Rotational Motion- 2 Practice Questions - DPP for NEET

 Page 1


(1) (b) As the mass of disc is negligible therefore only moment
of inertia of five particles will be considered.
I = 
2
mr
å
= 5 mr
2
 = 5 × 2 × (0.1)
2
 = 0.1 kg-m
2
(2) (a)
( )
2 22
11
I MR Rt pR
22
= = ´p ´´
4
IR Þµ (As t and p are same)
4
4
11
22
IR 0.21
I R 0.6 81
æö
æö
\= ==
ç÷
ç÷
èø
èø
(3) (a)
2 2 32
11
0.5 (0.1) 2.5 10
22
I MR kgm
-
= =´´ = ´-
(4) (a)
2
31.4
2.5
4
I kgm
t
= ==
ap
(5) (d) Let the mass of loop P (radius = r) = m
So the mass of loop Q (radius = nr) = nm
Q
P
r
Moment of inertia of loop P , I
P
 = mr
2
Moment of inertia of loop Q. I
Q
 = nm (nr)
2
 = n
3
 mr
2
3
82
Q
P
I
nn
I
\ = = Þ=
(6) (c) Moment of inertia of sphere about its tangent
22
77
55
MR MK KR = Þ=
(7) (a) Moment of inertia of system about point P
m m
m m
2
l
p
= 
2
2
42
2
l
m ml
æö
=
ç÷
èø
 and 4mK
2
 = 2ml
2
2
l
K \=
(8) (b)
22
1 2.5
1.76 cm
2 22
R
MR MKK = Þ ===
(9) (c) I = 2MR
2
 = 2 × 3 × (1)
2
 = 6 gm-cm
2
(10) (a)
2
5
I Mr
4
=
(11) (a)
R/2
y¢
y
Q
2R
Moment of inertia of the system about yy¢
I
yy¢ 
 = Moment of inertia of sphere P about yy¢
+ Moment of inertia of sphere Q about yy¢
Moment of inertia of sphere P about yy¢
2
2
2
()
52
R
M Mx
æö
=+
ç÷
èø
[Parallel axis theorem]
= 
2
2
2
(2)
52
R
M MR
æö
=+
ç÷
èø
 = 
2
2
4
10
MR
MR +
Moment of inertia of sphere Q about yy¢  is 
2
2
52
R
M
æö
ç÷
èø
Now 
2 2
22
2 21
4
10 5 25
yy
MRR
I MR M MR
¢
æö
= + +=
ç÷
èø
(12) (a) M.I. of system about the axis which passing through
m
1
m
1
m
2
m
3
a a
a/2 a/2
I
system
 = m
1
 (0)
2
 + 
22
23
22
aa
mm
æ ö æö
+
ç ÷ ç÷
è ø èø
I
system
 =
2
23
()
4
a
mm +
Page 2


(1) (b) As the mass of disc is negligible therefore only moment
of inertia of five particles will be considered.
I = 
2
mr
å
= 5 mr
2
 = 5 × 2 × (0.1)
2
 = 0.1 kg-m
2
(2) (a)
( )
2 22
11
I MR Rt pR
22
= = ´p ´´
4
IR Þµ (As t and p are same)
4
4
11
22
IR 0.21
I R 0.6 81
æö
æö
\= ==
ç÷
ç÷
èø
èø
(3) (a)
2 2 32
11
0.5 (0.1) 2.5 10
22
I MR kgm
-
= =´´ = ´-
(4) (a)
2
31.4
2.5
4
I kgm
t
= ==
ap
(5) (d) Let the mass of loop P (radius = r) = m
So the mass of loop Q (radius = nr) = nm
Q
P
r
Moment of inertia of loop P , I
P
 = mr
2
Moment of inertia of loop Q. I
Q
 = nm (nr)
2
 = n
3
 mr
2
3
82
Q
P
I
nn
I
\ = = Þ=
(6) (c) Moment of inertia of sphere about its tangent
22
77
55
MR MK KR = Þ=
(7) (a) Moment of inertia of system about point P
m m
m m
2
l
p
= 
2
2
42
2
l
m ml
æö
=
ç÷
èø
 and 4mK
2
 = 2ml
2
2
l
K \=
(8) (b)
22
1 2.5
1.76 cm
2 22
R
MR MKK = Þ ===
(9) (c) I = 2MR
2
 = 2 × 3 × (1)
2
 = 6 gm-cm
2
(10) (a)
2
5
I Mr
4
=
(11) (a)
R/2
y¢
y
Q
2R
Moment of inertia of the system about yy¢
I
yy¢ 
 = Moment of inertia of sphere P about yy¢
+ Moment of inertia of sphere Q about yy¢
Moment of inertia of sphere P about yy¢
2
2
2
()
52
R
M Mx
æö
=+
ç÷
èø
[Parallel axis theorem]
= 
2
2
2
(2)
52
R
M MR
æö
=+
ç÷
èø
 = 
2
2
4
10
MR
MR +
Moment of inertia of sphere Q about yy¢  is 
2
2
52
R
M
æö
ç÷
èø
Now 
2 2
22
2 21
4
10 5 25
yy
MRR
I MR M MR
¢
æö
= + +=
ç÷
èø
(12) (a) M.I. of system about the axis which passing through
m
1
m
1
m
2
m
3
a a
a/2 a/2
I
system
 = m
1
 (0)
2
 + 
22
23
22
aa
mm
æ ö æö
+
ç ÷ ç÷
è ø èø
I
system
 =
2
23
()
4
a
mm +
48
DPP/ P 16
(13) (a) M.I. of rod (1) about Z – axis I
1
 = 
2
3
Ml
1
2
3
M.I. of rod (2) about Z-axis, 
2
2
3
Ml
I =
M.I. of rod (3) about Z – axis, I
3
 = 0
Because this rod lies on Z-axis
\ I
system
 = I
1
 + I
2
 + I
3
 = 
2
2
3
Ml
(14) (c) Distribution of mass about BC axis is more than that
about AB axis i.e. radius of gyration about BC axis is
more than that about AB axis.
i.e. K
BC
 > K
AB
 \ I
BC
 > I
AB
 > I
CA
(15) (a)
22
2
0.12 1
0.01
12 12
Ml
I kgm
´
= = =-
(16) (c)
x
1
2
I
1
 = M.I. of ring about its diameter = 
2
1
2
mR
I
2
 = M.I. of ring about the axis normal to plane and
passing through centre = mR
2
Two rings are placed according to figure. Then
2 22
12
13
22
xx
I I I mR mR mR
¢
= + = +=
(17) (a) Mass of the centre disc would be 4M and its moment
of inertia about the given axis would be 
2
1
(4)
2
MR .
For the given section the moment of inertia about the
same axis would be one quarter of this i.e. 
2
1
.
2
MR
(18) (d) Mass per unit length of the wire = r
Mass of L length, M = rL
and since the wire of length L is bent in a or of circular
loop therefore 2pR = L 
2
L
R Þ=
p
Moment of inertia of loop about given axis 
2
3
2
MR =
= 
2 3
2
33
22
8
LL
L
r æö
r=
ç÷
èø p
p
(19) (b) M.I. of disc 
2
2
1 11
2 22
MM
MRM
tt
æö
= ==
ç÷
èø p r pr
2
2
As Therefore
MM
R
t
Rt
æö
r==
ç÷
èø pr
p
If mass and thickness are same then, 
1
I µ
r
12
21
3
.
1
I
I
r
\ ==
r
(20) (c) According to problem disc is melted and recasted into
a solid sphere so their volume will be same.
23
4
3
Disc Sphere Disc Sphere
V V R tR = Þp =p
33
4
63
Disc
Disc Sphere
R
RR
æö
Þp =p
ç÷
èø
 
,given
6
Disc
R
t
éù
=
êú
ëû
33
8
2
Disc
Disc Sphere Sphere
R
R RR Þ = Þ=
Moment of inertia of disc
2
Disc Disc
1
I MR I (given)
2
==
( )
2
Disc
M R 2I \=
Moment of inertia of sphere I
sphere
 = 
2
2
5
Sphere
MR
2
2
2 21
()
5 2 10 10 5
Disc
Discs
R MI
MR
æö
= = ==
ç÷
èø
(21) (d) Moment of inertia of system about YY'
I = I
1
 + I
2
 + I
3
= 
222
1 33
MR MR MR
222
++
Y
1
2 3
= 
2
7
MR
2
(22) (d) As C is the centre of mass, so, I
C
 will be minimum.
Also more mass is towards B so I
A
 > I
B
.
(23) (a) Applying the theorem of perpendicular axis,
I = I
1
 + I
2
 = I
3
 + I
4
Because of symmetry , we have I
1
 = I
2
 and I
3
 = I
4
 Hence
I = 2I
1
 = 2I
2
 = 2I
3
 = 2I
4
 or I
1
 = I
2
 = I
3
 = I
4
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.
Page 3


(1) (b) As the mass of disc is negligible therefore only moment
of inertia of five particles will be considered.
I = 
2
mr
å
= 5 mr
2
 = 5 × 2 × (0.1)
2
 = 0.1 kg-m
2
(2) (a)
( )
2 22
11
I MR Rt pR
22
= = ´p ´´
4
IR Þµ (As t and p are same)
4
4
11
22
IR 0.21
I R 0.6 81
æö
æö
\= ==
ç÷
ç÷
èø
èø
(3) (a)
2 2 32
11
0.5 (0.1) 2.5 10
22
I MR kgm
-
= =´´ = ´-
(4) (a)
2
31.4
2.5
4
I kgm
t
= ==
ap
(5) (d) Let the mass of loop P (radius = r) = m
So the mass of loop Q (radius = nr) = nm
Q
P
r
Moment of inertia of loop P , I
P
 = mr
2
Moment of inertia of loop Q. I
Q
 = nm (nr)
2
 = n
3
 mr
2
3
82
Q
P
I
nn
I
\ = = Þ=
(6) (c) Moment of inertia of sphere about its tangent
22
77
55
MR MK KR = Þ=
(7) (a) Moment of inertia of system about point P
m m
m m
2
l
p
= 
2
2
42
2
l
m ml
æö
=
ç÷
èø
 and 4mK
2
 = 2ml
2
2
l
K \=
(8) (b)
22
1 2.5
1.76 cm
2 22
R
MR MKK = Þ ===
(9) (c) I = 2MR
2
 = 2 × 3 × (1)
2
 = 6 gm-cm
2
(10) (a)
2
5
I Mr
4
=
(11) (a)
R/2
y¢
y
Q
2R
Moment of inertia of the system about yy¢
I
yy¢ 
 = Moment of inertia of sphere P about yy¢
+ Moment of inertia of sphere Q about yy¢
Moment of inertia of sphere P about yy¢
2
2
2
()
52
R
M Mx
æö
=+
ç÷
èø
[Parallel axis theorem]
= 
2
2
2
(2)
52
R
M MR
æö
=+
ç÷
èø
 = 
2
2
4
10
MR
MR +
Moment of inertia of sphere Q about yy¢  is 
2
2
52
R
M
æö
ç÷
èø
Now 
2 2
22
2 21
4
10 5 25
yy
MRR
I MR M MR
¢
æö
= + +=
ç÷
èø
(12) (a) M.I. of system about the axis which passing through
m
1
m
1
m
2
m
3
a a
a/2 a/2
I
system
 = m
1
 (0)
2
 + 
22
23
22
aa
mm
æ ö æö
+
ç ÷ ç÷
è ø èø
I
system
 =
2
23
()
4
a
mm +
48
DPP/ P 16
(13) (a) M.I. of rod (1) about Z – axis I
1
 = 
2
3
Ml
1
2
3
M.I. of rod (2) about Z-axis, 
2
2
3
Ml
I =
M.I. of rod (3) about Z – axis, I
3
 = 0
Because this rod lies on Z-axis
\ I
system
 = I
1
 + I
2
 + I
3
 = 
2
2
3
Ml
(14) (c) Distribution of mass about BC axis is more than that
about AB axis i.e. radius of gyration about BC axis is
more than that about AB axis.
i.e. K
BC
 > K
AB
 \ I
BC
 > I
AB
 > I
CA
(15) (a)
22
2
0.12 1
0.01
12 12
Ml
I kgm
´
= = =-
(16) (c)
x
1
2
I
1
 = M.I. of ring about its diameter = 
2
1
2
mR
I
2
 = M.I. of ring about the axis normal to plane and
passing through centre = mR
2
Two rings are placed according to figure. Then
2 22
12
13
22
xx
I I I mR mR mR
¢
= + = +=
(17) (a) Mass of the centre disc would be 4M and its moment
of inertia about the given axis would be 
2
1
(4)
2
MR .
For the given section the moment of inertia about the
same axis would be one quarter of this i.e. 
2
1
.
2
MR
(18) (d) Mass per unit length of the wire = r
Mass of L length, M = rL
and since the wire of length L is bent in a or of circular
loop therefore 2pR = L 
2
L
R Þ=
p
Moment of inertia of loop about given axis 
2
3
2
MR =
= 
2 3
2
33
22
8
LL
L
r æö
r=
ç÷
èø p
p
(19) (b) M.I. of disc 
2
2
1 11
2 22
MM
MRM
tt
æö
= ==
ç÷
èø p r pr
2
2
As Therefore
MM
R
t
Rt
æö
r==
ç÷
èø pr
p
If mass and thickness are same then, 
1
I µ
r
12
21
3
.
1
I
I
r
\ ==
r
(20) (c) According to problem disc is melted and recasted into
a solid sphere so their volume will be same.
23
4
3
Disc Sphere Disc Sphere
V V R tR = Þp =p
33
4
63
Disc
Disc Sphere
R
RR
æö
Þp =p
ç÷
èø
 
,given
6
Disc
R
t
éù
=
êú
ëû
33
8
2
Disc
Disc Sphere Sphere
R
R RR Þ = Þ=
Moment of inertia of disc
2
Disc Disc
1
I MR I (given)
2
==
( )
2
Disc
M R 2I \=
Moment of inertia of sphere I
sphere
 = 
2
2
5
Sphere
MR
2
2
2 21
()
5 2 10 10 5
Disc
Discs
R MI
MR
æö
= = ==
ç÷
èø
(21) (d) Moment of inertia of system about YY'
I = I
1
 + I
2
 + I
3
= 
222
1 33
MR MR MR
222
++
Y
1
2 3
= 
2
7
MR
2
(22) (d) As C is the centre of mass, so, I
C
 will be minimum.
Also more mass is towards B so I
A
 > I
B
.
(23) (a) Applying the theorem of perpendicular axis,
I = I
1
 + I
2
 = I
3
 + I
4
Because of symmetry , we have I
1
 = I
2
 and I
3
 = I
4
 Hence
I = 2I
1
 = 2I
2
 = 2I
3
 = 2I
4
 or I
1
 = I
2
 = I
3
 = I
4
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.
DPP/ P 16
49
(24) (d) Moment of inertia depends on all the three factors given
in (1), (2) & (3).
(25) (d) I = 
22
2
4 ( 2)
5
MR MR
éù
+
êú
ëû
= 
2
2
42
5
MR
éù
+
êú
ëû
= 
22
4 12 48
.
55
MR MR ´
=
(26) (b) Let a be the acceleration of centre of mass
Mg – T = 0 ... (i)
F.x = T.2x ... (ii)
M
F
x
(27) (c) remain the same
(28) (c) Radius of gyration of a body is not a constant quantity .
Its value changes with the change in location of the
axis of rotation. Radius of gyration of a body about a
given axis is given as
222
12
.....
n
rrr
K
n
+ ++
=
(29) (c) The moment of inertia of a particle about an axis of
rotation is given by the product of the mass of the
particle and the square of the perpendicular distance
of the particle from the axis of rotation. For different
axis, distance would be different, therefore moment of
inertia of a particle changes with the change in axis of
rotation.
(30) (a) When earth shrinks, it angular momentum remains
constant. i.e. 
2
22
5
L I mR
T
p
=w= ´= constant.
2
. TIR \ µµ
 It means if size of the earth changes
then its moment of inertia changes.
In the problem radius becomes half so time period
(Length of the day) will becomes 
1
4
 of the present value
i.e. 
24
6 hr.
4
=