JEE Test Paper 1 Solution - IT & Software

 Page 1


PART-1 : PHYSICS SOLUTION
HS-1/8
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575     info@allen.ac.in   www.allen.ac.in
SECTION-I
1. Ans. (A)
Sol. Let charge on 4 µF capacitor when switch
in position A is Q
Charge on 4µF capacitor after switch is
shifted to position B.
4 2Q
Q'Q
423
==
+
2Q/32
ratio
Q3
==
2. Ans. (C)
Sol. Velocity after collision will be
= 
ˆ ˆˆ
3i 10 j 10 k -+
Velocity after 1 second = 
ˆˆ
3i 10j -
3. Ans. (A)
Sol.
45
I 0.5A
90
==
4. Ans. (A)
Sol. Load on concrete = 
Y20A2
Y 20A 10Y A 3
×
=
× +×
5. Ans. (A)
Sol.
i ii
tan tan 3 60 q=mÞ q= Þq= °
also 1 sin 60° = 
3
 sin r Þ r = 30°
6. Ans. (A)
Sol. Initial current = 
21
A
10 20 15
=
+
Final current = 
21
A
20 10
=
7. Ans. (D)
Sol.
2
12
22 0
0
01
rms0
V
V t dt dt
7 2
VV
2 24
æö
+
ç÷
èø
==
òò
(0000CJA102119031) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE
JEE(Main)
AIOT
29-03-2020
ANSWER KEY
PART-1 : PHYSICS
PART -3 : MATHEMA TICS
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. A C A A A A D B C C
Q. 11 12 13 14 15 16 17 18 19 20
A. C A D C C C A D B A
Q. 1 2 3 4 5
A. 0.18 697.00 200.00 4.47 1316.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D B D B C
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A A D C B C C A
Q. 1 2 3 4 5
A. 324.00 –115.80 2.25 4.00 4.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B A B B C D B
Q. 11 12 13 14 15 16 17 18 19 20
A. A C D D B D C A D A
Q. 1 2 3 4 5
A. 7.00 1.00 –0.50 165.00 78.00
SECTION-I
SECTION-II
Page 2


PART-1 : PHYSICS SOLUTION
HS-1/8
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575     info@allen.ac.in   www.allen.ac.in
SECTION-I
1. Ans. (A)
Sol. Let charge on 4 µF capacitor when switch
in position A is Q
Charge on 4µF capacitor after switch is
shifted to position B.
4 2Q
Q'Q
423
==
+
2Q/32
ratio
Q3
==
2. Ans. (C)
Sol. Velocity after collision will be
= 
ˆ ˆˆ
3i 10 j 10 k -+
Velocity after 1 second = 
ˆˆ
3i 10j -
3. Ans. (A)
Sol.
45
I 0.5A
90
==
4. Ans. (A)
Sol. Load on concrete = 
Y20A2
Y 20A 10Y A 3
×
=
× +×
5. Ans. (A)
Sol.
i ii
tan tan 3 60 q=mÞ q= Þq= °
also 1 sin 60° = 
3
 sin r Þ r = 30°
6. Ans. (A)
Sol. Initial current = 
21
A
10 20 15
=
+
Final current = 
21
A
20 10
=
7. Ans. (D)
Sol.
2
12
22 0
0
01
rms0
V
V t dt dt
7 2
VV
2 24
æö
+
ç÷
èø
==
òò
(0000CJA102119031) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE
JEE(Main)
AIOT
29-03-2020
ANSWER KEY
PART-1 : PHYSICS
PART -3 : MATHEMA TICS
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. A C A A A A D B C C
Q. 11 12 13 14 15 16 17 18 19 20
A. C A D C C C A D B A
Q. 1 2 3 4 5
A. 0.18 697.00 200.00 4.47 1316.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D B D B C
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A A D C B C C A
Q. 1 2 3 4 5
A. 324.00 –115.80 2.25 4.00 4.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B A B B C D B
Q. 11 12 13 14 15 16 17 18 19 20
A. A C D D B D C A D A
Q. 1 2 3 4 5
A. 7.00 1.00 –0.50 165.00 78.00
SECTION-I
SECTION-II
ALLEN
LTS/HS-2/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
8. Ans. (B)
Sol.
1/2
t/t
t0
A A (0.5) =
1/2
t
1
t
11
22
æö æö
=
ç÷ ç÷
èø èø
\ 
1/2
tt=
A
t
 = A
0
e
–lt
ln A
t
 = ln A
0
 – lt
1/2
ln2
t
t
 = lt = ln A
0
 – ln A
t
()
1/2 2
1/2
ln2
dt
t -
 
0t
0
dA dA
t
AA
=+
0t
0t 1/2
1/2
dA dA
55
AA dt 1.5
1000 500
t ln2 ln2 ln2
+
+
= ==
9. Ans. (C)
Sol. Initially
mg
kx
0
Vg r
Þ Vrg = kx
0
 + mg ... (i)
Finally
m(g + a)
0
k(x + x)
0
V (g + a ) r
0
Þ Vr(g + a
0
) = k(x
0
 + x) + m(g + a
0
) ... (ii)
(ii) – (i)
Vra
0
 = kx + ma
0
0
V a ma
x
k
r-
=
0
ma V
x1
km
r æö
=-
ç÷
èø
0
m
ma V
x1
kV
æö r
=-
ç÷
r
èø
0
ma1
x1
k
æö
=-
ç÷
g
èø
10. Ans. (C)
Sol. Size of object = 
12
I I 2 4.5 3mm = ´=
11. Ans. (C)
Sol.
0
e
L
9
L
=
 & L
0
 + L
e
 = 100
L
0
 = 90
L
e
 = 10
Now 
1 1 1 90
u 9cm
90 u 10 10
-
- = Þ = =-
-
Tube length = 9 + 90 = 99 cm
Change = 1cm
12. Ans. (A)
Sol.
2
n n1
mvkk
v
r r mr
-
= Þ=
n1
2r 2r
T
v
k
mr
-
pp
==
or 
n1
2
Tr
+
µ
13. Ans. (D)
Sol. DQ
AB
 = 9 × 10
4
 J
DQ
BC
 = 0
55
CD
5 110 1 1102
Q nR
2 nR nR
éù ´´ ´´
D=-
êú
ëû
= –2.5 × 10
5
 J
55
DA
3 2.4 101 1101
Q nR
2 nR nR
éù ´ ´ ´´
D=-
êú
ëû
= 2.1 × 10
5
 J
W = SDQ = 5 × 10
4
 J
Page 3


PART-1 : PHYSICS SOLUTION
HS-1/8
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575     info@allen.ac.in   www.allen.ac.in
SECTION-I
1. Ans. (A)
Sol. Let charge on 4 µF capacitor when switch
in position A is Q
Charge on 4µF capacitor after switch is
shifted to position B.
4 2Q
Q'Q
423
==
+
2Q/32
ratio
Q3
==
2. Ans. (C)
Sol. Velocity after collision will be
= 
ˆ ˆˆ
3i 10 j 10 k -+
Velocity after 1 second = 
ˆˆ
3i 10j -
3. Ans. (A)
Sol.
45
I 0.5A
90
==
4. Ans. (A)
Sol. Load on concrete = 
Y20A2
Y 20A 10Y A 3
×
=
× +×
5. Ans. (A)
Sol.
i ii
tan tan 3 60 q=mÞ q= Þq= °
also 1 sin 60° = 
3
 sin r Þ r = 30°
6. Ans. (A)
Sol. Initial current = 
21
A
10 20 15
=
+
Final current = 
21
A
20 10
=
7. Ans. (D)
Sol.
2
12
22 0
0
01
rms0
V
V t dt dt
7 2
VV
2 24
æö
+
ç÷
èø
==
òò
(0000CJA102119031) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE
JEE(Main)
AIOT
29-03-2020
ANSWER KEY
PART-1 : PHYSICS
PART -3 : MATHEMA TICS
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. A C A A A A D B C C
Q. 11 12 13 14 15 16 17 18 19 20
A. C A D C C C A D B A
Q. 1 2 3 4 5
A. 0.18 697.00 200.00 4.47 1316.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D B D B C
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A A D C B C C A
Q. 1 2 3 4 5
A. 324.00 –115.80 2.25 4.00 4.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B A B B C D B
Q. 11 12 13 14 15 16 17 18 19 20
A. A C D D B D C A D A
Q. 1 2 3 4 5
A. 7.00 1.00 –0.50 165.00 78.00
SECTION-I
SECTION-II
ALLEN
LTS/HS-2/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
8. Ans. (B)
Sol.
1/2
t/t
t0
A A (0.5) =
1/2
t
1
t
11
22
æö æö
=
ç÷ ç÷
èø èø
\ 
1/2
tt=
A
t
 = A
0
e
–lt
ln A
t
 = ln A
0
 – lt
1/2
ln2
t
t
 = lt = ln A
0
 – ln A
t
()
1/2 2
1/2
ln2
dt
t -
 
0t
0
dA dA
t
AA
=+
0t
0t 1/2
1/2
dA dA
55
AA dt 1.5
1000 500
t ln2 ln2 ln2
+
+
= ==
9. Ans. (C)
Sol. Initially
mg
kx
0
Vg r
Þ Vrg = kx
0
 + mg ... (i)
Finally
m(g + a)
0
k(x + x)
0
V (g + a ) r
0
Þ Vr(g + a
0
) = k(x
0
 + x) + m(g + a
0
) ... (ii)
(ii) – (i)
Vra
0
 = kx + ma
0
0
V a ma
x
k
r-
=
0
ma V
x1
km
r æö
=-
ç÷
èø
0
m
ma V
x1
kV
æö r
=-
ç÷
r
èø
0
ma1
x1
k
æö
=-
ç÷
g
èø
10. Ans. (C)
Sol. Size of object = 
12
I I 2 4.5 3mm = ´=
11. Ans. (C)
Sol.
0
e
L
9
L
=
 & L
0
 + L
e
 = 100
L
0
 = 90
L
e
 = 10
Now 
1 1 1 90
u 9cm
90 u 10 10
-
- = Þ = =-
-
Tube length = 9 + 90 = 99 cm
Change = 1cm
12. Ans. (A)
Sol.
2
n n1
mvkk
v
r r mr
-
= Þ=
n1
2r 2r
T
v
k
mr
-
pp
==
or 
n1
2
Tr
+
µ
13. Ans. (D)
Sol. DQ
AB
 = 9 × 10
4
 J
DQ
BC
 = 0
55
CD
5 110 1 1102
Q nR
2 nR nR
éù ´´ ´´
D=-
êú
ëû
= –2.5 × 10
5
 J
55
DA
3 2.4 101 1101
Q nR
2 nR nR
éù ´ ´ ´´
D=-
êú
ëû
= 2.1 × 10
5
 J
W = SDQ = 5 × 10
4
 J
ALLEN
29032020 LTS/HS-3/8
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
14. Ans. (C)
Sol.
max
H
tan37
R/2
°=
22
2
3 u sin / 2g
4 u sin cos /g
q
=
qq
3
tan
2
Þ q=
15. Ans. (C)
Sol.
mv
R
qB
=
d = R cos 53°
Þ
53°
d
C
53°
v
v
Þ q = 90°
16. Ans. (C)
Sol. Top view of cylinder.
Rdq
dq
q
q
q
q
F
Change in momentum of one photon
2h
cos =q
l
No. of photons striking per second on small
area = N (Rdq cos q) × h
Þ Total change in momentum per second
striking small area
= (N(Rdq)cos q × h) 
2h
cosq
l
( )
2
2h
N Rhcosd
æö
= qq
ç÷
l
èø
Force applied = F cosq
3
2h
N Rhcosd
æö
= qq
ç÷
l
èø
3
2I
dF N Rh cos d
C
æö
= qq
ç÷
èø
òò
8IhR
3C
=
17. Ans. (A)
Sol. V = u + at
2
5
a m/s
4
Þ=
5
vt
4
Þ=
5
F ma
4
==
25
PFvt
16
Þ = ×=
18. Ans. (D)
Sol. Unit of µ
0
 = ohm sec/meter
C
2
 = 
0 0
1
Î m
Þ Î
0
 = 
0
2
C
1
m
C = velocity of light
x 
0
0
Î
m
 = 
2 2
0
C m = µ
0
C
Unit of C = metre/sec.
unit of 
0
0
Î
m
 = unit of µ
0
C = ohm sec/meter
× meter/sec = Ohm.
Page 4


PART-1 : PHYSICS SOLUTION
HS-1/8
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575     info@allen.ac.in   www.allen.ac.in
SECTION-I
1. Ans. (A)
Sol. Let charge on 4 µF capacitor when switch
in position A is Q
Charge on 4µF capacitor after switch is
shifted to position B.
4 2Q
Q'Q
423
==
+
2Q/32
ratio
Q3
==
2. Ans. (C)
Sol. Velocity after collision will be
= 
ˆ ˆˆ
3i 10 j 10 k -+
Velocity after 1 second = 
ˆˆ
3i 10j -
3. Ans. (A)
Sol.
45
I 0.5A
90
==
4. Ans. (A)
Sol. Load on concrete = 
Y20A2
Y 20A 10Y A 3
×
=
× +×
5. Ans. (A)
Sol.
i ii
tan tan 3 60 q=mÞ q= Þq= °
also 1 sin 60° = 
3
 sin r Þ r = 30°
6. Ans. (A)
Sol. Initial current = 
21
A
10 20 15
=
+
Final current = 
21
A
20 10
=
7. Ans. (D)
Sol.
2
12
22 0
0
01
rms0
V
V t dt dt
7 2
VV
2 24
æö
+
ç÷
èø
==
òò
(0000CJA102119031) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE
JEE(Main)
AIOT
29-03-2020
ANSWER KEY
PART-1 : PHYSICS
PART -3 : MATHEMA TICS
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. A C A A A A D B C C
Q. 11 12 13 14 15 16 17 18 19 20
A. C A D C C C A D B A
Q. 1 2 3 4 5
A. 0.18 697.00 200.00 4.47 1316.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D B D B C
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A A D C B C C A
Q. 1 2 3 4 5
A. 324.00 –115.80 2.25 4.00 4.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B A B B C D B
Q. 11 12 13 14 15 16 17 18 19 20
A. A C D D B D C A D A
Q. 1 2 3 4 5
A. 7.00 1.00 –0.50 165.00 78.00
SECTION-I
SECTION-II
ALLEN
LTS/HS-2/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
8. Ans. (B)
Sol.
1/2
t/t
t0
A A (0.5) =
1/2
t
1
t
11
22
æö æö
=
ç÷ ç÷
èø èø
\ 
1/2
tt=
A
t
 = A
0
e
–lt
ln A
t
 = ln A
0
 – lt
1/2
ln2
t
t
 = lt = ln A
0
 – ln A
t
()
1/2 2
1/2
ln2
dt
t -
 
0t
0
dA dA
t
AA
=+
0t
0t 1/2
1/2
dA dA
55
AA dt 1.5
1000 500
t ln2 ln2 ln2
+
+
= ==
9. Ans. (C)
Sol. Initially
mg
kx
0
Vg r
Þ Vrg = kx
0
 + mg ... (i)
Finally
m(g + a)
0
k(x + x)
0
V (g + a ) r
0
Þ Vr(g + a
0
) = k(x
0
 + x) + m(g + a
0
) ... (ii)
(ii) – (i)
Vra
0
 = kx + ma
0
0
V a ma
x
k
r-
=
0
ma V
x1
km
r æö
=-
ç÷
èø
0
m
ma V
x1
kV
æö r
=-
ç÷
r
èø
0
ma1
x1
k
æö
=-
ç÷
g
èø
10. Ans. (C)
Sol. Size of object = 
12
I I 2 4.5 3mm = ´=
11. Ans. (C)
Sol.
0
e
L
9
L
=
 & L
0
 + L
e
 = 100
L
0
 = 90
L
e
 = 10
Now 
1 1 1 90
u 9cm
90 u 10 10
-
- = Þ = =-
-
Tube length = 9 + 90 = 99 cm
Change = 1cm
12. Ans. (A)
Sol.
2
n n1
mvkk
v
r r mr
-
= Þ=
n1
2r 2r
T
v
k
mr
-
pp
==
or 
n1
2
Tr
+
µ
13. Ans. (D)
Sol. DQ
AB
 = 9 × 10
4
 J
DQ
BC
 = 0
55
CD
5 110 1 1102
Q nR
2 nR nR
éù ´´ ´´
D=-
êú
ëû
= –2.5 × 10
5
 J
55
DA
3 2.4 101 1101
Q nR
2 nR nR
éù ´ ´ ´´
D=-
êú
ëû
= 2.1 × 10
5
 J
W = SDQ = 5 × 10
4
 J
ALLEN
29032020 LTS/HS-3/8
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
14. Ans. (C)
Sol.
max
H
tan37
R/2
°=
22
2
3 u sin / 2g
4 u sin cos /g
q
=
qq
3
tan
2
Þ q=
15. Ans. (C)
Sol.
mv
R
qB
=
d = R cos 53°
Þ
53°
d
C
53°
v
v
Þ q = 90°
16. Ans. (C)
Sol. Top view of cylinder.
Rdq
dq
q
q
q
q
F
Change in momentum of one photon
2h
cos =q
l
No. of photons striking per second on small
area = N (Rdq cos q) × h
Þ Total change in momentum per second
striking small area
= (N(Rdq)cos q × h) 
2h
cosq
l
( )
2
2h
N Rhcosd
æö
= qq
ç÷
l
èø
Force applied = F cosq
3
2h
N Rhcosd
æö
= qq
ç÷
l
èø
3
2I
dF N Rh cos d
C
æö
= qq
ç÷
èø
òò
8IhR
3C
=
17. Ans. (A)
Sol. V = u + at
2
5
a m/s
4
Þ=
5
vt
4
Þ=
5
F ma
4
==
25
PFvt
16
Þ = ×=
18. Ans. (D)
Sol. Unit of µ
0
 = ohm sec/meter
C
2
 = 
0 0
1
Î m
Þ Î
0
 = 
0
2
C
1
m
C = velocity of light
x 
0
0
Î
m
 = 
2 2
0
C m = µ
0
C
Unit of C = metre/sec.
unit of 
0
0
Î
m
 = unit of µ
0
C = ohm sec/meter
× meter/sec = Ohm.
ALLEN
LTS/HS-4/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
PART–2 : CHEMISTRY SOLUTION
SECTION-I
1. Ans. (C)
2. Ans. (C)
3. Ans. (A)
4. Ans. (B)
5. Ans. (B)
6. Ans. (D)
7. Ans. (B)
8. Ans. (D)
9. Ans. (B)
10. Ans. (C)
11. Ans. (B)
12. Ans. (B)
13. Ans. (A)
19. Ans. (B)
Sol. Check truth table for corresponding
option.
20. Ans. (A)
Sol. v = f l
f = 10
5
 Hz
Þ Velocity of sound w.r.to ground
= 1447 m/s
5
app
1447
f 10
1447 10
Þ=´
-
= 100695.9 Hz
SECTION-II
1. Ans. 0.18
Sol.
10
3
D 6563 10 1m
d 3.6 10
-
-
l ´´
b==
´
b = 0.0182 cm
2. Ans. 697.00
Sol.
b
b
T mL
S nC n 697J/K
TT
D= += l
3. Ans. 200.00
Sol.
µN
N
60
100=Mg
80
F
µN = 60 Þ N = 120
Þ F = 200
4. Ans. 4.47
Sol.
V VV
ˆ ˆˆ
E i jk
x yz
¶¶¶ æö
=- ++
ç÷
¶ ¶¶
èø
( )
ˆ ˆˆ
2xi 2yj 2zk =- ++
( )
ˆ ˆˆ
E 2i 0j 4k =- ++
r
E 20 2 5 4.47 = ==
5. Ans. 1316.00
Sol.
2
mv
N mg
R
+=
( )
2
70 120
N 700
500
´
+=
N = 1316
14. Ans. (A)
15. Ans. (D)
16. Ans. (C)
17. Ans. (B)
18. Ans. (C)
19. Ans. (C)
20. Ans. (A)
SECTION-II
1. Ans. (324.00)
2. Ans. (–115.80)
3. Ans. (2.25)
4. Ans. (4.00)
5. Ans. (4.00)
Page 5


PART-1 : PHYSICS SOLUTION
HS-1/8
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575     info@allen.ac.in   www.allen.ac.in
SECTION-I
1. Ans. (A)
Sol. Let charge on 4 µF capacitor when switch
in position A is Q
Charge on 4µF capacitor after switch is
shifted to position B.
4 2Q
Q'Q
423
==
+
2Q/32
ratio
Q3
==
2. Ans. (C)
Sol. Velocity after collision will be
= 
ˆ ˆˆ
3i 10 j 10 k -+
Velocity after 1 second = 
ˆˆ
3i 10j -
3. Ans. (A)
Sol.
45
I 0.5A
90
==
4. Ans. (A)
Sol. Load on concrete = 
Y20A2
Y 20A 10Y A 3
×
=
× +×
5. Ans. (A)
Sol.
i ii
tan tan 3 60 q=mÞ q= Þq= °
also 1 sin 60° = 
3
 sin r Þ r = 30°
6. Ans. (A)
Sol. Initial current = 
21
A
10 20 15
=
+
Final current = 
21
A
20 10
=
7. Ans. (D)
Sol.
2
12
22 0
0
01
rms0
V
V t dt dt
7 2
VV
2 24
æö
+
ç÷
èø
==
òò
(0000CJA102119031) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Main) : LEADER TEST SERIES / JOINT PACKAGE COURSE
JEE(Main)
AIOT
29-03-2020
ANSWER KEY
PART-1 : PHYSICS
PART -3 : MATHEMA TICS
PART-2 : CHEMISTRY
Q. 1 2 3 4 5 6 7 8 9 10
A. A C A A A A D B C C
Q. 11 12 13 14 15 16 17 18 19 20
A. C A D C C C A D B A
Q. 1 2 3 4 5
A. 0.18 697.00 200.00 4.47 1316.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D B D B C
Q. 11 12 13 14 15 16 17 18 19 20
A. B B A A D C B C C A
Q. 1 2 3 4 5
A. 324.00 –115.80 2.25 4.00 4.00
SECTION-I
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10
A. C B D B A B B C D B
Q. 11 12 13 14 15 16 17 18 19 20
A. A C D D B D C A D A
Q. 1 2 3 4 5
A. 7.00 1.00 –0.50 165.00 78.00
SECTION-I
SECTION-II
ALLEN
LTS/HS-2/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
8. Ans. (B)
Sol.
1/2
t/t
t0
A A (0.5) =
1/2
t
1
t
11
22
æö æö
=
ç÷ ç÷
èø èø
\ 
1/2
tt=
A
t
 = A
0
e
–lt
ln A
t
 = ln A
0
 – lt
1/2
ln2
t
t
 = lt = ln A
0
 – ln A
t
()
1/2 2
1/2
ln2
dt
t -
 
0t
0
dA dA
t
AA
=+
0t
0t 1/2
1/2
dA dA
55
AA dt 1.5
1000 500
t ln2 ln2 ln2
+
+
= ==
9. Ans. (C)
Sol. Initially
mg
kx
0
Vg r
Þ Vrg = kx
0
 + mg ... (i)
Finally
m(g + a)
0
k(x + x)
0
V (g + a ) r
0
Þ Vr(g + a
0
) = k(x
0
 + x) + m(g + a
0
) ... (ii)
(ii) – (i)
Vra
0
 = kx + ma
0
0
V a ma
x
k
r-
=
0
ma V
x1
km
r æö
=-
ç÷
èø
0
m
ma V
x1
kV
æö r
=-
ç÷
r
èø
0
ma1
x1
k
æö
=-
ç÷
g
èø
10. Ans. (C)
Sol. Size of object = 
12
I I 2 4.5 3mm = ´=
11. Ans. (C)
Sol.
0
e
L
9
L
=
 & L
0
 + L
e
 = 100
L
0
 = 90
L
e
 = 10
Now 
1 1 1 90
u 9cm
90 u 10 10
-
- = Þ = =-
-
Tube length = 9 + 90 = 99 cm
Change = 1cm
12. Ans. (A)
Sol.
2
n n1
mvkk
v
r r mr
-
= Þ=
n1
2r 2r
T
v
k
mr
-
pp
==
or 
n1
2
Tr
+
µ
13. Ans. (D)
Sol. DQ
AB
 = 9 × 10
4
 J
DQ
BC
 = 0
55
CD
5 110 1 1102
Q nR
2 nR nR
éù ´´ ´´
D=-
êú
ëû
= –2.5 × 10
5
 J
55
DA
3 2.4 101 1101
Q nR
2 nR nR
éù ´ ´ ´´
D=-
êú
ëû
= 2.1 × 10
5
 J
W = SDQ = 5 × 10
4
 J
ALLEN
29032020 LTS/HS-3/8
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
14. Ans. (C)
Sol.
max
H
tan37
R/2
°=
22
2
3 u sin / 2g
4 u sin cos /g
q
=
qq
3
tan
2
Þ q=
15. Ans. (C)
Sol.
mv
R
qB
=
d = R cos 53°
Þ
53°
d
C
53°
v
v
Þ q = 90°
16. Ans. (C)
Sol. Top view of cylinder.
Rdq
dq
q
q
q
q
F
Change in momentum of one photon
2h
cos =q
l
No. of photons striking per second on small
area = N (Rdq cos q) × h
Þ Total change in momentum per second
striking small area
= (N(Rdq)cos q × h) 
2h
cosq
l
( )
2
2h
N Rhcosd
æö
= qq
ç÷
l
èø
Force applied = F cosq
3
2h
N Rhcosd
æö
= qq
ç÷
l
èø
3
2I
dF N Rh cos d
C
æö
= qq
ç÷
èø
òò
8IhR
3C
=
17. Ans. (A)
Sol. V = u + at
2
5
a m/s
4
Þ=
5
vt
4
Þ=
5
F ma
4
==
25
PFvt
16
Þ = ×=
18. Ans. (D)
Sol. Unit of µ
0
 = ohm sec/meter
C
2
 = 
0 0
1
Î m
Þ Î
0
 = 
0
2
C
1
m
C = velocity of light
x 
0
0
Î
m
 = 
2 2
0
C m = µ
0
C
Unit of C = metre/sec.
unit of 
0
0
Î
m
 = unit of µ
0
C = ohm sec/meter
× meter/sec = Ohm.
ALLEN
LTS/HS-4/8 29032020
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
PART–2 : CHEMISTRY SOLUTION
SECTION-I
1. Ans. (C)
2. Ans. (C)
3. Ans. (A)
4. Ans. (B)
5. Ans. (B)
6. Ans. (D)
7. Ans. (B)
8. Ans. (D)
9. Ans. (B)
10. Ans. (C)
11. Ans. (B)
12. Ans. (B)
13. Ans. (A)
19. Ans. (B)
Sol. Check truth table for corresponding
option.
20. Ans. (A)
Sol. v = f l
f = 10
5
 Hz
Þ Velocity of sound w.r.to ground
= 1447 m/s
5
app
1447
f 10
1447 10
Þ=´
-
= 100695.9 Hz
SECTION-II
1. Ans. 0.18
Sol.
10
3
D 6563 10 1m
d 3.6 10
-
-
l ´´
b==
´
b = 0.0182 cm
2. Ans. 697.00
Sol.
b
b
T mL
S nC n 697J/K
TT
D= += l
3. Ans. 200.00
Sol.
µN
N
60
100=Mg
80
F
µN = 60 Þ N = 120
Þ F = 200
4. Ans. 4.47
Sol.
V VV
ˆ ˆˆ
E i jk
x yz
¶¶¶ æö
=- ++
ç÷
¶ ¶¶
èø
( )
ˆ ˆˆ
2xi 2yj 2zk =- ++
( )
ˆ ˆˆ
E 2i 0j 4k =- ++
r
E 20 2 5 4.47 = ==
5. Ans. 1316.00
Sol.
2
mv
N mg
R
+=
( )
2
70 120
N 700
500
´
+=
N = 1316
14. Ans. (A)
15. Ans. (D)
16. Ans. (C)
17. Ans. (B)
18. Ans. (C)
19. Ans. (C)
20. Ans. (A)
SECTION-II
1. Ans. (324.00)
2. Ans. (–115.80)
3. Ans. (2.25)
4. Ans. (4.00)
5. Ans. (4.00)
ALLEN
29032020 LTS/HS-5/8
All India Open Test/Leader Test Series/Joint Package Course/JEE(Main) 0000CJA102119031
PART-3 : MATHEMATICS SOLUTION
SECTION–I
1. Ans. (C)
3sin
2
x – 7sinx + 2 < 0
Þ<£
1
sinx1
3
Þ sinx = 1
p pp
Þ=
59
x ,,
2 22
2. Ans. (B)
( )
éù = +-
ëû
n
270 2a n 1 2
2
Þ 270 = n(a +n – 1)
Þ 2 × 3
3
 × 5 = n(a +n – 1)
Number of possible value of n
= (1 + 1) (3 + 1) (1 + 1) – 1 – 1
= 14
3. Ans. (D)
n(n
2
 – 1) = (n – 1)n(n + 1)
put n = 2m + 1, m Î N
n(n
2
 – 1) = 4 m(2m + 1) (m + 1)
P(m) : m(2m + 1) (m + 1) is divisible by 6
P(1) is true
suppose P(k) is true, k Î N
Þ P(k + 1) is true
Þ P(m) is true for all m Î N
4. Ans. (B)
11
24
tan tan
1 4 2 1 84
--
q= ++
+ ´ +´
11
8 16
tan tan ...
1 16 8 1 32 16
--
++
+ ´ +´
11
4 2 84
tan tan
14 2 18 4
--
-- æ ö æö
= ++
ç÷ ç÷
+ ´ +´ èø èø
11
16 8 32 16
tan tan ...
1 16 8 1 32 16
--
-- æ ö æö
++
ç ÷ ç÷
+ ´ +´ è ø èø
= (tan
–1
4 – tan
–1
2) + (tan
–1
8 – tan
–1
4) +
(tan
–1
16 – tan
–1
8) + (tan
–1
32 – tan
–1
16) + ... to ¥
1
tan2
2
-
p
=-
( )
11
cot tan 2 cot cot 2 2
2
--
pæö
Þ - ==
ç÷
èø
5. Ans. (A)
D = N
2
 + 4 × 444N
= N(N + 4 × 444)
must be a perfect square, possible value of
N is 37 only
6. Ans. (B)
( ) ( ) ( ) ( ) Ù Ú Ù º Ú ÙÙ ~pq p ~q pq ~pq
( ) ( ) ( ) ( ) Þ ÙÚ Ù ÚÚ º Ù ~pq p ~q pq pq
( ) ( ) ( ) ( )
Þ Ù Ú Ù Ú Ú ºÙ ~p q p ~q p q ~p ~q
7. Ans. (B)
Co-ordinates of B & C 
A(2,3)
B C
are (0, 5) and (–6, –13)
equation of BC
Þ y = 3x + 5