Quiz: Mechanical Engineering 3rd July

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. The coefficient of restitution of a perfectly plastic impact is
(a) 0
(b) 1
(c) 2
(d) ∞

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q2. Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic energy stored in bolt A is 4 times that of the bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is
(a) 16
(b) 24
(c) 36
(d) 48

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q3. For A four bar linkage in toggle position, the value of mechanical advantage is
(a) 0.0
(b) 0.5
(c) 1.0
(d) ∞

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q4. Gear 2 rotates at 1200 rpm is counter clockwise direction and engages with Gear 5 engages with Gear 4. The numbers of teeth on Gears 2, 3, 4 and 5 are 20, 40, 15 and 30, respectively. The angular speed of gear 5 is

(a) 300 rpm counter clockwise
(b) 300 rpm clockwise
(c) 4800 rpm counter clockwise
(d) 4800 rpm clockwise

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q5. What is the natural frequency of the spring mass system shown below? The contact between the block and the inclined plane is frictionless. The mass of the block is denoted by m and the spring constants are denoted by k_1 and k_2 as shown below

(a) √((k_1+k_2)/2m )
(b) √((k_1+k_2)/4m )
(c) √((k_1-k_2)/m )
(d) √((k_1+k_2)/m )

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q6. A small element at the critical section of a component in biaxial state of stress with the two principal stress being 360 MPa and 140 MPa. The maximum working stress according to distortion energy theory is
(a) 220 MPa
(b) 110 MPa
(c) 314 MPa
(d) 330 MPa

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q7. A machine element is subjected to the following bi-axial state of stress σ_x=80 MPa; σ_y = 20 MPa; τ_xy = 40 MPa. If the shear strength of the materials is 100 MPa, the factor of safety as per Tresca’s maximum shear stress theory is
(a) 1.0
(b) 2.0
(c) 2.5
(d) 3.3

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q8. The velocity potential function for a source varies with the distance r as
(a) 1/r
(b) 1/r²
(c) e²
(d) ln r

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q9. Match List-I with List-II and select the correct answer using the codes given below the lists:
List- I
P. Centrifugal compressor
Q. Centrifugal pump
R. Pelton wheel
S. Kaplan turbine
List-II

  1. Axial flow
  2. Surging
  3. Priming
  4. Pure impulse
    Codes:
    P Q R S
    (a) 2 3 4 1
    (b) 2 3 1 4
    (c) 3 4 1 2
    (d) 1 2 3 4

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Q10. Considering the variation of static pressure and absolute velocity in an impulse steam turbine, across one row of moving blades
(a) both pressure and velocity decreases
(b) pressure decreases but velocity increases
(c) pressure remains constant, while velocity increases
(d) pressure remains constant, while velocity decreases

L1Difficulty 3
QTags Mechanical
QCereator Paper Maker 10

Solutions

S1. Ans.(a)
Sol. coefficient of restitution of a perfectly plastic impact is zero.

S2. Ans.(b)
Sol. Given
P_1 =P_2=P=(identical tensile load on bolt A& B)
L_1=L_2=L=(Same length)
d_A= 12 mm,
U_A=Strain energy in bolt A
U_B=Starin energy in bolt B
U_A=4U_B (Given)——-(1)
∵d_B= ?
Strain energy is given by U_A=1/2 P×S=1/2 (P^2 L)/AE
eq.(1) 1/2 (P_1^2 L_1)/(A_1 E)=4×1/2 (P_2^2 L_2)/(A_2 E)
1/A_1 =4×1/A_2
4/(π(12)^2 )=4×4/(π(d_B )^2 )
d_B=576 or d_B=24 mm [Diameter of bolt B]

S3. Ans.(d)
Sol.

ω_4 of the output link DC become zero at the extreme positions. The extreme positions of the linkage are known as “Toggle position”.
∴ mechanical advantage =ω_input/ω_output
∵ ω_output=0
∴ Mechanical advantage = ∞

S4. Ans.(a)
Sol.
Gears 2 3 4 5
Teeth on gears 20 40 15 30
Speed of gear in RPM -1200 (+20×1200)/40 (+20×1600)/40 (-20×1200)/40×15/30

∵ Angular speed of Gear 5
=(-20×1200×15)/(40×30)= -300 RPM
= 300 RPM (counter clockwise)

S5. Ans.(d)
Sol. mx ̈+(k_1+k_2 )x=0
x ̈+[(k_1+k_2)/m]x=0
ω_n=√((k_1+k_2)/m)

S6. Ans.(c)
Sol. According to distraction energy theory
τ=√(σ_1^2+σ_2^2-σ_1 σ_2 )=314 MPa

S7. Ans.(b)
Sol.
σ_1,2=1/2 [(σ_x+σ_y )±√((σ_x-σ_y )^2+4τ_xy^2 ) ]
=1/2 [(80+20)±√((60)^2+4×40^2 ) ]
σ_1=100 MPa
σ_2=0 MPa
τ_max=σ_1/2=50 MPa
f(factor os safety)=τ_ys/τ_max =100/50=2

S8. Ans.(d)
Sol.
u_r= -∂ϕ/∂r=-a/2πr
Or ϕ= -q/2π lnr+c

S9. Ans.(b)
Sol.
(P) Centrifugal compressor -surging
(Q) Centrifugal pump – priming
(R) Pelton wheel – Pure impulse
(S) Kaplan turbine – Axial flow.

S10. Ans.(d)
Sol. In impulse turbine, the steam pressure remains constant while it flows through the moving blades, where only the kinetic energy converts into mechanical energy.

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