Chapter 7 - Circular Motion (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 7 - Circular Motion (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


14. Let all the particles meet at time t
(seconds)
\ Distance travelled by B in t second : 
Distance travelled by C in t second
= 2.5t t :2
= 5 4 :   
Option (c) is correct.
JEE Corner
Assertion and Reason
1. For stopping car : 
Maximum retardation 
=
Maximum frictional force
m
     = = =
m m
m
N
m
mg
m
g
 v u as
2 2
2 = +
  0 2
2 2
= + - v g d ( )( ) m [ Q Initial velocity = v]
\      d
v
g
=
2
2m
For circular turn of car :
Centripetal force 
             = Maximum frictional force = m mg
\    
mv
d
mg
2
¢
= m
i.e., safe radius = ¢ d =
v
g
2
m
  Þ d d ¢ = 2 
Thus, Assertion and Reason are both
correct and Reason is the correct
explanation of the Assertion.
Option (a) is correct.
2. a
v v ®
® ®
=
-
av
B A
AB
t
=
®
BD
t
AB
i.e.,    | |
| |
a
BD ®
=
®
= av
t
v
t
AB AB
2
Now,    | |
| |
v
AB ®
=
®
= av
t
R
t
AB AB
2
\
| |
| |
a
v
®
®
= =
av
av
v
R
2
2
w (angular velocity)
Thus, assertion is correct.
In circular motion, when speed is
constant, the angular velocity will
obviously be constant; but this reason does 
not lead to the result as explained.
Option (b) is correct.
3. A frame moving in a circle with constant
speed can never be an inertial frame as
the frame is not moving with constant
velocity (due to change in direction).
Reason that the frame is having constant
acceleration is false.
Option (c) is correct.
4. If speed is constant angle between v
®
 and a
®
will always be 90°
\ v a
® ®
× = 0
150 | Mechanics-1
B
D
A
v
B
v
A
®
®
v
A
®
–1
2.5 ms
B
C A
–1
1 ms
–1
2 ms
a = a
c
w
q
®
® ®
Page 2


14. Let all the particles meet at time t
(seconds)
\ Distance travelled by B in t second : 
Distance travelled by C in t second
= 2.5t t :2
= 5 4 :   
Option (c) is correct.
JEE Corner
Assertion and Reason
1. For stopping car : 
Maximum retardation 
=
Maximum frictional force
m
     = = =
m m
m
N
m
mg
m
g
 v u as
2 2
2 = +
  0 2
2 2
= + - v g d ( )( ) m [ Q Initial velocity = v]
\      d
v
g
=
2
2m
For circular turn of car :
Centripetal force 
             = Maximum frictional force = m mg
\    
mv
d
mg
2
¢
= m
i.e., safe radius = ¢ d =
v
g
2
m
  Þ d d ¢ = 2 
Thus, Assertion and Reason are both
correct and Reason is the correct
explanation of the Assertion.
Option (a) is correct.
2. a
v v ®
® ®
=
-
av
B A
AB
t
=
®
BD
t
AB
i.e.,    | |
| |
a
BD ®
=
®
= av
t
v
t
AB AB
2
Now,    | |
| |
v
AB ®
=
®
= av
t
R
t
AB AB
2
\
| |
| |
a
v
®
®
= =
av
av
v
R
2
2
w (angular velocity)
Thus, assertion is correct.
In circular motion, when speed is
constant, the angular velocity will
obviously be constant; but this reason does 
not lead to the result as explained.
Option (b) is correct.
3. A frame moving in a circle with constant
speed can never be an inertial frame as
the frame is not moving with constant
velocity (due to change in direction).
Reason that the frame is having constant
acceleration is false.
Option (c) is correct.
4. If speed is constant angle between v
®
 and a
®
will always be 90°
\ v a
® ®
× = 0
150 | Mechanics-1
B
D
A
v
B
v
A
®
®
v
A
®
–1
2.5 ms
B
C A
–1
1 ms
–1
2 ms
a = a
c
w
q
®
® ®
If speed is increasing angle q between v
®
and a
®
 will be less than 90°.
\ v a
® ®
× will be positive.
If speed is decreasing angle between v
®
 and 
a
®
 will be greater than 90°.
\    v a
® ®
× will be negative.
Assertion is correct.
Rea son w
® ®
× = v 0 as both are
perpendicular to each other.
Reason is also true but not the correct
explanation of the assertion.
Option (b) is correct.
5. v i
®
= 2
^
 ms
-1
       a i j
®
= - +
^ ^
2  ms
-2
\      a j
®
-
= c 2
2 ^
ms
      a a
c c
® ®
= = | | 2 ms
-2
        v = =
®
| | v 2 ms
-1
       a
v
R
c
=
2
Þ        R
v
a
c
=
2
            = =
( ) 2
2
2
2
 m
         a i
T
®
= -
^
 ms
-2
Speed is decreasing (as a
T
®
 is -ive) at a
rate of 1 ms
-1
 per second i.e., 1 ms
-2
.
Both Assertion and the Reason are correct 
but reason has nothing to do with the
assertion.
Option (b) is correct.
6. a a a
® ® ®
= +
T c
 (Reason)
\ | | a
®
= +
v
r
g
2
2
Þ   | | a
®
> g (Assertion)
We see that both Assertion and the
Reason are correct and the reason is the
correct explanation of assertion.
Option (a) is correct.
7. At points A and C : The bob is
momentarily at rest.
i.e., v = 0 (Reason)
\          | | a
c
v
R
®
= =
2
0
but net acceleration is not zero (see figure) 
i.e., Assertion is false.
\ Option (d) is correct.
Circular Motion | 151
a
c
q
v a
T
a
® ®
® ®
R
2j
v = 2z
a
net
i
x
–
®
a
c
q
v a
T
a
® ®
®
®
a = g
T
a
c
a
A
® ®
®
B
g g
A C
® ®
Page 3


14. Let all the particles meet at time t
(seconds)
\ Distance travelled by B in t second : 
Distance travelled by C in t second
= 2.5t t :2
= 5 4 :   
Option (c) is correct.
JEE Corner
Assertion and Reason
1. For stopping car : 
Maximum retardation 
=
Maximum frictional force
m
     = = =
m m
m
N
m
mg
m
g
 v u as
2 2
2 = +
  0 2
2 2
= + - v g d ( )( ) m [ Q Initial velocity = v]
\      d
v
g
=
2
2m
For circular turn of car :
Centripetal force 
             = Maximum frictional force = m mg
\    
mv
d
mg
2
¢
= m
i.e., safe radius = ¢ d =
v
g
2
m
  Þ d d ¢ = 2 
Thus, Assertion and Reason are both
correct and Reason is the correct
explanation of the Assertion.
Option (a) is correct.
2. a
v v ®
® ®
=
-
av
B A
AB
t
=
®
BD
t
AB
i.e.,    | |
| |
a
BD ®
=
®
= av
t
v
t
AB AB
2
Now,    | |
| |
v
AB ®
=
®
= av
t
R
t
AB AB
2
\
| |
| |
a
v
®
®
= =
av
av
v
R
2
2
w (angular velocity)
Thus, assertion is correct.
In circular motion, when speed is
constant, the angular velocity will
obviously be constant; but this reason does 
not lead to the result as explained.
Option (b) is correct.
3. A frame moving in a circle with constant
speed can never be an inertial frame as
the frame is not moving with constant
velocity (due to change in direction).
Reason that the frame is having constant
acceleration is false.
Option (c) is correct.
4. If speed is constant angle between v
®
 and a
®
will always be 90°
\ v a
® ®
× = 0
150 | Mechanics-1
B
D
A
v
B
v
A
®
®
v
A
®
–1
2.5 ms
B
C A
–1
1 ms
–1
2 ms
a = a
c
w
q
®
® ®
If speed is increasing angle q between v
®
and a
®
 will be less than 90°.
\ v a
® ®
× will be positive.
If speed is decreasing angle between v
®
 and 
a
®
 will be greater than 90°.
\    v a
® ®
× will be negative.
Assertion is correct.
Rea son w
® ®
× = v 0 as both are
perpendicular to each other.
Reason is also true but not the correct
explanation of the assertion.
Option (b) is correct.
5. v i
®
= 2
^
 ms
-1
       a i j
®
= - +
^ ^
2  ms
-2
\      a j
®
-
= c 2
2 ^
ms
      a a
c c
® ®
= = | | 2 ms
-2
        v = =
®
| | v 2 ms
-1
       a
v
R
c
=
2
Þ        R
v
a
c
=
2
            = =
( ) 2
2
2
2
 m
         a i
T
®
= -
^
 ms
-2
Speed is decreasing (as a
T
®
 is -ive) at a
rate of 1 ms
-1
 per second i.e., 1 ms
-2
.
Both Assertion and the Reason are correct 
but reason has nothing to do with the
assertion.
Option (b) is correct.
6. a a a
® ® ®
= +
T c
 (Reason)
\ | | a
®
= +
v
r
g
2
2
Þ   | | a
®
> g (Assertion)
We see that both Assertion and the
Reason are correct and the reason is the
correct explanation of assertion.
Option (a) is correct.
7. At points A and C : The bob is
momentarily at rest.
i.e., v = 0 (Reason)
\          | | a
c
v
R
®
= =
2
0
but net acceleration is not zero (see figure) 
i.e., Assertion is false.
\ Option (d) is correct.
Circular Motion | 151
a
c
q
v a
T
a
® ®
® ®
R
2j
v = 2z
a
net
i
x
–
®
a
c
q
v a
T
a
® ®
®
®
a = g
T
a
c
a
A
® ®
®
B
g g
A C
® ®
8. v (speed) = - 4 12 t
For t < 3 (time unit) speed is negative,
which can’t describe a motion. Thus,
assertion is correct.
As speed can be changed linearly with
time, the reason is false.
Option (c) is correct.
9. In circular motion the acceleration
changes regularly where as in projectile
motion it is constant. Thus, in circular
motion we can't apply v u a
® ® ®
= + t directly,
whereas in projectile motion we can say
reason that in circular motion gravity has
no role is wrong.
Option (c) is correct.
10. N mg = cos q
Therefore, assertion is wrong.
Particle performs circular motion due to 
Nsinq 
\
mv
r
N
2
= sinq
Ncosq is balanced by mg (weight of
particle).
Acceleration is not along the surface of the 
funnel. It is along the centre O of the
circle. Thus, reason is true.
Option (d) is correct.
11. Centripetal force 
mv
r
N mg
2
æ
è
ç
ö
ø
÷ = +
i.e., Centripetal force (reason ) ³ wt mg ( ) of 
water
for N = 0, v gr =
If at the top of the circular path v gr ³
i.e., if bucket moved fast, the water will
not fall (Assertion).
As assertion and reason both are true and
reason is the correct explanation of the
assertion of the option would be (a).
Objective Questions (Level 2)
Single Correct Option
1. KE 
1
2
2
mv
æ
è
ç
ö
ø
÷ = Change in PE 
1
2
2
k x ( ) D
é
ë
ê
ù
û
ú
Dx = Length of spring (Collar at B)
- Length of spring (Collar at A)
= + + ( ) 7 5 5
2 2
m - 7m = 6 m
Thus,   
1
2
2
1
2
200 6
2 2
´ ´ = ´ ´ v
i.e.,          v
2
3600 =
\ Normal reaction =
mv
r
2
=
´ 2 3600
5
 = 1440 N
Option (a) is correct.
2. The particle will remain in equilibrium till 
w is constant. Any change in the value of w
will displace the particle up (if w increase)
and down (if w decreases). Thus, the
equilibrium is unstable.
Option (b) is correct.
3. Centripetal force = m mg
or        mr mg w m
2
=
or        
5
4
2
a
g w m =
or          w
2
4
15
=
g
a
 asm =
æ
è
ç
ö
ø
÷
1
3
Option (d) is correct.
152 | Mechanics-1
mg cos q
mg
N
O
N sinq
q
N
mg
Page 4


14. Let all the particles meet at time t
(seconds)
\ Distance travelled by B in t second : 
Distance travelled by C in t second
= 2.5t t :2
= 5 4 :   
Option (c) is correct.
JEE Corner
Assertion and Reason
1. For stopping car : 
Maximum retardation 
=
Maximum frictional force
m
     = = =
m m
m
N
m
mg
m
g
 v u as
2 2
2 = +
  0 2
2 2
= + - v g d ( )( ) m [ Q Initial velocity = v]
\      d
v
g
=
2
2m
For circular turn of car :
Centripetal force 
             = Maximum frictional force = m mg
\    
mv
d
mg
2
¢
= m
i.e., safe radius = ¢ d =
v
g
2
m
  Þ d d ¢ = 2 
Thus, Assertion and Reason are both
correct and Reason is the correct
explanation of the Assertion.
Option (a) is correct.
2. a
v v ®
® ®
=
-
av
B A
AB
t
=
®
BD
t
AB
i.e.,    | |
| |
a
BD ®
=
®
= av
t
v
t
AB AB
2
Now,    | |
| |
v
AB ®
=
®
= av
t
R
t
AB AB
2
\
| |
| |
a
v
®
®
= =
av
av
v
R
2
2
w (angular velocity)
Thus, assertion is correct.
In circular motion, when speed is
constant, the angular velocity will
obviously be constant; but this reason does 
not lead to the result as explained.
Option (b) is correct.
3. A frame moving in a circle with constant
speed can never be an inertial frame as
the frame is not moving with constant
velocity (due to change in direction).
Reason that the frame is having constant
acceleration is false.
Option (c) is correct.
4. If speed is constant angle between v
®
 and a
®
will always be 90°
\ v a
® ®
× = 0
150 | Mechanics-1
B
D
A
v
B
v
A
®
®
v
A
®
–1
2.5 ms
B
C A
–1
1 ms
–1
2 ms
a = a
c
w
q
®
® ®
If speed is increasing angle q between v
®
and a
®
 will be less than 90°.
\ v a
® ®
× will be positive.
If speed is decreasing angle between v
®
 and 
a
®
 will be greater than 90°.
\    v a
® ®
× will be negative.
Assertion is correct.
Rea son w
® ®
× = v 0 as both are
perpendicular to each other.
Reason is also true but not the correct
explanation of the assertion.
Option (b) is correct.
5. v i
®
= 2
^
 ms
-1
       a i j
®
= - +
^ ^
2  ms
-2
\      a j
®
-
= c 2
2 ^
ms
      a a
c c
® ®
= = | | 2 ms
-2
        v = =
®
| | v 2 ms
-1
       a
v
R
c
=
2
Þ        R
v
a
c
=
2
            = =
( ) 2
2
2
2
 m
         a i
T
®
= -
^
 ms
-2
Speed is decreasing (as a
T
®
 is -ive) at a
rate of 1 ms
-1
 per second i.e., 1 ms
-2
.
Both Assertion and the Reason are correct 
but reason has nothing to do with the
assertion.
Option (b) is correct.
6. a a a
® ® ®
= +
T c
 (Reason)
\ | | a
®
= +
v
r
g
2
2
Þ   | | a
®
> g (Assertion)
We see that both Assertion and the
Reason are correct and the reason is the
correct explanation of assertion.
Option (a) is correct.
7. At points A and C : The bob is
momentarily at rest.
i.e., v = 0 (Reason)
\          | | a
c
v
R
®
= =
2
0
but net acceleration is not zero (see figure) 
i.e., Assertion is false.
\ Option (d) is correct.
Circular Motion | 151
a
c
q
v a
T
a
® ®
® ®
R
2j
v = 2z
a
net
i
x
–
®
a
c
q
v a
T
a
® ®
®
®
a = g
T
a
c
a
A
® ®
®
B
g g
A C
® ®
8. v (speed) = - 4 12 t
For t < 3 (time unit) speed is negative,
which can’t describe a motion. Thus,
assertion is correct.
As speed can be changed linearly with
time, the reason is false.
Option (c) is correct.
9. In circular motion the acceleration
changes regularly where as in projectile
motion it is constant. Thus, in circular
motion we can't apply v u a
® ® ®
= + t directly,
whereas in projectile motion we can say
reason that in circular motion gravity has
no role is wrong.
Option (c) is correct.
10. N mg = cos q
Therefore, assertion is wrong.
Particle performs circular motion due to 
Nsinq 
\
mv
r
N
2
= sinq
Ncosq is balanced by mg (weight of
particle).
Acceleration is not along the surface of the 
funnel. It is along the centre O of the
circle. Thus, reason is true.
Option (d) is correct.
11. Centripetal force 
mv
r
N mg
2
æ
è
ç
ö
ø
÷ = +
i.e., Centripetal force (reason ) ³ wt mg ( ) of 
water
for N = 0, v gr =
If at the top of the circular path v gr ³
i.e., if bucket moved fast, the water will
not fall (Assertion).
As assertion and reason both are true and
reason is the correct explanation of the
assertion of the option would be (a).
Objective Questions (Level 2)
Single Correct Option
1. KE 
1
2
2
mv
æ
è
ç
ö
ø
÷ = Change in PE 
1
2
2
k x ( ) D
é
ë
ê
ù
û
ú
Dx = Length of spring (Collar at B)
- Length of spring (Collar at A)
= + + ( ) 7 5 5
2 2
m - 7m = 6 m
Thus,   
1
2
2
1
2
200 6
2 2
´ ´ = ´ ´ v
i.e.,          v
2
3600 =
\ Normal reaction =
mv
r
2
=
´ 2 3600
5
 = 1440 N
Option (a) is correct.
2. The particle will remain in equilibrium till 
w is constant. Any change in the value of w
will displace the particle up (if w increase)
and down (if w decreases). Thus, the
equilibrium is unstable.
Option (b) is correct.
3. Centripetal force = m mg
or        mr mg w m
2
=
or        
5
4
2
a
g w m =
or          w
2
4
15
=
g
a
 asm =
æ
è
ç
ö
ø
÷
1
3
Option (d) is correct.
152 | Mechanics-1
mg cos q
mg
N
O
N sinq
q
N
mg
4. Acceleration at B = Acceleration at A
\            
v
r
g
2
= sinq
or
2 1 gr
r
g
( cos )
sin
-
=
q
q
or 2 1 ( cos ) sin - = q q
or   [ ( cos )] cos 2 1 1
2 2
- = - q q
or ( cos ) (cos ) 5 3 1 0 q q - - =
As cos q = 1, i.e., q = ° 0 is not possible.
cos q =
3
5
i.e., q =
-
cos
1
3
5
Option (c) is correct.
5. At point P (for the circular motion)
mg N
mv
R
cos q - =
2
If at point P  skier leaves the hemisphere.
           N =0
\     mg
mv
R
cosq =
2
or    mg
m
R
g h
R
cosq = +
æ
è
ç
ö
ø
÷
2
4
or    mg
m
R
g R
R
cos ( cos ) q q = - +
é
ë
ê
ù
û
ú
2 1
4
Þ cos ( cos ) q q = - +
é
ë
ê
ù
û
ú
2 1
1
4
i.e.,        q =
-
cos
1
5
6
Option (c) is correct.
6. Velocity at B
        v gh = 2
         = - 2
2 1
gR(cos cos ) q q
         = ° - ° 2 37 53 gR(cos cos )
         = -
æ
è
ç
ö
ø
÷ 2
4
3
3
5
gR
         =
2
5
gR
For the circular motion at B, when block
just leaves the track
v
R
g
2
1
2
= cos q
or R
v
g
R
1
2
2 2
2
5
= =
cos cos q q
           =
× °
2
5 37
R
cos
           =
×
æ
è
ç
ö
ø
÷
2
5
4
5
R
           =
R
2
Option (c) is correct.
7.    mg
mv
a
cosq=
2
or    g
g
a
a
cosq =
+ × × 4 2
4
2
or  g
g
a
a
u g
a
a
-
=
+ × ×
4
2
4
2
or    g a u g
a
× = +
3
4
2
4
2
Þ        u
ag
=
2
Option (c) is correct.
8.      
v
r r
2
2
4
=
Þ      v
r
=
2
\    Momentum = mv
               =
2m
r
9.     N mr cosq w =
2
and  N mg sin q =
\       N m g r = +
2 2 4
w   
  = +
æ
è
ç
ö
ø
÷ m g r
T
2 2
4
2p
       = +
æ
è
ç
ö
ø
÷ 10 10
2
2 2
4
( ) 0.5
1.5
p
a
           =128 N
Option (b) is correct.
10.    q w a = +
0
2
1
2
t t
     = +
é
ë
ê
ù
û
ú
w a
0
1
2
t t
Circular Motion | 153
Page 5


14. Let all the particles meet at time t
(seconds)
\ Distance travelled by B in t second : 
Distance travelled by C in t second
= 2.5t t :2
= 5 4 :   
Option (c) is correct.
JEE Corner
Assertion and Reason
1. For stopping car : 
Maximum retardation 
=
Maximum frictional force
m
     = = =
m m
m
N
m
mg
m
g
 v u as
2 2
2 = +
  0 2
2 2
= + - v g d ( )( ) m [ Q Initial velocity = v]
\      d
v
g
=
2
2m
For circular turn of car :
Centripetal force 
             = Maximum frictional force = m mg
\    
mv
d
mg
2
¢
= m
i.e., safe radius = ¢ d =
v
g
2
m
  Þ d d ¢ = 2 
Thus, Assertion and Reason are both
correct and Reason is the correct
explanation of the Assertion.
Option (a) is correct.
2. a
v v ®
® ®
=
-
av
B A
AB
t
=
®
BD
t
AB
i.e.,    | |
| |
a
BD ®
=
®
= av
t
v
t
AB AB
2
Now,    | |
| |
v
AB ®
=
®
= av
t
R
t
AB AB
2
\
| |
| |
a
v
®
®
= =
av
av
v
R
2
2
w (angular velocity)
Thus, assertion is correct.
In circular motion, when speed is
constant, the angular velocity will
obviously be constant; but this reason does 
not lead to the result as explained.
Option (b) is correct.
3. A frame moving in a circle with constant
speed can never be an inertial frame as
the frame is not moving with constant
velocity (due to change in direction).
Reason that the frame is having constant
acceleration is false.
Option (c) is correct.
4. If speed is constant angle between v
®
 and a
®
will always be 90°
\ v a
® ®
× = 0
150 | Mechanics-1
B
D
A
v
B
v
A
®
®
v
A
®
–1
2.5 ms
B
C A
–1
1 ms
–1
2 ms
a = a
c
w
q
®
® ®
If speed is increasing angle q between v
®
and a
®
 will be less than 90°.
\ v a
® ®
× will be positive.
If speed is decreasing angle between v
®
 and 
a
®
 will be greater than 90°.
\    v a
® ®
× will be negative.
Assertion is correct.
Rea son w
® ®
× = v 0 as both are
perpendicular to each other.
Reason is also true but not the correct
explanation of the assertion.
Option (b) is correct.
5. v i
®
= 2
^
 ms
-1
       a i j
®
= - +
^ ^
2  ms
-2
\      a j
®
-
= c 2
2 ^
ms
      a a
c c
® ®
= = | | 2 ms
-2
        v = =
®
| | v 2 ms
-1
       a
v
R
c
=
2
Þ        R
v
a
c
=
2
            = =
( ) 2
2
2
2
 m
         a i
T
®
= -
^
 ms
-2
Speed is decreasing (as a
T
®
 is -ive) at a
rate of 1 ms
-1
 per second i.e., 1 ms
-2
.
Both Assertion and the Reason are correct 
but reason has nothing to do with the
assertion.
Option (b) is correct.
6. a a a
® ® ®
= +
T c
 (Reason)
\ | | a
®
= +
v
r
g
2
2
Þ   | | a
®
> g (Assertion)
We see that both Assertion and the
Reason are correct and the reason is the
correct explanation of assertion.
Option (a) is correct.
7. At points A and C : The bob is
momentarily at rest.
i.e., v = 0 (Reason)
\          | | a
c
v
R
®
= =
2
0
but net acceleration is not zero (see figure) 
i.e., Assertion is false.
\ Option (d) is correct.
Circular Motion | 151
a
c
q
v a
T
a
® ®
® ®
R
2j
v = 2z
a
net
i
x
–
®
a
c
q
v a
T
a
® ®
®
®
a = g
T
a
c
a
A
® ®
®
B
g g
A C
® ®
8. v (speed) = - 4 12 t
For t < 3 (time unit) speed is negative,
which can’t describe a motion. Thus,
assertion is correct.
As speed can be changed linearly with
time, the reason is false.
Option (c) is correct.
9. In circular motion the acceleration
changes regularly where as in projectile
motion it is constant. Thus, in circular
motion we can't apply v u a
® ® ®
= + t directly,
whereas in projectile motion we can say
reason that in circular motion gravity has
no role is wrong.
Option (c) is correct.
10. N mg = cos q
Therefore, assertion is wrong.
Particle performs circular motion due to 
Nsinq 
\
mv
r
N
2
= sinq
Ncosq is balanced by mg (weight of
particle).
Acceleration is not along the surface of the 
funnel. It is along the centre O of the
circle. Thus, reason is true.
Option (d) is correct.
11. Centripetal force 
mv
r
N mg
2
æ
è
ç
ö
ø
÷ = +
i.e., Centripetal force (reason ) ³ wt mg ( ) of 
water
for N = 0, v gr =
If at the top of the circular path v gr ³
i.e., if bucket moved fast, the water will
not fall (Assertion).
As assertion and reason both are true and
reason is the correct explanation of the
assertion of the option would be (a).
Objective Questions (Level 2)
Single Correct Option
1. KE 
1
2
2
mv
æ
è
ç
ö
ø
÷ = Change in PE 
1
2
2
k x ( ) D
é
ë
ê
ù
û
ú
Dx = Length of spring (Collar at B)
- Length of spring (Collar at A)
= + + ( ) 7 5 5
2 2
m - 7m = 6 m
Thus,   
1
2
2
1
2
200 6
2 2
´ ´ = ´ ´ v
i.e.,          v
2
3600 =
\ Normal reaction =
mv
r
2
=
´ 2 3600
5
 = 1440 N
Option (a) is correct.
2. The particle will remain in equilibrium till 
w is constant. Any change in the value of w
will displace the particle up (if w increase)
and down (if w decreases). Thus, the
equilibrium is unstable.
Option (b) is correct.
3. Centripetal force = m mg
or        mr mg w m
2
=
or        
5
4
2
a
g w m =
or          w
2
4
15
=
g
a
 asm =
æ
è
ç
ö
ø
÷
1
3
Option (d) is correct.
152 | Mechanics-1
mg cos q
mg
N
O
N sinq
q
N
mg
4. Acceleration at B = Acceleration at A
\            
v
r
g
2
= sinq
or
2 1 gr
r
g
( cos )
sin
-
=
q
q
or 2 1 ( cos ) sin - = q q
or   [ ( cos )] cos 2 1 1
2 2
- = - q q
or ( cos ) (cos ) 5 3 1 0 q q - - =
As cos q = 1, i.e., q = ° 0 is not possible.
cos q =
3
5
i.e., q =
-
cos
1
3
5
Option (c) is correct.
5. At point P (for the circular motion)
mg N
mv
R
cos q - =
2
If at point P  skier leaves the hemisphere.
           N =0
\     mg
mv
R
cosq =
2
or    mg
m
R
g h
R
cosq = +
æ
è
ç
ö
ø
÷
2
4
or    mg
m
R
g R
R
cos ( cos ) q q = - +
é
ë
ê
ù
û
ú
2 1
4
Þ cos ( cos ) q q = - +
é
ë
ê
ù
û
ú
2 1
1
4
i.e.,        q =
-
cos
1
5
6
Option (c) is correct.
6. Velocity at B
        v gh = 2
         = - 2
2 1
gR(cos cos ) q q
         = ° - ° 2 37 53 gR(cos cos )
         = -
æ
è
ç
ö
ø
÷ 2
4
3
3
5
gR
         =
2
5
gR
For the circular motion at B, when block
just leaves the track
v
R
g
2
1
2
= cos q
or R
v
g
R
1
2
2 2
2
5
= =
cos cos q q
           =
× °
2
5 37
R
cos
           =
×
æ
è
ç
ö
ø
÷
2
5
4
5
R
           =
R
2
Option (c) is correct.
7.    mg
mv
a
cosq=
2
or    g
g
a
a
cosq =
+ × × 4 2
4
2
or  g
g
a
a
u g
a
a
-
=
+ × ×
4
2
4
2
or    g a u g
a
× = +
3
4
2
4
2
Þ        u
ag
=
2
Option (c) is correct.
8.      
v
r r
2
2
4
=
Þ      v
r
=
2
\    Momentum = mv
               =
2m
r
9.     N mr cosq w =
2
and  N mg sin q =
\       N m g r = +
2 2 4
w   
  = +
æ
è
ç
ö
ø
÷ m g r
T
2 2
4
2p
       = +
æ
è
ç
ö
ø
÷ 10 10
2
2 2
4
( ) 0.5
1.5
p
a
           =128 N
Option (b) is correct.
10.    q w a = +
0
2
1
2
t t
     = +
é
ë
ê
ù
û
ú
w a
0
1
2
t t
Circular Motion | 153
       = + - ×
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
v
R R
v
R
R
v
t
1
2
1
4
4
2
p
p
       = -
é
ë
ê
ù
û
ú
v
R
v
R
R
v 2
4p
       = 4p
       =2 rev.
11.      H gT =
1
2
2
Þ     T
H
g
=
2
       w
p
p = =
2
2
2 T
g
H
        = p
2g
H
12. w (minute hand) =
-
2
3600
1
p
rad s
w (second hand) =
2
60
p
 rads
-1
For second hand to meet minute hand for
the first time.
2p + Angle moved by minute hand in t
second
= Angle moved by second hand in t second
or   2
2
3600
2
60
p
p p
+ = t t
 1
60 3600
= -
t t
1
60
59
60
= ´
t
  
Þ t =
3600
59
 s   
Option (d) is correct.
13.   ( ) ( ) ( ) PR QR PQ
2 2 2
+ =
( ) ( ) v v
A B
× + × = 2 2 30
2 2 2
        v v
A B
2 2
225 + = …(i)
Further, 
PR
PQ
= ° cos30
v
A
×
=
2
30
3
2
Þ v
A
=7.5 3 ms
-1
Substituting value of v
A
 in Eq. (i),
v
B
= 7.5  ms
-1
14. mg N
mv
r
cos q - =
2
When breaks off N = 0
\ mg
mv
r
cos q =
2
or g
gr
r
cos
( cos )
q
q
=
- 2 1
or cos ( cos ) q q = - 2 1
or         cosq =
2
3
Acceleration of particle when it leaves
sphere
= gsinq
=
g 5
3
   
Option (b) is correct.
154 | Mechanics-1
H
R
60° 30°
30 m
B'
v
B
t = 2 s
R
v
B
= AD
q
Acceleration
= g sin q
mg
N
O
mg sin q
q
Ö5
2
3
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