(a) g
L
12 /
(b)^3
4
g^12
L
/
(c)^33
2
12
g
L
/
(d)^3
2
g^12
L
/
- A solid sphere is rotating abouit a diameter at an
angular velocity Z. If it cools so that its radius
reduces to^1
n
of its original value, its angular velocity
becomes
(a)
Z
n
(b)
Z
n^2
(c) nZ (d) n^2 Z
- A cube of mass m and height H slides with a speed
v. It strikes the obstacle of height h=H
4
. e speed
of the CM of the cube just aer the collision is
H
m
v
h
(a)
5
6
v (b)
3
6
v (c)
5
4
v (d)
3
4
v
SOLUTIONS
- (d) : xCM = 0
or
mx mx mx mx
mmmm
11 22 33 44
1234
0
or (2m)(–a) + 4m(a) + m(a) + m 4 (–a) = 0
or m 4 = 3m
Similarly yCM = 0
or (2m)(–a) + 4m(–a) + m(a)
+ m 4 (a)= 0 or m 4 = 5m
Since, value of m 4 are dierent
to satisfy by both xCM = 0 and
yCM = 0
Hence, it is not possible.
- (d) : Iring = Ml^2
IDisc =
Ml^2
2
Isquare =
Ml() 2
6
2
=
2
3
Ml^2 (Using Ar axis theorem)
Ifour rods = 4
4
2
12 4
2
Ml()M()l 2
=
4
12
Ml^2
+ Ml^2
= Ml
2
3
+ Ml^2 =^4
3
Ml^2.
Thus, Ifour rods has the largest moment of inertia.
2 m
m 4
4 m
y m
a
a
x
aa
- (b) : Loss in PE = Mg l
2
sin
Gain in KE
=
1
2
Mv^2 +
1
2
IZ^2
For ring, I = MR^2 , Z = v
R
1
2
IZ^2 =
1
2
MR^2 ×
v
R
2
2 =
1
2
Mv^2
Hence gain in K.E. =^1
2
Mv^2 +^1
2
Mv^2 = Mv^2
Mv^2 = Mg
l
2
sin q v =
glsinT
2
.
- (c) : e acceleration of disc rolling down on an
inclined plane is given by,
aroll = g
I
MR
sinθ
1 + disc 2
= g
MR
MR
sin
/
1
(^22)
2
aroll =^2
3
g sin q
Also, aslide = g sinq
a
a
roll
slide
=
2
3
s =
1
2 1
()at^2
slide =
1
2 2
()at^2
roll^ ^ aslide × t 1
(^2) = a
roll × t 2
2
t
t
2
2
1
2 =
a
a
slide
roll
=^3
2
- (b) : vP = Zrc
Z
P
O
q
q/
q/
r
rc r
As is evident, rc = 2
2
rcos
vp = 2
2
rcos
=^2 cos 2
T(Zr)
= 2
2
cos
T
(vCM) vP = 2
2
vcos
T
q
q
l/
q/2r
rc q/2r
MONTHLY TUNE UP CLASS XII ANSWER KEY
- (b) 2. (b) 3. (c) 4. (b) 5. (d)
- (a) 7. (b) 8. (a) 9. (c) 10. (a)
- (b) 12. (b) 13. (b) 14. (a) 15. (b)
- (c) 17. (a) 18. (a) 19. (b) 20. (a,c,d)
- (a,b,d) 22. (a,b,c) 23. (a,c) 24. (30) 25. (8)
- (3.2) 27. (d) 28. (a) 29. (c) 30. (b)