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GATE Technical: GATE 2022 (ME Set-2) Mechanical Engineering questions with solutions

11 questions with worked solutions.

Questions

Q1. Equal sized circular regions are shaded in a square sheet of paper of 1 cm side length. Two cases, case M and case N, are considered as shown in the figures below. In the case M, four circles are shaded in the square sheet and in the case N, nine circles are shaded in the square sheet as shown. What is the ratio of the areas of unshaded regions of case M to that of case N?

  1. 2: 3
  2. 1: 1
  3. 3: 2
  4. 2: 1

Answer: 1: 1

In both cases, the total area of the square is 1 cm². In case M, four circles are shaded, each with a radius of 0.25 cm, giving a total shaded area of 4 × (π × (0.25)²) = π cm². In case N, nine circles are shaded, each with a radius of approximately 0.333 cm, resulting in a total shaded area of 9 × (π × (0.333)²) = π cm² as well. Since both cases have the same shaded area, the unshaded areas are equal, leading to a ratio of 1:1.

Q2. Which one of the following is the definition of ultimate tensile strength (UTS) obtained from a stress-strain test on a metal specimen?

  1. Stress value where the stress-strain curve transitions from elastic to plastic behavior
  2. The maximum load attained divided by the original cross-sectional area
  3. The maximum load attained divided by the corresponding instantaneous cross-sectional area
  4. Stress where the specimen fractures

Answer: The maximum load attained divided by the original cross-sectional area

Ultimate tensile strength (UTS) is defined as the maximum load a material can withstand divided by its original cross-sectional area, reflecting its ability to resist deformation and failure under tension.

Q3. A structural member under loading has a uniform state of plane stress which in usual notations is given by σx = 3P, σy = −2P and τxy = √2 P, where P > 0. The yield strength of the material is 350 MPa. If the member is designed using the maximum distortion energy theory, then the value of P at which yielding starts (according to the maximum distortion energy theory) is

  1. 70 MPa
  2. 90 MPa
  3. 120 MPa
  4. 75 MPa

Answer: 70 MPa

sigma_vm^2 = (3P)^2 - (3P)(-2P) + (-2P)^2 + 3(sqrt2 P)^2 = 9+6+4+6 = 25P^2, so sigma_vm = 5P. Setting 5P = 350 gives P = 70 MPa, not the stored 75 MPa.

Q4. Which one of the following CANNOT impart linear motion in a CNC machine?

  1. Linear motor
  2. Ball screw
  3. Lead screw
  4. Chain and sprocket

Answer: Chain and sprocket

Chain and sprocket systems are typically used for transferring rotary motion rather than directly imparting linear motion, making them less suitable for CNC applications compared to the other options.

Q5. Which one of the following is an intensive property of a thermodynamic system?

  1. Mass
  2. Density
  3. Energy
  4. Volume

Answer: Density

Density is an intensive property because it does not depend on the amount of substance present in the system; it remains constant regardless of the size or mass of the material.

Q6. Consider a steady flow through a horizontal divergent channel, as shown in the figure, with supersonic flow at the inlet. The direction of flow is from left to right. Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?

  1. Since volume flow rate is constant, velocity at B is lower than velocity at A
  2. Normal shock
  3. Viscous effect
  4. Boundary layer separation

Answer: Normal shock

A normal shock wave in a supersonic flow causes a sudden decrease in velocity and an increase in pressure. This phenomenon explains why the pressure at location B is higher than at location A, as the shock wave transitions the flow from supersonic to subsonic conditions.

Q7. Which of the following non-dimensional terms is an estimate of Nusselt number?

  1. Ratio of internal thermal resistance of a solid to the boundary layer thermal resistance
  2. Ratio of the rate at which internal energy is advected to the rate of conduction heat transfer
  3. Non-dimensional temperature gradient
  4. Non-dimensional velocity gradient multiplied by Prandtl number

Answer: Non-dimensional temperature gradient

Nusselt number is the non-dimensional temperature gradient at the surface (Nu = hL/k). The stored option (internal to boundary-layer thermal resistance ratio) describes the Biot number, not Nusselt.

Q8. Consider sand casting of a cube of edge length a. A cylindrical riser is placed at the top of the casting. Assume solidification time, tₛ ∝ V/A, where V is the volume and A is the total surface area dissipating heat. If the top of the riser is insulated, which of the following radius/radii of riser is/are acceptable?

  1. a/3
  2. a/2
  3. a/4
  4. a/6

Answer: a/2

The correct option, a/2, is acceptable because it provides a sufficient volume-to-surface area ratio that allows for effective heat dissipation during solidification, ensuring that the riser can maintain a liquid state longer than the casting, thus feeding it as it cools.

Q9. Parts P1-P7 are machined first on a milling machine and then polished at a separate machine. Using the information in the following table, the minimum total completion time required for carrying out both the operations for all 7 parts is ________ hours. Part | Milling (hours) | Polishing (hours) P1 | 8 | 6 P2 | 3 | 2 P3 | 3 | 4 P4 | 4 | 6 P5 | 5 | 7 P6 | 6 | 4 P7 | 2 | 1

  1. 31
  2. 33
  3. 30
  4. 32

Answer: 33

By Johnson's two-machine rule the optimal sequence is P3,P4,P5,P1,P6,P2,P7. Computing milling completion then polishing completion gives a total makespan of 33 hours, not 32.

Q10. The steady velocity field in an inviscid fluid of density 1.5 is given to be V = (y² - x²)î + (2xy)ĵ. Neglecting body forces, the pressure gradient at (x = 1, y = 1) is __________.

  1. 10ĵ
  2. 20î
  3. −6î − 6ĵ
  4. −4î − 4ĵ

Answer: −6î − 6ĵ

For steady inviscid flow, grad p = -rho*(V.grad)V. With V=(y^2-x^2, 2xy) the acceleration is (2x(x^2+y^2), 2y(x^2+y^2)); at (1,1) that is (4,4). Times -rho=-1.5 gives grad p = -6 i - 6 j, which is option 2, not -4i-4j.

Q11. In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is __________.

  1. 0.2
  2. 0.25
  3. 0.3
  4. 0.35

Answer: 0.25

The dryness fraction at the entry to the evaporator can be calculated using the specific enthalpy values. Given that the specific enthalpy of the saturated liquid is 50 kJ/kg and the enthalpy of the refrigerant entering the evaporator is 250 kJ/kg, the dryness fraction is determined by the formula: x = (h - hf) / (hg - hf), where hf is the enthalpy of the saturated liquid and hg is the enthalpy of the saturated vapor. This calculation yields a dryness fraction of 0.25, indicating that 25% of the refrigerant is in vapor form.

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