Engineering Jobs   »   Quiz: Mechanical Engineering 16th June

# Quiz: Mechanical Engineering 16th June

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. The cross-section of two solid bars made of the material are shown in the figure. The square cross-section has flexural (bending) rigidity I_1, while the circular cross-section has flexural rigidity I_2. Both sections have the same cross-sectional are. The ratio is I_1/I_2 is

(a) 1/π
(b) 2/π
(c) π/3
(d) π/6

Q2. The number of degrees of freedom in a planer mechanism having n links and j simple hinge joints is
(a) 3(n–3) – 2j
(b) 3(n–1) – 2j
(c) 3n–2j
(d) 2j–3n + 4

Q3. The INCORRECT statement about the characteristic of critical point of a pure substance is that
(a) there is no constant temperature vaporization process
(b) it has point of infection with zero slope
(c) the ice directly converts from solid phase to vapour phase
(d) saturated liquid and saturated vapour states are identical

Q4. A beam of length L is carrying a uniformly distributed load w per unit length. the flexural rigidity of the beams is EI. The reaction at the simple support at the right end is

(a) wL/2
(b) 3wL/8
(c) wL/4
(d) wL/8

Q5. In a meter forming operation when the material has just started yielding, the principal stresses are σ_1 = + 180 MPa, σ_2 = – 100MPa, σ_3 = 0. Following von Mises criterion, the yield stress is_________MPa.
(a) 245.76
(b) 240.12
(c) 248.57
(d) 251.98

Q6. The velocity components in the x and y direction of a two-dimensional potential flow are u and v, respectively. then ∂u/∂x, is equal to
(a) ∂V/∂x
(b) -∂V/∂x
(c) ∂V/∂y
(d) -∂V/∂y

Q7. Consider one-dimensional steady state heat conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The conductivity of the wall is given by k=k_o+bT; where k_O and b are positive constant, and T is temperature.

As x increases, the temperature gradient (dT/dx) will
(a) remains constant
(b) be zero
(c) increases
(d) decreases

Q8. Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temp is valid when
(a) Biot number < 0.1 (b) Biot number > 0.1
(c) Fourier number < 0.1 (d) Fourier number > 0.1

Q9. A diesel engine is usually more efficient than a spark ignition engine because
(a) diesel being a heavier hydrocarbon, releases more heat per kg than gasoline
(b) the air standard efficiency of diesel cycle is higher than the Otto cycle, at a fixed compression ration
(c) the compression ratio of a diesel engine is higher than that of an SI engine
(d) self-ignition temperature of temperature of diesel is higher than that of gasoline

Q10. With respect to metal working, match group A with Group B;

Group A
P. Defect in extrusion
Q. Defect in rolling
R. Product of skew rolling
S. Product of rolling through cluster mill

Group B
I. Alligatoring
II. Scab
III. fish tail
IV. seamless tube
V. thin sheet with tight tolerance
VI. semi-finished balls of ball bearing
(a) P-II, Q-III, R-VI, S-V
(b) P-III, Q-I, R-VI, S-V
(c) P-III, Q-I, R-IV, S-VI
(d) P-I, Q-II, R-V, S-V

Solutions
S1. Ans.(c)
Sol. Given that
Square cross section flexural rigidity = I_1
Circular cross section rigidity = I_2
Cross-section area A_1=A_2
Material is same, E_1=E_2
Flexural rigidity I_0 = EI
Where I – moment of inertia

A_1=A_2
a^2=πr^2
a=r √π
I_1=(E_1 I_1)/(E_2 I_2^1 )
I_1/I_2 =I_1/I_2 I_1^1=a^4/12
I_2^1=π/64 d^4=π/4 r^4
I_1/I_2 =(a^4/ 12)/(πr^4/4)
I_1/I_2 =π/3

S2. Ans.(b)
Sol. The number of degree of freedom in in a planning mechanism having n links and j simple hinge joint is given by grubler – kultbach equation
▭(F=2 (n-1)-1j)

S3. Ans.(c)
Sol. ICE converts directly into vapor, if it is heated at constant pursue which is less than triple point pressure. Therefore direct conversion of Ice into vapor does not involve critical point.

S4. Ans.(b)
Sol.

Let assume that it is a cantilever beam with UDL and we provide a support to avoid deflation at the free end of beam that means
deflection at free end due to UDL = deflation at support due to UDL
Y_1=(wL^4)/8EI
and deflection due to Reaction
Y_2=(RL^3)/3EI
Y_1=Y_2
(wL^4)/8EI = (RL^3)/3EI
R = 3wL/8

S5. Ans.(a)
Sol. We know that, the von-mises criteria is (σ1-σ_2 )^2+(σ_2-σ_3 )^2 (σ_3-σ_1 )^2=2 σ(m^2 )
Given, σ_1=180 MPa, σ_2 = –100 MPa, σ_3 = 0 MPa.
(180+100)² + (–100–0)² + (0–180)² = 2 σ_m^2
(280)² + (100)² + (180)² = 2 σ_m^2
78400 + 10000 + 32400 = 2 σ_m^2
120800 = 2 σ_m^2
60400 = σ_m^2
σ_m = 245.76 MPa

S6. Ans.(d)
Sol. for two dimensional flow, continuity equation should be satisfied.
i.e,
∂u/∂x+∂v/∂y=0
∵∂u/∂x= -∂v/∂y

S7. Ans.(d)
Sol.

Q= -kA dT/dx
Where k=k_0+bT
As x increases, T increases, then
K also increases.
i.e., K is +Ve
Equation (i)
Indicates dT/dx decreases.

S8. Ans.(a)
Sol. Biot no. should be less than 0.1.

S9. Ans.(c)
Sol. for same compression ratio and same heat supplied, Otto cycle is most efficient and diesel cycle is least efficient.
In practice, however the compression ratio of the diesel ranges between 14 and 25 where as that of the otto engine between 6 and 12. Because of is higher efficiency than the otto engine.

S10. Ans.(b)
Sol. defect in expression – fish tail
Defect in rolling – alligatoring
Product of skew rolling – semi finished bells of ball bearing
Product of rolling through cluster mill- thin sheet with High tolerance.

Sharing is caring!

### TOPICS:

[related_posts_view]