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30 Questions MCQ Test - Full Test 2 - EKT Mechanical

Full Test 2 - EKT Mechanical for AFCAT 2024 is part of AFCAT preparation. The Full Test 2 - EKT Mechanical questions and answers have been prepared according to the AFCAT exam syllabus.The Full Test 2 - EKT Mechanical MCQs are made for AFCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Full Test 2 - EKT Mechanical below.
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Full Test 2 - EKT Mechanical - Question 1

The order of differential equation 

Detailed Solution for Full Test 2 - EKT Mechanical - Question 1

Order of the differential equation is the highest derivative in a differential equation. Hence the order is 2. 

Full Test 2 - EKT Mechanical - Question 2

The system of linear equations
(4a – 1)x + y + z = 0
-y + z = 0
(4a – 1) z = 0
has a non-trivial solution, if a equals

Detailed Solution for Full Test 2 - EKT Mechanical - Question 2

The system of homogeneous linear equations has a non – trivial solution if

(4a – 1)2 = 0
a = 1/4 

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Full Test 2 - EKT Mechanical - Question 3

The value of 

Detailed Solution for Full Test 2 - EKT Mechanical - Question 3


Full Test 2 - EKT Mechanical - Question 4

If f(x, y)is a function satisfying Euler’ s theorem then

Detailed Solution for Full Test 2 - EKT Mechanical - Question 4

If u is a homogeneous function of degree n in x and y, then according to Euler’s Theorem:

Full Test 2 - EKT Mechanical - Question 5

If the centre of a circle is (-6, 8) and it passes through the origin, then equation to its tangent at the origin is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 5

The slope of the line joining the origin to the center is m1 (-8/6)
A tangent is perpendicular to the radius at the point of contact, so it’s slope will be m2 (-1/m) = (6/8)
Equation of line will be: y = mx

As, the line passes through the origin, so there’s no constant term c (recall y = mx + c)

Full Test 2 - EKT Mechanical - Question 6

A streamlined body is defined as a body about which ________.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 6

A stream lined body is defined as the body whose surface coincides with the stream - lines, when the body is placed in a flow. In that case the separation of flow will take place only at trailing edge (or rearmost part of the body).

Behind a stream lined body, wake formation zone will be very small and consequently the pressure drag will be very small. Thus the total drag on the stream lined body will be due to friction (shear) only.

Full Test 2 - EKT Mechanical - Question 7

A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω. Two objects each of mass ‘m’ are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity –

Detailed Solution for Full Test 2 - EKT Mechanical - Question 7

By conservation of angular momentum
I’ω’ = Iω
(M + 2m) r2ω’ = Mr2ω

Full Test 2 - EKT Mechanical - Question 8

One Coulomb passing a point in one second is one ______.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 8

The current at a point in the circuit is the amount of charge that passes that point in one second. A current of 1 A is flowing in a circuit if a charge of 1 coulomb passes any point in the circuit every second.

Full Test 2 - EKT Mechanical - Question 9

In an AC circuit, resonance occurs when

Detailed Solution for Full Test 2 - EKT Mechanical - Question 9

In an AC circuit, resonance occurs when capacitive reactance equals reactive reactance.
At resonance, XL = XC,Z = R

Full Test 2 - EKT Mechanical - Question 10

All the magnetic materials lose their magnetic properties when they are

Detailed Solution for Full Test 2 - EKT Mechanical - Question 10
  • For every material, there is a temperature specific to it known as its Curie temperature
  • Above curie temperature, they lose their magnetic properties
Full Test 2 - EKT Mechanical - Question 11

When setting up a mechanical drawing in AutoCAD the drafter should set the units to _________

Detailed Solution for Full Test 2 - EKT Mechanical - Question 11

In AutoCAD, to specify the drawing units in a new or existing drawing, the drafter should set the units to decimal. Decimal unit set is the default unit system in AutoCAD.

Full Test 2 - EKT Mechanical - Question 12

The energy absorb by brake is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 12

A brake is a device by means of which artificial frictional resistance is applied to a moving machine member,in order to retard or stop the motion of a machine.

In the process of performing this function, the brake absorbs either kinetic energy of the moving member or potential energy given up by objects being lowered by hoists, elevators etc.The energy absorbed by brakes is dissipated in the form of heat. This heat is dissipated in the surrounding air.

Full Test 2 - EKT Mechanical - Question 13

In the assembly design of shaft, pulley and key the weakest member is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 13

Key is the weakest member in the assembly of shaft, pulley and key. Key acts as a safety device, whenever there is excess load appears on the pulley key fails first and it keeps safer to shaft and pulley.

Full Test 2 - EKT Mechanical - Question 14

The most suitable bearing for carrying very heavy loads with slow speed is _____.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 14

A bearing is said to be hydrostatic bearing if it has a static fluid (liquid or air) along the surface of the shaft/journal where the fluid will be supplied externally and the pressure is generally maintained by an external pump; thus enabling non-contacting operation and the ability to support a load. Hydrostatic bearings can support large loads without journal rotation and provide large (accurate and controllable) direct stiffness as well as damping (energy dissipation) coefficients.

In case of hydrodynamic bearing the pressure is developed by the high speed journal at higher rpm, this helps the movement of thin film of lubricant. For Journal bearings mostly converging fluids are employed.

Full Test 2 - EKT Mechanical - Question 15

In the multiple disc clutch, if there are 6 discs on the driving shaft and 5 discs on the driven shaft, then the number of pairs of contact surfaces will be equal to

Detailed Solution for Full Test 2 - EKT Mechanical - Question 15

Number of pair of contact surfaces = n = n1 + n2 – 1
Where, n1 is the number of discs on driving shaft
N is the number of discs on driven shaft
N = 6 + 5 – 1 = 10

Full Test 2 - EKT Mechanical - Question 16

If a light and a heavy body have equal kinetic energy of translation, then ________.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 16

∴ the speed of the smaller object is higher than that of bigger.

Momentum of the smaller object:
 

Lighter body will have smaller momentum.

Full Test 2 - EKT Mechanical - Question 17

An elevator weighing 1000 kg attains an upward velocity of 4 m/sec in two seconds with uniform acceleration. The tension in the supporting cables will be:-

Detailed Solution for Full Test 2 - EKT Mechanical - Question 17

Force = T - W
ma = T - W
Acceleration is given as

Tension is given as
T = mg + ma = m (g + a)
T = 1000 (10 + 2)
T = 12000 N

Full Test 2 - EKT Mechanical - Question 18

A bar of length ‘L' meters extends by ‘l' mm under a tensile force of ‘P’. The strain produced in the bar is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 18

Whenever the bar is subjected to the axial tensile load, there will be increase in the length of the bar along the direction of the loading.

Longitudinal strain is defined as the ratio of increase in length of the bar in the direction of applied load to that of the original length.

Full Test 2 - EKT Mechanical - Question 19

When a point load on the free end of a cantilever beam is increased, failure will occur

Detailed Solution for Full Test 2 - EKT Mechanical - Question 19

A cantilever beam subjected to point load on free end will have maximum bending moment at fixed end and constant shear force throughout the length. So maximum stress will be at fixed end (σ = My/I) and failure will occur at that point.

Full Test 2 - EKT Mechanical - Question 20

The shear force and bending moment are zero at the free end of a cantilever beam, if it carries a

Detailed Solution for Full Test 2 - EKT Mechanical - Question 20


From SFD and BMD Diagram, we can conclude that shear force and Bending moment both are zero when cantilever beam is subjected to uniformly distributed load.

Full Test 2 - EKT Mechanical - Question 21

After reaching the yielding stage while testing a mild steel specimen strain

Detailed Solution for Full Test 2 - EKT Mechanical - Question 21

Salient points of Stress-Strain Diagram:

Point A (Limit of proportionality or the proportionality limit): The strain is proportional to strain or elongation is proportional to the load giving a st.line relationship. This law of proportionality is valid upto a point A.

Point B (Elastic Limit): For a short period beyond the point A, the material may still be elastic in the sense that the deformations are completely recovered when the load is removed.

Point (C) and (D) - Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when load is removed. These two points are termed as upper and lower yield points respectively.

Yield stress is defined as the stress after which material extension takes place more quickly with no or little increase in load.

Point E: A further increase in the load will cause marked deformation in the whole volume of the metal. The maximum load which the specimen can with stand without failure is called the load at the ultimate strength.

Point F: Beyond point E, the bar begins to forms neck. The load falling from the maximum until fracture occurs at F.

Full Test 2 - EKT Mechanical - Question 22

Isothermal compressibility of an ideal gas is 

Detailed Solution for Full Test 2 - EKT Mechanical - Question 22

Compressibility (τ):

Compressibility is thus inverse of bulk modulus. Hence compressibility can be defined as the incurred volummetric strain for unit change in pressure.
Isentropic compressibility: 
Isothermal compressibility:  
Since: PV=nRT

Hence, isothermal compressibility is

Full Test 2 - EKT Mechanical - Question 23

How much heat energy is gained when 5 kg of water at 20°C is brought to its boiling point (Specific heat of water = 4.2 kJ/kg oC)

Detailed Solution for Full Test 2 - EKT Mechanical - Question 23

Q  = mcΔT = 5 × 4.2 × (100−20) = 1680kJ

Full Test 2 - EKT Mechanical - Question 24

A Carnot engine receiving heat at 400 K has an efficiency of 25%. The COP of Carnot refrigerator , working between same temperature limits is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 24

Full Test 2 - EKT Mechanical - Question 25

The figure given below shows the variation of temperature across the thickness of materials with different thermal conductivities under steady states. Curve C will be applicable when thermal conductivity of the material______

Detailed Solution for Full Test 2 - EKT Mechanical - Question 25

The variation in thermal conductivity of a material with temperature in the temperature range of interest is given by k(T) = k0 (1 + αT) where α is called the temperature coefficient of thermal conductivity.

Curve B represents that α is less than zero
Curve D represents that α is zero
Curve A represents that α is greater than zero

Curve C represents that α is very large so the value of thermal conductivity is also very large. Curve C represents that there is almost no loss in heat transfer across the thickness which clearly indicates very large value of thermal conductivity.

Full Test 2 - EKT Mechanical - Question 26

Entropy change depends on ________.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 26

Absolute entropy (S) or entropy is a measure of energy dispersion in a system.

Following are basic features of entropy:

  • Entropy transfer is associated with heat transfer (direction of flow of entropy is same as that of heat flow)
  • Entropy is proportional to mass flow in an open system
  • Entropy does not transfer with work

Entropy linked with energy

Full Test 2 - EKT Mechanical - Question 27

The liquid is flowing separately through each of two pipes whose diameters are in the ratio of 2 : 1, if the ratio of the velocities of flow in the two pipes by 1 : 2, then the ratio of the amounts of the liquid flowing per sec through the pipe will be

Detailed Solution for Full Test 2 - EKT Mechanical - Question 27

Discharge through a pipe is Q = AV

Full Test 2 - EKT Mechanical - Question 28

A U tube manometer shown in figure is used to measure the gauge pressure of water of density ρ1 = 1000 kg/m3. If the density of manometer liquid ρ2 is 12000 kg/m3, h1 = 0.5 m & h2 = 1.0 m, gauge pressure at ‘A’, the centre of tube is (take g = 10 m/s2)

Detailed Solution for Full Test 2 - EKT Mechanical - Question 28

Full Test 2 - EKT Mechanical - Question 29

The reading of the pressure gauge fitted on a vessel is 25 bar. The atmospheric Pressure 1.03 bar and the value of g is 9.81 m/s2. The absolute pressure in the Vessel is

Detailed Solution for Full Test 2 - EKT Mechanical - Question 29

Absolute pressure = Gauge pressure + atmospheric pressure  = 25 + 1.03 = 26.03 bar

Full Test 2 - EKT Mechanical - Question 30

An impulse turbine is used for ________.

Detailed Solution for Full Test 2 - EKT Mechanical - Question 30

High head turbine: In this type of turbines, the net head varies from 150m to 2000m or even more, and these turbines require a small quantity of water. Example: Pelton wheel turbine.

Medium head turbine: The net head varies from 30m to 150m, and also these turbines require a moderate quantity of water. Example: Francis turbine.

Low head turbine: The net head is less than 30m and also these turbines require a large quantity of water. Example: Kaplan turbine.

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