6 KEY 1. B 2. C 3. C 4. D 5. D 6. C 7. C 8. A 9. A 10. A 11. C 12. B 13. C 14. B 15. B 16. A 17. B 18. B 19. C 20. B 21. B 22. B 23. B 24. B 25. B 26. B 27. C 28. A 29. B 30. C
1. B 2. C 3. C 4. D 5. D 6. C 7. C 8. A 9. A 10. A 11. C 12. B
13. C 14. B 15. B 16. A 17. B 18. B 19. C 20. B 21. B 22. B 23. B
24. B 25. B 26. B 27. C 28. A 29. B 30. C
7
SOLUTIONS 01. B 02. C 03. C 04. D 05. D F + f = ma .. (1)
F
f
( ) 22 aF f R MR5 R
= (2) By (1) and (2) we get 10Fa7m
=
06. C
Tcos mg = , 2mvTsin
r = .
07. C
xtan
y=
x y tan= Then differentiate it. 08. A mgh = 21 mv
2
09. A
2mvN mg cosr
= +
10. A 11. C 12. B t =
ddt = and at t = it stops
22 2
2
= = 13. C
.P =
. .
dP Id
=
2Pd I d = Integrating above relation 3
14. B Equate the tangential velocity at the rims. 15. B
8
In the absence of slipping, velocities of contacts points of
upper cylinders and lower cylinders are respectively
upw =32
ABv v
AB R=
lowerw = 2CDv v
CD R=
3uplower
w
w =
16. A If spool is not to translate cos ........1F f = If spool
is not to rotate . .........2F r FR= cosF f = And Fr F R=
1cos cosr r
R R = =
17. B In reference frame of truck angular Momentum is conserved
about P
20
2......1
5MvR Mv R MR= +
And for pure rolling 0 .......2v R=
1 and 2 057
v v=
Note that 0v is speed in truck frame, in ground frame velocity
is = 25
v v=
= 2
.
7v
18. B From fig (a) 1 2 .....1N N mg + = 1 2 .......2N N=
2 21mgN
=
+
310a
f mg =
From fig (b) 1 2 20 : 3bmgN N mg f N= = = =
910
a
b
ff =
19. C
2 2 2c
v v v v= + =
20. B ( )4 5 1 1 /A cmv v R m s= = = 21. B
9
2
2 2
12
1 12 2
IRotaionalKEfToatalKE
mv I
= =
+
22. B
2
2
sin 3.51
ga
KR
= =
+
And 12
s = 2 2at t s =
23. B
Fraction = 2
2
114
3 34
mv
mv
=
24. B
11 2 2 2
2
sinsin ; 3 / 2
1 /ag
a g ak R a
= = =+
25. B ( ) 2 21 1
2 2mg nR mv I mgR= + + Where 22
5I mR= and v R=
( )10 17
v gR n =
26. B
AD OA And AC OB
2CAD AOB = =
B
A
O
v
v
2
2
'O
2 2 2 2 cos 2Av v v vv = + + 2 cosAv v = 27. C
2
2
sin 5sin
71
g ga k
R
= =+
And 2 2 2v u as = 7s m = 28. A I fR = = Where 21
2I MR= 2 g
R
=
0 02 g
t tR
= + =
......2fa g v o at gtM
= = = + =
10
And for pure rolling 2og
v gt t RR =
0
3R
tg
=
29. B Let the cylinder move with linear velocity v . Then the
distance moved by centre of mass= l
1;t
v= As the velocity at the top most point of cylinder is 2 ,v the
length of the string pulled
