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# Test Paper 4 Solution MBBS Notes | EduRev

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## MBBS : Test Paper 4 Solution MBBS Notes | EduRev

``` Page 1

HINT – SHEET
1. Here q = charge at origin O=8 mC = 8 × 10
–3
C
q
0
= charge to be carried from P to Q via R
= –2×10
–9
C
\ r
1
= 3cm = 3 ×10
–2
m
r
2
= 4 cm = 4 × 10
–2
m
As electrostatic force are conservative forces, the
work done in moving q
0
is independent of the path
followed. Thus there is no relevance of the point
of the point R.
Z
Y O
(0,0,3cm) P
(0, 6cm, 9cm)
R
q=8µC
X
Q
(0,4cm,0)
(0999DMD310319004) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 04
18-08-2019
LTS/HS-1/8
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 - 01, 02 & 03
Let W
PQ
be the work done in moving q
0
from P to
Q, then using the relation,
PQ0
0 21
1 11
W qq
4 rr
æö
=-
ç÷
pe
èø
, we get
W
PQ
= 9 × 10
9
×(–2×10
–9
)
3
22
11
8 10
4 10 3 10
-
--
æö
´´-
ç÷
´´
èø
= 1.2 J
2. For ring,
q
Tsinq
T
N
mg
1
Tcosq
From force diagram of ring,
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 1 3 2 2 3 3 2 3 3 3 3 3 2 1 2 3 2 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.4 4 3 4 3 1 4 4 3 2 2 4 2 1 2 3 1 3 2 4
Que.41424344 4546 474849505152535455 5657 585960
Ans.3 3 2 3 1 4 1 3 2 2 3 2 4 1 3 1 4 4 4 1
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 2 2 3 1 1 2 2 2 3 2 1 1 2 1 2 2 2 1
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 4 4 1 3 4 3 2 1 1 3 1 1 1 2 1 2 1 2 2
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 2 2 3 2 3 3 3 2 4 3 3 1 1 2 4 4 1 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 3 3 4 3 4 4 1 1 2 3 3 3 2 1 3 3 3 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.3 4 3 1 2 2 4 4 3 2 4 4 1 3 3 3 4 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 1 1 4 4 2 2 3 3 2 1 2 4 2 2 4 1 3 1 2
Page 2

HINT – SHEET
1. Here q = charge at origin O=8 mC = 8 × 10
–3
C
q
0
= charge to be carried from P to Q via R
= –2×10
–9
C
\ r
1
= 3cm = 3 ×10
–2
m
r
2
= 4 cm = 4 × 10
–2
m
As electrostatic force are conservative forces, the
work done in moving q
0
is independent of the path
followed. Thus there is no relevance of the point
of the point R.
Z
Y O
(0,0,3cm) P
(0, 6cm, 9cm)
R
q=8µC
X
Q
(0,4cm,0)
(0999DMD310319004) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 04
18-08-2019
LTS/HS-1/8
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 - 01, 02 & 03
Let W
PQ
be the work done in moving q
0
from P to
Q, then using the relation,
PQ0
0 21
1 11
W qq
4 rr
æö
=-
ç÷
pe
èø
, we get
W
PQ
= 9 × 10
9
×(–2×10
–9
)
3
22
11
8 10
4 10 3 10
-
--
æö
´´-
ç÷
´´
èø
= 1.2 J
2. For ring,
q
Tsinq
T
N
mg
1
Tcosq
From force diagram of ring,
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 1 3 2 2 3 3 2 3 3 3 3 3 2 1 2 3 2 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.4 4 3 4 3 1 4 4 3 2 2 4 2 1 2 3 1 3 2 4
Que.41424344 4546 474849505152535455 5657 585960
Ans.3 3 2 3 1 4 1 3 2 2 3 2 4 1 3 1 4 4 4 1
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 2 2 3 1 1 2 2 2 3 2 1 1 2 1 2 2 2 1
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 4 4 1 3 4 3 2 1 1 3 1 1 1 2 1 2 1 2 2
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 2 2 3 2 3 3 3 2 4 3 3 1 1 2 4 4 1 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 3 3 4 3 4 4 1 1 2 3 3 3 2 1 3 3 3 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.3 4 3 1 2 2 4 4 3 2 4 4 1 3 3 3 4 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 1 1 4 4 2 2 3 3 2 1 2 4 2 2 4 1 3 1 2
LTS/HS-2/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
Tcosq = m
1
g ....(1)
For block B,  mg = T
T cosq = m
1
g
mg cosq=m
1
g
\
1
m 51
cog
m 102
q= ==
Þq=60°
3. Let at junction C, potential is V
Applying Kirchhoff's junction law at C
20V
A
2W 4W
5V
B
2W
S
i
1
i
2
C
i
3
D
0V
i
1
+ i
2
= i
3
or
A C B C CD
V V V V VV
2 42
- --
+=
or
20 V 5V V0
2 42
- --
+=
or 5V = 45
\ V = 9 volt
\ Current through switch S
3
9
i 4.5A
2
==
4.
1/2
dr 4 3 2t
ˆˆ ˆ
v t i j 2k
dt322
= =´ -+
r
r
=
1/2
ˆˆ ˆ
2t i t j 2k -+
1/2
dv
ˆˆ
a 2 t ij
dt2
-
1
==´-
r
r
At  t = 1 sec
2
ˆˆ
a i j 2 m/s = -=
r
5.
n1
2
dx xa
a sin
a
2axx
-
- æö
=
ç÷
èø -
ò
xa
a
-
= dimensions less
and
2
dx
2axx -
= dimensionless
so a
n
must be dimensionless \ n = 0
6.
m
ˆ
v 25i =
r
km/hr and
tm
ˆ
v 25 3j =
r
tm t m t tmm
v v v v vv = - Þ =+
r r r r rr
=
ˆˆ
25 3j 25i +
t
v 25 31 =+ = 50 km/hr
7.
20W
100V
10W
B
3µF
3µF
1µF
1µF
6µF 2µF
A B C
100V
AB
BC
V 21
V 63
==
\ V
AB
= 25 V and V
BC
= 75V
8.
3 5 15
y x yx
333 3
=-+Þ=-+
f
q
1
tan 30
3
f= Þf= °
\q = 180–f = 180°–30° = 150°
9.
t = 0
t = t
2
t = t
1
y = ut –
1
2
gt
2    Þ

1
2
gt
2
– ut + y = 0
t
1
t
2
=
y
g/2
=
2y
g
, as t
1
= 2, t
2
= 8
y =
g
2
t
1
t
2
=
10
2
× 2 × 8 = 80 m
Page 3

HINT – SHEET
1. Here q = charge at origin O=8 mC = 8 × 10
–3
C
q
0
= charge to be carried from P to Q via R
= –2×10
–9
C
\ r
1
= 3cm = 3 ×10
–2
m
r
2
= 4 cm = 4 × 10
–2
m
As electrostatic force are conservative forces, the
work done in moving q
0
is independent of the path
followed. Thus there is no relevance of the point
of the point R.
Z
Y O
(0,0,3cm) P
(0, 6cm, 9cm)
R
q=8µC
X
Q
(0,4cm,0)
(0999DMD310319004) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 04
18-08-2019
LTS/HS-1/8
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 - 01, 02 & 03
Let W
PQ
be the work done in moving q
0
from P to
Q, then using the relation,
PQ0
0 21
1 11
W qq
4 rr
æö
=-
ç÷
pe
èø
, we get
W
PQ
= 9 × 10
9
×(–2×10
–9
)
3
22
11
8 10
4 10 3 10
-
--
æö
´´-
ç÷
´´
èø
= 1.2 J
2. For ring,
q
Tsinq
T
N
mg
1
Tcosq
From force diagram of ring,
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 1 3 2 2 3 3 2 3 3 3 3 3 2 1 2 3 2 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.4 4 3 4 3 1 4 4 3 2 2 4 2 1 2 3 1 3 2 4
Que.41424344 4546 474849505152535455 5657 585960
Ans.3 3 2 3 1 4 1 3 2 2 3 2 4 1 3 1 4 4 4 1
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 2 2 3 1 1 2 2 2 3 2 1 1 2 1 2 2 2 1
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 4 4 1 3 4 3 2 1 1 3 1 1 1 2 1 2 1 2 2
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 2 2 3 2 3 3 3 2 4 3 3 1 1 2 4 4 1 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 3 3 4 3 4 4 1 1 2 3 3 3 2 1 3 3 3 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.3 4 3 1 2 2 4 4 3 2 4 4 1 3 3 3 4 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 1 1 4 4 2 2 3 3 2 1 2 4 2 2 4 1 3 1 2
LTS/HS-2/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
Tcosq = m
1
g ....(1)
For block B,  mg = T
T cosq = m
1
g
mg cosq=m
1
g
\
1
m 51
cog
m 102
q= ==
Þq=60°
3. Let at junction C, potential is V
Applying Kirchhoff's junction law at C
20V
A
2W 4W
5V
B
2W
S
i
1
i
2
C
i
3
D
0V
i
1
+ i
2
= i
3
or
A C B C CD
V V V V VV
2 42
- --
+=
or
20 V 5V V0
2 42
- --
+=
or 5V = 45
\ V = 9 volt
\ Current through switch S
3
9
i 4.5A
2
==
4.
1/2
dr 4 3 2t
ˆˆ ˆ
v t i j 2k
dt322
= =´ -+
r
r
=
1/2
ˆˆ ˆ
2t i t j 2k -+
1/2
dv
ˆˆ
a 2 t ij
dt2
-
1
==´-
r
r
At  t = 1 sec
2
ˆˆ
a i j 2 m/s = -=
r
5.
n1
2
dx xa
a sin
a
2axx
-
- æö
=
ç÷
èø -
ò
xa
a
-
= dimensions less
and
2
dx
2axx -
= dimensionless
so a
n
must be dimensionless \ n = 0
6.
m
ˆ
v 25i =
r
km/hr and
tm
ˆ
v 25 3j =
r
tm t m t tmm
v v v v vv = - Þ =+
r r r r rr
=
ˆˆ
25 3j 25i +
t
v 25 31 =+ = 50 km/hr
7.
20W
100V
10W
B
3µF
3µF
1µF
1µF
6µF 2µF
A B C
100V
AB
BC
V 21
V 63
==
\ V
AB
= 25 V and V
BC
= 75V
8.
3 5 15
y x yx
333 3
=-+Þ=-+
f
q
1
tan 30
3
f= Þf= °
\q = 180–f = 180°–30° = 150°
9.
t = 0
t = t
2
t = t
1
y = ut –
1
2
gt
2    Þ

1
2
gt
2
– ut + y = 0
t
1
t
2
=
y
g/2
=
2y
g
, as t
1
= 2, t
2
= 8
y =
g
2
t
1
t
2
=
10
2
× 2 × 8 = 80 m
AL LEN
0999DMD310319004 LTS/HS-3/8
10. Suppose resitance of ammeter is R', then equivalent
resistance from ammeter side
R
eq
= R' + R
\ Current, 4 =
20
R'R +
\ R + R' = 5
\ R = 5 – R'
So, R is less than 5W
12. Forces acting on the ball are as shown in figure. The
three concurrent forces are in equilibrium. Using
Lami's theorem.
12
TT 0
sin150 sin120 sin90
1
==
° °°
Þ
1 21
2
T TT 0 sin301
sin30 sin60 1 T sin60 3
1°
==Þ==
°°°
\T
1
= 10 sin30° = 10 × 0.5 = 5 N
and T
2
= 10 sin60° =
3
10
2
´ =
5 3N
13. For a fuse I
2
µ r
3
\
23
11
23
22
Ir
Ir
=
Þ
3
2
2
2
3 0.02
I 0.03
æö
=
ç÷
èø
Þ
3/2
2
3
I3A
2
æö
=´
ç÷
èø
14. The simple circuit is as under,
2µF
12 µF
2 µF
A
B
2µF
Þ
2F m
12F m 4F m
A B
Þ
4 12
23 25F
4 12
´
+ = + =m
+
15.
P QR +=
r rr
and
P QS -=
rr r
so (PQ)·(PQ) R·R + +=
rr r r rr
...(i)
also (P Q)·(P Q) S·S - -=
r r rr rr
...(ii)
eq. (i) + eq. (ii)  Þ
2 2 22
2(P Q ) (R S ) + =+
16.
1W
1W
I
I
I
I
The electric current through ideal voltmeter is zero.
According to loop rule,
E–1 × I – 1 × I = 0
I =
E2
1A
22
==
= V
A
– V
B
= {1×I} = {1×1}=1V
17.
2
max
u
R 500
g
==
2 22
uuu
S 500m
2a 2gsin30 g
= = ==
°
18. 3985 = 3.985 × 10
3
\ order of magnitude is 3
19. On introducing dielectric K in a parallel plate
capacitor, its capacity C' = KC
0
= 5C
0
energy stored
2
0
0
q
W
2C
=
\
22
0
qq
W'
2C' 2 5C
==
´
\
0
W 5
W'1
=
Þ  W' =
0
W
5
20.
60°
20
20 cos 60°
Distance travelled by the man
= 20 cos60° =
1
20 10m
2
´=
21. Maximum torque t
max
= pE = (ql)E
= (4 × 10
–8
× 2 × 10
–4
) × 4 × 10
8
= 32 × 10
–4
N–m
If q = 180°, then
W = pE(1–cos180°) = pE[1–(–1)]
W = 2pE = 2 × 32 × 10
–4
= 64 × 10
–4
J
Page 4

HINT – SHEET
1. Here q = charge at origin O=8 mC = 8 × 10
–3
C
q
0
= charge to be carried from P to Q via R
= –2×10
–9
C
\ r
1
= 3cm = 3 ×10
–2
m
r
2
= 4 cm = 4 × 10
–2
m
As electrostatic force are conservative forces, the
work done in moving q
0
is independent of the path
followed. Thus there is no relevance of the point
of the point R.
Z
Y O
(0,0,3cm) P
(0, 6cm, 9cm)
R
q=8µC
X
Q
(0,4cm,0)
(0999DMD310319004) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 04
18-08-2019
LTS/HS-1/8
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 - 01, 02 & 03
Let W
PQ
be the work done in moving q
0
from P to
Q, then using the relation,
PQ0
0 21
1 11
W qq
4 rr
æö
=-
ç÷
pe
èø
, we get
W
PQ
= 9 × 10
9
×(–2×10
–9
)
3
22
11
8 10
4 10 3 10
-
--
æö
´´-
ç÷
´´
èø
= 1.2 J
2. For ring,
q
Tsinq
T
N
mg
1
Tcosq
From force diagram of ring,
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 1 3 2 2 3 3 2 3 3 3 3 3 2 1 2 3 2 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.4 4 3 4 3 1 4 4 3 2 2 4 2 1 2 3 1 3 2 4
Que.41424344 4546 474849505152535455 5657 585960
Ans.3 3 2 3 1 4 1 3 2 2 3 2 4 1 3 1 4 4 4 1
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 2 2 3 1 1 2 2 2 3 2 1 1 2 1 2 2 2 1
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 4 4 1 3 4 3 2 1 1 3 1 1 1 2 1 2 1 2 2
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 2 2 3 2 3 3 3 2 4 3 3 1 1 2 4 4 1 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 3 3 4 3 4 4 1 1 2 3 3 3 2 1 3 3 3 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.3 4 3 1 2 2 4 4 3 2 4 4 1 3 3 3 4 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 1 1 4 4 2 2 3 3 2 1 2 4 2 2 4 1 3 1 2
LTS/HS-2/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
Tcosq = m
1
g ....(1)
For block B,  mg = T
T cosq = m
1
g
mg cosq=m
1
g
\
1
m 51
cog
m 102
q= ==
Þq=60°
3. Let at junction C, potential is V
Applying Kirchhoff's junction law at C
20V
A
2W 4W
5V
B
2W
S
i
1
i
2
C
i
3
D
0V
i
1
+ i
2
= i
3
or
A C B C CD
V V V V VV
2 42
- --
+=
or
20 V 5V V0
2 42
- --
+=
or 5V = 45
\ V = 9 volt
\ Current through switch S
3
9
i 4.5A
2
==
4.
1/2
dr 4 3 2t
ˆˆ ˆ
v t i j 2k
dt322
= =´ -+
r
r
=
1/2
ˆˆ ˆ
2t i t j 2k -+
1/2
dv
ˆˆ
a 2 t ij
dt2
-
1
==´-
r
r
At  t = 1 sec
2
ˆˆ
a i j 2 m/s = -=
r
5.
n1
2
dx xa
a sin
a
2axx
-
- æö
=
ç÷
èø -
ò
xa
a
-
= dimensions less
and
2
dx
2axx -
= dimensionless
so a
n
must be dimensionless \ n = 0
6.
m
ˆ
v 25i =
r
km/hr and
tm
ˆ
v 25 3j =
r
tm t m t tmm
v v v v vv = - Þ =+
r r r r rr
=
ˆˆ
25 3j 25i +
t
v 25 31 =+ = 50 km/hr
7.
20W
100V
10W
B
3µF
3µF
1µF
1µF
6µF 2µF
A B C
100V
AB
BC
V 21
V 63
==
\ V
AB
= 25 V and V
BC
= 75V
8.
3 5 15
y x yx
333 3
=-+Þ=-+
f
q
1
tan 30
3
f= Þf= °
\q = 180–f = 180°–30° = 150°
9.
t = 0
t = t
2
t = t
1
y = ut –
1
2
gt
2    Þ

1
2
gt
2
– ut + y = 0
t
1
t
2
=
y
g/2
=
2y
g
, as t
1
= 2, t
2
= 8
y =
g
2
t
1
t
2
=
10
2
× 2 × 8 = 80 m
AL LEN
0999DMD310319004 LTS/HS-3/8
10. Suppose resitance of ammeter is R', then equivalent
resistance from ammeter side
R
eq
= R' + R
\ Current, 4 =
20
R'R +
\ R + R' = 5
\ R = 5 – R'
So, R is less than 5W
12. Forces acting on the ball are as shown in figure. The
three concurrent forces are in equilibrium. Using
Lami's theorem.
12
TT 0
sin150 sin120 sin90
1
==
° °°
Þ
1 21
2
T TT 0 sin301
sin30 sin60 1 T sin60 3
1°
==Þ==
°°°
\T
1
= 10 sin30° = 10 × 0.5 = 5 N
and T
2
= 10 sin60° =
3
10
2
´ =
5 3N
13. For a fuse I
2
µ r
3
\
23
11
23
22
Ir
Ir
=
Þ
3
2
2
2
3 0.02
I 0.03
æö
=
ç÷
èø
Þ
3/2
2
3
I3A
2
æö
=´
ç÷
èø
14. The simple circuit is as under,
2µF
12 µF
2 µF
A
B
2µF
Þ
2F m
12F m 4F m
A B
Þ
4 12
23 25F
4 12
´
+ = + =m
+
15.
P QR +=
r rr
and
P QS -=
rr r
so (PQ)·(PQ) R·R + +=
rr r r rr
...(i)
also (P Q)·(P Q) S·S - -=
r r rr rr
...(ii)
eq. (i) + eq. (ii)  Þ
2 2 22
2(P Q ) (R S ) + =+
16.
1W
1W
I
I
I
I
The electric current through ideal voltmeter is zero.
According to loop rule,
E–1 × I – 1 × I = 0
I =
E2
1A
22
==
= V
A
– V
B
= {1×I} = {1×1}=1V
17.
2
max
u
R 500
g
==
2 22
uuu
S 500m
2a 2gsin30 g
= = ==
°
18. 3985 = 3.985 × 10
3
\ order of magnitude is 3
19. On introducing dielectric K in a parallel plate
capacitor, its capacity C' = KC
0
= 5C
0
energy stored
2
0
0
q
W
2C
=
\
22
0
qq
W'
2C' 2 5C
==
´
\
0
W 5
W'1
=
Þ  W' =
0
W
5
20.
60°
20
20 cos 60°
Distance travelled by the man
= 20 cos60° =
1
20 10m
2
´=
21. Maximum torque t
max
= pE = (ql)E
= (4 × 10
–8
× 2 × 10
–4
) × 4 × 10
8
= 32 × 10
–4
N–m
If q = 180°, then
W = pE(1–cos180°) = pE[1–(–1)]
W = 2pE = 2 × 32 × 10
–4
= 64 × 10
–4
J
LTS/HS-4/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
23. Comparing with the standard equation of projectile,
2
2
gx
y xtan
2u cos
= q-
q
We get, q = 45°
and
u 202 =
m/s
Time period of this projectile is 4s. Hence, after 4s
velocity vector will again make 45° with horizontal.
24.
Pitch
LC
Number of division on circular scale
=
Pitch
0.004 mm =
250
Þ Pitch = 1.0 mm
25. Angle of friction q = tan
–1
(µ)
or
1
1
tan 30º
3
-
æö
q==
ç÷
èø
a
F
Suppose the body is dragged by a force F acting at
an angle a with horizontal. Then,
N = mg – F sin a
and F cos a = µN = µ (mg – Fsin a)
\
mg
F
cos sin
m
=
a+ma
…(i)
F is minimum when denominator is maximum or,
d
(cos sin )0
d
a+m a=
a
Þ –sin a + µ cos a = 0
Þ
1
tan
3
a=m=
\ a = 30º, the angle of friction q
\ At a = 30º, force needed is minimum.
Substituting the values in Eq. (i) we have,
min
1
(25) (g)
3
F
( 3/2) (1/ 3)(1/2)
æö
ç÷
èø
=
+
= 12.5 g = 12.5 kgf
26. From the relation,
i
g
G = (i–i
g
)S  Þ S =
g
g
ii
G
i
æö -
ç÷
ç÷
èø
=
36
6
5 10 50 10
100
50 10
--
-
æö ´ -´
´
ç÷
´
èø
= 1W
a resistance of 1W should be connected in parallel
with galvanometer to convert it into the ammeter of
desired range
27. For 5s weight of the body is balanced by the given
force. Hence it will move in a straight line as shown.
5s
2
u sin2
R (ucos )(5)
g
q
= +q
=
2
(50) .sin60
(50 cos30 )(5)
10
°
+ ´°
=
250 3
m
28. As m =
velocityof lightin vaccum
velocityof lightin medium
,
hence, m is dimensionless. Thus, each term on the
R.H.S. of given equation should be dimensionless,
i.e.,
2
B
l
is dimensionless, i.e., B should have
dimension of l
2
, i.e., cm
2
, i.e., area.
29.
T T
W
30º
T
T = 60
W = T + T + T sin 30º
= 60 + 60 + 60 ×
1
2
= 150 N
30. In steady state no current will flow through capacitors.
10W 10W 10W
10W
10W 10W 10W
30W
A
B
100V
\
AB
100
V 50 50V
100
= ´=
\ q
s
= 5 × 10
–6
× 50 = 250 mC
Page 5

HINT – SHEET
1. Here q = charge at origin O=8 mC = 8 × 10
–3
C
q
0
= charge to be carried from P to Q via R
= –2×10
–9
C
\ r
1
= 3cm = 3 ×10
–2
m
r
2
= 4 cm = 4 × 10
–2
m
As electrostatic force are conservative forces, the
work done in moving q
0
is independent of the path
followed. Thus there is no relevance of the point
of the point R.
Z
Y O
(0,0,3cm) P
(0, 6cm, 9cm)
R
q=8µC
X
Q
(0,4cm,0)
(0999DMD310319004) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 04
18-08-2019
LTS/HS-1/8
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 - 01, 02 & 03
Let W
PQ
be the work done in moving q
0
from P to
Q, then using the relation,
PQ0
0 21
1 11
W qq
4 rr
æö
=-
ç÷
pe
èø
, we get
W
PQ
= 9 × 10
9
×(–2×10
–9
)
3
22
11
8 10
4 10 3 10
-
--
æö
´´-
ç÷
´´
èø
= 1.2 J
2. For ring,
q
Tsinq
T
N
mg
1
Tcosq
From force diagram of ring,
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 1 3 2 2 3 3 2 3 3 3 3 3 2 1 2 3 2 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.4 4 3 4 3 1 4 4 3 2 2 4 2 1 2 3 1 3 2 4
Que.41424344 4546 474849505152535455 5657 585960
Ans.3 3 2 3 1 4 1 3 2 2 3 2 4 1 3 1 4 4 4 1
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 2 2 3 1 1 2 2 2 3 2 1 1 2 1 2 2 2 1
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.3 4 4 1 3 4 3 2 1 1 3 1 1 1 2 1 2 1 2 2
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.4 4 2 2 3 2 3 3 3 2 4 3 3 1 1 2 4 4 1 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 3 3 4 3 4 4 1 1 2 3 3 3 2 1 3 3 3 3 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.3 4 3 1 2 2 4 4 3 2 4 4 1 3 3 3 4 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.3 1 1 4 4 2 2 3 3 2 1 2 4 2 2 4 1 3 1 2
LTS/HS-2/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
Tcosq = m
1
g ....(1)
For block B,  mg = T
T cosq = m
1
g
mg cosq=m
1
g
\
1
m 51
cog
m 102
q= ==
Þq=60°
3. Let at junction C, potential is V
Applying Kirchhoff's junction law at C
20V
A
2W 4W
5V
B
2W
S
i
1
i
2
C
i
3
D
0V
i
1
+ i
2
= i
3
or
A C B C CD
V V V V VV
2 42
- --
+=
or
20 V 5V V0
2 42
- --
+=
or 5V = 45
\ V = 9 volt
\ Current through switch S
3
9
i 4.5A
2
==
4.
1/2
dr 4 3 2t
ˆˆ ˆ
v t i j 2k
dt322
= =´ -+
r
r
=
1/2
ˆˆ ˆ
2t i t j 2k -+
1/2
dv
ˆˆ
a 2 t ij
dt2
-
1
==´-
r
r
At  t = 1 sec
2
ˆˆ
a i j 2 m/s = -=
r
5.
n1
2
dx xa
a sin
a
2axx
-
- æö
=
ç÷
èø -
ò
xa
a
-
= dimensions less
and
2
dx
2axx -
= dimensionless
so a
n
must be dimensionless \ n = 0
6.
m
ˆ
v 25i =
r
km/hr and
tm
ˆ
v 25 3j =
r
tm t m t tmm
v v v v vv = - Þ =+
r r r r rr
=
ˆˆ
25 3j 25i +
t
v 25 31 =+ = 50 km/hr
7.
20W
100V
10W
B
3µF
3µF
1µF
1µF
6µF 2µF
A B C
100V
AB
BC
V 21
V 63
==
\ V
AB
= 25 V and V
BC
= 75V
8.
3 5 15
y x yx
333 3
=-+Þ=-+
f
q
1
tan 30
3
f= Þf= °
\q = 180–f = 180°–30° = 150°
9.
t = 0
t = t
2
t = t
1
y = ut –
1
2
gt
2    Þ

1
2
gt
2
– ut + y = 0
t
1
t
2
=
y
g/2
=
2y
g
, as t
1
= 2, t
2
= 8
y =
g
2
t
1
t
2
=
10
2
× 2 × 8 = 80 m
AL LEN
0999DMD310319004 LTS/HS-3/8
10. Suppose resitance of ammeter is R', then equivalent
resistance from ammeter side
R
eq
= R' + R
\ Current, 4 =
20
R'R +
\ R + R' = 5
\ R = 5 – R'
So, R is less than 5W
12. Forces acting on the ball are as shown in figure. The
three concurrent forces are in equilibrium. Using
Lami's theorem.
12
TT 0
sin150 sin120 sin90
1
==
° °°
Þ
1 21
2
T TT 0 sin301
sin30 sin60 1 T sin60 3
1°
==Þ==
°°°
\T
1
= 10 sin30° = 10 × 0.5 = 5 N
and T
2
= 10 sin60° =
3
10
2
´ =
5 3N
13. For a fuse I
2
µ r
3
\
23
11
23
22
Ir
Ir
=
Þ
3
2
2
2
3 0.02
I 0.03
æö
=
ç÷
èø
Þ
3/2
2
3
I3A
2
æö
=´
ç÷
èø
14. The simple circuit is as under,
2µF
12 µF
2 µF
A
B
2µF
Þ
2F m
12F m 4F m
A B
Þ
4 12
23 25F
4 12
´
+ = + =m
+
15.
P QR +=
r rr
and
P QS -=
rr r
so (PQ)·(PQ) R·R + +=
rr r r rr
...(i)
also (P Q)·(P Q) S·S - -=
r r rr rr
...(ii)
eq. (i) + eq. (ii)  Þ
2 2 22
2(P Q ) (R S ) + =+
16.
1W
1W
I
I
I
I
The electric current through ideal voltmeter is zero.
According to loop rule,
E–1 × I – 1 × I = 0
I =
E2
1A
22
==
= V
A
– V
B
= {1×I} = {1×1}=1V
17.
2
max
u
R 500
g
==
2 22
uuu
S 500m
2a 2gsin30 g
= = ==
°
18. 3985 = 3.985 × 10
3
\ order of magnitude is 3
19. On introducing dielectric K in a parallel plate
capacitor, its capacity C' = KC
0
= 5C
0
energy stored
2
0
0
q
W
2C
=
\
22
0
qq
W'
2C' 2 5C
==
´
\
0
W 5
W'1
=
Þ  W' =
0
W
5
20.
60°
20
20 cos 60°
Distance travelled by the man
= 20 cos60° =
1
20 10m
2
´=
21. Maximum torque t
max
= pE = (ql)E
= (4 × 10
–8
× 2 × 10
–4
) × 4 × 10
8
= 32 × 10
–4
N–m
If q = 180°, then
W = pE(1–cos180°) = pE[1–(–1)]
W = 2pE = 2 × 32 × 10
–4
= 64 × 10
–4
J
LTS/HS-4/8 0999DMD310319004
Target : Pre-Medical 2020/NEET-UG/18-08-2019
AL LEN
23. Comparing with the standard equation of projectile,
2
2
gx
y xtan
2u cos
= q-
q
We get, q = 45°
and
u 202 =
m/s
Time period of this projectile is 4s. Hence, after 4s
velocity vector will again make 45° with horizontal.
24.
Pitch
LC
Number of division on circular scale
=
Pitch
0.004 mm =
250
Þ Pitch = 1.0 mm
25. Angle of friction q = tan
–1
(µ)
or
1
1
tan 30º
3
-
æö
q==
ç÷
èø
a
F
Suppose the body is dragged by a force F acting at
an angle a with horizontal. Then,
N = mg – F sin a
and F cos a = µN = µ (mg – Fsin a)
\
mg
F
cos sin
m
=
a+ma
…(i)
F is minimum when denominator is maximum or,
d
(cos sin )0
d
a+m a=
a
Þ –sin a + µ cos a = 0
Þ
1
tan
3
a=m=
\ a = 30º, the angle of friction q
\ At a = 30º, force needed is minimum.
Substituting the values in Eq. (i) we have,
min
1
(25) (g)
3
F
( 3/2) (1/ 3)(1/2)
æö
ç÷
èø
=
+
= 12.5 g = 12.5 kgf
26. From the relation,
i
g
G = (i–i
g
)S  Þ S =
g
g
ii
G
i
æö -
ç÷
ç÷
èø
=
36
6
5 10 50 10
100
50 10
--
-
æö ´ -´
´
ç÷
´
èø
= 1W
a resistance of 1W should be connected in parallel
with galvanometer to convert it into the ammeter of
desired range
27. For 5s weight of the body is balanced by the given
force. Hence it will move in a straight line as shown.
5s
2
u sin2
R (ucos )(5)
g
q
= +q
=
2
(50) .sin60
(50 cos30 )(5)
10
°
+ ´°
=
250 3
m
28. As m =
velocityof lightin vaccum
velocityof lightin medium
,
hence, m is dimensionless. Thus, each term on the
R.H.S. of given equation should be dimensionless,
i.e.,
2
B
l
is dimensionless, i.e., B should have
dimension of l
2
, i.e., cm
2
, i.e., area.
29.
T T
W
30º
T
T = 60
W = T + T + T sin 30º
= 60 + 60 + 60 ×
1
2
= 150 N
30. In steady state no current will flow through capacitors.
10W 10W 10W
10W
10W 10W 10W
30W
A
B
100V
\
AB
100
V 50 50V
100
= ´=
\ q
s
= 5 × 10
–6
× 50 = 250 mC
AL LEN
0999DMD310319004 LTS/HS-5/8
31. If two vectors are parallel their unit vector are same
to each other. Parallel vector is
( ) a·v
a
b
aa
=
r r r
r
rr
\
( ) ( )
222 22 2
ˆˆ ˆ ˆ ˆˆ
ˆ ˆˆ i j k · 6i 2j 2k
i jk
b
11 1 111
ìü
++ +-
++ ïï
=
íý
++ ++
ïï
îþ
r
( )
6 22
ˆ ˆˆ
i jk
3
+- æö
= ++
ç÷
èø

ˆ ˆˆ
2i 2 j 2k = ++
32. Net electric flux from a closed surface in uniform
electric field is always zero.
33. x
2
= t
2
+ 1  or
dx
2x. 2t
dt
=
or
dx
x.t
dt
=
\
2
dxtt
dtx
t1
==
+
2
2
2
2
2 2 2 3/23
t
t1
dx1
t1
dt (t 1) (t 1) x
+-
1
+
= ==
++
34.
A
4m
0.5m
4.5 m
B
a
common
= =
50 10
153
m/s
2
f
s, max
= mN =
1
5
× 50 = 10 N
f
required
=
10
3
× 5 =
50
3
N
Q f
s,max
< f
required
Þ blocks will move separately
N
N
G
f = f
= 10N
s,max

50
50
100 + N
A
B
f
\ a
A
=
10
5
= 2m/s
2
(right)
\ a
B
=
- 50 10
10
= 4m/s
2
(right)
a
rel
= a
B
– a
A
= 2m/s
2
Now \ s
rel
=
´
2
rel
1
at
2
4 =
1
2
× 2 × t
2
\t = 2s
35. Let a = maximum acceleration of A.
Under no slip condition acceleration of B is also a
FBD of A w.r.t. ground
N
a
y
x
mg
µN
45º
45º
45º
SF
y
= 0
\
NN
mg
22
m
=+
…(i)
SF
x
= ma
\
N µN
ma
22
+=
…(ii)
Solving these two equations, we get
1
ag
1
æö +m
=
ç÷
-m
èø
37. Component of
10
3
along the inclined plane is just
sufficient to balance component of weight along the
inclined plane. So there will be no tendency of
slipping. Therefore.
Friction = 0.
38.
wc
ˆˆ
v 20i,v 20j ==
rr
Here we have to look for velocity of wind w.r.t. car.
So
w/C wc
ˆˆ
v v v 20i 20 j = - =-
r rr
This is in south-east direction
39.
3
2
x
I sinxdx x dx cosx C
3
= + =- ++
òò
=
3
x
cosxC
3
-+
40. 2 × 100 cos q = 100
cos q =
1
2
Þ q = 60°
2q = 120°
q
q
V
1
W
1
```
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