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Vidyamandir Classes
VMC | JEE Main-2020 1 Solutions |8th January Evening
SOLUTIONS
JEE Main – 2020 | 8th January 2020 (Evening Shift) PHYSICS
SECTION – 1
1.(4) 3 2 1R R i.e. 4 32 1 0R R
1R can’t be solution so,
Remaining part is 3 2 1 1 0R R R R
i.e. R must satisfy 3 2 1 0R R R
2.(2) 00 2h
mv
00 0
eE tV V i V j km
02 2 2
2 20
12
hm v e E t
m v
3.(1) 10 1
100mA
R
.
9 9R . K 4.(1)
22 2
0 0 0 02
0
1 122
a adx a a aa a a ac lnd x d d d dd
5.(2) It is uniform circular motion. 6.(4) 2pI I I I cos
2
8 4.
12 12pI I
4mI I
112 0 85
2Ratio .
Vidyamandir Classes
VMC | JEE Main-2020 2 Solutions |8th January Evening
7.(3) T V
5
32
2 06 10 5 15 104
.T . N
8.(3)
9.(4) 2
242T g
g T
0.1 12 225 50
g Tg t T
100 0.4 4 4.4%gg
10.(4) 2 47 8 75 1010
K.E. mv . J
So, option is (4)
11.(3)
1
1 2
2 5 50 15 2 5 50 4
v
D
F g. .F g g .
12.(4) 1 t /i eR
0
20
11 1t / /E E E E ELQ e dt eR R R e R e eR
13.(1)
E i
15E BC
14. (4) 32
1 1 1 1 12
2 2 22 2
kQ / R R Q RR Q RkQ / R
21 1 1
22 2
kQ / R RRkQ / R
15.(2) 2 2V ax
Vidyamandir Classes
VMC | JEE Main-2020 3 Solutions |8th January Evening
16.(2) Total time to land (using C.O.M)
22 1 2 62 2 2 2
ghh h.T gT Tg
Time to collide 1
2 2h ht
ggh
So, time for combined mass to land 2 6 1 2 12 2 6 3
2 22 2 2 4T t
17.(4) 3 52 132 2
3 6v
R Rn. n RC mixn
196pRC mix ;
1913
Y
18.(4) fm
x f
f: mag of focal length
m = mag of magnification.
19.(2) 1
10H L
H
Q QQ
1 100
10 HH
W QQ
10H LQ Q
100 10 90L LQ Q
20.(4) 0 0 02 1
4 2 22 2outI I I
BR R R
SECTION – 2
21.(8.00) 211002
gt
21 181 ( )2 2
g t 10 59 0.5
t tt
2200 200 8
25g
t
22.(486) For Balmer series, Transition happen to 2n from higher orbits. For transition 3 2n to n
3 21
hcE E
2 2
1 113 66561Å3 2
hc. ……..(i)
For transition 4n to 2n
4 22
hcE E
2 22
1 113 64 2
hc.
…….(ii)
i / ii
2
1 1209 4
1 16561 2716 4
220 6561 4860Å 48627
nm
Vidyamandir Classes
VMC | JEE Main-2020 4 Solutions |8th January Evening
23.(50) Let containers have temperature 1 2T ,T and 3T C respectively. For case: 1 1 21 2 0w w w wc T T c T T
1 23 2 0T T T
1 22 180 60T T ....... i common temperature T
For case II
2 31 2 0w w w wc T T c T T
2 33 2 0T T T
2 32 90 30T T common temperature T in this csse …….(ii)
For case III:
1 32 1 0w w wc T T c T T
1 33 2 0T T T
1 32 180 60T T ......... iii common temperature T inthis case
For Case: -IV
1 2 31 1 1 0w w w w w wc T c T c T
1 2 33 0T T T ; 1 2 33
T T T
1 2 33 450i ii iii T T T C 1 2 3 450 503 9
T T T C
24.(16) From energy conservation i i f fU K U K
2 21 110 2 2
e ei f
GM m GM mm v mvR R
2 22 2 2 22 1 1 91 1 11 2 1210 10 10
ef i e i
GMv v v v .R
256 896 16 02fv . . km / s
25.(30) Current through the batteries;
10 10 20
30 3020 5 2530 30
iR RR R
P.d. across 20 internal resistance battery;
110 20 02
i i Or 20 1
30 22530
RR
3040 2530
RR
30 15 30R R ; 30R
Vidyamandir Classes
VMC | JEE Main-2020 5 Solutions |8th January Evening
CHEMISTRY SECTION – 1
1.(4)
2.(2) AgBr shows both Frenkel and Schottky defects.
3.(1) 2 3
2
FeO SiO FeSiO
1FeO Fe O2
These reactions don’t occur in the blast furnace during extraction of Fe
4.(4) Kjeldahl’s method can’t be used for Nitro compounds.
5.(4) Ea/RTK Ae
aElog k log A2.303RT
; aEslope2.303R
In the given graph, greater the slope, more is the aE Order c a d bE E E E
6.(1) Three isotopes of hydrogen are: Protinum Deuterium Tritium 1
1H 21 H 3
1 H
No. of neutrons in 11H 0
No. of neutrons in 21 H 1
No. of neutrons in 31 H 2
Total no. of neutrons 3
7.(4) Catalytic hydrogenation depends upon the extent of adsorption. Group 7-9 elements exhibit maximum adsorption property The reactants must get adsorbed reasonably strongly on to the catalyst to become active. However, they must not
get adsorbed so strongly that they are immobilized and other reactants are left with no space on the catalyst’s surface for adsorption.
Hence, assertion is true but reason is false.
8.(2)
Vidyamandir Classes
VMC | JEE Main-2020 6 Solutions |8th January Evening
9.(1) has greater boiling point than than because of extensive –H bonding.
10.(3) Greater the bond length, lesser is the bond energy. Order of bond length: C I C Br C Cl C F Order of bond energy: C F C Cl C Br C I
11.(1) More the no. of shells, greater is the atomic radius. Size of Br Size of Cl While moving along the period, the atomic size decreases due to increase in effective nuclear charge.
Size: C O F Overall order: Br Cl C O F
12.(4) Compound Geometry Geometrical isomerism
3 3Pt NH Cl
Square planar ×
3 5Pt NH Cl
Octahedral ×
3 22Pt NH ClNO Square Planar
23 2Pt NH ClBr
Octahedral
13.(3) Bohr’s radius 2
0naZ
For 2Li , n 2and z 3 04av3
14(4) In 4Ni CO , CO is SFL pairing occurs ; 0
Similarly for 24Ni CN
and 2 3 2PdCl PPh , under the effect of SFL, all e s are paired, hence 0
In 2
2 2 26 6Ni H O Cl or Ni H O
,
2H O is a weak field ligand,
2 8Ni : d configuration
Vidyamandir Classes
VMC | JEE Main-2020 7 Solutions |8th January Evening
15.(1)
16.(4)
17.(3) 2H O H OH Dissociation of 2H O is an endothermic reaction.
On increasing temperature, H ion concentration increases. pH decreases.
Vidyamandir Classes
VMC | JEE Main-2020 8 Solutions |8th January Evening
18.(3)
Basicity of 3 2H PO 1 (one replacable H ion per molecule)
19.(2)
Maltose is a disaccharide of two D glucose monomers.
20.(2) 3 23
2
Ag NHOZN/H OX A B
SECTION – 2 21.(20.00)
No. of 90° angles 8 No. of 90°angles 6 No. of 120° angles 0 No. of 120°angles 3
No. of 180° angles 2 No. of 180°angles 1 Total 20
Vidyamandir Classes
VMC | JEE Main-2020 9 Solutions |8th January Evening
22(13.00) A CH CH
Max no of atoms in one plane 13
23.(2130.00) 3 s s 2 g s sNaClO Fe O NaCl FeO
No. of moles of 2O no. of moles of 3NaClO
No. of moles of 2PV 1 492O n 20RT 0.082 300
Molar mass of 3NaClO 23 35.5 48 106.5
Mass of 3NaClO required 20 106.5 2130g
Ans. 2130
24.(2.15) 2 2Sn s | Sn aq,1M || Pb aq,1M | Pb s
Overall cell reaction 2 2
s aq aq sSn Pb Sn Pb
2
2
Sn0.06E E log2 Pb
E 0.13 0.14 0.01V
At equilibrium, E 0
2
2
Sn0 0.01 0.03log
Pb
2
2
Sn0.01 0.03log
Pb
2
2
Sn1 log3 Pb
2
1/32
Sn10 2.15
Pb
25.(6.25) vU nC T
v500 4 C 500 300
1 1v
5000C 6.25JK mol4 200
Vidyamandir Classes
VMC | JEE Main-2020 10 Solutions |8th January Evening
MATHEMATICS
SECTION – 1
1.(4)
2
2
1 21
22 3
1
xx ,
xf xx
, x ,x
2
2
2
2
(1 ) 0 (1, 2)(1 )
( )2(1 ) 0 [2,3)1
x xx
f xx xx
So, the function is decreasing in (1, 2) and [2, 3) therefore, range is 2 1 3 4
5 2 4 5, ,
2.(4 ) As given 10 20020
i
i
xx x
Similarly,
2 21
22 21100 4 2080
20 20
ix xx x
Final mean after correction 200 9 11 202
20 20
Final variance after correction
22 222080 9 11 202 2120
10 1 3 9920 20 20
. .
3.(2)
0
0
10x
x
t sin t dt
limx
;
0
100
1x
x sin xlim
4.(3) 0b c b a b c a
c a b
; a.c a.a b.a
; 0 6 4 ; 3
2
3
2c a i j k ; 3
2c.b a.b i j k i j k ;
34 3
2c.b
94
2c.b
; 9 1
42 2
c.b
5.(3) Given curve is 2 22 3 0x xy y
2 23 3 0x xy xy y
3 3 0x x y y x y
3 0x y x y
Equation of line to 0y x and passing through 2 2,
2 1 2 4y x x y
Its distance from origin 0 0 4
2 22
Vidyamandir Classes
VMC | JEE Main-2020 11 Solutions |8th January Evening
6.(2) Given 2
3 21 2 9 12 4
dxI
x x x
Say 3 22 9 12 4f x x x x
26 18 12f x x x
1 9 2 8f and f
f x decreases from 1 2x to x
So 3 2
1
2 9 12 4x x x
Increases from 1 2x ,to x
3 2
1 1 1
3 2 22 9 12 4x x x
2
3 21
1
3 2 22 9 12 4
dx dxdx
x x x
So 2 1 1
9 8I ,
7.(4) 2 2 1 09 4 0 1
A ,I
0A I characteristic equation
2 2
09 4
2 4 18 0
2 6 8 18 0
2 6 10 0
2 6 10 0A A I
16 10 0A I A
110 6A A I
8.(1) w ; clearly 3 21 and1 0
100 100
2 3
0 01 101k k
k ka w w , b b
202
2
11
1
wa w
w
2021 1
11 1
w wa
w w
Hence equation having roots 1 and 101
2 102 101 0x x
Vidyamandir Classes
VMC | JEE Main-2020 12 Solutions |8th January Evening
9.(Bonus) For constant function Which is continuous and differentiable in [0, 1] Option 2, 3, 4 are false
( ) (1) ( )1
f c f f dC
by LMVT ( , 1)d c
and we can’t compare ( ) and ( )f d f c None option is correct
10.(1) Given equation is 3 3 1 2 3 1 3 2x x x x
Let 3 0x t ,t
2 2 1 2t t t t
2 2y t t
1 2y t t
Point of intersections of the graphs of these two functions will be solution (for 0t ) 1B will intersect at only one point
2B will not intersect
For 3B
2 2 1 2t t t t
2 2 2 3t t t
2 3 5 0t t Roots are imaginary So solution set is a singleton
11.(3) 2 2 3 2x, y R : x y x
2 3 2x x
2 2 3 0x x 2 3 3 0x x x
3 1x ,x
Required area 11 1 3
2 2 2
3 3 3
3 2 3 2 33x
x x dx x x dx x x
1
3 1 9 9 93
32
3
12.(1) Given system of equations 2 2 5x y z
2 3 5 8x y z
4 6 10x y z
2 22 3 5 04 6
218 5 2 20 12 2 2 12 0
2 6 16 0
Vidyamandir Classes
VMC | JEE Main-2020 13 Solutions |8th January Evening
2 8and
Now, 1
5 2 28 3 5
10 2 6D
1 0D
Thus for 2 , given set of equations has no solution.
13.(2) Given family of curves is 2 4x b y b , b R .(i) Differentiating both sides w.r.t x
2 42
dy xx b b
dx dy dx ……..(ii)
On substituting the value of ‘b’ from (ii) to (i)
2 42 2
x xx y
dy dydx dx
We get 2
2dy dy
x y xdx dx
14.(2) Mid point of point and its image is 2 1 4
3 3 3, ,
Direction ratios of the normal 10 10 10
3 3 3, ,
So, the equation of plane is
10 2 10 1 10 4
03 3 3 3 3 3
x y z
1x y z
1 11, , ratio for equation of the plane
15.(3) 1
2P A B
2
25
P A P B P A B ……..(i)
1
2P A P B P A B …….(ii)
(i) – (ii)
2 1
5 2P A B
1
10P A B
1
10P A B
16.(1) ~ p q p q
~ ~ p q ~ p ~ q
Vidyamandir Classes
VMC | JEE Main-2020 14 Solutions |8th January Evening
17.(4) Let the equation of the hyperbola is 2 2
2 136
x y
b
Since it passes through 10 16,
22
100 2561 144
36b
b
So the equation of the hyperbola is 2 2
136 144
x y
Equation of normal 36 14410 36 16 144
x y
/ /
2 5 100x y
18.(2) 1
920
a d ……….(i)
1
1910
a d …….(ii)
1 1
1010 20
d ; 1
1020
d ; 1
200d
Now 9 1
200 20a ;
1 9
20 200a ;
1
200a
200200 2 1 201 1
199 100 1002 200 200 200 2
S
19.(3) 1 2L x y
Since y mx c is to 2x y So y mx c becomes 0y x c y x c
Now 0y x c is tangent to 2 23 1x y
3 0
1 3 22
cc
2 6 9 2c c ; 2 6 7 0c c
20.(4) 6 6 2 32 2 6 6 6 4 2 6 2 2 6 2
0 2 4 61 1 2 1 1 1x x x x C x C x x C x x C x
6 6 4 6 4 2 6 4 22 15 15 15 30 15 6 3 1x x x x x x x x x
6 4 2 6 4 22 32 48 18 1 16 96 36 2x x x x x x
96 36 96 36 132; ;
SECTION – 2 21.(2454) Given word is EXAMINATION 2A, 2I, 2N, 1E, 1X, 1M, 1T, 1O
Case (i) two pairs 32
4
2 2
!C
! !
Case (ii) one pair two diff 3 71 2
4
2
!C C
!
Case (iii) all diff 84 4C !
Total no of words 18 756 1680 2454
Vidyamandir Classes
VMC | JEE Main-2020 15 Solutions |8th January Evening
22.(3) Let 3 2 23 2f x ax bx cx d , f x ax bx c ; 6 2f x ax b 10a b c d …(i)
6a b c d …….(ii) 3 2 0a b c ……(iii) 6 2 0a b ….(iv) So, 1 3 9 5a ,b , c ,d
Also, for local minima 0f x 23 6 9 0x x 1 3x , So local minima at 3x
23(1) 2 1 1 2 1
01 2 7 2 10 2
sin cosand , ,
cos
2
2 1
71 2 1
sin
cos
1
07 2
sin,
cos
;
1
7tan
Now 21 1 2 1
2 10
sin
10
10 2sin , ,
110
sin , 1
3tan ;
2 3 2 5 32 1 8 41 3
9 9
/ /tan
Now
1 32 7 42 11 31 2 1
7 4
tan tantan
tan ,tan
24.(504)
7
1
1 2 1
4n
n n n
7
3 2
1
12 3
4 nn n n
; 21 7 8 7 8 15 7 8
2 34 2 6 2
25.(0.5) Area of 4OPQ
Slope of OP 2
t
2
2
0 0 11 1
0 1 42 4
1 11
4 2
t
t t
318
8t
3 64 4t t So 1
2m
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