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Mechanical engineering Questions and solutions set - 3

Mechanical engineering Questions and solutions set - 3
Question:-
Option (A)
0.4 radian
Option (B)
0.8 radian
Option (C)
1.6 radian
Option (D)
3.2 radian
Correct Option:
C
Question Solution:
qC/B = rotation of C w.r.t B
qB/A = ration of B w.r.t A
qC/B = [T(L/2)]/[G(p/32)(D/2)4]
qB/A = [T(L/2)]/[G(p/32)(D)4]
qC/B/qB/A
= [T(L/2)]/[G(p/32)(D/2)4] * [G(p/32)(D)4]/T(L/2)
= 16
qC/B = 16qB/A = 16 *0.1 = 1.6rad
Question:-
Option (A)
118 mm
Option (B)
161 mm
Option (C)
224 mm
Option (D)
312 mm
Correct Option:
C
Question Solution:
d/r = 100/0.4 = 250 > 20
\ D = Ö(d2 + 4dh)      (Q D>=20r D = Ö(d2 + 4dh))
= Ö(1002 +4*100*100)
=223.606mm
Question:-
Option (A)
0.425
Option (B)
2.25
Option (C)
0.225
Option (D)
4.25
Correct Option:
D
Question Solution:
This cylindrical cell m = 0.3
s1 = Longitudinal stress = PD/4t
s2 = hoop stress or circumferential stress = PD/2t
e1 = Longitudinal strain = s1/E - ms2/E
= PD/4tE - mPD/2te
= PD/4tE(1 - 2m)

e2
= circumferential strain
=s2/E - ms1/E
= PD/4tE(2 - m)

s2/s1
= (2 - m)/(1 - 2m)
= (2 - 0.3)/(1 - 2*0.3)
= 4.25
Question:-
Option (A)
25
Option (B)
30
Option (C)
48
Option (D)
62
Correct Option:
C
Question Solution:
ma3 X w3 = ma1 X w1 + ma2 X w2
=> w3 = (1*0.004+2*0.0051)/3 = 0.00473
j = (w/ws) * 100
= (0.00473/0.01)*100 = 47.33%
Question:-
Option (A)
Where the principle stress exist the shear stress is zero & where the shear stress is maximum the principle stress is zero
Option (B)
Where the principle stress exist the shear stress is zero & where the shear stress is zero the principle stress may or may not be zero
Option (C)
Where the principle stress exist the shear stress is maximum & where the shear stress is maximum the principle stress is maximum
Option (D)
Where the principle stress exist the shear stress is maximum & where the shear stress is maximum the principle stress is zero
Correct Option:
B
Question:-
Option (A)
-205.08 KJ
Option (B)
-225.08 KJ
Option (C)
-305.08 KJ
Option (D)
-105.08 KJ
Correct Option:
A
Question Solution:
dQ = CvDT
∫dQ = ∫125250ZT2dT
Q = Z*[T3/3]125250
Q = 0.045*[1253/3 - 2503/3]
= -205.078