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# Test Paper 5 Solution MBBS Notes | EduRev

## MBBS : Test Paper 5 Solution MBBS Notes | EduRev

``` Page 1

HINT – SHEET
1.
mg
2
mmg
1
m
1
a
a
acc.  a =
net pulling force
mass to be pulled
=
21
12
m g mg
mm
-m
+
as  v
2
= u
2
+ 2as and u = 0
21
12
2Lg(m m)
v 2La
mm
-m
==
+
OR
Work-energy method : The essence of the
process in terms of energy condsiderations is
that m
2
loses potential energy of amount m
2
gL
while coming down by L, and the loss of
potential energy of m
2
appears partly in m
1
and
m
2
in the form of kinetic energy and is partly
used up as work done against friction.
(0999DMD310319005) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 05
01-09-2019
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 Test # 04
m
2
gL =
2
121
1
(m m )v m gL
2
+ +m
Þ (m
2
– mm
1
)gL =
1
2
(m
1
+ m
2
)v
2
Þ v =
21
12
2(m m )gL
mm
-m
+
=
- ´´
=
+
2(2 0.5 2)10 5
5m /s
22
2. Since the wheels are connected with a common
belt, their linear velocity is constant, that is v
1
=v
2
w = w
0
+ wt  = 0 + 1.5 × 10 = 15 rad s
–1
r
1
w
1
= r
2
w
2
or w
2
= w
1

1
2
r
r
=
1
15
3
–1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.2 1 1 3 2 1 1 1 3 1 3 3 3 2 1 1 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.2 4 3 1 1 3 2 3 2 2 3 1 4 3 4 1 4 4 1 1
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.2 1 3 2 4 1 2 4 3 2 1 2 2 4 3 3 4 2 2 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.4 1 4 4 2 2 2 1 1 4 3 1 1 3 3 1 2 1 4 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 2 1 2 4 1 4 3 2 3 3 3 3 1 3 3 2 2 1
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 2 2 2 1 2 3 4 1 2 3 1 1 2 3 3 2 1 2
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 3 2 2 2 1 2 1 3 1 1 2 3 1 3 2 4 4 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.4 4 1 1 1 2 4 1 2 2 2 3 1 2 1 2 1 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 3 3 4 1 1 2 2 3 1 2 1 2 2 2 1 3 1 1
Page 2

HINT – SHEET
1.
mg
2
mmg
1
m
1
a
a
acc.  a =
net pulling force
mass to be pulled
=
21
12
m g mg
mm
-m
+
as  v
2
= u
2
+ 2as and u = 0
21
12
2Lg(m m)
v 2La
mm
-m
==
+
OR
Work-energy method : The essence of the
process in terms of energy condsiderations is
that m
2
loses potential energy of amount m
2
gL
while coming down by L, and the loss of
potential energy of m
2
appears partly in m
1
and
m
2
in the form of kinetic energy and is partly
used up as work done against friction.
(0999DMD310319005) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 05
01-09-2019
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 Test # 04
m
2
gL =
2
121
1
(m m )v m gL
2
+ +m
Þ (m
2
– mm
1
)gL =
1
2
(m
1
+ m
2
)v
2
Þ v =
21
12
2(m m )gL
mm
-m
+
=
- ´´
=
+
2(2 0.5 2)10 5
5m /s
22
2. Since the wheels are connected with a common
belt, their linear velocity is constant, that is v
1
=v
2
w = w
0
+ wt  = 0 + 1.5 × 10 = 15 rad s
–1
r
1
w
1
= r
2
w
2
or w
2
= w
1

1
2
r
r
=
1
15
3
–1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.2 1 1 3 2 1 1 1 3 1 3 3 3 2 1 1 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.2 4 3 1 1 3 2 3 2 2 3 1 4 3 4 1 4 4 1 1
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.2 1 3 2 4 1 2 4 3 2 1 2 2 4 3 3 4 2 2 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.4 1 4 4 2 2 2 1 1 4 3 1 1 3 3 1 2 1 4 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 2 1 2 4 1 4 3 2 3 3 3 3 1 3 3 2 2 1
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 2 2 2 1 2 3 4 1 2 3 1 1 2 3 3 2 1 2
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 3 2 2 2 1 2 1 3 1 1 2 3 1 3 2 4 4 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.4 4 1 1 1 2 4 1 2 2 2 3 1 2 1 2 1 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 3 3 4 1 1 2 2 3 1 2 1 2 2 2 1 3 1 1
LTS/HS-2/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
3. The field at P due to upper and lower squares will
cancel out. Resultant field at P will be due to the
square BCGF as shown in the figure. Solve to get,
B
P
=
0
4I
3a
m
p
A
D C
B
G
F
E
P
a/2
B
C
G
F
P
4. The observer moving with wedge will observe a
pseudo force ma
0
in the direction opposite to
0
a
r
.
a
0
l
h
F
pseudo

ps0
F ma =-
r
r
is a constant force
\ W
ps
=
ps
F .S
r r
= ma
0
l = 10 × 3 × 3 = 90 J
5. F = qvB =
0
qvi
2r
m
p
2
0
0
qvi mv 2 rmv
R
R 2r qi
m p
= Þ=
pm
6. 1
2
v 5gR 2 gR
5
==
222
1
v v 2gR v 4gR 2gR = - Þ =-
v 2gR Þ=
2
c
v
a 2g
R
==
v
B
a
t
a
q
a
c
a
t
= g
1 t
c
ag11
tan tan
a2g22
-
æö
a= = = Þa=
ç÷
èø
7. The horizontal components are (B
H
)
1
= B cos f
1
and (B
H
)
2
= B cos f
2
\
( )
()
H
1 1
H2
2
B
cos cos30
B cos cos 45
f°
==
f°
33
2
2 2
= ´=
8. Let R be the resistance force offered by the planks,
d
1
be the thickness of first plank and d
2
be the
thickness of second plank.
For first plank :
Loss in KE = work against resistance
2
22
11
1 1 4 19
mv m v Rd mv Rd
2 2 5 2 25
æ ö æö
- =Þ=
ç ÷ ç÷
è ø èø
(i)
For second plank :
2
2
22
1 4 1 16
m v 0 Rd mv Rd
2 5 2 25
æ ö æö
-=Þ=
ç ÷ ç÷
è ø èø
(ii)
Dividing eq. (i) by eq. (ii), we get
1
2
d 9
d 16
=
9.
2
2
2
22
nett
v 25
a a1
r 25
æö
æö
= +=+
ç÷ ç÷
èø
èø
=
–2
2 ms
F = m a
net
= 500 2 N.
10.
0 AB
2II
mg
4r
m
´=
p
l
where r = 2.5 cm
m = (0.100 g/cm)
l
Solve to get
B
250
IA
3
=
11. Power P = F.v  or  P = (ma)v
\ a =
P
mv
or  v .
dv
ds
=
P
mv
or  v
2
dv =
P
m
ds  or
P
m
òò
2
1
v s
2
0v
ds = v .dv
or
P
m
(s) =
1
3
(v
3
2
– v
1
3
)  or  s =
m
3P
(v
2
3
– v
1
3
)
Page 3

HINT – SHEET
1.
mg
2
mmg
1
m
1
a
a
acc.  a =
net pulling force
mass to be pulled
=
21
12
m g mg
mm
-m
+
as  v
2
= u
2
+ 2as and u = 0
21
12
2Lg(m m)
v 2La
mm
-m
==
+
OR
Work-energy method : The essence of the
process in terms of energy condsiderations is
that m
2
loses potential energy of amount m
2
gL
while coming down by L, and the loss of
potential energy of m
2
appears partly in m
1
and
m
2
in the form of kinetic energy and is partly
used up as work done against friction.
(0999DMD310319005) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 05
01-09-2019
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 Test # 04
m
2
gL =
2
121
1
(m m )v m gL
2
+ +m
Þ (m
2
– mm
1
)gL =
1
2
(m
1
+ m
2
)v
2
Þ v =
21
12
2(m m )gL
mm
-m
+
=
- ´´
=
+
2(2 0.5 2)10 5
5m /s
22
2. Since the wheels are connected with a common
belt, their linear velocity is constant, that is v
1
=v
2
w = w
0
+ wt  = 0 + 1.5 × 10 = 15 rad s
–1
r
1
w
1
= r
2
w
2
or w
2
= w
1

1
2
r
r
=
1
15
3
–1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.2 1 1 3 2 1 1 1 3 1 3 3 3 2 1 1 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.2 4 3 1 1 3 2 3 2 2 3 1 4 3 4 1 4 4 1 1
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.2 1 3 2 4 1 2 4 3 2 1 2 2 4 3 3 4 2 2 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.4 1 4 4 2 2 2 1 1 4 3 1 1 3 3 1 2 1 4 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 2 1 2 4 1 4 3 2 3 3 3 3 1 3 3 2 2 1
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 2 2 2 1 2 3 4 1 2 3 1 1 2 3 3 2 1 2
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 3 2 2 2 1 2 1 3 1 1 2 3 1 3 2 4 4 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.4 4 1 1 1 2 4 1 2 2 2 3 1 2 1 2 1 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 3 3 4 1 1 2 2 3 1 2 1 2 2 2 1 3 1 1
LTS/HS-2/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
3. The field at P due to upper and lower squares will
cancel out. Resultant field at P will be due to the
square BCGF as shown in the figure. Solve to get,
B
P
=
0
4I
3a
m
p
A
D C
B
G
F
E
P
a/2
B
C
G
F
P
4. The observer moving with wedge will observe a
pseudo force ma
0
in the direction opposite to
0
a
r
.
a
0
l
h
F
pseudo

ps0
F ma =-
r
r
is a constant force
\ W
ps
=
ps
F .S
r r
= ma
0
l = 10 × 3 × 3 = 90 J
5. F = qvB =
0
qvi
2r
m
p
2
0
0
qvi mv 2 rmv
R
R 2r qi
m p
= Þ=
pm
6. 1
2
v 5gR 2 gR
5
==
222
1
v v 2gR v 4gR 2gR = - Þ =-
v 2gR Þ=
2
c
v
a 2g
R
==
v
B
a
t
a
q
a
c
a
t
= g
1 t
c
ag11
tan tan
a2g22
-
æö
a= = = Þa=
ç÷
èø
7. The horizontal components are (B
H
)
1
= B cos f
1
and (B
H
)
2
= B cos f
2
\
( )
()
H
1 1
H2
2
B
cos cos30
B cos cos 45
f°
==
f°
33
2
2 2
= ´=
8. Let R be the resistance force offered by the planks,
d
1
be the thickness of first plank and d
2
be the
thickness of second plank.
For first plank :
Loss in KE = work against resistance
2
22
11
1 1 4 19
mv m v Rd mv Rd
2 2 5 2 25
æ ö æö
- =Þ=
ç ÷ ç÷
è ø èø
(i)
For second plank :
2
2
22
1 4 1 16
m v 0 Rd mv Rd
2 5 2 25
æ ö æö
-=Þ=
ç ÷ ç÷
è ø èø
(ii)
Dividing eq. (i) by eq. (ii), we get
1
2
d 9
d 16
=
9.
2
2
2
22
nett
v 25
a a1
r 25
æö
æö
= +=+
ç÷ ç÷
èø
èø
=
–2
2 ms
F = m a
net
= 500 2 N.
10.
0 AB
2II
mg
4r
m
´=
p
l
where r = 2.5 cm
m = (0.100 g/cm)
l
Solve to get
B
250
IA
3
=
11. Power P = F.v  or  P = (ma)v
\ a =
P
mv
or  v .
dv
ds
=
P
mv
or  v
2
dv =
P
m
ds  or
P
m
òò
2
1
v s
2
0v
ds = v .dv
or
P
m
(s) =
1
3
(v
3
2
– v
1
3
)  or  s =
m
3P
(v
2
3
– v
1
3
)
AL LEN
0999DMD310319005 LTS/HS-3/8
12. Let current in AB is I
1
and in DC, I
2
. Then
C D
A
I
E I
2
I
2
I
1
I
2
I
1
2
I 3
I1
=
. It is because resistance of AB will be one-
Now,
11
22
BI
3
BI
==
13. Total acceleration
2
2
2
T
v
aa
r
æö
=+
ç÷
èø
Friction will provide a
T
\
2
2
2
v
ga
r
æö
m=+
ç÷
èø
\
4
222
2
v
ga
r
m =+
v = [(m
2
g
2
– a
2
)r
2
]
1/4
14. I =
37
3
M M 7.5 10 9 10
0.09A / m
V m 75 10
-
-
r ´ ´´
===
´
15. Its initial speed v
1
=
1
36000
10ms
3600
-
=
If the car doubles its speed, finally its speed
becomes v
2
= 20 ms
–1
Change in KE of the car
= DKE = (1/2)
22
21
mv (1/ 2)mv -
During Dt = 1 min = 60 s
the power delivered by the engine,
P =
Work done
Time taken
=
| KE|
t
D
D
=
22
21
1
mvv
2
t
éù ´-
ëû
D
=
22
1 / 2 500[20 10 ]
60
´-
= 1250 W
16.
F FF
® ®®
==
\ Net force on frame
=
CD
3F (3) (2) (1) (4) 24N
®
==  (F = ilB)
17. v Rg =m Þ
2
v Rg =m
2
v 70 70
Rg 1000 9.8
´
m==
´
= 0.5
18. Let h be the height of the inclined plane
h = 5 sin 37º = 3 m
mgh =
2
1
mv 0 ( mgcos37º)s
2
æö
- +m
ç÷
èø
Þ
2
14
3mg mv (0.25) mg 5
25
= + ´ ´´
Þ
1
v 4g 6.26ms
-
==
19. From the figure it is clear that deviationis 180º if
x > R
× ×
× ×
×
× ×
× ×
l
v
0
R
v
0
×
×
×
×
×
x=1.5R
20. As they enter the magnetic field of the earth, they
are deflected away from the equator.
22. Net force on a current carrying loop in a uniform
magnetic field is zero. So magnetic force can not
balance its weight.
23. The radius of the horizontal circle is
r = l(1+sinq)
If T is tension in the string then
Tcosq = mg and Tsinq = mrw
2
\
2
r gtan
tan
g (1 sin)
wq
q= Þ w=
+q l
l sinq
l
q
T
Tcosq
m Tsinq
mg
Page 4

HINT – SHEET
1.
mg
2
mmg
1
m
1
a
a
acc.  a =
net pulling force
mass to be pulled
=
21
12
m g mg
mm
-m
+
as  v
2
= u
2
+ 2as and u = 0
21
12
2Lg(m m)
v 2La
mm
-m
==
+
OR
Work-energy method : The essence of the
process in terms of energy condsiderations is
that m
2
loses potential energy of amount m
2
gL
while coming down by L, and the loss of
potential energy of m
2
appears partly in m
1
and
m
2
in the form of kinetic energy and is partly
used up as work done against friction.
(0999DMD310319005) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 05
01-09-2019
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 Test # 04
m
2
gL =
2
121
1
(m m )v m gL
2
+ +m
Þ (m
2
– mm
1
)gL =
1
2
(m
1
+ m
2
)v
2
Þ v =
21
12
2(m m )gL
mm
-m
+
=
- ´´
=
+
2(2 0.5 2)10 5
5m /s
22
2. Since the wheels are connected with a common
belt, their linear velocity is constant, that is v
1
=v
2
w = w
0
+ wt  = 0 + 1.5 × 10 = 15 rad s
–1
r
1
w
1
= r
2
w
2
or w
2
= w
1

1
2
r
r
=
1
15
3
–1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.2 1 1 3 2 1 1 1 3 1 3 3 3 2 1 1 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.2 4 3 1 1 3 2 3 2 2 3 1 4 3 4 1 4 4 1 1
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.2 1 3 2 4 1 2 4 3 2 1 2 2 4 3 3 4 2 2 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.4 1 4 4 2 2 2 1 1 4 3 1 1 3 3 1 2 1 4 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 2 1 2 4 1 4 3 2 3 3 3 3 1 3 3 2 2 1
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 2 2 2 1 2 3 4 1 2 3 1 1 2 3 3 2 1 2
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 3 2 2 2 1 2 1 3 1 1 2 3 1 3 2 4 4 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.4 4 1 1 1 2 4 1 2 2 2 3 1 2 1 2 1 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 3 3 4 1 1 2 2 3 1 2 1 2 2 2 1 3 1 1
LTS/HS-2/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
3. The field at P due to upper and lower squares will
cancel out. Resultant field at P will be due to the
square BCGF as shown in the figure. Solve to get,
B
P
=
0
4I
3a
m
p
A
D C
B
G
F
E
P
a/2
B
C
G
F
P
4. The observer moving with wedge will observe a
pseudo force ma
0
in the direction opposite to
0
a
r
.
a
0
l
h
F
pseudo

ps0
F ma =-
r
r
is a constant force
\ W
ps
=
ps
F .S
r r
= ma
0
l = 10 × 3 × 3 = 90 J
5. F = qvB =
0
qvi
2r
m
p
2
0
0
qvi mv 2 rmv
R
R 2r qi
m p
= Þ=
pm
6. 1
2
v 5gR 2 gR
5
==
222
1
v v 2gR v 4gR 2gR = - Þ =-
v 2gR Þ=
2
c
v
a 2g
R
==
v
B
a
t
a
q
a
c
a
t
= g
1 t
c
ag11
tan tan
a2g22
-
æö
a= = = Þa=
ç÷
èø
7. The horizontal components are (B
H
)
1
= B cos f
1
and (B
H
)
2
= B cos f
2
\
( )
()
H
1 1
H2
2
B
cos cos30
B cos cos 45
f°
==
f°
33
2
2 2
= ´=
8. Let R be the resistance force offered by the planks,
d
1
be the thickness of first plank and d
2
be the
thickness of second plank.
For first plank :
Loss in KE = work against resistance
2
22
11
1 1 4 19
mv m v Rd mv Rd
2 2 5 2 25
æ ö æö
- =Þ=
ç ÷ ç÷
è ø èø
(i)
For second plank :
2
2
22
1 4 1 16
m v 0 Rd mv Rd
2 5 2 25
æ ö æö
-=Þ=
ç ÷ ç÷
è ø èø
(ii)
Dividing eq. (i) by eq. (ii), we get
1
2
d 9
d 16
=
9.
2
2
2
22
nett
v 25
a a1
r 25
æö
æö
= +=+
ç÷ ç÷
èø
èø
=
–2
2 ms
F = m a
net
= 500 2 N.
10.
0 AB
2II
mg
4r
m
´=
p
l
where r = 2.5 cm
m = (0.100 g/cm)
l
Solve to get
B
250
IA
3
=
11. Power P = F.v  or  P = (ma)v
\ a =
P
mv
or  v .
dv
ds
=
P
mv
or  v
2
dv =
P
m
ds  or
P
m
òò
2
1
v s
2
0v
ds = v .dv
or
P
m
(s) =
1
3
(v
3
2
– v
1
3
)  or  s =
m
3P
(v
2
3
– v
1
3
)
AL LEN
0999DMD310319005 LTS/HS-3/8
12. Let current in AB is I
1
and in DC, I
2
. Then
C D
A
I
E I
2
I
2
I
1
I
2
I
1
2
I 3
I1
=
. It is because resistance of AB will be one-
Now,
11
22
BI
3
BI
==
13. Total acceleration
2
2
2
T
v
aa
r
æö
=+
ç÷
èø
Friction will provide a
T
\
2
2
2
v
ga
r
æö
m=+
ç÷
èø
\
4
222
2
v
ga
r
m =+
v = [(m
2
g
2
– a
2
)r
2
]
1/4
14. I =
37
3
M M 7.5 10 9 10
0.09A / m
V m 75 10
-
-
r ´ ´´
===
´
15. Its initial speed v
1
=
1
36000
10ms
3600
-
=
If the car doubles its speed, finally its speed
becomes v
2
= 20 ms
–1
Change in KE of the car
= DKE = (1/2)
22
21
mv (1/ 2)mv -
During Dt = 1 min = 60 s
the power delivered by the engine,
P =
Work done
Time taken
=
| KE|
t
D
D
=
22
21
1
mvv
2
t
éù ´-
ëû
D
=
22
1 / 2 500[20 10 ]
60
´-
= 1250 W
16.
F FF
® ®®
==
\ Net force on frame
=
CD
3F (3) (2) (1) (4) 24N
®
==  (F = ilB)
17. v Rg =m Þ
2
v Rg =m
2
v 70 70
Rg 1000 9.8
´
m==
´
= 0.5
18. Let h be the height of the inclined plane
h = 5 sin 37º = 3 m
mgh =
2
1
mv 0 ( mgcos37º)s
2
æö
- +m
ç÷
èø
Þ
2
14
3mg mv (0.25) mg 5
25
= + ´ ´´
Þ
1
v 4g 6.26ms
-
==
19. From the figure it is clear that deviationis 180º if
x > R
× ×
× ×
×
× ×
× ×
l
v
0
R
v
0
×
×
×
×
×
x=1.5R
20. As they enter the magnetic field of the earth, they
are deflected away from the equator.
22. Net force on a current carrying loop in a uniform
magnetic field is zero. So magnetic force can not
balance its weight.
23. The radius of the horizontal circle is
r = l(1+sinq)
If T is tension in the string then
Tcosq = mg and Tsinq = mrw
2
\
2
r gtan
tan
g (1 sin)
wq
q= Þ w=
+q l
l sinq
l
q
T
Tcosq
m Tsinq
mg
LTS/HS-4/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
24.
m
m
.
0
i
B=2×
d
4
2
(sin 90º – sin 45º)
=
méù
Ä
êú
p
ëû
0
i 1
1–
2d2
25. mmgd =
2
1
mv mgh
2
-
Þ  d =
2
v 2gh
2g
-
m
=
36 2 10 1.1
1.17m
2 0.6 10
- ´´
=
´´
26.
30°
O
B
A
mgcosq mgsinq
mg
T
At A : cos T mg q = so acceleation = sin mg q
perpendicular to OA
27. Magnetic force
®
l
m
F =iB acts in the direction
shown in figure:
q
q
v
F=
m
iB l
Rod will move downwards with constant velocity
if net force on it
is zero.
or F
m
cos q = mg sin q
or ilB cos q = mg sin q
\ B =
éù
q
êú
ëû l
mg
tan
i
29.
2
2
q
w=   Þ
d2
d2
wq
=
q
d ddd
.
dt d dt d
w wqw
a = = =w
qq
=
2
.
2
q
q
3
2
q
a=
a
t
= ar =
()
3
2 .2
2
p
= 8p
3
30. E
mech
= K
A
+ U
gA
+ U
sA
= 0 + mgx
A
+
1
2
2
A
kx
= 25 × 10 × (–0.100) +
1
2
× 2.5 × 10
4
× (–0.100)
2
= –25 + 125 = 100 J
31. Let N be the number turns and R the radius of the
coil. Then l = 2pRN
or
l
R=
2pN
... (i)
Now magnetic moment of the coil is
M = NiA = Ni (pR
2
) =
p
p
ll
22
22
i
Ni×=
4pN 4N
Maximum value of M can be
M
max
=
p
l
2
i
4
at N = 1
\ t
max
= M
max
B sin 90º =
p
l
2
iB
4
34. Work done by pseudo-force = change in PE
ma
0
R = mgR
a
0
= g
35. At the lowest point : T
L
= mg sin37°+
2
L
mv
l
Þ 274 = 3×10 ×
2
L
3v 3
5 0.75
+ Þ v
L
= 8 ms
–1
36.
r
DA = – 2 cos 30º
ˆ
i – 2 sin 30º
ˆ
k
=
ˆ ˆ
(– 3 i – k)
®
ˆ
AB = 2j
\
® ®®
M = i (DA× AB)
=
éù
ëû
ˆˆ ˆ
1
(– 3 i – k) × (2j)
2
=
ˆ ˆ
– 3k+i
=
ˆ
2
(i – 3 k)A – m
38. Let the velocity at the top point is v
C
. Then for
projectile motion from C to A.
Range = v
c
c
2(height) 2 2R
3Rv
gg
´
Þ=
Þ
C
3
v Rg
2
=
To find velocity at B, apply conservation of
energy, i.e.,
2
B
1
mv
2
= mg2R +
2
C
1
mv
2
Þ  v
B
=
5
Rg
2
Page 5

HINT – SHEET
1.
mg
2
mmg
1
m
1
a
a
acc.  a =
net pulling force
mass to be pulled
=
21
12
m g mg
mm
-m
+
as  v
2
= u
2
+ 2as and u = 0
21
12
2Lg(m m)
v 2La
mm
-m
==
+
OR
Work-energy method : The essence of the
process in terms of energy condsiderations is
that m
2
loses potential energy of amount m
2
gL
while coming down by L, and the loss of
potential energy of m
2
appears partly in m
1
and
m
2
in the form of kinetic energy and is partly
used up as work done against friction.
(0999DMD310319005) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 05
01-09-2019
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 Test # 04
m
2
gL =
2
121
1
(m m )v m gL
2
+ +m
Þ (m
2
– mm
1
)gL =
1
2
(m
1
+ m
2
)v
2
Þ v =
21
12
2(m m )gL
mm
-m
+
=
- ´´
=
+
2(2 0.5 2)10 5
5m /s
22
2. Since the wheels are connected with a common
belt, their linear velocity is constant, that is v
1
=v
2
w = w
0
+ wt  = 0 + 1.5 × 10 = 15 rad s
–1
r
1
w
1
= r
2
w
2
or w
2
= w
1

1
2
r
r
=
1
15
3
–1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans.2 1 1 3 2 1 1 1 3 1 3 3 3 2 1 1 3 3 4 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans.2 4 3 1 1 3 2 3 2 2 3 1 4 3 4 1 4 4 1 1
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans.2 1 3 2 4 1 2 4 3 2 1 2 2 4 3 3 4 2 2 3
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans.4 1 4 4 2 2 2 1 1 4 3 1 1 3 3 1 2 1 4 2
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans.3 2 2 1 2 4 1 4 3 2 3 3 3 3 1 3 3 2 2 1
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 2 2 2 1 2 3 4 1 2 3 1 1 2 3 3 2 1 2
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.1 3 2 2 2 1 2 1 3 1 1 2 3 1 3 2 4 4 2 1
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.4 4 1 1 1 2 4 1 2 2 2 3 1 2 1 2 1 3 2 2
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans.4 4 3 3 4 1 1 2 2 3 1 2 1 2 2 2 1 3 1 1
LTS/HS-2/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
3. The field at P due to upper and lower squares will
cancel out. Resultant field at P will be due to the
square BCGF as shown in the figure. Solve to get,
B
P
=
0
4I
3a
m
p
A
D C
B
G
F
E
P
a/2
B
C
G
F
P
4. The observer moving with wedge will observe a
pseudo force ma
0
in the direction opposite to
0
a
r
.
a
0
l
h
F
pseudo

ps0
F ma =-
r
r
is a constant force
\ W
ps
=
ps
F .S
r r
= ma
0
l = 10 × 3 × 3 = 90 J
5. F = qvB =
0
qvi
2r
m
p
2
0
0
qvi mv 2 rmv
R
R 2r qi
m p
= Þ=
pm
6. 1
2
v 5gR 2 gR
5
==
222
1
v v 2gR v 4gR 2gR = - Þ =-
v 2gR Þ=
2
c
v
a 2g
R
==
v
B
a
t
a
q
a
c
a
t
= g
1 t
c
ag11
tan tan
a2g22
-
æö
a= = = Þa=
ç÷
èø
7. The horizontal components are (B
H
)
1
= B cos f
1
and (B
H
)
2
= B cos f
2
\
( )
()
H
1 1
H2
2
B
cos cos30
B cos cos 45
f°
==
f°
33
2
2 2
= ´=
8. Let R be the resistance force offered by the planks,
d
1
be the thickness of first plank and d
2
be the
thickness of second plank.
For first plank :
Loss in KE = work against resistance
2
22
11
1 1 4 19
mv m v Rd mv Rd
2 2 5 2 25
æ ö æö
- =Þ=
ç ÷ ç÷
è ø èø
(i)
For second plank :
2
2
22
1 4 1 16
m v 0 Rd mv Rd
2 5 2 25
æ ö æö
-=Þ=
ç ÷ ç÷
è ø èø
(ii)
Dividing eq. (i) by eq. (ii), we get
1
2
d 9
d 16
=
9.
2
2
2
22
nett
v 25
a a1
r 25
æö
æö
= +=+
ç÷ ç÷
èø
èø
=
–2
2 ms
F = m a
net
= 500 2 N.
10.
0 AB
2II
mg
4r
m
´=
p
l
where r = 2.5 cm
m = (0.100 g/cm)
l
Solve to get
B
250
IA
3
=
11. Power P = F.v  or  P = (ma)v
\ a =
P
mv
or  v .
dv
ds
=
P
mv
or  v
2
dv =
P
m
ds  or
P
m
òò
2
1
v s
2
0v
ds = v .dv
or
P
m
(s) =
1
3
(v
3
2
– v
1
3
)  or  s =
m
3P
(v
2
3
– v
1
3
)
AL LEN
0999DMD310319005 LTS/HS-3/8
12. Let current in AB is I
1
and in DC, I
2
. Then
C D
A
I
E I
2
I
2
I
1
I
2
I
1
2
I 3
I1
=
. It is because resistance of AB will be one-
Now,
11
22
BI
3
BI
==
13. Total acceleration
2
2
2
T
v
aa
r
æö
=+
ç÷
èø
Friction will provide a
T
\
2
2
2
v
ga
r
æö
m=+
ç÷
èø
\
4
222
2
v
ga
r
m =+
v = [(m
2
g
2
– a
2
)r
2
]
1/4
14. I =
37
3
M M 7.5 10 9 10
0.09A / m
V m 75 10
-
-
r ´ ´´
===
´
15. Its initial speed v
1
=
1
36000
10ms
3600
-
=
If the car doubles its speed, finally its speed
becomes v
2
= 20 ms
–1
Change in KE of the car
= DKE = (1/2)
22
21
mv (1/ 2)mv -
During Dt = 1 min = 60 s
the power delivered by the engine,
P =
Work done
Time taken
=
| KE|
t
D
D
=
22
21
1
mvv
2
t
éù ´-
ëû
D
=
22
1 / 2 500[20 10 ]
60
´-
= 1250 W
16.
F FF
® ®®
==
\ Net force on frame
=
CD
3F (3) (2) (1) (4) 24N
®
==  (F = ilB)
17. v Rg =m Þ
2
v Rg =m
2
v 70 70
Rg 1000 9.8
´
m==
´
= 0.5
18. Let h be the height of the inclined plane
h = 5 sin 37º = 3 m
mgh =
2
1
mv 0 ( mgcos37º)s
2
æö
- +m
ç÷
èø
Þ
2
14
3mg mv (0.25) mg 5
25
= + ´ ´´
Þ
1
v 4g 6.26ms
-
==
19. From the figure it is clear that deviationis 180º if
x > R
× ×
× ×
×
× ×
× ×
l
v
0
R
v
0
×
×
×
×
×
x=1.5R
20. As they enter the magnetic field of the earth, they
are deflected away from the equator.
22. Net force on a current carrying loop in a uniform
magnetic field is zero. So magnetic force can not
balance its weight.
23. The radius of the horizontal circle is
r = l(1+sinq)
If T is tension in the string then
Tcosq = mg and Tsinq = mrw
2
\
2
r gtan
tan
g (1 sin)
wq
q= Þ w=
+q l
l sinq
l
q
T
Tcosq
m Tsinq
mg
LTS/HS-4/8 0999DMD310319005
Target : Pre-Medical 2020/NEET-UG/01-09-2019
AL LEN
24.
m
m
.
0
i
B=2×
d
4
2
(sin 90º – sin 45º)
=
méù
Ä
êú
p
ëû
0
i 1
1–
2d2
25. mmgd =
2
1
mv mgh
2
-
Þ  d =
2
v 2gh
2g
-
m
=
36 2 10 1.1
1.17m
2 0.6 10
- ´´
=
´´
26.
30°
O
B
A
mgcosq mgsinq
mg
T
At A : cos T mg q = so acceleation = sin mg q
perpendicular to OA
27. Magnetic force
®
l
m
F =iB acts in the direction
shown in figure:
q
q
v
F=
m
iB l
Rod will move downwards with constant velocity
if net force on it
is zero.
or F
m
cos q = mg sin q
or ilB cos q = mg sin q
\ B =
éù
q
êú
ëû l
mg
tan
i
29.
2
2
q
w=   Þ
d2
d2
wq
=
q
d ddd
.
dt d dt d
w wqw
a = = =w
qq
=
2
.
2
q
q
3
2
q
a=
a
t
= ar =
()
3
2 .2
2
p
= 8p
3
30. E
mech
= K
A
+ U
gA
+ U
sA
= 0 + mgx
A
+
1
2
2
A
kx
= 25 × 10 × (–0.100) +
1
2
× 2.5 × 10
4
× (–0.100)
2
= –25 + 125 = 100 J
31. Let N be the number turns and R the radius of the
coil. Then l = 2pRN
or
l
R=
2pN
... (i)
Now magnetic moment of the coil is
M = NiA = Ni (pR
2
) =
p
p
ll
22
22
i
Ni×=
4pN 4N
Maximum value of M can be
M
max
=
p
l
2
i
4
at N = 1
\ t
max
= M
max
B sin 90º =
p
l
2
iB
4
34. Work done by pseudo-force = change in PE
ma
0
R = mgR
a
0
= g
35. At the lowest point : T
L
= mg sin37°+
2
L
mv
l
Þ 274 = 3×10 ×
2
L
3v 3
5 0.75
+ Þ v
L
= 8 ms
–1
36.
r
DA = – 2 cos 30º
ˆ
i – 2 sin 30º
ˆ
k
=
ˆ ˆ
(– 3 i – k)
®
ˆ
AB = 2j
\
® ®®
M = i (DA× AB)
=
éù
ëû
ˆˆ ˆ
1
(– 3 i – k) × (2j)
2
=
ˆ ˆ
– 3k+i
=
ˆ
2
(i – 3 k)A – m
38. Let the velocity at the top point is v
C
. Then for
projectile motion from C to A.
Range = v
c
c
2(height) 2 2R
3Rv
gg
´
Þ=
Þ
C
3
v Rg
2
=
To find velocity at B, apply conservation of
energy, i.e.,
2
B
1
mv
2
= mg2R +
2
C
1
mv
2
Þ  v
B
=
5
Rg
2
AL LEN
0999DMD310319005 LTS/HS-5/8
39. Point (0, 0, –a) lies on z-axis. Therefore, magnetic
field due to current along z-axis is zero and due
to rest two wires is
p
0
µ i
2a
in mutually
perpendicular directions along positive
y–direction and negative x–direction.
\
®
m
=
p
ˆˆ
0
i
B (j – i)
2a
40. r = 1 km = 1000m, v = 900 km
22
2
c
v (250)
a 62.5m / s
r 1000
===
41. Infinite segments 1 and 2. Then, induction at P is
12
B BB =+
r rr
.....(i)
y
x
P
a
a
2
1
q
1
q
2
'
1
B
r
=
0
1
µi
ˆ
(cos cos0)k
4a
q+
p
.....(ii)
2
B
r
=
' 0
1
µi
ˆ
(cos cos0)k
4a
q+
p
.....(iii)
Then,
B
r
=
0
12
µi
ˆ
(cos cos 2)k,
4a
q + q+
p
Where cosq
1
= cosq
2
=
1
2
Hence,
B
r
=
0
ˆ
µ ik
(2 2)
4a
+
p
42. Work done in lifting water = Gain in PE (potential
energy)
Work = 1000 g × 20 = 1.96 × 10
5
J min
–1
Work done (per minute) in giving it KE
=
2 2 51
11
mv (1000)(20) 2 10 J min
22
-
= =´
Power of the engine = Work done per second
=
53
1
(1.96 2) 10 J 6.6 10 W
60
+ ´ =´
Since 1 HP = 746 W, HP required = 8.85
43.
2
v
tan
Rg
q=

10 10
10 10
´
=
´
= 1
Þ q = 45°
44. use B.dl
ò
ur r
Ñ
= m
0
I
enclosed
Net current enclosed by path a is zero
Net current enclosed by path c is A
Net current enclosed by path d is 3 A
Net current enclosed by path b is 5 A
45. For equilibrium, potential energy has to be
minimum, maximum or constant.
46.
[ ] [ ]
252
d NO dO
1
2 dt dt
- =+
\
[ ]
25 -1
d NO
0.032
2 mol hour
dt 32
- =´
and
[ ]
25 –1
d NO
0.002 108 0.216 g hour
dt
- = ´=
47. Let _ rate = K[A]
x
[B]
y
[C]
z
Comparing (1) and (2) –
[ ] [ ] [ ]
[ ][ ][ ]
x yz
3
3 x yz
K 0.01 0.005 0.01
5 10
5 10 K 0.01 0.005 0.015
-
-
´
=
´
\
z
2
1
3
æö
=
ç÷
èø
or z = 0
Comparing (1) and (3) :-
[ ] [ ] [ ]
[ ][ ][ ]
x y0 3
2 x y0
K 0.01 0.005 0.01
5 10
1 10 K 0.01 0.01 0.01
-
-
´
=
´
\
y
11
y1
22
æö
= Þ=
ç÷
èø
Comparing (1) and (4)
[ ] [ ] [ ]
[ ][ ][ ]
x 1 0
3
x 1 0 3
K 0.01 0.005 0.01
5 10
2.5 10
K 0.005 0.005 0.01
-
-
´
=
´
\ 2 = (2)
x
or  x = 1
48.
[ ]
[ ]
25
25
d NO
K' NO
dt
-
=
and
[ ]
[ ]
25
25
d NO
1
K NO
2 dt
-=
\ K' = 2K
```
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