Rotational Motion - 3 Practice Questions - DPP for NEET

 Page 1


1. (a) Since disc is rolling (without slipping) about point O.
Hence
P
C
Q
w
O
OQ OC OP >> = Qvrw
\ >>
Q CP
v vv
2. ( d) Applying the theorem of perpendicular axis,
1 2 34
I I I II = + =+
Because of symmetry, we have
12
= II and 
34
= II
Hence 
1 234
2 2 22 = = == I I I II
or
1 2 34
= == I I II
i.e. sum of two moment of inertia of square plate about
any axis in a plane (Passing through centre) should be
equal to moment of inertia about the axis passing
through the centre and perpendicular to the plane of
the plate.
3. ( a) By the conservation of energy
l/2
a
P.E. of rod = Rotational K.E.
2
l1
mgsinI
22
a=w
2
2
l 1 ml
mg sin
2 23
Þ a=w 
3 sin a
Þw=
g
l
But in the problem length of the rod 2L is given
3 sin
2
a
\w=
g
L
4. (c) Graph should be parabola symmetric to I- axis, but it
should not pass from origin because there is a constant
value I
cm
 is present for 0 x = .
5. ( b)
2
2
2 24
1
3
1
1
2
= ==
+
+
gh gh
v gh
K
R
6. ( d)
2
2
sin sin / 2 5
2
7 / 5 14
1
1
5
g g gg
a
K
R
qq
= = ==
+
+
As 
30 q=
o
 and 
2
2
2
5
K
R
=
7. ( b) We know 
2
2
2
1
=
+
gh
v
k
r
22
2
\w==
+
v gh
r
rk
22 22
2mgh 2mgh 2mgh
mr mk mr I I mr
Þw= ==
+ ++
8. (a) Because its M.I. (or value of 
2
2
)
K
R
is minimum for
sphere.
9. (b) As body is moving on a frictionless surface. Its
mechanical energy is conserved. When body climbes
up the inclined plane it keeps on rotating with same
angular speed, as no friction force is present to provide
retarding torque so
2 22
1 11
I mv I mgh v 2gh
2 22
w + ³ w + Þ³
10. (a)
22
1
MR I MR 2I
2
=Þ=
Moment of inertia of disc about a tangent in a plane
2
5 55
(2)
4 42
= == MR II
11. ( d) Moment of inertia of system about YY’
1 23
I I II = ++
2 22
133
222
MR MR MR = ++ 
2
7
2
MR =
Y
1
2 3
Page 2


1. (a) Since disc is rolling (without slipping) about point O.
Hence
P
C
Q
w
O
OQ OC OP >> = Qvrw
\ >>
Q CP
v vv
2. ( d) Applying the theorem of perpendicular axis,
1 2 34
I I I II = + =+
Because of symmetry, we have
12
= II and 
34
= II
Hence 
1 234
2 2 22 = = == I I I II
or
1 2 34
= == I I II
i.e. sum of two moment of inertia of square plate about
any axis in a plane (Passing through centre) should be
equal to moment of inertia about the axis passing
through the centre and perpendicular to the plane of
the plate.
3. ( a) By the conservation of energy
l/2
a
P.E. of rod = Rotational K.E.
2
l1
mgsinI
22
a=w
2
2
l 1 ml
mg sin
2 23
Þ a=w 
3 sin a
Þw=
g
l
But in the problem length of the rod 2L is given
3 sin
2
a
\w=
g
L
4. (c) Graph should be parabola symmetric to I- axis, but it
should not pass from origin because there is a constant
value I
cm
 is present for 0 x = .
5. ( b)
2
2
2 24
1
3
1
1
2
= ==
+
+
gh gh
v gh
K
R
6. ( d)
2
2
sin sin / 2 5
2
7 / 5 14
1
1
5
g g gg
a
K
R
qq
= = ==
+
+
As 
30 q=
o
 and 
2
2
2
5
K
R
=
7. ( b) We know 
2
2
2
1
=
+
gh
v
k
r
22
2
\w==
+
v gh
r
rk
22 22
2mgh 2mgh 2mgh
mr mk mr I I mr
Þw= ==
+ ++
8. (a) Because its M.I. (or value of 
2
2
)
K
R
is minimum for
sphere.
9. (b) As body is moving on a frictionless surface. Its
mechanical energy is conserved. When body climbes
up the inclined plane it keeps on rotating with same
angular speed, as no friction force is present to provide
retarding torque so
2 22
1 11
I mv I mgh v 2gh
2 22
w + ³ w + Þ³
10. (a)
22
1
MR I MR 2I
2
=Þ=
Moment of inertia of disc about a tangent in a plane
2
5 55
(2)
4 42
= == MR II
11. ( d) Moment of inertia of system about YY’
1 23
I I II = ++
2 22
133
222
MR MR MR = ++ 
2
7
2
MR =
Y
1
2 3
DPP/ P 17
51
12. (b)
2
2
2 :1
1
2
==
Ring
Disc
I
MR
I
MR
13. (a)
14. (b) It follows from the theorem of parallel axes.
15. (a)
B A
l
O
P
Moment of inertia of Rod AB about point P and
perpendicular to the plane 
2
MI
12
=
M.I. of  rod AB about point ‘O’
2 22
MI I MI
M
12 23
æö
= +=
ç÷
èø
(By using parallel axis theorem)
but the system consists of four rods of similar type so
by the symmetry
2
system
Ml
I4
3
æö
=
ç÷
èø
16. (a)
17. (d)
l
O
b
M.I. of plate about O and parallel to length 
2
12
Mb
=
18. (d)
z xy
I II =+
x
I
y
I
z
I
2002
D Dd
I II = +=
100
D
I \=
2
gm cm ´
19. (a) M.I. of complete disc about ‘O’ point
2
1
(9)
2
Total
I MR -
O
R
2 /3 R
O
R
R/3
Radius of removed disc 
3
R
=
\ Mass of removed disc 
9
9
M
M ==
[As 
22
] M Rt MR =p \¥
M.I. of removed disc about its own axis
2 2
1
2 3 18
R MR
M
æö
==
ç÷
èø
Moment of inertia of removed disc about ‘O’
2 22
2
2
18 32
æö
=+= +=
ç÷
èø
removed disc cm
MR R MR
I ImxM
M. I. of complete disc can also be written as
Re Re Total moved disc maining disc
III =+
2
Re
2
Total maining disc
MR
II =+ ..... (ii)
Equating (i) and (ii) we get
22
Re
9
22
maining disc
MR MR
I+=
2 22
2
Re
98
4
2 22
maining disc
MR MR MR
I MR \ = - ==
Page 3


1. (a) Since disc is rolling (without slipping) about point O.
Hence
P
C
Q
w
O
OQ OC OP >> = Qvrw
\ >>
Q CP
v vv
2. ( d) Applying the theorem of perpendicular axis,
1 2 34
I I I II = + =+
Because of symmetry, we have
12
= II and 
34
= II
Hence 
1 234
2 2 22 = = == I I I II
or
1 2 34
= == I I II
i.e. sum of two moment of inertia of square plate about
any axis in a plane (Passing through centre) should be
equal to moment of inertia about the axis passing
through the centre and perpendicular to the plane of
the plate.
3. ( a) By the conservation of energy
l/2
a
P.E. of rod = Rotational K.E.
2
l1
mgsinI
22
a=w
2
2
l 1 ml
mg sin
2 23
Þ a=w 
3 sin a
Þw=
g
l
But in the problem length of the rod 2L is given
3 sin
2
a
\w=
g
L
4. (c) Graph should be parabola symmetric to I- axis, but it
should not pass from origin because there is a constant
value I
cm
 is present for 0 x = .
5. ( b)
2
2
2 24
1
3
1
1
2
= ==
+
+
gh gh
v gh
K
R
6. ( d)
2
2
sin sin / 2 5
2
7 / 5 14
1
1
5
g g gg
a
K
R
qq
= = ==
+
+
As 
30 q=
o
 and 
2
2
2
5
K
R
=
7. ( b) We know 
2
2
2
1
=
+
gh
v
k
r
22
2
\w==
+
v gh
r
rk
22 22
2mgh 2mgh 2mgh
mr mk mr I I mr
Þw= ==
+ ++
8. (a) Because its M.I. (or value of 
2
2
)
K
R
is minimum for
sphere.
9. (b) As body is moving on a frictionless surface. Its
mechanical energy is conserved. When body climbes
up the inclined plane it keeps on rotating with same
angular speed, as no friction force is present to provide
retarding torque so
2 22
1 11
I mv I mgh v 2gh
2 22
w + ³ w + Þ³
10. (a)
22
1
MR I MR 2I
2
=Þ=
Moment of inertia of disc about a tangent in a plane
2
5 55
(2)
4 42
= == MR II
11. ( d) Moment of inertia of system about YY’
1 23
I I II = ++
2 22
133
222
MR MR MR = ++ 
2
7
2
MR =
Y
1
2 3
DPP/ P 17
51
12. (b)
2
2
2 :1
1
2
==
Ring
Disc
I
MR
I
MR
13. (a)
14. (b) It follows from the theorem of parallel axes.
15. (a)
B A
l
O
P
Moment of inertia of Rod AB about point P and
perpendicular to the plane 
2
MI
12
=
M.I. of  rod AB about point ‘O’
2 22
MI I MI
M
12 23
æö
= +=
ç÷
èø
(By using parallel axis theorem)
but the system consists of four rods of similar type so
by the symmetry
2
system
Ml
I4
3
æö
=
ç÷
èø
16. (a)
17. (d)
l
O
b
M.I. of plate about O and parallel to length 
2
12
Mb
=
18. (d)
z xy
I II =+
x
I
y
I
z
I
2002
D Dd
I II = +=
100
D
I \=
2
gm cm ´
19. (a) M.I. of complete disc about ‘O’ point
2
1
(9)
2
Total
I MR -
O
R
2 /3 R
O
R
R/3
Radius of removed disc 
3
R
=
\ Mass of removed disc 
9
9
M
M ==
[As 
22
] M Rt MR =p \¥
M.I. of removed disc about its own axis
2 2
1
2 3 18
R MR
M
æö
==
ç÷
èø
Moment of inertia of removed disc about ‘O’
2 22
2
2
18 32
æö
=+= +=
ç÷
èø
removed disc cm
MR R MR
I ImxM
M. I. of complete disc can also be written as
Re Re Total moved disc maining disc
III =+
2
Re
2
Total maining disc
MR
II =+ ..... (ii)
Equating (i) and (ii) we get
22
Re
9
22
maining disc
MR MR
I+=
2 22
2
Re
98
4
2 22
maining disc
MR MR MR
I MR \ = - ==
52
DPP/ P 17
20. (d)
2 2
2
124
cm
MLL
IIMxM
æö
= + =+
ç÷
èø
I
L/4 L/4
cm
I
2 22
7
12 16 48
ML ML ML
= +=
21. (c)
2 2 22
0
2 042 w = w - aq Þ = p - aq n
2
2
2
1200
4
60
200
24
æö
p
ç÷
èø
q= =p
´
rad
2
2 200 100 314 nn \p-pÞ =p=
revolution
22. (b) Rotational K.E. = 
2
1
I
2
w
 &
22
11
T.E. I MV
22
= w+
2 22 22
1 11
I MR (I MR)
2 22
= w+ w= w+
For ring I = MR
2
2 22
1
T.E. (MR MR)
2
\ =w+ 
22
1
2MR
2
= w´
Rotational K.E. 
22
1
MR
2
=w
22
22
1
MR
1
2
1
2
2MR
2
w
\a==
w´
For a solid sphere I 
2
2
MR
5
=
2 22
12
T.E. MR MR
25
æö
\ =w+
ç÷
èø
 
22
17
MR
25
=w´
Rotational K.E.
22
12
MR
25
=´w
22
22
12
MR
2
25
17
7
MR
25
´w
b==
w´
23. (a) Time of descent 
2
2
K
R
µ . Time of descent depends upon
the value of radius of gyration (K) or moment of inertia
(I). Actually radius of gyration is a measure of moment
of inertia of the body.
24. (a)
22
22
TR
22
1 1 KK
K K mv mv 1
22
RR
æö
= Þ = Þ\=
ç÷
èø
This value of 
22
K /R match with hollow cylinder. .
25. (b) 26. (c)       27.    (d)
(i) Let acceleration of centre of mass of cylinder be a then
acceleration of block will be 2a.
For linear motion of cylinder T + f – 2mgsinq = 2m(a)
For rolling motion of cylinder
(T – f) R = Ia = 
2
2mRa
2R
æö
æö
ç÷ç÷
èø
èø
Þ T – f = ma
For linear motion of block
mg – T = m (2a) Þ a = 
2
(1 sin )g
7
-q
 ///////////////////////////////////////////
m
2m
R
q
mg
T
T
2mgsinq
f
2a
(ii)
2
T mg 2m g(1 sin)
7
æö
= - -q
ç÷
èø
 = 
3 4sin
mg
7
+q æö
ç÷
èø
(iii) F = T – ma = 
1 6sin
mg
7
+q æö
ç÷
èø
28. (c) The acceleration of a body rolling down an inclined
plane is given by  
2
g sin
a
I
1
MR
q
=
+
For hollow cylinder 
2
22
I MR
1
MR MR
==
For solid cylinder 
2
22
1
MR
I1
2
2
MR MR
==
Þ Acceleration  of solid cylinder is more than hollow
cylinder and therefore solid cylinder will reach the
bottom of the inclined plane first.
\ Statement -1 is false
• Statement - 2
In the case of rolling there will be no heat losses.
Therefore total mechanical energy remains conserved.
The potential energy therefore gets converted into
kinetic energy. In both the cases since the initial
potential energy is same, the final kinetic energy will
also be same. Therefore statement -2 is correct.
29. (b) Frictional force on an inclined plane
( )
1
sin for a disc .
3
= a g
30. (c) The moment of inertia about both the given axes shall
be same if they are parallel. Hence statement–1 is false.