NEET  >  DPP for NEET: Daily Practice Problems, Ch 7: Motion in a Plane- 2 (Solutions)

# DPP for NEET: Daily Practice Problems, Ch 7: Motion in a Plane- 2 (Solutions) - Physics Class 11

``` Page 1

(1) (b) We have angular displacement
=
linear displacement
Þ Dq =
S
r
D
Here, DS = n (2pr) = 1.5 (2p × 2 × 10
–2
) = 6p × 10
–2
\ Dq =
2
2
6 10
2 10
-
-
p´
´
(2) (a) We have
av
Total angular displacement
Total time
®
w=
For first one third part of circle,
angular displacement,   q
1
=
1
S
r
=
2 r/3
r
p
For second one third part of circle,
q
2
=
2r/32
r3
pp
Total angular displacement,
q  = q
1
+ q
2
Total time = 2 + 1 = 3 sec
\ av
®
w =
4 /3
3
p
4
6
p
=
2
3
p
(3) (c) Angular speed of hour hand,
w
1
=
t
Dq
D
=
2
12 60
p
´
Angular speed of minute hand,
w
2
=
2
60
p
2
1
12
1
w
=
w
(4) (d) We have q = w
0
t +
1
2
at
2

Þ
d
dt
q
= w
0
+ at
This is angular velocity at time t.
Now angular velocity at t = 2 sec will be
w =
t 2 sec
d
dt
=
q æö
ç÷
èø
= w
0
+ 2a  = 1 + 2 × 1.5 = 4 rad/sec
(5) (d) The distance covered in completing the circle is
2pr = 2p × 10 cm
The linear speed is v =
2r
t
p
=
2 10
4
p´
= 5p cm/s
The linear acceleration is,
a =
2
v
r
=
2
(5)
10
p
= 2.5 p
2
cm/s
2
This acceleration is directed towards the centre of the circle
(6) (d) We know that
Time period =
Circumference
Criticalspeed
=
2r
gr
p
=
2224
7 104
´´
´´
= 4 sec
(7) (b) V elocity =
Circumference
Timeof revolution
=
2r
60
p
=
21
60
p´
=
30
p
cm/s
Change in velocity Dv =
22
30 30
pp æö æö
+
ç÷ ç÷
èø èø
= 2
30
p
cm/s
(8) (a) Let the radius of the orbit be r and the number of
revolutions per second be n. Then the velocity of electron
is given by  v = 2pnr,
\ Acceleration a =
2
v
r
=
2 22
4 rn
r
p
=  4 p
2
r n
2
Substituting the given values, we have
a = 4 × (3.14)
2
× (5.3 × 10
–11
) (6.6 × 10
15
)
2
= 9.1 × 10
22
m/s
2
towards the nucleus.
The centripetal force is
F
C
= ma = (9.1 × 10
–31
) (9.1 × 10
22
)
= 8.3 × 10
–8
N towards the nucleus.
(9) (b) Given that radius of horizontal loop
r = 1 km = 1000 m
Speed v = 900 km/h =
9000 5
18
´
= 250 m/s
Centripetal acceleration a
c
=
2
v
r
=
250 250
1000
´
= 62.5 m/s
2
\
on accelerati nal Gravitatio
on accelerati l Centripeta
=
c
a
g
=
62.5
9.8
= 6.38 : 1
(10) (b) We know that, tan q =
2
v
rg
=
2
5
18
18
100 10
æö
´
ç÷
èø
´
=
1
40
Þ q = tan
–1
1
40
(11) (a) The angular velocity is  w =
v
r
Hence, v = 10 m/s
r = 20 cm = 0.2 m  \ w = 50 rad/s
Page 2

(1) (b) We have angular displacement
=
linear displacement
Þ Dq =
S
r
D
Here, DS = n (2pr) = 1.5 (2p × 2 × 10
–2
) = 6p × 10
–2
\ Dq =
2
2
6 10
2 10
-
-
p´
´
(2) (a) We have
av
Total angular displacement
Total time
®
w=
For first one third part of circle,
angular displacement,   q
1
=
1
S
r
=
2 r/3
r
p
For second one third part of circle,
q
2
=
2r/32
r3
pp
Total angular displacement,
q  = q
1
+ q
2
Total time = 2 + 1 = 3 sec
\ av
®
w =
4 /3
3
p
4
6
p
=
2
3
p
(3) (c) Angular speed of hour hand,
w
1
=
t
Dq
D
=
2
12 60
p
´
Angular speed of minute hand,
w
2
=
2
60
p
2
1
12
1
w
=
w
(4) (d) We have q = w
0
t +
1
2
at
2

Þ
d
dt
q
= w
0
+ at
This is angular velocity at time t.
Now angular velocity at t = 2 sec will be
w =
t 2 sec
d
dt
=
q æö
ç÷
èø
= w
0
+ 2a  = 1 + 2 × 1.5 = 4 rad/sec
(5) (d) The distance covered in completing the circle is
2pr = 2p × 10 cm
The linear speed is v =
2r
t
p
=
2 10
4
p´
= 5p cm/s
The linear acceleration is,
a =
2
v
r
=
2
(5)
10
p
= 2.5 p
2
cm/s
2
This acceleration is directed towards the centre of the circle
(6) (d) We know that
Time period =
Circumference
Criticalspeed
=
2r
gr
p
=
2224
7 104
´´
´´
= 4 sec
(7) (b) V elocity =
Circumference
Timeof revolution
=
2r
60
p
=
21
60
p´
=
30
p
cm/s
Change in velocity Dv =
22
30 30
pp æö æö
+
ç÷ ç÷
èø èø
= 2
30
p
cm/s
(8) (a) Let the radius of the orbit be r and the number of
revolutions per second be n. Then the velocity of electron
is given by  v = 2pnr,
\ Acceleration a =
2
v
r
=
2 22
4 rn
r
p
=  4 p
2
r n
2
Substituting the given values, we have
a = 4 × (3.14)
2
× (5.3 × 10
–11
) (6.6 × 10
15
)
2
= 9.1 × 10
22
m/s
2
towards the nucleus.
The centripetal force is
F
C
= ma = (9.1 × 10
–31
) (9.1 × 10
22
)
= 8.3 × 10
–8
N towards the nucleus.
(9) (b) Given that radius of horizontal loop
r = 1 km = 1000 m
Speed v = 900 km/h =
9000 5
18
´
= 250 m/s
Centripetal acceleration a
c
=
2
v
r
=
250 250
1000
´
= 62.5 m/s
2
\
on accelerati nal Gravitatio
on accelerati l Centripeta
=
c
a
g
=
62.5
9.8
= 6.38 : 1
(10) (b) We know that, tan q =
2
v
rg
=
2
5
18
18
100 10
æö
´
ç÷
èø
´
=
1
40
Þ q = tan
–1
1
40
(11) (a) The angular velocity is  w =
v
r
Hence, v = 10 m/s
r = 20 cm = 0.2 m  \ w = 50 rad/s
DPP/ P 07
19
(12) (b) Given that w = 1.5t – 3t
2
+ 2
a =
d
dt
w
= 1.5 – 6t
When,  a = 0
Þ 1.5 – 6t = 0
Þ t =
1.5
6
= 0.25 sec
(13) (c) Given v = 1.5 t
2
+ 2t
Linear acceleration a = dv/dt = 3t + 2
This is the linear acceleration at time t
Now angular acceleration at time t
a =
a
r
Þ a =
2
3t2
2 10
-
+
´
Angular acceleration at t = 2 sec
(a)
at t = 2sec
=
2
3 22
2 10
-
´+
´
=
8
2
× 10
2
= 4 × 10
2
2
(14) (a) Angular displacement after 4 sec is
q = w
0
t +
1
2
at
2
=
1
2
at
2
=
1
2
× 4 × 4
2
Angular velocity after 4 sec
w = w
0
+ at  = 0 + 4 × 4 = 16 rad/sec
(15) (a) Given a = 3t – t
2
Þ
d
dt
w
= 3t – t
2
Þ dw = (3t – t
2
)dt
Þ  w =
23
3tt
c
23
-+
At  t = 0, w = 0
\  c = 0, w =
23
3tt
23
-
Angular velocity at  t = 2 sec, (w)
t

= 2 sec
=
38
(4)
23
- =
10
3
Since there is no angular acceleration after 2 sec
\ The angular velocity after 6 sec remains the same.
(16) (d)
ˆ
ˆ ix yj + , x = r cos q,
y = r sin q where q = wt
r =
ˆ
i ( r cos wt) +
ˆ
j (r sin wt)
v = dr/dt = –
ˆ
i (wr sin wt) –
ˆ
j (w r cos wt)
a = d
2
r/dt
2
= – w
2
r
(17) (c) Let R be the normal reaction exerted by the road on the
car. At the highest point, we have
2
mv
(r h) +
= mg – R, R should not be negative.
Therefore v
2
£ (r + a)g =  (8.9 + 1.1) × 10
or v
2
£ 10 × 10v £ 10 m/sec
\  v
max
= 10 m/sec
(18) (a) Let W = Mg be the weight of the car. Friction
force  = 0.4 W
Centripetal force =
22
Mv Wv
r gr
=
0.4 W =
2
Wv
gr
Þ v
2
= 0.4 × g × r = 0.4 × 9.8 × 30 = 117.6
Þ v = 10.84 m/sec
(19) (c) Let v be the speed of earth's rotation.
We know that    W = mg
Hence
3
5
W = mg –
2
mv
r
or
3
5
mg = mg –
2
mv
r
\
2
3
mg =
2
mv
r
or v
2
=
2gr
5
Now   v
2
=
3
2 9.8 (6400 10 )
5
´ ´´
Solving, we get v = 5 × 10
9
m/sec,
w =
2g
5r
æö
ç÷
èø
= 7.8 × 10
4
(20) (a) Let  T be the tension produced in the stretched string.
The centripetal force required for the mass m to move in a
circle is provided by the tension T. The stretched length of
the spring is r (radius of the circle). Now ,
Elongation produced in the spring = (r – l
0
)
Tension produced in the spring,
T  = k (r – l
0
) ........ (1)
Where k is the force constant
Linear velocity of the motion v = 2p r n
\ Centripetal force  =
2
mv
r
=
2
m(2 rn)
r
p
= 4p
2
r n
2
m ........ (2)
Equating equation. (1) and (2), we get
k (r – l
0
) = 4p
2
r n
2
m (\ T = mv
2
/r)
Þ kr – k ll
0
= 4 p
2
r n
2
m
r (k – 4p
2
n
2
m) = k l
0
Þ  r =
0
22
k
(k 4 n m) -p
l
........(3)
Substituting the value of r in eqn. (1) we have
T = k
ú
û
ù
ê
ë
é
-
p -
0
2 2
0
) m n 4 k (
k
l
l
or T  =
22
0
22
4 nmk
(k 4 n m)
p
-p
l
(21) (c) Two types of acceleration are experienced by the car
(i) Radial acceleration due to circular path,
a
r
=
2
v
r
=
2
(30)
500
= 1.8 m/s
2
Page 3

(1) (b) We have angular displacement
=
linear displacement
Þ Dq =
S
r
D
Here, DS = n (2pr) = 1.5 (2p × 2 × 10
–2
) = 6p × 10
–2
\ Dq =
2
2
6 10
2 10
-
-
p´
´
(2) (a) We have
av
Total angular displacement
Total time
®
w=
For first one third part of circle,
angular displacement,   q
1
=
1
S
r
=
2 r/3
r
p
For second one third part of circle,
q
2
=
2r/32
r3
pp
Total angular displacement,
q  = q
1
+ q
2
Total time = 2 + 1 = 3 sec
\ av
®
w =
4 /3
3
p
4
6
p
=
2
3
p
(3) (c) Angular speed of hour hand,
w
1
=
t
Dq
D
=
2
12 60
p
´
Angular speed of minute hand,
w
2
=
2
60
p
2
1
12
1
w
=
w
(4) (d) We have q = w
0
t +
1
2
at
2

Þ
d
dt
q
= w
0
+ at
This is angular velocity at time t.
Now angular velocity at t = 2 sec will be
w =
t 2 sec
d
dt
=
q æö
ç÷
èø
= w
0
+ 2a  = 1 + 2 × 1.5 = 4 rad/sec
(5) (d) The distance covered in completing the circle is
2pr = 2p × 10 cm
The linear speed is v =
2r
t
p
=
2 10
4
p´
= 5p cm/s
The linear acceleration is,
a =
2
v
r
=
2
(5)
10
p
= 2.5 p
2
cm/s
2
This acceleration is directed towards the centre of the circle
(6) (d) We know that
Time period =
Circumference
Criticalspeed
=
2r
gr
p
=
2224
7 104
´´
´´
= 4 sec
(7) (b) V elocity =
Circumference
Timeof revolution
=
2r
60
p
=
21
60
p´
=
30
p
cm/s
Change in velocity Dv =
22
30 30
pp æö æö
+
ç÷ ç÷
èø èø
= 2
30
p
cm/s
(8) (a) Let the radius of the orbit be r and the number of
revolutions per second be n. Then the velocity of electron
is given by  v = 2pnr,
\ Acceleration a =
2
v
r
=
2 22
4 rn
r
p
=  4 p
2
r n
2
Substituting the given values, we have
a = 4 × (3.14)
2
× (5.3 × 10
–11
) (6.6 × 10
15
)
2
= 9.1 × 10
22
m/s
2
towards the nucleus.
The centripetal force is
F
C
= ma = (9.1 × 10
–31
) (9.1 × 10
22
)
= 8.3 × 10
–8
N towards the nucleus.
(9) (b) Given that radius of horizontal loop
r = 1 km = 1000 m
Speed v = 900 km/h =
9000 5
18
´
= 250 m/s
Centripetal acceleration a
c
=
2
v
r
=
250 250
1000
´
= 62.5 m/s
2
\
on accelerati nal Gravitatio
on accelerati l Centripeta
=
c
a
g
=
62.5
9.8
= 6.38 : 1
(10) (b) We know that, tan q =
2
v
rg
=
2
5
18
18
100 10
æö
´
ç÷
èø
´
=
1
40
Þ q = tan
–1
1
40
(11) (a) The angular velocity is  w =
v
r
Hence, v = 10 m/s
r = 20 cm = 0.2 m  \ w = 50 rad/s
DPP/ P 07
19
(12) (b) Given that w = 1.5t – 3t
2
+ 2
a =
d
dt
w
= 1.5 – 6t
When,  a = 0
Þ 1.5 – 6t = 0
Þ t =
1.5
6
= 0.25 sec
(13) (c) Given v = 1.5 t
2
+ 2t
Linear acceleration a = dv/dt = 3t + 2
This is the linear acceleration at time t
Now angular acceleration at time t
a =
a
r
Þ a =
2
3t2
2 10
-
+
´
Angular acceleration at t = 2 sec
(a)
at t = 2sec
=
2
3 22
2 10
-
´+
´
=
8
2
× 10
2
= 4 × 10
2
2
(14) (a) Angular displacement after 4 sec is
q = w
0
t +
1
2
at
2
=
1
2
at
2
=
1
2
× 4 × 4
2
Angular velocity after 4 sec
w = w
0
+ at  = 0 + 4 × 4 = 16 rad/sec
(15) (a) Given a = 3t – t
2
Þ
d
dt
w
= 3t – t
2
Þ dw = (3t – t
2
)dt
Þ  w =
23
3tt
c
23
-+
At  t = 0, w = 0
\  c = 0, w =
23
3tt
23
-
Angular velocity at  t = 2 sec, (w)
t

= 2 sec
=
38
(4)
23
- =
10
3
Since there is no angular acceleration after 2 sec
\ The angular velocity after 6 sec remains the same.
(16) (d)
ˆ
ˆ ix yj + , x = r cos q,
y = r sin q where q = wt
r =
ˆ
i ( r cos wt) +
ˆ
j (r sin wt)
v = dr/dt = –
ˆ
i (wr sin wt) –
ˆ
j (w r cos wt)
a = d
2
r/dt
2
= – w
2
r
(17) (c) Let R be the normal reaction exerted by the road on the
car. At the highest point, we have
2
mv
(r h) +
= mg – R, R should not be negative.
Therefore v
2
£ (r + a)g =  (8.9 + 1.1) × 10
or v
2
£ 10 × 10v £ 10 m/sec
\  v
max
= 10 m/sec
(18) (a) Let W = Mg be the weight of the car. Friction
force  = 0.4 W
Centripetal force =
22
Mv Wv
r gr
=
0.4 W =
2
Wv
gr
Þ v
2
= 0.4 × g × r = 0.4 × 9.8 × 30 = 117.6
Þ v = 10.84 m/sec
(19) (c) Let v be the speed of earth's rotation.
We know that    W = mg
Hence
3
5
W = mg –
2
mv
r
or
3
5
mg = mg –
2
mv
r
\
2
3
mg =
2
mv
r
or v
2
=
2gr
5
Now   v
2
=
3
2 9.8 (6400 10 )
5
´ ´´
Solving, we get v = 5 × 10
9
m/sec,
w =
2g
5r
æö
ç÷
èø
= 7.8 × 10
4
(20) (a) Let  T be the tension produced in the stretched string.
The centripetal force required for the mass m to move in a
circle is provided by the tension T. The stretched length of
the spring is r (radius of the circle). Now ,
Elongation produced in the spring = (r – l
0
)
Tension produced in the spring,
T  = k (r – l
0
) ........ (1)
Where k is the force constant
Linear velocity of the motion v = 2p r n
\ Centripetal force  =
2
mv
r
=
2
m(2 rn)
r
p
= 4p
2
r n
2
m ........ (2)
Equating equation. (1) and (2), we get
k (r – l
0
) = 4p
2
r n
2
m (\ T = mv
2
/r)
Þ kr – k ll
0
= 4 p
2
r n
2
m
r (k – 4p
2
n
2
m) = k l
0
Þ  r =
0
22
k
(k 4 n m) -p
l
........(3)
Substituting the value of r in eqn. (1) we have
T = k
ú
û
ù
ê
ë
é
-
p -
0
2 2
0
) m n 4 k (
k
l
l
or T  =
22
0
22
4 nmk
(k 4 n m)
p
-p
l
(21) (c) Two types of acceleration are experienced by the car
(i) Radial acceleration due to circular path,
a
r
=
2
v
r
=
2
(30)
500
= 1.8 m/s
2
20
DPP/ P 07
(ii) A tangential acceleration due to increase of tangential
speed given by
a
t
= Dv/Dt = 2 m/s
2
Radial and tangential acceleration are perpendicular to each
other.
Net acceleration of car
a  =
2 2 22
rt
a a (1.8) (2) +=+
= 2.7 m/s
2
(22) (a) For A :
Required centripetal force =
l 3
mv
2
A
(net force towards centre = T
1
)
This will provide required centripetal force
particle at A, \ T
1
=
l 3
mv
2
A
For B : Required centripetal force =
l 2
) v ( m
2
B
Remember w i.e. angular velocity , of all the particles is same
\ w =
A
v
3 l
=
B
v
2 l
=
C
v
l
When a system of particles rotates about an axis, the
angular velocity of all the particles will be same, but their
linear velocity will be different, because of different
distances from axis of rotation i.e. v = rw.
Thus for B, centripetal force =
2
A
2mv
9l
Net force towards the centre T
2
– T
1
=
2
A
2mv
9l
Þ T
2
=
2
A
2mv
9l
+ T
1
=
2
A
5mv
9 l
(Putting value of T
1
)
For C :
Centripetal force.
2
C
mv
3l
=
2
A
mv
9l
Net force towards centre  = T
3
– T
2
\ T
3
– T
2
=
2
A
mv
9 l
Þ   T
3
=
2
A
mv
9l
+ T
2
T
3
=
2
A
6mv
9 l
(on putting value of T
2
)
(23) (b) N cos q =
2
mv
r
and N sin q = mg
Þ
Nsin
Ncos
q
q
=
2
mg
mv /r
Þ tan q  =
2
rg
v
But  tan q =
r
h
\
r
h
=
2
rg
v
Þ  v  = hg =
2
9.8 9.8 10
-
´´ = 0.98 m/s
(24) (d) (1) Centripetal force is not a real force. It is only the
requirement for circular motion. It is not a new kind of
force. Any of the forces found in nature such as gravitational
force, electric friction force, tension in string, reaction force
may act as centripetal force.
(3) Work done by centripetal force is always zero.
(25) (a)  We know ,  a =
2
v
r
Hence  v = 10 m/s, r = 5 m \ a =
2
(10)
5
= 20 m/s
2
(26) (a) Given that the mass of the particle, m = 2 kg
Radius of circle = 3 m
Angular  velocity = 60 rev/minute
=
602
60
´p
Because the angle described during 1 revolution is 2p radian
The linear velocity v = rw  = 2p × 3 m/s = 6p m/s
The centripetal acceleration =
2
v
r
=
2
(6)
3
p
m/s
2
= 118.4 m/s
2
(27) (a) F =
2
mv
r
= mrw
2
Here m = 0.10 kg, r = 0.5 m
and  w =
2n
t
p
=
2 3.14 10
31.4
´´
F = 0.10 × 0.5 × (2)
2
= 0.2
(28) (a) In non-uniform circular motion acceleration vector
makes some angle with radius hence it is not
perpendicular to velocity vector.
(29) (c) If speed is increasing there is a tangential acceleration.
Net acceleration is not pointing towards centre.
(30) (b) Both statements are true but statement-2 is not correct
explanation for statement-1.
```

## Physics Class 11

116 videos|451 docs|188 tests

## Physics Class 11

116 videos|451 docs|188 tests Explore Courses for NEET exam ### How to Prepare for NEET

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