SOLUTIONS & ANSWERS FOR JEE MAINS-2021

25th February Shift 1

[PHYSICS, CHEMISTRY & MATHEMATICS]

PART – A – PHYSICS

SECTION A

Ans: 1

Sol:

( )n

0

Tt

0

2

N

2

NN ==

21 n

2

n

1

2

N

2

NN ==

Given n1 = 2 n2 = 6

( )

8222N

N 336nn

2

1 21 ==== −−

Ans: 1

Sol:

Current through resistance = A2

140

20 =

Voltage across capacitor = 10 V

Current through capacitor = A4

1

104

1

C

1=

F250C =

Voltage across inductor r = 20 V

Current through inductor = A4

1

H8.0L204

1L ==

Ans: 3 Sol: V2 = Vc

2 + 2aℓ --------(1) Vc

2 = u2 + 2aℓ --------(2)

(1) –(2) V2 – Vc2 = Vc

2 – u2

2

vuV

22

c+

A4

1 A4

1

~

C 10V

15 V

20 V L

A2

1

R=60 Ω

R’=40 Ω

Ans: 3 Sol: Maximum frequency of modulated signal = carrier wave frequency + signal frequency Minimum frequency of modulated signal = carrier wave frequency – signal frequency

Ans: 1

Sol: planetg

2T

=

22

2

2

planet s/m2T

4g =

=

Ans: 1

Sol:

mqv2

h=

22em

e2m4=

=

Ans: 2 Sol:

k4j3i2AO ++=→

→→→

+= BCABAC

++++

+++

++

→→→→→→→→→→

DECDBCABCDBCABBCABAB +

++++

→→→→→

EFDECDBCAB

++++++

→→→→→→

FGEFDECDBCAB

+++++++

→→→→→→→

GHFGEFDECDBCAB

−=

−=

−=

−=

→→

→→

→→

→→

DEHA

CDGH

BCFG

ABEF

+++

++++

+++

++

→→→→→→→→→→→→→

DECDBCDECDBCABCDBCABBCABAB

+

++

→→→

DEDECD

A B

C

D

E F

G

H

O

+++

→→→→

DECDBCAB4

k32j24i16 −+=

Ans: 4

Sol: A43.010

7.05i =

−=

Ans: 1 Sol: Option (1)

Ans: 3

Sol: f

VfV ==

504

336=

4

e =+

cm87.14=

11. Ans: 4

Sol:

( ) ( )( ) ( )

( ) ( )( ) ( )x221x2x221x2

x221x2x221x2

1x21x2

1x21x2

22

22

−++++

−+−++=

−++

−−+

1x2

x22

2x4

x24

+=

+

Ans: 1

Sol: 21

R9

mMGF =

212

R2

5

m8

GM

FF

−=22 R

GMm

509

41

50

1

9

1

R

GMm

−=

Ans: 3

Sol: T2

R5nTnCU v

==

TnRW =

2:7:5W:Q:U =

Ans: 3

Sol: ( )

m

q

m

BmvqqvBF ==

4

2:

2

1:

1

1F:F:F 321 =

m

1m

PVmVP

1:1:22:2:4

==

==

V1: V2 : V3 = 4

1:

2

1:

1

1 = 4 : 2 : 1

Ans: 2 Sol: At t = 0, inductor offers infinite resistance

At t = , inductor offers zero resistance

5 Ω 5 Ω

4 Ω 1 Ω

E

5 Ω 5 Ω

4 Ω 1 Ω

6 Ω 9 Ω I1

5 Ω

5 Ω

1 Ω

4 Ω

I2 E

5

4

2

5

5

14

2

5qRe +=

+=

10

33

10

825=

+=

Ans: 2 Sol:

= 1.287 × 10 3

Ans: 1

6400

600

200 kg

400 kg

1600 km

A

B

Sol: Lest count = 100

1

divisionofnumber

pitch=

100

18Error =

( )100

18

100

1721R2adingRe −+=

2R = 1.64

R = mm82.02

64.1=

Ans: 3

Sol: R

GM2Ve =

2

2

1

1

R

M

R

M=

Ans: 3 Sol: Option (3)

Ans: 1

Sol:

( ) 23

22

20

Rx2

NIRB

+

=

( )

( )

8

Rx2

NIR

Rx2

NIR

8B

B

23

222

20

23

221

20

2

1

=

+

+

=

(0.2)2 + R2 = 4 (0.052 + R2)

22

22

R4100

1R

100

4

R100100

254R

100

4

+=+

+

=+

m1.010

1RR3

100

3 2 ===

SECTION B

Ans: 50.00

Sol: = dV.KVdVPW 3

4

KVKV

4

VK

41

42

V

V

2 2

1

−=

=

( )124

142

114

122421

312

32

TTnRKVKV

VPKV,VPKV,PKV,PKV

−=−

====

( )( )

nR504

100300nR

TTnRW 12

=−

=

−=

Ans: 1.00

Sol: 3rr

U510−

−

=

( ) ( )

611 r

5

r

10

dr

dUF

−=

−=

−=

At equilibrium, F = 0

( )

611 r

5

r

10 =

b

a5

1

5

22r

5

10r

=

=

=

5

1

b

a=

5b,1a ==

Ans: 128.00 Sol: Let R → radius of bigger drop r → radius of small drop Vbig = 512 Vsmaller

R = 8r

V = r

qK

K

r2q

rq

K2 ==

Q = 512 q

Q = K

r1024

V128r8

K

r1024K

r

KQ'V =

==

Ans: 1.00

Sol: E ℓ E1 = K (380) E2 = K(760)

b

a

2

1

E

E

2

1 ==

1a =

Ans: 3600.00

Sol: TnCMVn2

1v

2 = TR2

3n =

R3

3600

R3

30304

R3

MVT

2

=

==

3600x =

Ans: 5.00 Sol:

A

B V

U

mg

T2

r

T1

Tmax – Tmin = 6 mg Given, Tmax/ Tmin = 5 On solving, we get

mg2

3T

mg2

15T

min

max

=

=

Ans: 10.00

Sol: 960

103VnnV

8=

==

At resonance,

C

1L

=

Ans: 144.00

Sol: dt

diLV =

=

=

4

0

4

0

dt2

t3i

dtL

Vi

124

t34

0

2

=

=

Ans: 15.00

Sol: 21 mm = (since the size of image is same)

21 uf

f

uf

f

+=

+

mg

21 uf

f

uf

f

+−=

+(since, one image is real and other is virtual)

f + u2= -f – u1 2f = -u2 – u1 2f = -(-10) – (-20) 2f = 10 + 20

f = 2

30= 15 cm

Sol: A.E= For Y – Z, plane, A1 = 0.2 i

0E5

6.0=

For X – Z plane, A2 = 0.3 i

( )j3.0jE5

4iE

5

3002

+= 0E

5

2.1=

2.1

6.0

1

2 =

1a

b

a

2

1

=

==

PART – B – CHEMISTRY

SECTION A

Ans: 2

Sol:

Ans: 1

Sol: (D) represent the correct plot with 2 radial nodes.

Ans: 3

Sol: For 4f series +3 oxidation state is more stable

Ce4+ acts as oxidizing agent while Eu2+ acts as a reducing agent

Ans: 1

Sol:

Number of moles of CxHyOz

Number of moles of CO2

Number of moles of H2O

Number of moles of CO2 =

x = z

The empirical formula is CxHyOz or CH2O

Ans: 4 Sol: Reducing smog consist of SO2 Photochemical smog consists of oxides of nitrogen

Ans: 3 Sol: Aromatisation reaction results in the formation of toluene

Ans: 3

Sol: AgCN

s' x’

Ans: 3

Sol: In the reaction 2I− + H2O2 + 2H+ → I2 + 2H2O hydrogen peroxide act as oxidizing agent

Ans: 2

Sol: The terminal B−H bonds have less p-character than bridged B−H bond.

Ans: 2

Sol: V2+ − [Ar] 3d3

Cr3+ − [Ar] 3d5

Mn2+ − [Ar] 3d5

Ni2+ − [Ar] 3d8

Cu+ − [Ar] 3d10

Fe2+ − [Ar] 3d6

Co+ − [Ar] 3d7 4s1

Ans: 2

Sol: The central metal ion in the given complexes are Mn2+ and Fe3+, both have 3d5 configuration, since

CN− is a strong field ligand both complexes will have d2sp3 hybridisation, odd number of d subshell electron makes then paramagnetic.

Ans: 4

Sol: Lactose is a disaccharide consists of galactose and glucose linked by C1-C4 glycosidic linkage.

Ans: 1 Sol: The synthesis of Buna-S needs nascent oxygen.

Ans: 2

Sol: Ellingham diagram is a plot between G (for the formation of metal oxide from element and one mole of oxygen) versus absolute temperature.

Ans: 4 Sol: There is no reaction between nitrobenzene and sodium borohydride

Ans: 2

Sol:

Ans: 1

Sol: Benzoic acid and picric acid are more acidic than carbonic acid, thus they react with NaHCO3 to form sodium salt and CO2

Ans: 2

Sol:

Ans: 2 Sol: The bond order of Be2 is zero

Ans: 4

Sol:

OCH3

H+

NO2

OCH3

Cl−

NO2

more stablecarbocation

OCH3

NaI

NO2A

dry acetone

Cl

OCH3

NO2

I

SECTION B

Ans: 70

Sol:

Ans: 4

Sol: 0.4 g (NaOH + Na2CO3 + Impurity)

Hph 17.5 mL 16 mL for NaOH MOH 1.5 mL 2HCl + Na2CO3 2 1.5 mL 0.1 1.5 m. moles 0.1 Na2CO3

Wt. = 1.5 10−3 106 0.1

= 159 10−3 g 0.1

Ans: 1 Sol: Only SF6 is inert towards water

Ans: 526

Sol:

T2 = 526 K

Ans: 4

Sol: = 4 10−1

Ans: 173 Sol: Equivalents of

+6 +3

→ n = 3

+2 +6

→ n = 8

V = 173 mL

Ans: 5576

Sol: Hf = 495.8 + (−325) + (−728.4)

= −557.6 kJ mol−1

= −5576 10−1 kJ mol−1

Ans: 7

Sol: 3CH3CHRed bbt Fe Co + HCl

AlCl3

O

HC

Ans: 375

Sol: I = 1 + (n −1)

= 1 + 4 0.6 = 3.4

Tb = 3.4 0.52 1 = 1.768 Boiling point = 373.15 + 1.768 = 375 K

Ans: 741

Sol: H = U + nRT = −742.24 + 0.5 8.314 10−3 298 = −741 kJ mol−1

Magnitude of H = 741

PART – C – MATHEMATICS

SECTION A

Ans: 2

Sol: ( ) ( ) 1nf1nf +=+

( ) ( )1f22f =

( ) ( )1f33f =

( ) ( )1f44f =

. . . . .

( ) ( )1nfnf =

f(x) is one-one

Generally fg is one-one f is one-one

fg is onto g is onto

Ans: 3

Sol: 0nm&0nm 222 =−+=−+

( ) 0mmmn222 =+−++=

0m2 =−

0m =

0m)or(0 ==

0= m = 0 0:1:1 1:0:1

2

1

22

100cos =

++=

2

3sin =

8

5

16

10

16

1

16

9cossin 44 ==+=+

Ans: 3

Sol:

=−

=+++=22

42

sin

1

cos1

1..............coscos1x

[infinite G.P series]

=−

=+++=22

42

cos

1

sin1

1.............sinsin1y

.................cossincossin1z 4422 +++=−

=22 cossin1

1

−

−−

=

y

11

x

111

1z

( )1yxxyxy

xyz

+−−−=

1yx

xyz

−+=

( ) zxyyxz +=+

Ans: 4

Sol: ( )11b

y

a

x 22

→=+

diff w.r.t ‘x’

0dx

dy

b

y2

a

x2=+

( )2ay

bx

dx

dy→

−=

( )31d

y

c

x 22

→=+

diff w.r.t ‘x’

1dx

dy

d

y2

c

x2=+

( )4cy

dx

dx

dy→

−=

1mm 21 −=

1cy

dx

ay

bx−=

−

−

( )5acybdx 22 →−=

( ) ( ) 0yd

1

b

1x

c

1

a

131 22 =

−+

−−

0xac

bd

bd

bdx

ac

ac 22 =

−

−+

−

( ) ( ) 0bdac =−−−

bdac −=−

badc −=−

Ans: 1 Sol: Given xyz = 2y

32xyz 3 =

s'2of.Noxi =

s'3of.Noyi =

3xxx 321 =++

Non-negative integral sol

1yyy 321 =++

13131 Cwaysof.No −−+= 2

3C=

Total no. of ways = 30CC 23

25 =

x3, y3

x y z

x1, y1 x2, y2

Ans: 4

Sol: Prob = 8

1

4

3.

3

24

=

Ans: 3

Sol: Given lines ( )( ) ( )( )iyxi2iyxi2 −+=+−

( ) ( ) ( ) ( )y2xiyx2xy2iyx2 −++=−++

y2xxy2 −=−

0y4x2 =−

( )10y2x →=−

( ) ( ) 0i4z2izi2 =−−++

( ) ( ) i4zi2zi2 +−=+

( )( ) ( )( ) i4iyxi2iyxi2 +−−=++

( ) ( ) ( ) ( )y2x4iyx2y2xiyx2 −−+−=++−

y2x4y2x −−=+

04y4x2 =−+

( )202y2x →=−+

Solving (1) & (2)

x − 2y = 0 x = 2y

From (2) 2y + 2y − 2 = 0

2y4 =

2

1y =

x = 1

2

1y,1x ==

0i1ziz =+++

( ) ( ) 0i1iyxiyxi =++−++

0i1iyxyix =++−+−

( ) 01yxor01yx =+−=+−

22

3

2

12

11

r =

+−

=

Ans: 4

Sol: Time to cover distance = v

y

13

20

−= ( )sec1310 +=

Ans: 4

Sol: ( ) 4bxaxxxf 23 −+−=

f(1) = f(2)

4b2a484ba1 −+−=−+−

•

x − y + 1 = 0

(1, ½)

( )17ba3 →=−

( ) bax2x3x'f 2 +−=

03

4'f =

0ba3

8

9

163 =+−

( )216b3a8 →−=+−

From (1) & (2) a = 5, b = 8

Ans: 4 Sol:

A B B→A A→(B→A)

T T T T

T F T T

F T F T

F F T T

Verify option (4)

A→ (AB)

A B A B A → (A B)

T T T T

T F T T

F T T T

F F F T

Ans: 4 Sol:

x − y + 1 = 0

(3, 5)

• (h, k)

•

( )

221111

ba

cbyax2

b

yk

a

xh

+

++−=

−=

−

( )

2

1532

1

5k

1

3h +−−=

−

−=

−

1k53h =−=−

4k,uh ==

( ) ( )4,4k,h =

Clearly (4, 4) lies on ( ) ( ) 44y2x22=−+−

Ans: 3

Sol: Put sin = t

cos d = dt

( ) dt6t3t2ttt 24246 ++++ ( ) ++++= dtt6t3t2ttt 24635

Let zt6t3t2 246 =++

( ) dzdtttt12 35 =++ = dzz12

1Cz

18

1 23

+= Csin6sin3sin218

1 23

246 ++=

( ) ( ) ( ) Ccos16cos13cos1218

1 23

22232 +

−+−+−=

Ccos2cos9cos181118

1 23

242 +−+−=

Ans: 3

Sol: 6,5,4,3,2,1c,b,a

n(S) = 63

equal roots D = 0

ac4b2 =

4

bac

2

=

If b = 2, ac = 1 a = 1, c = 1 If b = 4, ac = 4 a = 4, c = 1 a = 1, c = 4 a = 2, c = 2

If b = c, ac = 9 a = 3, c = 3

probability = 216

5

Ans: 4 Sol:

( ) 02cos12tan +

2

7,3

2

5,2

2

3,

2,02

4

7,

2

3

4

5,

4

3,

24,0

O 2

Ans: 3 Sol: Given point (0, 1, 2)

Line 2

1z

3

1y

2

1x

−

−=

+=

−

Verify option (3)

dr’s ( ) ( ) ( )( ) 0323432 =−++−

clearly passing through (0, 1, 2)

Ans: 1

Sol: 2x

8yx4x

dx

dy 2

−

++−=

( )( )2x

4y2x2

−

++−= ( )

2x

4y2x

−

++−=

Let x − 2 = t dx = dt

y + 4 = u dy = du

dt

du

dx

dy=

t

ut

dt

du+=

tt

u

dt

du=−

t

1eeF. tlog

dtt

1

=== −−

= dtt

1.t

t

1.u

Ctt

u+=

( ) C2x2x

4y+−=

−

+

Passing through (0, 0) C = 0

( )22x4y −=+

Option (1) (5, 5) satisfy the equation y + 4 = (x − 2)2

Ans: 3

Sol: Let

n

2n n

n

1..........

2

11

1limL

+++

+=→

almod1eLSo

n

n

1.........

3

1

2

11

imn

+++

→

=

++++

→

=2n n

1...............

n3

1

n2

1

n

1lim

e

0.............00e +++=

= e0 = 1

Ans: 2 Sol: D < 0

( ) ( ) 07k841k32 22−−−

( ) ( ) 07k841k6k94 22 −−+−

08k6k2 +−

( )( ) 02k4k −−

4k2 Then k = 3

Ans: 1

Sol: −

=

1

1

x2 dxexLet3

−

− +=

1

0

2

0

1

12 dxxdxex

1

0

30

1

3

3

x

e3

x

+

=

−

3

1

e3

1+=

Ans: 1

Sol: Equation of tangent y = m3

2mx +

Perpendicular slope of 2x + y = 1 is 2

1

Tangent is y = 32

x+

( )106y2x →=+−

(5, 4) does not lie in equation (1)

SECTION B

Ans: 11

Sol: acar =

( ) 0acr =−

acr =−

car +=

( ) ( ) 0cbba0br =+=

( ) 021 =+−

2=

ca2r +=

( ) ( )kjikj2i2 −−+−+=

k2j3i3 −+=

( ) ( )kj2ikj3i3ar −+−+= = 3 + 6 + 2 = 11

Ans: 144

Sol: ( ) 3456 xbxaxxxf +++=

( ) 2345 x3bx4ax5x6x'f +++=

Roots 1 & −1

03b4a56&03b4a56 =+−+−=+++

03b4a5&09b4a5 =−−=++

Solve we get 2

3b,

5

3a

−=

−=

( ) 3456 xx2

3x

5

3xxf +−−=

( )

+−−= 824

5

966452f5 = 144

Ans: 21

Sol:

322

213

11k

D3

−−

−=

( ) ( ) ( )26149143k0 −−++−+−=

813k0 −−=

k = 21

Ans: 2

Sol: ( )1034kykx3 →=−+

( )20k34yx3 →=−−

( ) ( )2k1 +

( ) 01k34kx32

0k34kykx3

034kykx3

2

2

=+−

=−−

=−+

( )3k

1k2x →

+=

From (2) ( )4kk

132y →

−=

From (3) & (4) kk

1

32

y,

k

1k

2

x−=+=

412

y

4

x 22

=−

148

y

16

x 22

=−

24

8

4

64

4

4816e ===

+=

Ans: 7

Sol: 32A −

=

100

010

001

yxz

xzy

zyx

yxz

xzy

zyx

Clearly 1zyx 222 =++

0zxyzxy =++

Now ( ) ( )zxyzxy2zyxzyx 2222+++++=++

x + y +z = 1

|||A| 2 =

1|A| =

( ) 1zyxxyz3 333 =++−

5)or(7123zyx 333 ==++

7zyx 333 =++

Ans: 3

Sol: ( ) |2xx||2x|3|1x2|xf 2 −+++−+=

|)2x()1x(||2x|3|1x2| +−++−+=

Non differentiable at 1,2,2

1x −

−=

Ans: 32

Sol: Divisible by ‘3’

Sum is 12 → 3, 4, 5 → 3! = 6

Sum is 9 → 2, 3, 4 → 3! = 6

Sum is 9 → 1, 3, 5 → 3! = 6

Sum is 6 → 1, 2, 3 → 3! = 6 24 Divisible by 5

32

5,1,3

5,3,1

5,3,4

5,4,3

41224noTotal

=

−+=

Ans: 9 Sol: Let a1, a2, a3…………an are length of sides of squares A1, A2,…………An 21 a2a =

2

aa 1

2 =

( )nn

1n

2

aa,e.i =+

Length of side A1 = 12

Length of side

( )12

2

12A =

Length of side

( )23

2

12A =

. . . .

Length of side

( ) 1nn

2

12A

−=

Area of

( )1

2

12A

2

1nn

=−

12

1441n=

−

n − 1 = 8 n = 9

3 ways

5

fixed

4 ways = 12

Ans: 64 Sol:

( )

−=

45

4

dxxcosxsinA 45

4

xsinxcos

−−=

+−

−−−=

2

1

2

1

2

1

2

122

2

4==

( ) 6422A44 ==

Ans: 13

Sol: ( )

( )

=

−=

10

01,

02

tan

2tan0

A 2

−=+

12

tan

2tan1

A2

−

=−

12

tan

2tan1

A2

( )

−

+=−

−

12

tan

2tan1

2tan1

1A

2

12

( )( )

−

−=−+

−

12

tan

2tan1

12

tan

2tan1

2cosAA 21

22

−

−−=

2tan1

2tan2

2tan2

2tan1

2cos

2

22

2tan1

2tan2

b,

2tan1

2tan1

a22

2

+

=+

−=

a = cos, b = sin

( ) ( ) 13sincos13ba13 2222 =+=+

45

4