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# Quiz: Mechanical Engineering 9th June

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. The figure shown a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of

(a) 0 Newton
(b) 490 Newtons in compression
(c) 981 Newtons in compression
(d) 981 Newtons in tension

Q2. A thin plate of uniform thickness is subject to pressure as shown in the figure below

Under the assumption of plane stress, which one of the following is correct?
(a) Normal stress is zero in the z-direction
(b) Normal stress is tensile in the z-direction
(c) Normal stress is compressive in the z-direction
(d) Normal stress varies in the z-direction

Q3. At a point in a stressed body the state of stress on two planes 45° apart is as shown below. Determine the two principal stresses in MPa.

(a) 8.242, 0.658
(b) 9.242, 0.758
(c) 9.242, 0.758
(d) 8.242, 0.758

Q4. A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are

(a) R_1=5qL/8,R_2=3qL/8,M=(qL^2)/8
(b) R_1=3qL/8,R_2=5qL/8,M=(qL^2)/8
(c) R_1=5qL/8,R_2=3qL/8,M=0
(d) R_1=3qL/8,R_2=5qL/8,M=0

Q5. The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is
(a) 1
(b) 2
(c) 3
(d) 4

Q6. List – I (Gear type)
Worm gears
Cross helical gears
Bevel gears
Spur gears
List – II (Application)
Parallel shafts
Nonparallel, intersecting shafts
Nonparallel, nonintersecting shafts
Large speed ratios
Codes:
A B C D
(a) 2 3 1 4
(b) 4 1 3 2
(c) 4 3 2 1
(d) 3 1 2 4

Q7. A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 Kg. the mass of the counterweight, for static balance, is

(a) 75 kg
(b) 150 kg
(c) 225 kg
(d) 300 kg

Q8. A Small steam whistle (Perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is
(a) 4 m/s
(b) 40 m/s
(c) 80 m/s
(d) 120 m/s

P. Influences the freezing range of the melt
Q. compensates the loss of fluidity of the melt
R. Facilitates top feeding of the melt
S. Avoids misrun
(a) P, R
(b) Q, S
(c) R, S
(d) R, S

Q10. A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the dia clearance is 3%. The required punch diameter is
(a) 19.88 mm
(b) 19.94 mm
(c) 20.06 mm
(d) 20.12 mm

Solutions
S1. Ans.(a)
Sol. Σf_H=0 & Σf_y=0
At joint ‘L’
∵ f_Lk -f_LM=0 (∵Σf_H=0 )
f_LN=0 (∵ Σf_y=0 )
Hence no force is cutting on the truss member L.N

S2. Ans.(a)
Sol. Thin plate of uniform thickness pertains to plane stress condition. So, stress out of plane would be zero.

S3. Ans.(b)
Sol.

2=((8+σ_y)/2)+((8-σ_y)/2) cos⁡〖90°〗-3 Sin90°
σ_y=2MPa
σ_1,2=(8+2)/2±√(((8-2)/2)^2+3^2 )
=5±4.242
σ_1,2=9.242,+0.758 MPa

S4. Ans.(a)
Sol.

the given propped cantilever bam can be assumed to be consisting of two types of loads. First part is simple cantilever with UDL and the second part is cantilever beam with point load of R_2 at end
Balancing the forces, we get
R_1+R_2=qL—–(1)
Balancing the deflection at the end point as net deflection at the end is zero.
Deflection at B due to the UDL alone,
δ_B=(qL^4)/8EI
Deflection at B due to R_2 alone,
δ_b^’=(R_2 L^3)/3EI
∵δ_B=δ_B’
⇒(qL^4)/8EI=(R_2 L^3)/3EI⇒R_2=3qL/8
From (1),
R_1=qL-R_2
=qL-3qL/8=5qL/8
Moment,
M=R_2 L-qL×L\/2
=(3qL^2)/8-(qL^2)/2
= (qL^2)/8

S5. Ans.(c)
Sol. No. of links l = 8
No. of revolute joints j = 9
No. of higher pair, h = 0
Number of degree of freedom,
m = 3 (L-1)-2j-h
=3(8-1)-2×9-0
=3

S6. Ans.(c)
Sol. Worm gear – large speed ratio
Cross helical gears – Non- parallel, nonintersecting shafts
Bevel gear – Non parallel, interesting shafts
Spur gear – Parallel shafts.

S7. Ans.(d)
Sol.

Let counterweight at R is m kg
For static condition
ΣM_Q=0
m×0.5=75×2
m=300 kg

S8. Ans.(b)
Sol. Given data: = ∆h = 0.8 kJ/kg
⇒ h_1-h_2 = 0.8 kJ/kg
h_1=(V_1^2)/2+gz_1+q=h_2+V_2^2+g2_2+w
Assumption
q=0,w=0 V_1≈0 and change in KE and P.E neglet
∵h_1=h_2+(V_2^2)/2
h_1=h_2+(V_2^2)/2000
⇒h_1-h_2=(V_2^2)/2000
0.8×2000=V_2^2
V_2^2=1600
V_2=40 m\/sec

S9. Ans.(b)
Sol. Proper gating design can reduces pouring time, which makes up for loss of fluidity and avoids misrun defects.

S10. Ans.(a)
Sol. Given punched diameter; d = 20 mm
Shut thickness, t = 2 mm
Clearance; C = 3% of t
Punch diameter,
D=d-2C
=20-2×3/100×2
=19.58 mm

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