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Vidyamandir Classes
VMC/JEE Mains-2018 1 Solutions
SOLUTIONS Joint Entrance Exam | IITJEE-2018
Paper Code - B 8th April 2017 | 9.30 AM – 12.30 PM
Vidyamandir Classes
VMC/JEE Mains-2018 2 Solutions
Joint Entrance Exam | JEE Mains 2018
PART-A PHYSICS
1.(3) For collision with deuterium:
1 22mv o mv mv (Conservation of momentum ) ......... (1)
2 1v v v ( 1e ) ......... (2)
By (1) and (2) 1 3vv
2 2
1
2
1 182 2 0.89
1 92
d
mv mvP
mv
For collision with carbon Nucleus
1 20 12mv mv mv (Conservation of momentum ) ......... (1)
2 1v v v ( 1e ) ......... (2)
By (1) and (2)
11113
v v
22
2
1 1 11482 2 13 0.281 169
2
mv m vPc
mv
2.(3) Change in momentum of a single molecule.
0 0 22
uP m
Total change in momentum per second
0 0. 2P n P n m u
Pressure 0 2nm uFA A
Substituting values: 3 22.35 10 / .P N m
3.(1) PB VV
V PV B
Also, 3V rV r
(As 24V r r )
3
r Pr B
3mgKa
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4.(4)
Equivalent circuit is
Where,
1 2 21 2 3eqr
1 2
1 2
1 2
12.331 1eq
V Vr r
v
r r
1010AB eq
eqV V
r
= 11.55 volts
5.(1) 22KUr
3dU KF r rdr r
2
3K mv
rr (As Force towards center
2mvr
)
K. E. = 22
12 2
KmVr
Total energy = KE + PE 2 22 2
K Kr r
= Zero
6.(4) For 1m to be at rest
5T g
For 2&m m to be at rest
5f T g
( )f N
20.15( )f m m g
23.33m kg
Amongst the options minimum mass that can be kept for no motion is 27.3 kg 7.(2) For Series limit of Lyman : 1 1n and 2n
2 1 11P RcZ
For Series limit of Pfund: 1 5n and 2n
2 1 125 25
LP RcZ
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8.(1) When an unpolarized light of intensity I passes through a polarizer for the 1st time, intensity of output is 2I
(irrespective of orientation of polarizer) So,
i.e., polarizers A and B have axes parallel to each other. Now let the axis of C make an angle with A, and with B.
4cos
2 8I I
Solving, 45
9 (3). 22 2
1 1 11n
RZn
1
2 21 11n
RZ n
Since n is very large, using binomial
2 21 11n
RZ n
2 2 21 1 1
nRZ RZ n
2nn
BA
As 2 2
2 22 12
4n
r n h nn nmZe
10 (1).
Voltage across Si diode in forward bias is 0.7 volts. Hence voltage across 200 resister is 3 – 0.7 = 2.3V
2.3200
I = 11.5 mA
11.(4) 2mv mKrqB qB
2 e
em K
reB
2 4
2pm K
reB
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2 p
pm K
reB
Comparing (1), (2) and (3) e pr r r
12.(3) iq CV
fq KCV
( 1)induced f iq q q K CV 125 1 90 10 20 1.23
nC
13.(3) Quality factor 02
Q
0LR
14.(1) Overall bandwidth use for transmission 10% of C
Number of telephonic channel Total bandwidth
Channel bandwidth
95
3
10 10 10100 2 10
5 10
15.(3) 12
c Yfl
10
31 9.27 10
2 0.6 2.7 10
71 9.27 10
1.2 2.7
34.88 10 Hz 5kHz
16.(2) 2 2
20 6 (2 )
2 2
MR MRI M R 2 21 553 242 2
MR MR
O is the centre of mass of the system. Applying parallel axis theorem between O & P.
2 2 20
557 (3 ) 632PI I M R MR MR 2181
2MR
17.(4) CA BB
kQkQ kQVb b c
2 2 2( )(4 ) ( )(4 ) ( )(4 )a b ck
b b c
2 2
0
1 44
a b cb b
2 2
0
a b cb
18.(4) Let potential difference per unit length of potentiometer wire be x. In case-I (52)( )x …(i)
In case-II
5
ir
(40)( )ir x
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405
r xr
5 405
xr
…(ii)
From (i) & (ii)
552 405
rx x
52 13 151 5 1 1.540 5 10 10
r r
19.(1) From wave equations :
In air: 22 , kc
In medium: , 2 kc k k
,
k kc c 2
2
cc
c c
2 2 1 10 0 0 0
1 1 12
r r r r
Medium and air are non-magnetic
1 21 ; 1r r
1
2 1 2
1 1 14 4
r
r r r
20.(3) Angular width of central maxima 2a
; (where a is slit width and is wavelength)
23a
… (i)
In YDSE, fringe width
D
d [where d is slit separation and D is distance of screen from slits)
6
D ad
6D ad
6
21 3.14 102 6 10
d
6100 104
25 m
21.(4) 12 1102
fT
122 10
2 km
22 3 1223
108 10 2 10 7 16 023 10
k m ..
22.(3) 2
2 29 232 2 3
RMM R RI M
2 2 2
29 42 18 9
MR MR MRI MR
23.(4)
It is given that final total kinetic energy has increased, so some internal energy of the system must have been
converted into kinetic energy.
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2 2 21 2 0
1 1 1mv mv 1.5 mv2 2 2
2 2 21 2 0v v 1.5v ...... (i)
Since, there is no external force, momentum can be conserved 1 2 0mv mv mv
1 2 0v v v ...... (ii)
From (i) & (ii)
2 21 0 1 0v v v 1.5v
2 21 0 1 02v 2v v 0.5v 0
Relative velocity 2 1v v Difference of roots 0D 2 va
24 (1). 201 1 1
1;
2
IB m I r
r
20
2 2 22
;2
IB m I r
r
22221 1
rm
m r
22
12 r
r
2
12r
r
1 2
2 12B r
B r
25 (1). 3M MV L
3M LM L
Maximum % error in density 1.5% 3 1% 4.5%
26.(1) Let the resistances in left and right slot be r and 1000 r respectively Initial: (100 ) (1000 )( )r x r x ..................(1)
After interchanging: (1000 )[100 ( 10)] ( 10)r x r x
(1000 )(110 ) ( 10)r x r x ..................(2)
From (1): 100 1000r rx x rx 10r x
From (2): (1000 ) 110 1010 10r rr r
2(1000 )(1100 ) 100r r r r
2 21000 1100 2100 100r r r r 1000 1100 5502000
r
27.(4) cosrms rmsP V I ; 4
100 20 12 2 2
P 1000
2
Wattless current, 20 1sin 102 2rmsI I
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28.(4) The (1), (2) and (3) graphs can represent the motion of a ball that is thrown in vertically upward direction. Initially speed decreases, becomes zero and then on the return trip, speed increases. Slope of graph in option (4) does not explain it.
29.(1) For mono atomic gas 53
Using 1 constantTV
2 23 3(300) 2V T V
2
3
300 1892
T K
3 32 8.314 189 300 27682 2
U n R T 2.7 kJ
30.(1) 1nF
R
2
nk mVF
RR 2
1nkV
mR
(1 )2n
V R
Now 1
2(1 )
2
2n
nR RT R
VR
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PART-B MATHEMATICS
31.(2) 7 62
yx 2 7 12x y 2 5x y 2 5 0x y
Also, centre of the circle is 8 6 , and the radius is 64 36 c
16 6 5 1005
c
5 100 95c c
32.(4) 2 2 3 2 0x y z
1 3 2 2 1 3 0x y z
221 3 2
2 1 3 2 3 3
2
So the equation of plane is 7 7 8 3 0x y z
Now, distance from origin equal to 2 2 2
3 13 27 7 8
33.(1) 2 1 0x x 21 32
x , (where and 2 are non-real cube roots of unity)
and 2
107101 2 101 214 2 1
34.(3) Equation of PQ, 4 0 3 36x y
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12y
Area of 1 15 6 5 45 52
TPQ
35.(2) 2 6yy'
1
6 32
y'y y
1 118 2 0x by y'
1 1 1 12 2
1 1 1 1
18 9 27 2712
x x x xy' bby by by y
21 1
962
y x b
36.(4) 1 33 2 02 4 3
kk
72
k
3 0x ky z …..(i)
3 2 0x ky z ….(ii)
2 4 3 0x y z …..(iii)
On solving (i) and (ii) 2 5 0x z …..(iv) On solving (iii) and (iv) 4 2y z
2 2
52 10
4
z zxzy z
37.(1) 2 3 6 6 0x x x
Case-I: 3x
2 3 6 6 0x x x 4 0x x 0 16x ,
As 9 16x x
Case-II: 3 2 6 6 6 0x x x x 8 12 0x x
6 2 0x x 36 4x ,
As, 3 4x x
There are exactly two elements in the given set.
38.(4) 2 2 18cos · cos sin 16 2
x x
23 18cos 1 cos 14 2
x x
28cos 4cos 1 2 14
x x
3cos3 4cos 3cosx x x
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2 cos3 1x
1cos32
x
3 0, 3x
3 , 2 , 23 3 3
x Sum = 13
9 .
39.(4)
Total probability 4 1 6 1 2. .
10 2 10 3 5
40.(2) Let 1g x x tx
211 0g' xx
20 0t R ; t ,
2
2 22
1 2 2 2f x x x t ,xx
f xh x
g x
2 2 2f x t tg x t t
Let 2h t tt
221h' tt
Local minimum value occurs at 2t
Local minimum value 22 2 2 22
h
41.(4) Since Set A is, | a 5 | 1 4 < a < 6
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and | b 5 | 1 4 < b < 6
Now B is 2 2(a 6) (b 5) 1
9 4
It can be seen that all vertices of rectangle lie inside the ellipse, therefore A B
42.(3) ( ) ( )p q p q
p q ( )p q p q ~ p
T F F F F T F F F F F T F T T F F T F T
43.(4) The equation of tangent at P
1y 16 (x 16)2
A ( 16, 0)
The normal is y 16 2(x 16)
B (24, 0) Since APB
2
AB is the diameter. Center of the circle C (4, 0)
Slope of 1PB 2 m
Slope of 24CP m3
2 1
2 1
m mtan 2
1 m m
44.(1) 2x 4 2x 2x2x x 4 2x (A Bx)(x A)2x 2x x 4
Put x = 0
34 0 0
0 4 0 A0 0 4
A 4
Put x = 1
23 2 2
2 3 2 (A B)(1 A)2 2 3
3(9 4) 2( 6 4) 2(4 6)
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15 20 20 ( 4 B)25
1 ( 4 B)
B = 5
45.(2) Let 3 1x y
5 5x y x y
5 5 5 4 5 5 5 5 4 50 1 5 0 1 55 5 C x C x y ....... C y C x C x y ........ C y
5 5 3 2 5 4 5 3 2 5 40 2 4 0 2 42 5 2 5 C x C x y C xy C x C x y C xy
25 3 3 3 5 6 3 6 32 10 1 5 1 2 10 10 5 1 2 x x x x x x x x x x x
5 6 3 7 42 10 10 5 5 10 2 1 10 5 5 2 x x x x x x
46.(1) 1 5 9 49 416a a a ........a 24 32a d ……(i)
9 43 66 25 33a a a d ….(ii)
From (i) and (ii) 1d and 8a
Now, 2 2 21 2 17 140a a ......a m
217
18 1 140
rr m
17
2
17 140 4760 140 34
rr m m m
47.(1) Let, ( , )R h k
(0, )P k
( ,0)Q h
Equation of line would be,
1x yh k . . . (i)
2 3 1h k
2 3k h hk
Locus of (h, k) is 2 3y x xy
48.(2) Given/ 2 2
/ 2
sin1 2x
x dx
2 22sin 2 (sin )( ) ( ) sin
1 2 1 2
x
x xx xf x f x x
/ 22
0
sin x dx
/ 2
2
0
sin4
x dx
49.(3) 2( ) cosg x x
( )f x x
( ( ) ) cosg f x x
Given, 2 218 9 0x x (6 ) (3 ) 0x x
,6 3
x
Area = / 3
/ 6
3 1cos2
x dx
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50.(1) 0
1 2 15lim ...........x
xx x x
0
1 1 2 2 15 15lim .......x
xx x x x x x
= 0 0
1 2 15lim (1 2 3 ........ 15) lim ........x x
xx x x
Now 0 1x x R 120
51.(1) Variance = 245 (1)9 5 1 4
Variance 2
52.(4) 2 2
5 3 2 3 2 5 2sin x cos x dx
(sin x cos x sin x sin x cos x cos x)
2 6
25 2 3
tan x sec xdx
tan x tan x tan x 1
Put tan x t 2 dtsec x
dx
2 2 2
3 2 2 2t (1 t ) dt
(t 1) (t 1)
3t 1 y
2 dy3tdt
21 dy 1 C3 3(y)y
31 C
3(tan x 1)
53.(3) Doubtful points for differentiability are 0 and
At x = 0
| || | ( 1) sin | | 0(0 ) limh
h o
h e hfh
( ) ( 1) sinlimh
h o
h e hh
0
sinhlim 1h h
and lim 1 0h
h oe
(0 ) 0 1 0f
| || | ( 1) sin | | 0(0 ) lim
h
h o
h e hfh
0
( ) ( 1) sinlimh
h
h e hh
0
sinhlim 1h h
and lim 1 0h
h oe
(0 ) 0 1 0f
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(0 ) (0 ) 0f f
Similarly ( ) ( ) 0f f
Hence ( )f x is differentiable x R
54.(1) 4 4dy y cot x x cosec x d y sin x xdxdx
Integrating both sides we get: 22y sin x x c
Also, 02
y
2
2c
2
222
y sin x x
286 9
y
55.(3) ( ) 0;u a b 0 u a and 24.u b
Let ˆ ˆ ˆ ˆ( ) ( )b b a a b u u
2 2 2ˆ ˆ| | ( ) ( )b b a b u
2
2 22ˆ( )ˆ| | ( )
ˆ| |b ub b a
u
2
22 (24)27 ˆ| |u
2| | 336u
56.(2) 5 1 41 1 1
x y z
5 1 4 P , ,
P is foot of perpendicular from A to plane 3 8 7
13
14 4 113 3 3
P , ,
4 1 31 1 1
x y z
4 1 3Q , ,
Q is foot of perpendicular from B to plane 3 6 7
13
13 2 103 3 3
Q , ,
1 4 1 6 23 3 3
PQ
57.(3) 13
hx
3x h
2 2200 3h h
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2 24 (200)h
24 40000h 100h
58.(3) 6 34 1 1 4!C C
6 5 3 24 45 24 10802
59.(4) 2 2 2 2 2 21 2 2 3 2 4 2 20A . . ........ A .
2 2 2 2 2 2 2 21 2 3 4 20 2 4 20......... .......
20 21 41 10 11 2146 6
2870 1540 4410 2870 1540 4410
40 41 81 4 20 21 416 6
B 540 41 41 280 41 820 33620
33620 8820 100 100 24800 248 60.(1)
2 3 3
3a
2 12 6a a
2 5 33
b 2 4 2b b
2 2(6 3) 3AC
Diameter 81 9AC 90
Radius 3 10 3 102 2 2
53
2
PART-C CHEMISTRY
61.(1) 3I is - 3sp d hybridised
- linear shape
62.(4) CH3COOK is a salt of a weak acid and a strong base
Most basic
63.(1)
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64.(1) Amidines, are stronger organic bases.
65.(1) Methyl orange is used for titration of strong acid and weak base. 66.(1) 67.(2) x y zC H O has z oxygen atom
x y 2 2 2y y
C H x O x CO H O4 2
O atoms required for combustiony
2 x4
1 y
z 2 x2 4
y
z x4
68.(1) During reduction 2 2 2H O H O
During oxidation 2 2 2H O O
69.(2)
Option (2) is correct [NCERT Class XII Part-II, Page No.-340]
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70.(1) 2 6 2 2 3 2B H 3O B O 3H O
2 627.66
nB H 127.66
2On required = 3
2 2 22H O 2 H O
n-factor for O2 = 4 Number of equivalent = 3 4 12F 12 96500C
i t 12 96500
12 96500
t s100
12 96500
h 3.2 hr100 3600
71.(3) G H T S RT nk H T S
H SnkRT R
Slope is HR
Since H is ve
Slope is positive.
72.(3) nr k A
n1 k 363 0.95 ….(i)
n0.5 k 363 0.67 ….(ii)
From (i) and (ii) n 2 73.(3)
Option (3) is correct [NCERT Class XII Part-II, Page No 405] 74.(4) Case - I
Case - II
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Two isomers (fac and mer) are produced if reactant complex ion is a cis isomer. Only one isomer (fac) is formed if reactant complex ion is a trans isomer.
75 (4).
76 (3).
OH OH
+ CO2NaOH H O3 +
COOH
(X)
OH OCOCH3
+ (CO CO) O3 2
H SO2 4
COOH
Aspirin
COOH
Cat
(X)
+ CH COOH3
[NCERT class XII part II/Page No. 330]
77.(1) 2 2 104 4 spBaSO (s) Ba (aq) SO (aq) K 10
22 4 4Na SO 2Na SO
Conc. of 24SO in final solution 50 1 0.1M
500
For final solution
2 2 104Ba SO 10 2 9Ba 10 M
i i f fM V M V
9 9C 450 10 500 C 1.1 10 M
78.(4) Kjeldahl method is not applicable to compounds containing nitrogen in nitro (NO2) and azo (N = N –) groups and nitrogen present in the ring (pyridine) as nitrogen of these compounds does not change to ammonium sulphate.
[NCERT Class XI part II/Page No. 358]
79.(1) NaOH23(M) white gel ppt
(X)
Al NaOH Al OH NaAlO
2 33Al OH Al O
2 3Al O is used in chromatography as an absorbent. (Refer NCERT Class XIth/Part-II, Page-352)
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80.(4) HCl H Cl 0.2M 0.2 M
2H S H HS 71K 10
2HS H S 132K 1.2 10
22H S 2H S 1 2K K K
201.2 10
2 2
2
H SK
H S
H 0.2M , 2H S 0.1
2 2
200.2 S
1.2 100.1
2 20S 3 10 M
81.(1) (Refer NCERT Class XIth Part-II, Page-407)
The F ions make the enamel on teeth much harder by converting hydroxyapatite, 4 2 23Ca PO Ca OH ,
into much harder fluorapatite i.e. 4 223Ca PO CaF
82.(2) 4 2 2 2NH NO N 2H O
4 4 3 2 42NH SO NH H SO
3 22Ba N Ba 3N
4 2 7 2 2 2 32NH Cr O N 4H O Cr O
83.(3) NCERT Class XII/Part-II, Page No. 443
84.(1) 2 36Cr H O Cl
x 0 3 0 x 3
6 6 2Cr C H
x 0 0 x 0
2 2 32 2K Cr CN O O NH
2 x 2 4 2 0 0 x 6
85.(1) Pressure of cation in interstitial sites is ‘Frenkel’ defect.
86.(2) 6 6 2 2 215C H ( ) O (g) 6CO (g) 3H O( )2
(g)3n2
(g)H U n RT
1.5 8.314 2983263.91000
3267.6 kJ / mol
87.(2) 3BCl and 3AlCl are e deficient and thus act as Lewis acid
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88.(1) KCl exist as K and Cl
89.(2) Depression in freezing pt
f fT i K m
Less the value of i,
Higher the value of freezing pt.
For (2) i = 1 (min)
90.(2) 22H does not exist as Bond order is zero
Electronic configuration of 2 2 22 1s 1sH : *
2 2B.O 02