ALLEN MOCK PAPER FOR AIIMS -- SOLUTION -- 3 NEET Notes | EduRev

NEET : ALLEN MOCK PAPER FOR AIIMS -- SOLUTION -- 3 NEET Notes | EduRev

 Page 1


HINT – SHEET
ANSWER KEY
1.
C A B ? ?
? ? ?
C
2
 = A
2
 + B
2
 + 2ABcos ?
x
2
 = x
2
 
+ B
2
 
+ 2(x)B
x
B
? ?
?
? ?
? ?
Let A = C = x
? ?
A
cos A C
B
?
? ? ?
? ?
?
B 2 x ?
So 
A x 1
cos
B 2x 2
?
? ? ? ? ? ?
3
145 rad
4
?
? ? ? ?
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ENTHUSIAST, LEADER & ACHIEVER COURSE
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 AIIMS
TEST DATE 
:
 26 - 04 - 2018
HS - 1/ 7 10 01 CM A305 31 70 30
TEST SYLLABUS : FULL SYLLABUS
2.
m 21m/s 2m 4m/s m 1m/s
A
B A
By COLM m(21) – 2m(4) = m(1) + 2mV
B
V
B
 = 6m/s
e =
6 1
0.2
21 ( 4)
?
?
? ?
( ?k)
loss
= 
2
2
1 2m
(25) [1 0.04]
2 3m
?
= 
m
625 0.96
3
? ?
 = 200 mJ
I = m(1) – m(21) = –20m
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
An s . 2 3 2 2 3 2 2 2 4 1 4 1 1 2 2 3 1 2 1 4
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
An s . 3 4 3 3 2 3 4 4 3 2 3 3 4 3 4 4 4 4 4 4
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
An s . 2 2 1 4 2 4 3 4 3 2 1 3 3 4 4 3 1 4 1 3
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
An s . 1 3 4 3 1 4 2 1 1 3 4 1 4 4 4 3 2 4 1 2
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
An s . 3 3 4 3 1 1 4 4 3 2 2 4 4 4 2 1 2 4 2 2
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
An s . 2 3 4 3 2 2 3 4 1 4 1 4 4 3 1 2 4 1 4 3
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
An s . 3 3 4 4 3 4 4 4 3 2 1 1 4 4 1 2 1 1 4 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
An s . 1 2 1 2 2 3 1 1 3 2 1 2 1 4 4 4 4 2 3 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
An s . 1 1 1 4 3 1 3 3 2 1 2 3 1 1 1 1 4 1 1 2
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
An s . 2 4 3 3 4 4 3 2 3 3 4 2 4 3 2 3 3 4 4 2
Page 2


HINT – SHEET
ANSWER KEY
1.
C A B ? ?
? ? ?
C
2
 = A
2
 + B
2
 + 2ABcos ?
x
2
 = x
2
 
+ B
2
 
+ 2(x)B
x
B
? ?
?
? ?
? ?
Let A = C = x
? ?
A
cos A C
B
?
? ? ?
? ?
?
B 2 x ?
So 
A x 1
cos
B 2x 2
?
? ? ? ? ? ?
3
145 rad
4
?
? ? ? ?
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ENTHUSIAST, LEADER & ACHIEVER COURSE
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 AIIMS
TEST DATE 
:
 26 - 04 - 2018
HS - 1/ 7 10 01 CM A305 31 70 30
TEST SYLLABUS : FULL SYLLABUS
2.
m 21m/s 2m 4m/s m 1m/s
A
B A
By COLM m(21) – 2m(4) = m(1) + 2mV
B
V
B
 = 6m/s
e =
6 1
0.2
21 ( 4)
?
?
? ?
( ?k)
loss
= 
2
2
1 2m
(25) [1 0.04]
2 3m
?
= 
m
625 0.96
3
? ?
 = 200 mJ
I = m(1) – m(21) = –20m
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
An s . 2 3 2 2 3 2 2 2 4 1 4 1 1 2 2 3 1 2 1 4
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
An s . 3 4 3 3 2 3 4 4 3 2 3 3 4 3 4 4 4 4 4 4
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
An s . 2 2 1 4 2 4 3 4 3 2 1 3 3 4 4 3 1 4 1 3
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
An s . 1 3 4 3 1 4 2 1 1 3 4 1 4 4 4 3 2 4 1 2
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
An s . 3 3 4 3 1 1 4 4 3 2 2 4 4 4 2 1 2 4 2 2
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
An s . 2 3 4 3 2 2 3 4 1 4 1 4 4 3 1 2 4 1 4 3
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
An s . 3 3 4 4 3 4 4 4 3 2 1 1 4 4 1 2 1 1 4 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
An s . 1 2 1 2 2 3 1 1 3 2 1 2 1 4 4 4 4 2 3 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
An s . 1 1 1 4 3 1 3 3 2 1 2 3 1 1 1 1 4 1 1 2
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
An s . 2 4 3 3 4 4 3 2 3 3 4 2 4 3 2 3 3 4 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
3.
2
8.75 10 m
4
?
?
? ?
2 2
v 350 35 10
n 1000Hz
4 8.75 10 35 10
? ?
?
? ? ? ?
? ? ? ?
4. energy stored in inductor = K.E. of conductry
rod
2 2
0
1 1
LI mv
2 2
?
   ? 
m
I
L
?
v
0
5.
2
1
S ut at
2
? ?
? ?
2
1
100 40t 10 t
2
? ? ?
    
+ive
–ive
40 m/s
100m
On solving t = –2 sec
t = 10 sec
6.
2T
g(8)
r
? ?
....(1)
2T
g(6)
r
? ?
....(2)
(1)/(2)
R 4
r 3
?
         
R
?
?
r 1 4
cos 3
?
?
3
cos
4
? ?
? ?
? ?
? ?
7. V
?
 = 
2 2
surface escape
V V ?
= 
2 2
escape escape
(3V ) V ?
    = 
escape
2 2 V =
2 2
×11.2 = 22.4
2
km/surface
8. I
2
R = P (R = resistance of choke coil)
also Heat produced = energy stored in choke coil
= 
2
1 1 P
LI L.
2 2 R
?
= 
1
P
2
?
9. a
rel
 = 2a from v
2
 = u
2
 + 2as
u
rel
 = 80 m/s 0 = (80)
2
 + 2(2a)(4×10
4
)
s
rel
 = 4 × 10
3
 m a = 0.8 m/s
2
10.
3
3
4
F 1.03 10 4 9.8 163N
0.2 10
? ?
? ? ? ? ? ?
? ?
?
? ?
11. q
1
 + q
2
 = Q and 
1
2
q
4 r ?
 = 
2
2
q
4 R ?
 (given)
q
1
 = 
2
2 2
Qr
R r ?
 and q
2
 = 
2
2 2
QR
R r ?
Potential at common centre
2 2
2 2 2 2
0
1 Qr QR
4 (R r )r (R r )R
? ?
?
? ?
? ? ? ?
? ?
 = 
2 2
0
Q(R r)
4 (R r )
?
? ? ?
12. Speed of light 
0
0
E
C
B
? C = ? ? ? = 
2
2
k
?? ?
?
?
?
0
0
E
B k
?
?
 ?  kE
0
 = ?B
0
13. F
?
 = ?
s
N = 0.7 × 80 = 56 N
10kg
37°
T
N
80N
mgsin = 60N ?
4kg
40N
T
µ=0.7
System will remain at rest
So T = 40 N
10kg
60N
T=40N
f =20N
s
and f
s
 = 20 N
14. Absorbing power 
2
1
r
?
Page 3


HINT – SHEET
ANSWER KEY
1.
C A B ? ?
? ? ?
C
2
 = A
2
 + B
2
 + 2ABcos ?
x
2
 = x
2
 
+ B
2
 
+ 2(x)B
x
B
? ?
?
? ?
? ?
Let A = C = x
? ?
A
cos A C
B
?
? ? ?
? ?
?
B 2 x ?
So 
A x 1
cos
B 2x 2
?
? ? ? ? ? ?
3
145 rad
4
?
? ? ? ?
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ENTHUSIAST, LEADER & ACHIEVER COURSE
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 AIIMS
TEST DATE 
:
 26 - 04 - 2018
HS - 1/ 7 10 01 CM A305 31 70 30
TEST SYLLABUS : FULL SYLLABUS
2.
m 21m/s 2m 4m/s m 1m/s
A
B A
By COLM m(21) – 2m(4) = m(1) + 2mV
B
V
B
 = 6m/s
e =
6 1
0.2
21 ( 4)
?
?
? ?
( ?k)
loss
= 
2
2
1 2m
(25) [1 0.04]
2 3m
?
= 
m
625 0.96
3
? ?
 = 200 mJ
I = m(1) – m(21) = –20m
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
An s . 2 3 2 2 3 2 2 2 4 1 4 1 1 2 2 3 1 2 1 4
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
An s . 3 4 3 3 2 3 4 4 3 2 3 3 4 3 4 4 4 4 4 4
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
An s . 2 2 1 4 2 4 3 4 3 2 1 3 3 4 4 3 1 4 1 3
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
An s . 1 3 4 3 1 4 2 1 1 3 4 1 4 4 4 3 2 4 1 2
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
An s . 3 3 4 3 1 1 4 4 3 2 2 4 4 4 2 1 2 4 2 2
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
An s . 2 3 4 3 2 2 3 4 1 4 1 4 4 3 1 2 4 1 4 3
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
An s . 3 3 4 4 3 4 4 4 3 2 1 1 4 4 1 2 1 1 4 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
An s . 1 2 1 2 2 3 1 1 3 2 1 2 1 4 4 4 4 2 3 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
An s . 1 1 1 4 3 1 3 3 2 1 2 3 1 1 1 1 4 1 1 2
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
An s . 2 4 3 3 4 4 3 2 3 3 4 2 4 3 2 3 3 4 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
3.
2
8.75 10 m
4
?
?
? ?
2 2
v 350 35 10
n 1000Hz
4 8.75 10 35 10
? ?
?
? ? ? ?
? ? ? ?
4. energy stored in inductor = K.E. of conductry
rod
2 2
0
1 1
LI mv
2 2
?
   ? 
m
I
L
?
v
0
5.
2
1
S ut at
2
? ?
? ?
2
1
100 40t 10 t
2
? ? ?
    
+ive
–ive
40 m/s
100m
On solving t = –2 sec
t = 10 sec
6.
2T
g(8)
r
? ?
....(1)
2T
g(6)
r
? ?
....(2)
(1)/(2)
R 4
r 3
?
         
R
?
?
r 1 4
cos 3
?
?
3
cos
4
? ?
? ?
? ?
? ?
7. V
?
 = 
2 2
surface escape
V V ?
= 
2 2
escape escape
(3V ) V ?
    = 
escape
2 2 V =
2 2
×11.2 = 22.4
2
km/surface
8. I
2
R = P (R = resistance of choke coil)
also Heat produced = energy stored in choke coil
= 
2
1 1 P
LI L.
2 2 R
?
= 
1
P
2
?
9. a
rel
 = 2a from v
2
 = u
2
 + 2as
u
rel
 = 80 m/s 0 = (80)
2
 + 2(2a)(4×10
4
)
s
rel
 = 4 × 10
3
 m a = 0.8 m/s
2
10.
3
3
4
F 1.03 10 4 9.8 163N
0.2 10
? ?
? ? ? ? ? ?
? ?
?
? ?
11. q
1
 + q
2
 = Q and 
1
2
q
4 r ?
 = 
2
2
q
4 R ?
 (given)
q
1
 = 
2
2 2
Qr
R r ?
 and q
2
 = 
2
2 2
QR
R r ?
Potential at common centre
2 2
2 2 2 2
0
1 Qr QR
4 (R r )r (R r )R
? ?
?
? ?
? ? ? ?
? ?
 = 
2 2
0
Q(R r)
4 (R r )
?
? ? ?
12. Speed of light 
0
0
E
C
B
? C = ? ? ? = 
2
2
k
?? ?
?
?
?
0
0
E
B k
?
?
 ?  kE
0
 = ?B
0
13. F
?
 = ?
s
N = 0.7 × 80 = 56 N
10kg
37°
T
N
80N
mgsin = 60N ?
4kg
40N
T
µ=0.7
System will remain at rest
So T = 40 N
10kg
60N
T=40N
f =20N
s
and f
s
 = 20 N
14. Absorbing power 
2
1
r
?
HS - 3/ 7
Enthusiast,Leader & Achiever Course/Phase-All/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
15. Force on l length of the wire 2 is
Q
l
?
2
?
1
R
F
2
 = QE
1
 = ( ?
2
l)
1
2k
R
?
?
2 1 2
F 2k
R
? ?
?
l
Also,  
1 2 1 2
F F 2k F
R
? ?
? ? ?
l l l
16.
B
C
A D
90º
75º
135º
60º
60º
P 60º
45º
45º
30º
60º
30º
60º
i
e
17.
Plank
2kg
? = 0.2
a=1 m/s
2
Their is no sliping between plane & Block so
f = ma = (2) (1)
= 2 N
18. T 2
g
? ?
?
?  
dT 1
T 2
?
?
?
  ?  
dT 1
T
T 2
? ? ??
?  
1
dT T T
2
? ? ? ? ? ?
= 
6
1
18 10 20 24 60 60
2
?
? ? ? ? ? ?
= 15.55 sec  
?
 15 sec.
19. From figure, it is clear that 
E
?
 at all points on
the y-axis is along 
ˆ
i
. Here 
E
?
 of all points on
x-axis cannot have the same direction.
Here electric potential at origin is zero so no
work is done in bringing a test charge from
infinity to origin.
• E=Ei
?
^
+q –q
(d, 0) (–d, 0)
Here dipole moment is in –x direction
(–q to +q).Hence only option (1) is correct
20. Final image is at infinity
M.P. = –
0
e
f
f
 = –5 ?   f
0
 = 5 f
e
L = f
0
 + f
e
 = 36
5f
e
 + f
e
 = 36 ? fe = 6 cm
f
0
 = 5f
e
 = 5 × 6 = 30 cm
21. T
1
 > T
2
30°
60°
T
1
T
2 V'
V
m
m
23. i = 2A
Applying KVL in loop
V
b
 = –4V
so V
a
 = +6V
24. ? ? ?? ? ? ? ?? ? ? ? ???Sperical wavefront
25. Kinetic energy = 
2
1
m
2
?
Further, ?
2
 = u
2
 + 2as = 0 + 2 ad
or ? ? ? ? ? ? ? ?
2
 = 
F
2 d
m
? ?
? ?
? ?
? KE = 
1 F
m 2 d Fd
2 m
? ?   ? KE ? d
Page 4


HINT – SHEET
ANSWER KEY
1.
C A B ? ?
? ? ?
C
2
 = A
2
 + B
2
 + 2ABcos ?
x
2
 = x
2
 
+ B
2
 
+ 2(x)B
x
B
? ?
?
? ?
? ?
Let A = C = x
? ?
A
cos A C
B
?
? ? ?
? ?
?
B 2 x ?
So 
A x 1
cos
B 2x 2
?
? ? ? ? ? ?
3
145 rad
4
?
? ? ? ?
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ENTHUSIAST, LEADER & ACHIEVER COURSE
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 AIIMS
TEST DATE 
:
 26 - 04 - 2018
HS - 1/ 7 10 01 CM A305 31 70 30
TEST SYLLABUS : FULL SYLLABUS
2.
m 21m/s 2m 4m/s m 1m/s
A
B A
By COLM m(21) – 2m(4) = m(1) + 2mV
B
V
B
 = 6m/s
e =
6 1
0.2
21 ( 4)
?
?
? ?
( ?k)
loss
= 
2
2
1 2m
(25) [1 0.04]
2 3m
?
= 
m
625 0.96
3
? ?
 = 200 mJ
I = m(1) – m(21) = –20m
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
An s . 2 3 2 2 3 2 2 2 4 1 4 1 1 2 2 3 1 2 1 4
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
An s . 3 4 3 3 2 3 4 4 3 2 3 3 4 3 4 4 4 4 4 4
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
An s . 2 2 1 4 2 4 3 4 3 2 1 3 3 4 4 3 1 4 1 3
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
An s . 1 3 4 3 1 4 2 1 1 3 4 1 4 4 4 3 2 4 1 2
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
An s . 3 3 4 3 1 1 4 4 3 2 2 4 4 4 2 1 2 4 2 2
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
An s . 2 3 4 3 2 2 3 4 1 4 1 4 4 3 1 2 4 1 4 3
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
An s . 3 3 4 4 3 4 4 4 3 2 1 1 4 4 1 2 1 1 4 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
An s . 1 2 1 2 2 3 1 1 3 2 1 2 1 4 4 4 4 2 3 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
An s . 1 1 1 4 3 1 3 3 2 1 2 3 1 1 1 1 4 1 1 2
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
An s . 2 4 3 3 4 4 3 2 3 3 4 2 4 3 2 3 3 4 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
3.
2
8.75 10 m
4
?
?
? ?
2 2
v 350 35 10
n 1000Hz
4 8.75 10 35 10
? ?
?
? ? ? ?
? ? ? ?
4. energy stored in inductor = K.E. of conductry
rod
2 2
0
1 1
LI mv
2 2
?
   ? 
m
I
L
?
v
0
5.
2
1
S ut at
2
? ?
? ?
2
1
100 40t 10 t
2
? ? ?
    
+ive
–ive
40 m/s
100m
On solving t = –2 sec
t = 10 sec
6.
2T
g(8)
r
? ?
....(1)
2T
g(6)
r
? ?
....(2)
(1)/(2)
R 4
r 3
?
         
R
?
?
r 1 4
cos 3
?
?
3
cos
4
? ?
? ?
? ?
? ?
7. V
?
 = 
2 2
surface escape
V V ?
= 
2 2
escape escape
(3V ) V ?
    = 
escape
2 2 V =
2 2
×11.2 = 22.4
2
km/surface
8. I
2
R = P (R = resistance of choke coil)
also Heat produced = energy stored in choke coil
= 
2
1 1 P
LI L.
2 2 R
?
= 
1
P
2
?
9. a
rel
 = 2a from v
2
 = u
2
 + 2as
u
rel
 = 80 m/s 0 = (80)
2
 + 2(2a)(4×10
4
)
s
rel
 = 4 × 10
3
 m a = 0.8 m/s
2
10.
3
3
4
F 1.03 10 4 9.8 163N
0.2 10
? ?
? ? ? ? ? ?
? ?
?
? ?
11. q
1
 + q
2
 = Q and 
1
2
q
4 r ?
 = 
2
2
q
4 R ?
 (given)
q
1
 = 
2
2 2
Qr
R r ?
 and q
2
 = 
2
2 2
QR
R r ?
Potential at common centre
2 2
2 2 2 2
0
1 Qr QR
4 (R r )r (R r )R
? ?
?
? ?
? ? ? ?
? ?
 = 
2 2
0
Q(R r)
4 (R r )
?
? ? ?
12. Speed of light 
0
0
E
C
B
? C = ? ? ? = 
2
2
k
?? ?
?
?
?
0
0
E
B k
?
?
 ?  kE
0
 = ?B
0
13. F
?
 = ?
s
N = 0.7 × 80 = 56 N
10kg
37°
T
N
80N
mgsin = 60N ?
4kg
40N
T
µ=0.7
System will remain at rest
So T = 40 N
10kg
60N
T=40N
f =20N
s
and f
s
 = 20 N
14. Absorbing power 
2
1
r
?
HS - 3/ 7
Enthusiast,Leader & Achiever Course/Phase-All/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
15. Force on l length of the wire 2 is
Q
l
?
2
?
1
R
F
2
 = QE
1
 = ( ?
2
l)
1
2k
R
?
?
2 1 2
F 2k
R
? ?
?
l
Also,  
1 2 1 2
F F 2k F
R
? ?
? ? ?
l l l
16.
B
C
A D
90º
75º
135º
60º
60º
P 60º
45º
45º
30º
60º
30º
60º
i
e
17.
Plank
2kg
? = 0.2
a=1 m/s
2
Their is no sliping between plane & Block so
f = ma = (2) (1)
= 2 N
18. T 2
g
? ?
?
?  
dT 1
T 2
?
?
?
  ?  
dT 1
T
T 2
? ? ??
?  
1
dT T T
2
? ? ? ? ? ?
= 
6
1
18 10 20 24 60 60
2
?
? ? ? ? ? ?
= 15.55 sec  
?
 15 sec.
19. From figure, it is clear that 
E
?
 at all points on
the y-axis is along 
ˆ
i
. Here 
E
?
 of all points on
x-axis cannot have the same direction.
Here electric potential at origin is zero so no
work is done in bringing a test charge from
infinity to origin.
• E=Ei
?
^
+q –q
(d, 0) (–d, 0)
Here dipole moment is in –x direction
(–q to +q).Hence only option (1) is correct
20. Final image is at infinity
M.P. = –
0
e
f
f
 = –5 ?   f
0
 = 5 f
e
L = f
0
 + f
e
 = 36
5f
e
 + f
e
 = 36 ? fe = 6 cm
f
0
 = 5f
e
 = 5 × 6 = 30 cm
21. T
1
 > T
2
30°
60°
T
1
T
2 V'
V
m
m
23. i = 2A
Applying KVL in loop
V
b
 = –4V
so V
a
 = +6V
24. ? ? ?? ? ? ? ?? ? ? ? ???Sperical wavefront
25. Kinetic energy = 
2
1
m
2
?
Further, ?
2
 = u
2
 + 2as = 0 + 2 ad
or ? ? ? ? ? ? ? ?
2
 = 
F
2 d
m
? ?
? ?
? ?
? KE = 
1 F
m 2 d Fd
2 m
? ?   ? KE ? d
HS - 4/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
26. In Real gas force between molecules are
attractive and potential energy is negative on
increasing the seperation potential energy
increases.
27.
28 ? 54 ?
8V 12V
6V
I
1
1
I
2
2
For loop (1)
8 + 6 = – 28 I
1
? I
1
 = 
1
2
?
amp
For loop (2)
12 + 6 = – 54 I
2
? I
2
 = 
1
3
?
I
3
 = I
1
 + I
2
 = 
1 1 5
2 3 6
? ? ? ?
amp
28. The total number of atoms neither remains
constant (as in option (1) not can ever increase
(as in option (2) and (3)). They will
continuously decrease with time. Therefore
option (4) is correct.
29. Moment of inertia of a cylinder about an axis
? to its circular face = 
2
MR
2
?
2 2
1 2
MR M(3R / 4)
I ;I
2 2
? ?
Hence,
2
1
2
2
I R 16
I (3R / 4) 9
? ?
30. v
2
 = ?
2
(A
2
 – b
2
) = 3 ?
2
b
2
  ? A
2
 = 4b
2
b = 
A
2
 = Asin ?t  ? t = 
6
?
?
Required time, t
1
 = 
T 2
t
4 4 6 3
? ? ?
? ? ? ?
? ? ?
31. B
1 
= 
? ?
? ?
2
0
3 2
2
2
2
2 2
/
I R
R R
?
? ?
?
? ?
? ?
B
2 
= 
? ?
? ? ? ?
2
0
3 2
2 2
2
2 2 2
/
I R
R R
?
? ?
?
? ?
? ?
  solve it
32. Voltage gain = 
Output voltage
Input voltage
? V
out
 = V
in
 × voltage gain
? V
out
 = V
in
 × current gain × resistance gain
= V
in
 × ? × 
L 3
BE
R 10
10 100 1V
R 1
?
? ? ? ?
33. K
R
 = 
2 2
1 1 L 1
I L
2 2 2
? ? ? ? ?
?
?
R
2K L
L , L'
4
? ?
?
34. The time period of simple harmonic motion is
T = 18 s.
–A A x = 0 3/2 cm
At t = 0, the particle is at x = 3/2 cm and
approaching positive extreme.
39s = 2 × 18 + 3 = 2T + 3 = 2T + T/6
Distance travelled by the particle in 2T is 8A
= 24 cm.
Distance travelled by the particle in further
T/6 is A/2 = 1.5 cm.
Total distance travelled = 25.5 cm
35.
1 2
P
?
2 ?
1
R
E
R
For long solenoid ?
1
 ? 0°
and for ?
2
 = cos ?
2
 =
R 1
2R 2
?  
?
2
R
R E P
2 R
2
?
?
? ?
2
 = 45°
B
p
 = 
0
1 2
µ ni
(cos cos )
2
? ? ?
 =
0
µ ni
2
(cos0°–cos45°)
            = 
0
µ ni 1
1
2 2
? ?
?
? ?
? ?
Page 5


HINT – SHEET
ANSWER KEY
1.
C A B ? ?
? ? ?
C
2
 = A
2
 + B
2
 + 2ABcos ?
x
2
 = x
2
 
+ B
2
 
+ 2(x)B
x
B
? ?
?
? ?
? ?
Let A = C = x
? ?
A
cos A C
B
?
? ? ?
? ?
?
B 2 x ?
So 
A x 1
cos
B 2x 2
?
? ? ? ? ? ?
3
145 rad
4
?
? ? ? ?
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
ENTHUSIAST, LEADER & ACHIEVER COURSE
ALL PHASE
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 AIIMS
TEST DATE 
:
 26 - 04 - 2018
HS - 1/ 7 10 01 CM A305 31 70 30
TEST SYLLABUS : FULL SYLLABUS
2.
m 21m/s 2m 4m/s m 1m/s
A
B A
By COLM m(21) – 2m(4) = m(1) + 2mV
B
V
B
 = 6m/s
e =
6 1
0.2
21 ( 4)
?
?
? ?
( ?k)
loss
= 
2
2
1 2m
(25) [1 0.04]
2 3m
?
= 
m
625 0.96
3
? ?
 = 200 mJ
I = m(1) – m(21) = –20m
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
An s . 2 3 2 2 3 2 2 2 4 1 4 1 1 2 2 3 1 2 1 4
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
An s . 3 4 3 3 2 3 4 4 3 2 3 3 4 3 4 4 4 4 4 4
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
An s . 2 2 1 4 2 4 3 4 3 2 1 3 3 4 4 3 1 4 1 3
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
An s . 1 3 4 3 1 4 2 1 1 3 4 1 4 4 4 3 2 4 1 2
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
An s . 3 3 4 3 1 1 4 4 3 2 2 4 4 4 2 1 2 4 2 2
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
An s . 2 3 4 3 2 2 3 4 1 4 1 4 4 3 1 2 4 1 4 3
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
An s . 3 3 4 4 3 4 4 4 3 2 1 1 4 4 1 2 1 1 4 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
An s . 1 2 1 2 2 3 1 1 3 2 1 2 1 4 4 4 4 2 3 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
An s . 1 1 1 4 3 1 3 3 2 1 2 3 1 1 1 1 4 1 1 2
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
An s . 2 4 3 3 4 4 3 2 3 3 4 2 4 3 2 3 3 4 4 2
HS - 2/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
3.
2
8.75 10 m
4
?
?
? ?
2 2
v 350 35 10
n 1000Hz
4 8.75 10 35 10
? ?
?
? ? ? ?
? ? ? ?
4. energy stored in inductor = K.E. of conductry
rod
2 2
0
1 1
LI mv
2 2
?
   ? 
m
I
L
?
v
0
5.
2
1
S ut at
2
? ?
? ?
2
1
100 40t 10 t
2
? ? ?
    
+ive
–ive
40 m/s
100m
On solving t = –2 sec
t = 10 sec
6.
2T
g(8)
r
? ?
....(1)
2T
g(6)
r
? ?
....(2)
(1)/(2)
R 4
r 3
?
         
R
?
?
r 1 4
cos 3
?
?
3
cos
4
? ?
? ?
? ?
? ?
7. V
?
 = 
2 2
surface escape
V V ?
= 
2 2
escape escape
(3V ) V ?
    = 
escape
2 2 V =
2 2
×11.2 = 22.4
2
km/surface
8. I
2
R = P (R = resistance of choke coil)
also Heat produced = energy stored in choke coil
= 
2
1 1 P
LI L.
2 2 R
?
= 
1
P
2
?
9. a
rel
 = 2a from v
2
 = u
2
 + 2as
u
rel
 = 80 m/s 0 = (80)
2
 + 2(2a)(4×10
4
)
s
rel
 = 4 × 10
3
 m a = 0.8 m/s
2
10.
3
3
4
F 1.03 10 4 9.8 163N
0.2 10
? ?
? ? ? ? ? ?
? ?
?
? ?
11. q
1
 + q
2
 = Q and 
1
2
q
4 r ?
 = 
2
2
q
4 R ?
 (given)
q
1
 = 
2
2 2
Qr
R r ?
 and q
2
 = 
2
2 2
QR
R r ?
Potential at common centre
2 2
2 2 2 2
0
1 Qr QR
4 (R r )r (R r )R
? ?
?
? ?
? ? ? ?
? ?
 = 
2 2
0
Q(R r)
4 (R r )
?
? ? ?
12. Speed of light 
0
0
E
C
B
? C = ? ? ? = 
2
2
k
?? ?
?
?
?
0
0
E
B k
?
?
 ?  kE
0
 = ?B
0
13. F
?
 = ?
s
N = 0.7 × 80 = 56 N
10kg
37°
T
N
80N
mgsin = 60N ?
4kg
40N
T
µ=0.7
System will remain at rest
So T = 40 N
10kg
60N
T=40N
f =20N
s
and f
s
 = 20 N
14. Absorbing power 
2
1
r
?
HS - 3/ 7
Enthusiast,Leader & Achiever Course/Phase-All/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
15. Force on l length of the wire 2 is
Q
l
?
2
?
1
R
F
2
 = QE
1
 = ( ?
2
l)
1
2k
R
?
?
2 1 2
F 2k
R
? ?
?
l
Also,  
1 2 1 2
F F 2k F
R
? ?
? ? ?
l l l
16.
B
C
A D
90º
75º
135º
60º
60º
P 60º
45º
45º
30º
60º
30º
60º
i
e
17.
Plank
2kg
? = 0.2
a=1 m/s
2
Their is no sliping between plane & Block so
f = ma = (2) (1)
= 2 N
18. T 2
g
? ?
?
?  
dT 1
T 2
?
?
?
  ?  
dT 1
T
T 2
? ? ??
?  
1
dT T T
2
? ? ? ? ? ?
= 
6
1
18 10 20 24 60 60
2
?
? ? ? ? ? ?
= 15.55 sec  
?
 15 sec.
19. From figure, it is clear that 
E
?
 at all points on
the y-axis is along 
ˆ
i
. Here 
E
?
 of all points on
x-axis cannot have the same direction.
Here electric potential at origin is zero so no
work is done in bringing a test charge from
infinity to origin.
• E=Ei
?
^
+q –q
(d, 0) (–d, 0)
Here dipole moment is in –x direction
(–q to +q).Hence only option (1) is correct
20. Final image is at infinity
M.P. = –
0
e
f
f
 = –5 ?   f
0
 = 5 f
e
L = f
0
 + f
e
 = 36
5f
e
 + f
e
 = 36 ? fe = 6 cm
f
0
 = 5f
e
 = 5 × 6 = 30 cm
21. T
1
 > T
2
30°
60°
T
1
T
2 V'
V
m
m
23. i = 2A
Applying KVL in loop
V
b
 = –4V
so V
a
 = +6V
24. ? ? ?? ? ? ? ?? ? ? ? ???Sperical wavefront
25. Kinetic energy = 
2
1
m
2
?
Further, ?
2
 = u
2
 + 2as = 0 + 2 ad
or ? ? ? ? ? ? ? ?
2
 = 
F
2 d
m
? ?
? ?
? ?
? KE = 
1 F
m 2 d Fd
2 m
? ?   ? KE ? d
HS - 4/ 7
Target : Pre-Medical 2018/Major/AIIMS/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
26. In Real gas force between molecules are
attractive and potential energy is negative on
increasing the seperation potential energy
increases.
27.
28 ? 54 ?
8V 12V
6V
I
1
1
I
2
2
For loop (1)
8 + 6 = – 28 I
1
? I
1
 = 
1
2
?
amp
For loop (2)
12 + 6 = – 54 I
2
? I
2
 = 
1
3
?
I
3
 = I
1
 + I
2
 = 
1 1 5
2 3 6
? ? ? ?
amp
28. The total number of atoms neither remains
constant (as in option (1) not can ever increase
(as in option (2) and (3)). They will
continuously decrease with time. Therefore
option (4) is correct.
29. Moment of inertia of a cylinder about an axis
? to its circular face = 
2
MR
2
?
2 2
1 2
MR M(3R / 4)
I ;I
2 2
? ?
Hence,
2
1
2
2
I R 16
I (3R / 4) 9
? ?
30. v
2
 = ?
2
(A
2
 – b
2
) = 3 ?
2
b
2
  ? A
2
 = 4b
2
b = 
A
2
 = Asin ?t  ? t = 
6
?
?
Required time, t
1
 = 
T 2
t
4 4 6 3
? ? ?
? ? ? ?
? ? ?
31. B
1 
= 
? ?
? ?
2
0
3 2
2
2
2
2 2
/
I R
R R
?
? ?
?
? ?
? ?
B
2 
= 
? ?
? ? ? ?
2
0
3 2
2 2
2
2 2 2
/
I R
R R
?
? ?
?
? ?
? ?
  solve it
32. Voltage gain = 
Output voltage
Input voltage
? V
out
 = V
in
 × voltage gain
? V
out
 = V
in
 × current gain × resistance gain
= V
in
 × ? × 
L 3
BE
R 10
10 100 1V
R 1
?
? ? ? ?
33. K
R
 = 
2 2
1 1 L 1
I L
2 2 2
? ? ? ? ?
?
?
R
2K L
L , L'
4
? ?
?
34. The time period of simple harmonic motion is
T = 18 s.
–A A x = 0 3/2 cm
At t = 0, the particle is at x = 3/2 cm and
approaching positive extreme.
39s = 2 × 18 + 3 = 2T + 3 = 2T + T/6
Distance travelled by the particle in 2T is 8A
= 24 cm.
Distance travelled by the particle in further
T/6 is A/2 = 1.5 cm.
Total distance travelled = 25.5 cm
35.
1 2
P
?
2 ?
1
R
E
R
For long solenoid ?
1
 ? 0°
and for ?
2
 = cos ?
2
 =
R 1
2R 2
?  
?
2
R
R E P
2 R
2
?
?
? ?
2
 = 45°
B
p
 = 
0
1 2
µ ni
(cos cos )
2
? ? ?
 =
0
µ ni
2
(cos0°–cos45°)
            = 
0
µ ni 1
1
2 2
? ?
?
? ?
? ?
HS - 5/ 7
Enthusiast,Leader & Achiever Course/Phase-All/26-04-2018
1 0 0 1C M A 3 0 5 3 17 0 3 0
37. By COLM
mv
mv 0 Mv'
2
? ? ?
M
? mv
MV'
2
?
mv
V' 5 g
2M
? ? ?
 
2M
V 5 g
m
? ?
39. r = 
?
mv m
qB q
(v, B same)
as r
2
 > r
1
 so 
2 1
m m
q q
? ? ? ?
?
? ? ? ?
? ? ? ?
40.
c
c
c
V 0.5
I 1mA
R 500
? ? ?
0.96
24
1 1 0.96
?
? ? ? ?
? ? ?
c
I 1
I mA
24
?
? ?
?
41. R = 0.529 × 
2
2
1
 ?(1)
x = 0.529 × 
2
3
11
 ?(2)
(1) ? (2)
R 4
11
x 9
? ?
?  x = 
9R
44
44. t
1/2
 = 
0.693
K
K = 
1/ 2
0.693
t
27°C       627°C
K
1
 = 0.01 K
2
 = 0.1
? 
? ?
? ?
? ?
? ?
2 a
1 1 2
K E 1 1
log
K 2.303 R T T
45.      4A + 2B + 3C — ? A
4
B
2
C
3
t 0 1 0.4 0.72
mole 1 0.4 0.72
S.C. 4 2 3
0.25 0.2 0.24
L.R. B
?
?
?
B A
4
B
2
C
3
S.C. 2 1
given moles 0.4 x
2x = 0.4 ? x = 0.2
46. 2 PCl
5
(s) ?PCl
4
+
+PCl
–
6
48. E = E
o
 – 
2 2
4
2
0.0591 [Fe ]
log
n [O ] [H ]
?
?
n = 4
[O
2
] = 
2
O
P 0.1 ?
[H
+
] = 10
–3
[Fe
2+
] = 10
–3
? E = 1.57 V
49. As value of a ?
Attractive force ? ? Ease of Liquefication ?
52.
2
a
C
K
(1 )
?
?
? ?
? = 
M
M
?
?
?
53. For free expansion of an ideal gas under
adiabatic condition
P
ext
 = 0 ? w = –P
ext
 ?V
w = 0
q = 0 (Adiabatic process)
By FLOT, ?E = ˆ ˆ q w ?
?E = 0
? ?E = nC
V
?T
? ?T = 0
54. CaCl
2
 reduces the melting point of the mixture.
57. By FLOT
?E = q + W
E
2
 – E
1
 = q + W
q = –450 J, W = +600 J
E
2
 = E
1
 –450 + 600
E
2
 = E
1
 + 150
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