← MES Jr. Grade
Past Questions
The value of the natural frequency obtained by Rayleigh’s method
A)
is always greater than the actual fundamental frequency
B)
is always less than the actual fundamental frequency
C)
depends upon the initial deflection curve chose and may be greater than or less than the actual fundamental frequency
D)
is independent of the initial deflection curve chosen
Option A is correct: is always greater than the actual fundamental frequency.
To determine hoop stress, efficiency of is to be considered.
A)
Construction joint
B)
Transverse joint
C)
Longitudinal joint
D)
Rivetjoint
Hoop stress (also known as circumferential stress) occurs in cylindrical pressure vessels due to internal pressure. The stress acts along the circumference of the cylinder, which tends to pull the cylinder walls apart along its length. In a manufactured pressure vessel, the cylinder is typically formed from a flat plate rolled into a cylinder and joined along its length. This joint is called the 'longitudinal joint'. Because the hoop stress is the primary stress acting to pull this joint apart, the efficiency of the longitudinal joint (the ratio of the strength of the joint to the strength of the solid plate) must be considered to determine the safe working pressure or the actual stress the vessel can withstand. Transverse joints (circumferential joints) are subjected to longitudinal stress, which is only half the magnitude of the hoop stress, making the longitudinal joint the critical factor for hoop stress calculations.
Maximum angular velocity of the connecting rod with a crank to connecting rod ratio 1:5 for a crank speed of 3000 rpm is around:
A)
300 rad/s
B)
60 rad/s
C)
30 rad/s
D)
3000 rad/s
Step-by-step calculation:
1. **Given Data**:
- Crank speed (N) = 3000 rpm
- Ratio of crank length (r) to connecting rod length (l) = 1:5. Therefore, n = l/r = 5.
2. **Calculate Angular Velocity of Crank (ω)**:
ω = (2 * π * N) / 60
ω = (2 * π * 3000) / 60 = 100 * π ≈ 314.16 rad/s
3. **Formula for Maximum Angular Velocity of Connecting Rod (ω_con)**:
In a slider-crank mechanism, the angular velocity of the connecting rod is given by:
ω_con = ω * (tan θ) / (n / cos φ) ≈ ω * (sin θ) / n
The maximum angular velocity of the connecting rod occurs approximately when the crank is perpendicular to the connecting rod. A widely accepted approximation for the maximum angular velocity of the connecting rod is:
ω_con_max ≈ ω * (sin θ_max) / n
For a large n, the maximum value occurs when sin θ is maximum (approx 1) or more accurately using the formula:
ω_con_max = ω * (sin θ) / sqrt(n^2 - sin^2 θ)
At the point of maximum angular velocity, sin θ ≈ 1/n (approximately), but for practical engineering calculations in competitive exams, the simplified maximum is often taken as:
ω_con_max ≈ ω / n
4. **Calculation**:
ω_con_max ≈ 314.16 / 5
ω_con_max ≈ 62.83 rad/s
5. **Conclusion**:
Comparing 62.83 rad/s with the given options, the closest value is 60 rad/s.
- Option (a) 300 rad/s is too high (close to the crank speed).
- Option (c) 30 rad/s is too low.
- Option (d) 3000 rad/s is incorrect as it matches the RPM value, not rad/s.
A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to
A)
41 MPa
B)
82 MPa
C)
164 MPa
D)
204 MPa
Step-by-step solution:
1. **Given Data:**
- Diameter of shaft (d) = 100 mm = 0.1 m
- Axial stress (σₓ) = 50 MPa = 50 × 10⁶ N/m²
- Torque (T) = 10 kNm = 10 × 10³ Nm
2. **Calculate Torsional Shear Stress (τ):**
The shear stress induced by torque in a solid circular shaft is given by:
τ = (16T) / (πd³)
τ = (16 × 10,000) / (π × 0.1³)
τ = 160,000 / (π × 0.001)
τ = 160,000,000 / π ≈ 50.93 MPa
3. **Calculate Maximum Principal Stress (σ₁):**
The shaft is subjected to a combined state of axial stress (σₓ) and shear stress (τ). The formula for the maximum principal stress is:
σ₁ = (σₓ / 2) + √((σₓ / 2)² + τ²)
σ₁ = (50 / 2) + √((50 / 2)² + 50.93²)
σ₁ = 25 + √(25² + 50.93²)
σ₁ = 25 + √(625 + 2593.86)
σ₁ = 25 + √3218.86
σ₁ = 25 + 56.73
σ₁ = 81.73 MPa
4. **Conclusion:**
The calculated maximum principal stress is approximately 81.73 MPa, which is closest to 82 MPa.
Comparing with options:
- (a) 41 MPa: Incorrect.
- (b) 82 MPa: Correct (closest value).
- (c) 164 MPa: Incorrect.
- (d) 204 MPa: Incorrect.
The following is not a friction clutch
A)
Centrifugal clutch
B)
Cone clutch
C)
Disc clutch
D)
Fluid clutch
A friction clutch is a device that transmits torque from one shaft to another through the application of friction between two surfaces.
1. Centrifugal clutch (Option a): Uses centrifugal force to push friction pads against a drum to engage the drive. It is a friction clutch.
2. Cone clutch (Option b): Uses conical surfaces that press against each other to transmit torque via friction. It is a friction clutch.
3. Disc clutch (Option c): Uses flat friction plates (discs) pressed together to transmit torque. It is a friction clutch.
4. Fluid clutch (Option d): Also known as a fluid coupling, it transmits torque through the hydrodynamic action of a fluid (usually oil) between an impeller and a runner. There is no physical contact or sliding friction between the driving and driven members. Therefore, it is a hydraulic/fluid coupling, not a friction clutch.
The ductility of a material is to the increase in percentage reduction in an area.
A)
inversely proportional
B)
directly proportional
C)
equal
D)
uniform
Ductility is a mechanical property of a material that allows it to undergo significant plastic deformation under tensile stress before rupture. In material science, ductility is quantitatively measured using two primary parameters: 1) Percentage Elongation and 2) Percentage Reduction in Area. The 'Percentage Reduction in Area' is calculated as [(Original Area - Final Area) / Original Area] * 100. A material that is highly ductile will experience a significant 'necking' effect, leading to a large reduction in the cross-sectional area at the point of fracture. Therefore, as the ductility of a material increases, the percentage reduction in area also increases. This positive correlation means that ductility is directly proportional to the increase in percentage reduction in area.
Transmission angle is the angle between
A)
Input link and coupler
B)
Input link and fixed link
C)
Output link and coupler
D)
Output link and fixed link
In the study of kinematics of machinery, specifically for a four-bar linkage, the transmission angle (denoted by μ) is defined as the angle between the coupler link and the output (follower) link.
1. **Purpose**: The transmission angle is a critical parameter used to evaluate the efficiency of motion transmission. It determines how effectively the force from the coupler is transmitted to the output link to produce rotation.
2. **Ideal Value**: For smooth operation and to avoid jamming or 'locking' of the mechanism, the transmission angle should ideally be as close to 90° as possible. Generally, it is recommended that the transmission angle does not deviate from 90° by more than 45° (i.e., it should stay between 45° and 135°).
3. **Analysis of Options**:
- (a) Input link and coupler: This angle changes but is not the 'transmission angle'.
- (b) Input link and fixed link: This is the input angle (θ).
- (c) Output link and coupler: This is the correct definition of the transmission angle.
- (d) Output link and fixed link: This is the output angle (ψ).
A beam AB of length 2 L having a concentrated load P at its mid-span is hinge supported at its two ends A and B on two identical cantilevers as shown in the given figure. The correct value of bending moment at A is
A)
Zero
B)
PL/2
C)
PL
D)
2PL
Option A is correct: Zero.
Belt slip may occur due to
A)
Heavy load
B)
Loose belt
C)
Driving pulley too small
D)
All of these
Belt slip occurs when the frictional force between the belt and the pulley is insufficient to transmit the required torque. The condition for no-slip is governed by the belt tension ratio formula: (T1/T2) ≤ e^(μθ), where T1 is the tight side tension, T2 is the slack side tension, μ is the coefficient of friction, and θ is the angle of wrap.
1. Heavy load (Option a): An increase in load increases the required torque, which increases the difference between T1 and T2. If T1/T2 exceeds e^(μθ), the belt will slip.
2. Loose belt (Option b): A loose belt reduces the initial tension (T0), which directly lowers the normal force acting on the pulley surface. Since frictional force is proportional to the normal force (F = μN), a loose belt significantly reduces the grip, leading to slip.
3. Driving pulley too small (Option c): A smaller pulley diameter reduces the angle of wrap (θ) if the center distance remains constant, or it increases the bending stress and reduces the effective contact area/pressure distribution. A smaller θ decreases the value of e^(μθ), making the system more prone to slipping.
Since all three factors contribute to the occurrence of belt slip, the correct option is 'All of these'.
The governor used in gramophone is of the following type
A)
Pickering
B)
Porter
C)
Hartnell
D)
Watt
A governor is a device used to maintain a constant speed of a machine regardless of the load. In the context of a gramophone (record player), it is critical to maintain a very precise and constant angular velocity to ensure the music plays at the correct pitch and tempo.
1. **Pickering Governor**: This is a centrifugal governor specifically designed for high precision and sensitivity in a compact space. It uses a spring-loaded mechanism that can react quickly to small changes in speed, making it ideal for the speed regulation required in gramophones.
2. **Porter Governor**: This is a dead-weight governor (a modification of the Watt governor) used primarily in large steam engines. It is too bulky and lacks the sensitivity required for a small electronic/mechanical device like a gramophone.
3. **Hartnell Governor**: This is a spring-loaded governor used in industrial applications where a wide range of speed control is needed. While it uses a spring, its design is intended for larger machinery rather than the precision instrumentation of a record player.
4. **Watt Governor**: This is the simplest form of a centrifugal governor (centrifugal balls on a rotating arm). It is highly inefficient at high speeds and lacks the precision needed for audio playback devices.
Therefore, the Pickering governor is the correct choice for the specific application of a gramophone.
Which of the following is a spring-controlled governor?
A)
Hartnell governor
B)
Hartung governor
C)
Wilson-Hartnell governor
D)
All of these
Governors are devices used to maintain the speed of an engine within specified limits regardless of the load. They are broadly classified into two types: Centrifugal Governors (which are further divided into Pendulum type and Fly-ball type) and Spring-controlled Governors.
1. Hartnell Governor: This is a classic example of a spring-loaded governor where a central spring provides the controlling force to counteract the centrifugal force of the rotating balls.
2. Hartung Governor: This is a specific type of spring-controlled governor where the spring is arranged such that the controlling force remains constant regardless of the lift of the sleeve.
3. Wilson-Hartnell Governor: This is a modification of the Hartnell governor that incorporates an additional spring to provide a more linear relationship between speed and lift, improving the sensitivity and stability of the governor.
Since all three listed governors (Hartnell, Hartung, and Wilson-Hartnell) utilize a spring to provide the controlling force, the correct option is 'All of these'.
In the case of an involute toothed gear, involute starts from
A)
addendum circle
B)
dedendum circle
C)
pitch circle
D)
base circle
An involute curve is defined as the path traced by a point on a straight line that rolls without slipping on a circle. In gear geometry, this 'rolling circle' is known as the base circle. Therefore, the involute profile of a gear tooth physically begins at the base circle and extends outward toward the addendum circle.
Analysis of options:
- a. Addendum circle: This is the outermost circle of the gear; the involute ends here.
- b. Dedendum circle: This is the bottom-most circle of the tooth space; the profile here is usually a radial fillet curve, not an involute.
- c. Pitch circle: This is the theoretical circle used to define the gear ratio and kinematics; while the involute passes through it, it does not start there.
- d. Base circle: This is the fundamental circle from which the involute is generated. The involute profile starts exactly at the base circle.
During tensile-testing of a specimen using a Universal Testing Machine, the parameters actually measured include
A)
True stress and true strain
B)
Poisson’s ratio and Young’s modulus
C)
Engineering stress and engineering strain
D)
Load and elongation
In a tensile test conducted using a Universal Testing Machine (UTM), the machine is equipped with sensors (such as a load cell for force and an extensometer or crosshead displacement sensor for length change). Therefore, the raw data 'actually measured' by the hardware are the applied Load (P) and the resulting Elongation (ΔL).
All other parameters mentioned in the options are 'calculated' or 'derived' values:
- Engineering Stress (σ = P/A₀) and Engineering Strain (ε = ΔL/L₀) are calculated by dividing the measured load and elongation by the original cross-sectional area and original length, respectively.
- True Stress and True Strain are derived from engineering values by accounting for the instantaneous change in area and length during deformation.
- Young's Modulus (E = σ/ε) and Poisson's ratio are material properties calculated from the slope of the stress-strain curve and the ratio of lateral to longitudinal strain, respectively.
Since the question asks for parameters 'actually measured', the correct answer is Load and elongation.
The longitudinal stress in the shell is
A)
pd/3t
B)
pd/4t
C)
pd/2t
D)
pd/ét
Option B is correct: pd/4t.
A cotter joint is used when no relative motion is permitted between the rods joined by the cotter. It is capable of transmitting
A)
twisting moment
B)
an axial tensile as well as compressive load
C)
the bending moment
D)
only compressive axial load
A cotter joint is a temporary fastening used to connect two co-axial rods. Its primary design purpose is to transmit axial forces.
1. **Axial Loads**: Because the cotter is driven into slots in the rods (socket and spigot), it creates a mechanical lock that can resist forces pulling the rods apart (tensile load) as well as forces pushing them together (compressive load). This makes it versatile for applications like piston rods in pumps.
2. **Twisting Moment (Option a)**: Cotter joints are not designed to transmit torque. A key or a splined shaft would be used for twisting moments.
3. **Bending Moment (Option c)**: The joint is designed for collinear axial loads. Applying a bending moment would create high stress concentrations at the edges of the cotter slot, likely leading to failure.
4. **Only Compressive Load (Option d)**: This is incorrect because the joint is specifically designed to prevent the rods from separating under tension as well.
The advantage of the piston valve over D-slide valve is that in the former case
A)
Wear is less
B)
Power absorbed is less
C)
Both wear and power absorbed are low
D)
The pressure developed being high provides tight sealing
In steam engines and compressors, the D-slide valve is a common mechanism, but it suffers from significant friction. Because the steam pressure acts directly on the face of the D-slide valve, it presses the valve firmly against the valve seat, creating high frictional resistance. This leads to two main problems: high power absorption (to overcome friction) and rapid wear of the valve surfaces.
In contrast, a piston valve is cylindrical and is 'balanced' against the steam pressure. Since the pressure acts equally around the circumference of the piston, there is no resultant force pushing the valve against the walls of the chest. This 'balanced' nature significantly reduces the friction between the valve and its guide. Consequently, the power required to operate the valve is much lower, and the mechanical wear is greatly reduced compared to the D-slide valve. Therefore, both wear and power absorption are low in the piston valve.
A fly wheel of moment of inertia 9.8 kgm² fluctuates by 30 rpm for a fluctuation in energy of 1936 Joules. The mean speed of the flywheel is (in rpm)
A)
600
B)
900
C)
968
D)
2940
Option A is correct: 600.
In a four stroke I.C. engine, the turning moment during the compression stroke is
A)
Negative throughout
B)
Positive throughout
C)
Negative during major portion of the stroke
D)
Positive during major portion of the stroke
In an Internal Combustion (IC) engine, the 'turning moment' (or torque) refers to the torque acting on the crankshaft.
1. **Positive Torque:** Occurs when the gas pressure in the cylinder pushes the piston down (during the power stroke), driving the crankshaft in its direction of rotation.
2. **Negative Torque:** Occurs when the crankshaft must do work to move the piston against the pressure of the gas (during the compression stroke). In this case, the gas pressure opposes the motion of the piston, and the energy is supplied by the flywheel.
During the compression stroke, the piston moves from Bottom Dead Center (BDC) to Top Dead Center (TDC), compressing the air-fuel mixture. Since the gas pressure is increasing and opposing the upward motion of the piston, the torque is negative for the vast majority of the stroke.
However, it is not 'negative throughout' because, near the dead centers (TDC and BDC), the geometry of the crank-slider mechanism results in a turning moment that approaches zero. Furthermore, in some specific engine geometries or due to inertia effects at very high speeds, there can be momentary fluctuations, but the dominant characteristic of the compression stroke is negative torque. Therefore, the most accurate description is that it is negative during the major portion of the stroke.
Deformation per unit length in the direction of force is known as
A)
Strain
B)
Lateral strain
C)
Linear strain
D)
Linear stress
Option A is correct: Strain.
A cantilever beam having square cross-section of side a is subjected to an end load. If a is increased by 19%, the tip deflection decreases approximately by
A)
19%
B)
29%
C)
41%
D)
50%
Option D is correct: 50%.
If the length of the cantilever beam is halved, then natural frequency of the mass M at the end of this cantilever beam of negligible mass is increased by a factor of
A)
2
B)
4
C)
√8
D)
8
1. **Identify the System**: The system is a mass-spring system where the 'spring' is a cantilever beam of length L. The natural frequency (f) of such a system is given by the formula: f = (1 / 2π) * √(k / M), where k is the stiffness of the beam and M is the mass at the end.
2. **Stiffness of a Cantilever Beam**: For a cantilever beam with a point load at the end, the stiffness (k) is defined as the load divided by the deflection. The formula for the stiffness of a cantilever beam is: k = (3EI / L³), where E is the Young's modulus, I is the area moment of inertia, and L is the length of the beam.
3. **Relationship between Frequency and Length**: Substituting k into the frequency formula: f ∝ √(k) ∝ √(1 / L³) ∝ L^(-3/2).
4. **Effect of Halving the Length**: Let the initial length be L₁ and the initial frequency be f₁. Let the new length be L₂ = L₁ / 2 and the new frequency be f₂.
f₂ / f₁ = (L₂ / L₁)^(-3/2)
f₂ / f₁ = (1 / 2)^(-3/2)
f₂ / f₁ = (2)^(3/2)
5. **Calculation**: (2)^(3/2) = √(2³) = √8.
Therefore, the natural frequency increases by a factor of √8 (which is approximately 2.828).
If the shear force acting at every section of a beam is of the same magnitude and of the same direction then it represents a
A)
Simply supported beam with a concentrated load at the centre.
B)
Overhung beam having equal overhang at both supports and carrying equal concentrated loads acting in the same direction at the free ends.
C)
Cantilever subjected to concentrated load at the free end.
D)
Simply supported beam having concentrated loads of equal magnitude and in the same direction acting at equal distances from the supports.
To determine the correct option, we must analyze the Shear Force (SF) distribution for each case. The condition given is that the shear force must be of the same magnitude and the same direction (constant SF) across every section of the beam.
1. **Option (a): Simply supported beam with a concentrated load at the center.** The SF is positive from the left support to the center and negative from the center to the right support. The magnitude is the same, but the direction changes. Thus, it is not constant.
2. **Option (b): Overhung beam with equal overhangs and equal loads at free ends.** The SF in the overhangs will be equal to the load, but the SF in the central span between the supports will be zero (since the loads at the ends cancel each other out in the internal section). Thus, it is not constant across the entire beam.
3. **Option (c): Cantilever subjected to a concentrated load at the free end.** For a cantilever of length L with a load P at the free end, the shear force at any section x (measured from the free end) is constant and equal to P. Since the load is acting in one direction, the SF remains the same in magnitude and direction throughout the entire length of the beam. This satisfies the condition.
4. **Option (d): Simply supported beam with equal loads at equal distances from supports.** The SF will be constant between the support and the load, but it will change magnitude (drop to zero or change sign) between the two loads. Thus, it is not constant across every section.
Therefore, only the cantilever beam with a point load at the free end maintains a constant shear force throughout its length.
A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield strength will
A)
Decrease
B)
Increase
C)
Remains same
D)
Becomes equal to ultimate tensile strength
This phenomenon is known as 'Strain Hardening' or 'Work Hardening'. When a ductile material (like mild steel) is stressed beyond its yield point, it undergoes plastic deformation. During this process, dislocations within the crystal lattice of the metal multiply and tangle, creating barriers to further dislocation movement. When the specimen is unloaded and then re-loaded, the material exhibits a higher yield strength than it did originally. The new yield point is effectively the stress level at which the material was previously unloaded. Therefore, the yield strength increases.
- Option (a) is incorrect because the material becomes stronger (harder), not weaker.
- Option (c) is incorrect because the internal structure of the material has been permanently altered by plastic deformation.
- Option (d) is incorrect because while the yield strength increases, it does not automatically jump to the ultimate tensile strength unless the material was stressed up to that specific limit before unloading.
The ratio of hoop stress to maximum shear stress is
A)
2
B)
3
C)
4
D)
6
In a thin-walled cylindrical pressure vessel subjected to internal pressure (P), there are two primary normal stresses:
1. Hoop Stress (Circumferential Stress), denoted as σ_h = Pd/2t
2. Longitudinal Stress, denoted as σ_l = Pd/4t
From these equations, it is evident that the hoop stress is twice the longitudinal stress (σ_h = 2σ_l).
To find the maximum shear stress (τ_max), we use the formula for maximum in-plane shear stress:
τ_max = (σ_max - σ_min) / 2
In the case of a thin cylinder, the maximum principal stress is the hoop stress (σ_h) and the minimum principal stress (considering the surface of the cylinder) is the radial stress, which is approximately zero (σ_r ≈ 0) for thin-walled vessels.
Therefore:
τ_max = (σ_h - 0) / 2 = σ_h / 2
Rearranging this to find the ratio of hoop stress to maximum shear stress:
Ratio = σ_h / τ_max = σ_h / (σ_h / 2) = 2
Thus, the ratio is 2.
A reed type tachometer uses the principle of
A)
torsional vibration
B)
longitudinal vibration
C)
transverse vibration
D)
damped free vibration
A reed-type tachometer operates on the principle of mechanical resonance. It consists of a set of steel reeds (cantilever beams) of varying lengths, each having a specific natural frequency of transverse vibration. A cam or a gear attached to the rotating shaft strikes these reeds at a frequency equal to the shaft's rotational speed. When the frequency of the striking cam matches the natural frequency of a particular reed, that reed enters resonance and vibrates with maximum amplitude (transverse vibration). The operator identifies the rotating speed by observing which reed is vibrating most vigorously.
- Torsional vibration (a) involves twisting of the shaft, which is not the mechanism here.
- Longitudinal vibration (b) involves axial movement, whereas the reeds move perpendicular to their length.
- Damped free vibration (d) refers to a system returning to equilibrium without external force, whereas the tachometer relies on forced vibration and resonance.