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DATE: 19-11-2019 JEE-MAIN MOCK TEST-1 XII
COURSE
NUCLEUS
TEST CODE
1 1 2 5 8
Q.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans
3
4
3
2
1
4
4
1
4
1
1
1
4
2
2
Q.No.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans
2
1
2
3
2
4
4
3
3
4
1
3
2
4
2
Q.No.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans
3
1
1
1
3
3
3
2
1
4
4
1
2
3
2
Q.No.
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans
2
2
1
4
1
2
2
2
1
1
2
4
2
3
3
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
Q.No.
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
Ans
2
2
4
4
2
4
4
3
2
3
2
2
4
2
2
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
PC
IOC
OC
Q.No.
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
Ans
1
2
2
3
2
3
1
3
2
1
2
3
1
4
4
HINTS & SOLUTIONS M AT H E M AT I C S Q.1
r a b c
taking dot with b c [ r b c ] = [a b c ] where a (0,1,1) ; b (1,1,1) and c ( 2,1, 0)
0 1 1 [a b c ] = 1 1 1 = 0 – (0 – 2) + 1(–1 + 2) = 3 2 1 0 and
x y z 1 1 1 = = [ r b c] 2 1 0
x(0 + 1) – y(0 – 2) + z(–1 + 2) = x + 2y + z hence equation of plane is x + 2y + z = 3;
p=
3 = 6
XII MT-1 [JEE Main]
3 Ans. ] 2
Q.2 a b V u d a [ a c d ] b (a . V ) b ( b . V ) a ( u . a ) d ( u . d ) a [ a c d ] b [ a c d ] b [ b c d ]a [ b c a ]d [ b c d ]a [ a c d ] b [ b c a ] d 0 a , b , c are coplanar ]
Q.3
Clearly minimum value of a2 + b2 + c2 O(0,0,0)
3x + 2y + z =7 P (a, b, c)
2
| (3(0) 2(0) (0) 7 | 49 7 = = = units. 2 2 2 2 (3) (2) (1) 14
(This is possible when P(a, b, c) is foot of perpendicular from O(0, 0, 0) on the plane.) Page # 1
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Downloaded from jeemain.guru Alternatively: Let V1 3î 2ˆj k̂ and V2 a î bˆj ck̂ Now V1 · V2 3a 2b c 7 V1 V2
E2 = 3, 4, 5, 6 on die
P(E2) =
2 3
E = (E E1) + (E E2) P(E) = P(E1) · P(E/E1) + P(E2) P(E/E2)
7 14 a 2 b 2 c 2 (a2 + b2 + c2) Hence
Q.4
w= =
a2
+
b2
+
49 14
c2|least
Using the law of total probabilities,
7 = Ans.] 2
P (red ball) =
3i (1 i) 2 3(1 i) 6 = 2i 2i
Q.9
(3 i)(2 i) 5 5i = =1–i 5 5
2 and amp. w = – 4 . ]
Hence | w | =
Q.5
Answer is 0 ]
Q.6
6! 6! 6! · 2! · 2! + · 2! + 3!· 3!· 2! 1!· 5! 2!· 4!
G1 G2 1 5 2 4 3 3 12 + 30 + 20 = 62 Alternatively: 1st tourist can go G1 or G2 in 2 ways |||ly all others. Hence required number of ways = 26 – 2 = 62 Ans. ]
Q.7
13C – 6C ; 5 5
Q.8
Urn A
6B / 13 \ ] 7G
9R 11W
;
Urn B
A : Coin randomly selected tossed 10 times, fell head wise 7 times B1 : coin was a fair coin P(B1) = 1/2 B2 : Coin was a weighted coin P(B2) = 1/2
XII MT-1 [JEE Main]
3
7
3
P(A/B1) =
1 1 1 · = 10C3 · 10 7· 2 2 2
P(A/B2) =
10C
47 4 1 10C · · · = 7 3 5 5 510 1 210
7
1 4 10 10 2 5
=
1 47 ·210 1 10 5
510 = 10 8 Ans. ] 5 8 Q.10
Let B1 : Taxi is black 0.85 B2 : Taxi is white 0.15 A: witness says that taxi involved in the hit and run accident was White. P(A/B1) = 0.2 P(A/B2) = 0.8
12R 3W
P(B1/A) =
E : event of drawing a red ball; E1 = 1 or 2 on die
7
10C
P(B1/A) =
(concept of grouping)
2 9 4 12 41 · + · = Ans. ] 6 20 6 15 60
1 P(E1) = 3
=
(0.85) (0.2) (0.85) (0.2) (0.15) (0.8)
0.170 17 17 = = Ans.] 0.170 0.120 17 12 29 Page # 2
Downloaded from jeemain.guru Q.11
Arrange the data in increasing order as,
O
7 5 1 1 p – , p – 3, p – , p – 2, p – , p + , 2 2 2 2 p + 4 , p + 5. As, number of observations = 8 So, median =
x
C 5m B 25m
(4 th observation ) (5 th observation ) p 5 Ans. = 4 2
tan2 =
C 128
16 160 24 56 40 216 B
Q.12
A
H
i 1
n
i 1
n
Hence, possible value of n = 18.
]
Equation of the line is r a t ( p q ) ....(1) now (1) intersects r · n d ....(2) substituting r from (1) in (2) a t (p q ) · n d a · n t[ p q n ] d (d a · n ) t= [p q n ] hence the position vector of R is (d a · n ) r a (p q) Ans. ] [p q n ]
Q.17
Vector perpendicular to 2î ˆj k̂ and
400 80 n 16 n n
=
3 . 2
Q.16
n
xi
2 3
x = 5 cot = 5.
Q.13 Since, root mean square arithmetic mean
x i2
2 tan 2 = 6tan tan2 = 2 3 1 tan
tan =
V = 800 Total playing game = 640 800 – 640 = 160 ]
n
30 tan 5
]
Q.14
î 3ˆj k̂ is
Q.15
î ĵ k̂ 2 1 1 2î ĵ 5k̂ 1 3 1
In OBC, we have tan =
5 x
30 x Dividing (ii) by(i), we have
Also, tan2 =
XII MT-1 [JEE Main]
(i) (ii)
Any general point on the line is (1 + 2, 1 + , 1 – ) at their point of intersection. This point satisfies equation of plane Page # 3
Downloaded from jeemain.guru (1 + 2) + 3(1 + ) – (1 – ) = 9 = 1 Point of intersection is (3, 2, 0). Hence required line is r 3î 2ˆj k 2î ˆj 5k̂
Q.22
4! 4! 4! = 2!· 2! + + 4! + 2!· 2! = 48 Ans. ] 2!
x 3 y2 z Ans.] 2 1 5
Q.23
P(W) = 1/3 ; P(B) = 2/3 p = 1/3 ; q = 2/3 and r = 4 or 5 and n = 5 Use P(r) = nCr pr qn r ]
Q.24
x>0;y>0 ;x<2;y<2 A : xy < 1 B : y < 2x ; n (S) = 4 n (A) =Area of shaded region.
Q.18 S = w + 2w2 + 3w3 + ..... + 9w9 Sw = + w2 + 2w3 + .......+ 8w9 + 9w10 where w9 = (cos 40° + sin40°)9 = 1 and | w | = 1 ———————————————— S(1 – w) = w + w2 + w3 + ..... + w9 – 9w10
S=– =
w (1 w ) – 9w = 0 – 9w 1 w
1 w 1 9w (using w9 = 1) ; = S 9w 1 w
1 | – 2 sin220° + 2i sin 20°cos 20° | 9
1 2 | 2 sin 20° i (cos 20° + i sin 20°) | = sin20° 9 9
2
Shaded area ABC = Answer is 6
Q.20
Treat W, B, G, R as beggar 0 0 0 0 0 0 Ø Ø Ø = 9C3 = 84 or co-eff. of x6 in (1 + x + x2 + x3 + x4 + x5 + x6)4 ;
Question No. 1 can
be printed in 5! ways |||ly
Question No.2 can be printed 5! ways and so on Total ways (5!)20 Ans. ] XII MT-1 [JEE Main]
1 2
15 i 2
Q.19
Q.21
1 1 1 . 2 2 2
1 | cos 40° + i sin40° – 1| 9
= =
2.
Shaded area OAB =
9
=
Possible cases if the product of four numbers a · b · c · d = 144 (1 a, b, c, d 6) 6, 6, 2, 2, ; 6, 6, 4, 1 ; 6, 4, 3, 2 and 4, 4, 3, 3
1 dx ln x 21 x
2
= ln2 + ln 2 + 3/2 ln2 Total area = p=
Q.25
3 ln 2 1 3 ln 2 1 2 2 2
3 ln 2 1 Ans. ] 8
n(A) = 3 Total number of relation in set A = 23×3 = 29 and maximum number of cartesian product = 9 out of which 3 ordered pair is necessary for reflexive . So, for remaining 6 ordered pair Number of ordered pair required = 6C0 + 6C1 + 6C2 + …… 6C6 = 26 ] Page # 4
Downloaded from jeemain.guru Q.26 p = T ~p=F (~ p q) = F ~r=F (~ p q) ~ r = F and p = T (~ p q) ~ r p = T ] Q.27
16C
3
– 2[2 ·
3C
3
+
4C
3]
P H YS I C S Q.31
Q.34
4C
–8· 3 = 16C3 – 12 – 32
1 = (µ2 – 1) 2 = –(µ1 – 1) = (µ2 – µ1) hc = + c . (3v0)
in case I
hc = + c. v0 2
in case II
hc where = (0 – threshold wavelength) 0 Q.35 560 – 44 = 516 Ans. ] Q.28
M (4, 5, 6) plane passes through 4, 5, 6 A(x – 4) + B(y – 5) + C(z – 5) = 0 Q.36
f f 10 1.8 f + 18 = f 18 = – 0.8 f f = – 22.5 cm in second case, u = – 50 Obj. is beyond C. Image is invereted and diminished. E B k̂
1.8 =
y
x
Q.29
A, B, C are 4, 4, 4 equation is x – 4 + y – 5 + z – 6 = 0 x + y + z = 15 ]
Answer is
z1
2
2
z2 ]
x4 – 2x3 – 2x2 + 4x + 3 (x – ) (x – ) (x – ) (x – ) substituting x = i 1 – 2i – 3 + 2i + 4 = (i – ) (i – ) (i – ) ( i – ) = 2 ....(1) substituting x = – i 1 + 2i – 3 – 2i + 4 = (– i – ) (– i – ) (– i – ) (– i – ) = 2 ....(2) Multiplying (1) and (2) (1+ 2) (1+ 2) (1+ 2) (1+ 2) = 4.
E
z
Q.37
O P
Q.30
XII MT-1 [JEE Main]
Q.38
k
P=
I
nhc t
n p e i = ex% = x% hc t
=
1.55 103 4 107 6.63 1034 3 108
Page # 5
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Downloaded from jeemain.guru
R=1m
Q.57
1.6
Q.39
A
O
I1
B
2m
2.0
1.6 1 1 .6 1 – = v1 = 16 v 2 1
2 .0 1 2 1 I2 – = v2 = 4 v 2 1 |v1 – v2| = 12 m
Q.44
d/2 – (µ – 1)t = 0 On screen = x = d D ( d / 2) x = d – (µ – 1) t = 0 D at centre on the screen
µ1 1
Q.45
Q.59
µ2 2
= Q.60
d2 2D
90° –
Q.61
2 1 = tan
2 sin C = = tan 1
C = sin–1 (tan )
Q.53
1. Due to emission of -particles mass will almost remain unchanged. 2. No. of -particles decayed = 3 × 1022 , so charge = 3 × 1022 × 1.6 × 10–19 = 4800C
After 2nd Polaroid, I =
I0 I0 cos2 60° = 2 8
After 3rd Polaroid, I =
I0 3I 0 cos2 30° = 8 32
12400 hc V= (inÅ)
1 34 / 34 M M Molarity = 2 2 Volume strength of the solution = 1 11.2V 5.6V 2
Q.62
Syn gas or water gas CO + H2
Q.63
Negative charged O-atom has more electron donating power than neutral O-atom therefore resonance energy.
Accordin to Malus law, I = I0 cos2
eV =
n 2 t
C H E M I S T RY
for C µ1 × sin C = µ2 × sin (90°)
Q.47
t = 4t
1 = e–4t 16 4ln2 = 4t
µ1 sin = µ1 × sin (90° – )
9 109 32 1.6 1019 = 96 × 105V 1.2 1015 (64)1 / 3
N0 dN = = N0e– × 4t 16 dt
Infront of upper slint
x = (µ – 1)t =
k 32 1.6 10 19 kQ V= = R 0 (40)1 / 3 R
O¯
Q.64
OH
Let, % of C be 7.5 x % and H be x % 7.5 x + x + 32 = 100 x=8 % of C = 60%, H = 8 % and O = 32 % E.F. = C 60 H 8 O 32 = C5H8O2 12
XII MT-1 [JEE Main]
1
16
Page # 6
Downloaded from jeemain.guru Q.65
O2 B.O. [O = O] = 2
(+M)
+
:
OH
H
: :
H2O2
O
O
Bond length
[due to resonance] B.O. = 1.5
: :
: :
O3 O
:: :
O
O
H
B.O. = 1
Q.72 Cl
(–I)
1 B.O.
H2O2 > O3 > O2 Q.73
CH3
Q.67
(i) / (ii)
Let, VO 2 = x mL, and
V = 3 eqn. (ii) T = 480
3.O2 (g) Ozonizer 2.O3(g) x = mL 0 (x–3u)mL (2u)mL= 600mL u = 300 x – 3 × 300 = 1100 x = 2000 Eqn (1) y = 1000
Q.74
In solvay process manufacture of sodium bicarbonate, the final biproduct is : NH 4 HCO 3 + NaCl NaHCO 3 + NH4Cl NH4Cl + CaO NH3 + CaCl2 + H2O
Q.69
O
Br
Q.70
d=
15 = 3 g/L = 3 kg/m3 5L
3P 3 10 4 vrms = = m/s = 100 m/s d 3 Q.71 Be and Mg does not impart colour to the flame due to their high Ionisation energy
Q.75
Cl Br
Chlorine atom lies at equatorial position because of its smaller size, bond length is shorter than Bromine to avoid 1,3 diaxial repulsion. Q.76
Theory based
Q.77
P4+ 3NaOH + 3H2OPH3 + 3NaH2PO2 White Phosphorus
CH3 is more stable than
CH2 due to back-bonding.
Due to inert pair effect as we more down the group stability of (+1) oxidation stateincreases.
Cl
VO 2 = 2 L and VN 2 = 1 L Ans. Q.68
5 V 5 / 300 = 1.5 / 240 V
hyperconjugation
VN 2 = y mL x + y = 3000 ....(i) t=0 t = tf
1 .5 L ( V ) L and 240K (T ) K ....(ii)
More number of – H, more will be
Q.66
5L (5 V )L 300K (T ) K ....(i)
Q.78
4 isomers with alcohol functional group (1) CH3 – CH2 – CH2 – CH2 – OH (2) CH 3 CH 2 CH CH 3 | OH CH3 | (3) CH 3 C OH | CH 3
(4) CH 3 CH CH 2 OH | CH 3 3 isomers with ether functional group XII MT-1 [JEE Main]
Page # 7
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Downloaded from jeemain.guru (1) CH3 – O – CH2 – CH2 – CH3
COOH OH
H
(2) CH 3 O CH CH 3 | CH 3
Q.87 H
OH COOH (I)
(3) CH3 – CH2 – O – CH2 – CH3 Q.79 Q.80
Theory based.
H
V2O5
2SO2(g) + O2 (g) 2SO3 (g) * Manufacture of H2SO4 by "Contact process"
HOOC H
COOH
H [Ca2+] = 400 ppm = 400 mg/L = 10 × 10–3 mol/L + [H ] = 20 × 10–3 mol/L n H = 20 mmol
H
H+ + NaOH Na+ + H2O 20 mmol 1M
Q.88
NH3 + 3Cl2 (excess) NCl3 + 3HCl CH3
Q.84
H3C–HC=CH
*
*
Q.89
OH
4-stereogeneic centres stereoisomers = 2n n = 3 23 = 8 Q.85
a P a Vm RT Z = 1 – 2 V Vm RT m
XeF6 (sp3d3) Distorted octahedral F
F
F
Xe F
F
F
XII MT-1 [JEE Main]
HO COOH
COOH OH
OH
OH
HO HO
H
COOH
H COOH
HO HO COOH
COOH
3 × n FeC2O4 = 5 × 50 × 0.1 n FeC2O4 =
25 mmol. 3
mFeC2O4 =
25 144 g = 1.2 g Ans. 3 1000
CaO Basic oxide CO2, SiO2 Acidic oxide SnO2 Amphoteric oxide
Q.90
(1) Gauche form of ethane-1,2-diol is most H
OH
OH
stable due to H-bonding H
96 Z =1– = 0.8 Ans. 20 0.08 300
Q.86
COOH
I and II are diastereomers
20 VNaOH = mL = 20 mL Ans. 1
Q.83
H
COOH
OH (II) H
HOOC
HO
COOH
OH
COOH COOH
At both position same groups are present Q.82
H
OH
OH
H Q.81
H
H
OH O | || (2) CH 3 C NH 2 CH 3 C NH Shows G .I.
(3) In methyl cyclohexane, methyl group lies at equatorial position than axial position to avoid 1,3-diaxial repulsion. Page # 8
Physics -ChronicleIAS Academy StudyMaterialforUPSCRailwaysSSC andStatePSCExamination(inPDF)
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