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1.1.1.1.1. (d)
2.2.2.2.2. (b)
3.3.3.3.3. (d)
4.4.4.4.4. (a)
5.5.5.5.5. (c)
6.6.6.6.6. (b)
13.13.13.13.13. (c)
14.14.14.14.14. (c)
15.15.15.15.15. (c)
16.16.16.16.16. (a)
17.17.17.17.17. (c)
18.18.18.18.18. (d)
19.19.19.19.19. (b)
20.20.20.20.20. (a)
21.21.21.21.21. (a)
22.22.22.22.22. (b)
23.23.23.23.23. (c)
24.24.24.24.24. (b)
25.25.25.25.25. (c)
26.26.26.26.26. (a)
27.27.27.27.27. (a)
28.28.28.28.28. (d)
29.29.29.29.29. (b)
30.30.30.30.30. (c)
7.7.7.7.7. (b)
8.8.8.8.8. (b)
9.9.9.9.9. (d)
10.10.10.10.10. (c)
11.11.11.11.11. (c)
12.12.12.12.12. (b)
Theory of MachineDate : 01/10/2015
Mechanical Engineering
CLASS TEST - 2015
Serial No. : B__ TOM_01102015_GH1
ME
ANSWERS
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CTME15 Delhi Noida Bhopal Hyderabad Jaipur
Lucknow Indore Pune Bhubaneswar Kolkata
EXPLANAEXPLANAEXPLANAEXPLANAEXPLANATIONSTIONSTIONSTIONSTIONS
11.11.11.11.11. (c)(c)(c)(c)(c)
= ( )( )
l
Area will be maximum when
A = 90i.e. PQR is a right angle triangle
l= tan 75 = 3.732
l = 3.732 r
12.12.12.12.12. (b)(b)(b)(b)(b)
=
=
16.16.16.16.16. (a)(a)(a)(a)(a)
=
+
=
+
fn
= ( )
= 5367 rpm
17.17.17.17.17. (c)(c)(c)(c)(c)
The logarithmic decrement represents the natural logarithm of the ratio of any two successiveamplitude on the same side of the mean position
=
+l
e = ( )
+= =
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19.19.19.19.19. (b)(b)(b)(b)(b)
=
Centre distance =
+
= ( )
+
= ( )
+
= 250 mm
20.20.20.20.20. (a)(a)(a)(a)(a) w = 10 rad/secV
C= 10 10 = 100 cm/sec= 1 m/sec
21.21.21.21.21. (a)(a)(a)(a)(a)
22.22.22.22.22. (b)(b)(b)(b)(b)
A spring is equivalent to 2 binary links connected by a turning pair.A cam pair is equivalent to 1 binary link with turning pairs at each end hence equivalent chain
23.23.23.23.23. (c)(c)(c)(c)(c)T
A= 80 T
B= 200
TB
=
+
TC = 60
+
y + x = 100y = 100 x
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and
= 50 or
=
x = 107.1 and y = 7.1
Thus speed of arm a = 7.1 (CCW)
24.24.24.24.24. (b)(b)(b)(b)(b)
Force in upper spring =
(W = mg)
Force in lower spring = W
Deflection of mass = deflection of upper spring + deflection of lower spring
= +
=
=
n =
=
25.25.25.25.25. (c)(c)(c)(c)(c)
=
6 106 = 7000 V2
V = 29.28 m/sec
29.28 =
d = 1.393 m
K.E at a = E
at b = E 28at c = E + 352
at d = E + 92
at e = E + 402
at f = E + 102
at g = E + 344
at h = E 36
at j = E + 229
at k = E
maximum energy = E + 402 (at e)
minimum energy = E 36 (at h)
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Max. fluctuation of energy
emax
=
= 22360 N-m
or 22.360 KJ
26. (a)26. (a)26. (a)26. (a)26. (a)
Work done per cycle = 300 103
= 200 103N.m
E = W CE =200 103 0.1= 20 103N.m
Cs
=
= = =
Cs = 0.01, I = mk2 = 4 m kg.m2
=
=
= 3
E = I 2 Cs
I =
=
=
= 22222.22 (210)
m =
= 5.5 103kgor 5.5 T
27.27.27.27.27. (a)(a)(a)(a)(a)
n
=
=
=
50 = e0.22 14.14 NTd
NTd = 126
Td =
= 0.455 sec
28. (d)28. (d)28. (d)28. (d)28. (d)
k =
=
=
= 555 n/cm or 55.5 kN/m
29.29.29.29.29. (b)(b)(b)(b)(b)
Let S = dist covered by the bicycle before it comes to rest
work done against friction = k.E. of the bicycle and the rider
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RnS =
0.06 80 S = ( )
S = 121.5 m
30.30.30.30.30. (c)(c)(c)(c)(c)
1
23
4
Number of links = 4
Number of lower pairs,j = 3
Number of higher pair, h = 1
dof = 3(n 1) 2j h
= 3(4 1) 2 3 1 = 3 3 6 1 = 2