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# Allen mock paper 4 for aiims solution NEET Notes | EduRev

## NEET : Allen mock paper 4 for aiims solution NEET Notes | EduRev

``` Page 1

HINT – SHEET
1 . Dot product of two perpendicular vector's
should be zero.
a + b = 0. Now put value from options.
2 .
m 2m
v
1kg
v
1
v
2
?
2m
m
By COLM
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern
:
AIIMS
TEST DATE
:
14 - 05 - 2018
HS - 1/ 7 10 01 CM A305 31 70 31
TEST SYLLABUS : FULL SYLLABUS
mv = 2mv
2
cos ? ? ? v
2
cos ? =
v
2
....(1)
2mv
2
sin ? = mv
1
? v
2
sin ? =
1
v
2
....(2)
By COKE
? ?
2 2 2
1 2
1 1 1
mv mv 2m v
2 2 2
? ?
2 2 2
1 2
v v 2v ? ?
2 2 2 2 2
2 2 2
4v cos 4v sin 2v ? ? ? ?
2cos
2
? = 2sin
2
? + 1
2cos
2
? = 2 – 2 cos
2
? + 1
4 cos
2
? = 3
cos ? =
3
2
? = 30°
Que . 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0
A ns . 4 2 3 1 3 3 4 2 3 1 3 2 3 4 3 1 2 2 2 4
Que . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0
A ns . 4 2 2 2 3 1 1 3 4 3 4 3 1 2 3 4 3 2 1 2
Que . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0
A ns . 3 3 2 1 2 3 2 2 3 1 2 4 3 3 4 2 2 2 3 1
Que . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0
A ns . 2 3 4 4 4 2 1 4 2 4 4 2 3 3 4 3 1 4 2 1
Que . 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 1 0 0
A ns . 4 3 1 3 3 1 2 4 3 2 4 3 2 2 3 2 3 1 1 4
Que . 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 2 0
A ns . 1 4 2 1 3 2 4 4 1 2 2 4 2 3 2 2 2 3 3 3
Que . 1 2 1 1 2 2 1 2 3 1 2 4 1 2 5 1 2 6 1 2 7 1 2 8 1 2 9 1 3 0 1 3 1 1 3 2 1 3 3 1 3 4 1 3 5 1 3 6 1 3 7 1 3 8 1 3 9 1 4 0
A ns . 4 1 1 2 4 4 3 1 1 2 3 1 1 3 4 1 2 3 1 1
Que . 1 4 1 1 4 2 1 4 3 1 4 4 1 4 5 1 4 6 1 4 7 1 4 8 1 4 9 1 5 0 1 5 1 1 5 2 1 5 3 1 5 4 1 5 5 1 5 6 1 5 7 1 5 8 1 5 9 1 6 0
A ns . 4 2 4 4 1 1 1 4 1 2 3 4 4 3 4 3 4 4 2 3
Que . 1 6 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 7 1 6 8 1 6 9 1 7 0 1 7 1 1 7 2 1 7 3 1 7 4 1 7 5 1 7 6 1 7 7 1 7 8 1 7 9 1 8 0
A ns . 2 1 4 3 1 2 1 2 2 2 3 4 1 3 3 1 4 2 1 3
Que . 1 8 1 1 8 2 1 8 3 1 8 4 1 8 5 1 8 6 1 8 7 1 8 8 1 8 9 1 9 0 1 9 1 1 9 2 1 9 3 1 9 4 1 9 5 1 9 6 1 9 7 1 9 8 1 9 9 2 0 0
A ns . 4 3 2 2 4 4 1 2 1 4 2 3 3 3 2 1 1 2 4 2
Page 2

HINT – SHEET
1 . Dot product of two perpendicular vector's
should be zero.
a + b = 0. Now put value from options.
2 .
m 2m
v
1kg
v
1
v
2
?
2m
m
By COLM
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern
:
AIIMS
TEST DATE
:
14 - 05 - 2018
HS - 1/ 7 10 01 CM A305 31 70 31
TEST SYLLABUS : FULL SYLLABUS
mv = 2mv
2
cos ? ? ? v
2
cos ? =
v
2
....(1)
2mv
2
sin ? = mv
1
? v
2
sin ? =
1
v
2
....(2)
By COKE
? ?
2 2 2
1 2
1 1 1
mv mv 2m v
2 2 2
? ?
2 2 2
1 2
v v 2v ? ?
2 2 2 2 2
2 2 2
4v cos 4v sin 2v ? ? ? ?
2cos
2
? = 2sin
2
? + 1
2cos
2
? = 2 – 2 cos
2
? + 1
4 cos
2
? = 3
cos ? =
3
2
? = 30°
Que . 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0
A ns . 4 2 3 1 3 3 4 2 3 1 3 2 3 4 3 1 2 2 2 4
Que . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0
A ns . 4 2 2 2 3 1 1 3 4 3 4 3 1 2 3 4 3 2 1 2
Que . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0
A ns . 3 3 2 1 2 3 2 2 3 1 2 4 3 3 4 2 2 2 3 1
Que . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0
A ns . 2 3 4 4 4 2 1 4 2 4 4 2 3 3 4 3 1 4 2 1
Que . 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 1 0 0
A ns . 4 3 1 3 3 1 2 4 3 2 4 3 2 2 3 2 3 1 1 4
Que . 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 2 0
A ns . 1 4 2 1 3 2 4 4 1 2 2 4 2 3 2 2 2 3 3 3
Que . 1 2 1 1 2 2 1 2 3 1 2 4 1 2 5 1 2 6 1 2 7 1 2 8 1 2 9 1 3 0 1 3 1 1 3 2 1 3 3 1 3 4 1 3 5 1 3 6 1 3 7 1 3 8 1 3 9 1 4 0
A ns . 4 1 1 2 4 4 3 1 1 2 3 1 1 3 4 1 2 3 1 1
Que . 1 4 1 1 4 2 1 4 3 1 4 4 1 4 5 1 4 6 1 4 7 1 4 8 1 4 9 1 5 0 1 5 1 1 5 2 1 5 3 1 5 4 1 5 5 1 5 6 1 5 7 1 5 8 1 5 9 1 6 0
A ns . 4 2 4 4 1 1 1 4 1 2 3 4 4 3 4 3 4 4 2 3
Que . 1 6 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 7 1 6 8 1 6 9 1 7 0 1 7 1 1 7 2 1 7 3 1 7 4 1 7 5 1 7 6 1 7 7 1 7 8 1 7 9 1 8 0
A ns . 2 1 4 3 1 2 1 2 2 2 3 4 1 3 3 1 4 2 1 3
Que . 1 8 1 1 8 2 1 8 3 1 8 4 1 8 5 1 8 6 1 8 7 1 8 8 1 8 9 1 9 0 1 9 1 1 9 2 1 9 3 1 9 4 1 9 5 1 9 6 1 9 7 1 9 8 1 9 9 2 0 0
A ns . 4 3 2 2 4 4 1 2 1 4 2 3 3 3 2 1 1 2 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
3 . For open pipe
f =
V
2 ?
For pipe dipped in water f ' =
v v
f
4 / 2 2
? ?
? ?
5 . ? = 2R cos ?
S = ut +
1
2
at
2
a = g cos ? ? ? 2R cos ? = 0 +
1
2
g cos ? t
2
? t =
R
2
g
6 . ?
w
gh =
1
2
?
w
v
2
;  v = 2gh
dV
dt
= Av = ( ?r
2
v)
7 . Acceleration due to gravity will be directed
towards centre of mass of earth.
8 .
d
e a 2at
dt
?
? ? ? ?
= a( ? – 2t)
H =
2
0
e
.dt
R
?
?
=
? ?
2
2 2 3
0
a 2t a
dt
R 3R
? ? ? ?
?
?
9 . From S = ut +
1
2
at
2
40 = u(5) +
1
2
a(5)
2
.....(1)
40 + 65 = u(10) +
1
2
a(10)
2
.....(2)
Solve equation (1) & (2)
then
u = 5.5 m/s
1 0 . W = 2T ?A
= (1.3 × 10
–4
) (3 × 10
–3
) × 2
= 7.8 × 10
–7
J
1 1 . C
eq
=
(3 3) (1 1) 6 2 5
1 1 µF
(3 3) (1 1) 6 2 2
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? Q = C × V =
5
2
× 100 = 250 µC
Charge in 6 µF branch = VC =
6 2
6 2
? ? ?
? ?
?
? ?
100
= 150 µF
•
100 V
1 µF
A C
2 µF 6 µF
B
1 3 .
12kg
8kg
a
a
12g
8g
T
1
T
2
T
1
– T
2
– 12 g = 12a ......(1)
T
2
– 8g = 8a ......(2)
Solve equation (1) & (2)
T
1
= 240 N, T
2
= 96 N
1 4 . Change in length for both rods should be same
? ?
1
= ? ?
2
? ?
1
?
1
?T = ?
2
?
2
?T ?? ?
1
?
1
= ?
2
?
2
1 6 .
2 1 2 1
v u R
? ? ? ? ?
? ?
?
? ?
? ?
1 1.5 1 1.5
1 u 2
u = –1.2 cm
Page 3

HINT – SHEET
1 . Dot product of two perpendicular vector's
should be zero.
a + b = 0. Now put value from options.
2 .
m 2m
v
1kg
v
1
v
2
?
2m
m
By COLM
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern
:
AIIMS
TEST DATE
:
14 - 05 - 2018
HS - 1/ 7 10 01 CM A305 31 70 31
TEST SYLLABUS : FULL SYLLABUS
mv = 2mv
2
cos ? ? ? v
2
cos ? =
v
2
....(1)
2mv
2
sin ? = mv
1
? v
2
sin ? =
1
v
2
....(2)
By COKE
? ?
2 2 2
1 2
1 1 1
mv mv 2m v
2 2 2
? ?
2 2 2
1 2
v v 2v ? ?
2 2 2 2 2
2 2 2
4v cos 4v sin 2v ? ? ? ?
2cos
2
? = 2sin
2
? + 1
2cos
2
? = 2 – 2 cos
2
? + 1
4 cos
2
? = 3
cos ? =
3
2
? = 30°
Que . 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0
A ns . 4 2 3 1 3 3 4 2 3 1 3 2 3 4 3 1 2 2 2 4
Que . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0
A ns . 4 2 2 2 3 1 1 3 4 3 4 3 1 2 3 4 3 2 1 2
Que . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0
A ns . 3 3 2 1 2 3 2 2 3 1 2 4 3 3 4 2 2 2 3 1
Que . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0
A ns . 2 3 4 4 4 2 1 4 2 4 4 2 3 3 4 3 1 4 2 1
Que . 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 1 0 0
A ns . 4 3 1 3 3 1 2 4 3 2 4 3 2 2 3 2 3 1 1 4
Que . 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 2 0
A ns . 1 4 2 1 3 2 4 4 1 2 2 4 2 3 2 2 2 3 3 3
Que . 1 2 1 1 2 2 1 2 3 1 2 4 1 2 5 1 2 6 1 2 7 1 2 8 1 2 9 1 3 0 1 3 1 1 3 2 1 3 3 1 3 4 1 3 5 1 3 6 1 3 7 1 3 8 1 3 9 1 4 0
A ns . 4 1 1 2 4 4 3 1 1 2 3 1 1 3 4 1 2 3 1 1
Que . 1 4 1 1 4 2 1 4 3 1 4 4 1 4 5 1 4 6 1 4 7 1 4 8 1 4 9 1 5 0 1 5 1 1 5 2 1 5 3 1 5 4 1 5 5 1 5 6 1 5 7 1 5 8 1 5 9 1 6 0
A ns . 4 2 4 4 1 1 1 4 1 2 3 4 4 3 4 3 4 4 2 3
Que . 1 6 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 7 1 6 8 1 6 9 1 7 0 1 7 1 1 7 2 1 7 3 1 7 4 1 7 5 1 7 6 1 7 7 1 7 8 1 7 9 1 8 0
A ns . 2 1 4 3 1 2 1 2 2 2 3 4 1 3 3 1 4 2 1 3
Que . 1 8 1 1 8 2 1 8 3 1 8 4 1 8 5 1 8 6 1 8 7 1 8 8 1 8 9 1 9 0 1 9 1 1 9 2 1 9 3 1 9 4 1 9 5 1 9 6 1 9 7 1 9 8 1 9 9 2 0 0
A ns . 4 3 2 2 4 4 1 2 1 4 2 3 3 3 2 1 1 2 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
3 . For open pipe
f =
V
2 ?
For pipe dipped in water f ' =
v v
f
4 / 2 2
? ?
? ?
5 . ? = 2R cos ?
S = ut +
1
2
at
2
a = g cos ? ? ? 2R cos ? = 0 +
1
2
g cos ? t
2
? t =
R
2
g
6 . ?
w
gh =
1
2
?
w
v
2
;  v = 2gh
dV
dt
= Av = ( ?r
2
v)
7 . Acceleration due to gravity will be directed
towards centre of mass of earth.
8 .
d
e a 2at
dt
?
? ? ? ?
= a( ? – 2t)
H =
2
0
e
.dt
R
?
?
=
? ?
2
2 2 3
0
a 2t a
dt
R 3R
? ? ? ?
?
?
9 . From S = ut +
1
2
at
2
40 = u(5) +
1
2
a(5)
2
.....(1)
40 + 65 = u(10) +
1
2
a(10)
2
.....(2)
Solve equation (1) & (2)
then
u = 5.5 m/s
1 0 . W = 2T ?A
= (1.3 × 10
–4
) (3 × 10
–3
) × 2
= 7.8 × 10
–7
J
1 1 . C
eq
=
(3 3) (1 1) 6 2 5
1 1 µF
(3 3) (1 1) 6 2 2
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? Q = C × V =
5
2
× 100 = 250 µC
Charge in 6 µF branch = VC =
6 2
6 2
? ? ?
? ?
?
? ?
100
= 150 µF
•
100 V
1 µF
A C
2 µF 6 µF
B
1 3 .
12kg
8kg
a
a
12g
8g
T
1
T
2
T
1
– T
2
– 12 g = 12a ......(1)
T
2
– 8g = 8a ......(2)
Solve equation (1) & (2)
T
1
= 240 N, T
2
= 96 N
1 4 . Change in length for both rods should be same
? ?
1
= ? ?
2
? ?
1
?
1
?T = ?
2
?
2
?T ?? ?
1
?
1
= ?
2
?
2
1 6 .
2 1 2 1
v u R
? ? ? ? ?
? ?
?
? ?
? ?
1 1.5 1 1.5
1 u 2
u = –1.2 cm
HS - 3/ 7
1 0 0 1C M A 3 0 5 3 17 0 3 1
1 7 . From constraints motion
v'cos (90 – ?) = v

V'
?
V
v
v'
sin
?
?
1 8 . ? = ?AT
4
....(1)
?
m
T = b
m
b
T ?
?
....(2)
from (1) & (2)
2
4 4
A r
? ? ?
? ?
1 9 . Electric field in the region 1, 3 and 5 is zero
i.e. E
1
= E
3
= E
5
slope of the line BC < slope
of the line DE
i.e. E
2
< E
4
1 2 3 4
V
A
B
C D
E
1 2 3 4 5
2 0 . For a normal eye, rays coming from infinity
should go the retina without effort when we look
at infinity, lens offers minimum power and
hence combination gives 40D + 20D = 60D.
Distance between the retina and the cornea eye
has must be equal to focal length.
f =
1
60
m = 1.67 cm
2 1 . Applying law of conservation of mechanical
energy between points A and B,
u
v
L
O B
A
2 2
1 1
mu mv mgL
2 2
? ?
?
2
v u 2gL ? ?
So, magnitude of change in velocity,
i.e.,
2 2
| v u | v u 2uv cos90 ? ? ? ? ?
? ?
?v =
2 2
v u ?
=
? ?
2
2 u gL ?
2 2 . Efficiency = 1 –
2
1
T
T
2 5 . From work-energy theorem,
?K.E = W
net
or K
f
– K
i
= Pdt
?
or
2
2 2
0
1 3
mv t dt
2 2
? ?
?
? ?
? ?
?
or v
2
=
2
3
0
t
2
? ?
? ?
? ?
? ?
? v = 2 m/s
28 . Number of lines in absorption spectrum = (n–1)
? 5 = n – 1 ? n = 6
? Number of bright lines in the emission
spectrum
n(n 1) 6(6 1)
15
2 2
? ?
? ? ?
.
Page 4

HINT – SHEET
1 . Dot product of two perpendicular vector's
should be zero.
a + b = 0. Now put value from options.
2 .
m 2m
v
1kg
v
1
v
2
?
2m
m
By COLM
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern
:
AIIMS
TEST DATE
:
14 - 05 - 2018
HS - 1/ 7 10 01 CM A305 31 70 31
TEST SYLLABUS : FULL SYLLABUS
mv = 2mv
2
cos ? ? ? v
2
cos ? =
v
2
....(1)
2mv
2
sin ? = mv
1
? v
2
sin ? =
1
v
2
....(2)
By COKE
? ?
2 2 2
1 2
1 1 1
mv mv 2m v
2 2 2
? ?
2 2 2
1 2
v v 2v ? ?
2 2 2 2 2
2 2 2
4v cos 4v sin 2v ? ? ? ?
2cos
2
? = 2sin
2
? + 1
2cos
2
? = 2 – 2 cos
2
? + 1
4 cos
2
? = 3
cos ? =
3
2
? = 30°
Que . 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0
A ns . 4 2 3 1 3 3 4 2 3 1 3 2 3 4 3 1 2 2 2 4
Que . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0
A ns . 4 2 2 2 3 1 1 3 4 3 4 3 1 2 3 4 3 2 1 2
Que . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0
A ns . 3 3 2 1 2 3 2 2 3 1 2 4 3 3 4 2 2 2 3 1
Que . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0
A ns . 2 3 4 4 4 2 1 4 2 4 4 2 3 3 4 3 1 4 2 1
Que . 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 1 0 0
A ns . 4 3 1 3 3 1 2 4 3 2 4 3 2 2 3 2 3 1 1 4
Que . 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 2 0
A ns . 1 4 2 1 3 2 4 4 1 2 2 4 2 3 2 2 2 3 3 3
Que . 1 2 1 1 2 2 1 2 3 1 2 4 1 2 5 1 2 6 1 2 7 1 2 8 1 2 9 1 3 0 1 3 1 1 3 2 1 3 3 1 3 4 1 3 5 1 3 6 1 3 7 1 3 8 1 3 9 1 4 0
A ns . 4 1 1 2 4 4 3 1 1 2 3 1 1 3 4 1 2 3 1 1
Que . 1 4 1 1 4 2 1 4 3 1 4 4 1 4 5 1 4 6 1 4 7 1 4 8 1 4 9 1 5 0 1 5 1 1 5 2 1 5 3 1 5 4 1 5 5 1 5 6 1 5 7 1 5 8 1 5 9 1 6 0
A ns . 4 2 4 4 1 1 1 4 1 2 3 4 4 3 4 3 4 4 2 3
Que . 1 6 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 7 1 6 8 1 6 9 1 7 0 1 7 1 1 7 2 1 7 3 1 7 4 1 7 5 1 7 6 1 7 7 1 7 8 1 7 9 1 8 0
A ns . 2 1 4 3 1 2 1 2 2 2 3 4 1 3 3 1 4 2 1 3
Que . 1 8 1 1 8 2 1 8 3 1 8 4 1 8 5 1 8 6 1 8 7 1 8 8 1 8 9 1 9 0 1 9 1 1 9 2 1 9 3 1 9 4 1 9 5 1 9 6 1 9 7 1 9 8 1 9 9 2 0 0
A ns . 4 3 2 2 4 4 1 2 1 4 2 3 3 3 2 1 1 2 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
3 . For open pipe
f =
V
2 ?
For pipe dipped in water f ' =
v v
f
4 / 2 2
? ?
? ?
5 . ? = 2R cos ?
S = ut +
1
2
at
2
a = g cos ? ? ? 2R cos ? = 0 +
1
2
g cos ? t
2
? t =
R
2
g
6 . ?
w
gh =
1
2
?
w
v
2
;  v = 2gh
dV
dt
= Av = ( ?r
2
v)
7 . Acceleration due to gravity will be directed
towards centre of mass of earth.
8 .
d
e a 2at
dt
?
? ? ? ?
= a( ? – 2t)
H =
2
0
e
.dt
R
?
?
=
? ?
2
2 2 3
0
a 2t a
dt
R 3R
? ? ? ?
?
?
9 . From S = ut +
1
2
at
2
40 = u(5) +
1
2
a(5)
2
.....(1)
40 + 65 = u(10) +
1
2
a(10)
2
.....(2)
Solve equation (1) & (2)
then
u = 5.5 m/s
1 0 . W = 2T ?A
= (1.3 × 10
–4
) (3 × 10
–3
) × 2
= 7.8 × 10
–7
J
1 1 . C
eq
=
(3 3) (1 1) 6 2 5
1 1 µF
(3 3) (1 1) 6 2 2
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? Q = C × V =
5
2
× 100 = 250 µC
Charge in 6 µF branch = VC =
6 2
6 2
? ? ?
? ?
?
? ?
100
= 150 µF
•
100 V
1 µF
A C
2 µF 6 µF
B
1 3 .
12kg
8kg
a
a
12g
8g
T
1
T
2
T
1
– T
2
– 12 g = 12a ......(1)
T
2
– 8g = 8a ......(2)
Solve equation (1) & (2)
T
1
= 240 N, T
2
= 96 N
1 4 . Change in length for both rods should be same
? ?
1
= ? ?
2
? ?
1
?
1
?T = ?
2
?
2
?T ?? ?
1
?
1
= ?
2
?
2
1 6 .
2 1 2 1
v u R
? ? ? ? ?
? ?
?
? ?
? ?
1 1.5 1 1.5
1 u 2
u = –1.2 cm
HS - 3/ 7
1 0 0 1C M A 3 0 5 3 17 0 3 1
1 7 . From constraints motion
v'cos (90 – ?) = v

V'
?
V
v
v'
sin
?
?
1 8 . ? = ?AT
4
....(1)
?
m
T = b
m
b
T ?
?
....(2)
from (1) & (2)
2
4 4
A r
? ? ?
? ?
1 9 . Electric field in the region 1, 3 and 5 is zero
i.e. E
1
= E
3
= E
5
slope of the line BC < slope
of the line DE
i.e. E
2
< E
4
1 2 3 4
V
A
B
C D
E
1 2 3 4 5
2 0 . For a normal eye, rays coming from infinity
should go the retina without effort when we look
at infinity, lens offers minimum power and
hence combination gives 40D + 20D = 60D.
Distance between the retina and the cornea eye
has must be equal to focal length.
f =
1
60
m = 1.67 cm
2 1 . Applying law of conservation of mechanical
energy between points A and B,
u
v
L
O B
A
2 2
1 1
mu mv mgL
2 2
? ?
?
2
v u 2gL ? ?
So, magnitude of change in velocity,
i.e.,
2 2
| v u | v u 2uv cos90 ? ? ? ? ?
? ?
?v =
2 2
v u ?
=
? ?
2
2 u gL ?
2 2 . Efficiency = 1 –
2
1
T
T
2 5 . From work-energy theorem,
?K.E = W
net
or K
f
– K
i
= Pdt
?
or
2
2 2
0
1 3
mv t dt
2 2
? ?
?
? ?
? ?
?
or v
2
=
2
3
0
t
2
? ?
? ?
? ?
? ?
? v = 2 m/s
28 . Number of lines in absorption spectrum = (n–1)
? 5 = n – 1 ? n = 6
? Number of bright lines in the emission
spectrum
n(n 1) 6(6 1)
15
2 2
? ?
? ? ?
.
HS - 4/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
2 9 .
B A
D C
AB
2
BC
?
? AB = DC =
3
l
6
l
Similarly, m
AB
= m
DC
=
m
3
and m
BC
= m
=
m
6
Now, I = 2I
AB
+ I
+ I
BC
=
2 2
m 1 m
2 0
3 3 3 6 3
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
l l
=
2 2
2 1
m m
81 54
? l l
=
2
7
m
162
l
3 0 . If m is mass, A is the amplitude of oscillation,
then maximum kinetic energy is
K = K
0
cos
2
?t
2 2
0 0
1
K m A
2
? ?
?
1/ 2
0
0
2
2K
A
m
? ?
?
? ?
?
? ?
Displacement x = A
0
sin ?t
x =
1
2
0
2
2K
sin t
m
? ?
?
? ?
?
? ?
3 1 .
B
1
= B
2
=
0
90 90
4 2
I
sin sin
d
? ? ? ? ? ?
? ? ? ?
? ? ? ?
?
? ? ? ?
2
d
sin
r
? ? ?
?
? ?
? ?
?
B
1
= B
2
=
0
1
2
4
2
I
cos
rsin
? ? ? ?
?
? ?
? ? ?
? ?
?
? ?
? ?
= B
B
0
= 2B   =
0
1
2
2
2
I cos
r sin
? ? ?
? ?
? ?
? ?
?
?
3 2 . The activity
dN
N
dt
? ?
? ? ?
? ?
? ?
?
1/ 2
e
dN T
N
dt log 2
? ? ? ?
? ?
? ?
? ?
? ?
? ?
Taking the ratio of this expression for
240
Pu to
this same expression for
243
Am,

Pu
1/ 2 Pu
pu
Am Am
1/ 2 Am
dN
(T )
N
(5µci) (6560y)
dt
1
dN N (4.45µci) (7370y)
1 (T )
dt
? ?
?
? ?
?
? ?
? ? ?
? ? ?
?
? ?
? ?
i.e. the two samples contains equal number of
nuclei.
3 3 . Work done,
W = ?d? = Fr d ?
= 200 × 3 ×
3
2
2
? ?
? ?
? ?
? ?
= 5652 J
Page 5

HINT – SHEET
1 . Dot product of two perpendicular vector's
should be zero.
a + b = 0. Now put value from options.
2 .
m 2m
v
1kg
v
1
v
2
?
2m
m
By COLM
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern
:
AIIMS
TEST DATE
:
14 - 05 - 2018
HS - 1/ 7 10 01 CM A305 31 70 31
TEST SYLLABUS : FULL SYLLABUS
mv = 2mv
2
cos ? ? ? v
2
cos ? =
v
2
....(1)
2mv
2
sin ? = mv
1
? v
2
sin ? =
1
v
2
....(2)
By COKE
? ?
2 2 2
1 2
1 1 1
mv mv 2m v
2 2 2
? ?
2 2 2
1 2
v v 2v ? ?
2 2 2 2 2
2 2 2
4v cos 4v sin 2v ? ? ? ?
2cos
2
? = 2sin
2
? + 1
2cos
2
? = 2 – 2 cos
2
? + 1
4 cos
2
? = 3
cos ? =
3
2
? = 30°
Que . 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0
A ns . 4 2 3 1 3 3 4 2 3 1 3 2 3 4 3 1 2 2 2 4
Que . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0
A ns . 4 2 2 2 3 1 1 3 4 3 4 3 1 2 3 4 3 2 1 2
Que . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0
A ns . 3 3 2 1 2 3 2 2 3 1 2 4 3 3 4 2 2 2 3 1
Que . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0
A ns . 2 3 4 4 4 2 1 4 2 4 4 2 3 3 4 3 1 4 2 1
Que . 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 1 0 0
A ns . 4 3 1 3 3 1 2 4 3 2 4 3 2 2 3 2 3 1 1 4
Que . 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 8 1 1 9 1 2 0
A ns . 1 4 2 1 3 2 4 4 1 2 2 4 2 3 2 2 2 3 3 3
Que . 1 2 1 1 2 2 1 2 3 1 2 4 1 2 5 1 2 6 1 2 7 1 2 8 1 2 9 1 3 0 1 3 1 1 3 2 1 3 3 1 3 4 1 3 5 1 3 6 1 3 7 1 3 8 1 3 9 1 4 0
A ns . 4 1 1 2 4 4 3 1 1 2 3 1 1 3 4 1 2 3 1 1
Que . 1 4 1 1 4 2 1 4 3 1 4 4 1 4 5 1 4 6 1 4 7 1 4 8 1 4 9 1 5 0 1 5 1 1 5 2 1 5 3 1 5 4 1 5 5 1 5 6 1 5 7 1 5 8 1 5 9 1 6 0
A ns . 4 2 4 4 1 1 1 4 1 2 3 4 4 3 4 3 4 4 2 3
Que . 1 6 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 7 1 6 8 1 6 9 1 7 0 1 7 1 1 7 2 1 7 3 1 7 4 1 7 5 1 7 6 1 7 7 1 7 8 1 7 9 1 8 0
A ns . 2 1 4 3 1 2 1 2 2 2 3 4 1 3 3 1 4 2 1 3
Que . 1 8 1 1 8 2 1 8 3 1 8 4 1 8 5 1 8 6 1 8 7 1 8 8 1 8 9 1 9 0 1 9 1 1 9 2 1 9 3 1 9 4 1 9 5 1 9 6 1 9 7 1 9 8 1 9 9 2 0 0
A ns . 4 3 2 2 4 4 1 2 1 4 2 3 3 3 2 1 1 2 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
3 . For open pipe
f =
V
2 ?
For pipe dipped in water f ' =
v v
f
4 / 2 2
? ?
? ?
5 . ? = 2R cos ?
S = ut +
1
2
at
2
a = g cos ? ? ? 2R cos ? = 0 +
1
2
g cos ? t
2
? t =
R
2
g
6 . ?
w
gh =
1
2
?
w
v
2
;  v = 2gh
dV
dt
= Av = ( ?r
2
v)
7 . Acceleration due to gravity will be directed
towards centre of mass of earth.
8 .
d
e a 2at
dt
?
? ? ? ?
= a( ? – 2t)
H =
2
0
e
.dt
R
?
?
=
? ?
2
2 2 3
0
a 2t a
dt
R 3R
? ? ? ?
?
?
9 . From S = ut +
1
2
at
2
40 = u(5) +
1
2
a(5)
2
.....(1)
40 + 65 = u(10) +
1
2
a(10)
2
.....(2)
Solve equation (1) & (2)
then
u = 5.5 m/s
1 0 . W = 2T ?A
= (1.3 × 10
–4
) (3 × 10
–3
) × 2
= 7.8 × 10
–7
J
1 1 . C
eq
=
(3 3) (1 1) 6 2 5
1 1 µF
(3 3) (1 1) 6 2 2
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
? Q = C × V =
5
2
× 100 = 250 µC
Charge in 6 µF branch = VC =
6 2
6 2
? ? ?
? ?
?
? ?
100
= 150 µF
•
100 V
1 µF
A C
2 µF 6 µF
B
1 3 .
12kg
8kg
a
a
12g
8g
T
1
T
2
T
1
– T
2
– 12 g = 12a ......(1)
T
2
– 8g = 8a ......(2)
Solve equation (1) & (2)
T
1
= 240 N, T
2
= 96 N
1 4 . Change in length for both rods should be same
? ?
1
= ? ?
2
? ?
1
?
1
?T = ?
2
?
2
?T ?? ?
1
?
1
= ?
2
?
2
1 6 .
2 1 2 1
v u R
? ? ? ? ?
? ?
?
? ?
? ?
1 1.5 1 1.5
1 u 2
u = –1.2 cm
HS - 3/ 7
1 0 0 1C M A 3 0 5 3 17 0 3 1
1 7 . From constraints motion
v'cos (90 – ?) = v

V'
?
V
v
v'
sin
?
?
1 8 . ? = ?AT
4
....(1)
?
m
T = b
m
b
T ?
?
....(2)
from (1) & (2)
2
4 4
A r
? ? ?
? ?
1 9 . Electric field in the region 1, 3 and 5 is zero
i.e. E
1
= E
3
= E
5
slope of the line BC < slope
of the line DE
i.e. E
2
< E
4
1 2 3 4
V
A
B
C D
E
1 2 3 4 5
2 0 . For a normal eye, rays coming from infinity
should go the retina without effort when we look
at infinity, lens offers minimum power and
hence combination gives 40D + 20D = 60D.
Distance between the retina and the cornea eye
has must be equal to focal length.
f =
1
60
m = 1.67 cm
2 1 . Applying law of conservation of mechanical
energy between points A and B,
u
v
L
O B
A
2 2
1 1
mu mv mgL
2 2
? ?
?
2
v u 2gL ? ?
So, magnitude of change in velocity,
i.e.,
2 2
| v u | v u 2uv cos90 ? ? ? ? ?
? ?
?v =
2 2
v u ?
=
? ?
2
2 u gL ?
2 2 . Efficiency = 1 –
2
1
T
T
2 5 . From work-energy theorem,
?K.E = W
net
or K
f
– K
i
= Pdt
?
or
2
2 2
0
1 3
mv t dt
2 2
? ?
?
? ?
? ?
?
or v
2
=
2
3
0
t
2
? ?
? ?
? ?
? ?
? v = 2 m/s
28 . Number of lines in absorption spectrum = (n–1)
? 5 = n – 1 ? n = 6
? Number of bright lines in the emission
spectrum
n(n 1) 6(6 1)
15
2 2
? ?
? ? ?
.
HS - 4/ 7
Target : Pre-Medical 2018/Major/AIIMS/14-05-2018
1 0 0 1C M A 3 0 5 3 17 0 3 1
2 9 .
B A
D C
AB
2
BC
?
? AB = DC =
3
l
6
l
Similarly, m
AB
= m
DC
=
m
3
and m
BC
= m
=
m
6
Now, I = 2I
AB
+ I
+ I
BC
=
2 2
m 1 m
2 0
3 3 3 6 3
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
l l
=
2 2
2 1
m m
81 54
? l l
=
2
7
m
162
l
3 0 . If m is mass, A is the amplitude of oscillation,
then maximum kinetic energy is
K = K
0
cos
2
?t
2 2
0 0
1
K m A
2
? ?
?
1/ 2
0
0
2
2K
A
m
? ?
?
? ?
?
? ?
Displacement x = A
0
sin ?t
x =
1
2
0
2
2K
sin t
m
? ?
?
? ?
?
? ?
3 1 .
B
1
= B
2
=
0
90 90
4 2
I
sin sin
d
? ? ? ? ? ?
? ? ? ?
? ? ? ?
?
? ? ? ?
2
d
sin
r
? ? ?
?
? ?
? ?
?
B
1
= B
2
=
0
1
2
4
2
I
cos
rsin
? ? ? ?
?
? ?
? ? ?
? ?
?
? ?
? ?
= B
B
0
= 2B   =
0
1
2
2
2
I cos
r sin
? ? ?
? ?
? ?
? ?
?
?
3 2 . The activity
dN
N
dt
? ?
? ? ?
? ?
? ?
?
1/ 2
e
dN T
N
dt log 2
? ? ? ?
? ?
? ?
? ?
? ?
? ?
Taking the ratio of this expression for
240
Pu to
this same expression for
243
Am,

Pu
1/ 2 Pu
pu
Am Am
1/ 2 Am
dN
(T )
N
(5µci) (6560y)
dt
1
dN N (4.45µci) (7370y)
1 (T )
dt
? ?
?
? ?
?
? ?
? ? ?
? ? ?
?
? ?
? ?
i.e. the two samples contains equal number of
nuclei.
3 3 . Work done,
W = ?d? = Fr d ?
= 200 × 3 ×
3
2
2
? ?
? ?
? ?
? ?
= 5652 J
HS - 5/ 7
1 0 0 1C M A 3 0 5 3 17 0 3 1
3 4 . T =
2
g
?
?
?
2
T ? ?
The time period of a simple pendulum is
independent of density of the material of the bol.
3 5 .
b
F
ab
F
bc
a c
? ? B
0
2B
0
90°
o
F
ab
= B
0
I
? ?
2R sin 90°
=
2
B
0
IR
F
bc
= 2B
0
I
? ?
2R sin90°
= 2
2
B
0
IR
F
net
=
2 2
ab bc
F F ? =
? ? ? ?
2 2
0 0
2B IR 2 2B IR ?
=
0
10B IR
3 6 . Y = A.B B.C ?
= A B B B ? ? ?
= A B C ? ? ? ? B B B ? ? ?
= A.B.C
3 7 . By COME
2
2 0
0
1 v 1
m mv mg mg
2 2 2 2
? ?
? ? ?
? ?
? ?
?
?
2 2
0 0
mv mv 3mg
8 2 2
? ?
?
2
0
5mv 3mg
8 2
?
?
0
12 g
v
5
?
?
3 8 .
f L
f L
? ?
?
? f =
L
f
L
?
?
= 1 ×
L
0.004 L
= 250 Hz
3 9 . For hollow cylinder
B =
2 2
0
2 2
µ I (r a )
2 r (b a )
?
? ?
,  2 2
I
J
(b a )
?
? ?
B =
2 2 2 2
0
2 2
µ J (b a )(r a )
2 r(b a )
? ? ?
? ?
=
2 2
0
µ J(r a )
2r
?
Put r = R/2, a = R/4
4 0 . I
B
=
C
I 1
mA
100
?
?
V
BB
= I
B
R
B
+ V
BE
=
? ?
3 3
1
10 100 10
100
? ? ?
? ? ?
? ?
? ?
+ 0.6 = 1.6 V
8 2 . NCERT Pg. # 243
8 4 . NCERT XII
th
Pg.#108,110,111,121
9 0 . NCERT Pg. # 157
9 2 . NCERT (XII) Pg. # 146,147 (E), 158(H)
1 0 2 . NCERT Pg # 279
1 0 3 . NCERT- (E) Pg # 128
NCERT- (H) Pg # 138
1 0 6 . NCERT Pg # 294
1 0 7 . NCERT- (E) Pg # 133
NCERT- (H) Pg # 143
1 1 1 . NCERT- (E) Pg # 137
NCERT- (H) Pg # 147
1 1 4 . NCERT Pg. # 321
1 1 8 . NCERT Pg. # 335
1 2 4 . Assertion is true by information. Reason is true
by formula but reson is not the explanation of
assertion.
```
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