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Vidyamandir Classes
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SOLUTIONS Joint Entrance Exam | IITJEE-2019
09th APRIL 2019 | Evening Session
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Joint Entrance Exam | JEE Mains 2019
PART-A PHYSICS
1.(1) Conservation of angular momentum
2.(3) We have confidence only till two places of decimal in actual measurement so the final answer accurate to the just 2 decimal place.
Note : It is actual measurement which using some device with least count.
3.(3)
4.(2) ,
, a = 0]
,
5.(4) Truth table can be formed as
Hence the equivalent is “OR” gate Method-2
OR (GATE)
6.(4) Limit of resolution
22 22
02 (0) 212 12 2ML ML Lm m
é ùé ù æ ö+ w + ´ wê úê ú ç ÷è øê ú ê úë û ë û
06
MM m
ww =
+
2mne
r =t
81.67 10 m-= ´ W
2 3( )x t at bt ct= + - 2( ) 2 3dx tv a bt ctdt
= = + -
2
2( ) 2 6d x ta b ct
dt= = -
3bgc
=2
2 33 3b bv a b cc c
æ ö æ ö= + -ç ÷ ç ÷è ø è ø
2
3bv ac
= +
Equivalent0 0 00 1 11 0 11 1 1
A B
1.22dl
=
9
21.22 600 10250 10
-
-´ ´
=´72.9 10 rad-= ´
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7.(3)
When moving away from each other
8.(1) Electric field at
Applying binomial approximation
9.(3) T = MB
10.(1) ,
,
0c
c
v vf fv vé ù-
= ê ú+ë û
0340 202000340 20
f -é ù= ê ú+ë û
0 2250f Hz=
1 1 1 22 cos 2 cosp E E= q - q
2 2 2 2 1/2 2 2 2 2 1/22 2
( ) ( ) [(2 ) ] [(2 ) ]Kq D Kq D
d D d D d D d D= ´ - ´
+ + + +
2 2 3/2 2 2 3/22 ( ) (4 )KqD d D d D- -é ù= + - +ë û3/2 3/22 2
3 2 22 41 1KqD d dD D D
- -é ùæ ö æ öê ú= + - +ç ÷ ç ÷ç ÷ ç ÷ê úè ø è øë ûd D\ <<
2 2
3 2 22 3 3 41 1
2 2KqD d dD D D
é ùæ ö´= - - -ê úç ÷ç ÷ê úè øë û
2 2
3 2 22 12 3
2 2KqD d dD D D
é ù= -ê ú
ê úë û2
49KqdD
=
C NIABf =6
34 3
10 10175 10 10
B T-
-- -´p
= =´ ´
2( ) Krr
r =2
2Mv GMR R
¢=
22
02
( )4R
G r r dnMvR R
r p
=ò 2
22 4GK Rm RT Rp pæ ö =ç ÷
è ø
constantTR=
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11.(2)
12.(2)
13.(1) L = 2.0 m f = 240 Hz
,
v = 320,
14.(2)
15.(1) ,
16.(1)
Linear momentum conservation
, c
17.(2) , b
At t = 0 18.(4) Potential Energy of spring = Total Head Energy
( 2 )IP a rC
= +
2 34 4
IC
æ ö= ´ +ç ÷è ø
85 20 8 104IPC
-= = × ´
!2 215 (4 20 )r t i t j®= + -!
!2
2 30 ( 40)dt i jdt
®
= + -!
250 / seca m=
3 2402vfL
= =3 2404v=
0 802vf HzL
= =
2 13 ( ) ( )3
dQ KA KAddt d d
q -q q-q= =
1 2910 10q q
q = +
3R = W 2R lµ22 3 12R¢ = ´ = W
15 106
R Rp ¢= ´ = Wp
2 26
R Rp ¢= ´ = Wp
1 2
1 2
53
R RRR R
= = W+
2 2 0 22vm v mv m m¢
+ = ´ + ´ 2 2v v¢ =
2outer 0( )4tnKte R-af µ p [1 ]td Ce t
dt-a- f
e = = -a
induced[1 ]
(Resistance)
tCe ati-a- -
=
inducedi ve= -
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,
19.(3)
t = 2 sec
20.(4)
n = 2, z = 2
21.(3) Magnification is 2
If image is real,
If image is virtual,
22.(3)
When is filled between lens and mirror
23.(3)
21 1 2 2
12kA m s m s= Dq+ Dq
2
1 1 2 2
12kA
m s m sDq =
+
2
5
2400100 3.6 10
(4184 200)k-
æ ö´ç ÷è øDq = = ´+
220 / secrada =
0 tw =w +a
20tw =
21 12002Iw =
21 1.5 (20 ) 12002
t´ ´ =
2
213.6nzEn
= -
2
2213.6 13.62
E eV= - ´ = -
132fx =
2 2fx =
1
23:1x
x=
1
1 1 2 12 18 18f
= ´ =
1
2
1 ( 1)18f
µ --
-
1µ
11
2 2 2 2 2( 1)18 18 18
P - µ += - µ - =
1
182mF
æ ö= = -ç ÷-µè ø
12 6 3= - µ
13 4µ =
143
µ =
1 2P P P+ =
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24.(3)
25.(2) The physical size of antenna of receiver and transmitter both inversely proportional to carrier frequency.
26.(1) By conservation of L momentum
Conservation of M.E.
,
27.(4)
28.(4) Initially
After pouring of oil
29.(1) Q = Equivalent
x y
l l l- =
l l l
x y
y x
l ll =
l -l
( )g gI I S I G- ´ =
0.002 50 0.20.5 0.002
S ´æ ö= = Wç ÷-è ø
0 5mv mv¢=
0 05 5mv vvm
¢ = =
2 20
1 1 5 ( )2 2mv m v mgH¢= +
2025vHg
=
22 30 1
2 2 3/2 102( )q
i R rR x
-µf = ´p =
+2
20 12 2 3/22(r )pi r Rx
µf = ´p
+2 2 3/2
22 2 3/2 3
1
( )( ) 10
P P
Q
i R xi r x -
f + f= × =
f +
323 10
P-
f=
46.67 10P-f = ´
45V g mgrw =
02 2V Vmg g g= rw +r
0 42 2 5
rrw+ = rw
0 4 12 5 2r é ù= rw -ê úë û
0 8 52 10r -é ù= rwê úë û
035
r = rw
eqQ C V=
( 1)Q n CV= +
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After intersection dielectric equivalent
Conservation of charge
30.(4) For A
For B
PART-B CHEMISTRY 1.(2) Kieselghur is an amorphous form of slica.
2.(1) CO is diamagnetic according to molecular orbital theory.
3.(2) By passing 0.1 Faraday electricity, 0.1 gm-equivalents of will be discharged. Number of gm-equivalent = (n – factor) × number of moles 0.1 = 2 × number of moles
Number of moles
4.(2)
More the intensity of +ve charge on carbon atom of carbonyl group, more is the reactivity towards nucleophilic addition reaction. So propanal is more reactivity than acetone.
5.(2)
eqC
eqC KC nC¢ = +
( )eqC C n K¢ = +
1) 1( )n CV nV Vn n K C n K( + +æ ö= = ç ÷+ +è ø
7p vR C C= - =
4422 6.32 7vfRC f= = Þ = =
9p vR C C= - =
42212 9vfRC f= = Þ =
2Ni+
Þ
Þ0.1 0.052
= =
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6.(3)
7.(1) Noradrenaline is one of the neurotransmitters that plays a role in mood change
8.(1) Assertion is correct as Haematite ore is used for extraction of Fe. Haematite is an oxide ore, so reason is incorrect.
9.(1) Neptunium (Np) and plutonium (Pu) show maximum number of oxidation states starting from +3 to +7.
10.(2) 20% (mass/mass) means 100 gm of solution has 20 gm of KI 80 gm of solvent contains 20 gm of KI
Molality,
= 1.506
11.(3) According to the postulates of VBT we can’t predict the magnetic properties of complexes, it can’t distinguish ligands as weak and strong field thus is not able to explain colour of complexes.
12.(1) for inert gases
‘b’ depends on size of the gas atom the gas atom therefore steepest increase in plot of is maximum for Xenon.
13.(1) Boron trioxide is acidic
and amphoteric
and are basic
14.(2) Initially solution is of NaOH therefore pH is high then it starts decreasing with addition of HCl. After equivalence point pH will become constant and in acidic range due to presence of strong acid.
15.(1) Due to intermolecular hydrogen bonding HF has highest b.p. among hydrogen halides.
16.(1)
17.(3) As the distance from nucleus increases total energy increases and is minimum at distance
18.(1) Hinsberg’s reagent is benzenesulphonylchloride.
19.(2) Ceric ammonium nitrate test is for alcohols and carbylamines test is for amines. Both these functional groups are present in peptide formed by serine and lysine.
20.(4) More stable is the carbocation formed more is the reactivity for reactions.
Þ20 1000m166 80
= ´
1.51m»
Pbz 1RT
= +
z b,µ z
2 3(B O )
2 3Al O 2 3Ga O
2 3In O 2 3Tl O
U q WD = +
U ( 2) 10 8kJD = - + =
0a .
NS 1
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21.(3) i = 3
= 3 × 4 × 0.03 = 0.36 K
22.(4) According to given graph, activation enthalpy to from C is greater than that to from D.
23.(1) Bauxide , Malachite
Siderite , Calamine
24.(3) Stratosphere lies between 10 to 50 km from sea level.
25.(4) Number of particles
Total area = Area covered by one particle × number of particle 0.24 =
(a) edge length
26.(1) Common CN of transition elements = 6 Common CN of inner transition elements = 8 to 12.
27.(1) It is acid catalyzed intermolecular esterification reaction
28.(4) is a two step reaction in which first step is rate determining step and stable carbocation is formed. So, peak of first step is higher than second step.
29.(2) Electrophilic substitution is more in phenol ring.
30.(4) In vapour phase, exit as dimer
In solid state, has polymer chain structure
PART-C MATHEMATICS
22 4 4K SO 2K SO+ -= +
f fAl i K m= ´ ´
2 3Al O® 3 2CuCO .Cu(OH)®
3FeCO® 3ZnCO®
3
A310 10 N10
- ´=
323
310 10Area 6 1010
- ´´ ´ ´
20 2Area 4 10 cm-= ´102 10 cm 2pm-= ´ =
NS 1
2BeCl
2BeCl
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1.(4) ,
,
,
,
2.(4) ... (i)
... (ii)
Clearly circle (i) & circle (ii) touches internally because
Equation of common tangent Point (6, –2) passes through the tangent.
3.(1)
Put
Ans.
4.(3) Let number of people in city = 100
( )55 1
zwz
+=
-5 5 5w wz z- = +
5 55 3wzw-
=+
13/5
wzw
-=
+
1 13/5
wzw
-= <
+315
w w- < +
( ) 1Re5
w > ( )5Re 1w >
2 2 4x y+ =2 2 6 8 24 0x y x y+ + + - =
( ) ( )2 23 4 49x y+ + + =
1 2 1 2c c r r= -
1 2 0s s- =
1 2 6 8 20 0s s x y- = + - =
3 4 10x y+ =
( )1
1 2 4
0
cot 1I x x x dx-= × - +ò2x t=2xdx dt=
( )1
1 2
0
1 cot 12
I t t dt-= - +ò
( )
11
0
1 1tan2 1 1
dtt t
- æ ö= ç ÷ç ÷+ -è øò
( )( )
11
0
11 tan2 1 1
t tdt
t t- æ ö- -
= ç ÷ç ÷+ × -è øò
( ){ }1
1 1
0
1 tan tan 12
t t dt- -= + -ò
\ ( )1 1
1 1
0 0
tan tan 1t dt t dt- -= -ò ò1
1
0
tanI t dt-= ò111
200
tan1tt t dtt
-= × -+ò ( )
12
0
1 14 2
n tp= - +!
1 24 2
np= - !
17A BÇ =
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30% of
40% of
50% of
Total = 5.1 + 4.8 + 4 = 13.9
5.(4) & ... (i)
... (ii)
Using (i) & (ii)
6.(1)
,
Equation (i)
Equation (ii)
An adding
7.(4)
Let We know that
,
,
8.(1)
( ) 5.1A BÇ =
12A BÇ =
( ) 4.8A BÇ =
8A BÇ =
( ) 4A BÇ =
2 4x ¹ 3 0x x- >2, 2x ¹ -
( )( )1 1 0x x x× - + >
( ) ( )1, 0 1,xÎ - È ¥
( ) ( ) ( )1, 0 1, 2 2,xÎ - È È ¥
2 3 11 2 02 1 1
D k-
= - =-
( )2 2 3 5 2 1 0k k- - ´ + + =
4 10k = 92
k =
2 3 0x zy y× + - =
2 1 0x zy y× - + =
12
xy
-=
2zy=
14
xz= -
12
x x z ky z y+ + + =
1ˆ cos3 2
a i p× = = =!
"
1ˆ cos4 2
a j mp× = = =!
! cosa k n× = = q"
2 2 2 1m n+ + =!21 1 1
4 2n+ + =
1 cos2
n = ± = q2,
3 3p p
q =
( ) ( )2199 2
2n n
n+
+ = -
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n = 19
No. Of balls used for the triangle = = 190.
9.(2) Ar(ABC)
.
10.(3)
11.(4)
.
12.(2) Let a point lying on the parabola is
Equation of tangent at to the parabola is
If this line also touches the ellipse then
(–ive sign rejected)
13.(2) upto 11th term
20192
´
12BC AN= ´ ´
sinAN AD= qˆˆ ˆ3 2 2AD i j k= - +
!!!"
ˆˆ3 4i k= +!"
sinAD AD lq = ´!!!"
#
( ) ( )ˆ ˆˆ ˆ ˆ3 2 2 3 4
5
i j k i k- + ´ +=
ˆˆ ˆ8 6 6 1365 5
i j k- + += =
( ) 1 13652 5
Ar ABC = ´ ´ 34=
5 1 5 3a bp- + = - p +
( ) ( )5 1 5 3a b- p + = - p +
( )( )5 2a b- - p =
25
a b- =- p
5tan15d
° =
Þ52 3d
- =
Þ5
2 3d =
-( )5 2 3= +
2y x= ( )2,b b
( )2,b b 2y x=2
2xy +b
b =
Þ2 2xy b
= +b
2 21
1 1/ 2x y
+ = 2 2 2 2c a m b= +
2
21 11
4 24b
= × +b
Þ 4 21 2b = + b Þ 4 22 1 0b - b - =
( )22 1 2b - = Þ 2 1 2b - = ±
2 1 2b = ±
Þ 2 1 2b = + Þ 1 2a = +
1 2 3 3 5 4 7 ....+ ´ + ´ + ´ +(2 1)nT n n= -
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= 946
14.(1)
Sum of roots
is max. when m = 0
= 9 – 2 = 7
15.(2)
Given that
(Using L.H. Rule)
16.(4)
Differentiating both sides w.r.t. x
17.(3) Equation of required plane is
Equation of plane
Distance
18.(1) Mid point of AC lies on diameter
11 11 112
111 1 1
2nn n n
S T n n= = =
= = -å å å 11 12 23 11 1226 2
´ ´ ´= ´ -
( ) ( )22 2 21 3 1 0m x x m+ - + + =
223 , 11
mm
= = a +b ab = ++
a +b
Þ 3, 1a +b = ab = Þ ( )22 2 2a +b = a +b - ab
( )( )3 3 2 2a -b = a -b a +ab+b ( ) ( )2 2 24= a +b - ab a +ab+b
( )9 4 7 1 8 5= - + =
( )
26
2lim2
f x
x
t dtx® -ò
( )2 6f =
( )
( )6
2
2
lim2
f x
x
t dt
x®=
-
ò 0 form0æ öç ÷è ø
( ) ( )2
2 ' 0lim
1 0x
f x f x®
-=
-
( )2 6 ' 2f= ´
( )12 ' 2f=
( ) ( )( ) ( )sec 2 secsec tan sec tan secx xe x x f x x x x dx e f x c+ + = +ò
Þ ( )sec 2sec tan ( ) sec tan secxe x x f x x x x+ + ( ) ( )sec sec' sec tanx xe f x f x e x x= + ×
Þ ( ) 2' sec tan secf x x x x= + Þ ( ) sec tanf x x x K= + +
1 2 0P P+ l =
( )2 3 5 6 0x y z x y z+ + + + l + + - =
( ) ( ) ( ) ˆˆ ˆ2 3 1i j kl = + l + + l + + l!
! 0kl × ="
1l = -2 11 0x y= + + =
2 2
11
1 2=
+
115
=
151,2y+æ ö-ç ÷
è ø3 7y x= +
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Area of rectangle = 14 × 6 = 84
19.(1)
Solve and we get (8, 4) & (2, –2)
Shaded area = 18
Ans. (1)
20.(3)
h = 2r
21.(3)
,
lines are and point of intersection of lines is
Distance from origin .
22.(3)
I.F.
,
,
,
Þ 153 1 72y+æ ö = - +ç ÷
è øÞ 1 1y = -
2 2 , 4y x x y£ £ +2 2y x= 4;x y= +
( )4 2
2
42yy dy
-
æ ö= + -ç ÷ç ÷
è øò
1 1tan2
-q =
1tan2
rh
q = = Þ
213
V r h= p
2 313 4 12h hV h p
= p × =
2312
dv dhhdt dt
p= ×
Þ 5 1004
dhdt
p= × ´ Þ
15
dhdt
=p
( )1 1x a y+ - =
22 1x a y+ = { }0, 1a RÎ -
1 2 1m m = -
21 2 11a a-
- × = --
3 22 a a- = -
3 2 2 0a a- + = Þ 1a = -
\ 2 1x y- = 2 1x y+ = \3 1,5 5æ ö-ç ÷è ø
9 1 10 225 25 25 5
= + = =
6tan ; 0,cos 2
dy xy x xdx x
pæ ö- = Îç ÷è ø
tan log cos cosdx xe e x
- += = =ò6cos coscosxy x xdxx
× = ×ò 2cos 3y x x c= +
, 03
y pæ öç ÷è ø
20 3
9C
æ öp= +ç ÷ç ÷
è ø2
3C p= -
22cos 3
3y x x p
= -
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,
23.(1)
d = 4, – 4
24.(1)
... (i)
r = 2.
Average
25.(3) Slope of normal at P = slope of CP
Slope of normal
2 2 2 2 2 23cos 36 36 3 12 3 12 4
yæ öp p p p p p p
= - = - = - = -ç ÷ç ÷è ø
2 224 3 2 3
y p -p= - × =
, , Termsa d a a d- + ®
3 33 11a a= Þ =
( )( ) 1155a a d a d- + =
( )211 121 1155d- =
2121 105d- =2 16d = Þ
( )( )11 15 10 4 40 15 25T = + - = - + = -
1 1: : 2 :15 : 70n n nr r rC C C- + =
12: ,15
n nr rC C- =
1
152
nr
nr
CC -
=
1 152
n rr
- +=
2 2 2 15n r r- + =
2 17 2n r- = -
( )1 1 170 1415 1 3
nr
nr
n rCrC
+ - + += Þ =
+
141 3
n rr-
=+
Þ 3 3 14 14n r r- = +3 17 14 ...( )2 17 2 ...( )
16
n r iin r i
n
- =- = -
- + +=
48 17 14r- =17 34r =
Þ
16 16 16 171 2 3 3
17.16.151616 16 680 6961.2.3 2323 3 3 3 3
C C C C ++ + + += = = = = =
2 4y x=
2 ' 4yy =
2'yy
=
2 12-
= = -
211
b-- =
a -1 2-a + = b-
3b = -a
( ) ( )2 2 21 2a - + b- = b
2 2 22 4 5a +b - a - b+ = b
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... (i)
Radius
Area
26.(1) Verify with options
27.(4)
28.(2) 10, 22, 26, 29, 34, x, 42, 67, 70, y
Mean
... (i)
Median x = 36 y = 84
29.(1)
2 2 4 5 0a - a - b+ =2 2 4 5 0a - a - b+ =2 2 12 4 5 0a - a - + a + =2 2 7 0a + a - =
( )21 8a + =
1 2 2a + = ±
1 2 2, 1 2 2a = - - - +
\ ( )3 1 2 2b = - - +
4 2 2b = -
( )2= p b
( ) ( ) ( ) ( )2 24 2 2 4 2 2 4 4 2 4 2 8 3 2 2= p - = p - = p + - = p -
( )p q rÞ È
; ;p T q F r F= = =
( )T F FÞ È
T FÞ
sin10 sin30 sin50 sin70° ° ° °1cos80 cos40 cos202
æ ö° ° °ç ÷è ø
1 cos20 cos40 cos802
° ° °
( )3
3
sin 2 2012 2 sin 20
× °×
°
1 sin16016 sin 20
°°
( )sin 180 201 sin 20 116 sin 20 16sin 20 16
°- ° °= = =
° °
10 22 26 29 34 42 67 7010x y+ + + + + + + + +
=
3004210x y+ +
= Þ 420 300 x y= + +
120x y+ =
34 352
x += = \
84 736 3
yx= =
33TA A I=
0 2 2 0 2 1 1 0 02 2 1 3 0 1 01 1 1 2 1 0 0 1
x x yy y y R y
x y
é ù é ù é ùê ú ê ú ê ú- - =ê ú ê ú ê úê ú ê ú ê ú- - +ë û ë û ë û
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;
Ans. 4
30.(4)
is continuous at x = 4.
2
2
8 0 0 3 0 00 6 0 0 3 00 0 3 0 0 3
x
y
é ù é ùê ú ê úê ú = ê úê ú ê úë ûê úë û28 3x =
38
x = ± 26 3y = Þ12
y = ±
( ) [ ]4xF x x é ù= - ê úë û
( ) ( ) [ ] ( )0 0 04
4lim lim 4 lim 4 lim 4 1 34h h hx
hF x F h h+ ® ® ®®
æ ö+é ù= + = + - = - =ç ÷ê úë ûè ø
( ) ( ) [ ] ( )0 0 04
4lim lim 4 lim 4 lim 3 0 34h h hx
hF x F h h- ® ® ®®
æ ö-é ù= - = - - = - =ç ÷ê úë ûè ø
( ) [ ] 44 4 4 1 34
F é ù= - = - =ê úë û
( )F x
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