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# ALLEN MOCK PAPER SOLUTION FOR AIIMS -- 1 NEET Notes | EduRev

## NEET : ALLEN MOCK PAPER SOLUTION FOR AIIMS -- 1 NEET Notes | EduRev

``` Page 1

HINT – SHEET
DISTANCE LEARNING PROGRAMME
(Academic Session : 2017 - 2018)
0000CMA303117002 LATS/HS - 1/7
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
1 . Here, u = 56 m s
–1
Let ? be the angle of projection with the
horizontal to have maximum range, with
maximum height = 40 m
Maximum height, H =
2 2
u sin
2g
?
40 =
? ?
2
2
56 sin
2 9.8
?
?
sin
2
? =
? ?
2
2 9.8 40 1
4
56
? ?
?
or sin ? =
1
2
? = sin
–1

1
2
? ?
? ?
? ?
= 30º
2 . Field at centre of curvature of circular
arc =
0
Q 1
. sin
2 R 2
? ? ?
? ?
? ? ? ?
Which is greatest for option (ii)
3 .
F
F
?
?
? =
2
? ?
? ?
? ?
?
F sin ? +
2
? ?
? ?
? ?
?
F sin ?
Test Type : MAJOR  Test Pattern
:
AIIMS
ALL INDIA OPEN TEST # 02 TEST DATE
:
22 - 04 - 2018
Q u e . 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 n s . 2 2 3 1 3 2 1 1 4 2 2 4 2 2 4 1 2 3 1 1
Q u e . 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 n s . 3 1 4 2 3 2 1 2 2 2 2 4 1 2 1 4 1 4 3 2
Q u e . 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 n s . 1 1 2 4 1 4 4 2 3 2 3 2 1 3 4 3 2 2 2 4
Q u e . 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 n s . 3 2 4 4 4 4 3 1 1 3 2 3 3 2 4 4 3 4 1 4
Q u e . 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 n s . 3 3 1 2 1 1 4 1 4 2 4 1 3 3 3 3 1 2 1 3
Q u e . 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 n s . 3 2 3 3 3 4 4 3 2 2 3 3 4 2 2 2 2 2 4 2
Q u e . 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 n s . 1 1 3 1 4 4 1 3 4 1 2 1 1 1 1 1 1 4 2 1
Q u e . 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 n s . 2 3 2 1 1 1 4 3 1 4 1 1 1 3 4 3 1 3 1 1
Q u e . 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 n s . 3 2 1 1 4 3 4 3 1 2 2 1 2 2 2 3 1 4 1 4
Q u e . 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 n s . 2 3 4 2 1 4 3 1 1 4 4 2 2 2 1 2 3 3 4 3
Page 2

HINT – SHEET
DISTANCE LEARNING PROGRAMME
(Academic Session : 2017 - 2018)
0000CMA303117002 LATS/HS - 1/7
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
1 . Here, u = 56 m s
–1
Let ? be the angle of projection with the
horizontal to have maximum range, with
maximum height = 40 m
Maximum height, H =
2 2
u sin
2g
?
40 =
? ?
2
2
56 sin
2 9.8
?
?
sin
2
? =
? ?
2
2 9.8 40 1
4
56
? ?
?
or sin ? =
1
2
? = sin
–1

1
2
? ?
? ?
? ?
= 30º
2 . Field at centre of curvature of circular
arc =
0
Q 1
. sin
2 R 2
? ? ?
? ?
? ? ? ?
Which is greatest for option (ii)
3 .
F
F
?
?
? =
2
? ?
? ?
? ?
?
F sin ? +
2
? ?
? ?
? ?
?
F sin ?
Test Type : MAJOR  Test Pattern
:
AIIMS
ALL INDIA OPEN TEST # 02 TEST DATE
:
22 - 04 - 2018
Q u e . 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 n s . 2 2 3 1 3 2 1 1 4 2 2 4 2 2 4 1 2 3 1 1
Q u e . 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 n s . 3 1 4 2 3 2 1 2 2 2 2 4 1 2 1 4 1 4 3 2
Q u e . 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 n s . 1 1 2 4 1 4 4 2 3 2 3 2 1 3 4 3 2 2 2 4
Q u e . 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 n s . 3 2 4 4 4 4 3 1 1 3 2 3 3 2 4 4 3 4 1 4
Q u e . 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 n s . 3 3 1 2 1 1 4 1 4 2 4 1 3 3 3 3 1 2 1 3
Q u e . 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 n s . 3 2 3 3 3 4 4 3 2 2 3 3 4 2 2 2 2 2 4 2
Q u e . 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 n s . 1 1 3 1 4 4 1 3 4 1 2 1 1 1 1 1 1 4 2 1
Q u e . 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 n s . 2 3 2 1 1 1 4 3 1 4 1 1 1 3 4 3 1 3 1 1
Q u e . 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 n s . 3 2 1 1 4 3 4 3 1 2 2 1 2 2 2 3 1 4 1 4
Q u e . 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 n s . 2 3 4 2 1 4 3 1 1 4 4 2 2 2 1 2 3 3 4 3
0000CMA303117002 HS - 2/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
? = F ? sin ?
If axis is changed
F
a
b
F
? = b F sin ? + aF sin ?
= (a + b) F sin ?
a + b = ?
? = ??F sin ?
4 . y
1
= a sin ( ?t + kx + 0.57)
y
2
= a cos ( ?t + kx) = a sin
t kx
2
? ? ?
? ? ?
? ?
? ?
? Phase difference
2
?
?
– 0.57 = 1.57 – 0.57
5 . F = 49 N
N = W = mg = 10 × 9.8 N
So µ =
F
N
=
49
10 9.8 ?
= 0.5
6 . When  S
1
closed,
V
1
=
E 3
4
When S
2
closed
V
2
=
6 E
7
When S
1
and S
2
are closed
V
3
=
2 E
3
7 .
L
h = L sin ?
?
1
2
mv
2
2
2
K
1
R
? ?
?
? ?
? ?
= mgh = mgL sin ?
v =
2
2
2gLsin
K
1
R
? ?
?
? ?
? ?
? ? ?
? ?
For disc
2
2
K
R
=
1
2
v =
4gLsin
3
?
8 . The motion of the particle is oscillatory.
Its time period = 4
2h / sin 4 2h
gsin sin g
?
?
? ?
9 . Here, m = 10 kg, v
i
= 10 m s
–1
Initial kinetic energy of the block is
K
i
= ? ? ? ?
2
2 1
i
1 1
mv 10kg 10ms 500J
2 2
?
? ? ? ?
Work done by retarding force
W =
30
30 2
r
20 20
x
F dx 0.1x dx 0.1
2
? ?
? ? ? ?
? ?
? ?
? ?
= –0.1
900 400
2
? ? ?
? ?
? ?
= –25 J
According to work energy theorem
W = K
f
– K
i
K
f
= W + K
i
= –25 J + 500 J = 475 J
1 0 . I =
6
30
A
Reading = 4 + 20I = 8 volt
1 1 . T
2
? r
3
2
1
T
T
=
3/ 2
2
1
r
r
? ?
? ?
? ?
=
3 / 2
1
1
4r
r
? ?
? ?
? ?
= 8
T
2
= 8T
1
T
1
= 1 day for communication satellite.
1 2 . Shift = t
1
1
? ?
?
? ?
?
? ?
= 1 cm
Ray is falling normally, image coincides with
object
Page 3

HINT – SHEET
DISTANCE LEARNING PROGRAMME
(Academic Session : 2017 - 2018)
0000CMA303117002 LATS/HS - 1/7
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
1 . Here, u = 56 m s
–1
Let ? be the angle of projection with the
horizontal to have maximum range, with
maximum height = 40 m
Maximum height, H =
2 2
u sin
2g
?
40 =
? ?
2
2
56 sin
2 9.8
?
?
sin
2
? =
? ?
2
2 9.8 40 1
4
56
? ?
?
or sin ? =
1
2
? = sin
–1

1
2
? ?
? ?
? ?
= 30º
2 . Field at centre of curvature of circular
arc =
0
Q 1
. sin
2 R 2
? ? ?
? ?
? ? ? ?
Which is greatest for option (ii)
3 .
F
F
?
?
? =
2
? ?
? ?
? ?
?
F sin ? +
2
? ?
? ?
? ?
?
F sin ?
Test Type : MAJOR  Test Pattern
:
AIIMS
ALL INDIA OPEN TEST # 02 TEST DATE
:
22 - 04 - 2018
Q u e . 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 n s . 2 2 3 1 3 2 1 1 4 2 2 4 2 2 4 1 2 3 1 1
Q u e . 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 n s . 3 1 4 2 3 2 1 2 2 2 2 4 1 2 1 4 1 4 3 2
Q u e . 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 n s . 1 1 2 4 1 4 4 2 3 2 3 2 1 3 4 3 2 2 2 4
Q u e . 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 n s . 3 2 4 4 4 4 3 1 1 3 2 3 3 2 4 4 3 4 1 4
Q u e . 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 n s . 3 3 1 2 1 1 4 1 4 2 4 1 3 3 3 3 1 2 1 3
Q u e . 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 n s . 3 2 3 3 3 4 4 3 2 2 3 3 4 2 2 2 2 2 4 2
Q u e . 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 n s . 1 1 3 1 4 4 1 3 4 1 2 1 1 1 1 1 1 4 2 1
Q u e . 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 n s . 2 3 2 1 1 1 4 3 1 4 1 1 1 3 4 3 1 3 1 1
Q u e . 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 n s . 3 2 1 1 4 3 4 3 1 2 2 1 2 2 2 3 1 4 1 4
Q u e . 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 n s . 2 3 4 2 1 4 3 1 1 4 4 2 2 2 1 2 3 3 4 3
0000CMA303117002 HS - 2/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
? = F ? sin ?
If axis is changed
F
a
b
F
? = b F sin ? + aF sin ?
= (a + b) F sin ?
a + b = ?
? = ??F sin ?
4 . y
1
= a sin ( ?t + kx + 0.57)
y
2
= a cos ( ?t + kx) = a sin
t kx
2
? ? ?
? ? ?
? ?
? ?
? Phase difference
2
?
?
– 0.57 = 1.57 – 0.57
5 . F = 49 N
N = W = mg = 10 × 9.8 N
So µ =
F
N
=
49
10 9.8 ?
= 0.5
6 . When  S
1
closed,
V
1
=
E 3
4
When S
2
closed
V
2
=
6 E
7
When S
1
and S
2
are closed
V
3
=
2 E
3
7 .
L
h = L sin ?
?
1
2
mv
2
2
2
K
1
R
? ?
?
? ?
? ?
= mgh = mgL sin ?
v =
2
2
2gLsin
K
1
R
? ?
?
? ?
? ?
? ? ?
? ?
For disc
2
2
K
R
=
1
2
v =
4gLsin
3
?
8 . The motion of the particle is oscillatory.
Its time period = 4
2h / sin 4 2h
gsin sin g
?
?
? ?
9 . Here, m = 10 kg, v
i
= 10 m s
–1
Initial kinetic energy of the block is
K
i
= ? ? ? ?
2
2 1
i
1 1
mv 10kg 10ms 500J
2 2
?
? ? ? ?
Work done by retarding force
W =
30
30 2
r
20 20
x
F dx 0.1x dx 0.1
2
? ?
? ? ? ?
? ?
? ?
? ?
= –0.1
900 400
2
? ? ?
? ?
? ?
= –25 J
According to work energy theorem
W = K
f
– K
i
K
f
= W + K
i
= –25 J + 500 J = 475 J
1 0 . I =
6
30
A
Reading = 4 + 20I = 8 volt
1 1 . T
2
? r
3
2
1
T
T
=
3/ 2
2
1
r
r
? ?
? ?
? ?
=
3 / 2
1
1
4r
r
? ?
? ?
? ?
= 8
T
2
= 8T
1
T
1
= 1 day for communication satellite.
1 2 . Shift = t
1
1
? ?
?
? ?
?
? ?
= 1 cm
Ray is falling normally, image coincides with
object
HS - 3/7 0000CMA303117002
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
1 4 . V = E – Ir
I = 0,  E = V = 2 volt
V = 0, r =
E 2
I 5
?
= 0.4 ?
1 5 . Take ? = density of coin
? = density of liquid ( ? < ?)
A = area of wooden block
m = mass of coin, M = mass of wooden block.
Initially volume displaced by wooden block
V
0
= A ?, ?A ?g = (m + M)g
V
0
= A ? =
m
?
+
M
?
.... (1)
When coin falls volume displaced
by wooden block + coin V
1
=
M
?
+
m
?
...(2)
M
?
+
m
?
<
m
?
+
M
?
(So V
1
< V
0
)
So liquid falls by
This volume difference (V
0
– V
1
).
So h decrease volume displaced by wooden
block
A ?
1
=
M
?
( ?
1
< ?)
Both ? & h decrease
1 6 .
F 10 1
m
F u 10 10 2
? ? ?
? ?
image is virtual
so, sense of motion remains same radius of
image =
1
2
cm
Speed =
2 r
t 2
? ?
?
cm/sec
1 7 . The relation between linear velocity v
?
and
angular velocity ?
? ?
is
v r ? ? ?
? ? ? ?
1 8 . C
PR
=
C C 5C
3 2 6
? ?
C
PQ
=
C 5C
C
4 4
? ?
C
PR
/ C
PQ
=
2
3
1 9 .
T
0
Q ms dt ?
?
=
20
4
0
mA
T
4
? ?
? ?
= 4 × 10
4
mA
2 0 . As D >>d & ? << d.
Hence we can use ? ? =
D
d
?
so distance between 5
th
bright fringe and 3
rd
dark fringe = 5 ? – (2 ? + ?/2)
=
5
2
?
=
–7
3
5 6.5 10 1
2 10
?
? ?
?
= 1.625 mm
2 1 . Weight of body on the surface of the earth
= mg = 72 N
Acceleration due to gravity at height h is

? ?
2 2
E E
h 2
E
E
E
gR gR 4
g g
R 9
R h
R
2
? ? ?
? ?
?
?
? ?
? ?

E
R
h
2
? ?
?
? ?
? ?
?
Gravitational force on body at height h is
F = mg
h
= m ×
4 4 4
g mg 72 32N
9 9 9
? ? ? ? ?
2 2 . Field due to lower straight current = 0
field due to square loop = 0
field due to upper straight current =
0
2I 1
1
4 a 2
? ? ?
?
? ?
?
? ?
2 3 .
? ?
1 2
KA T T d
dt L
? ?
?
? Rate of melting
Page 4

HINT – SHEET
DISTANCE LEARNING PROGRAMME
(Academic Session : 2017 - 2018)
0000CMA303117002 LATS/HS - 1/7
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
1 . Here, u = 56 m s
–1
Let ? be the angle of projection with the
horizontal to have maximum range, with
maximum height = 40 m
Maximum height, H =
2 2
u sin
2g
?
40 =
? ?
2
2
56 sin
2 9.8
?
?
sin
2
? =
? ?
2
2 9.8 40 1
4
56
? ?
?
or sin ? =
1
2
? = sin
–1

1
2
? ?
? ?
? ?
= 30º
2 . Field at centre of curvature of circular
arc =
0
Q 1
. sin
2 R 2
? ? ?
? ?
? ? ? ?
Which is greatest for option (ii)
3 .
F
F
?
?
? =
2
? ?
? ?
? ?
?
F sin ? +
2
? ?
? ?
? ?
?
F sin ?
Test Type : MAJOR  Test Pattern
:
AIIMS
ALL INDIA OPEN TEST # 02 TEST DATE
:
22 - 04 - 2018
Q u e . 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 n s . 2 2 3 1 3 2 1 1 4 2 2 4 2 2 4 1 2 3 1 1
Q u e . 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 n s . 3 1 4 2 3 2 1 2 2 2 2 4 1 2 1 4 1 4 3 2
Q u e . 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 n s . 1 1 2 4 1 4 4 2 3 2 3 2 1 3 4 3 2 2 2 4
Q u e . 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 n s . 3 2 4 4 4 4 3 1 1 3 2 3 3 2 4 4 3 4 1 4
Q u e . 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 n s . 3 3 1 2 1 1 4 1 4 2 4 1 3 3 3 3 1 2 1 3
Q u e . 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 n s . 3 2 3 3 3 4 4 3 2 2 3 3 4 2 2 2 2 2 4 2
Q u e . 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 n s . 1 1 3 1 4 4 1 3 4 1 2 1 1 1 1 1 1 4 2 1
Q u e . 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 n s . 2 3 2 1 1 1 4 3 1 4 1 1 1 3 4 3 1 3 1 1
Q u e . 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 n s . 3 2 1 1 4 3 4 3 1 2 2 1 2 2 2 3 1 4 1 4
Q u e . 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 n s . 2 3 4 2 1 4 3 1 1 4 4 2 2 2 1 2 3 3 4 3
0000CMA303117002 HS - 2/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
? = F ? sin ?
If axis is changed
F
a
b
F
? = b F sin ? + aF sin ?
= (a + b) F sin ?
a + b = ?
? = ??F sin ?
4 . y
1
= a sin ( ?t + kx + 0.57)
y
2
= a cos ( ?t + kx) = a sin
t kx
2
? ? ?
? ? ?
? ?
? ?
? Phase difference
2
?
?
– 0.57 = 1.57 – 0.57
5 . F = 49 N
N = W = mg = 10 × 9.8 N
So µ =
F
N
=
49
10 9.8 ?
= 0.5
6 . When  S
1
closed,
V
1
=
E 3
4
When S
2
closed
V
2
=
6 E
7
When S
1
and S
2
are closed
V
3
=
2 E
3
7 .
L
h = L sin ?
?
1
2
mv
2
2
2
K
1
R
? ?
?
? ?
? ?
= mgh = mgL sin ?
v =
2
2
2gLsin
K
1
R
? ?
?
? ?
? ?
? ? ?
? ?
For disc
2
2
K
R
=
1
2
v =
4gLsin
3
?
8 . The motion of the particle is oscillatory.
Its time period = 4
2h / sin 4 2h
gsin sin g
?
?
? ?
9 . Here, m = 10 kg, v
i
= 10 m s
–1
Initial kinetic energy of the block is
K
i
= ? ? ? ?
2
2 1
i
1 1
mv 10kg 10ms 500J
2 2
?
? ? ? ?
Work done by retarding force
W =
30
30 2
r
20 20
x
F dx 0.1x dx 0.1
2
? ?
? ? ? ?
? ?
? ?
? ?
= –0.1
900 400
2
? ? ?
? ?
? ?
= –25 J
According to work energy theorem
W = K
f
– K
i
K
f
= W + K
i
= –25 J + 500 J = 475 J
1 0 . I =
6
30
A
Reading = 4 + 20I = 8 volt
1 1 . T
2
? r
3
2
1
T
T
=
3/ 2
2
1
r
r
? ?
? ?
? ?
=
3 / 2
1
1
4r
r
? ?
? ?
? ?
= 8
T
2
= 8T
1
T
1
= 1 day for communication satellite.
1 2 . Shift = t
1
1
? ?
?
? ?
?
? ?
= 1 cm
Ray is falling normally, image coincides with
object
HS - 3/7 0000CMA303117002
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
1 4 . V = E – Ir
I = 0,  E = V = 2 volt
V = 0, r =
E 2
I 5
?
= 0.4 ?
1 5 . Take ? = density of coin
? = density of liquid ( ? < ?)
A = area of wooden block
m = mass of coin, M = mass of wooden block.
Initially volume displaced by wooden block
V
0
= A ?, ?A ?g = (m + M)g
V
0
= A ? =
m
?
+
M
?
.... (1)
When coin falls volume displaced
by wooden block + coin V
1
=
M
?
+
m
?
...(2)
M
?
+
m
?
<
m
?
+
M
?
(So V
1
< V
0
)
So liquid falls by
This volume difference (V
0
– V
1
).
So h decrease volume displaced by wooden
block
A ?
1
=
M
?
( ?
1
< ?)
Both ? & h decrease
1 6 .
F 10 1
m
F u 10 10 2
? ? ?
? ?
image is virtual
so, sense of motion remains same radius of
image =
1
2
cm
Speed =
2 r
t 2
? ?
?
cm/sec
1 7 . The relation between linear velocity v
?
and
angular velocity ?
? ?
is
v r ? ? ?
? ? ? ?
1 8 . C
PR
=
C C 5C
3 2 6
? ?
C
PQ
=
C 5C
C
4 4
? ?
C
PR
/ C
PQ
=
2
3
1 9 .
T
0
Q ms dt ?
?
=
20
4
0
mA
T
4
? ?
? ?
= 4 × 10
4
mA
2 0 . As D >>d & ? << d.
Hence we can use ? ? =
D
d
?
so distance between 5
th
bright fringe and 3
rd
dark fringe = 5 ? – (2 ? + ?/2)
=
5
2
?
=
–7
3
5 6.5 10 1
2 10
?
? ?
?
= 1.625 mm
2 1 . Weight of body on the surface of the earth
= mg = 72 N
Acceleration due to gravity at height h is

? ?
2 2
E E
h 2
E
E
E
gR gR 4
g g
R 9
R h
R
2
? ? ?
? ?
?
?
? ?
? ?

E
R
h
2
? ?
?
? ?
? ?
?
Gravitational force on body at height h is
F = mg
h
= m ×
4 4 4
g mg 72 32N
9 9 9
? ? ? ? ?
2 2 . Field due to lower straight current = 0
field due to square loop = 0
field due to upper straight current =
0
2I 1
1
4 a 2
? ? ?
?
? ?
?
? ?
2 3 .
? ?
1 2
KA T T d
dt L
? ?
?
? Rate of melting
0000CMA303117002 HS - 4/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
2 4 . tan (i
p
) = µ
tan (i
p
) =
3
i
p
= 60°
2 5 . Gravitational potential energy at any point at
a distance r from the centre of the earth is
E
GM m
U
r
? ?
where M
E
and m be masses of earth and body
repectively.
At the surface of the earth, r = R
E
? U
1
= –
E
E
GM m
R
At a height h from the surface,
r = R
E
+ h = R
E
+ nR
E
= (1 + n) R
E
? U
2
= –
? ?
E
E
GM m
n 1 R ?
Change in potential energy is
?U = U
2
– U
1
= –
? ?
E E
E E
GM m GM m
n 1 R R
? ?
? ?
? ?
?
? ?
=
? ? ? ?
E E
E E
GM m 1 GM mn
1
R n 1 n 1 R
? ?
? ?
? ?
? ?
? ?
? ?
=
? ?
E
E 2
E
n GM
mgR g
n 1 R
? ?
?
? ?
?
? ?
?
2 6 . Field due to long straight current
=
–6 0
2I
ˆ ˆ
k 10 k
4 d
?
?
?
wb/m
2
B
net
=
6
2 10
?
?
wb/m
2
2 7 . ? ?
0
dT
T T
dt
? ?
? ( ??)
n
n = 1
2 8 . K
max
(eV) = 12375
0
1 1
(Å) (Å)
? ?
?
? ?
? ?
? ?
= 12375
1 1
1000 2000
? ?
?
? ?
? ?
= 6.2 eV
2 9 . Work done , W =
1
2
× F × ?L
For a given F
W
?
?L
and ?L =
FL
AY
As F, A and L are constants
?
1
L
Y
? ?
From (i) and (ii), we get
1
W
Y
?
?
copper
steel
copper steel
Y
W
W Y
?
As Y
copper
< Y
steel
? W
steel
< W
copper
So, less work is done on steel spring.
3 0 . Due to motion along direction (3) these is no
change in flux, so emf = 0
3 1 .
P, V
300 K
V, P
150 K
2
Q
1
Q
2
300 K
P
2
, 2V,
Q
1
= nC
v
(–150)
Q
2
= nC
p
(150)
Q = n(150) (C
p
– C
v
)
= 300 R
Page 5

HINT – SHEET
DISTANCE LEARNING PROGRAMME
(Academic Session : 2017 - 2018)
0000CMA303117002 LATS/HS - 1/7
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
1 . Here, u = 56 m s
–1
Let ? be the angle of projection with the
horizontal to have maximum range, with
maximum height = 40 m
Maximum height, H =
2 2
u sin
2g
?
40 =
? ?
2
2
56 sin
2 9.8
?
?
sin
2
? =
? ?
2
2 9.8 40 1
4
56
? ?
?
or sin ? =
1
2
? = sin
–1

1
2
? ?
? ?
? ?
= 30º
2 . Field at centre of curvature of circular
arc =
0
Q 1
. sin
2 R 2
? ? ?
? ?
? ? ? ?
Which is greatest for option (ii)
3 .
F
F
?
?
? =
2
? ?
? ?
? ?
?
F sin ? +
2
? ?
? ?
? ?
?
F sin ?
Test Type : MAJOR  Test Pattern
:
AIIMS
ALL INDIA OPEN TEST # 02 TEST DATE
:
22 - 04 - 2018
Q u e . 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 n s . 2 2 3 1 3 2 1 1 4 2 2 4 2 2 4 1 2 3 1 1
Q u e . 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 n s . 3 1 4 2 3 2 1 2 2 2 2 4 1 2 1 4 1 4 3 2
Q u e . 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 n s . 1 1 2 4 1 4 4 2 3 2 3 2 1 3 4 3 2 2 2 4
Q u e . 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 n s . 3 2 4 4 4 4 3 1 1 3 2 3 3 2 4 4 3 4 1 4
Q u e . 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 n s . 3 3 1 2 1 1 4 1 4 2 4 1 3 3 3 3 1 2 1 3
Q u e . 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 n s . 3 2 3 3 3 4 4 3 2 2 3 3 4 2 2 2 2 2 4 2
Q u e . 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 n s . 1 1 3 1 4 4 1 3 4 1 2 1 1 1 1 1 1 4 2 1
Q u e . 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 n s . 2 3 2 1 1 1 4 3 1 4 1 1 1 3 4 3 1 3 1 1
Q u e . 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 n s . 3 2 1 1 4 3 4 3 1 2 2 1 2 2 2 3 1 4 1 4
Q u e . 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 n s . 2 3 4 2 1 4 3 1 1 4 4 2 2 2 1 2 3 3 4 3
0000CMA303117002 HS - 2/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
? = F ? sin ?
If axis is changed
F
a
b
F
? = b F sin ? + aF sin ?
= (a + b) F sin ?
a + b = ?
? = ??F sin ?
4 . y
1
= a sin ( ?t + kx + 0.57)
y
2
= a cos ( ?t + kx) = a sin
t kx
2
? ? ?
? ? ?
? ?
? ?
? Phase difference
2
?
?
– 0.57 = 1.57 – 0.57
5 . F = 49 N
N = W = mg = 10 × 9.8 N
So µ =
F
N
=
49
10 9.8 ?
= 0.5
6 . When  S
1
closed,
V
1
=
E 3
4
When S
2
closed
V
2
=
6 E
7
When S
1
and S
2
are closed
V
3
=
2 E
3
7 .
L
h = L sin ?
?
1
2
mv
2
2
2
K
1
R
? ?
?
? ?
? ?
= mgh = mgL sin ?
v =
2
2
2gLsin
K
1
R
? ?
?
? ?
? ?
? ? ?
? ?
For disc
2
2
K
R
=
1
2
v =
4gLsin
3
?
8 . The motion of the particle is oscillatory.
Its time period = 4
2h / sin 4 2h
gsin sin g
?
?
? ?
9 . Here, m = 10 kg, v
i
= 10 m s
–1
Initial kinetic energy of the block is
K
i
= ? ? ? ?
2
2 1
i
1 1
mv 10kg 10ms 500J
2 2
?
? ? ? ?
Work done by retarding force
W =
30
30 2
r
20 20
x
F dx 0.1x dx 0.1
2
? ?
? ? ? ?
? ?
? ?
? ?
= –0.1
900 400
2
? ? ?
? ?
? ?
= –25 J
According to work energy theorem
W = K
f
– K
i
K
f
= W + K
i
= –25 J + 500 J = 475 J
1 0 . I =
6
30
A
Reading = 4 + 20I = 8 volt
1 1 . T
2
? r
3
2
1
T
T
=
3/ 2
2
1
r
r
? ?
? ?
? ?
=
3 / 2
1
1
4r
r
? ?
? ?
? ?
= 8
T
2
= 8T
1
T
1
= 1 day for communication satellite.
1 2 . Shift = t
1
1
? ?
?
? ?
?
? ?
= 1 cm
Ray is falling normally, image coincides with
object
HS - 3/7 0000CMA303117002
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
1 4 . V = E – Ir
I = 0,  E = V = 2 volt
V = 0, r =
E 2
I 5
?
= 0.4 ?
1 5 . Take ? = density of coin
? = density of liquid ( ? < ?)
A = area of wooden block
m = mass of coin, M = mass of wooden block.
Initially volume displaced by wooden block
V
0
= A ?, ?A ?g = (m + M)g
V
0
= A ? =
m
?
+
M
?
.... (1)
When coin falls volume displaced
by wooden block + coin V
1
=
M
?
+
m
?
...(2)
M
?
+
m
?
<
m
?
+
M
?
(So V
1
< V
0
)
So liquid falls by
This volume difference (V
0
– V
1
).
So h decrease volume displaced by wooden
block
A ?
1
=
M
?
( ?
1
< ?)
Both ? & h decrease
1 6 .
F 10 1
m
F u 10 10 2
? ? ?
? ?
image is virtual
so, sense of motion remains same radius of
image =
1
2
cm
Speed =
2 r
t 2
? ?
?
cm/sec
1 7 . The relation between linear velocity v
?
and
angular velocity ?
? ?
is
v r ? ? ?
? ? ? ?
1 8 . C
PR
=
C C 5C
3 2 6
? ?
C
PQ
=
C 5C
C
4 4
? ?
C
PR
/ C
PQ
=
2
3
1 9 .
T
0
Q ms dt ?
?
=
20
4
0
mA
T
4
? ?
? ?
= 4 × 10
4
mA
2 0 . As D >>d & ? << d.
Hence we can use ? ? =
D
d
?
so distance between 5
th
bright fringe and 3
rd
dark fringe = 5 ? – (2 ? + ?/2)
=
5
2
?
=
–7
3
5 6.5 10 1
2 10
?
? ?
?
= 1.625 mm
2 1 . Weight of body on the surface of the earth
= mg = 72 N
Acceleration due to gravity at height h is

? ?
2 2
E E
h 2
E
E
E
gR gR 4
g g
R 9
R h
R
2
? ? ?
? ?
?
?
? ?
? ?

E
R
h
2
? ?
?
? ?
? ?
?
Gravitational force on body at height h is
F = mg
h
= m ×
4 4 4
g mg 72 32N
9 9 9
? ? ? ? ?
2 2 . Field due to lower straight current = 0
field due to square loop = 0
field due to upper straight current =
0
2I 1
1
4 a 2
? ? ?
?
? ?
?
? ?
2 3 .
? ?
1 2
KA T T d
dt L
? ?
?
? Rate of melting
0000CMA303117002 HS - 4/7
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
2 4 . tan (i
p
) = µ
tan (i
p
) =
3
i
p
= 60°
2 5 . Gravitational potential energy at any point at
a distance r from the centre of the earth is
E
GM m
U
r
? ?
where M
E
and m be masses of earth and body
repectively.
At the surface of the earth, r = R
E
? U
1
= –
E
E
GM m
R
At a height h from the surface,
r = R
E
+ h = R
E
+ nR
E
= (1 + n) R
E
? U
2
= –
? ?
E
E
GM m
n 1 R ?
Change in potential energy is
?U = U
2
– U
1
= –
? ?
E E
E E
GM m GM m
n 1 R R
? ?
? ?
? ?
?
? ?
=
? ? ? ?
E E
E E
GM m 1 GM mn
1
R n 1 n 1 R
? ?
? ?
? ?
? ?
? ?
? ?
=
? ?
E
E 2
E
n GM
mgR g
n 1 R
? ?
?
? ?
?
? ?
?
2 6 . Field due to long straight current
=
–6 0
2I
ˆ ˆ
k 10 k
4 d
?
?
?
wb/m
2
B
net
=
6
2 10
?
?
wb/m
2
2 7 . ? ?
0
dT
T T
dt
? ?
? ( ??)
n
n = 1
2 8 . K
max
(eV) = 12375
0
1 1
(Å) (Å)
? ?
?
? ?
? ?
? ?
= 12375
1 1
1000 2000
? ?
?
? ?
? ?
= 6.2 eV
2 9 . Work done , W =
1
2
× F × ?L
For a given F
W
?
?L
and ?L =
FL
AY
As F, A and L are constants
?
1
L
Y
? ?
From (i) and (ii), we get
1
W
Y
?
?
copper
steel
copper steel
Y
W
W Y
?
As Y
copper
< Y
steel
? W
steel
< W
copper
So, less work is done on steel spring.
3 0 . Due to motion along direction (3) these is no
change in flux, so emf = 0
3 1 .
P, V
300 K
V, P
150 K
2
Q
1
Q
2
300 K
P
2
, 2V,
Q
1
= nC
v
(–150)
Q
2
= nC
p
(150)
Q = n(150) (C
p
– C
v
)
= 300 R
HS - 5/7 0000CMA303117002
ALL INDIA OPEN TEST/Pre-Medical /AIIMS/22-04-2018
3 2 . ?-particle cannot be attracted by the nucleus.
3 3 . According to equation of continuity A
1
v
1
=
A
2
v
2
2
2
1 2 2 2
2
2 1 1 1
v A D / 4 D
v A D / 4 D
? ? ?
? ? ?
? ?
?
? ?
Here, D
1
= 2.5 cm, D
2
= 3.75 cm
?
2 2
1
2
v 3.75 3 9
v 2.5 2 4
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
3 4 .
L
R
? ?
? ? ? ?   L = R ?
Power factor
2 2
R
R (L )
?
? ?
2
1
1 ( )
?
? ??
3 5 .
V
C
??
?
?
V =
??
?
C = 5 × 10
5
m/s
3 6 . E = ?m.c
2
? E =
0.3
1000
× (3×10
8
)
2
= 2.7 × 10
13
J
=
13
6
2.7 10
3.6 10
?
?
= 7.5 × 10
6
kWh.
3 7 .
F
B
F
A
F
C
A C
B
2
B 2
KQ
F 3
L
?
2
A C 2
KQ
F F
L
? ?
3 8 . Magnetic moment of diamagnetic atom = zero,
due to paired electron and for ferro and para
magnetic it is not zero due to unpaired
electrons.
3 9 . For a normal person max. audible frequency = 20
kHz
= 20000 Hz If n represent the max. number of
overtones generated by COP then.
20000
2n 1 n 6
1500
? ? ? ?
8 1 . NCERT-XI, Pg.# 71 fig. 5.9 (b)
8 3 . NCERT Pg.# 338
8 5 . NCERT-XI, Pg.# 74, 75
8 6 . NCERT XI
th
Pg.# 246
8 7 . NCERT XII
th
Pg # 35, 36 (2.4.2)
9 0 . NCERT Pg.# 233
9 1 . NCERT XII
th
Pg # 35 (2.4.2)
9 2 . NCERT XII
th
Pg # 131, Para.-7.3
9 4 . NCERT Pg.# 223
9 5 . NCERT XII
th
Pg # 36 (2.4.3)
9 6 . NCERT XII
th
Pg # 131,133,136, Para.-7.3 to
7.6
9 7 . NCERT XI
th
Pg # 36,38,40
9 9 . NCERT XII
th
Pg # 61 para 4.2
1 0 1 . NCERT XI
th
Pg # 23
1 0 3 . NCERT XII
th
Pg # 48 Para 3.3
1 0 9 . NCERT XI
th
Pg # 104 fig. 7.1.3
1 1 3 . NCERT XI
th
Pg.# 135, 136
1 1 7 . NCERT XI
th
Pg.# 133 - 8.5.3.1
1 2 1 . Distance is the total length of path covered by
the particle.
Mag. of displacement is the shortest path
between initial and final points.
eg. A = initial point = final point
disp = 0
A
r
distance ? 0
distance = 2 ?r
1 2 2 . Surface tension decreases as temperature
increases.
```
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