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: