ALLEN MOCK PAPER SOLUTION -- FULL SYLLABUS -- 4 NEET Notes | EduRev

NEET : ALLEN MOCK PAPER SOLUTION -- FULL SYLLABUS -- 4 NEET Notes | EduRev

 Page 1


HINT – SHEET
ANSWER KEY
1 . Among the given physical quantities angle has a
unit but no dimensions. Angle = [M
0
L
0
T
0
]
The SI unit of angle is radian
2 . The object will slip if centripetal force ? force of
friction
mr ?
2
 ? µmg
r ?
2
 ? µg
r ?
2
 ? constant, or 
2
1 2
2 1
r
r
? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
2
2
4cm 2
r
? ? ?
?
? ?
?
? ?
    ? r
2 
= 1 cm
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
LEADER & ACHIEVER COURSE
PHASE : MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 NEET(UG)
TEST DATE 
:
 17 - 04 - 2018
HS -1 /7 1001 CM D30 5 11 7061
TEST SYLLABUS : FULL SYLLABUS
3 . No. of beats per second = 
9
3
 = 3 s
–1
No. of beats per second = ?
1
 – ?
2
1 2 1 2
v v 1 1
3 v
? ?
? ? ? ?
? ?
? ? ? ?
? ?
2
3 1 1
300 2
? ?
?
2
1 1 1 50 1 49
2 100 100 100
?
? ? ? ?
?
?
2
 = 
49
100
 = 2.04 m
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 . 4 1 2 1 2 1 3 3 2 1 4 4 1 4 4 1 2 3 3 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 . 4 2 1 3 2 1 4 2 4 4 1 3 3 4 1 1 2 1 1 1
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 4 3 3 3 2 3 4 1 2 3 4 1 2 4 2 4 4 3 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
A n s . 4 3 2 2 2 1 2 4 2 2 1 2 3 1 3 3 2 2 4 3
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 2 1 2 2 2 3 2 4 4 2 3 2 3 4 1 4 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 4 1 4 1 2 1 4 4 2 4 1 1 2 3 4 1 2 1 4
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 . 2 4 4 1 1 4 4 3 4 3 1 2 2 4 3 4 4 4 2 2
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 . 1 2 1 2 2 1 4 2 3 2 1 1 3 1 1 3 2 4 4 2
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 . 1 1 3 2 2 2 3 3 3 2 4 1 2 1 1 4 2 4 4 2
Page 2


HINT – SHEET
ANSWER KEY
1 . Among the given physical quantities angle has a
unit but no dimensions. Angle = [M
0
L
0
T
0
]
The SI unit of angle is radian
2 . The object will slip if centripetal force ? force of
friction
mr ?
2
 ? µmg
r ?
2
 ? µg
r ?
2
 ? constant, or 
2
1 2
2 1
r
r
? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
2
2
4cm 2
r
? ? ?
?
? ?
?
? ?
    ? r
2 
= 1 cm
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
LEADER & ACHIEVER COURSE
PHASE : MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 NEET(UG)
TEST DATE 
:
 17 - 04 - 2018
HS -1 /7 1001 CM D30 5 11 7061
TEST SYLLABUS : FULL SYLLABUS
3 . No. of beats per second = 
9
3
 = 3 s
–1
No. of beats per second = ?
1
 – ?
2
1 2 1 2
v v 1 1
3 v
? ?
? ? ? ?
? ?
? ? ? ?
? ?
2
3 1 1
300 2
? ?
?
2
1 1 1 50 1 49
2 100 100 100
?
? ? ? ?
?
?
2
 = 
49
100
 = 2.04 m
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 . 4 1 2 1 2 1 3 3 2 1 4 4 1 4 4 1 2 3 3 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 . 4 2 1 3 2 1 4 2 4 4 1 3 3 4 1 1 2 1 1 1
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 4 3 3 3 2 3 4 1 2 3 4 1 2 4 2 4 4 3 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
A n s . 4 3 2 2 2 1 2 4 2 2 1 2 3 1 3 3 2 2 4 3
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 2 1 2 2 2 3 2 4 4 2 3 2 3 4 1 4 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 4 1 4 1 2 1 4 4 2 4 1 1 2 3 4 1 2 1 4
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 . 2 4 4 1 1 4 4 3 4 3 1 2 2 4 3 4 4 4 2 2
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 . 1 2 1 2 2 1 4 2 3 2 1 1 3 1 1 3 2 4 4 2
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 . 1 1 3 2 2 2 3 3 3 2 4 1 2 1 1 4 2 4 4 2
HS -2 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
4 .
C
N
D
S
N
F S
B
1
E
B
2
d
d
Magnetic induction at point E due to magnet at
F (axial point) is
0
1
3
µ 2m
B
4
d
?
?
It acts along EF.
Magnetic induction at point E due to magnet at
D (equatorial point) is
0
2
3
µ m
B
4
d
?
?
It acts along FE.
Resultant magnetic induction at point E is
B = B
1
 – B
2
 = 
0
3
µ m
4 d ?
5 . Fresnel distance, 
? ?
2
3
2
F
9
4 10
a
z
500 10
?
?
?
? ?
?
?
? z
F
 = 32 m
6 . Here, a = g – bv
When an object falls with constant speed v
c
, its
acceleration becomes zero.
? g – bv
c
 = 0 or v
c
 = 
g
b
7 . Gravitational potential on the surface of the shell
is V = Gravitational potential due to particle (V
1
)
+ Gravitational potential due to shell itself (V
2
)
Gm G3m 4Gm
R R R
? ?
? ? ? ? ?
? ?
? ?
8 . Here, distance of point from the centre of the
sphere, r = 20 cm = 0.2 m
Electric field, E = –1.2 × 10
3
 N C
–1
As E = 
2
0
q
4 r ? ?
? q = (4 ??
0
r
2
) E = 
2 3
9
(0.2) ( 1.2 10 )
9 10
? ? ?
?
= –5.3 × 10
–9 
C
9 . When a wire of irregular shape turns into a circular
loop, are of the loop tends to increase. Thereforre,
magnetic flux linked with the loop increases.
According to Lenz's law, the direction of induced
current mus oppose the magnetic flux, for which
induced current should flow along adcba.
1 0. For a source S
1
, Wavelength, ?
1
 = 5000Å
Number of photons emitted per second, N
1
 = 10
15
Energy of each photon, E
1 
= 
1
hc
?
Power of source S
1
, P
1
 = E
1
N
1
 = 
1
1
N hc
?
For a source S
2
, Wavelength ?
2
 = 5100Å
Number of photons emitted per second,
N
2
 = 1.02 × 10
15
Energy of each photons, E
2
 = 
2
hc
?
Power of source S
2
, P
2
 = N
2
E
2 
= 
2
2
N hc
?
?
2
2 2 2 1 2
1
1 1 1 2
1
N hc
Power of S P N
N hc
Power of S P N
? ?
? ? ?
?
?
    = 
? ? ? ?
? ? ? ?
15 10
15 10
1.02 10 photons /s 5000 10
51
1
51
10 photons / s 5100 10
?
?
? ? ?
? ?
? ?
Page 3


HINT – SHEET
ANSWER KEY
1 . Among the given physical quantities angle has a
unit but no dimensions. Angle = [M
0
L
0
T
0
]
The SI unit of angle is radian
2 . The object will slip if centripetal force ? force of
friction
mr ?
2
 ? µmg
r ?
2
 ? µg
r ?
2
 ? constant, or 
2
1 2
2 1
r
r
? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
2
2
4cm 2
r
? ? ?
?
? ?
?
? ?
    ? r
2 
= 1 cm
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
LEADER & ACHIEVER COURSE
PHASE : MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 NEET(UG)
TEST DATE 
:
 17 - 04 - 2018
HS -1 /7 1001 CM D30 5 11 7061
TEST SYLLABUS : FULL SYLLABUS
3 . No. of beats per second = 
9
3
 = 3 s
–1
No. of beats per second = ?
1
 – ?
2
1 2 1 2
v v 1 1
3 v
? ?
? ? ? ?
? ?
? ? ? ?
? ?
2
3 1 1
300 2
? ?
?
2
1 1 1 50 1 49
2 100 100 100
?
? ? ? ?
?
?
2
 = 
49
100
 = 2.04 m
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 . 4 1 2 1 2 1 3 3 2 1 4 4 1 4 4 1 2 3 3 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 . 4 2 1 3 2 1 4 2 4 4 1 3 3 4 1 1 2 1 1 1
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 4 3 3 3 2 3 4 1 2 3 4 1 2 4 2 4 4 3 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
A n s . 4 3 2 2 2 1 2 4 2 2 1 2 3 1 3 3 2 2 4 3
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 2 1 2 2 2 3 2 4 4 2 3 2 3 4 1 4 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 4 1 4 1 2 1 4 4 2 4 1 1 2 3 4 1 2 1 4
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 . 2 4 4 1 1 4 4 3 4 3 1 2 2 4 3 4 4 4 2 2
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 . 1 2 1 2 2 1 4 2 3 2 1 1 3 1 1 3 2 4 4 2
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 . 1 1 3 2 2 2 3 3 3 2 4 1 2 1 1 4 2 4 4 2
HS -2 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
4 .
C
N
D
S
N
F S
B
1
E
B
2
d
d
Magnetic induction at point E due to magnet at
F (axial point) is
0
1
3
µ 2m
B
4
d
?
?
It acts along EF.
Magnetic induction at point E due to magnet at
D (equatorial point) is
0
2
3
µ m
B
4
d
?
?
It acts along FE.
Resultant magnetic induction at point E is
B = B
1
 – B
2
 = 
0
3
µ m
4 d ?
5 . Fresnel distance, 
? ?
2
3
2
F
9
4 10
a
z
500 10
?
?
?
? ?
?
?
? z
F
 = 32 m
6 . Here, a = g – bv
When an object falls with constant speed v
c
, its
acceleration becomes zero.
? g – bv
c
 = 0 or v
c
 = 
g
b
7 . Gravitational potential on the surface of the shell
is V = Gravitational potential due to particle (V
1
)
+ Gravitational potential due to shell itself (V
2
)
Gm G3m 4Gm
R R R
? ?
? ? ? ? ?
? ?
? ?
8 . Here, distance of point from the centre of the
sphere, r = 20 cm = 0.2 m
Electric field, E = –1.2 × 10
3
 N C
–1
As E = 
2
0
q
4 r ? ?
? q = (4 ??
0
r
2
) E = 
2 3
9
(0.2) ( 1.2 10 )
9 10
? ? ?
?
= –5.3 × 10
–9 
C
9 . When a wire of irregular shape turns into a circular
loop, are of the loop tends to increase. Thereforre,
magnetic flux linked with the loop increases.
According to Lenz's law, the direction of induced
current mus oppose the magnetic flux, for which
induced current should flow along adcba.
1 0. For a source S
1
, Wavelength, ?
1
 = 5000Å
Number of photons emitted per second, N
1
 = 10
15
Energy of each photon, E
1 
= 
1
hc
?
Power of source S
1
, P
1
 = E
1
N
1
 = 
1
1
N hc
?
For a source S
2
, Wavelength ?
2
 = 5100Å
Number of photons emitted per second,
N
2
 = 1.02 × 10
15
Energy of each photons, E
2
 = 
2
hc
?
Power of source S
2
, P
2
 = N
2
E
2 
= 
2
2
N hc
?
?
2
2 2 2 1 2
1
1 1 1 2
1
N hc
Power of S P N
N hc
Power of S P N
? ?
? ? ?
?
?
    = 
? ? ? ?
? ? ? ?
15 10
15 10
1.02 10 photons /s 5000 10
51
1
51
10 photons / s 5100 10
?
?
? ? ?
? ?
? ?
HS -3 /7
Leader & Achiever Course/Phase-MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP/17-04-2018
1001 CM D30 5 11 7061
1 1. Figure shows conditions of the question.
Child Father
Stationary
observer
Mother
+ve
4 kmh
–1
Moving belt
In this case,
Speed of belt w.r.t. ground ? ? v
BC
 = 4 km h
–1
Speed of child w.r.t. belt   ? ? ? v
CB
 = 9 km h
–1
? For an observer on a stationary platform, speed
of child running in the direction of motion of the
belt is
v
CG
 = v
CB
 + v
BG
 = 9 km h
–1 
+ 4 km h
–1
 = 13km h
–1
1 2. Bulk modulus, B = 
P
V / V ?
? Fractional change in volume, 
V P
V B
?
?
Here, P = 10 atm = 10 × 1 × 10
5
 N m
–2
B = 37 × 10
9
 N m
–2
?
6 2
9 2
V 1 10 N m
V
37 10 N m
?
?
? ?
?
?
 = 0.027 × 10
–3
= 2.7 × 10
–5
1 3. Potential energy of system
U = 
1 2
0
1 q q
4 r ? ?
0.5 × 10
–6
 = 
9 9 9
2
9 10 5 10 ( 2) 10
(x 2) 10
? ?
?
? ? ? ? ? ?
? ?
? x = 20 cm.
1 4. Here, V
P
 = 11000 V, V
S
 = 220 V
N
P
 = 6000, ? = 60%; P
O 
= 9 kW = 9 × 10
3 
W
Efficiency, ? = 
O
i
Output power P
Input power P
?
? P
i
 = 
3
O
P 9 10
60 /100
?
?
?
 = 1.5 × 10
4
 = 15 kW
1 5. Radius of the circular path of a charged particle
in a magnetic field is given by
R = 
mv
Bq
 or mv = RBq
Here, R = 0.83 cm = 0.83 × 10
–2
m
B = 0.25 Wb m
–2
q = 2e = 2 × 1.6 × 10
–19
C
? mv = (0.83 × 10
–2
)(0.25)(2 × 1.6 × 10
–19
)
de Broglie wavelength,
34
2 19
h 6.6 10
mv
0.83 10 0.25 2 1.6 10
? ?
?
? ? ?
? ? ? ? ?
= 0.01Å
1 7 . The velocity of outflow of water remains unchanged
because it depends upon the height of water level
and is independent of the size of the hole. The volume
depends directly on the size of the hole.
1 8 . Charge on capacitor plates without the dielectric is
Q = CV = (5 × 10
–6
F) × 1 V = 5 × 10
–6
C = 5 µC
The capacitance after the dielectric is introduced is
C' = 
0 0
A A / d
t t
d t t
K K
1
d
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
?
? ?
? ?
? ?
? ?
= 
C 5µF
t 4cm
t 4cm
K 4
1 1
d 6cm
?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
= 
5µF
4 1
1
6
? ? ?
?
? ?
? ?
 = 10 µF
? Charge on capacitor plates now will be
Q' = C'V = 10 µF × 1 V = 10 µC
Additional charge transferred
= Q' – Q = 10 µC – 5 µC = 5 µC
1 9. Resistance of the circuit,
R = R
1
 + R
2
 + 40 ? + 40 ? = 80 ?
Impedance of the circuit,
? ? ? ? ? ?
2 2 2
2
L C
Z R X X 80 100 40 ? ? ? ? ? ?
= 
? ? ? ?
2 2
80 60 ? = 100 ?
Power factor, cos ? = 
R 80
0.8
Z 100
? ?
Page 4


HINT – SHEET
ANSWER KEY
1 . Among the given physical quantities angle has a
unit but no dimensions. Angle = [M
0
L
0
T
0
]
The SI unit of angle is radian
2 . The object will slip if centripetal force ? force of
friction
mr ?
2
 ? µmg
r ?
2
 ? µg
r ?
2
 ? constant, or 
2
1 2
2 1
r
r
? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
2
2
4cm 2
r
? ? ?
?
? ?
?
? ?
    ? r
2 
= 1 cm
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
LEADER & ACHIEVER COURSE
PHASE : MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 NEET(UG)
TEST DATE 
:
 17 - 04 - 2018
HS -1 /7 1001 CM D30 5 11 7061
TEST SYLLABUS : FULL SYLLABUS
3 . No. of beats per second = 
9
3
 = 3 s
–1
No. of beats per second = ?
1
 – ?
2
1 2 1 2
v v 1 1
3 v
? ?
? ? ? ?
? ?
? ? ? ?
? ?
2
3 1 1
300 2
? ?
?
2
1 1 1 50 1 49
2 100 100 100
?
? ? ? ?
?
?
2
 = 
49
100
 = 2.04 m
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 . 4 1 2 1 2 1 3 3 2 1 4 4 1 4 4 1 2 3 3 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 . 4 2 1 3 2 1 4 2 4 4 1 3 3 4 1 1 2 1 1 1
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 4 3 3 3 2 3 4 1 2 3 4 1 2 4 2 4 4 3 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
A n s . 4 3 2 2 2 1 2 4 2 2 1 2 3 1 3 3 2 2 4 3
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 2 1 2 2 2 3 2 4 4 2 3 2 3 4 1 4 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 4 1 4 1 2 1 4 4 2 4 1 1 2 3 4 1 2 1 4
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 . 2 4 4 1 1 4 4 3 4 3 1 2 2 4 3 4 4 4 2 2
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 . 1 2 1 2 2 1 4 2 3 2 1 1 3 1 1 3 2 4 4 2
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 . 1 1 3 2 2 2 3 3 3 2 4 1 2 1 1 4 2 4 4 2
HS -2 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
4 .
C
N
D
S
N
F S
B
1
E
B
2
d
d
Magnetic induction at point E due to magnet at
F (axial point) is
0
1
3
µ 2m
B
4
d
?
?
It acts along EF.
Magnetic induction at point E due to magnet at
D (equatorial point) is
0
2
3
µ m
B
4
d
?
?
It acts along FE.
Resultant magnetic induction at point E is
B = B
1
 – B
2
 = 
0
3
µ m
4 d ?
5 . Fresnel distance, 
? ?
2
3
2
F
9
4 10
a
z
500 10
?
?
?
? ?
?
?
? z
F
 = 32 m
6 . Here, a = g – bv
When an object falls with constant speed v
c
, its
acceleration becomes zero.
? g – bv
c
 = 0 or v
c
 = 
g
b
7 . Gravitational potential on the surface of the shell
is V = Gravitational potential due to particle (V
1
)
+ Gravitational potential due to shell itself (V
2
)
Gm G3m 4Gm
R R R
? ?
? ? ? ? ?
? ?
? ?
8 . Here, distance of point from the centre of the
sphere, r = 20 cm = 0.2 m
Electric field, E = –1.2 × 10
3
 N C
–1
As E = 
2
0
q
4 r ? ?
? q = (4 ??
0
r
2
) E = 
2 3
9
(0.2) ( 1.2 10 )
9 10
? ? ?
?
= –5.3 × 10
–9 
C
9 . When a wire of irregular shape turns into a circular
loop, are of the loop tends to increase. Thereforre,
magnetic flux linked with the loop increases.
According to Lenz's law, the direction of induced
current mus oppose the magnetic flux, for which
induced current should flow along adcba.
1 0. For a source S
1
, Wavelength, ?
1
 = 5000Å
Number of photons emitted per second, N
1
 = 10
15
Energy of each photon, E
1 
= 
1
hc
?
Power of source S
1
, P
1
 = E
1
N
1
 = 
1
1
N hc
?
For a source S
2
, Wavelength ?
2
 = 5100Å
Number of photons emitted per second,
N
2
 = 1.02 × 10
15
Energy of each photons, E
2
 = 
2
hc
?
Power of source S
2
, P
2
 = N
2
E
2 
= 
2
2
N hc
?
?
2
2 2 2 1 2
1
1 1 1 2
1
N hc
Power of S P N
N hc
Power of S P N
? ?
? ? ?
?
?
    = 
? ? ? ?
? ? ? ?
15 10
15 10
1.02 10 photons /s 5000 10
51
1
51
10 photons / s 5100 10
?
?
? ? ?
? ?
? ?
HS -3 /7
Leader & Achiever Course/Phase-MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP/17-04-2018
1001 CM D30 5 11 7061
1 1. Figure shows conditions of the question.
Child Father
Stationary
observer
Mother
+ve
4 kmh
–1
Moving belt
In this case,
Speed of belt w.r.t. ground ? ? v
BC
 = 4 km h
–1
Speed of child w.r.t. belt   ? ? ? v
CB
 = 9 km h
–1
? For an observer on a stationary platform, speed
of child running in the direction of motion of the
belt is
v
CG
 = v
CB
 + v
BG
 = 9 km h
–1 
+ 4 km h
–1
 = 13km h
–1
1 2. Bulk modulus, B = 
P
V / V ?
? Fractional change in volume, 
V P
V B
?
?
Here, P = 10 atm = 10 × 1 × 10
5
 N m
–2
B = 37 × 10
9
 N m
–2
?
6 2
9 2
V 1 10 N m
V
37 10 N m
?
?
? ?
?
?
 = 0.027 × 10
–3
= 2.7 × 10
–5
1 3. Potential energy of system
U = 
1 2
0
1 q q
4 r ? ?
0.5 × 10
–6
 = 
9 9 9
2
9 10 5 10 ( 2) 10
(x 2) 10
? ?
?
? ? ? ? ? ?
? ?
? x = 20 cm.
1 4. Here, V
P
 = 11000 V, V
S
 = 220 V
N
P
 = 6000, ? = 60%; P
O 
= 9 kW = 9 × 10
3 
W
Efficiency, ? = 
O
i
Output power P
Input power P
?
? P
i
 = 
3
O
P 9 10
60 /100
?
?
?
 = 1.5 × 10
4
 = 15 kW
1 5. Radius of the circular path of a charged particle
in a magnetic field is given by
R = 
mv
Bq
 or mv = RBq
Here, R = 0.83 cm = 0.83 × 10
–2
m
B = 0.25 Wb m
–2
q = 2e = 2 × 1.6 × 10
–19
C
? mv = (0.83 × 10
–2
)(0.25)(2 × 1.6 × 10
–19
)
de Broglie wavelength,
34
2 19
h 6.6 10
mv
0.83 10 0.25 2 1.6 10
? ?
?
? ? ?
? ? ? ? ?
= 0.01Å
1 7 . The velocity of outflow of water remains unchanged
because it depends upon the height of water level
and is independent of the size of the hole. The volume
depends directly on the size of the hole.
1 8 . Charge on capacitor plates without the dielectric is
Q = CV = (5 × 10
–6
F) × 1 V = 5 × 10
–6
C = 5 µC
The capacitance after the dielectric is introduced is
C' = 
0 0
A A / d
t t
d t t
K K
1
d
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
?
? ?
? ?
? ?
? ?
= 
C 5µF
t 4cm
t 4cm
K 4
1 1
d 6cm
?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
= 
5µF
4 1
1
6
? ? ?
?
? ?
? ?
 = 10 µF
? Charge on capacitor plates now will be
Q' = C'V = 10 µF × 1 V = 10 µC
Additional charge transferred
= Q' – Q = 10 µC – 5 µC = 5 µC
1 9. Resistance of the circuit,
R = R
1
 + R
2
 + 40 ? + 40 ? = 80 ?
Impedance of the circuit,
? ? ? ? ? ?
2 2 2
2
L C
Z R X X 80 100 40 ? ? ? ? ? ?
= 
? ? ? ?
2 2
80 60 ? = 100 ?
Power factor, cos ? = 
R 80
0.8
Z 100
? ?
HS -4 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
2 0. We know that r
n
 ? n
2
 or 
n
1
r
r
? ?
? ?
? ?
= n
2
So, log
n
1
r
r
? ?
? ?
? ?
= 2logn
Hence, the graph between log
n
1
r
r
? ?
? ?
? ?
 and logn will
be a straight line passing through origin.
The positive slope is given by tan ? = 2.
2 1.
B
A
F
µ
A
µ
B
Here, m
A
 = 
m
2
, m
B
 = m
µ
A
 = 0.2, µ
B
 = 0.1
Let both the blocks are moving with common
acceleration a. Then,
a = 
A A
A
µ m g
m
 = µ
A
g = 0.2 g
and F – µ
B
(m
B
 + m
A
)g = (m
B
 + m
A
)a
F = (m
B
 + m
A
)a + µ
B
(m
B
 + m
A
)g
= ? ? ? ?
m m
m 0.2g 0.1 m g
2 2
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
= ? ? ? ?
3 3 0.9
m 0.2g m 0.1g mg
2 2 2
? ? ? ?
? ?
? ? ? ?
? ? ? ?
= 0.45 mg
2 2. Let T be the temperature at the junction.
Let L and A be the length and area of cross-section
of each rod respectively.
90ºC
90ºC
0ºC
X
Y
Z
T
W
? Heat current from Y to X is
H
1
 = 
KA(90 C T)
L
? ?
Heat current from Z to X is
H
2
 = 
KA(90 C T)
L
? ?
Heat current from X to W is
H
3
 = 
KA(T 0 C)
L
? ?
At the junction X,
H
1
 + H
2
 = H
3
? 90°C – T + 90°C – T = T
or 3T = 180°C or T = 60°C
2 3.
40 cm
100 cm
10 mV
G
B A
C
2V
(  )
R
The current in the potentiometer wire AC is
I = 
2
10 R ?
The potential difference across the potentiometer
wire is V = current × resistance
= 
2
10
10 R
?
?
The length of the wire is l = 100 cm
So, the potential gradient along the wire is
V 2 10
K
t 10 R 100
? ?
? ? ?
? ?
?
? ?
...(i)
The source of emf 10 mV is balanced against a
length of 40 cm of the potentiometer wire
i.e. 10 × 10
–3
 = k × 40
or 10 × 10
–3
 = 
? ?
2 40
10 R 10
?
?
      (Using (i))
or R = 790 ?.
2 4. In an electromagnetic wave both electric and
magnetic vectors are perpendicular to each other
as well as perpendicular to the direction of
propagation of wave.
2 6. Volume of wind flowing per second = Av
Mass of wind flowing per second = Av ?
Mass of air passing in time t is = Av ?t
Page 5


HINT – SHEET
ANSWER KEY
1 . Among the given physical quantities angle has a
unit but no dimensions. Angle = [M
0
L
0
T
0
]
The SI unit of angle is radian
2 . The object will slip if centripetal force ? force of
friction
mr ?
2
 ? µmg
r ?
2
 ? µg
r ?
2
 ? constant, or 
2
1 2
2 1
r
r
? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
2
2
4cm 2
r
? ? ?
?
? ?
?
? ?
    ? r
2 
= 1 cm
CLASSROOM CONTACT PROGRAMME
(Academic Session : 2017 - 2018)
LEADER & ACHIEVER COURSE
PHASE : MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP
TARGET : PRE-MEDICAL 2018
Test Type : MAJOR  Test Pattern 
:
 NEET(UG)
TEST DATE 
:
 17 - 04 - 2018
HS -1 /7 1001 CM D30 5 11 7061
TEST SYLLABUS : FULL SYLLABUS
3 . No. of beats per second = 
9
3
 = 3 s
–1
No. of beats per second = ?
1
 – ?
2
1 2 1 2
v v 1 1
3 v
? ?
? ? ? ?
? ?
? ? ? ?
? ?
2
3 1 1
300 2
? ?
?
2
1 1 1 50 1 49
2 100 100 100
?
? ? ? ?
?
?
2
 = 
49
100
 = 2.04 m
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 . 4 1 2 1 2 1 3 3 2 1 4 4 1 4 4 1 2 3 3 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 . 4 2 1 3 2 1 4 2 4 4 1 3 3 4 1 1 2 1 1 1
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 4 3 3 3 2 3 4 1 2 3 4 1 2 4 2 4 4 3 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
A n s . 4 3 2 2 2 1 2 4 2 2 1 2 3 1 3 3 2 2 4 3
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 2 1 2 2 2 3 2 4 4 2 3 2 3 4 1 4 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 4 1 4 1 2 1 4 4 2 4 1 1 2 3 4 1 2 1 4
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 . 2 4 4 1 1 4 4 3 4 3 1 2 2 4 3 4 4 4 2 2
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 . 1 2 1 2 2 1 4 2 3 2 1 1 3 1 1 3 2 4 4 2
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 . 1 1 3 2 2 2 3 3 3 2 4 1 2 1 1 4 2 4 4 2
HS -2 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
4 .
C
N
D
S
N
F S
B
1
E
B
2
d
d
Magnetic induction at point E due to magnet at
F (axial point) is
0
1
3
µ 2m
B
4
d
?
?
It acts along EF.
Magnetic induction at point E due to magnet at
D (equatorial point) is
0
2
3
µ m
B
4
d
?
?
It acts along FE.
Resultant magnetic induction at point E is
B = B
1
 – B
2
 = 
0
3
µ m
4 d ?
5 . Fresnel distance, 
? ?
2
3
2
F
9
4 10
a
z
500 10
?
?
?
? ?
?
?
? z
F
 = 32 m
6 . Here, a = g – bv
When an object falls with constant speed v
c
, its
acceleration becomes zero.
? g – bv
c
 = 0 or v
c
 = 
g
b
7 . Gravitational potential on the surface of the shell
is V = Gravitational potential due to particle (V
1
)
+ Gravitational potential due to shell itself (V
2
)
Gm G3m 4Gm
R R R
? ?
? ? ? ? ?
? ?
? ?
8 . Here, distance of point from the centre of the
sphere, r = 20 cm = 0.2 m
Electric field, E = –1.2 × 10
3
 N C
–1
As E = 
2
0
q
4 r ? ?
? q = (4 ??
0
r
2
) E = 
2 3
9
(0.2) ( 1.2 10 )
9 10
? ? ?
?
= –5.3 × 10
–9 
C
9 . When a wire of irregular shape turns into a circular
loop, are of the loop tends to increase. Thereforre,
magnetic flux linked with the loop increases.
According to Lenz's law, the direction of induced
current mus oppose the magnetic flux, for which
induced current should flow along adcba.
1 0. For a source S
1
, Wavelength, ?
1
 = 5000Å
Number of photons emitted per second, N
1
 = 10
15
Energy of each photon, E
1 
= 
1
hc
?
Power of source S
1
, P
1
 = E
1
N
1
 = 
1
1
N hc
?
For a source S
2
, Wavelength ?
2
 = 5100Å
Number of photons emitted per second,
N
2
 = 1.02 × 10
15
Energy of each photons, E
2
 = 
2
hc
?
Power of source S
2
, P
2
 = N
2
E
2 
= 
2
2
N hc
?
?
2
2 2 2 1 2
1
1 1 1 2
1
N hc
Power of S P N
N hc
Power of S P N
? ?
? ? ?
?
?
    = 
? ? ? ?
? ? ? ?
15 10
15 10
1.02 10 photons /s 5000 10
51
1
51
10 photons / s 5100 10
?
?
? ? ?
? ?
? ?
HS -3 /7
Leader & Achiever Course/Phase-MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP/17-04-2018
1001 CM D30 5 11 7061
1 1. Figure shows conditions of the question.
Child Father
Stationary
observer
Mother
+ve
4 kmh
–1
Moving belt
In this case,
Speed of belt w.r.t. ground ? ? v
BC
 = 4 km h
–1
Speed of child w.r.t. belt   ? ? ? v
CB
 = 9 km h
–1
? For an observer on a stationary platform, speed
of child running in the direction of motion of the
belt is
v
CG
 = v
CB
 + v
BG
 = 9 km h
–1 
+ 4 km h
–1
 = 13km h
–1
1 2. Bulk modulus, B = 
P
V / V ?
? Fractional change in volume, 
V P
V B
?
?
Here, P = 10 atm = 10 × 1 × 10
5
 N m
–2
B = 37 × 10
9
 N m
–2
?
6 2
9 2
V 1 10 N m
V
37 10 N m
?
?
? ?
?
?
 = 0.027 × 10
–3
= 2.7 × 10
–5
1 3. Potential energy of system
U = 
1 2
0
1 q q
4 r ? ?
0.5 × 10
–6
 = 
9 9 9
2
9 10 5 10 ( 2) 10
(x 2) 10
? ?
?
? ? ? ? ? ?
? ?
? x = 20 cm.
1 4. Here, V
P
 = 11000 V, V
S
 = 220 V
N
P
 = 6000, ? = 60%; P
O 
= 9 kW = 9 × 10
3 
W
Efficiency, ? = 
O
i
Output power P
Input power P
?
? P
i
 = 
3
O
P 9 10
60 /100
?
?
?
 = 1.5 × 10
4
 = 15 kW
1 5. Radius of the circular path of a charged particle
in a magnetic field is given by
R = 
mv
Bq
 or mv = RBq
Here, R = 0.83 cm = 0.83 × 10
–2
m
B = 0.25 Wb m
–2
q = 2e = 2 × 1.6 × 10
–19
C
? mv = (0.83 × 10
–2
)(0.25)(2 × 1.6 × 10
–19
)
de Broglie wavelength,
34
2 19
h 6.6 10
mv
0.83 10 0.25 2 1.6 10
? ?
?
? ? ?
? ? ? ? ?
= 0.01Å
1 7 . The velocity of outflow of water remains unchanged
because it depends upon the height of water level
and is independent of the size of the hole. The volume
depends directly on the size of the hole.
1 8 . Charge on capacitor plates without the dielectric is
Q = CV = (5 × 10
–6
F) × 1 V = 5 × 10
–6
C = 5 µC
The capacitance after the dielectric is introduced is
C' = 
0 0
A A / d
t t
d t t
K K
1
d
? ?
?
? ? ? ?
? ? ?
? ? ? ?
? ?
?
? ?
? ?
? ?
? ?
= 
C 5µF
t 4cm
t 4cm
K 4
1 1
d 6cm
?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
= 
5µF
4 1
1
6
? ? ?
?
? ?
? ?
 = 10 µF
? Charge on capacitor plates now will be
Q' = C'V = 10 µF × 1 V = 10 µC
Additional charge transferred
= Q' – Q = 10 µC – 5 µC = 5 µC
1 9. Resistance of the circuit,
R = R
1
 + R
2
 + 40 ? + 40 ? = 80 ?
Impedance of the circuit,
? ? ? ? ? ?
2 2 2
2
L C
Z R X X 80 100 40 ? ? ? ? ? ?
= 
? ? ? ?
2 2
80 60 ? = 100 ?
Power factor, cos ? = 
R 80
0.8
Z 100
? ?
HS -4 /7
Target : Pre-Medical 2018/Major/17-04-2018
1001 CM D30 5 11 7061
2 0. We know that r
n
 ? n
2
 or 
n
1
r
r
? ?
? ?
? ?
= n
2
So, log
n
1
r
r
? ?
? ?
? ?
= 2logn
Hence, the graph between log
n
1
r
r
? ?
? ?
? ?
 and logn will
be a straight line passing through origin.
The positive slope is given by tan ? = 2.
2 1.
B
A
F
µ
A
µ
B
Here, m
A
 = 
m
2
, m
B
 = m
µ
A
 = 0.2, µ
B
 = 0.1
Let both the blocks are moving with common
acceleration a. Then,
a = 
A A
A
µ m g
m
 = µ
A
g = 0.2 g
and F – µ
B
(m
B
 + m
A
)g = (m
B
 + m
A
)a
F = (m
B
 + m
A
)a + µ
B
(m
B
 + m
A
)g
= ? ? ? ?
m m
m 0.2g 0.1 m g
2 2
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
= ? ? ? ?
3 3 0.9
m 0.2g m 0.1g mg
2 2 2
? ? ? ?
? ?
? ? ? ?
? ? ? ?
= 0.45 mg
2 2. Let T be the temperature at the junction.
Let L and A be the length and area of cross-section
of each rod respectively.
90ºC
90ºC
0ºC
X
Y
Z
T
W
? Heat current from Y to X is
H
1
 = 
KA(90 C T)
L
? ?
Heat current from Z to X is
H
2
 = 
KA(90 C T)
L
? ?
Heat current from X to W is
H
3
 = 
KA(T 0 C)
L
? ?
At the junction X,
H
1
 + H
2
 = H
3
? 90°C – T + 90°C – T = T
or 3T = 180°C or T = 60°C
2 3.
40 cm
100 cm
10 mV
G
B A
C
2V
(  )
R
The current in the potentiometer wire AC is
I = 
2
10 R ?
The potential difference across the potentiometer
wire is V = current × resistance
= 
2
10
10 R
?
?
The length of the wire is l = 100 cm
So, the potential gradient along the wire is
V 2 10
K
t 10 R 100
? ?
? ? ?
? ?
?
? ?
...(i)
The source of emf 10 mV is balanced against a
length of 40 cm of the potentiometer wire
i.e. 10 × 10
–3
 = k × 40
or 10 × 10
–3
 = 
? ?
2 40
10 R 10
?
?
      (Using (i))
or R = 790 ?.
2 4. In an electromagnetic wave both electric and
magnetic vectors are perpendicular to each other
as well as perpendicular to the direction of
propagation of wave.
2 6. Volume of wind flowing per second = Av
Mass of wind flowing per second = Av ?
Mass of air passing in time t is = Av ?t
HS -5 /7
Leader & Achiever Course/Phase-MLF ,G,H,J,P ,SP ,M,MAZA,ZB,ZI,ZJ,ZQ,ZR,ZS,Z T,ZX,ZY & MAP/17-04-2018
1001 CM D30 5 11 7061
2 7. Here, ?
1 
= 1 – 
2
1
T
T
or 0.25 = 1 – 
2
1
T
T
 ? 
2
1
1 T
1
4 T
? ?
2
1
T 1 3
1
T 4 4
? ? ? ...(i)
According to question,
?
2
 = 2 ?
1
, and T
2
 = T
2 
– 58ºC
?
? ?
2
1
T 58 C 1
2 1
4 T
? ?
? ? ?
?
2
1
1 T 58 C
1
2 T
? ?
? ?
2
1 1 1
1 T 58 3 1 58
2 T T 4 2 T
?
? ? ? ? ?
? T
1
 = 232°C
2 8 . Let ? be emf and r be internal resistance of the battery.
In first case, 12 = ? – 2r ...(i)
In second case, 15 = ? + 3r ...(ii)
Substract (ii) from (i), we get, r = 
3
5
?
Putting this value of r in equation (i), we get
? = 12 + 
2 3 60 6 66
13.2V
5 5 5
? ?
? ? ?
2 9. Apparent depth of the dot
1 2 3 1 2 3
h h h h 1 1 1
3µ 3µ 3µ 3 µ µ µ
? ?
? ? ? ? ? ?
? ?
? ?
3 0. Let the number of fissions per second be n.
Energy released per second
= n × 200 MeV = n × 200 × 1.6 × 10
–13
J
Energy required per second = power × time
= 1 kW × 1 s = 1000 J
? n × 200 × 1.6 × 10
–13
 = 1000
or n = 
13 13
11
1000 10
10 3.125 10
3.2
3.2 10
?
? ? ? ?
?
3 1. Choosing the x and y axes as shown in the figure.
The coordinates of the vertices of the L-shaped
lamina is as shown in the figure. Divide the
L-shape lamina into three squares each of side 1m
and mass 1 kg ( ? the lamina is uniform).
By symmetry, the centres of mass C
1
, C
2
 and C
3
of the squares are their geometric centres and have
coordinates
y
x
C
3
C
1
C
2
A(2,0)
(0,0)
m
D(1,1) B(2,1)
E(1,2)
2m
F(0,2)
    
1 2
1 1 3 1
C , ,C ,
2 2 2 2
? ? ? ?
? ? ? ?
? ? ? ?
 and 
3
1 3
C ,
2 2
? ?
? ?
? ?
 respectively.
The coordinates of the centre of mass of the
L-shaped lamina is
1 1 2 2 3 3
CM
1 2 3
m x m x m x
X
m m m
? ?
?
? ?
= 
1 3 1
1 1 1
5
2 2 2
m
1 1 1 6
? ? ? ? ?
?
? ?
1 1 2 2 3 3
CM
1 2 3
m y m y m y
Y
m m m
? ?
?
? ?
= 
1 1 3
1 1 1
5
2 2 2
m
1 1 1 6
? ? ? ? ?
?
? ?
3 2. For a monatomic gas like helium ?
He
 = 
5
3
For a diatomic gas like oxygen ?
O
2
 = 
7
5
?
? ?
O He
2
mix
3 2
3 2
? ? ? ? ?
? ?
?
= 
7 5 21 10
3 2
113
5 3 5 3
1.5
5 5 15 5
? ? ? ?
? ? ?
?
3 3. In absence of magnetic field the weight added in
one pan balances the rectangular coil in the other
pan of balance,
? Mgl = W
coil
 l or W
coil
 = Mg = 0.5 × 9.8 N
When current I is passed through the coil and the
magnetic field is switched on.
Let m mass be added in the first pan to regain the
balance
Then Mgl + mgl = W
coil
 l + IBL sin90°l
mgl = IBLl
or m = 
2
IBL 9.8 0.4 1.5 10
g 9.8
?
? ? ?
?
= 0.6 × 10
–2
 kg = 6 × 10
–3
 kg = 6 g
Read More
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