Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
SOLUTIONS & ANSWERS FOR JEE MAINS-2021
26th February Shift 1
[PHYSICS, CHEMISTRY & MATHEMATICS]
PART – A – PHYSICS
SECTION A
Ans: 4 Sol: Option (4)
Ans: 4 Sol:
T2'T
2K'K
=
=
m
m
T=2πK
m
K
K K
K
m22
2K
M2
'K
M2'T ===
Ans: 3 Sol: Option (3)
Ans: 1 Sol:
E
2
L3
2L
2L
3
1
23.L
2L
tan ==
= 30°
( )2L3
KQ230sin30sin
L2
L3
KQE =+
=
2
0 L32
QE
=
Ans: NTA key is (2) Sol: In sufficient data
Ans: 4
Sol: It is a balanced wheatstone’s bridge
R2
R2R eq ==
Ans: 4 Sol:
8
1tan =
( )( )
=== 2
R3
.M.Gdm
3
8dFcoscosdFF
22 R
GMm
27
8dm
R27
GM8==
OR
( ) ( ) 23
2223
22 R8R
R8GmMx
xR
GMxMMEF
+
=
+
==
2R27
GMm8=
B
R
R R
R
A R
8 R
R
M
dF
F cos
Ans: 2
Sol: r
mVF
2
=
m
K
r
1V
Vmr
K
r
mV
r
K 2
2
2
3
=
==
2r
m
K
r
1
r2
V
r2T
=
=
2rT
Ans: 3
Sol: 2
2
2
1
RZ16
15
4
11RZ
1=
−=
−=
22
2
2 4
1
3
1RZ
1
135
7
2
1 =
Ans: 3
Sol: 2xKT =
ML2 T-2 = [] [L2]
[] = MT-2
2 = energy = ML2 T-2
2 = L2
[] = [L]
Ans: 4 Sol: I = I1 + I2 + I3 + I4 [ I1 = I2, I3 = I4] = 2I1 + 2I3
= (2 × ma2) + 2( ma2 + mb2)
= ma2 + 2mb2.
Ans: 4
Sol: As image has same orientation as that of the object, image is virtual.
m =
So, virtual and diminished image will form by only convex mirror on opposite side.
Ans: 1
Sol: Areal velocity = = constant
Ans: 2
Sol:
as v = constant
Ans: 3 Sol: R3 = nr3
A = n × 4r2 - 4R2
JH = A × T
Heat per unit vol =
Ans: 2
Sol:
tan =
Ans: 2 Sol: F = mg’
mg’ = (OB)m
FR = F cos =
=
xR
GMa
3R =
T =
2R
x
B
F
O
Fres
Ans: 3
Sol: 2
2
1
1
RRH
−=
−=
21
2112
21
2
2
1
1
RR
RR
R
1
R
1
RR
+
+=
+
+
=
Ans: 3
Sol: Eg
hchcEg =
=
19
834834
106.19.1
1031063.6
eV9.1
1031063.6
−
−−
=
=
= 654 nm and red colour
Ans: 3
Sol: 3
9
102
110500
d
D
−
−
=
=
= 250 × 10-6 m
= 250 m = 0.25 mm
SECTION B
Ans: 400.00
Sol: R
LQ
=
422R
R
L
'L
Q
'Q
1
===
Q’ = 4 Q = 4 × 100 = 400
Ans: 500.00
Sol: j10i20F +=
j5i102
j10i20
m
Fa +=
+==
22
x 10102
1ta
2
1x == = 500
Ans: 25.00 Sol: By energy conservation
( )RTnn2
fRTn
2
fRTn
2
f212211 +=+
( )212211 VVP2
fVP
2
fVP
2
f+=+ [PV = nRT]
10
5.535.42
VV
VPVPP
21
2211 +=
+
+= = 25.5 × 10-1 atm
Ans: 1215.00
Sol: = 0.135 g/ cm = m/kg10
10135.0
2
3
−
−= 135 × 10-4 kg/m
301
30
KV ==
=
90010135rTT
V 42 ==
= −= 1215 × 10-2 N
Ans: 20.00 Sol: R = 130 + 50 + 120 = 300 Ω
mA20A1020300
6
R
Vi 3 ==== −
Ans: 300.00
Sol: = QV = 15 × 20 = 300 J
Ans: 10.00 Sol:
mg2
F3N +=
For no slipping, maxf2
F
+
+ g3F
2
3
33
1mgF
2
3
2
F
N
m
2F
F2
3
f
3
g
3
F
3g
6
F
2
F
3g
6
F
2
F
−
+
3
10x10F =
Ans: 33.00
Sol: 2
AAA minmax
m
−=
3
1
24
8
816
816
A
Am
2
AAA
c
mi
minmaxc
==+
−==
+=
= 33 ×10-2
Ans: 492.00 Sol: mg – N = 1.8 × m N = m (10 – 1.80) = 60 × 8.2 = 492 N
Ans: 137.00
Sol: ( ) Efficiency24
PowerEc
2
1I
2
200
==
( )( )
40
125E
109
103
2
1
100
1
4
25.11000Ec4
2
1
209
8
200
=
=
1
0 109.136E −= = 137 × 10-1 v/m
PART – B – CHEMISTRY
SECTION A
Ans: 4
Sol:
CN
CH2−CH3
Br2
UV light
CN
CH−Br
CH3
Ans: 4
Sol: CH3−C−CH2−CH3
Cl
ClHydrolysis
373 KCH3−C−CH2−CH3
O
(A)
(B)
Ans: 3
Sol:
OH
(i) CHCl3 / NaOH
(ii) H3O+
(B)
CHO
OH
Br2 in CS2
OH
(A)Br
Ans: 1 Sol: A is false but R is true
Dipole-Dipole interactions are non-covalent in nature but ion-dipole interactions can also result in Hydrogen bond formation.
Ans: 1 Sol: Yb does not form MO2 type oxide
Ans: 1
Sol: o-Nitrophenol has low melting point (44.8C). The melting point depend on packing efficiency and not hydrogen bonding.
Ans: 4
Sol: No. of radial nodes = n − − 1 No. of angular nodes = For 5d orbital, radial nodes = 5 − 2 − 1 = 2 angular nodes = 2
Ans: 4
Sol: Kernite − Na2B4O7.4H2O
Cassiterite − SnO2
Calamine − ZnCO3
Cryolite − Na3AlF6
Ans: 4 Sol: O3 generate photochemical smog in troposphere
Ans: 1 Sol: Both statements are true
Ans: 2 Sol: Neoprene is a polymer of 2-chloro-1,3-butadiene (chloroprene)
−CH2−C=CH−CH2−
Cl n
CH2=C−CH=CH2
polymer isation
ClNeopreneChloroprene
(2-chloro-1,3-butadiene)
Ans : 4
Sol: CH=CH−Br
Br
NaNH2
CCH
CH3
Red hot iron tube
873 K
CH3
CH3 CH3(A)
Ans: 1
Sol: NH3 + H2O + CO2 (NH4)2CO3(A)
(NH4)2CO3 + H2O + CO2 NH4HCO3(B)
NH4HCO3 + NaCl NH4Cl + NaHCO3(C)
Ans: 3
Sol: Viscosity of heavy water, D2O (1.07 centipoise) is greater than ordinary water, H2O ( 0.89 centipoise)
Ans: 4
Sol: Compound+ dil.H2SO4 → SO2 (X)
SO2 + K2Cr2O7 (Y)
Ans: 4
Sol: 1s2 2s2 − Be
1s2 2s2 2p1− B
1s2 2s2 2p3− N
1s2 2s2 2p4− O The correct increasing order of the first ionization enthalpy among these elements is of the order B < Be < O < N
Ans: 2 Sol: Carius method is used for the estimation of halogens and sulphur in the organic compounds.
Ans: 3 Sol: Vitamin K is helpful in delaying the clotting of blood.
Ans: 4 Sol: PbO2 is an amphoteric oxide
Ans: 2
Sol: 2 amine react with benzene sulphonyl chloride giving a product which is insoluble in alkali.
R−NH−CH2−CH3 can be prepared by ammonolysis of ethyl halide.
SECTION B
Ans: 2.00 Sol: PV = nRT
1.2 g Pt metal absorb 2.4 L oxygen Volume adsorbed per gram = 2L
Ans: 200
Sol: At eqbm, G = 0 H = TS
80 103 J mol−1 = T 2T K−1 mol−1
2T2 = 80 103 T = 200 K
Ans: 0 Sol: The structure of Mn2(CO)10 is
Mn
CO
CO
CO CO
Mn
COCO
CO
CO
CO CO
Ans: 6 Sol:
Oxidation number of Cr in is +6
Ans: 73
Sol: AB2 A + 2B
1 0 0Initial no. of moles1- 2At. ebqm
PV = nRT
1.9 25 = n 0.0821 300 n = 1.93
1 + 2 = 1.93
2 = 0.93
= 0.465
At eqbm,
Ans: 24 Sol:
24 − x = 0.24
X = 23.76
lowering of vapour pressure, P = 0.24 = 24 10−2 mm of Hg
Ans: 25 Sol: 1 mole required 5F
5 mole required 25 F
Ans: 1
Sol:
X = 1
Ans: 8
Sol: 50000.020 10−3 Number of significant figures = 8
Ans: 50
Sol:
−20 = 30 −
PART – C – MATHEMATICS
SECTION A
Ans: 4
Sol:
Sum of possible values =
Ans: 3
Sol:
Let f(x) = 2x3 – 15x2 + 36 x – 19
f(x) = 6x2 – 30x + 36 = 0 x2 – 5x + 6 = 0 x = 2, 3
f (x) = 12x – 30
f(x) < 0 for x = 2 At x = 2 y = 8 – 40 + 72 – 38 y = 72 – 70 = 2
(2, 2)
Ans: 3
Sol:
f(y) = c
c = 1
f(x) = 1
Ans: 4
Sol:
Ans: 4
Sol:
Put r2 =u
Sum of 4th, 6th and 8th terms is = 5[1 + 2 + 4] = 35
Ans: 3
Sol:
S =
Ans: 3 Sol: One of possibility is : 1, 1, 1, 1, 1, 2, 3
Number of ways of arranging =
Second possibility is I : 1, 1, 1, 1, 2, 2, 2
Number of ways of arranging =
Total number of ways = 42 + 35 = 77
Ans: 1
Sol: Let
Ans: 1 Sol: The vertices of the triangle formed by the lines is (3,0), (1,1) and (2,2). The lengths of the sides are
. Therefore, the triangle is an isosceles triangle.
Ans: 1 Sol: c1 → c1 – c2, c2 → c2 – c3
R2 → R2 – R1 and R3 → R3 – R1
= 6 – 8 = -2
Ans: 2 Sol: The direction ratios of P1 and P3 are proportional. Therefore P1 and P3 are parallel.
Ans: 4
Sol:
Ans: 1
Sol: Let A
= a2 + 2b2 + c2 = 1 a = 1, b = 0, c = 0 a = 0, b = 0, c = 1 a = -1, b = 0, c = 0 c = -1, b = 0, a = 0
Ans: 1
Sol:
=
= (e – 1) + (e – 1) + (e – 1) + ……… (e – 1) = 100 (e – 1)
Ans: 3 Sol:
Ans: 3
Sol:
Ans: 4
Sol:
Ans: 4
Sol:
Sol:
Ans: 3 Sol: Applying componendo dividend,
Now,
Ans: 3
Sol:
Put t = 2, and x = 1200
Now
SECTION B
Ans: 1
Sol:
Solution is given by,
Ans: 8
Sol:
Ans: 2
Sol:
Here, order is 1 and degree 3. Therefore, difference is 2.
Ans: 1
Sol:
Therefore only 1 solution
Ans: 2
Sol:
n = 1.00
Ans: 45
Sol:
Ans: 1
Sol: Solving we get cosx =1 and cosx =
Therefore, number of solutions is 1.
Ans: 11
Sol:
Ans: 3 Sol:
Ans: 4
Sol: