromite Dolomite b A refractory material is a material that retains its strength and form at high temperature.
Refractory material are used in furnaces, kilns, in nuclear and reactors.
The oxides of aluminium (alumina), silicon (silica) and magnesium (magnesia) are the most important. Materials used in the making of refractories, natural fractural.
Common examples of materials are alumina, chromite, granites etc.
• Karnelite is evaporites material.		 Comments Active
48 ____________ is better suited for lighter duty bearings. Phosphor Bronze Plastics White metal Monel metal a White metal or Babbit metal:
• The white metal is used for making bearings that are subjected to heavy load.
• The white metal is an alloy of Lead and Lithium metal.
• These alloys usually consist of relatively hard crystals embedded in a softer matrix, a structure important for machine bearings.
• They are composed primarily of tin, copper, and antimony, with traces of other metals added in some cases and lead substituted for tin in others.
• The alkali metals are quite useful.
Phosphorous bronze:
• Phosphorous bronze contains 90% copper, 9.7% tin and 0.3% phosphorus.
• The box nut or replaceable nut of a bench vice is made up of phosphorous bronze.
Monel metal:
• Monel metal is an important alloy of nickel and copper.
• It contains 68% nickel, 29% copper, and 3% other constituents like iron, manganese, silicon, and carbon.
• It resembles nickel in appearance and is strong, ductile, and tough. It is superior to brass and bronze in corrosion-resisting properties.
• It is used for making propellers, pump fittings, condenser tubes, steam turbine blades, seawater exposed parts, tanks, and chemical and food handling plants.		 Comments Active
49 The pH value of neutral solution is Equal to 7 Less than 7 Greater than 7 None of these a A neutral solution is one that has a pH value equal to 7.
If 0 pH < 7 For acidic solution \(≤\) \(→\)
pH = 7 For neutral solution \(→\)
7 < pH 14 For basic solution (alkaline). \(≤\) \(→\)		 Comments Active
50 In 18-4-1 HSS (High speed steel) the percentage of chromium is 1% 4% 18% 20% b • High-speed steel is special alloy steel which is obtained by alloying tungsten,
Chromium, Vanadium, Cobalt and molybdenum with steel
• HSS is an alloy of 18% tungsten, 4% chromium and 1% Vanadium
• Stellite is an alloy of 30% chromium, 20% tungsten and 1 to 4% carbon and the remaining consists of cobalt.		 Comments Active
51 Water flows through a turbine in which due to friction is a temperature rise from 30áµ’ to 35áµ’C. If there is no heat transfer taking place during the process. What is the change in the entropy of water? 0.077 0.079 0.406 0.496 b Given
\(T_{1}=30+273=303K\)
\(T_{2}=35+273=308K\)
We know, dQ = TdS
dS = \(\frac{mC_{v}dT}{T}\)
\(S_{2}-S_{1}=mC_{v}ln\frac{T_{2}}{T_{1}}=1×4.187ln\frac{308}{303}\)
J/kg \(S_{2}-S_{1}=0.06852\)		 Comments Active
52 All the processes in an air standard cycle are assumed to be _______. Adiabatic Irreversible Isothermal Reversible d Air – standard cycle, which is based on the following assumptions:
The working medium is assumed to be a perfect gas and follows the relation,
pV = mRT or p = ρRT
There is no change in the mass of the working medium.
All the processes that constitute the cycle are reversible.
Heat is assumed to be supplied from a constant high – temperature source and not from chemical reactions during the cycle.
The working medium has constant specific heats throughout the cycle.		 Comments Active
53 How is the thermal efficiency of an I.C engine related to compression ratio? Increase Decrease May increase or decrease Remains same c The compression ratio is defined as the ratio of the maximum volume of the combustion chamber to the minimum volume of the combustion chamber in an internal combustion engine. The compression ratio has a direct impact on the thermal efficiency of the engine.
When the compression ratio in an IC engine is increases, the thermal efficiency also increases. This is because the higher compression ration results in higher combustion temperatures and pressures, leading to better combustion of the fuel – air mixture. As a result, more energy is extracted from the fuel and converted into useful work.
However, there is a limit to how much the compression ration can be increased. If the compression ratio becomes too high, it can lead to knocking or detonation, which can damage the engine. At this point, the thermal efficiency will decrease due to incomplete combustion of the fuel – air mixture.		 Comments Active
54 What is the temperature at which a system goes under a reversible isothermal process without heat transfer? Absolute zero temperature Critical temperature Reversible temperature Boiling temperature a the temperature at which a system undergoes a reversible isothermal process without transfer of heat is called as absolute zero. Comments Active
55 What processes does Carnot cycle consist of? Two isothermal and two adiabatic processes Two isothermal and two constant volume processes Two isothermal and two constant pressure processes Two constant pressure and two constant volume processes a The Carnot cycle is composed of four reversible processes. Carnot cycle consists of 2 isothermal and 2 adiabatic processes. An isothermal process is a very slow process and the adiabatic process is a very fast process and the combination of a slow process and fast process is very difficult.
 Comments Active
56 What is the ratio W/Q (W = work transfer from cycle, Q = heat transfer to the cycle) known as? Air standard efficiency Specific consumption Specific work transfer Work ratio a \(Efficiency=\frac{Net work output}{Heat supplied}\)
\(W_{net}=Q_{in}-Q_{out}\)
\(η=\frac{W_{net}}{Q_{in}}=\frac{W_{net}}{Q_{supplied}}\)		 Comments Active
57 Tones of refrigeration mean ___. The weight of the machine is one tone. The weight of the refrigerant is one tone. The rate of the heat extraction is one tone. None of these c One ton of refrigeration is defined as the rate of heat removal required to freeze 1 ton (2000 lb) of water at 0°C into ice at 0°C in 24 hours. Comments Active
58 Which of the below stated are properties of a PMM – 2?
1. When network is equal to the heat absorbed and work efficiency is 100%
2. Heat is exchanged from one heat reservoir only
3. It violates Kelvin – Planck statement
4. It is a hypothetical machine
Options		 1, 2 and 4 1, 3 and 4 2, 3 and 4 1, 2, 3 and 4 d In conclusion, the properties of a PMM-2 include the equality of network and heat absorbed with 100% work efficiency, heat exchange from one reservoir only, violation of the Kelvin-Planck statement, and its hypothetical nature. Comments Active
59 If a Carnot refrigerator has a COP of 6. What is the ratio of the lower to the higher absolute temperature? 1-6 7-8 6-7 1-7 c Given
COP of refrigerator =6
COP of refrigerator == \(\frac{T_{L}}{T_{H}-T_{L}}\) \(6\)
\((T_{H}-T_{L})=\frac{T_{L}}{6} \)
\(\frac{T_{L}}{T_{H}}=6/7\)		 Comments Active
60 Which factor contributed to the efficiency of the Otto cycle? Compression ratio Operating pressure Operating temperature None of these a Efficiency of otto cycle is given by –
\(η_{otto}=1-\frac{1}{(r)^{γ-1}}\)
Where r = compression ration
γ = adiabatic index
On increasing compression ratio and adiabatic index efficiency of otto cycle increases.		 Comments Active
61 What value of the dryness fraction (in %) represents the dry saturated state of vapour? 0 25 50 100 d The dryness fraction of saturated steam or dry saturated steam is 1. Wet steam has dryness fraction less than 1, depending on amount of water vapor present in steam. Comments Active
62 What is the state at which phase change occurs without a change in pressure and temperature is known as? Critical state Saturation state Equilibrium state None of these b At a saturation state, any additional heat or mass transfer to the system would cause a phase change without changing the temperature or pressure. Comments Active
63 The mean effective pressure of Otto cycle is given by _____. \(P_{m}=\frac{p_{1}r_{k}(r_{p}-1)(rkγ-1-1)}{(γ-1)(r_{k}-1)}\) \(P_{m}=\frac{p_{1}r_{k}(r_{p}-1)(rkγ+1-1)}{(γ+1)(r_{k}-1)}\) \( P_{m}=\frac{p_{1}r_{k}(r_{p}+1)(rkγ-1-1)}{(γ-1)(r_{k}+1)}\) \(P_{m}=\frac{p_{1}r_{k}(r_{p}-1)(rkγ-1+1)}{(γ-1)(r_{k}+1)}\) a
\(W_{net}=Q_{1}-Q_{2}=mC_v((T_{3}-T_{2})-(T_{4}-T_{1}))\)
\(V_{s}=V_{1}-V_{2}=V_{1}(1-\frac{V_{2}}{V_{1}})=V_1(1-\frac{1}{r})\)
\(V_{s}=V_{1}(\frac{r-1}{r})=\frac{mRT_{1}}{P_{1}}(\frac{r-1}{r})=\frac{mC_{v}(r-1)T_{1}}{P_{1}}(\frac{r-1}{r})\)
\(mep=(\frac{P_{1}}{T_{1}})(\frac{1}{γ-1})(\frac{r}{r-1})((T_{3}-T_{2})-(T_{4}-T_{1})) \)
\(T_{2}=T_{1}(r)^{γ-1}\)
\(r_{p}=\frac{P_{3}}{P_{2}}=\frac{T_{3}}{T_{2}}\)
\(T_{3}=T_{2}r_{p}=T_{1}(r)^{γ-1}r_{p}\)
\(T_{4}=T_{3}(\frac{1}{r})^{γ-1}=T_{1}r_{p}\)
\(((T_{3}-T_{2})-(T_{4}-T_{1}))=T_{1}(r^{γ-1}-1)-(r_{p}-1)\)] \(T_{1}((r^{γ-1}-1)(r_{p}-1))\)
\(mep=(\frac{P_{1}}{T_{1}})(\frac{1}{γ-1})(\frac{r}{r-1})T_{1}((r^{γ-1}-1)(r_{p}-1))\)
\(mep=P_{1}r(\frac{r^{γ-1}-1)(r_{p}-1)}{(r-1)(γ-1)})\)
Increasing pressure ratio will increase the mean effective pressure and which increases the efficiency. \((r_{p})\)		 Comments Active
64 For the same operating maximum pressure and temperature, which cycle will have the highest efficiency? Diesel cycle Dual cycle Otto cycle None of these a
P – V and T – s diagram of otto cycle, diesel cycle and dual cycle are:
1 – 6 – 4 – 5 – Otto cycle
1 – 2 – 3 – 4 – 5 – Dual cycle
1 – 7 – 4 – 5 - Diesel cycle
As we know that,
Efficiency of a cycle is given by:
\(η=\frac{work done}{heat input}\)
As heat input is same in all three cycles and work done is equal to area under P – V curve.
As seen from the diagram area under 1 – 7 – 4 – 5 (diesel cycle) is maximum, therefore efficiency of diesel cycle will be maximum.		 Comments Active
65 In an Otto cycle, r is the compression ratio and γ is the adiabatic index. Which relation defines the air standard efficiency? \(η=1-\frac{1}{r^{γ-1}}\) \(η=1-\frac{1}{r^{γ}}\) \(η=1-\frac{1}{r^{γ}+1}\) \(η=1-\frac{1}{r^{\frac{γ}{γ-1}}}\) a Thermal efficiency of Otto cycle:
\(η_{otto}=1-\frac{1}{r^{γ-1}}\)
Compression ratio \(r=V_{1}/V_{2}\)
\(\frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}})^{\frac{γ-1}{γ}}=(\frac{V_{2}}{V_{1}})^{γ-1}\)		 Comments Active
66 In a surrounding, the amount of irreversibility of a process undergone by a system is determined by _______. Entropy change of the system Entropy change of the surrounding Entropy increase of the universe Entropy decrease of the universe c Extent of irreversibility of any process is determined by the entropy increase of the universe. Comments Active
67 A carnot engine operates between T1 and T2. If there is an increase in source temperature by t and a decrease in sink temperature by t. How the efficiency in both cases is related? \(η_{1}>η_{2}\) \(η_{1}<η_{2}\) \(η_{1}=η_{2}\) None of these b If we decrease the temperature of sink, its efficiency will increase. Therefore . \(η_{1}<η_{2}\)
Efficiency \(η=1-\frac{T_{2}}{T_{1}}\)		 Comments Active
68 The entropy always increases for an isolated system and when the equilibrium is reached, it is _____. Maximum Same as the initial starting state More than initial starting state Zero a The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium. Comments Active
69 What is the CORRECT order of decrease in entropy?
1. Solid phase
2. Liquid phase
3. Gaseous phase
Options		 (1) > (2) > (3) (3) > (2) > (1) (3) > (1) > (2) (1) > (3) > (2) b The entropy of a system is actually the measure of the randomness of a system. Among solid liquid and gas, as the randomness follows the order gas> liquid> solid, entropy also follows the same order. Comments Active
70 For a pure substance in gaseous phase to be liquefied, the gas has to undergo initial cooling below the ________. Critical state Saturated liquid line Saturated vapor line Triple point line a The gases should be cooled down below their critical temperatures for their liquification. Upon cooling, the movement of molecules slow down. Therefore, the inter-molecular forces hold the slowly moving molecules together and thus, the gas liquifies. Comments Active
71 A steel bar of dimension 10 mm × 1 mm × 1 mm is free to expand is heated from 15ᵒC to 25ᵒC. What stress shall be developed? Tensile stress No stress Shear stress Compressive stress b In case of free expansion there will be no stress developed in the bar. Comments Active
72 A beam of diameter 75 mm has a span length of 10 m is subjected to uniform distributed load of w kN/m. The maximum value of bending stress produced is 6.75 kN-m. What is the value of distributed load, if the beam is simply supported? 1.5 0.54 2 5.4 b
Given Mmax = 6.75 kN-m
D= 0.075 mm
L = 10 m
\(M_{max}=\frac{wl^{2}}{8}\)
\(6.75=\frac{w×10^{2}}{8}\)
w = 0.54 kN/m		 Comments Active
73 Which is the correct option for the stress experienced by the shaft when the power transmission in shaft takes place through the gears? Varying bending and constant torsional stresses Constant bending and varying torsional stresses Torsional stress Bending stress only a When power is transmitted through gears mounted on a shaft, the shaft experiences two primary types of stresses:
Torsional Stress: Due to the torque applied to the shaft to transmit power from one gear to another. This torsional stress is constant as long as the transmitted torque is constant.
Bending Stress: When gears are mounted on a shaft, they exert radial forces (gear tooth forces) which act perpendicular to the shaft. These forces create bending moments on the shaft, which vary along its length, especially at gear locations and supports.
Therefore, bending stress is varying.		 Comments Active
74 A power transmission shaft of diameter d rotating at the speed of N rpm. The power P is related to diameter d and N as ______. \(P∝(Nd^{3})\) \(P∝(Nd^{3})^{\frac{1}{2}}\) \(P∝(Nd^{3})^{\frac{2}{3}}\) \(P∝(Nd^{3})^{\frac{1}{3}}\) a Power Transmitted by a shaft P = \(\frac{2πNT}{60}\)
From torsion equation
\(\frac{T}{J}=\frac{Ï„}{r}\)
is the allowable shear stress in the shaft, J = Polar moment of inertia, T is the torque acting on the shaft, r is the radius of the shaft \(Ï„\)
Now from torsion equation
\(T=\frac{Ï€}{16}d^{3} Ï„\)
P = \(\frac{2πN}{60}×\frac{π}{16}d^{3} τ\)
\(P∝(Nd^{3})\)		 Comments Active
75 A metal pipe is subjected to internal pressure of 10 kgf/cm2. If the permissible tensile stress in the metal is 200 kgf/cm2 and the thickness of the pipe is 2.5 cm. What is the diameter of the metal pipe? 10 100 200 20 b permissible tensile stress or hoop stress= \(\frac{Pd}{2t}\)
Where P=internal pressure, t=thickness, = stress \(σ_{h}\)
Diameter of pipe d=cm. \(\frac{200×2×2.5}{10}=100\)		 Comments Active
76 The thin walled cylindrical vessel has the wall thickness t and diameter d is subjected to the gauge pressure of p. If the diameter of the vessel is doubled, then what is the ratio of hoop stress as compared to initial value? 1:1 2:1 1:2 1:4 c Hoop Stress=Pd/2t
If the diameter of vessel is double, than hoop stress will be=P2d/2t
\(=\frac{\frac{Pd}{2t}}{\frac{P2d}{2t}}=1:2\)		 Comments Active
77 What is the ratio of the Euler’s buckling loads of column having (i) both ends fixed and (ii) one end fixed and other end free is? 4:1 16:1 1:4 2:1 b \(\frac{\frac{4π^{2EI}}{L^{2}}}{\frac{π^{2EI}}{4L^{2}}}=16:1\)
Ends Conditions
Le
Buckling Load
Both ends hinged
L
\(Ï€^{2}EI/L^{2}\)
Both ends fixed
L/2
\(4Ï€^{2}EI/L^{2}\)
One end fixed and another end is free
2L
\(Ï€^{2}EI/4L^{2}\)
One end fixed and other end hinged
\(L/2\)
\(2Ï€^{2}EI/L^{2}\)		 Comments Active
78 The expression for the Rankine’s crippling load is given as _______. \(P=\frac{σ_{c}A}{1-K(\frac{l_{e}}{K})^{2}}\) \(P=\frac{σ_{c}A}{1+K(\frac{l_{e}}{K})^{2}}\) \(P=\frac{σ_{c}A}{1-2K(\frac{l_{e}}{K})^{2}}\) \(P=\frac{σ_{c}A}{1+2K(\frac{l_{e}}{K})^{2}}\) b The Rankine’s Gordon formula is applicable for both long and short column. Comments Active
79 The expression for the slenderness Ration of the columns is given as _______. \((\frac{l_{e}}{K_{min }})^{2}\) \((\frac{2l_{e}}{K_{min }})^{2}\) \((\frac{l_{e}}{K_{min }})^{ }\) \((\frac{2l_{e}}{K_{min }})^{ }\) c Slenderness raio is the ratio of efective length() to the least radius of gyration(k). \(l_{e}\) Comments Active
80 The equivalent length of the column, when both the ends are hinged is ____. l l/2 l/4 2l a 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
81 Maximum principal stress theory was postulated by _____. ST Venant Mohr Rankine Tresca c Maximum Principal stress theory was postulated by Rankine. It is suitable for brittle materials.
Maximum Shear stress theory was postulated by Tresca. This theory is suitable for ductile materials
Maximum Principal strain theory was postulated by St Venant. This theory is not accurate for brittle and ductile materials both.
Maximum shear strain energy theory was postulated by Von-mises. Its results in case of pure shear are the accurate for ductile materials		 Comments Active
82 What is the S.I. unit of Poisson’s ratio? kN/mm2 N/mm2 mm Unitless d Poisson’s ratio is the ratio of lateral strain to longitudinal strain in the direction of the stretching force. Comments Active
83 What is the direction of application of friction force with respect to the direction of motion of an object? Same Perpendicular Opposite Downward c Friction is the resistance to motion of one object moving relative to another.
It produces heat.
It opposes relative motion between two object which are in contact. Hence friction force always acts opposite to direction of motion.
It leads to decrease in velocity of object.
It depends on the nature of the surface.		 Comments Active
84 Which of the following term can also be used to state the dynamic equilibrium? Translatory equilibrium Static equilibrium Kinetic equilibrium Rotational equilibrium c We can use kinetic energy equation for dynamic equilibrium in rotation as well as translation both. Thus, kinetic equilibrium is a broader and synonymous term for dynamic equilibrium, covering both translational and rotational motion where net force and torque are zero. Comments Active
85 When a body is said to be in dynamic equilibrium? When the body is moving with non uniform accelerations When the body is in uniform motion along a circular path When the body is in uniform motion along a straight line When the body is moving with instantaneous velocity c When acceleration is zero then body is said to be in dynamic equilibrium. When the body is in uniform motion along a straight line. Comments Active
86 Which of the following options is correct about the motion of the follower? Uniform velocity Simple harmonic motion Uniform acceleration All option are correct d Uniform Velocity motion: The abrupt changes of velocity, which result in large forces at the beginning and the end of the stroke. These forces are undesirable, especially when the cam rotates at high velocity. The constant velocity motion is therefore only of theoretical interest.
Simple Harmonic Motion (SHM): There is an abrupt change of acceleration from zero to maximum at the beginning of the follower motion and also at the end of the follower motion. The same pattern is repeated during descent. This leads to Jerk, vibration and noises etc. Therefore, SHM should be adopted only for low and moderate cam speeds.
Cycloidal motion: It may be observed that there are no abrupt changes in the velocity and acceleration at any stage of the motion. Therefore, cycloidal is most ideal for high speed follower motion.		 Comments Active
87 The types of failure involved in the analysis of the riveted joints are?
1. Shear failure of rivet
2. Tensile failure of the plate
3. Crushing failure of the plate
Which of the following statements are is correct for the analysis of the riveted joints?		 1 and 2 only 2 and 3 only 1, 2 and 3 1 and 3 only c According to conventional theory, the failure of the riveted joint may occur in any one or more of the following ways:
1. Shear failure of the rivet
2. Tensile failure of the plate between two consecutive rivets
3. Crushing or bearing failure of the plate
4. Shear failure of the plate in the margin area
5. Tearing of plate in the margin area		 Comments Active
88 Which of the following is NOT the type of loaded type governor? Dead weight governor Spring controlled governor Watt governor Porter governor c Comments Active
89 The profile of gears having module of 12 mm and length of arc of contact is 60 mm. What is the contact ratio of the gear profile? 2 1.2 1.6 5 c Given
M = 12 mm
AOC = 60 mm
PC = \(π×m\)
\(=3.14×12=37.68\)
Contact Ratio = \(\frac{AOC}{P_{C}}\)
\(=\frac{60}{37.68}\)
\(=1.59≈1.6\)		 Comments Active
90 Which of the following statement is TRUE about the length of arc of contact? Inversely proportional to the module Varies directly to the length of the path of contact Inversely proportional to the circular pitch All option are correct b The longer the arc of contact, the more teeth are in contact at any given moment. Therefore, the contact ratio is directly proportional to the length of the arc of contact.
Arc of contact = \(\frac{Path of contact}{cos∅}\)
This means that as the length of the arc of contact increases, the contact ratio also increases.		 Comments Active
91 The equation for the calculation of torque transmitting capacity in the conical clutch as per uniform wear theory is _____. \(M_{t}=\frac{μP}{4sinα}(D+d)\) \(M_{t}=\frac{μP}{4sinα}(D-d)\) \(M_{t}=\frac{μP}{sinα}(D-d)\) \(M_{t}=\frac{2μP}{sinα}(D-d)\) a Torque transmitting capacity as per uniform pressure theory is :
\(M_{t}=\frac{μP}{3sinα}\frac{(D^{3}-d^{3})}{(D^{2}-d^{2})}\)
μ is coefficient is coeffcient of friction
P is normal pressure density
D is outer diameter of cone
d is inner diameter of cone
is semi- cone angle \(α\)		 Comments Active
92 Which of the following theory is used in the design of clutches, when the friction linings get warn out? Uniform pressure theory Uniform wear theory Uniform friction theory None of these b The uniform pressure theory is applicable only when friction lining is new.
The uniform wear theory is applicable when the friction lining gets worn out, a major portion of the life of the friction lining comes under the uniform wear criterion.
The torque transmitting capacity of the new clutch is slightly more than worn-out clutches because the friction radius of the new clutch is slightly more than worn-out clutches.		 Comments Active
93 Which of the following is NOT the inversion of single slider crank chain? Pendulum pump Whitworth quick return motion mechanism Oscillating cylinder engine Elliptical trammel d Elliptical trammel are used for drawing ellipse. They can be used to draw smaller ellipse but only draw on half at a time, having to be reverse to draw the complete ellipse. The elliptical trammel is simple mechanism which can trace and exact elliptical path. Comments Active
94 The scotch yoke mechanism is the inversion of __________. Four bar link chain Double slider crank chain Single slider crank chain None of these b The reciprocating mechanism that changes rotational motions into reciprocating motions or vice versa is called the Scotch Yoke Mechanism. As mentioned above, it is an inversion of the four-bar linkage mechanism. This inversion is obtained when one of the sliding links of a double slider-crank chain is fixed Scotch yoke mechanism is the one of the mechanism of Double Slider crank mechanism. Comments Active
95 The power transmission takes place from a pulley of diameter 1 m running at the speed of 250 rpm to a pulley of 2.50m diameter by means of a belt. Determine the speed (in rpm) of the driven pulley? 110 150 100 625 c Given
\(D_{1}-1m\)
\(N_{1}=250 rpm \)
\(D_{2}=2.50 m\)
\(\frac{N_{1}}{N_{2}}=\frac{D_{2}}{D_{1}}\)
\(N_{2}=\frac{N_{1}D_{1}}{D_{2}}\)
\(=\frac{250×1}{2.50}\)
\(N_{2}=100 rpm\)		 Comments Active
96 The angular acceleration and the moment of inertia of the flywheel is 0.6 rad/s2 and 2500 kg. m2 respectively. What is the kinetic energy (in kN-m) of the flywheel after 10 seconds from the start? 45 450 4.5 4500 a Given
\(α=0.6 rad/s^{2}\)
\(I=2500 kg/m^{2}\)
\(t=10 sec\)
\(w=αt\)
\(=0.6×10\)
\(=6 rad/s\)
\(K.E=\frac{1}{2}Iw^{2}\)
\(K.E=\frac{1}{2}×2500×6^{2}\)
\(K.E=45,000 N-m\)
\(K.E=45 KN-m\)		 Comments Active
97 Which of the following statement is TRUE regarding work done per cycle? Work done per cycle is directly proportional to the mean torque Work done per cycle is directly proportional to the angle turned in one revolution Work done per cycle is inversely proportional to the angle turned in one revolution Both work done per cycle is directly proportional to the mean torque and work done per cycle is directly proportional to the angle turned in one revolution d The formula for calculating work on a flywheel is
W = Tmean x θ,
where W is the work, T is the torque, and θ is the angular displacement of the flywheel in radians.
Both work done per cycle is directly proportional to the mean torque and work done per cycle is directly proportional to the angle turned in one revolution		 Comments Active
98 What is the degree of freedom of the mechanism shown below?
 1 2 3 4 a L = 6
j = 7
h = 0
F = 3 (L-1) -2j –h
\(=3(6-1)-2×7-0\)
\(=3×5-14\)
F = 1		 Comments Active
99 Which of the following will lead to the formation of higher pair? Sliding pair Turning pair Rolling pair None of these c A pair of toothed gearing, belt and rope drives, ball and roller bearings and cam and follower are the examples of higher pairs. In Belt and pulley due to line contact, they form a higher pair. Comments Active
100 The number of joints (j) which constitutes a kinematic chain can be expressed in terms of number of links (l) as _____________. \(j=\frac{3}{4}l-2\) \(j=\frac{3}{2}l+2\) \(j=\frac{3}{2}l-2 \) \(j=l-\frac{2}{3}\) c The relation between the number of link and number of kinematic pair is given by,
L = 2p - 4
where, L = number of links, p = number of kinematic pair.
The above formula is used to check whether a given chain is a kinematic chain or not.
The relation between the number of joints and the number of links is given by,
\(J+\frac{H}{2}=\frac{3L}{2}-2\)
J = Number of binary joints, H = Number of higher pairs		 Comments Active