(a) g
L12 /
(b)^3
4g^12
L/(c)^33
212
g
L/
(d)^3
2g^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
Hm
v
h(a)5
6
v (b)3
6
v (c)5
4
v (d)3
4
vSOLUTIONS- (d) : xCM = 0
ormx mx mx mx
mmmm11 22 33 44
12340
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
62
=2
3
Ml^2 (Using Ar axis theorem)Ifour rods = 4
42
12 4
2
Ml()M()l 2
=
4
12
Ml^2
+ Ml^2= Ml23+ Ml^2 =^4
3Ml^2.
Thus, Ifour rods has the largest moment of inertia.2 mm 44 my maaxaa- (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
R2
2 =1
2
Mv^2Hence gain in K.E. =^1
2Mv^2 +^1
2Mv^2 = Mv^2 Mv^2 = Mgl
2sin q v =glsinT
2.
- (c) : e acceleration of disc rolling down on an
inclined plane is given by,
aroll = g
I
MRsinθ1 + disc 2= g
MR
MRsin
/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
aslide
roll=^3
2
- (b) : vP = Zrc
ZP
Oq
q/
q/rrc rAs is evident, rc = 2
2rcos
vp = 2
2 rcos
=^2 cos 2T(Zr)= 2
2
cosT
(vCM) vP = 2
2vcosT
qql/q/2r
rc q/2rMONTHLY 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)