Test Paper 6 Solution MBBS Notes | EduRev

MBBS : Test Paper 6 Solution MBBS Notes | EduRev

 Page 1


HINT – SHEET
ANSWER KEY
(0999DMD310319006) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 06
15-09-2019
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 - 05
1. Since you are decreasing current in the circuit by
increasing resistance of the circuit.
Induced emf across inductor will support 20 V
battery.
Hence, net emf of the circuit is greater than 20 V.
Or current in the circuit is more than 4 A.
2. At maximum compression (x
m
) velocity of both
the blocks is same, say it is v. Applying
conservation of linear momentum, we have,
(m
A
 + m
B
)v = m
B
v
0
or (1.0 + 1.0)v = (1.0) v
0
or
= ==
0
v 2.0
v 1.0 m/s
22
Using conservation of mechanical energy, we
have
( ) = ++
2 22
B0 ABm
111
m v m m v kx
222
Substituting the values, we get
()() ´´
2 1
1 2.0
2
( )() () =´+´ +´ ´
2
2
m
11
1.0 1.0 1.0 200 x
22
or 2 = 1.0 + 100 x
m
2
orx
m
 = 0.1 m = 10.0 cm
3. ()
dB
E2rA
dt
p=
( )
20
d nI
R
dt
m
=p
2
0
dI
Rn
dt
=m
( )
2
27
2
6 10 4 10 2000 0.01
2 8 10
--
-
´ ´p´ ´´
=
´´
7
5.6 10 V / m
-
=´
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 3 4 3 4 2 2 1 4 4 4 2 3 1 4 2 3 3 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.1 3 3 2 2 4 4 4 4 1 4 1 4 1 3 2 2 2 4 2
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 2 1 1 1 1 2 2 4 1 2 4 4 4 1 4 1 3 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 3 3 3 4 2 4 2 4 2 4 2 1 2 4 4 2 2 2
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.1 1 3 4 1 2 4 1 3 1 4 2 3 4 3 1 4 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 3 4 2 4 2 2 3 2 2 3 2 2 4 1 2 3 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 1 3 4 4 4 3 2 2 2 3 1 1 4 4 3 4 4 2
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 2 1 3 2 4 1 1 2 3 3 3 2 2 3 1 2 2 3 2
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 4 2 3 1 3 3 3 3 4 2 1 1 3 3 2 1 4
Page 2


HINT – SHEET
ANSWER KEY
(0999DMD310319006) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 06
15-09-2019
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 - 05
1. Since you are decreasing current in the circuit by
increasing resistance of the circuit.
Induced emf across inductor will support 20 V
battery.
Hence, net emf of the circuit is greater than 20 V.
Or current in the circuit is more than 4 A.
2. At maximum compression (x
m
) velocity of both
the blocks is same, say it is v. Applying
conservation of linear momentum, we have,
(m
A
 + m
B
)v = m
B
v
0
or (1.0 + 1.0)v = (1.0) v
0
or
= ==
0
v 2.0
v 1.0 m/s
22
Using conservation of mechanical energy, we
have
( ) = ++
2 22
B0 ABm
111
m v m m v kx
222
Substituting the values, we get
()() ´´
2 1
1 2.0
2
( )() () =´+´ +´ ´
2
2
m
11
1.0 1.0 1.0 200 x
22
or 2 = 1.0 + 100 x
m
2
orx
m
 = 0.1 m = 10.0 cm
3. ()
dB
E2rA
dt
p=
( )
20
d nI
R
dt
m
=p
2
0
dI
Rn
dt
=m
( )
2
27
2
6 10 4 10 2000 0.01
2 8 10
--
-
´ ´p´ ´´
=
´´
7
5.6 10 V / m
-
=´
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 3 4 3 4 2 2 1 4 4 4 2 3 1 4 2 3 3 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.1 3 3 2 2 4 4 4 4 1 4 1 4 1 3 2 2 2 4 2
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 2 1 1 1 1 2 2 4 1 2 4 4 4 1 4 1 3 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 3 3 3 4 2 4 2 4 2 4 2 1 2 4 4 2 2 2
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.1 1 3 4 1 2 4 1 3 1 4 2 3 4 3 1 4 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 3 4 2 4 2 2 3 2 2 3 2 2 4 1 2 3 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 1 3 4 4 4 3 2 2 2 3 1 1 4 4 3 4 4 2
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 2 1 3 2 4 1 1 2 3 3 3 2 2 3 1 2 2 3 2
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 4 2 3 1 3 3 3 3 4 2 1 1 3 3 2 1 4
LTS/HS-2/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
4.
( )( ) ( ) ( )( )( )
( ) ( ) ()
22
cm 22
20 0 46
X
20 n4
s +-sp
=
s -s
x
1
O 8cm
2cm
     = 
( )( ) 166
400 16
-p
-p
    = 
301.44
349.76
-
    = – 0.86 cm
0.86cm left of O
5. Here, v
1
 = –v, v
2
 = 2v, m
1
 = m and m
2
 = 2m.
2m
m
2v
v
+ve
Substituting these values in Eqs. (iii) and (iv)
we get
() ()
1
m 2m 4m
v ' v 2v
m 2m m 2m
- æö æö
= -+
ç÷ ç÷
++
èø èø
or
1
v 8v
v ' 3v
33
= +=
and 
() ()
2
2m m 2m
v' 2vv
m 2m m 2m
- æö æö
= +-
ç÷ ç÷
++
èø èø
or 2
22
v' v v0
33
æö
= -=
ç÷
èø
2m
m
3v
i.e., the second particle (of mass 2m) comes to a
rest while the first (of mass m) moves with velocity
3v in the direction shown in figure.
6. Magnetic lines on circular loop, due to current in
wire AB are tangnetial to its plane.
\
d
0 or 0
dt
f
f==
 or induced emf = 0.
7. Let velocity of I ball and II ball after collision be
v
1
 and v
2
Q
v
2 
– v
1
 = e(u
1
 – u
2
)
\ v
2
 – v
1
 = 0.5 × 10 ...(i)
and mv
2
 + mv
1
 = m × 10 ...(ii)
Þ v
2
 + v
1
 = 10
Solving eqs. (i) and (ii)
v
1
 = 2.5 m/s, v
2
 = 7.5 m/s
Ball II after moving 10 m collides with ball III
elastically and stops. But ball I moves towards ball
II. Time taken between two consecutive collisions.
10
t 4s
2.5
==
8. Terminal velocity is attained when magnetic force
is equal to mg sin q
\ F
m
 = mg sinq    
mg sinq
q
Fm
or 
iB l
 = mg sinq
or 
æö
=q
ç÷
èø
l
e
B mgsin
R
or 
( ) n
´
l
l
T
B
B
R
 = mg sin q
22 T
mgRsin
B
q
\n=
l
9. As inward B increases; so ACW current in bigger
loop & CW current in smaller loop.
10.
211
2 21
21 12
m m 2m
v'vv
m m mm
æöæö -
=+
ç÷ç÷
++
èøèø
v
1
 = 0
\
2 21
2 21
v mm
v' mm
æö +
=
ç÷
-
èø
(Q v
1
 = 0)
      
m 2m
3
m 2m
+ æö
= =-
ç÷
-
èø
\
2
22
22
Kv
9
K' v'
æö
==
ç÷
èø
11.
=
PD
I
Resistance
\ PD = 2 × 12 = 24 V = B v l
Here l = AD sin 37° = 
´=
3
0.3 0.18 m
5
\
===
´
24 24 100
v m/s
B 4 0.18 3 l
Page 3


HINT – SHEET
ANSWER KEY
(0999DMD310319006) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 06
15-09-2019
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 - 05
1. Since you are decreasing current in the circuit by
increasing resistance of the circuit.
Induced emf across inductor will support 20 V
battery.
Hence, net emf of the circuit is greater than 20 V.
Or current in the circuit is more than 4 A.
2. At maximum compression (x
m
) velocity of both
the blocks is same, say it is v. Applying
conservation of linear momentum, we have,
(m
A
 + m
B
)v = m
B
v
0
or (1.0 + 1.0)v = (1.0) v
0
or
= ==
0
v 2.0
v 1.0 m/s
22
Using conservation of mechanical energy, we
have
( ) = ++
2 22
B0 ABm
111
m v m m v kx
222
Substituting the values, we get
()() ´´
2 1
1 2.0
2
( )() () =´+´ +´ ´
2
2
m
11
1.0 1.0 1.0 200 x
22
or 2 = 1.0 + 100 x
m
2
orx
m
 = 0.1 m = 10.0 cm
3. ()
dB
E2rA
dt
p=
( )
20
d nI
R
dt
m
=p
2
0
dI
Rn
dt
=m
( )
2
27
2
6 10 4 10 2000 0.01
2 8 10
--
-
´ ´p´ ´´
=
´´
7
5.6 10 V / m
-
=´
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 3 4 3 4 2 2 1 4 4 4 2 3 1 4 2 3 3 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.1 3 3 2 2 4 4 4 4 1 4 1 4 1 3 2 2 2 4 2
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 2 1 1 1 1 2 2 4 1 2 4 4 4 1 4 1 3 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 3 3 3 4 2 4 2 4 2 4 2 1 2 4 4 2 2 2
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.1 1 3 4 1 2 4 1 3 1 4 2 3 4 3 1 4 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 3 4 2 4 2 2 3 2 2 3 2 2 4 1 2 3 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 1 3 4 4 4 3 2 2 2 3 1 1 4 4 3 4 4 2
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 2 1 3 2 4 1 1 2 3 3 3 2 2 3 1 2 2 3 2
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 4 2 3 1 3 3 3 3 4 2 1 1 3 3 2 1 4
LTS/HS-2/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
4.
( )( ) ( ) ( )( )( )
( ) ( ) ()
22
cm 22
20 0 46
X
20 n4
s +-sp
=
s -s
x
1
O 8cm
2cm
     = 
( )( ) 166
400 16
-p
-p
    = 
301.44
349.76
-
    = – 0.86 cm
0.86cm left of O
5. Here, v
1
 = –v, v
2
 = 2v, m
1
 = m and m
2
 = 2m.
2m
m
2v
v
+ve
Substituting these values in Eqs. (iii) and (iv)
we get
() ()
1
m 2m 4m
v ' v 2v
m 2m m 2m
- æö æö
= -+
ç÷ ç÷
++
èø èø
or
1
v 8v
v ' 3v
33
= +=
and 
() ()
2
2m m 2m
v' 2vv
m 2m m 2m
- æö æö
= +-
ç÷ ç÷
++
èø èø
or 2
22
v' v v0
33
æö
= -=
ç÷
èø
2m
m
3v
i.e., the second particle (of mass 2m) comes to a
rest while the first (of mass m) moves with velocity
3v in the direction shown in figure.
6. Magnetic lines on circular loop, due to current in
wire AB are tangnetial to its plane.
\
d
0 or 0
dt
f
f==
 or induced emf = 0.
7. Let velocity of I ball and II ball after collision be
v
1
 and v
2
Q
v
2 
– v
1
 = e(u
1
 – u
2
)
\ v
2
 – v
1
 = 0.5 × 10 ...(i)
and mv
2
 + mv
1
 = m × 10 ...(ii)
Þ v
2
 + v
1
 = 10
Solving eqs. (i) and (ii)
v
1
 = 2.5 m/s, v
2
 = 7.5 m/s
Ball II after moving 10 m collides with ball III
elastically and stops. But ball I moves towards ball
II. Time taken between two consecutive collisions.
10
t 4s
2.5
==
8. Terminal velocity is attained when magnetic force
is equal to mg sin q
\ F
m
 = mg sinq    
mg sinq
q
Fm
or 
iB l
 = mg sinq
or 
æö
=q
ç÷
èø
l
e
B mgsin
R
or 
( ) n
´
l
l
T
B
B
R
 = mg sin q
22 T
mgRsin
B
q
\n=
l
9. As inward B increases; so ACW current in bigger
loop & CW current in smaller loop.
10.
211
2 21
21 12
m m 2m
v'vv
m m mm
æöæö -
=+
ç÷ç÷
++
èøèø
v
1
 = 0
\
2 21
2 21
v mm
v' mm
æö +
=
ç÷
-
èø
(Q v
1
 = 0)
      
m 2m
3
m 2m
+ æö
= =-
ç÷
-
èø
\
2
22
22
Kv
9
K' v'
æö
==
ç÷
èø
11.
=
PD
I
Resistance
\ PD = 2 × 12 = 24 V = B v l
Here l = AD sin 37° = 
´=
3
0.3 0.18 m
5
\
===
´
24 24 100
v m/s
B 4 0.18 3 l
Leader Test Series/Joint Package Course/NEET-UG/15-09-2019
AL LEN
0999DMD310319006 LTS/HS-3/7
12. Both the balls in air have acceleration g downwards.
Hence, the acceleration of their centre of mass will
also be g downwards.
13.
v
v
m
m
q
2m
v
2
mv cos
v
mv cos 2m
2 22
qq
+=
1
cos
22
q
=
60º 120º
2
q
= Þq=
14.
=-
di
EL
dt
or induced emf µ – slope of i-t graph.
15. 0 + 2mV = mv + 0  Þ V = 
v
2
m
v
2m
Þ
m 2m
v/2
Rest
Relative velocity of separation
e
Relative velocity of approach
=
v/21
v2
==
16.
-
æö
==´´=
ç÷
èø
2
26
0
115
U Li 50 10 J 2.5 mJ
2 2 0.5
Now, after switching off the battery this energy
is dissipated in coil (resistance = 0.5) and
resistance (10 W) in the ratio of their resistances
(H = i
2
Rt or H µ R). Therefore, heat generated in
the coil.
æö
==
ç÷
+
èø
0.5
H 2.5 mJ 0.12 mJ
0.5 10
17. ( ) ( )
CM
m O m PQ m PR
X from P = m=1kg
m mm
´ +´ +´
++
P Q R
or
CM
PQ PR
X
3
+
=
18. \ Initial momentum of each ball
p
i
 = mv
   = (0.05) (6)
   = 0.30 kg ms
–1
As after collision, the direction of velocity of each
ball is reversed on rebounding.
\ Final momentum of each ball = m(–v)
p
f
 = 0.05 (–6)
   = –0.30 kg ms
–1
\ Impulse imparted of each ball
= p
f
 – p
i
= change in momentum of each ball
= – 0.30 – (0.30)
= –0.60 kg ms
–1
or magnitude of impulse imparted by one ball due
to collision with the other = 0.6 kg ms
–1
. The two
impulses are opposite in direction.
19. M
12
 = M
21
, when rate of charge of current is
doubled, induced emf also become two times.
20. h
n
 = he
2n
 = 
4
1
32
2
æö
ç÷
èø
 
32
2m
16
==
(here n = 2, e = 1/2)
21.
=+
22
0
1 11
CV LI CV
2 22
Þ
( )
-
=
22
0
CVV
I
L
( )
-
-
´-
=
´
6 22
3
2 10 12 6
0.6 10
= 0.6 A
23. Nf = LI
L = 
f ´´
=
–4
N 200 8 10
I4
 = 4 × 10
–2
 H
24. e = M
dI
dt
straight line
I = Kt
e = constant
25. Time constant 
t=
1
L
R
= 2 × 10
–3
.....(1)
t=
+
2
L
R 90
 = 0.5 × 10
–3
.....(2)
Page 4


HINT – SHEET
ANSWER KEY
(0999DMD310319006) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 06
15-09-2019
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 - 05
1. Since you are decreasing current in the circuit by
increasing resistance of the circuit.
Induced emf across inductor will support 20 V
battery.
Hence, net emf of the circuit is greater than 20 V.
Or current in the circuit is more than 4 A.
2. At maximum compression (x
m
) velocity of both
the blocks is same, say it is v. Applying
conservation of linear momentum, we have,
(m
A
 + m
B
)v = m
B
v
0
or (1.0 + 1.0)v = (1.0) v
0
or
= ==
0
v 2.0
v 1.0 m/s
22
Using conservation of mechanical energy, we
have
( ) = ++
2 22
B0 ABm
111
m v m m v kx
222
Substituting the values, we get
()() ´´
2 1
1 2.0
2
( )() () =´+´ +´ ´
2
2
m
11
1.0 1.0 1.0 200 x
22
or 2 = 1.0 + 100 x
m
2
orx
m
 = 0.1 m = 10.0 cm
3. ()
dB
E2rA
dt
p=
( )
20
d nI
R
dt
m
=p
2
0
dI
Rn
dt
=m
( )
2
27
2
6 10 4 10 2000 0.01
2 8 10
--
-
´ ´p´ ´´
=
´´
7
5.6 10 V / m
-
=´
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 3 4 3 4 2 2 1 4 4 4 2 3 1 4 2 3 3 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.1 3 3 2 2 4 4 4 4 1 4 1 4 1 3 2 2 2 4 2
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 2 1 1 1 1 2 2 4 1 2 4 4 4 1 4 1 3 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 3 3 3 4 2 4 2 4 2 4 2 1 2 4 4 2 2 2
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.1 1 3 4 1 2 4 1 3 1 4 2 3 4 3 1 4 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 3 4 2 4 2 2 3 2 2 3 2 2 4 1 2 3 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 1 3 4 4 4 3 2 2 2 3 1 1 4 4 3 4 4 2
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 2 1 3 2 4 1 1 2 3 3 3 2 2 3 1 2 2 3 2
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 4 2 3 1 3 3 3 3 4 2 1 1 3 3 2 1 4
LTS/HS-2/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
4.
( )( ) ( ) ( )( )( )
( ) ( ) ()
22
cm 22
20 0 46
X
20 n4
s +-sp
=
s -s
x
1
O 8cm
2cm
     = 
( )( ) 166
400 16
-p
-p
    = 
301.44
349.76
-
    = – 0.86 cm
0.86cm left of O
5. Here, v
1
 = –v, v
2
 = 2v, m
1
 = m and m
2
 = 2m.
2m
m
2v
v
+ve
Substituting these values in Eqs. (iii) and (iv)
we get
() ()
1
m 2m 4m
v ' v 2v
m 2m m 2m
- æö æö
= -+
ç÷ ç÷
++
èø èø
or
1
v 8v
v ' 3v
33
= +=
and 
() ()
2
2m m 2m
v' 2vv
m 2m m 2m
- æö æö
= +-
ç÷ ç÷
++
èø èø
or 2
22
v' v v0
33
æö
= -=
ç÷
èø
2m
m
3v
i.e., the second particle (of mass 2m) comes to a
rest while the first (of mass m) moves with velocity
3v in the direction shown in figure.
6. Magnetic lines on circular loop, due to current in
wire AB are tangnetial to its plane.
\
d
0 or 0
dt
f
f==
 or induced emf = 0.
7. Let velocity of I ball and II ball after collision be
v
1
 and v
2
Q
v
2 
– v
1
 = e(u
1
 – u
2
)
\ v
2
 – v
1
 = 0.5 × 10 ...(i)
and mv
2
 + mv
1
 = m × 10 ...(ii)
Þ v
2
 + v
1
 = 10
Solving eqs. (i) and (ii)
v
1
 = 2.5 m/s, v
2
 = 7.5 m/s
Ball II after moving 10 m collides with ball III
elastically and stops. But ball I moves towards ball
II. Time taken between two consecutive collisions.
10
t 4s
2.5
==
8. Terminal velocity is attained when magnetic force
is equal to mg sin q
\ F
m
 = mg sinq    
mg sinq
q
Fm
or 
iB l
 = mg sinq
or 
æö
=q
ç÷
èø
l
e
B mgsin
R
or 
( ) n
´
l
l
T
B
B
R
 = mg sin q
22 T
mgRsin
B
q
\n=
l
9. As inward B increases; so ACW current in bigger
loop & CW current in smaller loop.
10.
211
2 21
21 12
m m 2m
v'vv
m m mm
æöæö -
=+
ç÷ç÷
++
èøèø
v
1
 = 0
\
2 21
2 21
v mm
v' mm
æö +
=
ç÷
-
èø
(Q v
1
 = 0)
      
m 2m
3
m 2m
+ æö
= =-
ç÷
-
èø
\
2
22
22
Kv
9
K' v'
æö
==
ç÷
èø
11.
=
PD
I
Resistance
\ PD = 2 × 12 = 24 V = B v l
Here l = AD sin 37° = 
´=
3
0.3 0.18 m
5
\
===
´
24 24 100
v m/s
B 4 0.18 3 l
Leader Test Series/Joint Package Course/NEET-UG/15-09-2019
AL LEN
0999DMD310319006 LTS/HS-3/7
12. Both the balls in air have acceleration g downwards.
Hence, the acceleration of their centre of mass will
also be g downwards.
13.
v
v
m
m
q
2m
v
2
mv cos
v
mv cos 2m
2 22
qq
+=
1
cos
22
q
=
60º 120º
2
q
= Þq=
14.
=-
di
EL
dt
or induced emf µ – slope of i-t graph.
15. 0 + 2mV = mv + 0  Þ V = 
v
2
m
v
2m
Þ
m 2m
v/2
Rest
Relative velocity of separation
e
Relative velocity of approach
=
v/21
v2
==
16.
-
æö
==´´=
ç÷
èø
2
26
0
115
U Li 50 10 J 2.5 mJ
2 2 0.5
Now, after switching off the battery this energy
is dissipated in coil (resistance = 0.5) and
resistance (10 W) in the ratio of their resistances
(H = i
2
Rt or H µ R). Therefore, heat generated in
the coil.
æö
==
ç÷
+
èø
0.5
H 2.5 mJ 0.12 mJ
0.5 10
17. ( ) ( )
CM
m O m PQ m PR
X from P = m=1kg
m mm
´ +´ +´
++
P Q R
or
CM
PQ PR
X
3
+
=
18. \ Initial momentum of each ball
p
i
 = mv
   = (0.05) (6)
   = 0.30 kg ms
–1
As after collision, the direction of velocity of each
ball is reversed on rebounding.
\ Final momentum of each ball = m(–v)
p
f
 = 0.05 (–6)
   = –0.30 kg ms
–1
\ Impulse imparted of each ball
= p
f
 – p
i
= change in momentum of each ball
= – 0.30 – (0.30)
= –0.60 kg ms
–1
or magnitude of impulse imparted by one ball due
to collision with the other = 0.6 kg ms
–1
. The two
impulses are opposite in direction.
19. M
12
 = M
21
, when rate of charge of current is
doubled, induced emf also become two times.
20. h
n
 = he
2n
 = 
4
1
32
2
æö
ç÷
èø
 
32
2m
16
==
(here n = 2, e = 1/2)
21.
=+
22
0
1 11
CV LI CV
2 22
Þ
( )
-
=
22
0
CVV
I
L
( )
-
-
´-
=
´
6 22
3
2 10 12 6
0.6 10
= 0.6 A
23. Nf = LI
L = 
f ´´
=
–4
N 200 8 10
I4
 = 4 × 10
–2
 H
24. e = M
dI
dt
straight line
I = Kt
e = constant
25. Time constant 
t=
1
L
R
= 2 × 10
–3
.....(1)
t=
+
2
L
R 90
 = 0.5 × 10
–3
.....(2)
LTS/HS-4/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
By solving eq (1) & (2)
L = 60 mH, R = 30W
26.
1 1 22
CM
12
my my
y
mm
+
=
+
or
( )( ) ( )
2
10 7 30 y
1
10 30
+
+=
+
\ y
2
 = –1 cm
27.
ind
d
6t4
dt
f
e = =+
at = 2sec    e
ind 
= 6 × 2 + 4 = 16 volt
28. After striking at the floor the ball cannot return
with double the velocity because there will be
some loss of KE of the ball after the collision
which will appear in the form of sound energy,
heat energy, etc.
30. C
1
 ® Position of centre of mass of rods AB and
CD (nearer to CD, as it is heavy)
D
C
B A
C
1
C
C
2
C
2
 ® Position of centre of mass of rods BC and
DA.
C ® Overall centre of mass of all four rods.
32. Let the velocity of block = v’
Then from conservation of linear momentum
mu = mv + mnv’ or v’ = 
u–v
n
éù
êú
ëû
\ Velocity of bullet relative to block will be v
r
= v – v’ = v – 
u–v
n
éù
êú
ëû
 = 
(1 n)v–u
n
+
33. At t = 0 centre of mass is at mid-point or at
(2.25 m, 0)
Velocity of centre of mass is zero. Hence, centre
of mass will remain at this position all the time.
34. C
net 
= C/2 (in series) and L
net
 = 2 L (in series)
net net
1
f LC
2
=´
p
1 C1
2L
22 2 LC
= ´=
p p
35. Induced emf, when rod rotates in a vertical plane
perpendicular to magnetic field
=w
2
1
eB
2
l
  
() =´ ´´
2 1
0.3 2 100
2
 = 60 V
36. After 1 s, coordinates of first particle will become
(4m, 4m, 6m) and co-ordinates of second particle
will become (6m, 4m, 8m).
\
CM
46
X 5 m
2
+
==
CM
44
Y 4 m
2
+
==
and 
CM
68
Z 7 m
2
+
==
Note : In this problem answer is independent of
the fact whether the first particle has velocity v
1
or v
2
. Think why?
37. Growth of current in the circuit is given by
i = i
0
(1 – e
–Rt/L
)
where, i
0
 is peak value of current and
==
0
5
i 1A
5
\ i = 1(1 – e
–5×2/10
)
= (1 – e
–1
) A
38. Velocity on hitting the surface
2 9.8 4.9 = ´´
= 9.8 m/s
Velocity after first bounce
3
v 9.8
4
=´
Time taken from first bounce to the second bounce
= 
2v
g
Page 5


HINT – SHEET
ANSWER KEY
(0999DMD310319006) Test Pattern
DIST ANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE
NEET(UG)
MINOR TEST # 06
15-09-2019
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 - 05
1. Since you are decreasing current in the circuit by
increasing resistance of the circuit.
Induced emf across inductor will support 20 V
battery.
Hence, net emf of the circuit is greater than 20 V.
Or current in the circuit is more than 4 A.
2. At maximum compression (x
m
) velocity of both
the blocks is same, say it is v. Applying
conservation of linear momentum, we have,
(m
A
 + m
B
)v = m
B
v
0
or (1.0 + 1.0)v = (1.0) v
0
or
= ==
0
v 2.0
v 1.0 m/s
22
Using conservation of mechanical energy, we
have
( ) = ++
2 22
B0 ABm
111
m v m m v kx
222
Substituting the values, we get
()() ´´
2 1
1 2.0
2
( )() () =´+´ +´ ´
2
2
m
11
1.0 1.0 1.0 200 x
22
or 2 = 1.0 + 100 x
m
2
orx
m
 = 0.1 m = 10.0 cm
3. ()
dB
E2rA
dt
p=
( )
20
d nI
R
dt
m
=p
2
0
dI
Rn
dt
=m
( )
2
27
2
6 10 4 10 2000 0.01
2 8 10
--
-
´ ´p´ ´´
=
´´
7
5.6 10 V / m
-
=´
Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920
Ans.3 2 3 4 3 4 2 2 1 4 4 4 2 3 1 4 2 3 3 1
Que.21222324 2526 272829303132333435 3637 383940
Ans.1 3 3 2 2 4 4 4 4 1 4 1 4 1 3 2 2 2 4 2
Que.41424344 4546 474849505152535455 5657 585960
Ans.1 4 2 1 1 1 1 2 2 4 1 2 4 4 4 1 4 1 3 2
Que.61626364 6566 676869707172737475 7677 787980
Ans.3 4 3 3 3 4 2 4 2 4 2 4 2 1 2 4 4 2 2 2
Que.81828384 8586 878889909192939495 9697 9899 100
Ans.1 1 3 4 1 2 4 1 3 1 4 2 3 4 3 1 4 3 3 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans.3 4 3 4 2 4 2 2 3 2 2 3 2 2 4 1 2 3 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans.3 2 1 3 4 4 4 3 2 2 2 3 1 1 4 4 3 4 4 2
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans.1 2 1 3 2 4 1 1 2 3 3 3 2 2 3 1 2 2 3 2
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 4 2 3 1 3 3 3 3 4 2 1 1 3 3 2 1 4
LTS/HS-2/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
4.
( )( ) ( ) ( )( )( )
( ) ( ) ()
22
cm 22
20 0 46
X
20 n4
s +-sp
=
s -s
x
1
O 8cm
2cm
     = 
( )( ) 166
400 16
-p
-p
    = 
301.44
349.76
-
    = – 0.86 cm
0.86cm left of O
5. Here, v
1
 = –v, v
2
 = 2v, m
1
 = m and m
2
 = 2m.
2m
m
2v
v
+ve
Substituting these values in Eqs. (iii) and (iv)
we get
() ()
1
m 2m 4m
v ' v 2v
m 2m m 2m
- æö æö
= -+
ç÷ ç÷
++
èø èø
or
1
v 8v
v ' 3v
33
= +=
and 
() ()
2
2m m 2m
v' 2vv
m 2m m 2m
- æö æö
= +-
ç÷ ç÷
++
èø èø
or 2
22
v' v v0
33
æö
= -=
ç÷
èø
2m
m
3v
i.e., the second particle (of mass 2m) comes to a
rest while the first (of mass m) moves with velocity
3v in the direction shown in figure.
6. Magnetic lines on circular loop, due to current in
wire AB are tangnetial to its plane.
\
d
0 or 0
dt
f
f==
 or induced emf = 0.
7. Let velocity of I ball and II ball after collision be
v
1
 and v
2
Q
v
2 
– v
1
 = e(u
1
 – u
2
)
\ v
2
 – v
1
 = 0.5 × 10 ...(i)
and mv
2
 + mv
1
 = m × 10 ...(ii)
Þ v
2
 + v
1
 = 10
Solving eqs. (i) and (ii)
v
1
 = 2.5 m/s, v
2
 = 7.5 m/s
Ball II after moving 10 m collides with ball III
elastically and stops. But ball I moves towards ball
II. Time taken between two consecutive collisions.
10
t 4s
2.5
==
8. Terminal velocity is attained when magnetic force
is equal to mg sin q
\ F
m
 = mg sinq    
mg sinq
q
Fm
or 
iB l
 = mg sinq
or 
æö
=q
ç÷
èø
l
e
B mgsin
R
or 
( ) n
´
l
l
T
B
B
R
 = mg sin q
22 T
mgRsin
B
q
\n=
l
9. As inward B increases; so ACW current in bigger
loop & CW current in smaller loop.
10.
211
2 21
21 12
m m 2m
v'vv
m m mm
æöæö -
=+
ç÷ç÷
++
èøèø
v
1
 = 0
\
2 21
2 21
v mm
v' mm
æö +
=
ç÷
-
èø
(Q v
1
 = 0)
      
m 2m
3
m 2m
+ æö
= =-
ç÷
-
èø
\
2
22
22
Kv
9
K' v'
æö
==
ç÷
èø
11.
=
PD
I
Resistance
\ PD = 2 × 12 = 24 V = B v l
Here l = AD sin 37° = 
´=
3
0.3 0.18 m
5
\
===
´
24 24 100
v m/s
B 4 0.18 3 l
Leader Test Series/Joint Package Course/NEET-UG/15-09-2019
AL LEN
0999DMD310319006 LTS/HS-3/7
12. Both the balls in air have acceleration g downwards.
Hence, the acceleration of their centre of mass will
also be g downwards.
13.
v
v
m
m
q
2m
v
2
mv cos
v
mv cos 2m
2 22
qq
+=
1
cos
22
q
=
60º 120º
2
q
= Þq=
14.
=-
di
EL
dt
or induced emf µ – slope of i-t graph.
15. 0 + 2mV = mv + 0  Þ V = 
v
2
m
v
2m
Þ
m 2m
v/2
Rest
Relative velocity of separation
e
Relative velocity of approach
=
v/21
v2
==
16.
-
æö
==´´=
ç÷
èø
2
26
0
115
U Li 50 10 J 2.5 mJ
2 2 0.5
Now, after switching off the battery this energy
is dissipated in coil (resistance = 0.5) and
resistance (10 W) in the ratio of their resistances
(H = i
2
Rt or H µ R). Therefore, heat generated in
the coil.
æö
==
ç÷
+
èø
0.5
H 2.5 mJ 0.12 mJ
0.5 10
17. ( ) ( )
CM
m O m PQ m PR
X from P = m=1kg
m mm
´ +´ +´
++
P Q R
or
CM
PQ PR
X
3
+
=
18. \ Initial momentum of each ball
p
i
 = mv
   = (0.05) (6)
   = 0.30 kg ms
–1
As after collision, the direction of velocity of each
ball is reversed on rebounding.
\ Final momentum of each ball = m(–v)
p
f
 = 0.05 (–6)
   = –0.30 kg ms
–1
\ Impulse imparted of each ball
= p
f
 – p
i
= change in momentum of each ball
= – 0.30 – (0.30)
= –0.60 kg ms
–1
or magnitude of impulse imparted by one ball due
to collision with the other = 0.6 kg ms
–1
. The two
impulses are opposite in direction.
19. M
12
 = M
21
, when rate of charge of current is
doubled, induced emf also become two times.
20. h
n
 = he
2n
 = 
4
1
32
2
æö
ç÷
èø
 
32
2m
16
==
(here n = 2, e = 1/2)
21.
=+
22
0
1 11
CV LI CV
2 22
Þ
( )
-
=
22
0
CVV
I
L
( )
-
-
´-
=
´
6 22
3
2 10 12 6
0.6 10
= 0.6 A
23. Nf = LI
L = 
f ´´
=
–4
N 200 8 10
I4
 = 4 × 10
–2
 H
24. e = M
dI
dt
straight line
I = Kt
e = constant
25. Time constant 
t=
1
L
R
= 2 × 10
–3
.....(1)
t=
+
2
L
R 90
 = 0.5 × 10
–3
.....(2)
LTS/HS-4/7 0999DMD310319006
Target : Pre-Medical 2020/NEET-UG/15-09-2019
AL LEN
By solving eq (1) & (2)
L = 60 mH, R = 30W
26.
1 1 22
CM
12
my my
y
mm
+
=
+
or
( )( ) ( )
2
10 7 30 y
1
10 30
+
+=
+
\ y
2
 = –1 cm
27.
ind
d
6t4
dt
f
e = =+
at = 2sec    e
ind 
= 6 × 2 + 4 = 16 volt
28. After striking at the floor the ball cannot return
with double the velocity because there will be
some loss of KE of the ball after the collision
which will appear in the form of sound energy,
heat energy, etc.
30. C
1
 ® Position of centre of mass of rods AB and
CD (nearer to CD, as it is heavy)
D
C
B A
C
1
C
C
2
C
2
 ® Position of centre of mass of rods BC and
DA.
C ® Overall centre of mass of all four rods.
32. Let the velocity of block = v’
Then from conservation of linear momentum
mu = mv + mnv’ or v’ = 
u–v
n
éù
êú
ëû
\ Velocity of bullet relative to block will be v
r
= v – v’ = v – 
u–v
n
éù
êú
ëû
 = 
(1 n)v–u
n
+
33. At t = 0 centre of mass is at mid-point or at
(2.25 m, 0)
Velocity of centre of mass is zero. Hence, centre
of mass will remain at this position all the time.
34. C
net 
= C/2 (in series) and L
net
 = 2 L (in series)
net net
1
f LC
2
=´
p
1 C1
2L
22 2 LC
= ´=
p p
35. Induced emf, when rod rotates in a vertical plane
perpendicular to magnetic field
=w
2
1
eB
2
l
  
() =´ ´´
2 1
0.3 2 100
2
 = 60 V
36. After 1 s, coordinates of first particle will become
(4m, 4m, 6m) and co-ordinates of second particle
will become (6m, 4m, 8m).
\
CM
46
X 5 m
2
+
==
CM
44
Y 4 m
2
+
==
and 
CM
68
Z 7 m
2
+
==
Note : In this problem answer is independent of
the fact whether the first particle has velocity v
1
or v
2
. Think why?
37. Growth of current in the circuit is given by
i = i
0
(1 – e
–Rt/L
)
where, i
0
 is peak value of current and
==
0
5
i 1A
5
\ i = 1(1 – e
–5×2/10
)
= (1 – e
–1
) A
38. Velocity on hitting the surface
2 9.8 4.9 = ´´
= 9.8 m/s
Velocity after first bounce
3
v 9.8
4
=´
Time taken from first bounce to the second bounce
= 
2v
g
Leader Test Series/Joint Package Course/NEET-UG/15-09-2019
AL LEN
0999DMD310319006 LTS/HS-5/7
31
2 9.8
4 9.8
=´ ´´
 = 1.5 s
OR
h' = he
2
2
2h'2
T 2 2 he
gg
==
2h 3 2 4.9
2e2
g 4 9.8
´
= = ´´
= 1.5 s
40. Induced emf in the circuit
( )() = ==
di
e M 0.2 5 1 V
dt
42. Distance between bullet and block at this instant
is (D – d)
mx = M(D-d-x)
Distance from bullet x = ( )
M
Dd
Mm
-
+
And distance from block = ( )
m
Dd
Mm
-
+
43. From conservation of linear momentum
1
111
v
mv m 5glm
3
=+
     Þ 1
1
3m
v 5gl
2m
=
45. Centre of mass will remain at height h.
\
CM
m 0 MH
hh
mM
´+
==
+
\
m
H h1
M
æö
=+
ç÷
èø
46. Atomic weight of He = 4 amu
1 amu = 1.66 × 10
–24
 g
\ mass of one He atom in gram
= 4 × 1.66 × 10
–24
 g
= 6.64 × 10
–24
 g
47. Atomic weight of H atom = 1 amu
so number of H atom in 50 amu is 50
48. mole (n) = 
A
weight(w) No. of atom(N)
atomic wt. N
=
in 8g O
2
 No. of atoms
A
A
8NN
N
16N2
= Þ=
49. C
4
H
10
 + 2
13
O
2
 ¾® 4CO
2
 + 5H
2
O
1 mole    
13
2
mole
58 g   ¾®   208 g
1 kg   ¾®   
208
58
 = 3.58 kg
50. 1 Mg + 2 H
2
O ¾® Mg(OH)
2
 + 1 H
2
Q 1 mole Mg produce 1 mol H
2
51. 2KClO
3
 ¾® 2KCl + 3O
2
Q 3 × 22.4 L O
2
 is formed by 2 mol KClO
3
\ 5.6 L O
2
 is formed by 
25.61
3 22.4 6
´
=
´
mol KClO
3
52. Molecular weight of MCl
x
80 = x(4.5 + 35.5)
80 = 40x
x = 2
Atomic weight = Equivalent weight × Valency
= 4.5 × 2 = 9
53. 1 molecule of CO
2
 contains 2 atoms of oxygen
12.04 × 10
22
 molecules contain
= 12.04 × 10
22
 × 2
= 24.08 × 10
22
 atoms of oxygen
54. Number of oxygen atoms
           = 
23
210
6.02 10 2
22400
´ ´´
            = 11.28 × 10
21
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