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# ALLEN : QUES 7 ANS NEET Notes | EduRev

## NEET : ALLEN : QUES 7 ANS NEET Notes | EduRev

``` Page 1

HINT â€“ SHEET
LTS/HS-1/7 0999DMD310318007
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test                Test # 07 Test Pattern
:
NEET-UG
TEST DATE
:
07 - 10 - 2018
1. In a rigid body, angular velocity of any point
on the rigid body w.r.t any other point on the
rigid body is constant and equal to angular
velocity of rigid body.
3. Since K.E. of rotation =
1
2
Iw
2
, I =
2
5
MR
2
\ K.E. =
22
12
1 (.3) (50) 45 J
25
´´´ ´=
4.
22
1 122
22
I mr ,I mr
53
==
I
1
= I
2
Þ
22
1 21
2
r r r5
5 3 r3
= Þ=
5.
2
22
2
1 K 12
KE m1 v m1 v
2 25 R
æö
æö
= + =+
ç÷ ç÷
èø
èø
=
2
17
50 (5) 875 erg
25
´ ´´=
6. Time taken by the rolling body to reach the
bottom
t
rolling
=
1
sinq

2
2
2hK
1
gR
æö
+
ç÷
èø
Here, q, h and g are constant
\ t
rolling
µ
2
2
K
1
R
+
Que. 1 2 3 4 5 6 7 8 9 101112 1314151617 181920
Ans.2 1 4 3 3 3 2 2 1 3 3 2 2 2 2 3 2 3 4 1
Que.2122 2324252627 2829303132 3334353637 383940
Ans.2 2 2 1 3 4 1 2 1 3 3 2 3 4 2 1 2 3 2 4
Que.4142 4344454647 4849505152 5354555657 585960
Ans.3 2 2 3 1 2 3 1 3 3 2 3 3 4 1 4 3 4 3 3
Que.6162 6364656667 6869707172 7374757677 787980
Ans.2 3 2 1 1 1 1 1 3 3 2 4 4 4 4 4 4 3 2 3
Que.8182 8384858687 8889909192 9394959697 9899 100
Ans.3 3 4 4 2 1 4 2 1 3 1 1 3 3 4 4 1 2 3 4
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 3 3 3 4 1 3 2 4 1 3 1 1 3 3 1 2 2 3 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 3 3 2 3 2 3 3 4 3 2 2 1 4 2 2 2 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 2 2 4 1 2 3 2 4 2 3 4 1 1 3 2 4 4 3
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 4 1 1 2 4 3 3 3 4 3 1 1 4 2 4 3 1 2
Page 2

HINT â€“ SHEET
LTS/HS-1/7 0999DMD310318007
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test                Test # 07 Test Pattern
:
NEET-UG
TEST DATE
:
07 - 10 - 2018
1. In a rigid body, angular velocity of any point
on the rigid body w.r.t any other point on the
rigid body is constant and equal to angular
velocity of rigid body.
3. Since K.E. of rotation =
1
2
Iw
2
, I =
2
5
MR
2
\ K.E. =
22
12
1 (.3) (50) 45 J
25
´´´ ´=
4.
22
1 122
22
I mr ,I mr
53
==
I
1
= I
2
Þ
22
1 21
2
r r r5
5 3 r3
= Þ=
5.
2
22
2
1 K 12
KE m1 v m1 v
2 25 R
æö
æö
= + =+
ç÷ ç÷
èø
èø
=
2
17
50 (5) 875 erg
25
´ ´´=
6. Time taken by the rolling body to reach the
bottom
t
rolling
=
1
sinq

2
2
2hK
1
gR
æö
+
ç÷
èø
Here, q, h and g are constant
\ t
rolling
µ
2
2
K
1
R
+
Que. 1 2 3 4 5 6 7 8 9 101112 1314151617 181920
Ans.2 1 4 3 3 3 2 2 1 3 3 2 2 2 2 3 2 3 4 1
Que.2122 2324252627 2829303132 3334353637 383940
Ans.2 2 2 1 3 4 1 2 1 3 3 2 3 4 2 1 2 3 2 4
Que.4142 4344454647 4849505152 5354555657 585960
Ans.3 2 2 3 1 2 3 1 3 3 2 3 3 4 1 4 3 4 3 3
Que.6162 6364656667 6869707172 7374757677 787980
Ans.2 3 2 1 1 1 1 1 3 3 2 4 4 4 4 4 4 3 2 3
Que.8182 8384858687 8889909192 9394959697 9899 100
Ans.3 3 4 4 2 1 4 2 1 3 1 1 3 3 4 4 1 2 3 4
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 3 3 3 4 1 3 2 4 1 3 1 1 3 3 1 2 2 3 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 3 3 2 3 2 3 3 4 3 2 2 1 4 2 2 2 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 2 2 4 1 2 3 2 4 2 3 4 1 1 3 2 4 4 3
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 4 1 1 2 4 3 3 3 4 3 1 1 4 2 4 3 1 2
LTS/HS-2/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
22
22
ring disc
K K1
1,
R R2
æö æö
==
ç÷ ç÷
èø èø
;
22
22
sphere shell
K 2K2
,
R 5R3
æ ö æö
==
ç ÷ ç÷
è ø èø
\t
sphere
< t
disc
< t
shell
< t
ring
7. The moment of inertia of each of the sphere A
and B about an axis passing through their centres
is
2
5
Ma
2
. Similarly the moment of inertia of C
and D about their central axes =
2
5
ma
2
A B
C D
b
b b
b
Now the spheres C and D are at distance b
apart, therefore by the theorem of parallel axes
Moment of inertia of C or D =
22
2
Ma Mb
5
æö
+
ç÷
èø
Therefore the moment of inertia of all the four
2 2 2 22
228
2 Ma Ma Mb Ma 2Mb
555
éù æö
= + + =+
ç÷
êú
èø ëû
8. Given
1
2
(1.2) × w
2
= 1500
\
3000
1.2
w=
t
w
a=
Þ
t
w
=
a
2
)
\
50
t 2sec
25
==
9. L = Iw = 0.6 × 2p ×
1
2
= 0.6p kg-m
2
/sec
10. The linear acceleration of the bodies rolling
down an inclined plane,
2
2
1
a
K
1
R
µ
+
Hence for sphere
1
5
a
7
= and for disc
2
2
a
3
=
from relation s =
1
2
at
2
Þ t
2
=
2s
a
\
12
21
ta 2/3
t a 5/7
==
14 : 15 =
11. I depends on mass distribution, mass is more
away from the axis of rotation the MI will be
more
I
CA
< I
AB
< I
BC
12. Rotational K.E. =
1
2
Iw
2
2
2
1 1 702
72 (0.5) 240J
2 2 60
´p é ùæö
=´´ ´=
ç÷
êú
ë ûèø
13.
2
g
ML
I
12
Now I = I
g
+ Ma
2

L LL
a
2 36
éù
= -=
êú
ëû
\
2 22
ML ML ML
I
12 369
= +=
14. I =
2
MR
4 I
I =
2
(1) (0.1)
4
= 2.5 × 10
â€“3
kg-m
2
15. By perpendicular axes theorem
zz' xx' yy'
I II =+
16.
R
r
V
w
x
y
r =R
1
O
Angular momentum of disc about origin is :
L = Iw +mV(r
1
)
Page 3

HINT â€“ SHEET
LTS/HS-1/7 0999DMD310318007
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test                Test # 07 Test Pattern
:
NEET-UG
TEST DATE
:
07 - 10 - 2018
1. In a rigid body, angular velocity of any point
on the rigid body w.r.t any other point on the
rigid body is constant and equal to angular
velocity of rigid body.
3. Since K.E. of rotation =
1
2
Iw
2
, I =
2
5
MR
2
\ K.E. =
22
12
1 (.3) (50) 45 J
25
´´´ ´=
4.
22
1 122
22
I mr ,I mr
53
==
I
1
= I
2
Þ
22
1 21
2
r r r5
5 3 r3
= Þ=
5.
2
22
2
1 K 12
KE m1 v m1 v
2 25 R
æö
æö
= + =+
ç÷ ç÷
èø
èø
=
2
17
50 (5) 875 erg
25
´ ´´=
6. Time taken by the rolling body to reach the
bottom
t
rolling
=
1
sinq

2
2
2hK
1
gR
æö
+
ç÷
èø
Here, q, h and g are constant
\ t
rolling
µ
2
2
K
1
R
+
Que. 1 2 3 4 5 6 7 8 9 101112 1314151617 181920
Ans.2 1 4 3 3 3 2 2 1 3 3 2 2 2 2 3 2 3 4 1
Que.2122 2324252627 2829303132 3334353637 383940
Ans.2 2 2 1 3 4 1 2 1 3 3 2 3 4 2 1 2 3 2 4
Que.4142 4344454647 4849505152 5354555657 585960
Ans.3 2 2 3 1 2 3 1 3 3 2 3 3 4 1 4 3 4 3 3
Que.6162 6364656667 6869707172 7374757677 787980
Ans.2 3 2 1 1 1 1 1 3 3 2 4 4 4 4 4 4 3 2 3
Que.8182 8384858687 8889909192 9394959697 9899 100
Ans.3 3 4 4 2 1 4 2 1 3 1 1 3 3 4 4 1 2 3 4
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 3 3 3 4 1 3 2 4 1 3 1 1 3 3 1 2 2 3 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 3 3 2 3 2 3 3 4 3 2 2 1 4 2 2 2 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 2 2 4 1 2 3 2 4 2 3 4 1 1 3 2 4 4 3
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 4 1 1 2 4 3 3 3 4 3 1 1 4 2 4 3 1 2
LTS/HS-2/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
22
22
ring disc
K K1
1,
R R2
æö æö
==
ç÷ ç÷
èø èø
;
22
22
sphere shell
K 2K2
,
R 5R3
æ ö æö
==
ç ÷ ç÷
è ø èø
\t
sphere
< t
disc
< t
shell
< t
ring
7. The moment of inertia of each of the sphere A
and B about an axis passing through their centres
is
2
5
Ma
2
. Similarly the moment of inertia of C
and D about their central axes =
2
5
ma
2
A B
C D
b
b b
b
Now the spheres C and D are at distance b
apart, therefore by the theorem of parallel axes
Moment of inertia of C or D =
22
2
Ma Mb
5
æö
+
ç÷
èø
Therefore the moment of inertia of all the four
2 2 2 22
228
2 Ma Ma Mb Ma 2Mb
555
éù æö
= + + =+
ç÷
êú
èø ëû
8. Given
1
2
(1.2) × w
2
= 1500
\
3000
1.2
w=
t
w
a=
Þ
t
w
=
a
2
)
\
50
t 2sec
25
==
9. L = Iw = 0.6 × 2p ×
1
2
= 0.6p kg-m
2
/sec
10. The linear acceleration of the bodies rolling
down an inclined plane,
2
2
1
a
K
1
R
µ
+
Hence for sphere
1
5
a
7
= and for disc
2
2
a
3
=
from relation s =
1
2
at
2
Þ t
2
=
2s
a
\
12
21
ta 2/3
t a 5/7
==
14 : 15 =
11. I depends on mass distribution, mass is more
away from the axis of rotation the MI will be
more
I
CA
< I
AB
< I
BC
12. Rotational K.E. =
1
2
Iw
2
2
2
1 1 702
72 (0.5) 240J
2 2 60
´p é ùæö
=´´ ´=
ç÷
êú
ë ûèø
13.
2
g
ML
I
12
Now I = I
g
+ Ma
2

L LL
a
2 36
éù
= -=
êú
ëû
\
2 22
ML ML ML
I
12 369
= +=
14. I =
2
MR
4 I
I =
2
(1) (0.1)
4
= 2.5 × 10
â€“3
kg-m
2
15. By perpendicular axes theorem
zz' xx' yy'
I II =+
16.
R
r
V
w
x
y
r =R
1
O
Angular momentum of disc about origin is :
L = Iw +mV(r
1
)
LTS/HS-3/7 0999DMD310318007
=
2
MR
mV(R)
2
w+
=
2
MR
M( R)(R)
2
w+w
=
2
3
MR
2
w
17. According to the law of conservation of
angular momentum
Mr
2
w = (Mr
2
+ 2mr
2
) w' Þ
M
'
M 2m
w
w=
+
18.
N
A
N
B
l/3 l/6 3l/10 l/5
B A C
By taking torque about point C
N
A
3
10
æö
ç÷
èø
l
= N
B
6
æö
ç÷
èø
l

A
B
N5
N9
=
19. as there is no external torque working on the
system so angular momentum will remain same
20.
2
2
ML
MK
12
= Þ
L
K
12
=
22.
2
Ring
2
Disc
I
MR
2:1
1
I
MR
2
==
23.
22
1
MR MK
2
=
R 2.5
K 1.76
22
= ==
cm
24.
1.05m
B
F
1800kg
N
2
N
1 1.8 m
0.75m
N
1
×1.8 = 1800g × 0.75
N
1
× 1.8 =
1800 10 0.75
1.8
´´
N
1
= 7500 N
Force exerted by ground on each front wheel
=
1
N
3750 N
2
=
25. rms
100 2
V 100 volt
2
==
V
2
RMS
= V
R
2
+ V
L
2
Þ V
L
2
= (100)
2
â€“ (60)
2
= 6400
V
L
= 80 volt
26. P
R
= 10 W    Þ R =
22
V (60)
360
P 10
= =W
V
R
= 60 V      V
C
2
+ V
R
2
= V
S
2
Þ V
C
= 80V
~
X
C
+80Vâ€“
+60Vâ€“
I I
360W
100 V
I =
V 601
R 360 6
==
Applying Ohm's law for capacitor
V
C
= IX
C
X
C
=
80
480
1/6
=W
27. When connected to DC source
V = IR
Resistance of the coil R =
100
4
25
=W
~
(100V , 50 Hz)
X
L
R=4W
I=20A
with A.C. Z =
V 100
5
I 20
==
Page 4

HINT â€“ SHEET
LTS/HS-1/7 0999DMD310318007
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test                Test # 07 Test Pattern
:
NEET-UG
TEST DATE
:
07 - 10 - 2018
1. In a rigid body, angular velocity of any point
on the rigid body w.r.t any other point on the
rigid body is constant and equal to angular
velocity of rigid body.
3. Since K.E. of rotation =
1
2
Iw
2
, I =
2
5
MR
2
\ K.E. =
22
12
1 (.3) (50) 45 J
25
´´´ ´=
4.
22
1 122
22
I mr ,I mr
53
==
I
1
= I
2
Þ
22
1 21
2
r r r5
5 3 r3
= Þ=
5.
2
22
2
1 K 12
KE m1 v m1 v
2 25 R
æö
æö
= + =+
ç÷ ç÷
èø
èø
=
2
17
50 (5) 875 erg
25
´ ´´=
6. Time taken by the rolling body to reach the
bottom
t
rolling
=
1
sinq

2
2
2hK
1
gR
æö
+
ç÷
èø
Here, q, h and g are constant
\ t
rolling
µ
2
2
K
1
R
+
Que. 1 2 3 4 5 6 7 8 9 101112 1314151617 181920
Ans.2 1 4 3 3 3 2 2 1 3 3 2 2 2 2 3 2 3 4 1
Que.2122 2324252627 2829303132 3334353637 383940
Ans.2 2 2 1 3 4 1 2 1 3 3 2 3 4 2 1 2 3 2 4
Que.4142 4344454647 4849505152 5354555657 585960
Ans.3 2 2 3 1 2 3 1 3 3 2 3 3 4 1 4 3 4 3 3
Que.6162 6364656667 6869707172 7374757677 787980
Ans.2 3 2 1 1 1 1 1 3 3 2 4 4 4 4 4 4 3 2 3
Que.8182 8384858687 8889909192 9394959697 9899 100
Ans.3 3 4 4 2 1 4 2 1 3 1 1 3 3 4 4 1 2 3 4
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 3 3 3 4 1 3 2 4 1 3 1 1 3 3 1 2 2 3 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 3 3 2 3 2 3 3 4 3 2 2 1 4 2 2 2 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 2 2 4 1 2 3 2 4 2 3 4 1 1 3 2 4 4 3
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 4 1 1 2 4 3 3 3 4 3 1 1 4 2 4 3 1 2
LTS/HS-2/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
22
22
ring disc
K K1
1,
R R2
æö æö
==
ç÷ ç÷
èø èø
;
22
22
sphere shell
K 2K2
,
R 5R3
æ ö æö
==
ç ÷ ç÷
è ø èø
\t
sphere
< t
disc
< t
shell
< t
ring
7. The moment of inertia of each of the sphere A
and B about an axis passing through their centres
is
2
5
Ma
2
. Similarly the moment of inertia of C
and D about their central axes =
2
5
ma
2
A B
C D
b
b b
b
Now the spheres C and D are at distance b
apart, therefore by the theorem of parallel axes
Moment of inertia of C or D =
22
2
Ma Mb
5
æö
+
ç÷
èø
Therefore the moment of inertia of all the four
2 2 2 22
228
2 Ma Ma Mb Ma 2Mb
555
éù æö
= + + =+
ç÷
êú
èø ëû
8. Given
1
2
(1.2) × w
2
= 1500
\
3000
1.2
w=
t
w
a=
Þ
t
w
=
a
2
)
\
50
t 2sec
25
==
9. L = Iw = 0.6 × 2p ×
1
2
= 0.6p kg-m
2
/sec
10. The linear acceleration of the bodies rolling
down an inclined plane,
2
2
1
a
K
1
R
µ
+
Hence for sphere
1
5
a
7
= and for disc
2
2
a
3
=
from relation s =
1
2
at
2
Þ t
2
=
2s
a
\
12
21
ta 2/3
t a 5/7
==
14 : 15 =
11. I depends on mass distribution, mass is more
away from the axis of rotation the MI will be
more
I
CA
< I
AB
< I
BC
12. Rotational K.E. =
1
2
Iw
2
2
2
1 1 702
72 (0.5) 240J
2 2 60
´p é ùæö
=´´ ´=
ç÷
êú
ë ûèø
13.
2
g
ML
I
12
Now I = I
g
+ Ma
2

L LL
a
2 36
éù
= -=
êú
ëû
\
2 22
ML ML ML
I
12 369
= +=
14. I =
2
MR
4 I
I =
2
(1) (0.1)
4
= 2.5 × 10
â€“3
kg-m
2
15. By perpendicular axes theorem
zz' xx' yy'
I II =+
16.
R
r
V
w
x
y
r =R
1
O
Angular momentum of disc about origin is :
L = Iw +mV(r
1
)
LTS/HS-3/7 0999DMD310318007
=
2
MR
mV(R)
2
w+
=
2
MR
M( R)(R)
2
w+w
=
2
3
MR
2
w
17. According to the law of conservation of
angular momentum
Mr
2
w = (Mr
2
+ 2mr
2
) w' Þ
M
'
M 2m
w
w=
+
18.
N
A
N
B
l/3 l/6 3l/10 l/5
B A C
By taking torque about point C
N
A
3
10
æö
ç÷
èø
l
= N
B
6
æö
ç÷
èø
l

A
B
N5
N9
=
19. as there is no external torque working on the
system so angular momentum will remain same
20.
2
2
ML
MK
12
= Þ
L
K
12
=
22.
2
Ring
2
Disc
I
MR
2:1
1
I
MR
2
==
23.
22
1
MR MK
2
=
R 2.5
K 1.76
22
= ==
cm
24.
1.05m
B
F
1800kg
N
2
N
1 1.8 m
0.75m
N
1
×1.8 = 1800g × 0.75
N
1
× 1.8 =
1800 10 0.75
1.8
´´
N
1
= 7500 N
Force exerted by ground on each front wheel
=
1
N
3750 N
2
=
25. rms
100 2
V 100 volt
2
==
V
2
RMS
= V
R
2
+ V
L
2
Þ V
L
2
= (100)
2
â€“ (60)
2
= 6400
V
L
= 80 volt
26. P
R
= 10 W    Þ R =
22
V (60)
360
P 10
= =W
V
R
= 60 V      V
C
2
+ V
R
2
= V
S
2
Þ V
C
= 80V
~
X
C
+80Vâ€“
+60Vâ€“
I I
360W
100 V
I =
V 601
R 360 6
==
Applying Ohm's law for capacitor
V
C
= IX
C
X
C
=
80
480
1/6
=W
27. When connected to DC source
V = IR
Resistance of the coil R =
100
4
25
=W
~
(100V , 50 Hz)
X
L
R=4W
I=20A
with A.C. Z =
V 100
5
I 20
==
LTS/HS-4/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
Z =
2 2 2 22
LL
X R (5) X (4) + Þ =+
ÞX
L
= 3 W
28. It is condition of resonance
So,  Z = R    Þ   I =
220
2.2A
100
=
X
L
â€“ X
C
Þ  V
R
= V = 220 V
29. Q
dIq
L
dtC
-=
\
dIq
dt LC
=
So,
0
max
q dI
dt LC
=
30.
rms rms
150 150
; E ,I
3 22
p
f= ==
P = E
rm
I
rm
cosf
=
150 150
cos 60º
22
´
=
150 150
5625W
4
´
=
31. R
net
= 40 + 40 = 80 W
X
L
= 100 W
Z =
22
(80) (100 40) +-
X
C
= 40 W           Z= 100 W
Power factor = cos f =
R 80
0.8
Z 100
==
32. V = 50 × 2sin(100pt) cos(100pt)
= 50 sin (200 pt)
V
0
= 50 and w = 200 p
f =
200
22
wp
=
pp
= 100 Hz
33. I
rms
= 2A
Wattless current = I
rms
sinf
3 2 sin =f
3
sin 60º
2
f= Þf=
\ power factor = cos f = cos 60º =
1
2
34. When only capacitor is removed
Z
R
60°
Þ Z =
R
2R 200
cos60
= =W
22
L
R X 200 +=
X
L
= 173.2 W
When only indictor is removed
R
Z
60°
Þ X
C
= 173.2 W
So, this is resonance condition.
Power =
2
rms
V 40000
400W
R 100
==
OR
LC
XX
tan
R
-
f=
when X
C
= 0,  tan f
1
=
L
X
R
(f
1
= 60º)
when L = 0,  tan f
1
=
C
X
R
(f
2
= 60º)
\ X
L
= X
C
and f = 0
it is condition of resonance and hence |Z| = R
= 100 W
\
2
rms
V 200 200
P 400W
R 100
´
===
35. f =
50
Hz
p
R = 1W   L = 10 mH
X
L
= wLX
L
=
100 10
1
1000
´
=W
Power factor =
22
L
R
RX +
=
22
11
2
11
=
+
36.
36
11
LC
8 10 20 10
--
w==
´ ´´
=
10000
4
=
0
0
E 220 2
I 5 2A
R 44
===
Page 5

HINT â€“ SHEET
LTS/HS-1/7 0999DMD310318007
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test                Test # 07 Test Pattern
:
NEET-UG
TEST DATE
:
07 - 10 - 2018
1. In a rigid body, angular velocity of any point
on the rigid body w.r.t any other point on the
rigid body is constant and equal to angular
velocity of rigid body.
3. Since K.E. of rotation =
1
2
Iw
2
, I =
2
5
MR
2
\ K.E. =
22
12
1 (.3) (50) 45 J
25
´´´ ´=
4.
22
1 122
22
I mr ,I mr
53
==
I
1
= I
2
Þ
22
1 21
2
r r r5
5 3 r3
= Þ=
5.
2
22
2
1 K 12
KE m1 v m1 v
2 25 R
æö
æö
= + =+
ç÷ ç÷
èø
èø
=
2
17
50 (5) 875 erg
25
´ ´´=
6. Time taken by the rolling body to reach the
bottom
t
rolling
=
1
sinq

2
2
2hK
1
gR
æö
+
ç÷
èø
Here, q, h and g are constant
\ t
rolling
µ
2
2
K
1
R
+
Que. 1 2 3 4 5 6 7 8 9 101112 1314151617 181920
Ans.2 1 4 3 3 3 2 2 1 3 3 2 2 2 2 3 2 3 4 1
Que.2122 2324252627 2829303132 3334353637 383940
Ans.2 2 2 1 3 4 1 2 1 3 3 2 3 4 2 1 2 3 2 4
Que.4142 4344454647 4849505152 5354555657 585960
Ans.3 2 2 3 1 2 3 1 3 3 2 3 3 4 1 4 3 4 3 3
Que.6162 6364656667 6869707172 7374757677 787980
Ans.2 3 2 1 1 1 1 1 3 3 2 4 4 4 4 4 4 3 2 3
Que.8182 8384858687 8889909192 9394959697 9899 100
Ans.3 3 4 4 2 1 4 2 1 3 1 1 3 3 4 4 1 2 3 4
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 3 3 3 4 1 3 2 4 1 3 1 1 3 3 1 2 2 3 1
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 3 3 2 3 2 3 3 4 3 2 2 1 4 2 2 2 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 2 2 4 1 2 3 2 4 2 3 4 1 1 3 2 4 4 3
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 3 4 1 1 2 4 3 3 3 4 3 1 1 4 2 4 3 1 2
LTS/HS-2/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
22
22
ring disc
K K1
1,
R R2
æö æö
==
ç÷ ç÷
èø èø
;
22
22
sphere shell
K 2K2
,
R 5R3
æ ö æö
==
ç ÷ ç÷
è ø èø
\t
sphere
< t
disc
< t
shell
< t
ring
7. The moment of inertia of each of the sphere A
and B about an axis passing through their centres
is
2
5
Ma
2
. Similarly the moment of inertia of C
and D about their central axes =
2
5
ma
2
A B
C D
b
b b
b
Now the spheres C and D are at distance b
apart, therefore by the theorem of parallel axes
Moment of inertia of C or D =
22
2
Ma Mb
5
æö
+
ç÷
èø
Therefore the moment of inertia of all the four
2 2 2 22
228
2 Ma Ma Mb Ma 2Mb
555
éù æö
= + + =+
ç÷
êú
èø ëû
8. Given
1
2
(1.2) × w
2
= 1500
\
3000
1.2
w=
t
w
a=
Þ
t
w
=
a
2
)
\
50
t 2sec
25
==
9. L = Iw = 0.6 × 2p ×
1
2
= 0.6p kg-m
2
/sec
10. The linear acceleration of the bodies rolling
down an inclined plane,
2
2
1
a
K
1
R
µ
+
Hence for sphere
1
5
a
7
= and for disc
2
2
a
3
=
from relation s =
1
2
at
2
Þ t
2
=
2s
a
\
12
21
ta 2/3
t a 5/7
==
14 : 15 =
11. I depends on mass distribution, mass is more
away from the axis of rotation the MI will be
more
I
CA
< I
AB
< I
BC
12. Rotational K.E. =
1
2
Iw
2
2
2
1 1 702
72 (0.5) 240J
2 2 60
´p é ùæö
=´´ ´=
ç÷
êú
ë ûèø
13.
2
g
ML
I
12
Now I = I
g
+ Ma
2

L LL
a
2 36
éù
= -=
êú
ëû
\
2 22
ML ML ML
I
12 369
= +=
14. I =
2
MR
4 I
I =
2
(1) (0.1)
4
= 2.5 × 10
â€“3
kg-m
2
15. By perpendicular axes theorem
zz' xx' yy'
I II =+
16.
R
r
V
w
x
y
r =R
1
O
Angular momentum of disc about origin is :
L = Iw +mV(r
1
)
LTS/HS-3/7 0999DMD310318007
=
2
MR
mV(R)
2
w+
=
2
MR
M( R)(R)
2
w+w
=
2
3
MR
2
w
17. According to the law of conservation of
angular momentum
Mr
2
w = (Mr
2
+ 2mr
2
) w' Þ
M
'
M 2m
w
w=
+
18.
N
A
N
B
l/3 l/6 3l/10 l/5
B A C
By taking torque about point C
N
A
3
10
æö
ç÷
èø
l
= N
B
6
æö
ç÷
èø
l

A
B
N5
N9
=
19. as there is no external torque working on the
system so angular momentum will remain same
20.
2
2
ML
MK
12
= Þ
L
K
12
=
22.
2
Ring
2
Disc
I
MR
2:1
1
I
MR
2
==
23.
22
1
MR MK
2
=
R 2.5
K 1.76
22
= ==
cm
24.
1.05m
B
F
1800kg
N
2
N
1 1.8 m
0.75m
N
1
×1.8 = 1800g × 0.75
N
1
× 1.8 =
1800 10 0.75
1.8
´´
N
1
= 7500 N
Force exerted by ground on each front wheel
=
1
N
3750 N
2
=
25. rms
100 2
V 100 volt
2
==
V
2
RMS
= V
R
2
+ V
L
2
Þ V
L
2
= (100)
2
â€“ (60)
2
= 6400
V
L
= 80 volt
26. P
R
= 10 W    Þ R =
22
V (60)
360
P 10
= =W
V
R
= 60 V      V
C
2
+ V
R
2
= V
S
2
Þ V
C
= 80V
~
X
C
+80Vâ€“
+60Vâ€“
I I
360W
100 V
I =
V 601
R 360 6
==
Applying Ohm's law for capacitor
V
C
= IX
C
X
C
=
80
480
1/6
=W
27. When connected to DC source
V = IR
Resistance of the coil R =
100
4
25
=W
~
(100V , 50 Hz)
X
L
R=4W
I=20A
with A.C. Z =
V 100
5
I 20
==
LTS/HS-4/7 0999DMD310318007
Target : Pre-Medical 2019/NEET-UG/07-10-2018
Z =
2 2 2 22
LL
X R (5) X (4) + Þ =+
ÞX
L
= 3 W
28. It is condition of resonance
So,  Z = R    Þ   I =
220
2.2A
100
=
X
L
â€“ X
C
Þ  V
R
= V = 220 V
29. Q
dIq
L
dtC
-=
\
dIq
dt LC
=
So,
0
max
q dI
dt LC
=
30.
rms rms
150 150
; E ,I
3 22
p
f= ==
P = E
rm
I
rm
cosf
=
150 150
cos 60º
22
´
=
150 150
5625W
4
´
=
31. R
net
= 40 + 40 = 80 W
X
L
= 100 W
Z =
22
(80) (100 40) +-
X
C
= 40 W           Z= 100 W
Power factor = cos f =
R 80
0.8
Z 100
==
32. V = 50 × 2sin(100pt) cos(100pt)
= 50 sin (200 pt)
V
0
= 50 and w = 200 p
f =
200
22
wp
=
pp
= 100 Hz
33. I
rms
= 2A
Wattless current = I
rms
sinf
3 2 sin =f
3
sin 60º
2
f= Þf=
\ power factor = cos f = cos 60º =
1
2
34. When only capacitor is removed
Z
R
60°
Þ Z =
R
2R 200
cos60
= =W
22
L
R X 200 +=
X
L
= 173.2 W
When only indictor is removed
R
Z
60°
Þ X
C
= 173.2 W
So, this is resonance condition.
Power =
2
rms
V 40000
400W
R 100
==
OR
LC
XX
tan
R
-
f=
when X
C
= 0,  tan f
1
=
L
X
R
(f
1
= 60º)
when L = 0,  tan f
1
=
C
X
R
(f
2
= 60º)
\ X
L
= X
C
and f = 0
it is condition of resonance and hence |Z| = R
= 100 W
\
2
rms
V 200 200
P 400W
R 100
´
===
35. f =
50
Hz
p
R = 1W   L = 10 mH
X
L
= wLX
L
=
100 10
1
1000
´
=W
Power factor =
22
L
R
RX +
=
22
11
2
11
=
+
36.
36
11
LC
8 10 20 10
--
w==
´ ´´
=
10000
4
=
0
0
E 220 2
I 5 2A
R 44
===
LTS/HS-5/7 0999DMD310318007
37. At resonance condition power delivered is
maximum and at resonance
X
L
= X
C
wL =
1
C w
38. When f > f
R
then X
L
> X
C
Circuit is inductive.
39. I
2
= 4t
<I
2
> =
44 22
2
22
4(4 2)
I dt 4t dt 12
22
-
= ==
´
òò
I
rms
=
2
I 12 2 3A <>==
40.
6
V V 3 1.8V
37
- = ´=
+
AB
6
V V 2 2V
24
- = ´=
+
\ V
D
â€“ V
B
= 2â€“1.8 = 0.2 V
potential difference across the capacitor = 0.2V
\
2 62
11
U CV 3 10 (0.2)
22
-
= =´´´
= 60 × 10
â€“9
= 60 nJ
41. Energy stored in capacitor
=
2 6 21
11
CV 5 10 (200) 10 J
22
--
=´´ ´=
The fraction lost in 500 W is
8
5
of total so
H =
8
5
×
10
1
=
12
JJ
16 32
=
42. nâ€“1 capacitance are in series combination.
1
C
C
n1
=
-
A B
C
C
nâ€“1
eq
C C (n 1)C Cn
CC
n1 n1 n1
+-
=+==
- --
43. V
R
= e.e
â€“t/RC
V
C
= e[1 â€“ e
â€“t/RC
]
At t = 100 ms
V
R
= V
C
Þ e
â€“t/RC
= 1 â€“ e
â€“t/RC
Þ 2e
â€“t/RC
= 1
Þ e
â€“t/RC
= 1/2 Þ e
t/RC
= 2 Þ
t
n2
RC
= l
Þ
100 100
n 2 RC 145.45 ms
RC n(2)
= Þ == l
l
(as ln2 = 0.693 ; 0.7)
44. Both the capacitors are in series. Therefore
charge stored on them will be same.
Net capacity =
C 2C 2C
C 2C3
´
=
+
=
2
6 4µF
3
´=
Potential difference = 10 V
q = CV = 4 × 10
â€“6
× 10 = 40 µC
45. q
1
= CE
q
2
=
E
C3
23
´´
+
= CE ×
3
5
47. BF     +    NH
33
sp
2
sp
3
Trigonal
planar
Tetrahedral
B N H
H
H
F
F
F
sp
3
sp
3
Tetrahedral Tetrahedral
reqular geometry is charged
48. In BF
3
, pp-pp back bonding is present.
49.
3
3.5
SiO
-
pyro sillicate
50. PCl
5
(s) ¾® PCl
4
+
+ PCl
6
â€“
51. Al
4
C
3
+ 12Hâ€“OH ¾® 4Al(OH)
3
+ 3CH
4
Be
2
C + 4Hâ€“OH ¾® 2Be(OH)
2
+ CH
4
53. PCl  +  HO  HPO
5 2 34
¾®
(+5) (+5)
Same O.S.
54.
242 23
( 4)
N O H O HNO HNO
+
+ ¾¾®+
Cl
2
O
6
+ H
2
O ¾® HClO
3
+ HClO
4
55.
2 34
( 1)
( 3) ( 7) ( 5)
HOCl HClO HClO HClO
+
++ +
< <<
Acidic nature µ EN µ + O.S.
```
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