Test Paper 1 solution MBBS Notes | EduRev

MBBS : Test Paper 1 solution MBBS Notes | EduRev

 Page 1


HINT – SHEET
LTS/HS-1/8 0999DMD310319001
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2020
Test Type : Unit Test             Test # 01 Test Pattern 
:
 NEET-UG
TEST DATE 
:
 07 - 07 - 2019
ANSWER KEY
1. Electric field E = –
dV
dr
æö
ç÷
èø
–30V +80V
(1)
20V –10V +70V +90V
(2) (3)
For the set of plates as in figure (I)
E
1
 = –
1
300 ( 80) 110
Vm
dr dr
-
- -++ éù
=
êú
ëû
Similarly E
II
 = –
1
20 ( 70) 50
Vm
dr dr
-
+-++ éù
=
êú
ëû
And E
III
 = –
1
10 ( 90) 100
Vm
dr dr
-
- -+ éù
=
êú
ëû
\  E
1
 > E
2
 > E
3
The order of the pair of plates (for E
r
 is
descending order) is I, III and II.
2.
dq
i
dt
=
Q /2
00
dq idt
p
=
òò
Q = () ( )
/2
/2
0
0
4
4sin 2t dt cos2t
2
p
p
=-
ò
= 
[ ]
4
cos cos0
2
- p- = 
[ ]
4
1 1 4coulomb
2
- --=
Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920
Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3
Que.21222324 25262728 29 303132 33 34353637 383940
Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3
Que.41424344 45464748 49 505152 53 54555657 585960
Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1
Que.61626364 65666768 69 707172 73 74757677 787980
Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2
Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2
Page 2


HINT – SHEET
LTS/HS-1/8 0999DMD310319001
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2020
Test Type : Unit Test             Test # 01 Test Pattern 
:
 NEET-UG
TEST DATE 
:
 07 - 07 - 2019
ANSWER KEY
1. Electric field E = –
dV
dr
æö
ç÷
èø
–30V +80V
(1)
20V –10V +70V +90V
(2) (3)
For the set of plates as in figure (I)
E
1
 = –
1
300 ( 80) 110
Vm
dr dr
-
- -++ éù
=
êú
ëû
Similarly E
II
 = –
1
20 ( 70) 50
Vm
dr dr
-
+-++ éù
=
êú
ëû
And E
III
 = –
1
10 ( 90) 100
Vm
dr dr
-
- -+ éù
=
êú
ëû
\  E
1
 > E
2
 > E
3
The order of the pair of plates (for E
r
 is
descending order) is I, III and II.
2.
dq
i
dt
=
Q /2
00
dq idt
p
=
òò
Q = () ( )
/2
/2
0
0
4
4sin 2t dt cos2t
2
p
p
=-
ò
= 
[ ]
4
cos cos0
2
- p- = 
[ ]
4
1 1 4coulomb
2
- --=
Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920
Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3
Que.21222324 25262728 29 303132 33 34353637 383940
Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3
Que.41424344 45464748 49 505152 53 54555657 585960
Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1
Que.61626364 65666768 69 707172 73 74757677 787980
Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2
Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2
LTS/HS-2/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
3. For balance mg = eE Þ E = 
mg
e
Also m = 
3 73
4 4 22
r d (10 ) 1000kg
3 37
-
p=´´´
Þ E = 
73
19
4 22
(10 ) 1000 10
37
260N / C
1.6 10
-
-
´ ´ ´´
=
´
(g = 10 newton/kg, e = 1.6 × 10
–19
 coulomb)
4. Let A and B be two forces
Greatest resultant = A + B = 10 .....(1)
Least resultant = A – B  = 6 .... (2)
Solving (1) & (2) we get A = 8N and B = 2N
when each force is increased by 3N
A' = A + 3 = 8 + 3 = 11 N
B' = B + 3 = 2 + 3 = 5 N
As the forces are acting at an angle of 90º
So  R' = 
( ) ()
22
22
A' B' 115 + =+ = 146 N
5. Flux through curved surface will be equal to
flux through flat surface.
6. In case of spherical metal conductor the charge
quickly spreads uniformly overthe entire
surface because of which charges stay for
longer time on the spherical surface. While in
case of non-spherical surface, the charge
concentration is different at different points due
to which the charges do not stay on the surface
or longer time.
7. Trailing zeroes will be counted if unit placed.
8. q
y
80V 120V
a
O
–q
x
1cm 1cm
It may be that charges of same magnitude but
opposite sign may have placed on y-axis in
such a manner that potential due to them on
x-axis is zero, but they will produce electric
field in addition to 
120 80
20 V / cm
2
-
=
9. The dimensions of
( )( )
( )
2
22 21
2
522
5 132
MLT MLT
EJ
1
MG
M M LT
--
--
==
So, it represents dimensions quantity like angle.
10. The net flux through each closed surface is
determined by the net charge inside
1
S
00
( Q 3Q) 2Q +--
f==
ee
2
S
0
( Q 2Q 3Q)
0
++-
f==
e
3
S
00
(2Q3Q)Q +-
f = =-
ee
4
S
0
Q
f =+
e
11. The cork floats motionless if the weight of the
cork is equal to electrostatic force due to
uniformly charged plane of earth's surface
mg = F
e
 = QE
Since earth is considered as infinite that
E = 
0
2
s
e
mg = 
0
2
s
e
Q
123
0
8
2 mg 2 8.85 10 10 9.8
Q 1 10
--
-
e ´ ´ ´´
s==
´
= 17.3 × 10
–8
 C/m
2
12. cos
–
1
2
æö
- =q
ç÷
èø
cos q = 
1
2
-
q > 90°
q = 120°
as cos (120) = –sin(90+30) = –sin30 = 
1
2
-
13. The potential energy of the system of charges
relative to infinite separation is
U =
B C AC AB
0
q q qq qq 1
4 AB BC AC
éù
++
êú
pe
ëû
=8.55×10
–4
J
A B C
1×10 C
–8
2×10 C
–8
3×10 C
–8
x=0 x=1cm x=2cm x=3cm
U = 
9 16
2
9 10 10 1 2 2 3 13
1 12 10
-
-
´ ´ ´ ´´ æö
++
ç÷
èø
= 9 × 10
–5
 (9.5) = 85.5 × 10
–5
 = 8.55 × 10
–4
 J
Page 3


HINT – SHEET
LTS/HS-1/8 0999DMD310319001
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2020
Test Type : Unit Test             Test # 01 Test Pattern 
:
 NEET-UG
TEST DATE 
:
 07 - 07 - 2019
ANSWER KEY
1. Electric field E = –
dV
dr
æö
ç÷
èø
–30V +80V
(1)
20V –10V +70V +90V
(2) (3)
For the set of plates as in figure (I)
E
1
 = –
1
300 ( 80) 110
Vm
dr dr
-
- -++ éù
=
êú
ëû
Similarly E
II
 = –
1
20 ( 70) 50
Vm
dr dr
-
+-++ éù
=
êú
ëû
And E
III
 = –
1
10 ( 90) 100
Vm
dr dr
-
- -+ éù
=
êú
ëû
\  E
1
 > E
2
 > E
3
The order of the pair of plates (for E
r
 is
descending order) is I, III and II.
2.
dq
i
dt
=
Q /2
00
dq idt
p
=
òò
Q = () ( )
/2
/2
0
0
4
4sin 2t dt cos2t
2
p
p
=-
ò
= 
[ ]
4
cos cos0
2
- p- = 
[ ]
4
1 1 4coulomb
2
- --=
Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920
Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3
Que.21222324 25262728 29 303132 33 34353637 383940
Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3
Que.41424344 45464748 49 505152 53 54555657 585960
Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1
Que.61626364 65666768 69 707172 73 74757677 787980
Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2
Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2
LTS/HS-2/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
3. For balance mg = eE Þ E = 
mg
e
Also m = 
3 73
4 4 22
r d (10 ) 1000kg
3 37
-
p=´´´
Þ E = 
73
19
4 22
(10 ) 1000 10
37
260N / C
1.6 10
-
-
´ ´ ´´
=
´
(g = 10 newton/kg, e = 1.6 × 10
–19
 coulomb)
4. Let A and B be two forces
Greatest resultant = A + B = 10 .....(1)
Least resultant = A – B  = 6 .... (2)
Solving (1) & (2) we get A = 8N and B = 2N
when each force is increased by 3N
A' = A + 3 = 8 + 3 = 11 N
B' = B + 3 = 2 + 3 = 5 N
As the forces are acting at an angle of 90º
So  R' = 
( ) ()
22
22
A' B' 115 + =+ = 146 N
5. Flux through curved surface will be equal to
flux through flat surface.
6. In case of spherical metal conductor the charge
quickly spreads uniformly overthe entire
surface because of which charges stay for
longer time on the spherical surface. While in
case of non-spherical surface, the charge
concentration is different at different points due
to which the charges do not stay on the surface
or longer time.
7. Trailing zeroes will be counted if unit placed.
8. q
y
80V 120V
a
O
–q
x
1cm 1cm
It may be that charges of same magnitude but
opposite sign may have placed on y-axis in
such a manner that potential due to them on
x-axis is zero, but they will produce electric
field in addition to 
120 80
20 V / cm
2
-
=
9. The dimensions of
( )( )
( )
2
22 21
2
522
5 132
MLT MLT
EJ
1
MG
M M LT
--
--
==
So, it represents dimensions quantity like angle.
10. The net flux through each closed surface is
determined by the net charge inside
1
S
00
( Q 3Q) 2Q +--
f==
ee
2
S
0
( Q 2Q 3Q)
0
++-
f==
e
3
S
00
(2Q3Q)Q +-
f = =-
ee
4
S
0
Q
f =+
e
11. The cork floats motionless if the weight of the
cork is equal to electrostatic force due to
uniformly charged plane of earth's surface
mg = F
e
 = QE
Since earth is considered as infinite that
E = 
0
2
s
e
mg = 
0
2
s
e
Q
123
0
8
2 mg 2 8.85 10 10 9.8
Q 1 10
--
-
e ´ ´ ´´
s==
´
= 17.3 × 10
–8
 C/m
2
12. cos
–
1
2
æö
- =q
ç÷
èø
cos q = 
1
2
-
q > 90°
q = 120°
as cos (120) = –sin(90+30) = –sin30 = 
1
2
-
13. The potential energy of the system of charges
relative to infinite separation is
U =
B C AC AB
0
q q qq qq 1
4 AB BC AC
éù
++
êú
pe
ëû
=8.55×10
–4
J
A B C
1×10 C
–8
2×10 C
–8
3×10 C
–8
x=0 x=1cm x=2cm x=3cm
U = 
9 16
2
9 10 10 1 2 2 3 13
1 12 10
-
-
´ ´ ´ ´´ æö
++
ç÷
èø
= 9 × 10
–5
 (9.5) = 85.5 × 10
–5
 = 8.55 × 10
–4
 J
LTS/HS-3/8 0999DMD310319001
Leader Test Series/Joint Package Course/NEET-UG/07-07-2019
14.
T  =cos30° 
T
T =sin30°  A
0
A
0
30°
W
A
0
0
A /2
T 30 =
0
A
T cos30 W °=
0
A
T 60N =
3
60W
2
=
W = 30 3N
15. a
x
 = 
67
qE 10 2 10
m2
-
´´
= = 10 m/s
2
u=10m/s
45º
Y
X
Time of flight T = 
1
2 10
2u sin
2
2s
g 10
´´
q
==
Hence, Range R = u
x
T + 
1
2
a
x
T
2
R = 10 cos 45º × T + 
1
2
a
x
T
2
101
2 10 2 20m
2 2
´ + ´ ´=
(use g = 10 m/s
2
)
16. Initially, force between A and C
F = 
2
2
Q
k
r
C
B
A
+Q +Q –Q
r/2 r/2
r
FF
A C
When a similar sphere B having charge +Q is
kept at the mid point of line joining A and C,
then
Net force on B is F
net
 = F
A
 + F
C
= k
22
22
Q kQ
(r / 2) (r / 2)
+
= 8 
2
2
kQ
8F
r
=
(direction is shown in figure)
17. because dimension of resistance is
[ML
2
T
–3
A
–2
]
18. Force of repulsion between the charges,
F = 
9 12
2
9 10 40 40 10 160
N
9 (0.9)
-
´ ´´´
=
Since F > mg, so string always remains tight at
any position even if velocity of bob is zero.So
minimum speed at lowest point will be such
that it is sufficient to take particle to highest
point, i.e., velocity becomes zero at highest
point.
For this   u
min
 = 4gl 4 10 0.9 6m / s = ´ ´=
20. Pont P lies at equatorial positions of dipole 1
and 2 and axial osition of dipole 3.
Hence field at P due to dipole 1
E
1
 = 
3
k.p
x
 (towards left) due to dipole 2
E
2
 = 
2
k.p
x
 (towards left) due to dipole 3
E
3
 = 
3
k(2p)
x
 (toward righ)
So net field at P will be zero.
+Q –Q
+Q –Q
+Q –Q
3
2
1
E
3
E
1
E
2
P
21. If 
a
r
 is perpendicular to 
b
r
a.b0 =
r
r
3 + l + 6 = 0
l = –9
22.
2
00
mv q 2kq
qv
r 2r 2mm
l ll
= Þ==
pe pe
T = 
2rm
2r
v 2kq
p
=p
l
23. The electric field E at any point on an
equipotential surface acts perpendicular to it.
Therefore, it cannot have any component
parallel (tangential) to the surface.
Page 4


HINT – SHEET
LTS/HS-1/8 0999DMD310319001
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2020
Test Type : Unit Test             Test # 01 Test Pattern 
:
 NEET-UG
TEST DATE 
:
 07 - 07 - 2019
ANSWER KEY
1. Electric field E = –
dV
dr
æö
ç÷
èø
–30V +80V
(1)
20V –10V +70V +90V
(2) (3)
For the set of plates as in figure (I)
E
1
 = –
1
300 ( 80) 110
Vm
dr dr
-
- -++ éù
=
êú
ëû
Similarly E
II
 = –
1
20 ( 70) 50
Vm
dr dr
-
+-++ éù
=
êú
ëû
And E
III
 = –
1
10 ( 90) 100
Vm
dr dr
-
- -+ éù
=
êú
ëû
\  E
1
 > E
2
 > E
3
The order of the pair of plates (for E
r
 is
descending order) is I, III and II.
2.
dq
i
dt
=
Q /2
00
dq idt
p
=
òò
Q = () ( )
/2
/2
0
0
4
4sin 2t dt cos2t
2
p
p
=-
ò
= 
[ ]
4
cos cos0
2
- p- = 
[ ]
4
1 1 4coulomb
2
- --=
Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920
Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3
Que.21222324 25262728 29 303132 33 34353637 383940
Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3
Que.41424344 45464748 49 505152 53 54555657 585960
Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1
Que.61626364 65666768 69 707172 73 74757677 787980
Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2
Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2
LTS/HS-2/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
3. For balance mg = eE Þ E = 
mg
e
Also m = 
3 73
4 4 22
r d (10 ) 1000kg
3 37
-
p=´´´
Þ E = 
73
19
4 22
(10 ) 1000 10
37
260N / C
1.6 10
-
-
´ ´ ´´
=
´
(g = 10 newton/kg, e = 1.6 × 10
–19
 coulomb)
4. Let A and B be two forces
Greatest resultant = A + B = 10 .....(1)
Least resultant = A – B  = 6 .... (2)
Solving (1) & (2) we get A = 8N and B = 2N
when each force is increased by 3N
A' = A + 3 = 8 + 3 = 11 N
B' = B + 3 = 2 + 3 = 5 N
As the forces are acting at an angle of 90º
So  R' = 
( ) ()
22
22
A' B' 115 + =+ = 146 N
5. Flux through curved surface will be equal to
flux through flat surface.
6. In case of spherical metal conductor the charge
quickly spreads uniformly overthe entire
surface because of which charges stay for
longer time on the spherical surface. While in
case of non-spherical surface, the charge
concentration is different at different points due
to which the charges do not stay on the surface
or longer time.
7. Trailing zeroes will be counted if unit placed.
8. q
y
80V 120V
a
O
–q
x
1cm 1cm
It may be that charges of same magnitude but
opposite sign may have placed on y-axis in
such a manner that potential due to them on
x-axis is zero, but they will produce electric
field in addition to 
120 80
20 V / cm
2
-
=
9. The dimensions of
( )( )
( )
2
22 21
2
522
5 132
MLT MLT
EJ
1
MG
M M LT
--
--
==
So, it represents dimensions quantity like angle.
10. The net flux through each closed surface is
determined by the net charge inside
1
S
00
( Q 3Q) 2Q +--
f==
ee
2
S
0
( Q 2Q 3Q)
0
++-
f==
e
3
S
00
(2Q3Q)Q +-
f = =-
ee
4
S
0
Q
f =+
e
11. The cork floats motionless if the weight of the
cork is equal to electrostatic force due to
uniformly charged plane of earth's surface
mg = F
e
 = QE
Since earth is considered as infinite that
E = 
0
2
s
e
mg = 
0
2
s
e
Q
123
0
8
2 mg 2 8.85 10 10 9.8
Q 1 10
--
-
e ´ ´ ´´
s==
´
= 17.3 × 10
–8
 C/m
2
12. cos
–
1
2
æö
- =q
ç÷
èø
cos q = 
1
2
-
q > 90°
q = 120°
as cos (120) = –sin(90+30) = –sin30 = 
1
2
-
13. The potential energy of the system of charges
relative to infinite separation is
U =
B C AC AB
0
q q qq qq 1
4 AB BC AC
éù
++
êú
pe
ëû
=8.55×10
–4
J
A B C
1×10 C
–8
2×10 C
–8
3×10 C
–8
x=0 x=1cm x=2cm x=3cm
U = 
9 16
2
9 10 10 1 2 2 3 13
1 12 10
-
-
´ ´ ´ ´´ æö
++
ç÷
èø
= 9 × 10
–5
 (9.5) = 85.5 × 10
–5
 = 8.55 × 10
–4
 J
LTS/HS-3/8 0999DMD310319001
Leader Test Series/Joint Package Course/NEET-UG/07-07-2019
14.
T  =cos30° 
T
T =sin30°  A
0
A
0
30°
W
A
0
0
A /2
T 30 =
0
A
T cos30 W °=
0
A
T 60N =
3
60W
2
=
W = 30 3N
15. a
x
 = 
67
qE 10 2 10
m2
-
´´
= = 10 m/s
2
u=10m/s
45º
Y
X
Time of flight T = 
1
2 10
2u sin
2
2s
g 10
´´
q
==
Hence, Range R = u
x
T + 
1
2
a
x
T
2
R = 10 cos 45º × T + 
1
2
a
x
T
2
101
2 10 2 20m
2 2
´ + ´ ´=
(use g = 10 m/s
2
)
16. Initially, force between A and C
F = 
2
2
Q
k
r
C
B
A
+Q +Q –Q
r/2 r/2
r
FF
A C
When a similar sphere B having charge +Q is
kept at the mid point of line joining A and C,
then
Net force on B is F
net
 = F
A
 + F
C
= k
22
22
Q kQ
(r / 2) (r / 2)
+
= 8 
2
2
kQ
8F
r
=
(direction is shown in figure)
17. because dimension of resistance is
[ML
2
T
–3
A
–2
]
18. Force of repulsion between the charges,
F = 
9 12
2
9 10 40 40 10 160
N
9 (0.9)
-
´ ´´´
=
Since F > mg, so string always remains tight at
any position even if velocity of bob is zero.So
minimum speed at lowest point will be such
that it is sufficient to take particle to highest
point, i.e., velocity becomes zero at highest
point.
For this   u
min
 = 4gl 4 10 0.9 6m / s = ´ ´=
20. Pont P lies at equatorial positions of dipole 1
and 2 and axial osition of dipole 3.
Hence field at P due to dipole 1
E
1
 = 
3
k.p
x
 (towards left) due to dipole 2
E
2
 = 
2
k.p
x
 (towards left) due to dipole 3
E
3
 = 
3
k(2p)
x
 (toward righ)
So net field at P will be zero.
+Q –Q
+Q –Q
+Q –Q
3
2
1
E
3
E
1
E
2
P
21. If 
a
r
 is perpendicular to 
b
r
a.b0 =
r
r
3 + l + 6 = 0
l = –9
22.
2
00
mv q 2kq
qv
r 2r 2mm
l ll
= Þ==
pe pe
T = 
2rm
2r
v 2kq
p
=p
l
23. The electric field E at any point on an
equipotential surface acts perpendicular to it.
Therefore, it cannot have any component
parallel (tangential) to the surface.
LTS/HS-4/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
24.
ˆˆ
b ij =+
r
component of a along b
r
r
a cos q
ˆ
b = 
a.b
ˆ
b
b
æö
ç÷
èø
r
r
22 22
ˆ ˆ ˆˆ ˆˆ
(2i 3j).(i j) (i j)
.
1 1 11
+ ++
=
++
= 
23
ˆˆ
(i j)
2
+
+
= 
5
ˆˆ
(i j)
2
+
25.
mgsinq
mgcosq
qEcosq
q
qEsinq
qE
mg
N = mg cos q + qE sin q
mg sin q = qE cos q + m (mg cos q + qE sin q)
Þ q = 
mg(1)
E(1)
-m
+m
 (for q = 45º, sin q = cos q)
= 
2
1 10(1 0.5) 10 0.5
3.3 10 C
100(1 0.5) 100 1.5
-
´-´
= =´
+´
26. Potential at large distances (¥) is V¥ = 0.
Potential at the surface of the sphere V
s
=
0
1Q
4R pe
.
The potential at the centre of the spherical shell
V
centre
 = 
0
3 1Q
24R
´
pe
Kinetic energy at the surface = –q (V
s
 – V
¥
)
i.e., (KE)
s
  = –q ×
0
1Q
4R
´
pe
Kinetic energy at the centre,
(KE)
C
 = –q(V
C
–V
¥
) = –q 
0
3 1Q
24R
´
pe
2
centre c
2
surface
(KE) (1/ 2)mv 33
(KE) 22 (1/ 2)mv
==
Þ v
c
 = 1.5v
27. MI I = MR
2
I M 2R
100 100 100
I MR
D DD æö
´ =± ´ +´
ç÷
èø
= ± (1 + 2 × 0.5)
= ± 2%
28. No field inside the hollow conducting sphere.
29. Zero error :
Least count of circular scale 
1
0.02 mm
50
==
Reading of main scale = 0.0 cm
Number of division coincied = 4
reading of circular scale = 4 × 0.02
Zero error = 0.08 mm = +0.008 cm
For diameter sphere :
Reading of main scale = 1.1 cm
Number of division coincied = 14
reading of circular scale = 14 × 0.02 = 0.28 mm
        = 0.028 cm
Reading = 1.128 cm
Diameter = Reading – zero error
      = 1.128 – 0.008 = 1.120 cm
30. A = 5t
2 
+ 4t + 8
dA
10t4
dt
=+
t = 3 sec t
dA
dt
= 10 × 3 + 4 = 34
31.
Q Q/2 –Q
4cm
x
2 22
kQ kQ kQ
0
2 4 x 2(x 4)
- -=
´-
Þ
111
0
8 x 2(x 4)
--=
-
Þ x(x – 4) – 8x(x – 4) – 4x = 0
Þ x
2
 – 16x + 32 = 0  Þ  x ; 13 cm
Page 5


HINT – SHEET
LTS/HS-1/8 0999DMD310319001
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2020
Test Type : Unit Test             Test # 01 Test Pattern 
:
 NEET-UG
TEST DATE 
:
 07 - 07 - 2019
ANSWER KEY
1. Electric field E = –
dV
dr
æö
ç÷
èø
–30V +80V
(1)
20V –10V +70V +90V
(2) (3)
For the set of plates as in figure (I)
E
1
 = –
1
300 ( 80) 110
Vm
dr dr
-
- -++ éù
=
êú
ëû
Similarly E
II
 = –
1
20 ( 70) 50
Vm
dr dr
-
+-++ éù
=
êú
ëû
And E
III
 = –
1
10 ( 90) 100
Vm
dr dr
-
- -+ éù
=
êú
ëû
\  E
1
 > E
2
 > E
3
The order of the pair of plates (for E
r
 is
descending order) is I, III and II.
2.
dq
i
dt
=
Q /2
00
dq idt
p
=
òò
Q = () ( )
/2
/2
0
0
4
4sin 2t dt cos2t
2
p
p
=-
ò
= 
[ ]
4
cos cos0
2
- p- = 
[ ]
4
1 1 4coulomb
2
- --=
Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920
Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3
Que.21222324 25262728 29 303132 33 34353637 383940
Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3
Que.41424344 45464748 49 505152 53 54555657 585960
Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1
Que.61626364 65666768 69 707172 73 74757677 787980
Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4
Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4
Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2
Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3
Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2
LTS/HS-2/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
3. For balance mg = eE Þ E = 
mg
e
Also m = 
3 73
4 4 22
r d (10 ) 1000kg
3 37
-
p=´´´
Þ E = 
73
19
4 22
(10 ) 1000 10
37
260N / C
1.6 10
-
-
´ ´ ´´
=
´
(g = 10 newton/kg, e = 1.6 × 10
–19
 coulomb)
4. Let A and B be two forces
Greatest resultant = A + B = 10 .....(1)
Least resultant = A – B  = 6 .... (2)
Solving (1) & (2) we get A = 8N and B = 2N
when each force is increased by 3N
A' = A + 3 = 8 + 3 = 11 N
B' = B + 3 = 2 + 3 = 5 N
As the forces are acting at an angle of 90º
So  R' = 
( ) ()
22
22
A' B' 115 + =+ = 146 N
5. Flux through curved surface will be equal to
flux through flat surface.
6. In case of spherical metal conductor the charge
quickly spreads uniformly overthe entire
surface because of which charges stay for
longer time on the spherical surface. While in
case of non-spherical surface, the charge
concentration is different at different points due
to which the charges do not stay on the surface
or longer time.
7. Trailing zeroes will be counted if unit placed.
8. q
y
80V 120V
a
O
–q
x
1cm 1cm
It may be that charges of same magnitude but
opposite sign may have placed on y-axis in
such a manner that potential due to them on
x-axis is zero, but they will produce electric
field in addition to 
120 80
20 V / cm
2
-
=
9. The dimensions of
( )( )
( )
2
22 21
2
522
5 132
MLT MLT
EJ
1
MG
M M LT
--
--
==
So, it represents dimensions quantity like angle.
10. The net flux through each closed surface is
determined by the net charge inside
1
S
00
( Q 3Q) 2Q +--
f==
ee
2
S
0
( Q 2Q 3Q)
0
++-
f==
e
3
S
00
(2Q3Q)Q +-
f = =-
ee
4
S
0
Q
f =+
e
11. The cork floats motionless if the weight of the
cork is equal to electrostatic force due to
uniformly charged plane of earth's surface
mg = F
e
 = QE
Since earth is considered as infinite that
E = 
0
2
s
e
mg = 
0
2
s
e
Q
123
0
8
2 mg 2 8.85 10 10 9.8
Q 1 10
--
-
e ´ ´ ´´
s==
´
= 17.3 × 10
–8
 C/m
2
12. cos
–
1
2
æö
- =q
ç÷
èø
cos q = 
1
2
-
q > 90°
q = 120°
as cos (120) = –sin(90+30) = –sin30 = 
1
2
-
13. The potential energy of the system of charges
relative to infinite separation is
U =
B C AC AB
0
q q qq qq 1
4 AB BC AC
éù
++
êú
pe
ëû
=8.55×10
–4
J
A B C
1×10 C
–8
2×10 C
–8
3×10 C
–8
x=0 x=1cm x=2cm x=3cm
U = 
9 16
2
9 10 10 1 2 2 3 13
1 12 10
-
-
´ ´ ´ ´´ æö
++
ç÷
èø
= 9 × 10
–5
 (9.5) = 85.5 × 10
–5
 = 8.55 × 10
–4
 J
LTS/HS-3/8 0999DMD310319001
Leader Test Series/Joint Package Course/NEET-UG/07-07-2019
14.
T  =cos30° 
T
T =sin30°  A
0
A
0
30°
W
A
0
0
A /2
T 30 =
0
A
T cos30 W °=
0
A
T 60N =
3
60W
2
=
W = 30 3N
15. a
x
 = 
67
qE 10 2 10
m2
-
´´
= = 10 m/s
2
u=10m/s
45º
Y
X
Time of flight T = 
1
2 10
2u sin
2
2s
g 10
´´
q
==
Hence, Range R = u
x
T + 
1
2
a
x
T
2
R = 10 cos 45º × T + 
1
2
a
x
T
2
101
2 10 2 20m
2 2
´ + ´ ´=
(use g = 10 m/s
2
)
16. Initially, force between A and C
F = 
2
2
Q
k
r
C
B
A
+Q +Q –Q
r/2 r/2
r
FF
A C
When a similar sphere B having charge +Q is
kept at the mid point of line joining A and C,
then
Net force on B is F
net
 = F
A
 + F
C
= k
22
22
Q kQ
(r / 2) (r / 2)
+
= 8 
2
2
kQ
8F
r
=
(direction is shown in figure)
17. because dimension of resistance is
[ML
2
T
–3
A
–2
]
18. Force of repulsion between the charges,
F = 
9 12
2
9 10 40 40 10 160
N
9 (0.9)
-
´ ´´´
=
Since F > mg, so string always remains tight at
any position even if velocity of bob is zero.So
minimum speed at lowest point will be such
that it is sufficient to take particle to highest
point, i.e., velocity becomes zero at highest
point.
For this   u
min
 = 4gl 4 10 0.9 6m / s = ´ ´=
20. Pont P lies at equatorial positions of dipole 1
and 2 and axial osition of dipole 3.
Hence field at P due to dipole 1
E
1
 = 
3
k.p
x
 (towards left) due to dipole 2
E
2
 = 
2
k.p
x
 (towards left) due to dipole 3
E
3
 = 
3
k(2p)
x
 (toward righ)
So net field at P will be zero.
+Q –Q
+Q –Q
+Q –Q
3
2
1
E
3
E
1
E
2
P
21. If 
a
r
 is perpendicular to 
b
r
a.b0 =
r
r
3 + l + 6 = 0
l = –9
22.
2
00
mv q 2kq
qv
r 2r 2mm
l ll
= Þ==
pe pe
T = 
2rm
2r
v 2kq
p
=p
l
23. The electric field E at any point on an
equipotential surface acts perpendicular to it.
Therefore, it cannot have any component
parallel (tangential) to the surface.
LTS/HS-4/8 0999DMD310319001
Target : Pre-Medical 2020/NEET-UG/07-07-2019
24.
ˆˆ
b ij =+
r
component of a along b
r
r
a cos q
ˆ
b = 
a.b
ˆ
b
b
æö
ç÷
èø
r
r
22 22
ˆ ˆ ˆˆ ˆˆ
(2i 3j).(i j) (i j)
.
1 1 11
+ ++
=
++
= 
23
ˆˆ
(i j)
2
+
+
= 
5
ˆˆ
(i j)
2
+
25.
mgsinq
mgcosq
qEcosq
q
qEsinq
qE
mg
N = mg cos q + qE sin q
mg sin q = qE cos q + m (mg cos q + qE sin q)
Þ q = 
mg(1)
E(1)
-m
+m
 (for q = 45º, sin q = cos q)
= 
2
1 10(1 0.5) 10 0.5
3.3 10 C
100(1 0.5) 100 1.5
-
´-´
= =´
+´
26. Potential at large distances (¥) is V¥ = 0.
Potential at the surface of the sphere V
s
=
0
1Q
4R pe
.
The potential at the centre of the spherical shell
V
centre
 = 
0
3 1Q
24R
´
pe
Kinetic energy at the surface = –q (V
s
 – V
¥
)
i.e., (KE)
s
  = –q ×
0
1Q
4R
´
pe
Kinetic energy at the centre,
(KE)
C
 = –q(V
C
–V
¥
) = –q 
0
3 1Q
24R
´
pe
2
centre c
2
surface
(KE) (1/ 2)mv 33
(KE) 22 (1/ 2)mv
==
Þ v
c
 = 1.5v
27. MI I = MR
2
I M 2R
100 100 100
I MR
D DD æö
´ =± ´ +´
ç÷
èø
= ± (1 + 2 × 0.5)
= ± 2%
28. No field inside the hollow conducting sphere.
29. Zero error :
Least count of circular scale 
1
0.02 mm
50
==
Reading of main scale = 0.0 cm
Number of division coincied = 4
reading of circular scale = 4 × 0.02
Zero error = 0.08 mm = +0.008 cm
For diameter sphere :
Reading of main scale = 1.1 cm
Number of division coincied = 14
reading of circular scale = 14 × 0.02 = 0.28 mm
        = 0.028 cm
Reading = 1.128 cm
Diameter = Reading – zero error
      = 1.128 – 0.008 = 1.120 cm
30. A = 5t
2 
+ 4t + 8
dA
10t4
dt
=+
t = 3 sec t
dA
dt
= 10 × 3 + 4 = 34
31.
Q Q/2 –Q
4cm
x
2 22
kQ kQ kQ
0
2 4 x 2(x 4)
- -=
´-
Þ
111
0
8 x 2(x 4)
--=
-
Þ x(x – 4) – 8x(x – 4) – 4x = 0
Þ x
2
 – 16x + 32 = 0  Þ  x ; 13 cm
LTS/HS-5/8 0999DMD310319001
Leader Test Series/Joint Package Course/NEET-UG/07-07-2019
32.
ˆˆ ˆ
i jk
r F 3 23
2 34
t= ´=
-
r
r r
= 
ˆˆ
[(2 4) – (3 3)]i [(2 3)– (3 4)]j ´ ´- +´´
ˆˆ ˆˆ
[(3 3)– (2 2)]k 17i 6 j 13k + ´- ´ = --
33. The facing surfaces have equal and opposite
charges.
Q
O
A
B
C
Q
O
A
B
C
–Q
+Q
–Q
+Q
34.
1.26
2.3
3.56
+
Rounding off to 1 decimal place we get 3.6.
In addition or subtraction the number of decimal
places in the result should be equal to the number
of decimal places of that term in the operation
which contain lesser number of decimal places.
e.g. 12.587 – 12.5 = 0.087 = 0.1 (Q second term
contain lesser i.e. one decimal place)
35. Let T µ P
a
 d
b
 E
c
Writing dimensions on both sides.
[M
0
L
0
T] = [ML
–1
T
–2
]
a
[ML
–3
]
b
[ML
2
T
–2
]
c
[M
0
L
0
T] = M
a+b+c
 L
–a–3b+2c
T
–2a–2c
Thus,
a + b + c = 0, –a – 3b + 2c = 0, –2a – 2c = 1
On solving these equation, we get
a = 
5
6
-
,  b = 
1
2
 and c = 
1
3
36. because electric field applies the force on
electron in the direction opposite to its motion.
37. 38.7
If the digit to be rounded off is more than
5, then the preceding digit is increased by
one. e.g. 6.87» 6.9.
If the digit to be rounded off is less than 5,
then the preceding digit is  left unchanged.
e.g. 3.94 » 3.9
38. Total enclosed charge q = 100 Q coulomb
f
E 
= 
00
q 100Q
=
ee
39. The electric field at a point x is
E(x) = 
2
dV(x)d
[4(1 x )] 8x
dx dx
- =- + =-
The electric field and hence force on a positive
charge of 1C, is in negative x-axis direction.
(Q F = qE)
40.
AB ´
r r
 is ^ to plane of AandB
r r
41. Metal plate acts as an equipotential surface,
therefore the field lines should enter normal to
the surface of the metal plate.
42. g = 
2
2
4
T
p l
g
100
g
D
´ = 
T
100 2 100
T
DD
´ +´
l
l
Dl and DT are minimum for option (4) and also
maximum number of observations are taken in
option (4) only.
43. Distance between A(1, 2, 4) and B(3, 2, 1)
2 22
r (3 1) (2 2) (1 4) 13 = - + - +-=
V = 
98
0
1 Q 9 10 2 10
50volt
4r 13
-
´ ´´
==
pÎ
E = 
98
2
0
1 Q 9 10 2 10 180
N/C
4 13 13 r
-
´ ´´
==
pÎ
Unit vector along AB, 
ˆ ˆ
2i 3k
ˆ
r
13
-
=
180
ˆ ˆ
ˆ E Er (2i 3k)
13 13
==-
r
44. At t = 1
x
A
 = 4, x
B
 = 7
y
A
 = 3, y
B
 = 7
distance = 
22
(7–4) (7–3) + = 5
45. V = 4 + 5x
2
(i) x = 1, V
1
 = 9
   x = –2, V
2
 = 24
   V
2
 – V
1
 = 15 V, (i) is O.K.
(ii) E
x
 = –
dV
dx
 = –10x
    at x = –1m, E = 10 NC
    F = qE = 1 × 10 = 10N, (ii) is O.K.
(iii) 
ˆˆ
F qE 10xi 10 iN = =-=
rr
, (iii) is O.K.
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