System of Particles & Rotational Motion Practice Questions - DPP for JEE

 Page 1


1. (b) I = 1.2  kg m
2
, E
r
 = 1500 J,
a = 25 rad/sec
2
, ?
1
 = 0, t  = ?
As E
r
 
From 
50 = 0 + 25 t, ?   t = 2 seconds
2. (a) F = 20t – 5t
2
?
?
?
? (as ? = 0 at t = 0, 6s)
? ? = 36 rad
? 
3. (b) Since no external torque act on gymnast, so angular momentum
(L=I ) is conserved. After pulling her arms & legs, the angular
velocity increases but moment of inertia of gymnast, decreases in,
such a way that angular momentum remains constant.
4. (b) The M.I. about the axis of rotation is not constant as the
perpendicular distance of the bead with the axis of rotation
increases.
Also since no external torque is acting.
Page 2


1. (b) I = 1.2  kg m
2
, E
r
 = 1500 J,
a = 25 rad/sec
2
, ?
1
 = 0, t  = ?
As E
r
 
From 
50 = 0 + 25 t, ?   t = 2 seconds
2. (a) F = 20t – 5t
2
?
?
?
? (as ? = 0 at t = 0, 6s)
? ? = 36 rad
? 
3. (b) Since no external torque act on gymnast, so angular momentum
(L=I ) is conserved. After pulling her arms & legs, the angular
velocity increases but moment of inertia of gymnast, decreases in,
such a way that angular momentum remains constant.
4. (b) The M.I. about the axis of rotation is not constant as the
perpendicular distance of the bead with the axis of rotation
increases.
Also since no external torque is acting.
? t
ext
 =   L = constant I? = constant
Since, I increases, ? decreases.
5. (b) ....... (1)
....... (2)
a
1
 = a ....... (3)
Acceleration of point b = acceleration of point a
ra
1
 = a
cm
 – ra ....... (4)
Hence, 2ra = a
cm
6. (b) Velocity of the tennis ball on the surface of the earth or ground
 ( where k = radius of gyration of spherical shell = 
)
Horizontal range AB = 
Solving we get AB = 2.08 m
7. (d) Initially centre of mass is at the centre. When sand is poured it will
Page 3


1. (b) I = 1.2  kg m
2
, E
r
 = 1500 J,
a = 25 rad/sec
2
, ?
1
 = 0, t  = ?
As E
r
 
From 
50 = 0 + 25 t, ?   t = 2 seconds
2. (a) F = 20t – 5t
2
?
?
?
? (as ? = 0 at t = 0, 6s)
? ? = 36 rad
? 
3. (b) Since no external torque act on gymnast, so angular momentum
(L=I ) is conserved. After pulling her arms & legs, the angular
velocity increases but moment of inertia of gymnast, decreases in,
such a way that angular momentum remains constant.
4. (b) The M.I. about the axis of rotation is not constant as the
perpendicular distance of the bead with the axis of rotation
increases.
Also since no external torque is acting.
? t
ext
 =   L = constant I? = constant
Since, I increases, ? decreases.
5. (b) ....... (1)
....... (2)
a
1
 = a ....... (3)
Acceleration of point b = acceleration of point a
ra
1
 = a
cm
 – ra ....... (4)
Hence, 2ra = a
cm
6. (b) Velocity of the tennis ball on the surface of the earth or ground
 ( where k = radius of gyration of spherical shell = 
)
Horizontal range AB = 
Solving we get AB = 2.08 m
7. (d) Initially centre of mass is at the centre. When sand is poured it will
fall and again after a limit, centre of mass will rise.
8. (a) Here 
Now, 
  
Moment of inertia of the cube about the given axis,  
9. (d) Minimum velocity for a body rolling without slipping 
For solid sphere, 
?
10. (c)
From conservation of angular momentum about any fix point on the
Page 4


1. (b) I = 1.2  kg m
2
, E
r
 = 1500 J,
a = 25 rad/sec
2
, ?
1
 = 0, t  = ?
As E
r
 
From 
50 = 0 + 25 t, ?   t = 2 seconds
2. (a) F = 20t – 5t
2
?
?
?
? (as ? = 0 at t = 0, 6s)
? ? = 36 rad
? 
3. (b) Since no external torque act on gymnast, so angular momentum
(L=I ) is conserved. After pulling her arms & legs, the angular
velocity increases but moment of inertia of gymnast, decreases in,
such a way that angular momentum remains constant.
4. (b) The M.I. about the axis of rotation is not constant as the
perpendicular distance of the bead with the axis of rotation
increases.
Also since no external torque is acting.
? t
ext
 =   L = constant I? = constant
Since, I increases, ? decreases.
5. (b) ....... (1)
....... (2)
a
1
 = a ....... (3)
Acceleration of point b = acceleration of point a
ra
1
 = a
cm
 – ra ....... (4)
Hence, 2ra = a
cm
6. (b) Velocity of the tennis ball on the surface of the earth or ground
 ( where k = radius of gyration of spherical shell = 
)
Horizontal range AB = 
Solving we get AB = 2.08 m
7. (d) Initially centre of mass is at the centre. When sand is poured it will
fall and again after a limit, centre of mass will rise.
8. (a) Here 
Now, 
  
Moment of inertia of the cube about the given axis,  
9. (d) Minimum velocity for a body rolling without slipping 
For solid sphere, 
?
10. (c)
From conservation of angular momentum about any fix point on the
surface,
mr
2
?
0
 = 2mr
2
?
?    ?     
11. (a) When n = 0, x = k where k is a constant. This means that the
linear mass density is constant. In this case the centre of mass will
be at the middle of the rod i.e., at L/2. Therefore (c) is ruled out
  n is positive and as its value increases, the rate of increase of
linear mass density with increase in x increases. This shows that
the centre of mass will shift towards that end of the rod where n =
L as the value of n increases. Therefore graph (b) is ruled out.
 The linear mass density 
With increase in the value of n, the centre of mass shift towards the end
x = L such that first the shifting is at a higher rate with increase in
the value of n and then the rate decreases with the value of n.
These characteristics are represented by graph (a). 
 
Page 5


1. (b) I = 1.2  kg m
2
, E
r
 = 1500 J,
a = 25 rad/sec
2
, ?
1
 = 0, t  = ?
As E
r
 
From 
50 = 0 + 25 t, ?   t = 2 seconds
2. (a) F = 20t – 5t
2
?
?
?
? (as ? = 0 at t = 0, 6s)
? ? = 36 rad
? 
3. (b) Since no external torque act on gymnast, so angular momentum
(L=I ) is conserved. After pulling her arms & legs, the angular
velocity increases but moment of inertia of gymnast, decreases in,
such a way that angular momentum remains constant.
4. (b) The M.I. about the axis of rotation is not constant as the
perpendicular distance of the bead with the axis of rotation
increases.
Also since no external torque is acting.
? t
ext
 =   L = constant I? = constant
Since, I increases, ? decreases.
5. (b) ....... (1)
....... (2)
a
1
 = a ....... (3)
Acceleration of point b = acceleration of point a
ra
1
 = a
cm
 – ra ....... (4)
Hence, 2ra = a
cm
6. (b) Velocity of the tennis ball on the surface of the earth or ground
 ( where k = radius of gyration of spherical shell = 
)
Horizontal range AB = 
Solving we get AB = 2.08 m
7. (d) Initially centre of mass is at the centre. When sand is poured it will
fall and again after a limit, centre of mass will rise.
8. (a) Here 
Now, 
  
Moment of inertia of the cube about the given axis,  
9. (d) Minimum velocity for a body rolling without slipping 
For solid sphere, 
?
10. (c)
From conservation of angular momentum about any fix point on the
surface,
mr
2
?
0
 = 2mr
2
?
?    ?     
11. (a) When n = 0, x = k where k is a constant. This means that the
linear mass density is constant. In this case the centre of mass will
be at the middle of the rod i.e., at L/2. Therefore (c) is ruled out
  n is positive and as its value increases, the rate of increase of
linear mass density with increase in x increases. This shows that
the centre of mass will shift towards that end of the rod where n =
L as the value of n increases. Therefore graph (b) is ruled out.
 The linear mass density 
With increase in the value of n, the centre of mass shift towards the end
x = L such that first the shifting is at a higher rate with increase in
the value of n and then the rate decreases with the value of n.
These characteristics are represented by graph (a). 
 
12. (b) Applying angular momentum conservation
mV
0
R
0
 = (m) (V
1
) 
? v
1
 = 2V
0              
Therefore, new KE = m (2V
0
)
2
 = 
13. (b)
14. (c) Kinetic energy 
(rotational)
 K
R
 = 
Kinetic energy 
(translational) 
K
T
 = (v = R?)
M.I.
(initial)
 I
ring
 = MR
2
; ?
initial
 = ?
M.I.
(new)
 I'
(system)
 = 
?'
(system)
? 
=
Solving we get loss in K.E. = 
15. (c) After collision velocity of COM of A becomes zero and that of B
becomes equal to initial velocity of COM of A. But angular
velocity of A remains unchanged as the two spheres are smooth.
16. (c) Velocity of 
17. (b) Couple produces purely rotational motion.