Engineering Jobs   »   Quiz: Mechanical Engineering 21 April 2021

Quiz: Mechanical Engineering 21 April 2021

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 15 Minutes

Q1. The polytropic process is represented by a straight line in the following figure. What is X and Y respectively?

Quiz: Mechanical Engineering 21 April 2021_30.1

(a) In V and in P
(b) V and P
(c) In P and in V
(d) P and V

Q2. What is minimum amount of work input to a refrigerator which convert 1 kg of water at 20 °C into ice at -5 °C while max. COP of refrigerator is 10?
(a) 40 kJ
(b) 50 kJ
(c) 42.91 kJ
(d) 52.91 kJ

Q3. Q8. An empty tank is charged through a pipeline. The fluid in the pipeline is an ideal gas (γ = 1.67) having a temperature T = 300 K. the final temperature of the tank when fully charged is (consider the process as adiabatic and quasistatic)
(a) 180 K
(b) 501 K
(c) 501°C
(d) 228 K

Q4. Formation of inter-granular cavities or voids by process of nucleation in sheet metal forming is known as…….
(a) Super plasticity
(b) Spinnability
(c) Cavitation
(d) Intrinsic voids

Q5. In a condenser of a power plant, the steam condenses at a temperature of 60°C. The cooling water enters at 30°C and leaves at 45°C. The logarithmic mean temperature Difference (LMTD) of the condenser is
(a) 16.2°C
(b) 21.6°C
(c) 30°C
(d) 37.5°C

Q6. A rod is subjected to a uni-axial load within linear elastic limit. When the change in the stress is 200 MPa, the change in the strain is 0.001 if the poison’s ratio of the rod is 0.3, the modulus of rigidity (in GPa) is __
(a) 76.9230 GPa
(b) 89.9230 GPa
(c) 79.9230 GPa
(d) 87.9230 GPa

Q7. The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 rpm is 3000 hours. What is the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm.
(a) Unchanged
(b) 12000 hours
(c) 1500 hours
(d) 6000 hours

Q8. A planar closed kinematic chain is formed with rigid links PQ = 2.0 m, QR =3.0 m, RS = 2.5 m and SP = 2.7 m with all revolute joints. The link to be fixed to obtain rocker (rocker – rocker) mechanism is
(a) PQ
(b) QR
(c) RS
(d) SP

Q9. A column has a rectangular cross-section of 10 mm × 20 mm and a length of 1m. the slenderness ratio of the column is close to
(a) 200
(b) 346
(c) 477
(d) 1000

Q10. Semi brittle materials can be extruded by
(a) impact extrusion
(b) closed cavity extrusion
(c) hydrostatic extrusion
(d) backward extrusion


S1. Ans.(c)
Sol. PV^n=C (polytropic process)
InP + n InV = C
X + nY = C

X/C+Y/((C\/n) )=1 (Equation of straight line)

S2. Ans.(c)
Sol. Q = m_w C_pW (0-20)+m_i C_pi (-0.5-0)-325
= -1 × 4.18 (20) – 1 × 2.09(5) – 335
= -429.05 kJ
∵W_min=Q/(COP)_max =429.05/10=42.905
= 42.91 kJ

S3. Ans.(b)
Sol. T2=γT1=1.67×300=501 K

S4. Cavitation is the phenomenon which is related to the formation of inter-granular cavities or voids by process of nucleation in sheet metal forming. One can easily correlate with cavitation phenomenon of fluid particle where pressure becomes lower than the vapour pressure of fluid.

S5. Ans.(b)
Sol. LMTD = (∆T1-∆T2)/(ln [(∆T1)/(∆T2 )] )

Quiz: Mechanical Engineering 21 April 2021_40.1

=(30-15)/ln⁡(2) =21.64°C

S6. Ans.(a)
Sol. with in the linear elastic limit

Quiz: Mechanical Engineering 21 April 2021_50.1

E-slope of σ Vs ε curve
E=200 GPa
G=E/2(1+μ) =200/2(1+0.3) =100/1.3
G=76.9230 GPa

S7. Ans.(b)
L1/L2 =(P2/P1 )^3
⇒(1000×60×3000)/(2000×60×t2 )=[4900/9800]^3
t2=12000 hrs

S8. Ans.(c)
Sol. PQ = 2.0 m, QR = 3.0 m, RS = 2.5 m, SP = 2.7 m
(i) If shortest link PQ = 2.0 m is fixed, then we will get double crank.
(ii) If link adjacent to shortest is fixed, then we will get crank rocker.
(iii) If we fix coupler ‘RS’, we will get double rocker maximum

S9. Ans.(b)
Sol. Assuming both ends of column to be hinged
Given: Leq = 1m
Slenderness ratio = Leq/r min
r_min=√(I_min/A)=√((1/12×(ω)^2×200×10^16)/(10×20×10^(-6) ))
Slenderness=1/(2.88×10^(-3) )=347.22≈346

S10. Ans.(c)
Sol. Hydrostatic pressure increases the ductility of the material and also there is no friction to overcome along the container walls

Sharing is caring!

Leave a comment

Your email address will not be published. Required fields are marked *