Solutions to Mock JEE Main - 1 |JEE-2021

Solutions | Mock JEE Main-1 | JEE 2021

Solutions to Mock JEE Main - 1 |JEE-2021

SECTION-1 1.(A) /F Adv dx

1 1timearea

dx MVF m adv A A

2.(B) Applying conservation of momentum, ˆ ˆ ˆ ˆ( 2 ) 3 (0) 4 ( )m ui uj m m vi v j

/ 4v u and / 2v u

3.(C) 2 21 1 313.6 13.61 4 4

hc Z Z ; 4

3k

line of Paschen

2 1 113.69 25

hc Z

9 2516

k ; 3 9 254 16

3 2

63 5

2

4.(D) 0.5m

c

AA

{ 20 volt, 40 volt}m cA A

( ) sinm mm t A t 6{ 2 10 }c

( ) sinc cc t A t 3{ 2 10 10 }m

( ) ( sin )sinm c m m cC t A A t t {1 sin }sinc m cA t t

5.(C) 11 2

LRa

; 22 23

LRa

Both are in parallel 1 2 1 22

1 2 1 2(3 )eqR R LR

R R a

6.(A) 24pIr

… (i)

and 20 0

12

I E c … (ii)

From (i) and (ii)

20 02

124

p E cr

0 20

24

pEr c

; 9

30 6 8

9 10 2 15 2 10 V/m15 15 10 3 10

E

7.(C) RTvM

; 1 1

2 2

V TV T

; 1363 340 1.1 340 374 m/s300

V

8.(D) Fact. Due to decrease in intermolecular forces

9.(B) Time taken by trains to collide 300 2hrs150

Speed of bird = 200 km/hr Total distance travelled by bird 200 2 400 km

PHYSICS


10.(B) At minima, 20 0 0( 25 ) 16I I I I

At maxima, 20 0 0( 25 ) 36I I I I Ratio

16 436 9

11.(C) The time spent in magnetic field (t) is independent of velocity of charge. 1 24

mtqB

12.(C) At second surface, 25i 25r

50i r

13.(A) Workdone by gas = area enclosed in P-V diagram 2(2) lt-atm

14.(C) Taking torque at top pt, sin30 / 2 cos30 / 2 3mg L F L F mg

15.(B) For permanent magnetic Y is good choice, As it has very high coercivity (comparitively)

16.(C) Let the origin be at the CM of the particles, let the initial positions of the particles be2ax

and2ax and let the instantaneous positions of the particles be x r and x r

Let the instantaneous velocity of each particle be v

Let the time after which the distance between the particles has reduced to 2a be T

Then, for the particle that was initially at 2ax ,

2

22Gm v dvm

drr 2

4Gm v dvdrr

20 /24

v r

a

drvdv Gmr

2 1 22

Gmvr a

1/21 22

Gmvr a

[v is negative because the velocity is towards the –X direction]

1/21 22

dr Gmdt r a

/4

/2 02 2a T

a

r Gmdr dta r a

/4

/2 2 2a

a

r Gmdr Ta r a

… (i)

Let us now evaluate the integral 2

rI dra r

Let 2sin

2ar

sin cosdr a d

Therefore, 2 1sin 1 cos sin22 2 2 2 2

a a aI d d

Since 2sin

2ar

, 1 2sin ra

and 8 22 2sin 2 2sin cos 2 1

r a rr ra a a


So, 1 12 2 22 2sin sin

22 2 2 2

r a r r a ra r a rIa a a

Therefore, from equation (i),

/4

1

/2

22 2 2 1sin2 4 22 2 2 2 2 2

a

a

r a ra a r a a aTGm a Gm

Hence, 32

8 4 8a a a aT

Gm Gm

17.(B) In y axis 0yu v 0 /ya E q m

0,ys 21 02y yu t a t 0

0

22 /y yv mt u aE q

x coordinate at that time xv t 0

00

2v m vE q

20

0

2v mE q

18.(D) Pitch 0.5 mmp Device has +ve zero error

6 28 450 50p pR p 246

50pp ; 243mm 0.5

50 3.24 mm

19.(A) Centripetal acceleration is provided by friction force. Maximum friction force should be equal to or

greater that 2 / .mv R 2mvmg

R ;

100100 10

; 1/10 ; 0.1

20.(C) Using parallel axis theorem for the side BC, the moment of inertia of the triangle about an axis passing through A and perpendicular to the plane of the triangle,

22 2 21 1 3 32

3 12 2 2AaI ma ma m ma

If the angular velocity of the triangle at any instant is , the velocity of the vertex B at that instant is a Therefore, the velocity of B is maximum at the instant the angular velocity is maximum, i.e. when the side BC becomes horizontal Let the angular velocity at this instant be MAX Then, by conservation of energy Gain in kinetic energy = Loss in potential energy

21 3 Loss in height of CM of triangle2 A MAXI mg

2 21 3 332 2 23 2 3MAX

a ama mg mga 2 2

3MAXga

Therefore, the maximum velocity of the vertex B is given by 2 2 2 23MAX MAX

gav a


SECTION-2 21.(7) In case 1, photon 2E eV ; metal surface photon max 1E KE eV

In case 2, photon12400 81550

E eV ; max . 8 1KE eV eV 7eV

22.(3.33) Charge initially on capacitor 1 10 C

Charge initially on capacitor 2 20 C

Finally the potential difference across

both capacitor will be equal.

Let us assume that to be V.

1 2 10V V

10 / 3 3.33V V

23.(70) When relative sliding stops, both move with same velocity. Only friction acts in horizontal direction. for system no external force, we can apply conservation of momentum.

1 12 5 0 6 v 2 m/sv

Now work done by friction on block = change in kinetic energy of block

2 21 (2 12 )2

w m 1 (1)(4 144)2

1 ( 140)2

70 J

24.(1.60) 82Rx ;

52Ry ;

8 1.65

xy

25.(8) EMF induced 4 2 1BLV 8V

Current produced BLVR

8 81

A

26.(175) 1/2 ,t half life log 2 3 se

Initially 200 nuclei were there After 3 half-lives, 9 seconds,

No. of nuclei = 31 200 252

Therefore, the number of nuclei that decayed = 200 – 25 = 175

27.(5) Gold band is equal to 5% tolerance 28.(21) 0 300Hzf

3rd overtone 07 2100 Hzf

29.(800) H L

H L

Q QT T

300 1600 800 J600LQ

work done 1600 800 800 JH LQ Q

30.(100) 1 0.5 1.5R

1 1R 1000 10010

R


SECTION-1 1.(B) Here cyclohexanone is formed as follows

2.(D) If the carbonyl compound is sterically crowded, then it will be reluctant to undergo addition reaction. Moreover, attachment of bulkier alkyl group with the carbonyl carbon decreases the partial positive

charge resulting into the minimization of attack by R from RMgBr. So, the order is

3.(B) 2

23 2 3 3 2 4 3 2 2H O(P) (Q) (R)

CH CH COOH NH CH CH CO NH CH CH CONH

2KOH Br3 2 2 2CH CH NH H O

4.(A) (A) 2SiO has three-dimensional network structure of Si O bonds; while carbon dioxide

consists of discrete 2CO molecules. 2SiO is solid, whereas 2CO is a gas

(B) Because of its great affinity for oxygen, Si always occurs as the oxide, silica 2(SiO ) or in the

form of silicates, which are the compounds of 2SiO with other metal oxides.

(C) 2SiO 2Mg Si 2MgO

(D) The reluctance of silicon to form p p bonds to itself is clearly shown by the fact thatsilicon does not exist in graphite-like structure, but only in diamond like structure.

5.(D) h ,p

there is inverse relation between de-Broglie wavelength and momentum, hence the graph will

be rectangular hyperbola.

6.(D) To stop bleeding, 3FeCl is applied locally because 3FeCl causes denaturation of proteins present in blood. It is a case of coagulation.

7.(A) i. 2 2 2Zn 2NaOH Na ZnO H

ii. 2 2 24Au 8NaCN O 2H O 4Na[Au(CN) ] 4NaOH

iii. 3 3 2 2 2(conc.)

Cu 4HNO Cu(NO ) 2NO 2H O

iv. Formation of passive layer of 2 3Fe O on the surface of Fe and 2NO gas is evolved

8.(D) 33 6[Co(NH ) ] is diamagnetic 6 0

2g(t eg ) and is inner orbital complex.

Chemistry


9.(A) When Ferric chloride reacts with potassium thiocyanate a blood red colour of Ferric thiocyanate is formed

3 3FeCl 3KSCN Fe(SCN) 3KCl

10.(D) As 2 2 3Zn(OH) , BeO, Al O are amphoteric, so they can react with both HCl and NaOH.

11.(D) In NS 2 reaction, inversion takes place, so configuration get reversed.

12.(A) 5 4 3 2 1

3 2

2 5

CH C CH CH COOH|C O C H||O

4-ethoxycarbonylpent-3-enoic acid

13.(D) The given value of (spin only)

2.84 n(n 2) BM, So, n 2

Among the given configurations, 4d system in strong field ligand will have 2-unpaired e in 2gt set of orbitals as shown below.

14.(C) The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy.

15.(D) Nitrogen, sulphur and halogens are tested in an organic compound by lassaigne’s test. The organic compound is fused with sodium metal as to convert these elements into ionisable inorganic substances.

Na C N NaCN

22Na S Na S

22Na X 2NaX The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.

16.(A) The second ionization energy of K is maximum, because the second electron is removed from fully

filled 63p subshell. The second ionization energies of group 2 elements decrease down the group. Hence, second I.E. of Ca second I.E. of Ba.

17.(A) All monosaccharides which differ in configuration at 1C and 2C gives the same osazone. Since,

glucose and fructose differ from each other only in configuration at 1C and 2C therefore, they give the same osazone. All other options given in the questions do not satisfy this condition and hence, do not from the same osazone.

18.(B) 2 22CO(g) O (g) 2CO (g)

gn 2 (2 1) 1

gH E n RT or H E 1RT

i.e. H E


19.(D) Sometimes, when maximum covalency is obtained, the halides become inert to water, thus 6SF (or

similarly 4CCl ) is stable. This is because 6SF is coordinately saturated and sterically hindered. Thus,

6SF is inert to water, because of kinetic rather than thermodynamic factor.

A. 5 2 3 4PCl 4H O H PO 5HCl

B. 4 2 4SiCl 4H O Si(OH) 4HCl

C. 3 2 3BCl 3H O B(OH) 3HCl

20.(B) One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process G 0. As phase transformation of water is an equilibrium process and at equilibrium, free energy change is always zero.

SECTION-2

21.(5) ATP has phosphate group, 34PO

So x 4( 2) 3 or x 5

22.(16)

The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the

number of stereoisomers 42 16.

23.(836)

rxn f 2 f 2 f 2 fH H (N O, g) 3 H (CO ,g) (2 H NO , g) 3 H (CO, g) 81 3 ( 393) 2 34 3( 110)

81 1179 68 330 836kJ

24.(0) As 3f-orbital does not exist as value of cannot be equal to n, hence no electron can be accommodated.

25.(400) As 500g of tooth-paste has 0.2g of F

So, 610 g of toothpaste has60.2g 10 g F 400g of F

500g

Hence, concentration in ppm is 400.


26.(150) 1P 1atm

240P 1atm 0.40atm

100

31V 100cm

2V ?

At constant temperature, 1 1 2 2P V P V

So 3

31 12

2

P V 1atm 100cmV 250cmP 0.40atm

Hence, the volume of bulb 3 3B (250 100)cm 150cm

27.(4145.40)

pGº 2.303RT log K 32.303 2 300log10

12.303 2 300 ( 3) 4145.4cal mol

28.(1.93) To decompose 1 mole of 2H O, we need two moles of electrons i.e. 2 faraday

To decompose 4 moles of 2H O, we need 8 faraday.

Now, Q i t 58 96500t 1.93 10 sec4

29.(0.50) Number of mol of A reacted 5 51 10 10

Number of molecule of 5 23A 10 6 10

Quantum efficiency 18

196 10 0.50

1.2 10

30.(64) Temperature coefficient is 2 that means rate of reaction doubles at every 10°C rise in temperature.

Thus, 690 C

30 C

k 2 64k


SECTION-1 1.(C) (10 ) .1p hr ( /10 ) .8p s hr

(7 ) .2p hr ( / 7 ) .6p s hr (4 ) .7p hr ( / 4 ) .4p s hr ( ) .1 .8 .2 .6 .7 .4p s

.7 .4 28 28 7(4 / ).1 .8 .2 .6 .7 .4 8 12 28 48 12

p hr s

2.(B) 0 1 1

1 1 01 1 1

1(1 1) ( 1) 1( 1) 0 22 1 0

2 1 1, 1 2 2 2a b

3.(C) 1 2 3, ,b b b 2, , ........a ar ar

11(1 )

a ar ar

12 2

1kk

abr

21 2

1 r

2 1 12 2

r r

2102

b r 1 2 2a b

4.(D) 2 2( ) ( )( ) 2 ( 2 )I M I M I M I M M I M 3 2( ) ( 2 )( ) 3 2 ( 3 )I M I M I M I M M I M 50( ) 50I M I M

50det(( ) 50 ) det( ) 1I M M I

5.(C) 2 4y x 2 2( 3) 9x y

1y mxm

(3, 0), 3r

2

133

1

mm

m

Mathematics


2 22

19 6 9 9m mm

21 13

3m

m

13

m in first quadrant

3 3 33

xy y x

6.(B) 1 13

r

r

TT

1 2 3, ,T T T …………G.P.

71

243T

6 1243

ar

6 51 3

3 3a a

Sum of infinite series 1 2 2 3 3 4T T T T T T …………..

1 22

13 3 2731 81 19

T Tr

7.(A) ~ ( (~ )p p q

~ (~ (~ ))p p q ~ ( ~ )p p q

8.(B) 3 sin 22

y

sinx e 3cos2 3 (cos sin )

(cos sin )dydx e e

9.(C) 2( ) ( )ar New ar Oldv k v 4 5 20

10.(B) 3 3

3 2

0 0

( 1)( 2)( 3) 6 11 6x x x dx x x x dx

1 2 33 2 3 2 3 2

0 1 2

( 6 11 6) ( 6 11 6) ( 6 11 6)A x x x dx x x x dx x x x dx 114

A


11.(A) 2 22 2z z

0x Hence z minimum ‘0’

12.(C) 2( ) ( 2)f x x

212 (2) 3 2( ) 3( 2)f x x ; 2(6) 3 4 48f

13.(B) 1tan ( )x t 2 4 3 0t t

( , 1) (3, )t

1tan ( ) , 12

x

( , tan1)x

Largest integral 1x 14.(A) log x t

122 1t

t

12 2 121

1( )r

rr rT C t

t

12 24 3 ( 1)r rrC t

For constant term 8r & coefficient 128C

12 11 10 9 45 11 49524

15.(B) 221dy dx

xy

1 1sin ( )2

y c cx

1sin ( )3 2 6

y

12

y

16.(C) 5 4

( ) 520 12x xf x

4 33 1'( )

4 3 4 3x x xf x x

3 2"( )f x x x 2 ( 1)x x 1x point of inflection


17.(C) '( ) (1 )f x x y

1dy xdx

y

2

ln( 1)2xy c

(0, 0) 0c

2 2

2 21 2x x

y e e 2

(ln 2 0)2x k

2x k 18.(D) For point of intersection

1 3 3 2 1 2 5 5 1, 1 Point of intersection (4, 3, 5) For greatest distance from origin perpendicular from meet plane at point of intersection Hence equation ˆˆ ˆ(4 3 5 ) 50r i j k

19.(D) Differentiable continuous Continuous at 2 2 0x a b Continuous at 3x 0 9 3 1p q Differentiable at 2x

2 2 5 1a a Differentiable at 3x 2 3 5 2 3 6 1p q p q

4 51, 2, ,9 3

a b p q

20.(C) 2 22 3 35a b 0,a b I

21, 11a b b I

2, 9 3 or 3a b b 2 173,

3a b b I

24, 1 1a b b

SECTION-2 21.(570) All even + 2 odd 1 even

10 10 103 2 1C C C

10 9 8 10 9 106 2

120 450 570


22.(12) Family of circle through (2, 2) and (9, 9), ( 2)( 9) ( 2)( 9) ( ) 0x x y y y x

Touches 0y (x-axis)

( 2)( 9) 18 0x x x 2 11 36 0x x x 2 (11 ) 36 0x x has repeated roots

0 1, 23D

6 or 6x x Difference = 12

23.(2) 2 223 4 9 16 24a b a b a b

But 0,a b a b k

3 4 5a b k

4 3 5a b k

10 20 2k k a b

24.(8) / 2

0

cos 22x x dx

/ 4 /2

0 /4

cos 2 cos 22 2x xx dx x dx

/4 /2/4 /2

0 /40 /4

sin 2 sin 2 sin 2 sin 22 2 4 2 2 4x x x x x xdx dx

/4 /2

0 /4

cos 2 cos 216 8 16 8

x x

1 18 8 8 8

25.(4) 2

20

coslim 1sin

x

x

ae b x cx x

x

2

20 0

cos 1lim limsin

x

x x

ae b x cxx

x

For limit to exist 0,a b c

1 2, 22c c b

4a b c

26.(220) 4 4 51 1r r rC C C

4 5 41 1r r rC C C


38 4 8 4 8 4 8 4 8 4

1 0 1 1 2 2 3 3 40

r rr

C C C C C C C C C C

8 4 8 4 8 4 8 4

8 1 7 2 6 3 5 4C C C C C C C C

129

12 11 10 2206

C

27.(0) 2 4 4 0A A I 2 2 2 4 0A AI A I

2( 2 ) 0A I 2 0A I

297 594B A I 297( 2 )A I

28.(16) 4 5 0 2 50 1 2 51 (1 ) (1 ) (1 ) .......... (1 )x x a x a x a x a x

Differentiate 3 4 2

1 2 34 5 2 (1 ) 3 (1 )x x a a x a x ………. 2 3

2 312 20 2 6 (1 )x x a a x ………. Put 1x

2 212 20 2 ............ 16a a 29.(9)

5, 2, 1c a b

30.(8) PQ is focal chord. Quadrilateral PTQR is square

Area 82

PQ TR

Solutions to Mock JEE Main - 1 |JEE-2021

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