JEE Test Paper 2 solution IT & Software Notes | EduRev

IT & Software : JEE Test Paper 2 solution IT & Software Notes | EduRev

 Page 1


HS-1/7
Corporate Office : AL L EN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
SOLUTION
PART-1 : PHYSICS ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B C B B A,C C,D A,B,C C,D A B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6
A. 8 2 4 4 1 3
A B C D A B C D
Q R,S R,S R,S R T P T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
(0999DJA110319001) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Advanced)
MINOR TEST # 01
07-07-2019
JEE(Main + Advanced) : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit Test # 01
SECTION-I
1. Ans. (B)
Sol.
() 31
180
p æö
´
ç÷
èø
2. Ans. (C)
3. Ans. (B)
Sol. In new system,
Force unit = 
( )( )
( )
2
10kg 5m
1sec
 50 kgm/s
2
 = 50 N
4. Ans. (B)
Sol. ( )
a b 122
LML MLTC
-
=
a + c = 0, b – 2c = 0
2c = 1 
11
c ,a
22
Þ = =-
b = 1
5. Ans. (A, C)
Sol. W = F.r
r
r
36 = 3C + 8 + C
2
solving we get, C = 4, –7
6. Ans. (C, D)
7. Ans. (A,B,C)
Sol. Mass = 
Force
Acceleration
 = 
61
244
=
,
Time = 
Velocity
Acceleration
 = 
41
246
=
Energy = (mass) (velocity)
2
 = 
()
2 1
4
4
æö
ç÷
èø
          = 4 joule,
Momentum = force × time = 6 × 
1
6
 = 1Ns
Power = 
Energy
time
 = 
4
1/6
 = 24 watt
8. Ans. (C,D)
Sol. From a to b E = 
2
kq
x
 3q
b
–q
q
a
c
v
in
 = 
k 3q kq kq
0
c ba
´
- +¹
E
inside
 conductor = 0
E
out
 = 
2
3kq
r
9. Ans. (A)
Sol. use polygon law
10. Ans. (B)
Sol. use polygon law
Page 2


HS-1/7
Corporate Office : AL L EN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
SOLUTION
PART-1 : PHYSICS ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B C B B A,C C,D A,B,C C,D A B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6
A. 8 2 4 4 1 3
A B C D A B C D
Q R,S R,S R,S R T P T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
(0999DJA110319001) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Advanced)
MINOR TEST # 01
07-07-2019
JEE(Main + Advanced) : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit Test # 01
SECTION-I
1. Ans. (B)
Sol.
() 31
180
p æö
´
ç÷
èø
2. Ans. (C)
3. Ans. (B)
Sol. In new system,
Force unit = 
( )( )
( )
2
10kg 5m
1sec
 50 kgm/s
2
 = 50 N
4. Ans. (B)
Sol. ( )
a b 122
LML MLTC
-
=
a + c = 0, b – 2c = 0
2c = 1 
11
c ,a
22
Þ = =-
b = 1
5. Ans. (A, C)
Sol. W = F.r
r
r
36 = 3C + 8 + C
2
solving we get, C = 4, –7
6. Ans. (C, D)
7. Ans. (A,B,C)
Sol. Mass = 
Force
Acceleration
 = 
61
244
=
,
Time = 
Velocity
Acceleration
 = 
41
246
=
Energy = (mass) (velocity)
2
 = 
()
2 1
4
4
æö
ç÷
èø
          = 4 joule,
Momentum = force × time = 6 × 
1
6
 = 1Ns
Power = 
Energy
time
 = 
4
1/6
 = 24 watt
8. Ans. (C,D)
Sol. From a to b E = 
2
kq
x
 3q
b
–q
q
a
c
v
in
 = 
k 3q kq kq
0
c ba
´
- +¹
E
inside
 conductor = 0
E
out
 = 
2
3kq
r
9. Ans. (A)
Sol. use polygon law
10. Ans. (B)
Sol. use polygon law
LTS-2/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
11. Ans. (C)
Sol. For 30 C charge, angle Î (5°, 9°) Þ 7°
12. Ans. (C)
Sol. In (iii) most of positive charge run away to
the metal knob.
So due to less charge on the leaves, the
leaves will come closer than before.
SECTION-III
1. Ans. 8
Sol.
i
ˆˆ
r 1i 2j =+
r
, 
1
ˆ ˆˆ
2i j 2k
ˆv
3
++
=
( ) ( )
fi11
ˆˆ r r v9 3 v6 4 =+ ´-´
rr
2. Ans. 2
Sol. Energy = M
1
L
2
T
–2
Now, according to questions
       [ ]
22
111
2 kg M sec
21
-
é ùéù
Þ
ê úêú
ë ûëû
1 22
1
2 kg m sec
4
-
Þ´
122
1
kg m sec
2
-
Þ
3. Ans. 4
Sol. Q = net charge in sphere of radius 3R
Q = 
3
4
R
3
æö
rp
ç÷
èø
 –  
() ( )
rp-
3
3
4
2 2RR
3
                            
() () ( )
+rp-
33 4
3 3R 2R
3
then E = 
()
2
kQ
3R
4. Ans. 4
Sol. ()
a
b
K.E.
kdv
Volume
=
( )( )
22
ab
31
3
MLT
ML LT
L
-
--
=
\ a = 1, b = 2
\ when velocity is doubled,
kinetic energy/volume becomes 4 times
5. Ans. 1
Sol. qE = kx Þ x = 10
–5
 m
qEx – 
22
11
kx mv
22
=
Þ v = 10
–3
 m/s = 1 mm/s
6. Ans. 3
SECTION-IV
1. Ans. (A) ® (Q); (B) ® (R,S); (C) ® (R,S);
(D) ® (R,S)
2. Ans. (A) ® (R); (B) ® (T); (C) ® (P);
(D) ® (T)
Sol. For (A) : Q[Acceleration] = LT
–2
 \ unit of
acceleration = (1km) (1min)
–2
 = 
5
18
m/s
2
For (B) : Q[Kinetic energy] = ML
2
T
–2
\ unit of kinetic energy
             = (1 Quintal) (1km)
2
(1min)
–2
              = 
5
18
 × 10
5
 kgm
2
s
–2
For (C) : Q[Pressure] = ML
–1
T
–2
   \ unit of pressure = (1 Quintal)
(1km)
–1
(1min)
–2
 = 
5
18
 × 10
–4
 kgm
–1
s
–2
For (D) : Q[Work] = ML
2
T
–2
   \ unit of work = (1 Quintal)
(1km)
2
(1min)
–2
 = 
5
18
 × 10
5
 kgm
2
s
–2
Page 3


HS-1/7
Corporate Office : AL L EN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
SOLUTION
PART-1 : PHYSICS ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B C B B A,C C,D A,B,C C,D A B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6
A. 8 2 4 4 1 3
A B C D A B C D
Q R,S R,S R,S R T P T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
(0999DJA110319001) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Advanced)
MINOR TEST # 01
07-07-2019
JEE(Main + Advanced) : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit Test # 01
SECTION-I
1. Ans. (B)
Sol.
() 31
180
p æö
´
ç÷
èø
2. Ans. (C)
3. Ans. (B)
Sol. In new system,
Force unit = 
( )( )
( )
2
10kg 5m
1sec
 50 kgm/s
2
 = 50 N
4. Ans. (B)
Sol. ( )
a b 122
LML MLTC
-
=
a + c = 0, b – 2c = 0
2c = 1 
11
c ,a
22
Þ = =-
b = 1
5. Ans. (A, C)
Sol. W = F.r
r
r
36 = 3C + 8 + C
2
solving we get, C = 4, –7
6. Ans. (C, D)
7. Ans. (A,B,C)
Sol. Mass = 
Force
Acceleration
 = 
61
244
=
,
Time = 
Velocity
Acceleration
 = 
41
246
=
Energy = (mass) (velocity)
2
 = 
()
2 1
4
4
æö
ç÷
èø
          = 4 joule,
Momentum = force × time = 6 × 
1
6
 = 1Ns
Power = 
Energy
time
 = 
4
1/6
 = 24 watt
8. Ans. (C,D)
Sol. From a to b E = 
2
kq
x
 3q
b
–q
q
a
c
v
in
 = 
k 3q kq kq
0
c ba
´
- +¹
E
inside
 conductor = 0
E
out
 = 
2
3kq
r
9. Ans. (A)
Sol. use polygon law
10. Ans. (B)
Sol. use polygon law
LTS-2/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
11. Ans. (C)
Sol. For 30 C charge, angle Î (5°, 9°) Þ 7°
12. Ans. (C)
Sol. In (iii) most of positive charge run away to
the metal knob.
So due to less charge on the leaves, the
leaves will come closer than before.
SECTION-III
1. Ans. 8
Sol.
i
ˆˆ
r 1i 2j =+
r
, 
1
ˆ ˆˆ
2i j 2k
ˆv
3
++
=
( ) ( )
fi11
ˆˆ r r v9 3 v6 4 =+ ´-´
rr
2. Ans. 2
Sol. Energy = M
1
L
2
T
–2
Now, according to questions
       [ ]
22
111
2 kg M sec
21
-
é ùéù
Þ
ê úêú
ë ûëû
1 22
1
2 kg m sec
4
-
Þ´
122
1
kg m sec
2
-
Þ
3. Ans. 4
Sol. Q = net charge in sphere of radius 3R
Q = 
3
4
R
3
æö
rp
ç÷
èø
 –  
() ( )
rp-
3
3
4
2 2RR
3
                            
() () ( )
+rp-
33 4
3 3R 2R
3
then E = 
()
2
kQ
3R
4. Ans. 4
Sol. ()
a
b
K.E.
kdv
Volume
=
( )( )
22
ab
31
3
MLT
ML LT
L
-
--
=
\ a = 1, b = 2
\ when velocity is doubled,
kinetic energy/volume becomes 4 times
5. Ans. 1
Sol. qE = kx Þ x = 10
–5
 m
qEx – 
22
11
kx mv
22
=
Þ v = 10
–3
 m/s = 1 mm/s
6. Ans. 3
SECTION-IV
1. Ans. (A) ® (Q); (B) ® (R,S); (C) ® (R,S);
(D) ® (R,S)
2. Ans. (A) ® (R); (B) ® (T); (C) ® (P);
(D) ® (T)
Sol. For (A) : Q[Acceleration] = LT
–2
 \ unit of
acceleration = (1km) (1min)
–2
 = 
5
18
m/s
2
For (B) : Q[Kinetic energy] = ML
2
T
–2
\ unit of kinetic energy
             = (1 Quintal) (1km)
2
(1min)
–2
              = 
5
18
 × 10
5
 kgm
2
s
–2
For (C) : Q[Pressure] = ML
–1
T
–2
   \ unit of pressure = (1 Quintal)
(1km)
–1
(1min)
–2
 = 
5
18
 × 10
–4
 kgm
–1
s
–2
For (D) : Q[Work] = ML
2
T
–2
   \ unit of work = (1 Quintal)
(1km)
2
(1min)
–2
 = 
5
18
 × 10
5
 kgm
2
s
–2
LTS-3/7 0999DJA110319001
Leader Test Series/Joint Package Course/07-07-2019
ALLEN
SOLUTION
PART-2 : CHEMISTRY ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B D B D A,B,C A,B,D A,C,D A,B,D D C
Q. 11 12
A. B A
Q. 1 2 3 4 5 6
A. 1 8 7 7 3 8
A B C D A B C D
P,R P,R Q,S,T Q,S P,Q P,S P,Q,R P,S,T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
SECTION-I
1. Ans. (B)
CaC
2
    + H
2
O ® CaO + C
2
H
2
3
20 10
312.5
64
´
=      312.5
C
2
H
2
 + H
2
  ®  C
2
H
4
312.5      312.5
n(C
2
H
4
)      (C
2
H
4
)
n
mass of C
2
H
4
 = 312.5 × 28
= mass of polyethene = 8.75 kg
2. Ans. (D)
22
15 11 11
R RR
36 49 23
æö æö
= = =´ - -
ç÷ ç÷
l èø èø
2
1R 11
R
4 2
2
æö
== -
ç÷
l è ¥ø
l - l' = 
H
36 41 36
3.2/R 4
5R RR 5
éù
-==-
êú
ëû
3. Ans. (B)
4. Ans. (D)
5. Ans. (A,B,C)
          % of C by mass = 
massofC
100
Total mass of compound
´
6. Ans.(A,B,D)
7. Ans.(A,C,D)
Sol. In presence of Na metal N
2
 gas will be
eliminate from the above compounds.]
8. Ans.(A,B,D)
Sol. (A) 
OH
2 5
4 3
Cl
Me
      3-Chloro-4-methyl cyclopentanol-1
(B) 
C ? C ? C ? C ?  ? C C ? NH
2
Br
O
6 5 4 3 2
       3-bromo hexanamide
(C) 
1
COOH
Cl
2
3
4
4-Chloro-3-methylcyclohexane
carboxylic acid
(D) C ? C ? C ? C ? C ? O H ? C 
Br
6 5 4 3 2
C
7
1
      4-Bromoheptan-2-ol
9. Ans. (D)
10. Ans. (C)
11. Ans. (B)
12. Ans. (A)
SECTION-III
1. Ans. (1)
Sol. l = 
h
3mkT
 , l µ m
–1/2 
 T
–1/2
Page 4


HS-1/7
Corporate Office : AL L EN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
SOLUTION
PART-1 : PHYSICS ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B C B B A,C C,D A,B,C C,D A B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6
A. 8 2 4 4 1 3
A B C D A B C D
Q R,S R,S R,S R T P T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
(0999DJA110319001) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Advanced)
MINOR TEST # 01
07-07-2019
JEE(Main + Advanced) : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit Test # 01
SECTION-I
1. Ans. (B)
Sol.
() 31
180
p æö
´
ç÷
èø
2. Ans. (C)
3. Ans. (B)
Sol. In new system,
Force unit = 
( )( )
( )
2
10kg 5m
1sec
 50 kgm/s
2
 = 50 N
4. Ans. (B)
Sol. ( )
a b 122
LML MLTC
-
=
a + c = 0, b – 2c = 0
2c = 1 
11
c ,a
22
Þ = =-
b = 1
5. Ans. (A, C)
Sol. W = F.r
r
r
36 = 3C + 8 + C
2
solving we get, C = 4, –7
6. Ans. (C, D)
7. Ans. (A,B,C)
Sol. Mass = 
Force
Acceleration
 = 
61
244
=
,
Time = 
Velocity
Acceleration
 = 
41
246
=
Energy = (mass) (velocity)
2
 = 
()
2 1
4
4
æö
ç÷
èø
          = 4 joule,
Momentum = force × time = 6 × 
1
6
 = 1Ns
Power = 
Energy
time
 = 
4
1/6
 = 24 watt
8. Ans. (C,D)
Sol. From a to b E = 
2
kq
x
 3q
b
–q
q
a
c
v
in
 = 
k 3q kq kq
0
c ba
´
- +¹
E
inside
 conductor = 0
E
out
 = 
2
3kq
r
9. Ans. (A)
Sol. use polygon law
10. Ans. (B)
Sol. use polygon law
LTS-2/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
11. Ans. (C)
Sol. For 30 C charge, angle Î (5°, 9°) Þ 7°
12. Ans. (C)
Sol. In (iii) most of positive charge run away to
the metal knob.
So due to less charge on the leaves, the
leaves will come closer than before.
SECTION-III
1. Ans. 8
Sol.
i
ˆˆ
r 1i 2j =+
r
, 
1
ˆ ˆˆ
2i j 2k
ˆv
3
++
=
( ) ( )
fi11
ˆˆ r r v9 3 v6 4 =+ ´-´
rr
2. Ans. 2
Sol. Energy = M
1
L
2
T
–2
Now, according to questions
       [ ]
22
111
2 kg M sec
21
-
é ùéù
Þ
ê úêú
ë ûëû
1 22
1
2 kg m sec
4
-
Þ´
122
1
kg m sec
2
-
Þ
3. Ans. 4
Sol. Q = net charge in sphere of radius 3R
Q = 
3
4
R
3
æö
rp
ç÷
èø
 –  
() ( )
rp-
3
3
4
2 2RR
3
                            
() () ( )
+rp-
33 4
3 3R 2R
3
then E = 
()
2
kQ
3R
4. Ans. 4
Sol. ()
a
b
K.E.
kdv
Volume
=
( )( )
22
ab
31
3
MLT
ML LT
L
-
--
=
\ a = 1, b = 2
\ when velocity is doubled,
kinetic energy/volume becomes 4 times
5. Ans. 1
Sol. qE = kx Þ x = 10
–5
 m
qEx – 
22
11
kx mv
22
=
Þ v = 10
–3
 m/s = 1 mm/s
6. Ans. 3
SECTION-IV
1. Ans. (A) ® (Q); (B) ® (R,S); (C) ® (R,S);
(D) ® (R,S)
2. Ans. (A) ® (R); (B) ® (T); (C) ® (P);
(D) ® (T)
Sol. For (A) : Q[Acceleration] = LT
–2
 \ unit of
acceleration = (1km) (1min)
–2
 = 
5
18
m/s
2
For (B) : Q[Kinetic energy] = ML
2
T
–2
\ unit of kinetic energy
             = (1 Quintal) (1km)
2
(1min)
–2
              = 
5
18
 × 10
5
 kgm
2
s
–2
For (C) : Q[Pressure] = ML
–1
T
–2
   \ unit of pressure = (1 Quintal)
(1km)
–1
(1min)
–2
 = 
5
18
 × 10
–4
 kgm
–1
s
–2
For (D) : Q[Work] = ML
2
T
–2
   \ unit of work = (1 Quintal)
(1km)
2
(1min)
–2
 = 
5
18
 × 10
5
 kgm
2
s
–2
LTS-3/7 0999DJA110319001
Leader Test Series/Joint Package Course/07-07-2019
ALLEN
SOLUTION
PART-2 : CHEMISTRY ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B D B D A,B,C A,B,D A,C,D A,B,D D C
Q. 11 12
A. B A
Q. 1 2 3 4 5 6
A. 1 8 7 7 3 8
A B C D A B C D
P,R P,R Q,S,T Q,S P,Q P,S P,Q,R P,S,T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
SECTION-I
1. Ans. (B)
CaC
2
    + H
2
O ® CaO + C
2
H
2
3
20 10
312.5
64
´
=      312.5
C
2
H
2
 + H
2
  ®  C
2
H
4
312.5      312.5
n(C
2
H
4
)      (C
2
H
4
)
n
mass of C
2
H
4
 = 312.5 × 28
= mass of polyethene = 8.75 kg
2. Ans. (D)
22
15 11 11
R RR
36 49 23
æö æö
= = =´ - -
ç÷ ç÷
l èø èø
2
1R 11
R
4 2
2
æö
== -
ç÷
l è ¥ø
l - l' = 
H
36 41 36
3.2/R 4
5R RR 5
éù
-==-
êú
ëû
3. Ans. (B)
4. Ans. (D)
5. Ans. (A,B,C)
          % of C by mass = 
massofC
100
Total mass of compound
´
6. Ans.(A,B,D)
7. Ans.(A,C,D)
Sol. In presence of Na metal N
2
 gas will be
eliminate from the above compounds.]
8. Ans.(A,B,D)
Sol. (A) 
OH
2 5
4 3
Cl
Me
      3-Chloro-4-methyl cyclopentanol-1
(B) 
C ? C ? C ? C ?  ? C C ? NH
2
Br
O
6 5 4 3 2
       3-bromo hexanamide
(C) 
1
COOH
Cl
2
3
4
4-Chloro-3-methylcyclohexane
carboxylic acid
(D) C ? C ? C ? C ? C ? O H ? C 
Br
6 5 4 3 2
C
7
1
      4-Bromoheptan-2-ol
9. Ans. (D)
10. Ans. (C)
11. Ans. (B)
12. Ans. (A)
SECTION-III
1. Ans. (1)
Sol. l = 
h
3mkT
 , l µ m
–1/2 
 T
–1/2
LTS-4/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
PART -3 : MATHEMA TICS ANSWER KEY
SOLUTION
Q. 1 2 3 4 5 6 7 8 9 10
A. C C D A A,B B,C,D A,B,C A,B,C,D D A
Q. 11 12
A. B A
Q. 1 2 3 4 5 6
A. 6 1 4 1 9 4
A B C D A B C D
Q,R,S R R,S T T P Q,R Q,S
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
SECTION–I
1. Ans. (C)
Sol. a = log
2
x, b = log
4
x, g = log
6
x
abg = ab + bg + ga
1 11
10
æö
abg ++- =
ç÷
a bg
èø
Þ either a = 0 or b = 0 or g = 0 or
1 11
10 + + -=
a bg
2. Ans. (C)
Sol. We know that (1 + tan k°)(1 + tan(45 – k)°)
= 2
Þ (1 + tan1°)(1 + tan2°).......
      (1 + tan44°)(1 + tan45°)
= 2{(1 + tan1°)(1 + tan44°)}.{(1 + tan2°)
(1 + tan43°)}.......{(1 + tan22°)(1 + tan23°)}
= 2.2
22
 = 2
23
\ n = 23
3. Ans. (D)
Sol. ByC
1
 ® C
1
 – (C
2
 + C
3
)
g(x) = 0   Þ g(0) + .... + g(10) = 0
2. Ans.(8)
3. Ans.(7)
Let assume 
34
Cl has x gram and 
38
Cl has
(7– x) gram
moles of 
34
Cl = x/34 and moles of 
38
Cl
= (7 – x) / 38
M
avg.
 = 
7
(x/34) (7 x)/38 +-
put the value of M
avg.
 find the value of x, so
x = 5.1 g
moles of 
34
Cl = 5.1/34 = 0.15
and moles of 
38
Cl = (7 – 5.1)/38 =.05
n = 0.15 × 17 + 0.05 × 21 = 3.6
p = 0.15 × 17 + 0.05 × 17 = 3.4
\ p + n = 7
4. Ans. (7)
5. Ans. (3)
6. Ans. (8)
SECTION-IV
1. Ans. (A)-(P,R); (B)-(P,R); (C)-(Q,S,T);
(D)–(S,Q)
2. Ans. (A)-(P,Q); (B)-(P,S); (C)-(P,Q,R);
(D)-(P,S,T)
Sol.
OH
O
H
Acidic hydrogen
DU = 4
DU = 3  
O O
OH
O
HO
HO
ester
H
Acidic
hydrogen
DU = 3  
N
Can act as base
:
Page 5


HS-1/7
Corporate Office : AL L EN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005
  +91-744-2757575      dlp@allen.ac.in   www.allen.ac.in
SOLUTION
PART-1 : PHYSICS ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B C B B A,C C,D A,B,C C,D A B
Q. 11 12
A. C C
Q. 1 2 3 4 5 6
A. 8 2 4 4 1 3
A B C D A B C D
Q R,S R,S R,S R T P T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
(0999DJA110319001) Test Pattern
DISTANCE LEARNING PROGRAMME
(Academic Session : 2019 - 2020)
JEE(Advanced)
MINOR TEST # 01
07-07-2019
JEE(Main + Advanced) : LEADER TEST SERIES / JOINT PACKAGE COURSE
Test Type : Unit Test # 01
SECTION-I
1. Ans. (B)
Sol.
() 31
180
p æö
´
ç÷
èø
2. Ans. (C)
3. Ans. (B)
Sol. In new system,
Force unit = 
( )( )
( )
2
10kg 5m
1sec
 50 kgm/s
2
 = 50 N
4. Ans. (B)
Sol. ( )
a b 122
LML MLTC
-
=
a + c = 0, b – 2c = 0
2c = 1 
11
c ,a
22
Þ = =-
b = 1
5. Ans. (A, C)
Sol. W = F.r
r
r
36 = 3C + 8 + C
2
solving we get, C = 4, –7
6. Ans. (C, D)
7. Ans. (A,B,C)
Sol. Mass = 
Force
Acceleration
 = 
61
244
=
,
Time = 
Velocity
Acceleration
 = 
41
246
=
Energy = (mass) (velocity)
2
 = 
()
2 1
4
4
æö
ç÷
èø
          = 4 joule,
Momentum = force × time = 6 × 
1
6
 = 1Ns
Power = 
Energy
time
 = 
4
1/6
 = 24 watt
8. Ans. (C,D)
Sol. From a to b E = 
2
kq
x
 3q
b
–q
q
a
c
v
in
 = 
k 3q kq kq
0
c ba
´
- +¹
E
inside
 conductor = 0
E
out
 = 
2
3kq
r
9. Ans. (A)
Sol. use polygon law
10. Ans. (B)
Sol. use polygon law
LTS-2/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
11. Ans. (C)
Sol. For 30 C charge, angle Î (5°, 9°) Þ 7°
12. Ans. (C)
Sol. In (iii) most of positive charge run away to
the metal knob.
So due to less charge on the leaves, the
leaves will come closer than before.
SECTION-III
1. Ans. 8
Sol.
i
ˆˆ
r 1i 2j =+
r
, 
1
ˆ ˆˆ
2i j 2k
ˆv
3
++
=
( ) ( )
fi11
ˆˆ r r v9 3 v6 4 =+ ´-´
rr
2. Ans. 2
Sol. Energy = M
1
L
2
T
–2
Now, according to questions
       [ ]
22
111
2 kg M sec
21
-
é ùéù
Þ
ê úêú
ë ûëû
1 22
1
2 kg m sec
4
-
Þ´
122
1
kg m sec
2
-
Þ
3. Ans. 4
Sol. Q = net charge in sphere of radius 3R
Q = 
3
4
R
3
æö
rp
ç÷
èø
 –  
() ( )
rp-
3
3
4
2 2RR
3
                            
() () ( )
+rp-
33 4
3 3R 2R
3
then E = 
()
2
kQ
3R
4. Ans. 4
Sol. ()
a
b
K.E.
kdv
Volume
=
( )( )
22
ab
31
3
MLT
ML LT
L
-
--
=
\ a = 1, b = 2
\ when velocity is doubled,
kinetic energy/volume becomes 4 times
5. Ans. 1
Sol. qE = kx Þ x = 10
–5
 m
qEx – 
22
11
kx mv
22
=
Þ v = 10
–3
 m/s = 1 mm/s
6. Ans. 3
SECTION-IV
1. Ans. (A) ® (Q); (B) ® (R,S); (C) ® (R,S);
(D) ® (R,S)
2. Ans. (A) ® (R); (B) ® (T); (C) ® (P);
(D) ® (T)
Sol. For (A) : Q[Acceleration] = LT
–2
 \ unit of
acceleration = (1km) (1min)
–2
 = 
5
18
m/s
2
For (B) : Q[Kinetic energy] = ML
2
T
–2
\ unit of kinetic energy
             = (1 Quintal) (1km)
2
(1min)
–2
              = 
5
18
 × 10
5
 kgm
2
s
–2
For (C) : Q[Pressure] = ML
–1
T
–2
   \ unit of pressure = (1 Quintal)
(1km)
–1
(1min)
–2
 = 
5
18
 × 10
–4
 kgm
–1
s
–2
For (D) : Q[Work] = ML
2
T
–2
   \ unit of work = (1 Quintal)
(1km)
2
(1min)
–2
 = 
5
18
 × 10
5
 kgm
2
s
–2
LTS-3/7 0999DJA110319001
Leader Test Series/Joint Package Course/07-07-2019
ALLEN
SOLUTION
PART-2 : CHEMISTRY ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B D B D A,B,C A,B,D A,C,D A,B,D D C
Q. 11 12
A. B A
Q. 1 2 3 4 5 6
A. 1 8 7 7 3 8
A B C D A B C D
P,R P,R Q,S,T Q,S P,Q P,S P,Q,R P,S,T
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
SECTION-I
1. Ans. (B)
CaC
2
    + H
2
O ® CaO + C
2
H
2
3
20 10
312.5
64
´
=      312.5
C
2
H
2
 + H
2
  ®  C
2
H
4
312.5      312.5
n(C
2
H
4
)      (C
2
H
4
)
n
mass of C
2
H
4
 = 312.5 × 28
= mass of polyethene = 8.75 kg
2. Ans. (D)
22
15 11 11
R RR
36 49 23
æö æö
= = =´ - -
ç÷ ç÷
l èø èø
2
1R 11
R
4 2
2
æö
== -
ç÷
l è ¥ø
l - l' = 
H
36 41 36
3.2/R 4
5R RR 5
éù
-==-
êú
ëû
3. Ans. (B)
4. Ans. (D)
5. Ans. (A,B,C)
          % of C by mass = 
massofC
100
Total mass of compound
´
6. Ans.(A,B,D)
7. Ans.(A,C,D)
Sol. In presence of Na metal N
2
 gas will be
eliminate from the above compounds.]
8. Ans.(A,B,D)
Sol. (A) 
OH
2 5
4 3
Cl
Me
      3-Chloro-4-methyl cyclopentanol-1
(B) 
C ? C ? C ? C ?  ? C C ? NH
2
Br
O
6 5 4 3 2
       3-bromo hexanamide
(C) 
1
COOH
Cl
2
3
4
4-Chloro-3-methylcyclohexane
carboxylic acid
(D) C ? C ? C ? C ? C ? O H ? C 
Br
6 5 4 3 2
C
7
1
      4-Bromoheptan-2-ol
9. Ans. (D)
10. Ans. (C)
11. Ans. (B)
12. Ans. (A)
SECTION-III
1. Ans. (1)
Sol. l = 
h
3mkT
 , l µ m
–1/2 
 T
–1/2
LTS-4/7 0999DJA110319001
Target : JEE (Main + Advanced) 2020/07-07-2019
ALLEN
PART -3 : MATHEMA TICS ANSWER KEY
SOLUTION
Q. 1 2 3 4 5 6 7 8 9 10
A. C C D A A,B B,C,D A,B,C A,B,C,D D A
Q. 11 12
A. B A
Q. 1 2 3 4 5 6
A. 6 1 4 1 9 4
A B C D A B C D
Q,R,S R R,S T T P Q,R Q,S
Q.1 Q.2
SECTION-I
SECTION-III
SECTION-IV
SECTION–I
1. Ans. (C)
Sol. a = log
2
x, b = log
4
x, g = log
6
x
abg = ab + bg + ga
1 11
10
æö
abg ++- =
ç÷
a bg
èø
Þ either a = 0 or b = 0 or g = 0 or
1 11
10 + + -=
a bg
2. Ans. (C)
Sol. We know that (1 + tan k°)(1 + tan(45 – k)°)
= 2
Þ (1 + tan1°)(1 + tan2°).......
      (1 + tan44°)(1 + tan45°)
= 2{(1 + tan1°)(1 + tan44°)}.{(1 + tan2°)
(1 + tan43°)}.......{(1 + tan22°)(1 + tan23°)}
= 2.2
22
 = 2
23
\ n = 23
3. Ans. (D)
Sol. ByC
1
 ® C
1
 – (C
2
 + C
3
)
g(x) = 0   Þ g(0) + .... + g(10) = 0
2. Ans.(8)
3. Ans.(7)
Let assume 
34
Cl has x gram and 
38
Cl has
(7– x) gram
moles of 
34
Cl = x/34 and moles of 
38
Cl
= (7 – x) / 38
M
avg.
 = 
7
(x/34) (7 x)/38 +-
put the value of M
avg.
 find the value of x, so
x = 5.1 g
moles of 
34
Cl = 5.1/34 = 0.15
and moles of 
38
Cl = (7 – 5.1)/38 =.05
n = 0.15 × 17 + 0.05 × 21 = 3.6
p = 0.15 × 17 + 0.05 × 17 = 3.4
\ p + n = 7
4. Ans. (7)
5. Ans. (3)
6. Ans. (8)
SECTION-IV
1. Ans. (A)-(P,R); (B)-(P,R); (C)-(Q,S,T);
(D)–(S,Q)
2. Ans. (A)-(P,Q); (B)-(P,S); (C)-(P,Q,R);
(D)-(P,S,T)
Sol.
OH
O
H
Acidic hydrogen
DU = 4
DU = 3  
O O
OH
O
HO
HO
ester
H
Acidic
hydrogen
DU = 3  
N
Can act as base
:
LTS-5/7 0999DJA110319001
Leader Test Series/Joint Package Course/07-07-2019
ALLEN
4. Ans. (A)
Sol. det((3AC)(2B)
–1
) = |3AC|.|(2B)
–1
|
= 9|A|.|C|.
1
1
.B
2
-
 = 9|A|.|C|.
1
4
.
1
B
= 
1 11
9..2..
4 49
=
1
8
5. Ans. (A,B)
Sol. A + A
2
 + A
3
+........¥
= 
1/3 00
0 1/20
0 0 1/2
æö
ç÷
ç÷
ç÷
èø
+
2
2
2
1/3 00
0 1/20
0 0 1/2
æö
ç÷
ç÷
ç÷
èø
       +...=
1/2 00
0 10
0 01
æö
ç÷
ç÷
ç÷
èø
(B
2
.B
2
.B
2
.B
2
.....¥) = I
Given equation becomes
1/2 00
0 10
0 01
æö
ç÷
ç÷
ç÷
èø
x 1/2
y4
z6
æöæö
ç÷ç÷
=
ç÷ç÷
ç÷ç÷
èøèø
Þ x = 1, y = 4, z = 6
6. Ans. (B,C,D)
Sol.
0 01
A 0 10
1 00
éù
êú
=
êú
êú
ëû
adjA = 
0 01
0 10
1 00
- éù
êú
-
êú
êú -
ëû
A
T
 = 
0 01
0 10
1 00
éù
êú
êú
êú
ëû
A
2
 = 
1 00
0 10
0 01
éù
êú
êú
êú
ëû
hence A
T
 = A
A
2
 = I and A
2
 = I
Now, A
2
 = I Þ A
3
 = A so A is also
periodic matrix.
7. Ans. (A,B,C)
Sol. c
2 
® c
2
 – c
1
x
      
2
1 11 1 1 11 1
22
2 22 2 2 22 2
2
3 33 3 3 33 3
a bxb(1 x) c a bxb c
a bxb(1 x) c (1 x)a bxb c
a bx b (1 x ) c a bx b c
+-+
+ - =-+
+-+
c
1
 ® c
1
 – c
2
x 
1 11
2
2 22
3 33
a bc
(1x)a b c0
a bc
=-=
8. Ans. (A,B,C,D)
Sol.
3 3 m 3m
m
(3sin1 4sin 1)(3sin3 4sin 3) (3sin3 4sin 3 )
...
sin1 sin3 sin(3)
- --
nn
n1
sin3 sin9 sin(3 ) sin(3 )
.....
sin1 sin3 sin(1) sin(3 )
-
´ ´´=
3
n
 + 3 = 4, 0, 8, 2 as last digits.
9. Ans. (D)
Consider D = 
abc
bca
c ab
= 
( )
( )( )( ) ( )
222 a bc
a b bc ca
2
++
- + - +-
if  a ¹ b ¹ c  &  a + b + c ¹ 0, then D ¹ 0
lines are not concurrent ( la xk eh)  & no
solution.
10. Ans. (A)
a = b = c makes all the lines coincident
a(x + y + 1) = 0
a ¹ 0 because a + b + c ¹ 0
x + y + 1 = 0  Þ   infinite solutions, all lying
on the line x + y + 1 = 0.
11. Ans. (B)
    
e
log sec 3232
2
æö p æö
ç÷ ç÷
èø èø
= 
e
log sec(1616 ) p = log
e
1 = 0
12. Ans. (A)
p æö
ç÷
èø
p æö
ç÷
èø
cos
4
395
log sin
4
 = 
æö
ç÷
èø
1
2
1
log
2
= 1
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