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# Allen minor test 22/07/2018 (DLP) Solutions NEET Notes | EduRev

## NEET : Allen minor test 22/07/2018 (DLP) Solutions NEET Notes | EduRev

``` Page 1

HINT – SHEET
LT S/H S- 1/ 7 0 99 9D M D3 10 3 18 00 2
DISTANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test             Test # 02 Test Pattern
:
NEET-UG
TEST DATE
:
22 - 07 - 2018
1 .
40cm
50cm
u=0
2h
t
g
?
? t
2
– t
1
=
2 90 2 40
980 980
? ?
?
9 4
4a 49
?
=
3 2 1
s
7 7 7
? ?
2 . h =
2
1
gt
2
...(i)
and v = gt ..(ii)
h' =
2
t 1 t
v g
2 2 2
? ?
?
? ?
? ?
= gt
2
t 1 1
gt
2 4 2
? ?
= h –
1 3
h h
4 4
?
3 . initial velocity of food packet u = 4 ms
–1
(upward)
ita acceleration = – g downwards
v = +4 – 9.8 × 3 = –25.4 ms
–1
–ive means downwards
Q u e. 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 14 1 5 16 1 7 1 8 1 9 2 0
A n s. 3 3 2 2 3 3 3 2 3 4 3 1 3 2 1 2 3 4 1 4
Q u e. 2 1 22 2 3 24 2 5 2 6 2 7 2 8 29 3 0 3 1 3 2 3 3 34 3 5 36 3 7 3 8 3 9 4 0
A n s. 1 2 1 4 3 3 3 3 3 4 3 3 3 4 4 4 1 2 3 3
Q u e. 4 1 42 4 3 44 4 5 4 6 4 7 4 8 49 5 0 5 1 5 2 5 3 54 5 5 56 5 7 5 8 5 9 6 0
A n s. 2 1 2 3 2 3 2 1 2 4 2 4 2 2 1 4 2 1 4 2
Q u e. 6 1 62 6 3 64 6 5 6 6 6 7 6 8 69 7 0 7 1 7 2 7 3 74 7 5 76 7 7 7 8 7 9 8 0
A n s. 4 4 1 3 4 1 3 3 1 2 1 3 4 1 3 4 2 1 3 2
Q u e. 8 1 82 8 3 84 8 5 8 6 8 7 8 8 89 9 0 9 1 9 2 9 3 94 9 5 96 9 7 9 8 9 9 10 0
A n s. 4 1 1 3 3 3 1 2 1 2 3 4 3 3 2 4 3 3 4 1
Q u e. 1 0 1 1 0 2 10 3 1 0 4 1 05 1 0 6 1 0 7 10 8 1 0 9 11 0 1 1 1 1 12 1 1 3 1 1 4 11 5 1 1 6 1 17 1 1 8 1 19 12 0
A n s. 4 4 3 1 1 2 4 2 1 3 4 1 2 3 1 1 3 1 4 1
Q u e. 1 2 1 1 2 2 12 3 1 2 4 1 25 1 2 6 1 2 7 12 8 1 2 9 13 0 1 3 1 1 32 1 3 3 1 3 4 13 5 1 3 6 1 37 1 3 8 1 39 14 0
A n s. 3 1 1 3 1 3 4 4 3 2 1 2 2 3 2 3 3 4 3 3
Q u e. 1 4 1 1 4 2 14 3 1 4 4 1 45 1 4 6 1 4 7 14 8 1 4 9 15 0 1 5 1 1 52 1 5 3 1 5 4 15 5 1 5 6 1 57 1 5 8 1 59 16 0
A n s. 2 2 2 4 2 2 1 3 3 4 2 2 3 3 4 4 2 2 2 1
Q u e. 1 6 1 1 6 2 16 3 1 6 4 1 65 1 6 6 1 6 7 16 8 1 6 9 17 0 1 7 1 1 72 1 7 3 1 7 4 17 5 1 7 6 1 77 1 7 8 1 79 18 0
A n s. 2 4 2 4 3 3 1 1 2 4 3 4 3 2 2 3 4 4 2 3
Page 2

HINT – SHEET
LT S/H S- 1/ 7 0 99 9D M D3 10 3 18 00 2
DISTANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test             Test # 02 Test Pattern
:
NEET-UG
TEST DATE
:
22 - 07 - 2018
1 .
40cm
50cm
u=0
2h
t
g
?
? t
2
– t
1
=
2 90 2 40
980 980
? ?
?
9 4
4a 49
?
=
3 2 1
s
7 7 7
? ?
2 . h =
2
1
gt
2
...(i)
and v = gt ..(ii)
h' =
2
t 1 t
v g
2 2 2
? ?
?
? ?
? ?
= gt
2
t 1 1
gt
2 4 2
? ?
= h –
1 3
h h
4 4
?
3 . initial velocity of food packet u = 4 ms
–1
(upward)
ita acceleration = – g downwards
v = +4 – 9.8 × 3 = –25.4 ms
–1
–ive means downwards
Q u e. 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 14 1 5 16 1 7 1 8 1 9 2 0
A n s. 3 3 2 2 3 3 3 2 3 4 3 1 3 2 1 2 3 4 1 4
Q u e. 2 1 22 2 3 24 2 5 2 6 2 7 2 8 29 3 0 3 1 3 2 3 3 34 3 5 36 3 7 3 8 3 9 4 0
A n s. 1 2 1 4 3 3 3 3 3 4 3 3 3 4 4 4 1 2 3 3
Q u e. 4 1 42 4 3 44 4 5 4 6 4 7 4 8 49 5 0 5 1 5 2 5 3 54 5 5 56 5 7 5 8 5 9 6 0
A n s. 2 1 2 3 2 3 2 1 2 4 2 4 2 2 1 4 2 1 4 2
Q u e. 6 1 62 6 3 64 6 5 6 6 6 7 6 8 69 7 0 7 1 7 2 7 3 74 7 5 76 7 7 7 8 7 9 8 0
A n s. 4 4 1 3 4 1 3 3 1 2 1 3 4 1 3 4 2 1 3 2
Q u e. 8 1 82 8 3 84 8 5 8 6 8 7 8 8 89 9 0 9 1 9 2 9 3 94 9 5 96 9 7 9 8 9 9 10 0
A n s. 4 1 1 3 3 3 1 2 1 2 3 4 3 3 2 4 3 3 4 1
Q u e. 1 0 1 1 0 2 10 3 1 0 4 1 05 1 0 6 1 0 7 10 8 1 0 9 11 0 1 1 1 1 12 1 1 3 1 1 4 11 5 1 1 6 1 17 1 1 8 1 19 12 0
A n s. 4 4 3 1 1 2 4 2 1 3 4 1 2 3 1 1 3 1 4 1
Q u e. 1 2 1 1 2 2 12 3 1 2 4 1 25 1 2 6 1 2 7 12 8 1 2 9 13 0 1 3 1 1 32 1 3 3 1 3 4 13 5 1 3 6 1 37 1 3 8 1 39 14 0
A n s. 3 1 1 3 1 3 4 4 3 2 1 2 2 3 2 3 3 4 3 3
Q u e. 1 4 1 1 4 2 14 3 1 4 4 1 45 1 4 6 1 4 7 14 8 1 4 9 15 0 1 5 1 1 52 1 5 3 1 5 4 15 5 1 5 6 1 57 1 5 8 1 59 16 0
A n s. 2 2 2 4 2 2 1 3 3 4 2 2 3 3 4 4 2 2 2 1
Q u e. 1 6 1 1 6 2 16 3 1 6 4 1 65 1 6 6 1 6 7 16 8 1 6 9 17 0 1 7 1 1 72 1 7 3 1 7 4 17 5 1 7 6 1 77 1 7 8 1 79 18 0
A n s. 2 4 2 4 3 3 1 1 2 4 3 4 3 2 2 3 4 4 2 3
LT S/H S- 2/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
4 . time of flight for each ball = 1s
??h =
1
2
gt
2
=
1
2
× 10 × (1)
2
= 5m
5 . Speed of the child w.r.t. moving belt = 9km/h.
As the belt is moving with a speed of 4 km/h,
hence w.r.t. observer on the platform, speed of
child in the direction of motion of the belt
= 9 + 4 = 13 km/h
6 .
x
h
5m
time for 5m fall = 1 s
it means each drop will released after every
1
2
s.
x =
2
1
2
gt
=
2
1 1 5
10 m
2 2 4
? ?
? ? ?
? ?
? ?
h = 5 –
5 15
3.75m
4 4
? ?
7 .
v
time
t s
60 km/hr
t s 8t s
Ave. speed =
totaldistance area under v t graph
total time total time
?
?
=
? ?
1
8t 10t 60
2
54km / hr
10t
? ?
?
o r
average speed for A to B =
0 60
30km / hr
2
?
?
average speed for C to D =
60 0
30km / hr
2
?
?
average speed for A to D
v
ave
=
30 t 60 8t 30 t
t 8t t
? ? ? ? ?
? ?
=
30 480 30 540
54 km / hr
10 10
? ?
? ?
8 . Given : u = 0
2
2
2
1
1
a(20)
S
2
4
1
S
a(10)
2
? ?
? ? ? ? ?  S
2
= 4 S
1
9 . During the first 5 seconds of the motion, the
acceleration is –ve and during the next 5
seconds it becomes positive. (Example: a stone
thrown upwards, coming to momentary rest at
the highest point.) The distance covered
remains same during the two intervals of time.
10 . Relative velocity of policeman w.r.t. the thief is
10 – 9 = 1 ms
–1
. Since, the separation between them
is 100 m, hence the time taken will be 100 s.
1 1 . In 60 sec the tip of seconds hand travels a
distance 2 ? ?,
i.e.,
2
60
? ?
= 1.05
or
0 1 05 7 63 7
2 22 44
. ? ? ? ?
? ?
?
?
=
441
10
44
cm ?
1 2 . v
rel.
= 45 + 36 = 81 km/h
S = v
rel.
× t = 81 ×
5 81
6 75
60 12
. km ? ?
1 3 . It is only possible when particle goes upto
maximum height in 5
th
and come back in 6
th
.
so v = u – gt ? ?0 = u – 9.8 × 5 ? ?4 = 49 m/s
1 4 .
2
2
6 4
3
ˆ ˆ
dv F t i tj
a m /s
dt m
?
? ? ?
?
?
?
as acceleration depending on time so its not a
case of constant acceleration hence
t
0
v a dt ?
?
? ?
?
2
3
0
6 4
3 3
t t
ˆ ˆ
v i j dt
? ?
? ?
? ?
? ?
?
?
=
3
3 2
0
6 4
18 16
9 6
t t
ˆ ˆ ˆ ˆ
i j i j
? ?
? ? ?
? ?
? ?
1 5 . F.s. =
2 3 2
2
2
1 mu 25 10 (200)
mu F
2 2s 2 5 10
?
?
? ?
? ? ?
? ?
= 10 × 10
3
N = 10 kN
Page 3

HINT – SHEET
LT S/H S- 1/ 7 0 99 9D M D3 10 3 18 00 2
DISTANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test             Test # 02 Test Pattern
:
NEET-UG
TEST DATE
:
22 - 07 - 2018
1 .
40cm
50cm
u=0
2h
t
g
?
? t
2
– t
1
=
2 90 2 40
980 980
? ?
?
9 4
4a 49
?
=
3 2 1
s
7 7 7
? ?
2 . h =
2
1
gt
2
...(i)
and v = gt ..(ii)
h' =
2
t 1 t
v g
2 2 2
? ?
?
? ?
? ?
= gt
2
t 1 1
gt
2 4 2
? ?
= h –
1 3
h h
4 4
?
3 . initial velocity of food packet u = 4 ms
–1
(upward)
ita acceleration = – g downwards
v = +4 – 9.8 × 3 = –25.4 ms
–1
–ive means downwards
Q u e. 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 14 1 5 16 1 7 1 8 1 9 2 0
A n s. 3 3 2 2 3 3 3 2 3 4 3 1 3 2 1 2 3 4 1 4
Q u e. 2 1 22 2 3 24 2 5 2 6 2 7 2 8 29 3 0 3 1 3 2 3 3 34 3 5 36 3 7 3 8 3 9 4 0
A n s. 1 2 1 4 3 3 3 3 3 4 3 3 3 4 4 4 1 2 3 3
Q u e. 4 1 42 4 3 44 4 5 4 6 4 7 4 8 49 5 0 5 1 5 2 5 3 54 5 5 56 5 7 5 8 5 9 6 0
A n s. 2 1 2 3 2 3 2 1 2 4 2 4 2 2 1 4 2 1 4 2
Q u e. 6 1 62 6 3 64 6 5 6 6 6 7 6 8 69 7 0 7 1 7 2 7 3 74 7 5 76 7 7 7 8 7 9 8 0
A n s. 4 4 1 3 4 1 3 3 1 2 1 3 4 1 3 4 2 1 3 2
Q u e. 8 1 82 8 3 84 8 5 8 6 8 7 8 8 89 9 0 9 1 9 2 9 3 94 9 5 96 9 7 9 8 9 9 10 0
A n s. 4 1 1 3 3 3 1 2 1 2 3 4 3 3 2 4 3 3 4 1
Q u e. 1 0 1 1 0 2 10 3 1 0 4 1 05 1 0 6 1 0 7 10 8 1 0 9 11 0 1 1 1 1 12 1 1 3 1 1 4 11 5 1 1 6 1 17 1 1 8 1 19 12 0
A n s. 4 4 3 1 1 2 4 2 1 3 4 1 2 3 1 1 3 1 4 1
Q u e. 1 2 1 1 2 2 12 3 1 2 4 1 25 1 2 6 1 2 7 12 8 1 2 9 13 0 1 3 1 1 32 1 3 3 1 3 4 13 5 1 3 6 1 37 1 3 8 1 39 14 0
A n s. 3 1 1 3 1 3 4 4 3 2 1 2 2 3 2 3 3 4 3 3
Q u e. 1 4 1 1 4 2 14 3 1 4 4 1 45 1 4 6 1 4 7 14 8 1 4 9 15 0 1 5 1 1 52 1 5 3 1 5 4 15 5 1 5 6 1 57 1 5 8 1 59 16 0
A n s. 2 2 2 4 2 2 1 3 3 4 2 2 3 3 4 4 2 2 2 1
Q u e. 1 6 1 1 6 2 16 3 1 6 4 1 65 1 6 6 1 6 7 16 8 1 6 9 17 0 1 7 1 1 72 1 7 3 1 7 4 17 5 1 7 6 1 77 1 7 8 1 79 18 0
A n s. 2 4 2 4 3 3 1 1 2 4 3 4 3 2 2 3 4 4 2 3
LT S/H S- 2/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
4 . time of flight for each ball = 1s
??h =
1
2
gt
2
=
1
2
× 10 × (1)
2
= 5m
5 . Speed of the child w.r.t. moving belt = 9km/h.
As the belt is moving with a speed of 4 km/h,
hence w.r.t. observer on the platform, speed of
child in the direction of motion of the belt
= 9 + 4 = 13 km/h
6 .
x
h
5m
time for 5m fall = 1 s
it means each drop will released after every
1
2
s.
x =
2
1
2
gt
=
2
1 1 5
10 m
2 2 4
? ?
? ? ?
? ?
? ?
h = 5 –
5 15
3.75m
4 4
? ?
7 .
v
time
t s
60 km/hr
t s 8t s
Ave. speed =
totaldistance area under v t graph
total time total time
?
?
=
? ?
1
8t 10t 60
2
54km / hr
10t
? ?
?
o r
average speed for A to B =
0 60
30km / hr
2
?
?
average speed for C to D =
60 0
30km / hr
2
?
?
average speed for A to D
v
ave
=
30 t 60 8t 30 t
t 8t t
? ? ? ? ?
? ?
=
30 480 30 540
54 km / hr
10 10
? ?
? ?
8 . Given : u = 0
2
2
2
1
1
a(20)
S
2
4
1
S
a(10)
2
? ?
? ? ? ? ?  S
2
= 4 S
1
9 . During the first 5 seconds of the motion, the
acceleration is –ve and during the next 5
seconds it becomes positive. (Example: a stone
thrown upwards, coming to momentary rest at
the highest point.) The distance covered
remains same during the two intervals of time.
10 . Relative velocity of policeman w.r.t. the thief is
10 – 9 = 1 ms
–1
. Since, the separation between them
is 100 m, hence the time taken will be 100 s.
1 1 . In 60 sec the tip of seconds hand travels a
distance 2 ? ?,
i.e.,
2
60
? ?
= 1.05
or
0 1 05 7 63 7
2 22 44
. ? ? ? ?
? ?
?
?
=
441
10
44
cm ?
1 2 . v
rel.
= 45 + 36 = 81 km/h
S = v
rel.
× t = 81 ×
5 81
6 75
60 12
. km ? ?
1 3 . It is only possible when particle goes upto
maximum height in 5
th
and come back in 6
th
.
so v = u – gt ? ?0 = u – 9.8 × 5 ? ?4 = 49 m/s
1 4 .
2
2
6 4
3
ˆ ˆ
dv F t i tj
a m /s
dt m
?
? ? ?
?
?
?
as acceleration depending on time so its not a
case of constant acceleration hence
t
0
v a dt ?
?
? ?
?
2
3
0
6 4
3 3
t t
ˆ ˆ
v i j dt
? ?
? ?
? ?
? ?
?
?
=
3
3 2
0
6 4
18 16
9 6
t t
ˆ ˆ ˆ ˆ
i j i j
? ?
? ? ?
? ?
? ?
1 5 . F.s. =
2 3 2
2
2
1 mu 25 10 (200)
mu F
2 2s 2 5 10
?
?
? ?
? ? ?
? ?
= 10 × 10
3
N = 10 kN
LT S/H S- 3/ 7 0 99 9D M D3 10 3 18 00 2
1 6 .
t=u/a
a=–10ms
–2
u=60ms
–1
O
a=     =
f
m
10
1
=10m/s
2
T =
2u 60
2 12s
a 10
? ? ?
1 7 . Velocity of bullet w.r. to jeep v
bj
=
5
72
18
? =20m/s
velocity of bullet w.r. to ground v
bg
=20+5=25m/s
?? velocity of bullet w.r. to thief's car
bg th
v v 25 10 15m / s ? ? ? ?
1 8 . time for one side t =
4
2s
2
?
0 = u – gt ??u = gt = 10 × 2 = 20 m/s
o r
time of flight T =
2u gT 10 4
u 20m / s
g 2 2
?
? ? ? ?
1 9 . f =
dv 0.1 (10 0)
m 10N
dt 0.1
? ?
? ?
2 0 . x = ?t
3
and y = ?t
3
2 2
x y
dx dy
v 3 t and v 3 t
dt dt
? ? ? ? ? ?
2 2
x y
v v v ? ?
=
2 2 2 2
3 3 ( t ) ( t ) ? ? ? = 3t
2
2 2
? ? ?
2 1 . Velocity of paratrooper after bailing out = 0
after falling 50m, v
2
= 2ah = 2 × 9.8 × 50 = 980
let he cover distance h, with retardation of 2m/s
2
then 3
2
= v
2
– 2 × 2 × h
1
? 9 = 980 – 4h
1
or h
1
= 242.75 m
? Total height = 242.75 + 50 ? 293 m
2 2 . Velocity, v =
dx
x
dt
? ?
or
dx
dt
x
? ?
??
dx
dt
x
? ?
? ?
or 2x
1/2
= ?t or x ? t
2
2 3 .
d d
eE
v v E
m
? ? ? ?
2 4 .
R
A
? ?
l
and m = Ald   ??A =
m
d ?
?
2
R
m
?
l
? R
1
: R
2
: R
3
=
25 9 1
: :
1 3 5
= 125 : 15 : 1
2 5 . When the cells are connected in series, current
I
1
is given by :
1
nE
I
R nr
?
?
....(i)
When the cells connected in parallel, current
I
2
is given by :
2
E nE
I
r
nR r
R
n
? ?
?
?
....(ii)
As I
1
= I
2
So,
nE nE
R nr nR r
?
? ?
? R + nr = nR + r
or (n – 1)r = (n – 1)R or   r = R
2 6 . I = neAv
d
A = ?r
2
=
2
d
2
? ?
?
? ?
? ?
2 7 .
A B
2
R R
A t 2 t t
? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ?
?? ?R
A
= R
B
2 8 .
? ?
E E
I
r nr r n 1
? ?
? ?
V = E – Ir = E –
? ?
E nE V n
r
r n 1 n 1 E n 1
? ? ?
? ? ?
2 9 . In parallel connection
i
i
eq
i
E
r
E
1
r
?
?
?
??
1 2
1 2 1 2 2 1
eq
1 2
1 2
E E
r r E r E r
E
1 1
r r
r r
?
?
?
?
?
eq i 1 2
1 1 1 1
r r r r
? ? ? ?
? r
eq
=
1 2
1 2
r r
r r ?
3 0 . Total resistance of the circuit
R = 1 ? + 2 ? + 3 ? = 6 ?
Current,
10 4
I 1 amp
6
?
? ?
The direction of the current would be from a
to b via e.
Page 4

HINT – SHEET
LT S/H S- 1/ 7 0 99 9D M D3 10 3 18 00 2
DISTANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test             Test # 02 Test Pattern
:
NEET-UG
TEST DATE
:
22 - 07 - 2018
1 .
40cm
50cm
u=0
2h
t
g
?
? t
2
– t
1
=
2 90 2 40
980 980
? ?
?
9 4
4a 49
?
=
3 2 1
s
7 7 7
? ?
2 . h =
2
1
gt
2
...(i)
and v = gt ..(ii)
h' =
2
t 1 t
v g
2 2 2
? ?
?
? ?
? ?
= gt
2
t 1 1
gt
2 4 2
? ?
= h –
1 3
h h
4 4
?
3 . initial velocity of food packet u = 4 ms
–1
(upward)
ita acceleration = – g downwards
v = +4 – 9.8 × 3 = –25.4 ms
–1
–ive means downwards
Q u e. 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 14 1 5 16 1 7 1 8 1 9 2 0
A n s. 3 3 2 2 3 3 3 2 3 4 3 1 3 2 1 2 3 4 1 4
Q u e. 2 1 22 2 3 24 2 5 2 6 2 7 2 8 29 3 0 3 1 3 2 3 3 34 3 5 36 3 7 3 8 3 9 4 0
A n s. 1 2 1 4 3 3 3 3 3 4 3 3 3 4 4 4 1 2 3 3
Q u e. 4 1 42 4 3 44 4 5 4 6 4 7 4 8 49 5 0 5 1 5 2 5 3 54 5 5 56 5 7 5 8 5 9 6 0
A n s. 2 1 2 3 2 3 2 1 2 4 2 4 2 2 1 4 2 1 4 2
Q u e. 6 1 62 6 3 64 6 5 6 6 6 7 6 8 69 7 0 7 1 7 2 7 3 74 7 5 76 7 7 7 8 7 9 8 0
A n s. 4 4 1 3 4 1 3 3 1 2 1 3 4 1 3 4 2 1 3 2
Q u e. 8 1 82 8 3 84 8 5 8 6 8 7 8 8 89 9 0 9 1 9 2 9 3 94 9 5 96 9 7 9 8 9 9 10 0
A n s. 4 1 1 3 3 3 1 2 1 2 3 4 3 3 2 4 3 3 4 1
Q u e. 1 0 1 1 0 2 10 3 1 0 4 1 05 1 0 6 1 0 7 10 8 1 0 9 11 0 1 1 1 1 12 1 1 3 1 1 4 11 5 1 1 6 1 17 1 1 8 1 19 12 0
A n s. 4 4 3 1 1 2 4 2 1 3 4 1 2 3 1 1 3 1 4 1
Q u e. 1 2 1 1 2 2 12 3 1 2 4 1 25 1 2 6 1 2 7 12 8 1 2 9 13 0 1 3 1 1 32 1 3 3 1 3 4 13 5 1 3 6 1 37 1 3 8 1 39 14 0
A n s. 3 1 1 3 1 3 4 4 3 2 1 2 2 3 2 3 3 4 3 3
Q u e. 1 4 1 1 4 2 14 3 1 4 4 1 45 1 4 6 1 4 7 14 8 1 4 9 15 0 1 5 1 1 52 1 5 3 1 5 4 15 5 1 5 6 1 57 1 5 8 1 59 16 0
A n s. 2 2 2 4 2 2 1 3 3 4 2 2 3 3 4 4 2 2 2 1
Q u e. 1 6 1 1 6 2 16 3 1 6 4 1 65 1 6 6 1 6 7 16 8 1 6 9 17 0 1 7 1 1 72 1 7 3 1 7 4 17 5 1 7 6 1 77 1 7 8 1 79 18 0
A n s. 2 4 2 4 3 3 1 1 2 4 3 4 3 2 2 3 4 4 2 3
LT S/H S- 2/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
4 . time of flight for each ball = 1s
??h =
1
2
gt
2
=
1
2
× 10 × (1)
2
= 5m
5 . Speed of the child w.r.t. moving belt = 9km/h.
As the belt is moving with a speed of 4 km/h,
hence w.r.t. observer on the platform, speed of
child in the direction of motion of the belt
= 9 + 4 = 13 km/h
6 .
x
h
5m
time for 5m fall = 1 s
it means each drop will released after every
1
2
s.
x =
2
1
2
gt
=
2
1 1 5
10 m
2 2 4
? ?
? ? ?
? ?
? ?
h = 5 –
5 15
3.75m
4 4
? ?
7 .
v
time
t s
60 km/hr
t s 8t s
Ave. speed =
totaldistance area under v t graph
total time total time
?
?
=
? ?
1
8t 10t 60
2
54km / hr
10t
? ?
?
o r
average speed for A to B =
0 60
30km / hr
2
?
?
average speed for C to D =
60 0
30km / hr
2
?
?
average speed for A to D
v
ave
=
30 t 60 8t 30 t
t 8t t
? ? ? ? ?
? ?
=
30 480 30 540
54 km / hr
10 10
? ?
? ?
8 . Given : u = 0
2
2
2
1
1
a(20)
S
2
4
1
S
a(10)
2
? ?
? ? ? ? ?  S
2
= 4 S
1
9 . During the first 5 seconds of the motion, the
acceleration is –ve and during the next 5
seconds it becomes positive. (Example: a stone
thrown upwards, coming to momentary rest at
the highest point.) The distance covered
remains same during the two intervals of time.
10 . Relative velocity of policeman w.r.t. the thief is
10 – 9 = 1 ms
–1
. Since, the separation between them
is 100 m, hence the time taken will be 100 s.
1 1 . In 60 sec the tip of seconds hand travels a
distance 2 ? ?,
i.e.,
2
60
? ?
= 1.05
or
0 1 05 7 63 7
2 22 44
. ? ? ? ?
? ?
?
?
=
441
10
44
cm ?
1 2 . v
rel.
= 45 + 36 = 81 km/h
S = v
rel.
× t = 81 ×
5 81
6 75
60 12
. km ? ?
1 3 . It is only possible when particle goes upto
maximum height in 5
th
and come back in 6
th
.
so v = u – gt ? ?0 = u – 9.8 × 5 ? ?4 = 49 m/s
1 4 .
2
2
6 4
3
ˆ ˆ
dv F t i tj
a m /s
dt m
?
? ? ?
?
?
?
as acceleration depending on time so its not a
case of constant acceleration hence
t
0
v a dt ?
?
? ?
?
2
3
0
6 4
3 3
t t
ˆ ˆ
v i j dt
? ?
? ?
? ?
? ?
?
?
=
3
3 2
0
6 4
18 16
9 6
t t
ˆ ˆ ˆ ˆ
i j i j
? ?
? ? ?
? ?
? ?
1 5 . F.s. =
2 3 2
2
2
1 mu 25 10 (200)
mu F
2 2s 2 5 10
?
?
? ?
? ? ?
? ?
= 10 × 10
3
N = 10 kN
LT S/H S- 3/ 7 0 99 9D M D3 10 3 18 00 2
1 6 .
t=u/a
a=–10ms
–2
u=60ms
–1
O
a=     =
f
m
10
1
=10m/s
2
T =
2u 60
2 12s
a 10
? ? ?
1 7 . Velocity of bullet w.r. to jeep v
bj
=
5
72
18
? =20m/s
velocity of bullet w.r. to ground v
bg
=20+5=25m/s
?? velocity of bullet w.r. to thief's car
bg th
v v 25 10 15m / s ? ? ? ?
1 8 . time for one side t =
4
2s
2
?
0 = u – gt ??u = gt = 10 × 2 = 20 m/s
o r
time of flight T =
2u gT 10 4
u 20m / s
g 2 2
?
? ? ? ?
1 9 . f =
dv 0.1 (10 0)
m 10N
dt 0.1
? ?
? ?
2 0 . x = ?t
3
and y = ?t
3
2 2
x y
dx dy
v 3 t and v 3 t
dt dt
? ? ? ? ? ?
2 2
x y
v v v ? ?
=
2 2 2 2
3 3 ( t ) ( t ) ? ? ? = 3t
2
2 2
? ? ?
2 1 . Velocity of paratrooper after bailing out = 0
after falling 50m, v
2
= 2ah = 2 × 9.8 × 50 = 980
let he cover distance h, with retardation of 2m/s
2
then 3
2
= v
2
– 2 × 2 × h
1
? 9 = 980 – 4h
1
or h
1
= 242.75 m
? Total height = 242.75 + 50 ? 293 m
2 2 . Velocity, v =
dx
x
dt
? ?
or
dx
dt
x
? ?
??
dx
dt
x
? ?
? ?
or 2x
1/2
= ?t or x ? t
2
2 3 .
d d
eE
v v E
m
? ? ? ?
2 4 .
R
A
? ?
l
and m = Ald   ??A =
m
d ?
?
2
R
m
?
l
? R
1
: R
2
: R
3
=
25 9 1
: :
1 3 5
= 125 : 15 : 1
2 5 . When the cells are connected in series, current
I
1
is given by :
1
nE
I
R nr
?
?
....(i)
When the cells connected in parallel, current
I
2
is given by :
2
E nE
I
r
nR r
R
n
? ?
?
?
....(ii)
As I
1
= I
2
So,
nE nE
R nr nR r
?
? ?
? R + nr = nR + r
or (n – 1)r = (n – 1)R or   r = R
2 6 . I = neAv
d
A = ?r
2
=
2
d
2
? ?
?
? ?
? ?
2 7 .
A B
2
R R
A t 2 t t
? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ?
?? ?R
A
= R
B
2 8 .
? ?
E E
I
r nr r n 1
? ?
? ?
V = E – Ir = E –
? ?
E nE V n
r
r n 1 n 1 E n 1
? ? ?
? ? ?
2 9 . In parallel connection
i
i
eq
i
E
r
E
1
r
?
?
?
??
1 2
1 2 1 2 2 1
eq
1 2
1 2
E E
r r E r E r
E
1 1
r r
r r
?
?
?
?
?
eq i 1 2
1 1 1 1
r r r r
? ? ? ?
? r
eq
=
1 2
1 2
r r
r r ?
3 0 . Total resistance of the circuit
R = 1 ? + 2 ? + 3 ? = 6 ?
Current,
10 4
I 1 amp
6
?
? ?
The direction of the current would be from a
to b via e.
LT S/H S- 4/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
3 1 .
10 10 10
20
20
10
10
10
10
10
10
10
10
=30 ?
3 2 . Current I through CBD = (2/15) amp
Current I through CDA = (2/15) amp
C B
2
V V 10 volt
15
? ? ?
C A
2
V V 5 volt
15
? ? ?
? V
A
– V
B
= (V
C
– V
B
) – (V
C
– V
A
)
? ?
2 2
10 5 volt
15 3
? ?
? ? ?
? ?
? ?
3 3 . Current in each branch =
10
2.5A
4
?
V
C
– V
A
= 2.5 × 1 ...(i)
V
C
– V
B
= 2.5 × 3 ...(ii)
eq. (ii) – (i)
V
A
– V
B
= 2.5 (3 – 1) = 5V
3 4 .
i
3
i
2
i
1
i –i 2 3
i
2
i –i
2 3
i
1
1
1
1
1
1 1
1
8/3
2
1 ?
2/3 ?
1 ?
2 ?
1? || 2 ? =
1 2 2
1 2 3
?
? ?
?
8
2
16 8
3
8
14 7
2
3
?
? ? ?
?
3 5 . R
1
= R
o1
(1 + ?
1
t
1
)
R
2
= R
o2
(1 + ?
2
t
2
)
As R
1
= R
2
and R
o1
= R
o2
?
? ?
? ?
1 1
1
2 2 2
1 t
R
1
R 1 t
? ?
? ?
? ?
1 + ?
1
t
1
= 1 + ?
2
t
2
?
1
t
1
= ?
2
t
2
or
1 2
2 1
t
t
?
?
?
3 6 . Let R
0
be the initial resistance of both conductors.
? At temperature ? their resistance will be,
R
1
= R
0
(1 + ?
1
?)
and R
2
= R
0
(1 + ?
2
?)
For, series combination, R
s
= R
1
+ R
2
R
s0
(1 + ?
s
?) = R
0
(1 + ?
1
?) + R
0
(1 + ?
2
?)
where, R
s0
= R
0
+ R
0
= 2R
0
? 2R
0
(1 + ?
s
?) = 2R
0
+ R
0
?( ?
1
+ ?
2
)
or
1 2
s
2
? ? ?
? ?
3 7 .
2V
V
A
V
C
–    +
1V
2V
+
–
1A
V
D
V
B
V
B
– V
A
= (V
B
– V
D
) + (V
D
– V
C
) + (V
C
– V
A
)
= –2 + 2 + 1 = +1 volt
3 8 . V = IR ??
dI 1
dV R
?

dI
slope
dV
?
dI 1
i.e.R
dV R
? ? ?
it T ??then R ?? ? T
1
< T
2
3 9 . Resistance of bulb =
? ?
? ?
2
Rated voltage
Rated power
when the bulbs are connected in series
? ? ? ?
1 2
2 2
s B B
220 220
R R R
40 60
? ? ? ?
? ? ? ?
? ?
2
2 2
220
1 1 100
220 220
40 60 2400 24
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
?
? ?
? ?
2
2
s
1 2
s
V 24
P 220 24 W
R
220
? ? ? ?
P
2
= 60 + 40 = 100 W
Page 5

HINT – SHEET
LT S/H S- 1/ 7 0 99 9D M D3 10 3 18 00 2
DISTANCE LEARNING PROGRAMME
(Academic Session : 2018 - 2019)
LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2019
Test Type : Unit Test             Test # 02 Test Pattern
:
NEET-UG
TEST DATE
:
22 - 07 - 2018
1 .
40cm
50cm
u=0
2h
t
g
?
? t
2
– t
1
=
2 90 2 40
980 980
? ?
?
9 4
4a 49
?
=
3 2 1
s
7 7 7
? ?
2 . h =
2
1
gt
2
...(i)
and v = gt ..(ii)
h' =
2
t 1 t
v g
2 2 2
? ?
?
? ?
? ?
= gt
2
t 1 1
gt
2 4 2
? ?
= h –
1 3
h h
4 4
?
3 . initial velocity of food packet u = 4 ms
–1
(upward)
ita acceleration = – g downwards
v = +4 – 9.8 × 3 = –25.4 ms
–1
–ive means downwards
Q u e. 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 14 1 5 16 1 7 1 8 1 9 2 0
A n s. 3 3 2 2 3 3 3 2 3 4 3 1 3 2 1 2 3 4 1 4
Q u e. 2 1 22 2 3 24 2 5 2 6 2 7 2 8 29 3 0 3 1 3 2 3 3 34 3 5 36 3 7 3 8 3 9 4 0
A n s. 1 2 1 4 3 3 3 3 3 4 3 3 3 4 4 4 1 2 3 3
Q u e. 4 1 42 4 3 44 4 5 4 6 4 7 4 8 49 5 0 5 1 5 2 5 3 54 5 5 56 5 7 5 8 5 9 6 0
A n s. 2 1 2 3 2 3 2 1 2 4 2 4 2 2 1 4 2 1 4 2
Q u e. 6 1 62 6 3 64 6 5 6 6 6 7 6 8 69 7 0 7 1 7 2 7 3 74 7 5 76 7 7 7 8 7 9 8 0
A n s. 4 4 1 3 4 1 3 3 1 2 1 3 4 1 3 4 2 1 3 2
Q u e. 8 1 82 8 3 84 8 5 8 6 8 7 8 8 89 9 0 9 1 9 2 9 3 94 9 5 96 9 7 9 8 9 9 10 0
A n s. 4 1 1 3 3 3 1 2 1 2 3 4 3 3 2 4 3 3 4 1
Q u e. 1 0 1 1 0 2 10 3 1 0 4 1 05 1 0 6 1 0 7 10 8 1 0 9 11 0 1 1 1 1 12 1 1 3 1 1 4 11 5 1 1 6 1 17 1 1 8 1 19 12 0
A n s. 4 4 3 1 1 2 4 2 1 3 4 1 2 3 1 1 3 1 4 1
Q u e. 1 2 1 1 2 2 12 3 1 2 4 1 25 1 2 6 1 2 7 12 8 1 2 9 13 0 1 3 1 1 32 1 3 3 1 3 4 13 5 1 3 6 1 37 1 3 8 1 39 14 0
A n s. 3 1 1 3 1 3 4 4 3 2 1 2 2 3 2 3 3 4 3 3
Q u e. 1 4 1 1 4 2 14 3 1 4 4 1 45 1 4 6 1 4 7 14 8 1 4 9 15 0 1 5 1 1 52 1 5 3 1 5 4 15 5 1 5 6 1 57 1 5 8 1 59 16 0
A n s. 2 2 2 4 2 2 1 3 3 4 2 2 3 3 4 4 2 2 2 1
Q u e. 1 6 1 1 6 2 16 3 1 6 4 1 65 1 6 6 1 6 7 16 8 1 6 9 17 0 1 7 1 1 72 1 7 3 1 7 4 17 5 1 7 6 1 77 1 7 8 1 79 18 0
A n s. 2 4 2 4 3 3 1 1 2 4 3 4 3 2 2 3 4 4 2 3
LT S/H S- 2/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
4 . time of flight for each ball = 1s
??h =
1
2
gt
2
=
1
2
× 10 × (1)
2
= 5m
5 . Speed of the child w.r.t. moving belt = 9km/h.
As the belt is moving with a speed of 4 km/h,
hence w.r.t. observer on the platform, speed of
child in the direction of motion of the belt
= 9 + 4 = 13 km/h
6 .
x
h
5m
time for 5m fall = 1 s
it means each drop will released after every
1
2
s.
x =
2
1
2
gt
=
2
1 1 5
10 m
2 2 4
? ?
? ? ?
? ?
? ?
h = 5 –
5 15
3.75m
4 4
? ?
7 .
v
time
t s
60 km/hr
t s 8t s
Ave. speed =
totaldistance area under v t graph
total time total time
?
?
=
? ?
1
8t 10t 60
2
54km / hr
10t
? ?
?
o r
average speed for A to B =
0 60
30km / hr
2
?
?
average speed for C to D =
60 0
30km / hr
2
?
?
average speed for A to D
v
ave
=
30 t 60 8t 30 t
t 8t t
? ? ? ? ?
? ?
=
30 480 30 540
54 km / hr
10 10
? ?
? ?
8 . Given : u = 0
2
2
2
1
1
a(20)
S
2
4
1
S
a(10)
2
? ?
? ? ? ? ?  S
2
= 4 S
1
9 . During the first 5 seconds of the motion, the
acceleration is –ve and during the next 5
seconds it becomes positive. (Example: a stone
thrown upwards, coming to momentary rest at
the highest point.) The distance covered
remains same during the two intervals of time.
10 . Relative velocity of policeman w.r.t. the thief is
10 – 9 = 1 ms
–1
. Since, the separation between them
is 100 m, hence the time taken will be 100 s.
1 1 . In 60 sec the tip of seconds hand travels a
distance 2 ? ?,
i.e.,
2
60
? ?
= 1.05
or
0 1 05 7 63 7
2 22 44
. ? ? ? ?
? ?
?
?
=
441
10
44
cm ?
1 2 . v
rel.
= 45 + 36 = 81 km/h
S = v
rel.
× t = 81 ×
5 81
6 75
60 12
. km ? ?
1 3 . It is only possible when particle goes upto
maximum height in 5
th
and come back in 6
th
.
so v = u – gt ? ?0 = u – 9.8 × 5 ? ?4 = 49 m/s
1 4 .
2
2
6 4
3
ˆ ˆ
dv F t i tj
a m /s
dt m
?
? ? ?
?
?
?
as acceleration depending on time so its not a
case of constant acceleration hence
t
0
v a dt ?
?
? ?
?
2
3
0
6 4
3 3
t t
ˆ ˆ
v i j dt
? ?
? ?
? ?
? ?
?
?
=
3
3 2
0
6 4
18 16
9 6
t t
ˆ ˆ ˆ ˆ
i j i j
? ?
? ? ?
? ?
? ?
1 5 . F.s. =
2 3 2
2
2
1 mu 25 10 (200)
mu F
2 2s 2 5 10
?
?
? ?
? ? ?
? ?
= 10 × 10
3
N = 10 kN
LT S/H S- 3/ 7 0 99 9D M D3 10 3 18 00 2
1 6 .
t=u/a
a=–10ms
–2
u=60ms
–1
O
a=     =
f
m
10
1
=10m/s
2
T =
2u 60
2 12s
a 10
? ? ?
1 7 . Velocity of bullet w.r. to jeep v
bj
=
5
72
18
? =20m/s
velocity of bullet w.r. to ground v
bg
=20+5=25m/s
?? velocity of bullet w.r. to thief's car
bg th
v v 25 10 15m / s ? ? ? ?
1 8 . time for one side t =
4
2s
2
?
0 = u – gt ??u = gt = 10 × 2 = 20 m/s
o r
time of flight T =
2u gT 10 4
u 20m / s
g 2 2
?
? ? ? ?
1 9 . f =
dv 0.1 (10 0)
m 10N
dt 0.1
? ?
? ?
2 0 . x = ?t
3
and y = ?t
3
2 2
x y
dx dy
v 3 t and v 3 t
dt dt
? ? ? ? ? ?
2 2
x y
v v v ? ?
=
2 2 2 2
3 3 ( t ) ( t ) ? ? ? = 3t
2
2 2
? ? ?
2 1 . Velocity of paratrooper after bailing out = 0
after falling 50m, v
2
= 2ah = 2 × 9.8 × 50 = 980
let he cover distance h, with retardation of 2m/s
2
then 3
2
= v
2
– 2 × 2 × h
1
? 9 = 980 – 4h
1
or h
1
= 242.75 m
? Total height = 242.75 + 50 ? 293 m
2 2 . Velocity, v =
dx
x
dt
? ?
or
dx
dt
x
? ?
??
dx
dt
x
? ?
? ?
or 2x
1/2
= ?t or x ? t
2
2 3 .
d d
eE
v v E
m
? ? ? ?
2 4 .
R
A
? ?
l
and m = Ald   ??A =
m
d ?
?
2
R
m
?
l
? R
1
: R
2
: R
3
=
25 9 1
: :
1 3 5
= 125 : 15 : 1
2 5 . When the cells are connected in series, current
I
1
is given by :
1
nE
I
R nr
?
?
....(i)
When the cells connected in parallel, current
I
2
is given by :
2
E nE
I
r
nR r
R
n
? ?
?
?
....(ii)
As I
1
= I
2
So,
nE nE
R nr nR r
?
? ?
? R + nr = nR + r
or (n – 1)r = (n – 1)R or   r = R
2 6 . I = neAv
d
A = ?r
2
=
2
d
2
? ?
?
? ?
? ?
2 7 .
A B
2
R R
A t 2 t t
? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ?
?? ?R
A
= R
B
2 8 .
? ?
E E
I
r nr r n 1
? ?
? ?
V = E – Ir = E –
? ?
E nE V n
r
r n 1 n 1 E n 1
? ? ?
? ? ?
2 9 . In parallel connection
i
i
eq
i
E
r
E
1
r
?
?
?
??
1 2
1 2 1 2 2 1
eq
1 2
1 2
E E
r r E r E r
E
1 1
r r
r r
?
?
?
?
?
eq i 1 2
1 1 1 1
r r r r
? ? ? ?
? r
eq
=
1 2
1 2
r r
r r ?
3 0 . Total resistance of the circuit
R = 1 ? + 2 ? + 3 ? = 6 ?
Current,
10 4
I 1 amp
6
?
? ?
The direction of the current would be from a
to b via e.
LT S/H S- 4/ 7 09 99 D MD3 10 3 18 0 02
Target : Pre-Medical 2019/NEET-UG/22-07-2018
3 1 .
10 10 10
20
20
10
10
10
10
10
10
10
10
=30 ?
3 2 . Current I through CBD = (2/15) amp
Current I through CDA = (2/15) amp
C B
2
V V 10 volt
15
? ? ?
C A
2
V V 5 volt
15
? ? ?
? V
A
– V
B
= (V
C
– V
B
) – (V
C
– V
A
)
? ?
2 2
10 5 volt
15 3
? ?
? ? ?
? ?
? ?
3 3 . Current in each branch =
10
2.5A
4
?
V
C
– V
A
= 2.5 × 1 ...(i)
V
C
– V
B
= 2.5 × 3 ...(ii)
eq. (ii) – (i)
V
A
– V
B
= 2.5 (3 – 1) = 5V
3 4 .
i
3
i
2
i
1
i –i 2 3
i
2
i –i
2 3
i
1
1
1
1
1
1 1
1
8/3
2
1 ?
2/3 ?
1 ?
2 ?
1? || 2 ? =
1 2 2
1 2 3
?
? ?
?
8
2
16 8
3
8
14 7
2
3
?
? ? ?
?
3 5 . R
1
= R
o1
(1 + ?
1
t
1
)
R
2
= R
o2
(1 + ?
2
t
2
)
As R
1
= R
2
and R
o1
= R
o2
?
? ?
? ?
1 1
1
2 2 2
1 t
R
1
R 1 t
? ?
? ?
? ?
1 + ?
1
t
1
= 1 + ?
2
t
2
?
1
t
1
= ?
2
t
2
or
1 2
2 1
t
t
?
?
?
3 6 . Let R
0
be the initial resistance of both conductors.
? At temperature ? their resistance will be,
R
1
= R
0
(1 + ?
1
?)
and R
2
= R
0
(1 + ?
2
?)
For, series combination, R
s
= R
1
+ R
2
R
s0
(1 + ?
s
?) = R
0
(1 + ?
1
?) + R
0
(1 + ?
2
?)
where, R
s0
= R
0
+ R
0
= 2R
0
? 2R
0
(1 + ?
s
?) = 2R
0
+ R
0
?( ?
1
+ ?
2
)
or
1 2
s
2
? ? ?
? ?
3 7 .
2V
V
A
V
C
–    +
1V
2V
+
–
1A
V
D
V
B
V
B
– V
A
= (V
B
– V
D
) + (V
D
– V
C
) + (V
C
– V
A
)
= –2 + 2 + 1 = +1 volt
3 8 . V = IR ??
dI 1
dV R
?

dI
slope
dV
?
dI 1
i.e.R
dV R
? ? ?
it T ??then R ?? ? T
1
< T
2
3 9 . Resistance of bulb =
? ?
? ?
2
Rated voltage
Rated power
when the bulbs are connected in series
? ? ? ?
1 2
2 2
s B B
220 220
R R R
40 60
? ? ? ?
? ? ? ?
? ?
2
2 2
220
1 1 100
220 220
40 60 2400 24
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
?
? ?
? ?
2
2
s
1 2
s
V 24
P 220 24 W
R
220
? ? ? ?
P
2
= 60 + 40 = 100 W
LT S/H S- 5/ 7 0 99 9D M D3 10 3 18 00 2
4 0 . Bulbs A and B are in parallel, their effective
power is
P' = P
A
+ P
B
= 200 W + 200 W = 400 W
P' and bulb C are in series, the resultant power
of the combination is
C
R
C
P'  P 400 W × 400 W
P 200 W
P' + P 400 W + 400 W
?
? ? ?
4 1 . = 3A
2 ?
4 ?
1 ? 5 ?
i
1
i
2
i
3
A
C
E
D
F
B
V
2?, 4 ? and (1 ? + 5 ?) are in parallel so their
potential difference are the same.
V = 3 × 2 = 6 V
3
6
i 1 A
1 5
? ?
?
P
S
=
2 2
3
i 5 (1) 5 5W ? ? ? ?
4 2 . Q = at – bt
2
dQ
I a 2bt
dt
? ? ?
I = 0 for t = t
0
= (a/2b)
Current flows from t = 0 to t = t
0
.
Heat produced =
0
t
2
0
I R dt
?
=
? ?
0
t
2
0
a 2bt R dt ?
?
? ?
0
t
2 2 2
0
a R 4abRt 4b Rt dt ? ?
?
Solving above equation and putting t
0
= (a/2b),
we get
Heat produced =
3
a R
6b
? ?
? ?
? ?
4 3 . P
6
= I
6
2
× R ??6 = I
6
2
× 6 ??I
6
= 1
I
4
× 4 = 1 × 6 ??I
4
= 1.5
P
4
= (1.5)
2
× 4 = 2.25 × 4 = 9W
4 5 . 150 W on two  so 75W on one
? ?
2
V 15 15
75 R 3
R 75
?
? ? ? ? ?
4 8 .
58
Ce =
54
[Xe]4f
2
6s
2
. f-block element placed in
IIIB group.
5 1 . Representative elements are s & p-block
element.
5 2 .
O
P S Cl
5 5 . Max unpaired e
–
s in p
–
Subshell belong to 'N'
family.
5 6 . 5d>3d>4d- due to Lanthanoid contraction
5 7 . For isoelectronic species size order is
cation < Neutral < Anion
6 0 . 2
nd
I.E. of alkali Metal is max. because after
loosing an el
??
it attains inert gas configuration.
6 1 . 2
nd
IP ??? ???????????? ????????? ???? ?? ?? ??? ????? ?? ??el
–
??????? ??? ????? ? ??? ?? ??? ? ??????? ??? ??? ??? ????? ?? ?
6 3 . Cr = [Ar] 3d
5
4s
1
attains stable configuration
after removal of 1
st
el
–
.
6 5 . Polarity of Bond ? ?EN
6 6 . ??? ?????? ??? ??? ? ? ????? ???? ??? ???? ??< ?????????<
??? ????? ??? ?? ???
6 9 . M
eq
of conc. HCl = M
eq
of dil. HCl
? ? ?
1
10 10 V
10
V = 1000 ml
Thus, 990 ml of water should be added to
10 ml of conc. HCl to get decinormal solution.
7 0 .
? ? ?
moleof NaCl 1
m 1
Weight of solvent in kg. 1
weight of solvent = weight of solution – weight
of NaCl
= 1.0585 × 1000 – 58.5
= 1058.5 – 58.5 = 1000 g = 1 kg
7 1 . O.P. is colligative property which depends on
number of particles
7 2 . ?T
f
= 1.86 × 0.05 = 0.093
? T
f
= 0 – 0.093 = –0.093°C
7 3 . C
2
H
5
OH shows hydrogen bonding as well as
polarity both.
```
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