NEET Exam  >  NEET Notes  >  Physics Class 11  >  HC Verma Solutions: Chapter 7 - Circular Motion

HC Verma Solutions: Chapter 7 - Circular Motion | Physics Class 11 - NEET PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


7.1
SOLUTIONS TO CONCEPTS     circular motion;;
CHAPTER 7
1. Distance between Earth & Moon 
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6 
sec
v = 
T
r 2 ?
= 
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a = 
r
v
2
= 
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V = 
T
R 2 ?
= 
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a = 
R
V
2
= 
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a = 
r
v
2
= 
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a = 
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg, 
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is 
r
mv
2
= 
30
) 10 ( 150
2
?
= 
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? = 
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? = 
rmg
mv
2
= 
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? = 
rg
v
2
= 
10 30
100
?
= (1/3) 
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? = 
rg
v
2
? ? = tan 
–1
rg
v
2
= tan
–1
100
25
= tan 
–1
(1/4)
R
mv
2
/R
mg
Page 2


7.1
SOLUTIONS TO CONCEPTS     circular motion;;
CHAPTER 7
1. Distance between Earth & Moon 
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6 
sec
v = 
T
r 2 ?
= 
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a = 
r
v
2
= 
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V = 
T
R 2 ?
= 
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a = 
R
V
2
= 
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a = 
r
v
2
= 
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a = 
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg, 
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is 
r
mv
2
= 
30
) 10 ( 150
2
?
= 
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? = 
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? = 
rmg
mv
2
= 
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? = 
rg
v
2
= 
10 30
100
?
= (1/3) 
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? = 
rg
v
2
? ? = tan 
–1
rg
v
2
= tan
–1
100
25
= tan 
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So 
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30° 
Radius = r = 50m
tan ? = 
rg
v
2
? tan 30° = 
rg
v
2
?
3
1
= 
rg
v
2
? v
2
= 
3
rg
= 
3
10 50 ?
? v = 
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre. 
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
= 
2
2
r
q
k ? v
2
= 
rm
kq
2
= 
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13 
= 4.7 × 10
12
? v = 
12
10 7 . 4 ? = 2.2 × 10
6 
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2 
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle 
also exerts a force of 14.8N on the blade along its surface. 
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 
3
1
33 rpm.
n = 
3
1
33 rpm = 
60 3
100
?
rps
?? = 2 ? n = 2 ? × 
180
100
= 
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? = 
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Page 3


7.1
SOLUTIONS TO CONCEPTS     circular motion;;
CHAPTER 7
1. Distance between Earth & Moon 
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6 
sec
v = 
T
r 2 ?
= 
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a = 
r
v
2
= 
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V = 
T
R 2 ?
= 
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a = 
R
V
2
= 
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a = 
r
v
2
= 
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a = 
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg, 
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is 
r
mv
2
= 
30
) 10 ( 150
2
?
= 
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? = 
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? = 
rmg
mv
2
= 
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? = 
rg
v
2
= 
10 30
100
?
= (1/3) 
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? = 
rg
v
2
? ? = tan 
–1
rg
v
2
= tan
–1
100
25
= tan 
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So 
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30° 
Radius = r = 50m
tan ? = 
rg
v
2
? tan 30° = 
rg
v
2
?
3
1
= 
rg
v
2
? v
2
= 
3
rg
= 
3
10 50 ?
? v = 
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre. 
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
= 
2
2
r
q
k ? v
2
= 
rm
kq
2
= 
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13 
= 4.7 × 10
12
? v = 
12
10 7 . 4 ? = 2.2 × 10
6 
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2 
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle 
also exerts a force of 14.8N on the blade along its surface. 
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 
3
1
33 rpm.
n = 
3
1
33 rpm = 
60 3
100
?
rps
?? = 2 ? n = 2 ? × 
180
100
= 
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? = 
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? = 
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
= 
rmg
mv
2
? tan ? = 
rg
v
2
? ? = tan 
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg + 
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg + 
r
mv
2
= 
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N 
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram, 
T – mg cos ? = 
R
mv
2
? T = 
R
mv
2
+ mg cos ?
? T = 
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 × 
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? = 
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
= 
mg
) r m mg ( mg
2
? ? ?
= 
g
2
?
= 
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T = 
?
? 2
= 
8 . 9
10 6400 2
2
3
? ?
? ? sec = 
49
10 64
2
6
?
? ? sec = 
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos 
? ?
T
m ?
2
/R
mg
R
Page 4


7.1
SOLUTIONS TO CONCEPTS     circular motion;;
CHAPTER 7
1. Distance between Earth & Moon 
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6 
sec
v = 
T
r 2 ?
= 
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a = 
r
v
2
= 
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V = 
T
R 2 ?
= 
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a = 
R
V
2
= 
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a = 
r
v
2
= 
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a = 
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg, 
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is 
r
mv
2
= 
30
) 10 ( 150
2
?
= 
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? = 
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? = 
rmg
mv
2
= 
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? = 
rg
v
2
= 
10 30
100
?
= (1/3) 
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? = 
rg
v
2
? ? = tan 
–1
rg
v
2
= tan
–1
100
25
= tan 
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So 
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30° 
Radius = r = 50m
tan ? = 
rg
v
2
? tan 30° = 
rg
v
2
?
3
1
= 
rg
v
2
? v
2
= 
3
rg
= 
3
10 50 ?
? v = 
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre. 
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
= 
2
2
r
q
k ? v
2
= 
rm
kq
2
= 
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13 
= 4.7 × 10
12
? v = 
12
10 7 . 4 ? = 2.2 × 10
6 
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2 
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle 
also exerts a force of 14.8N on the blade along its surface. 
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 
3
1
33 rpm.
n = 
3
1
33 rpm = 
60 3
100
?
rps
?? = 2 ? n = 2 ? × 
180
100
= 
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? = 
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? = 
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
= 
rmg
mv
2
? tan ? = 
rg
v
2
? ? = tan 
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg + 
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg + 
r
mv
2
= 
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N 
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram, 
T – mg cos ? = 
R
mv
2
? T = 
R
mv
2
+ mg cos ?
? T = 
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 × 
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? = 
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
= 
mg
) r m mg ( mg
2
? ? ?
= 
g
2
?
= 
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T = 
?
? 2
= 
8 . 9
10 6400 2
2
3
? ?
? ? sec = 
49
10 64
2
6
?
? ? sec = 
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos 
? ?
T
m ?
2
/R
mg
R
Chapter 7
7.4
18. Given, v = 36km/hr = 10m/s, r = 20m, ? = 0.4
The road is banked with an angle,
? = tan 
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan 
–1
?
?
?
?
?
?
?10 20
100
= tan
–1 ?
?
?
?
?
?
2
1
or tan ? = 0.5
When the car travels at max. speed so that it slips upward, ?R
1
acts downward as shown in Fig.1
So, R
1
– mg cos ? –
r
mv
2
1
sin ? = 0 ..(i)
And ?R
1
+ mg sin ? –
r
mv
2
1
cos ? = 0 ..(ii)
Solving the equation we get, 
V
1
= 
? ? ?
? ? ?
tan 1
tan
rg = 
2 . 1
1 . 0
10 20 ? ? = 4.082 m/s = 14.7 km/hr
So, the possible speeds are between 14.7 km/hr and 54km/hr. 
19. R = radius of the bridge
L = total length of the over bridge
a) At the highest pt.
mg = 
R
mv
2
? v
2
= Rg ? v = Rg
b) Given, v = Rg
2
1
suppose it loses contact at B. So, at B, mg cos ? = 
R
mv
2
? v
2 
= Rg cos ?
?
2
2
Rv
?
?
?
?
?
?
?
?
= Rg cos ? ?
2
Rg
= Rg cos ? ? cos ? = 1/2 ? ? = 60° = ?/3
? = 
r
?
? l = r ? = 
3
R ?
So, it will lose contact at distance 
3
R ?
from highest point
c) Let the uniform speed on the bridge be v.
The chances of losing contact is maximum at the end of the bridge for which ? = 
R 2
L
.
So, 
R
mv
2
= mg cos ? ? v = ?
?
?
?
?
?
R 2
L
cos gR
?
20. Since the motion is nonuniform, the acceleration has both radial & tangential 
component 
a
r
= 
r
v
2
a
t
= 
dt
dv
=a
Resultant magnitude = 
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
Now ?N = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ? mg = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ?
2
g
2
= 
2
4
a
2 r
v
?
?
?
?
?
?
?
?
?
? v
4
= ( ?
2
g
2
– a
2
) r
2
? v = [( ?
2
g
2
– a
2
) r
2
]
1/4
? ?
?R 1
? ?
mv
1
2
/r
mg
R 1
? ?
mv
2
2
/r
?R 2
? ?
mg
R 2
? ?
2 ? ?
mv
2
/R
mg
2 ?= L/R ?
? ?
mv
2
/R
mg
2 ? ?
2 ?= L/R ?
? ?
mv
2
/R
2 ? ?
2 ?= L/R ?
mv
2
/R ? mg
m dv/dt
m
N
mv
2
/R
m 
Page 5


7.1
SOLUTIONS TO CONCEPTS     circular motion;;
CHAPTER 7
1. Distance between Earth & Moon 
r = 3.85 × 10
5
km = 3.85 × 10
8
m
T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10
6 
sec
v = 
T
r 2 ?
= 
6
8
10 36 . 2
10 85 . 3 14 . 3 2
?
? ? ?
= 1025.42m/sec
a = 
r
v
2
= 
8
2
10 85 . 3
) 42 . 1025 (
?
= 0.00273m/sec
2
= 2.73 × 10
–3
m/sec
2
2. Diameter of earth = 12800km
Radius R = 6400km = 64 × 10
5
m
V = 
T
R 2 ?
= 
3600 24
10 64 14 . 3 2
5
?
? ? ?
m/sec = 465.185
a = 
R
V
2
= 
5
2
10 64
) 5185 . 46 (
?
= 0.0338m/sec
2
3. V = 2t, r = 1cm
a) Radial acceleration at t = 1 sec.
a = 
r
v
2
= 
1
2
2
= 4cm/sec
2
b) Tangential acceleration at t = 1sec.
a = 
dt
dv
= ) t 2 (
dt
d
= 2cm/sec
2
c) Magnitude of acceleration at t = 1sec
a = 
2 2
2 4 ? = 20 cm/sec
2
4. Given that m = 150kg, 
v= 36km/hr = 10m/sec, r = 30m
Horizontal force needed is 
r
mv
2
= 
30
) 10 ( 150
2
?
= 
30
100 150 ?
= 500N
5. in the diagram
R cos ? = mg ..(i)
R sin ? = 
r
mv
2
..(ii)
Dividing equation (i) with equation (ii)
Tan ? = 
rmg
mv
2
= 
rg
v
2
v = 36km/hr = 10m/sec, r = 30m
Tan ? = 
rg
v
2
= 
10 30
100
?
= (1/3) 
? ? = tan
–1
(1/3) ?
6. Radius of Park = r = 10m
speed of vehicle = 18km/hr = 5 m/sec
Angle of banking tan ? = 
rg
v
2
? ? = tan 
–1
rg
v
2
= tan
–1
100
25
= tan 
–1
(1/4)
R
mv
2
/R
mg
Chapter 7
7.2
7. The road is horizontal (no banking)
R
mv
2
= ?N
and N = mg
So 
R
mv
2
= ? mg v = 5m/sec, R = 10m
?
10
25
= ?g ? ? =
100
25
= 0.25 ?
8. Angle of banking = ? = 30° 
Radius = r = 50m
tan ? = 
rg
v
2
? tan 30° = 
rg
v
2
?
3
1
= 
rg
v
2
? v
2
= 
3
rg
= 
3
10 50 ?
? v = 
3
500
= 17m/sec.
9. Electron revolves around the proton in a circle having proton at the centre. 
Centripetal force is provided by coulomb attraction.
r = 5.3 ?t 10
–11
m m = mass of electron = 9.1 × 10
–3
kg.
charge of electron = 1.6 × 10
–19
c.
r
mv
2
= 
2
2
r
q
k ? v
2
= 
rm
kq
2
= 
31 11
38 9
10 1 . 9 10 3 . 5
10 6 . 1 6 . 1 10 9
? ?
?
? ? ?
? ? ? ?
= 
13
10
23 . 48
04 . 23
?
? v
2
= 0.477 × 10
13 
= 4.7 × 10
12
? v = 
12
10 7 . 4 ? = 2.2 × 10
6 
m/sec
10. At the highest point of a vertical circle
R
mv
2
= mg
? v
2 
= Rg ? v = Rg
11. A celling fan has a diameter = 120cm.
?Radius = r = 60cm = 0/6m
Mass of particle on the outer end of a blade is 1g.
n = 1500 rev/min = 25 rev/sec
? = 2 ??n = 2 ? ×25 = 157.14
Force of the particle on the blade = Mr ?
2
= (0.001) × 0.6 × (157.14) = 14.8N
The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle 
also exerts a force of 14.8N on the blade along its surface. 
12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 
3
1
33 rpm.
n = 
3
1
33 rpm = 
60 3
100
?
rps
?? = 2 ? n = 2 ? × 
180
100
= 
9
10 ?
rad/sec
r = 10cm =0.1m, g = 10m/sec
2
?mg ? mr ?
2
? ? = 
g
r
2
?
?
10
9
10
1 . 0
2
?
?
?
?
?
? ?
?
? ? ?
81
2
?
?
R
mg
??g mv
2
/R
Chapter 7
7.3
13. A pendulum is suspended from the ceiling of a car taking a turn
r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec
2
From the figure T sin ? = 
r
mv
2
..(i)
T cos ? = mg ..(ii)
?
?
?
cos
sin
= 
rmg
mv
2
? tan ? = 
rg
v
2
? ? = tan 
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan
–1
10 10
100
?
= tan
–1
(1) ? ? = 45° ?
14. At the lowest pt.
T = mg + 
r
mv
2
Here m = 100g = 1/10 kg, r = 1m, v = 1.4 m/sec
T = mg + 
r
mv
2
= 
10
) 4 . 1 (
8 . 9
10
1
2
? ? = 0.98 + 0.196 = 1.176 = 1.2 N 
15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian.
m = 100g = 0.1kg, r = 1m, v = 1.4m/sec.
From the diagram, 
T – mg cos ? = 
R
mv
2
? T = 
R
mv
2
+ mg cos ?
? T = 
?
?
?
?
?
?
?
?
?
? ? ? ?
?
2
1 8 . 9 ) 1 . 0 (
1
) 4 . 1 ( 1 . 0
2 2
? T = 0.196 + 9.8 × 
?
?
?
?
?
?
?
?
?
2
) 2 (.
1
2
( ? cos ? = 
2
1
2
?
? for small ?)
? T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N ? 1.16 N ?
16. At the extreme position, velocity of the pendulum is zero.
So there is no centrifugal force.
So T = mg cos ?
o
?
17. a) Net force on the spring balance.
R = mg – m ?
2
r
So, fraction less than the true weight (3mg) is
= 
mg
) r m mg ( mg
2
? ? ?
= 
g
2
?
= 
10
10 6400
3600 24
2
3
2
?
? ?
?
?
?
?
?
?
?
= 3.5 × 10
–3
b) When the balance reading is half the true weight, ?
mg
) r m mg ( mg
2
? ? ?
= 1/2
?
2
r = g/2 ? ????
r 2
g
???
3
10 6400 2
10
? ?
rad/sec
? Duration of the day is
T = 
?
? 2
= 
8 . 9
10 6400 2
2
3
? ?
? ? sec = 
49
10 64
2
6
?
? ? sec = 
3600 7
8000 2
?
? ?
hr = 2hr
mv
2
/R
mg
? ?
mg ?
mv
2
/r ?
T
mg sin ? ?
mg cos ? ?
T
mg sin ? ?
mg cos 
? ?
T
m ?
2
/R
mg
R
Chapter 7
7.4
18. Given, v = 36km/hr = 10m/s, r = 20m, ? = 0.4
The road is banked with an angle,
? = tan 
–1
?
?
?
?
?
?
?
?
rg
v
2
= tan 
–1
?
?
?
?
?
?
?10 20
100
= tan
–1 ?
?
?
?
?
?
2
1
or tan ? = 0.5
When the car travels at max. speed so that it slips upward, ?R
1
acts downward as shown in Fig.1
So, R
1
– mg cos ? –
r
mv
2
1
sin ? = 0 ..(i)
And ?R
1
+ mg sin ? –
r
mv
2
1
cos ? = 0 ..(ii)
Solving the equation we get, 
V
1
= 
? ? ?
? ? ?
tan 1
tan
rg = 
2 . 1
1 . 0
10 20 ? ? = 4.082 m/s = 14.7 km/hr
So, the possible speeds are between 14.7 km/hr and 54km/hr. 
19. R = radius of the bridge
L = total length of the over bridge
a) At the highest pt.
mg = 
R
mv
2
? v
2
= Rg ? v = Rg
b) Given, v = Rg
2
1
suppose it loses contact at B. So, at B, mg cos ? = 
R
mv
2
? v
2 
= Rg cos ?
?
2
2
Rv
?
?
?
?
?
?
?
?
= Rg cos ? ?
2
Rg
= Rg cos ? ? cos ? = 1/2 ? ? = 60° = ?/3
? = 
r
?
? l = r ? = 
3
R ?
So, it will lose contact at distance 
3
R ?
from highest point
c) Let the uniform speed on the bridge be v.
The chances of losing contact is maximum at the end of the bridge for which ? = 
R 2
L
.
So, 
R
mv
2
= mg cos ? ? v = ?
?
?
?
?
?
R 2
L
cos gR
?
20. Since the motion is nonuniform, the acceleration has both radial & tangential 
component 
a
r
= 
r
v
2
a
t
= 
dt
dv
=a
Resultant magnitude = 
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
Now ?N = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ? mg = m
2
2
2
a
r
v
?
?
?
?
?
?
?
?
?
? ?
2
g
2
= 
2
4
a
2 r
v
?
?
?
?
?
?
?
?
?
? v
4
= ( ?
2
g
2
– a
2
) r
2
? v = [( ?
2
g
2
– a
2
) r
2
]
1/4
? ?
?R 1
? ?
mv
1
2
/r
mg
R 1
? ?
mv
2
2
/r
?R 2
? ?
mg
R 2
? ?
2 ? ?
mv
2
/R
mg
2 ?= L/R ?
? ?
mv
2
/R
mg
2 ? ?
2 ?= L/R ?
? ?
mv
2
/R
2 ? ?
2 ?= L/R ?
mv
2
/R ? mg
m dv/dt
m
N
mv
2
/R
m 
Chapter 7
7.5
21. a) When the ruler makes uniform circular motion in the horizontal 
plane, (fig–a)
? mg = m ?
?
?
L
?
?
?
??
L
g ?
?
b) When the ruler makes uniformly accelerated circular motion,(fig–b)
? mg = 
2 2 2
2
) mL ( ) L m ( ? ? ? ? ?
2
4
+ ?
2
= 
2
2 2
L
g ?
? ?
2
= 
4 / 1
2
2
L
g
?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
? ?
(When viewed from top)
22. Radius of the curves = 100m 
Weight = 100kg
Velocity = 18km/hr = 5m/sec
a) at B mg –
R
mv
2
= N ? N = (100 × 10) –
100
25 100 ?
= 1000 – 25 = 975N
At d, N = mg + 
R
mv
2
= 1000 + 25 = 1025 N
b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero.
At ‘C’, mg sin ? = F ? F = 1000 × 
2
1
= 707N
c) (i) Before ‘C’ mg cos ? – N = 
R
mv
2
? N = mg cos ? –
R
mv
2
= 707 – 25 = 683N
(ii) N – mg cos ? = 
R
mv
2
? N = 
R
mv
2
+ mg cos ? = 25 + 707 = 732N
d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where 
N is minimum)
Now, ? N = mg sin ? ? ? × 682 = 707
So, ? = 1.037 ?
23. d = 3m ? R = 1.5m
R = distance from the centre to one of the kids
N = 20 rev per min = 20/60 = 1/3 rev per sec
??= 2 ?r = 2 ?/3
m = 15kg
? Frictional force F = mr ?
2
= 15 × (1.5) × 
9
) 2 (
2
?
= 5 × (0.5) × 4 ?
2
= 10 ?
2
? Frictional force on one of the kids is 10 ?
2
?
24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force 
acts downward.
Here, r = R sin ?
From FBD –1 
R
1
– mg cos ? – m ?
?
2
(R sin ?) sin ? = 0 ..(i) [because r = R sin ?]
and ?R
1
mg sin ? – m ?
1
2
(R sin ?) cos ? = 0 ..(ii)
Substituting the value of R
1
from Eq (i) in Eq(ii), it can be found out that
?
1
= 
2 / 1
) sin (cos sin R
) cos (sin g
?
?
?
?
?
?
? ? ? ? ?
? ? ? ?
Again, for minimum speed, the frictional force 
?R
2
acts upward. From FBD–2, it can be proved 
that,
??? 1
2
L ? mg
mg
? ? ? ?
R
L
(Fig–a)
(Fig–b)
m ? 2
2
L ? mg
mL ? ?
?R 1
? ?
m ? 1
2
r
R 1  
(FBD – 1)
?R 2
? ?
m ? 2
2
r
R 2  
(FBD – 2)
?
E
D
C
B
A
mv
2
/R
mg
N
B
15kg
mg
m ?
2
r ? F
15kg
Read More
102 videos|411 docs|121 tests

Top Courses for NEET

FAQs on HC Verma Solutions: Chapter 7 - Circular Motion - Physics Class 11 - NEET

1. What is circular motion?
Ans. Circular motion is the movement of an object in a circular path around a fixed point or axis. In this type of motion, the object constantly changes its direction but maintains a constant distance from the center.
2. What are the two types of circular motion?
Ans. The two types of circular motion are uniform circular motion and non-uniform circular motion. In uniform circular motion, the object moves at a constant speed along a circular path, while in non-uniform circular motion, the object's speed varies along the path.
3. How is centripetal force related to circular motion?
Ans. Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and is responsible for continuously changing the object's direction. The magnitude of centripetal force is directly proportional to the mass of the object and the square of its velocity, and inversely proportional to the radius of the circular path.
4. What is the difference between linear and circular motion?
Ans. Linear motion refers to the movement of an object in a straight line, while circular motion involves the movement of an object in a circular path. In linear motion, the object's speed and direction are constant, whereas in circular motion, the object's speed remains constant, but the direction continuously changes.
5. How does the concept of angular velocity relate to circular motion?
Ans. Angular velocity is a measure of how quickly an object rotates around a fixed point or axis in circular motion. It is defined as the rate of change of angular displacement with respect to time. The magnitude of angular velocity depends on the object's rotational speed and the radius of the circular path it follows.
102 videos|411 docs|121 tests
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

practice quizzes

,

MCQs

,

Semester Notes

,

Previous Year Questions with Solutions

,

HC Verma Solutions: Chapter 7 - Circular Motion | Physics Class 11 - NEET

,

Viva Questions

,

shortcuts and tricks

,

Free

,

Summary

,

ppt

,

Objective type Questions

,

Important questions

,

video lectures

,

HC Verma Solutions: Chapter 7 - Circular Motion | Physics Class 11 - NEET

,

pdf

,

Exam

,

past year papers

,

Sample Paper

,

study material

,

HC Verma Solutions: Chapter 7 - Circular Motion | Physics Class 11 - NEET

,

Extra Questions

,

mock tests for examination

;