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Mechanical engineering Questions and solutions set - 2

Mechanical engineering Questions and solutions set - 2
Question:-
Option (A)
60m
Option (B)
90m
Option (C)
120m
Option (D)
150m
Correct Option:
C
Question Solution:
head developed by a pump will be in the ratio

H1 / H2 =(N1 / N2)

H1 =30m ,N1 =1000rpm

H2 = ? , N2 =2000rpm

H2 = (N2/N1)2 ´ H1 = (2000/1000)2 ´ 30

H2 = 120m
Question:-
Option (A)
double of that of the original spring
Option (B)
same as that of the original spring
Option (C)
half of that of the original spring
Option (D)
one-fourth of that of the original spring
Correct Option:
B
Question Solution:
For series spring connection :- Equivalvent spring const

W/keq = W/k1 +W/k2

1/keq = 1/2k + 1/2k =2/2k [\k1 = k2 =2k]

\Keq =k
Question:-
Option (A)
Cup-and cone fracture
Option (B)
Fracture along a plane normal to the axis of the specimen
Option (C)
Fracture along a helix of approximately 45°
Option (D)
Fracture along a plane inclined at 60° to the anis
Correct Option:
C
Question Solution:
Cast Iron is a brittle metal,

Materials when subjected to torsion is a case of pure shear. Under pure shear the max. stress occurs approximately at 45°. Hence the specimen will fail along a helix at 45°.


Question:-
Option (A)
1.25 m/sec.
Option (B)
0.5 m/sec.
Option (C)
0.4 m/sec.
Option (D)
0.2 m/sec
Correct Option:
D
Question Solution:
Given VB = 0.5 m/sec

VB = w3 ´ [I23 to I13] Þ VB = w3 ´ 0.25

0.5 = w3 ´ 0.25

W3 = 2 rad/sec

To find VC

W3 ´ [I34 – I13] = w4 ×[I14 – I34]

2 ´ 0.1 = VC

[V = rw]

Vc =0.2 m/s


Question:-
Option (A)
Stirling cycle
Option (B)
Atkinson cycle
Option (C)
Ericsson cycle
Option (D)
Brayton cycle
Correct Option:
C
Question Solution:
Normal Brayton Cycle consists of Two reversible adiabatic and two constant pressure processes when the cycle is divided into infinite no. of stages during compression (inter cooling) expansion (Reheating). The Brayton Cycle approaches Ericsson Cycle i.e Two reversible is isotherms and Two reversible is isobars.


Question:-
Option (A)
0.04
Option (B)
0.05
Option (C)
0.10
Option (D)
0.12
Correct Option:
C
Question Solution:
For clearance Fit.

Minimum clearance = Lower Limit of Hole – Upper limit of shaft.

0.01 = 40 – Upper limit of shaft

\ Upper limit of shaft = 40 – 0.01 = 39.99 mm

Tolerance on shaft = 0.04 mm

\Lower Limit of shaft = 39.99 – 0.04 = 39.95 mm

Max clearance = Upper limit of Hole – Lower limit of shaft

Upper limit of Hole = 40 + 0.05 = 40.05 mm.

\ Max clearance = 40.05 – 39.95 = 0.1 mm