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# Test Paper 7 Solution MBBS Notes | EduRev

## MBBS : Test Paper 7 Solution MBBS Notes | EduRev

``` Page 1

HINT – SHEET
(0999DMD310319007) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 07
29-09-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit - 06
1. The given AC circuit is the combination of two
pure parallel circuits with the applied voltage. In
which I
2
is in phase with V and I
1
2. Work done by the retarding torque = initial KE
i.e., tq = K
Since t is same, hence q or the number of
revolutions will also be same.
3. X
C
> X
L
. Hence, current will lead the voltage.
Z =
22
CL
R (X –X) +
= 102W
\ I
rms
=
rms
V
Z
=
400 / 2
102
= 20 A
cos f =
R1
Z 2
=
V
R
= I
rms
R = (20) (10) = 200 V
4. Mw
2
L = T
0
Þ
0
T
ML
w=
0
1T
f
2 2 ML
w
==
pp
5. tan f =
L 2 50 0.21
R 12
w p´´
=
= 5.5
as tan 45º = 1 and tan 90º = ¥
so best answer is only in between 45º and 90º
which is f = 80º
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 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 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3
Page 2

HINT – SHEET
(0999DMD310319007) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 07
29-09-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit - 06
1. The given AC circuit is the combination of two
pure parallel circuits with the applied voltage. In
which I
2
is in phase with V and I
1
2. Work done by the retarding torque = initial KE
i.e., tq = K
Since t is same, hence q or the number of
revolutions will also be same.
3. X
C
> X
L
. Hence, current will lead the voltage.
Z =
22
CL
R (X –X) +
= 102W
\ I
rms
=
rms
V
Z
=
400 / 2
102
= 20 A
cos f =
R1
Z 2
=
V
R
= I
rms
R = (20) (10) = 200 V
4. Mw
2
L = T
0
Þ
0
T
ML
w=
0
1T
f
2 2 ML
w
==
pp
5. tan f =
L 2 50 0.21
R 12
w p´´
=
= 5.5
as tan 45º = 1 and tan 90º = ¥
so best answer is only in between 45º and 90º
which is f = 80º
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 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 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3
LTS/HS-2/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
6. The average value of energy density (energy/volume)
is given by
u
av
=
2
00
1
E
2
e
Total volume of the cylinder V = A·1
\ Total energy contained in the cylinder,
U = (u
av
) (V)
=
2
00
1
E (Al)
2
æö
e
ç÷
èø
Substituting the values, we have
U =
1
2
× (8.86×10
–12
) (50)
2
(10×10
–4
) (50×10
–2
) J
= 5.5 × 10
–12
J
7. KE in both cases should be same and equal to the
loss in potential energy (according to law of
conservation of energy). Hence,
2
2
22
2
1 1 v 1 5v
Mv MkM
2 2 R 24
æö
+=
ç÷
èø
Þ k =
3R
4
8. I
0
=
0
V 100
R 20
=
= 5 A
\ I
rms
=
0
I5
A
22
=
9. E
0
= B
0
c = 2 × 10
–7
× 3 × 10
8
= 60 N/C
EB ^
rr
and as direction of propogation is along
x so E is along z.
10. The cylinder will just begin to slip and will not
topple, if
Moment of force Mg sin q about A
= moment of N about A
45º
G
h/2
Mg sin q
Mg cos q
Mg
r
N
A
45º
Mg sin q ×
h
2
= Mg cos q × r
Þ r =
h
tan
2
q
for q = 45º, r =
h
2
11. e =
22 12
12
2 2 22
1 2 12
esin t ecost
ee
e e ee
æö
ww
ç÷ ++
ç÷
++
èø
Let      cos f =
1
22
12
e
ee +
sin f =
2
22
12
e
ee +
then     e =
22
12
ee + sin(wt + f)
\      e
rms
=
22 22
12 12
ee ee
2 2
+ +
=
12. Intensity I =
2
00
1
EC
2
Î
Þ  E
0
=
0
2I
C Î
=
128
24
8.8 10 3 10
-
´
´ ´´
= 54.87 N/C
and B
0
=
0
E
c
=
8
54.87
3 10 ´
= 1.83 × 10
–7
T
13. According to law of conservation of linear
momentum
mv = MV .......(1)
According to law of conservation of angular
momentum,
mv·
2
L ML
2 12
=w .......(2)
Since the collision is elastic, therefore
2 22
1 11
mv MVI
2 22
= +w
Putting the values of w and V in above equation,
m =
M
4
Page 3

HINT – SHEET
(0999DMD310319007) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 07
29-09-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit - 06
1. The given AC circuit is the combination of two
pure parallel circuits with the applied voltage. In
which I
2
is in phase with V and I
1
2. Work done by the retarding torque = initial KE
i.e., tq = K
Since t is same, hence q or the number of
revolutions will also be same.
3. X
C
> X
L
. Hence, current will lead the voltage.
Z =
22
CL
R (X –X) +
= 102W
\ I
rms
=
rms
V
Z
=
400 / 2
102
= 20 A
cos f =
R1
Z 2
=
V
R
= I
rms
R = (20) (10) = 200 V
4. Mw
2
L = T
0
Þ
0
T
ML
w=
0
1T
f
2 2 ML
w
==
pp
5. tan f =
L 2 50 0.21
R 12
w p´´
=
= 5.5
as tan 45º = 1 and tan 90º = ¥
so best answer is only in between 45º and 90º
which is f = 80º
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 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 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3
LTS/HS-2/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
6. The average value of energy density (energy/volume)
is given by
u
av
=
2
00
1
E
2
e
Total volume of the cylinder V = A·1
\ Total energy contained in the cylinder,
U = (u
av
) (V)
=
2
00
1
E (Al)
2
æö
e
ç÷
èø
Substituting the values, we have
U =
1
2
× (8.86×10
–12
) (50)
2
(10×10
–4
) (50×10
–2
) J
= 5.5 × 10
–12
J
7. KE in both cases should be same and equal to the
loss in potential energy (according to law of
conservation of energy). Hence,
2
2
22
2
1 1 v 1 5v
Mv MkM
2 2 R 24
æö
+=
ç÷
èø
Þ k =
3R
4
8. I
0
=
0
V 100
R 20
=
= 5 A
\ I
rms
=
0
I5
A
22
=
9. E
0
= B
0
c = 2 × 10
–7
× 3 × 10
8
= 60 N/C
EB ^
rr
and as direction of propogation is along
x so E is along z.
10. The cylinder will just begin to slip and will not
topple, if
Moment of force Mg sin q about A
= moment of N about A
45º
G
h/2
Mg sin q
Mg cos q
Mg
r
N
A
45º
Mg sin q ×
h
2
= Mg cos q × r
Þ r =
h
tan
2
q
for q = 45º, r =
h
2
11. e =
22 12
12
2 2 22
1 2 12
esin t ecost
ee
e e ee
æö
ww
ç÷ ++
ç÷
++
èø
Let      cos f =
1
22
12
e
ee +
sin f =
2
22
12
e
ee +
then     e =
22
12
ee + sin(wt + f)
\      e
rms
=
22 22
12 12
ee ee
2 2
+ +
=
12. Intensity I =
2
00
1
EC
2
Î
Þ  E
0
=
0
2I
C Î
=
128
24
8.8 10 3 10
-
´
´ ´´
= 54.87 N/C
and B
0
=
0
E
c
=
8
54.87
3 10 ´
= 1.83 × 10
–7
T
13. According to law of conservation of linear
momentum
mv = MV .......(1)
According to law of conservation of angular
momentum,
mv·
2
L ML
2 12
=w .......(2)
Since the collision is elastic, therefore
2 22
1 11
mv MVI
2 22
= +w
Putting the values of w and V in above equation,
m =
M
4
AL LEN
0999DMD310319007 LTS/HS-3/7
14. In the given graph current is leading the voltage
by 45º. Therefore circuit is C–R
tan f =
C
X
R
or   X
C
= R
16. The cylinder rotating under gravity has both
translational and rotational motions. Let v be the
linear velocity of its centre of mass and w its
angular velocity about the axis of rotation. We can
consider as it is case of inclined plane with angle
of inclination a = 90º
a =
2
2
gsin
K
1
R
a
+

T
T
T
T
M
=
gsin90º
1
1
2
+
=
2
g
3
17. For capacitive circuits
X
C
=
1
C w
\ i =
C
V
X
= VwC  Þ i µ w
18. I = MR
2
8 × (0.2)
2
= 0.32 kg × m
2
t = Ia     0.32 × 3 = 0.96 Nm
But F =
0.96
4.8N
R 0.2
t
==
19. Inductive reactance
X
L
= Lw = 2pfL
also,
E
I
= Lw = X
L
E = IX
L
E =
5
1256
1000
´
E = 6.28 volt (rms)
20. As v =
2
2gh
I
1
MR
+
hence velocity is independent of the inclination
of the plane and depends only on height h through
which body descends
But because t =
2
1 2hI
1
sin g MR
æö
+
ç÷
q
èø
depends on
the inclination also, hence greater the inclination
lesser will be the time of descend. Hence in present
case, the speeds will be same (because h is same)
but time of descend will be different (because of
different inclinations).
21. tan f =
L
X
R
æö
ç÷
èø
and X
L
= wL = (2pfL)
= (2p) (50) (0.01) = pW
also R = 1 W
f = tan
–1
(p)
22. t = Ia Þ a =
I
t
=
500
100
2
w = w
0
+ at
(w – w
0
) = at = 5 × 2 = 10 rad/s
23. I =
2
sin [wt +
42
pp
+
]
Hence phase difference between V and I =
2
p
P = 0
24. By defination, angular momentum
x yz
ˆ ˆˆ
i jk
L r p mx yz
v vv
= ´=
r
rr
Now for motion in xy-plant z = 0, v
z
= 0
So L
r
=
yx
xy
ˆ ˆˆ
i jk
ˆ
m x y 0 mk(xv yv)
vv0
=-
25. I
rms
=
rms
L
E
X
where X
L
= wL = 2 pfL is the reactance of the
inductor
I
rms
=
3
220
22
2 50 44 10
7
-
´ ´ ´´
=
10 1000
100 6
´
´
= 15.9 A
Page 4

HINT – SHEET
(0999DMD310319007) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 07
29-09-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit - 06
1. The given AC circuit is the combination of two
pure parallel circuits with the applied voltage. In
which I
2
is in phase with V and I
1
2. Work done by the retarding torque = initial KE
i.e., tq = K
Since t is same, hence q or the number of
revolutions will also be same.
3. X
C
> X
L
. Hence, current will lead the voltage.
Z =
22
CL
R (X –X) +
= 102W
\ I
rms
=
rms
V
Z
=
400 / 2
102
= 20 A
cos f =
R1
Z 2
=
V
R
= I
rms
R = (20) (10) = 200 V
4. Mw
2
L = T
0
Þ
0
T
ML
w=
0
1T
f
2 2 ML
w
==
pp
5. tan f =
L 2 50 0.21
R 12
w p´´
=
= 5.5
as tan 45º = 1 and tan 90º = ¥
so best answer is only in between 45º and 90º
which is f = 80º
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 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 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3
LTS/HS-2/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
6. The average value of energy density (energy/volume)
is given by
u
av
=
2
00
1
E
2
e
Total volume of the cylinder V = A·1
\ Total energy contained in the cylinder,
U = (u
av
) (V)
=
2
00
1
E (Al)
2
æö
e
ç÷
èø
Substituting the values, we have
U =
1
2
× (8.86×10
–12
) (50)
2
(10×10
–4
) (50×10
–2
) J
= 5.5 × 10
–12
J
7. KE in both cases should be same and equal to the
loss in potential energy (according to law of
conservation of energy). Hence,
2
2
22
2
1 1 v 1 5v
Mv MkM
2 2 R 24
æö
+=
ç÷
èø
Þ k =
3R
4
8. I
0
=
0
V 100
R 20
=
= 5 A
\ I
rms
=
0
I5
A
22
=
9. E
0
= B
0
c = 2 × 10
–7
× 3 × 10
8
= 60 N/C
EB ^
rr
and as direction of propogation is along
x so E is along z.
10. The cylinder will just begin to slip and will not
topple, if
Moment of force Mg sin q about A
= moment of N about A
45º
G
h/2
Mg sin q
Mg cos q
Mg
r
N
A
45º
Mg sin q ×
h
2
= Mg cos q × r
Þ r =
h
tan
2
q
for q = 45º, r =
h
2
11. e =
22 12
12
2 2 22
1 2 12
esin t ecost
ee
e e ee
æö
ww
ç÷ ++
ç÷
++
èø
Let      cos f =
1
22
12
e
ee +
sin f =
2
22
12
e
ee +
then     e =
22
12
ee + sin(wt + f)
\      e
rms
=
22 22
12 12
ee ee
2 2
+ +
=
12. Intensity I =
2
00
1
EC
2
Î
Þ  E
0
=
0
2I
C Î
=
128
24
8.8 10 3 10
-
´
´ ´´
= 54.87 N/C
and B
0
=
0
E
c
=
8
54.87
3 10 ´
= 1.83 × 10
–7
T
13. According to law of conservation of linear
momentum
mv = MV .......(1)
According to law of conservation of angular
momentum,
mv·
2
L ML
2 12
=w .......(2)
Since the collision is elastic, therefore
2 22
1 11
mv MVI
2 22
= +w
Putting the values of w and V in above equation,
m =
M
4
AL LEN
0999DMD310319007 LTS/HS-3/7
14. In the given graph current is leading the voltage
by 45º. Therefore circuit is C–R
tan f =
C
X
R
or   X
C
= R
16. The cylinder rotating under gravity has both
translational and rotational motions. Let v be the
linear velocity of its centre of mass and w its
angular velocity about the axis of rotation. We can
consider as it is case of inclined plane with angle
of inclination a = 90º
a =
2
2
gsin
K
1
R
a
+

T
T
T
T
M
=
gsin90º
1
1
2
+
=
2
g
3
17. For capacitive circuits
X
C
=
1
C w
\ i =
C
V
X
= VwC  Þ i µ w
18. I = MR
2
8 × (0.2)
2
= 0.32 kg × m
2
t = Ia     0.32 × 3 = 0.96 Nm
But F =
0.96
4.8N
R 0.2
t
==
19. Inductive reactance
X
L
= Lw = 2pfL
also,
E
I
= Lw = X
L
E = IX
L
E =
5
1256
1000
´
E = 6.28 volt (rms)
20. As v =
2
2gh
I
1
MR
+
hence velocity is independent of the inclination
of the plane and depends only on height h through
which body descends
But because t =
2
1 2hI
1
sin g MR
æö
+
ç÷
q
èø
depends on
the inclination also, hence greater the inclination
lesser will be the time of descend. Hence in present
case, the speeds will be same (because h is same)
but time of descend will be different (because of
different inclinations).
21. tan f =
L
X
R
æö
ç÷
èø
and X
L
= wL = (2pfL)
= (2p) (50) (0.01) = pW
also R = 1 W
f = tan
–1
(p)
22. t = Ia Þ a =
I
t
=
500
100
2
w = w
0
+ at
(w – w
0
) = at = 5 × 2 = 10 rad/s
23. I =
2
sin [wt +
42
pp
+
]
Hence phase difference between V and I =
2
p
P = 0
24. By defination, angular momentum
x yz
ˆ ˆˆ
i jk
L r p mx yz
v vv
= ´=
r
rr
Now for motion in xy-plant z = 0, v
z
= 0
So L
r
=
yx
xy
ˆ ˆˆ
i jk
ˆ
m x y 0 mk(xv yv)
vv0
=-
25. I
rms
=
rms
L
E
X
where X
L
= wL = 2 pfL is the reactance of the
inductor
I
rms
=
3
220
22
2 50 44 10
7
-
´ ´ ´´
=
10 1000
100 6
´
´
= 15.9 A
LTS/HS-4/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
26. In electromagnetic waves, electric field E
r
and
magnetic field B
r
are perpendicular to the direction
of wave. So direction of wave propagation is given
by
EB ´
rr
.
27. r 3L =
60º
r
O
L
The desired moment of inertia about O is
I = 6 × I
one side
=
2
2
m(2L)
6 mr
12
éù
+
êú
ëû
=
2
2
mL
6 3mL
3
éù
+
êú
ëû
= 20mL
2
28. T =
11
s
f 60
=
w = 2pf = 377 rad/s
X
L
= wL = (377) (0.040)
= 15.08 W
Z =
22
L
xR + = 25.05 W
Amplitudes (maximum value) are,
0
0
V 150
i 6A
Z 25.05
= =»
(V
0
)
R
= i
0
R = 120 V
(V
0
)
L
= i
0
X
L
= 90.5 V
29. I = I
c
+ mR
2
= mR
2
+ mR
2
= 2mR
2
= 2 × 3 × (1)
2
= 6 gm cm
2
30.
3 /2
m
/2
I sin t dt
i
3
22
pw
pw
w
< >=
pp
-
ww
ò
=
3 /2
m
/2
cost
I
0
pw
pw
w æö
-
ç÷
w
èø
=
p
w
31.
k 2 11
2c
p
= =-
w l ´ pn ln
32. I' = 2I
1
= 2kg m
2
I'w
2
= Iw
1
Þ 2Iw
2
= Iw
1
Þw
2
=
1
100
22
w
==
DKE =
22
1 1 22
11
II
22
w-w
=
22
1
1 (100) 2 (50)
2
éù ´ -´
ëû
=
2
1
(50) (4 2)
2
´- = 2500 J = 2.5 kJ
33. cos f =
R
Z
=
R
V IR
IZV
=
22
8 84
105
8 (12 6)
==
+-
35.
2kg
10 N
T T
a
a
For block : 10 – T = 2a ...(i)
For pulley : TR = Ia
T × 0.25 = 2a Þ T = 8a ...(ii)
aRa
4
a
= aÞ=
...(iii)
1082
42
aa
-aÞ ´=
17
10
2
a
Þ = Þa=
20
17
2
37.
CL
XX
tan
R
-
f=
1
2 fL
2 fC
tan45
R
-p
p
°=
( )
1
C
2f2 fL R
éù
=
êú
p p+
êú
ëû
Page 5

HINT – SHEET
(0999DMD310319007) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 07
29-09-2019
LTS/HS-1/7
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
+91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
Test Type : Unit - 06
1. The given AC circuit is the combination of two
pure parallel circuits with the applied voltage. In
which I
2
is in phase with V and I
1
2. Work done by the retarding torque = initial KE
i.e., tq = K
Since t is same, hence q or the number of
revolutions will also be same.
3. X
C
> X
L
. Hence, current will lead the voltage.
Z =
22
CL
R (X –X) +
= 102W
\ I
rms
=
rms
V
Z
=
400 / 2
102
= 20 A
cos f =
R1
Z 2
=
V
R
= I
rms
R = (20) (10) = 200 V
4. Mw
2
L = T
0
Þ
0
T
ML
w=
0
1T
f
2 2 ML
w
==
pp
5. tan f =
L 2 50 0.21
R 12
w p´´
=
= 5.5
as tan 45º = 1 and tan 90º = ¥
so best answer is only in between 45º and 90º
which is f = 80º
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 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 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3
LTS/HS-2/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
6. The average value of energy density (energy/volume)
is given by
u
av
=
2
00
1
E
2
e
Total volume of the cylinder V = A·1
\ Total energy contained in the cylinder,
U = (u
av
) (V)
=
2
00
1
E (Al)
2
æö
e
ç÷
èø
Substituting the values, we have
U =
1
2
× (8.86×10
–12
) (50)
2
(10×10
–4
) (50×10
–2
) J
= 5.5 × 10
–12
J
7. KE in both cases should be same and equal to the
loss in potential energy (according to law of
conservation of energy). Hence,
2
2
22
2
1 1 v 1 5v
Mv MkM
2 2 R 24
æö
+=
ç÷
èø
Þ k =
3R
4
8. I
0
=
0
V 100
R 20
=
= 5 A
\ I
rms
=
0
I5
A
22
=
9. E
0
= B
0
c = 2 × 10
–7
× 3 × 10
8
= 60 N/C
EB ^
rr
and as direction of propogation is along
x so E is along z.
10. The cylinder will just begin to slip and will not
topple, if
Moment of force Mg sin q about A
= moment of N about A
45º
G
h/2
Mg sin q
Mg cos q
Mg
r
N
A
45º
Mg sin q ×
h
2
= Mg cos q × r
Þ r =
h
tan
2
q
for q = 45º, r =
h
2
11. e =
22 12
12
2 2 22
1 2 12
esin t ecost
ee
e e ee
æö
ww
ç÷ ++
ç÷
++
èø
Let      cos f =
1
22
12
e
ee +
sin f =
2
22
12
e
ee +
then     e =
22
12
ee + sin(wt + f)
\      e
rms
=
22 22
12 12
ee ee
2 2
+ +
=
12. Intensity I =
2
00
1
EC
2
Î
Þ  E
0
=
0
2I
C Î
=
128
24
8.8 10 3 10
-
´
´ ´´
= 54.87 N/C
and B
0
=
0
E
c
=
8
54.87
3 10 ´
= 1.83 × 10
–7
T
13. According to law of conservation of linear
momentum
mv = MV .......(1)
According to law of conservation of angular
momentum,
mv·
2
L ML
2 12
=w .......(2)
Since the collision is elastic, therefore
2 22
1 11
mv MVI
2 22
= +w
Putting the values of w and V in above equation,
m =
M
4
AL LEN
0999DMD310319007 LTS/HS-3/7
14. In the given graph current is leading the voltage
by 45º. Therefore circuit is C–R
tan f =
C
X
R
or   X
C
= R
16. The cylinder rotating under gravity has both
translational and rotational motions. Let v be the
linear velocity of its centre of mass and w its
angular velocity about the axis of rotation. We can
consider as it is case of inclined plane with angle
of inclination a = 90º
a =
2
2
gsin
K
1
R
a
+

T
T
T
T
M
=
gsin90º
1
1
2
+
=
2
g
3
17. For capacitive circuits
X
C
=
1
C w
\ i =
C
V
X
= VwC  Þ i µ w
18. I = MR
2
8 × (0.2)
2
= 0.32 kg × m
2
t = Ia     0.32 × 3 = 0.96 Nm
But F =
0.96
4.8N
R 0.2
t
==
19. Inductive reactance
X
L
= Lw = 2pfL
also,
E
I
= Lw = X
L
E = IX
L
E =
5
1256
1000
´
E = 6.28 volt (rms)
20. As v =
2
2gh
I
1
MR
+
hence velocity is independent of the inclination
of the plane and depends only on height h through
which body descends
But because t =
2
1 2hI
1
sin g MR
æö
+
ç÷
q
èø
depends on
the inclination also, hence greater the inclination
lesser will be the time of descend. Hence in present
case, the speeds will be same (because h is same)
but time of descend will be different (because of
different inclinations).
21. tan f =
L
X
R
æö
ç÷
èø
and X
L
= wL = (2pfL)
= (2p) (50) (0.01) = pW
also R = 1 W
f = tan
–1
(p)
22. t = Ia Þ a =
I
t
=
500
100
2
w = w
0
+ at
(w – w
0
) = at = 5 × 2 = 10 rad/s
23. I =
2
sin [wt +
42
pp
+
]
Hence phase difference between V and I =
2
p
P = 0
24. By defination, angular momentum
x yz
ˆ ˆˆ
i jk
L r p mx yz
v vv
= ´=
r
rr
Now for motion in xy-plant z = 0, v
z
= 0
So L
r
=
yx
xy
ˆ ˆˆ
i jk
ˆ
m x y 0 mk(xv yv)
vv0
=-
25. I
rms
=
rms
L
E
X
where X
L
= wL = 2 pfL is the reactance of the
inductor
I
rms
=
3
220
22
2 50 44 10
7
-
´ ´ ´´
=
10 1000
100 6
´
´
= 15.9 A
LTS/HS-4/7 0999DMD310319007
Target : Pre-Medical 2020/NEET-UG/29-09-2019
AL LEN
26. In electromagnetic waves, electric field E
r
and
magnetic field B
r
are perpendicular to the direction
of wave. So direction of wave propagation is given
by
EB ´
rr
.
27. r 3L =
60º
r
O
L
The desired moment of inertia about O is
I = 6 × I
one side
=
2
2
m(2L)
6 mr
12
éù
+
êú
ëû
=
2
2
mL
6 3mL
3
éù
+
êú
ëû
= 20mL
2
28. T =
11
s
f 60
=
w = 2pf = 377 rad/s
X
L
= wL = (377) (0.040)
= 15.08 W
Z =
22
L
xR + = 25.05 W
Amplitudes (maximum value) are,
0
0
V 150
i 6A
Z 25.05
= =»
(V
0
)
R
= i
0
R = 120 V
(V
0
)
L
= i
0
X
L
= 90.5 V
29. I = I
c
+ mR
2
= mR
2
+ mR
2
= 2mR
2
= 2 × 3 × (1)
2
= 6 gm cm
2
30.
3 /2
m
/2
I sin t dt
i
3
22
pw
pw
w
< >=
pp
-
ww
ò
=
3 /2
m
/2
cost
I
0
pw
pw
w æö
-
ç÷
w
èø
=
p
w
31.
k 2 11
2c
p
= =-
w l ´ pn ln
32. I' = 2I
1
= 2kg m
2
I'w
2
= Iw
1
Þ 2Iw
2
= Iw
1
Þw
2
=
1
100
22
w
==
DKE =
22
1 1 22
11
II
22
w-w
=
22
1
1 (100) 2 (50)
2
éù ´ -´
ëû
=
2
1
(50) (4 2)
2
´- = 2500 J = 2.5 kJ
33. cos f =
R
Z
=
R
V IR
IZV
=
22
8 84
105
8 (12 6)
==
+-
35.
2kg
10 N
T T
a
a
For block : 10 – T = 2a ...(i)
For pulley : TR = Ia
T × 0.25 = 2a Þ T = 8a ...(ii)
aRa
4
a
= aÞ=
...(iii)
1082
42
aa
-aÞ ´=
17
10
2
a
Þ = Þa=
20
17
2
37.
CL
XX
tan
R
-
f=
1
2 fL
2 fC
tan45
R
-p
p
°=
( )
1
C
2f2 fL R
éù
=
êú
p p+
êú
ëû
AL LEN
0999DMD310319007 LTS/HS-5/7
38.
w
M
y
O x
R
Angular momentum of disc about origin
L = mvR + Iw
and v = wR,
So L = MwR
2
+
2
MR
2
w
=
2
3
MR
2
w
39. At the time of maximum speed of point P, the rod
should be vertical
DK + DU = 0
Þ
2
2
1m
0 mg0
232
éù æö
éù
w-+-=
êú ç÷
êú
ëû
èø ëû
ll
Þ
3g
w=
l
Velocity of point P V
P
= w l = 3g l .
42. I = I
larger
+ I
smaller
I = 4M(4R)
2
+ [MR
2
+ M(5R)
2
] = 90 MR
2
44. Net current :
I 68 2sint =+w
Now find
( )
T
2
0
v
6 8 2 sin t dt
I 10A
T
+w
==
ò
45.
2
2
mr
I mr
2
=+      L = 2pr m = r×L
2
3
mr
2
=
r =
L
2p
23
22
3 L 3L
L
2 48
r
=´r´=
pp
46. 2NO
2
+ H
2
O ¾¾® HNO
2
+ HNO
3
(X)       (Y)
(X) & (Y) both are acidic
47. Reactivity order. White P > Red P > Black P
P P
P
> P P
P
P
– >
P
P
P P P
P P
P P
P P
P
P
–
P
due to angle strain
48. R–SiCl
3

HOH
¾¾¾®
R–Si(OH)
3

Polymerisation
49. Al
4
C
3
+ 12 HOH ¾¾®  4Al(OH)
3
+ 3CH
4
Be
2
C + 4HOH  ¾¾®  2 Be (OH)
2
+ CH
4
50. I
2
Cl
6
®
I
Cl
Cl
I
Cl
Cl
Cl
Cl
Hybridisation – sp
3
d
2
(4s + 2LP) – planar
51. Pyrosulphuric acid  – H
2
S
2
O
7
S
O
O
O H O S
O
O
O H
all Bond length are not equal
52. Acidic strength µ +ive oxidation state
53. BF
3
®
F B
F
F
54. In pseudo halide 'N' must be present.
55. H
2
S

+ KMnO
4
®     MnO + S  + H O
22
¯
Colloidal sulphur
```
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