answers & hints...2020/10/18 · test - 1 (code-a)(answers) all india aakash test series for...

JEE (Main)-2022

(Divisions of Aakash Educational Services Limited)

A18-10-2020

Test Booklet Code

(XI Studying Students)

TEST No. 1

ANSWERS & HINTS

Test - 1 (Code-A)(Answers) All India Aakash Test Series for JEE (Main)-2022

All India Aakash Test Series for JEE (Main)-2022

Test Date : 18/10/2020

ANSWERS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

TEST -1 (Code-A)

PHYSICS CHEMISTRY MATHEMATICS

1. (1)

2. (2)

3. (2)

4. (3)

5. (4)

6. (3)

7. (3)

8. (4)

9. (2)

10. (3)

11. (3)

12. (4)

13. (4)

14. (2)

15. (1)

16. (4)

17. (1)

18. (3)

19. (2)

20. (4)

21. (02.00)

22. (55.00)

23. (01.00)

24. (08.00)

25. (06.00)

26. (2)

27. (2)

28. (1)

29. (4)

30. (2)

31. (3)

32. (3)

33. (3)

34. (1)

35. (3)

36. (2)

37. (3)

38. (4)

39. (3)

40. (2)

41. (1)

42. (1)

43. (2)

44. (2)

45. (2)

46. (04.00)

47. (20.00)

48. (01.50)

49. (14.00)

50. (10.00)

51. (3)

52. (4)

53. (4)

54. (1)

55. (3)

56. (2)

57. (2)

58. (3)

59. (4)

60. (3)

61. (3)

62. (3)

63. (2)

64. (2)

65. (1)

66. (3)

67. (3)

68. (2)

69. (2)

70. (4)

71. (05.00)

72. (02.00)

73. (05.50)

74. (01.00)

75. (04.00)

All India Aakash Test Series for JEE (Main)-2022 Test - 1 (Code-A)(Hints & Solutions)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 2/10

PART - A (PHYSICS) 1. Answer (1)

Hint. : 3v r

v r∆ ∆

=

Sol. 343

v r= π

3 0.2100 100 3 100 11%5.2

v rv r

∆ ∆ ∴ × = × = ×

2. Answer (2)

Hint. : v = distance

time

Sol. s v s tvt v s t

∆ ∆ ∆= ⇒ = +

0.2 0.3

13.8 4.0= +

13.8 0.2 0.3 0.3 m/s4 13.8 4.0

v ∴ ∆ = =+

3. Answer (2) Hint. : All non-zero digits are significant. Sol. For a number less than 1, all the zeroes to

the right of the decimal point but to the left of first non-zero digit are not significant.

4. Answer (3) Hint. : v = λαρβgγ Sol. [LT–1] = [L]α[ML–3]β[LT–2]γ

1, 02

⇒ α = γ = β =

Comparing the powers on both sides, we get α, β, & γ.

5. Answer (4) Hint. : v(t) = 4 – 2t Sol. v = u + at 0 = 4 – 2t ⇒ t = 2 s 6. Answer (3)

Hint. : 2 – 22tv gh g ∆

=

Sol. 212

h ut gt= − or 21 02

gt ut h− + =

1 22ut tg

+ = and 1 22ht tg

=

Now ∆t = t2 – t1

2 22 1 1 2( ) ( ) 4t t t t t⇒ ∆ = + −

21 8 ( )2

u gh tg⇒ = + ∆

7. Answer (3)

Hint. : distancespeed =

time

Sol. Remaining distance = 300 – [100 + 40] = 160 km

Time remaining = 3.25 – [1 + 1] = 1.25 h

∴ Least speed = 160 1281.25

= km/h

8. Answer (4) Hint. : Distance travelled during last second of

upward journey is independent of velocity. Sol. Distance travelled in last second of upward

journey = distance covered in 1st second of downward journey and is independent of velocity of projection.

9. Answer (2)

Hint. : h = 21 1

1–2

ut gt+

h = + ut2 + 22

12

gt

Sol. For 1st projection

21 1

12

h ut gt= − + ...(i)

(u : speed of projection) and taking downward as positive)

For 2nd projection

22 2

12

h ut gt= − + ...(ii)

h : height of that point above the ground If it is dropped u = 0

212

h gt= ...(iii)

(i) × t2 + (ii) × t1 gives

1 2 1 2 1 21( ) ( )2

h t t gt t t t+ = +

i.e. 1 2 1 212

h gt t t t t= ⇒ =

Test - 1 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2022


10. Answer (3) Hint. : 2xv = 2t

v = tx

Sol. x2 = 1 + t2 or x = (1 + t2)1/2

2 1/2 2 1/21 (1 ) 2 (1 )2

dx t t t tdt

− −∴ = + ⋅ = +

2

2 1/2 2 3/22

1(1 ) 1 (1 ) 22

d xa t t t tdt

− −− = = + ⋅ + + ⋅

2

31 tx x

= −

11. Answer (3) Hint. : Draw v – t graph and area under v – t

graph gives distance.

Sol. 2 2

2 2v vsa b

= +

Again v v abt v ta b a b

= + ⇒ = +

2 1 1 1

2abs t

a b a b ⇒ = + +

21 1

2t

a b s⇒ + =

12. Answer (4)

Hint. : Average velocity = Total distanceTotal time

Sol. 1/34 12

s st = = (s : total distance)

0 02 2 4( )3s t kt= + or 0

23(2 4 )

stk

=+

Avg. velocity = 1 0 0

6(2 4 ) 35 6

s kt t t k k

+= =

+ + +

12

k⇒ =

∴ Required time = 0

2t

13. Answer (4) Hint. : BD is dimensionless. Sol. From the relation AD = C ln(BD), we know

that BD must be dimensionless, so AD has same dimensions as C.

(i) Now ABD = BC ln(BD) But [BD] = [I] ⇒ [A] = [BC] possible

(ii) [ ] [ ]A AD CB

= = possible

(iii) [ ] andA CABD D

= ⇒ [AD] = [C] possible

(iv) [ ]C CBD

= and [AD] = [C]

⇒ [AD2] = [CD] not possible 14. Answer (2) Hint. : Draw v – t graph and area of v – t graph

gives displacement.

Sol.

20

12

s at=

and 20

1( )2

s at t at− = −

(t : time to return after acceleration changes)

Adding 2 20 0

1 1 02 2

at at t at+ − =

( ) 01 2t t⇒ = +

∴ Total time = ( ) 02 2 t+

15. Answer (1)

Hint. : –vdv C Dxdx

=

( )–vdv C Dx dx=∫ ∫

Sol. a = C – DX

⇒ dvv C DXdx

= − or vdv = (C – DX)dx

Integrating v2 = 2CX – DX2

For v(or v2) to be maximum, 2

0dvdx

=

⇒ CXD

=

maxCvD

∴ =

Also v(or v2) = 0 at X = D and 2CXD

=



16. Answer (4)

Hint. : a = vdvdx

Sol. dva vds

= = |v(slope of the v – s graph)|

= 4 tan60° = 4 3 m/s2

17. Answer (1) Hint. : In frame of train average velocity is zero. Sol. : With respect to train, displacement = 0 ∴ Average velocity w.r.t. train = 0 With respect to ground, displacement of bird =

displacement of train With respect to ground, the bird always moves

with speed v2 18. Answer (3) Hint. : Area under a – t curve gives change in

velocity. Sol. :

Area under the a – t graph gives change in

velocity. The final velocity will be same as initial if the area under the graph is zero.

⇒ 0 00 0 0

( 10)1 15 5 ( 10)2 2 5

t aa a t −× + × × = × − ×

Hence t0 is the sought time

⇒ t0 = 10 5 3+ s

19. Answer (2)

Hint. : a = vdvdx

Sol. : dva vds

= = v × slope of v – s graph

1slope of normal to curve

v −= ×

14

4/1−

= × =−

1 m/s2

20. Answer (4)

Hint. : t = Relative Velocity

h

Sol. : For first stone 21

12

s gt=

For second stone 22

12

s ut gt= −

Where 2

4 82u h u ghg

= ⇒ =

Also s1 + s2 = h (when they cross)

2 21 12 2

gt ut gt h⇒ + − =

8htg

⇒ =

21. Answer (02.00) Hint. : Use concept of relative velocity. Sol. : aball relative to lift = 2 + 10 = 12 m/s2 in

downward direction

rel

rel

2 2 12 212

ut sa

×∴ = = =

22. Answer (55.00)

Hint. : h = 110

0

vdt∫

Sol. : Maximum height will be obtained at t = 110 s after that the velocity becomes

negative i.e. the rocket starts to come back. The maximum height will be area under the graph

1 110 1000 55 km2

= × × =

23. Answer (01.00)

Hint. : ∆v = ( )1 22g h h+

Sol. : aavg = ( )1 22g h hv

t t

+∆=

∆ ∆

3 230 10 m/s0.03avga = =

24. Answer (08.00) Hint. : Apply equation of uniformly accelerated

motion.

Sol. : t1 = t2 – t, 2 21 1 2 2

1 12 2

s a t a t= =

v1 = v2 + v v1 = a1t1 and v2 = a2t2 2 2 1 1 1 2( )a t v a t a t t⇒ + = = −



12

1 2

v a tta a

+⇒ =

−

Also 2 1 1 2

1 2 2 1

( )1 1a t t a ata t t v a t

−= = − = −

+

1 2( )v a a t⇒ = = 8 m/s

25. Answer (06.00)

Hint. : S = 212

at -

Sol. : 212

h at=

1 3 4 6 m2

= × × =

PART - B (CHEMISTRY)

26. Answer (2)

Hint : 104.5% means, 100 g oleum sample requires 4.5 g H2O to produce 104.5 g H2SO4.

Sol. : 10 g sample water required = 0.45 g

SO3 + H2O → H2SO4

0.45 g

Free SO3 = 80 0.45 2 g18

× =

Now, 0.09 g H2O is added which react with SO3

SO3 + H2O → H2SO4

2 g 0.09 g

Mole = 280

0.0918

After reaction 0.02 0 0.005 (mole) Wt. 1.6 g 0 0.49 g ______________________________ Now wt. of H2SO4 =8 + 0.49 = 8.49

% free SO3 = 1.6 1008.49 1.6

×+

= 15.86%

27. Answer (2) Hint : Meq. of NaOH initially taken = 100 × 1 = 100 Meq. of NaOH consumed H2SO4 = 60 × 1

= 60

Sol. : Meq. of NaOH reacting with salt

= 100 – 60

= 40

Acc. to law of equivalence

Meq. of salt = Meq. of NH3

Meq. of NH3 = 40

3NH40 17W 0.68 g1000

×= =

% NH3 = 0.68 100 60.18%1.13

×=

28. Answer (1)

Hint : Balanced reaction is

2X + 3Y2 → 2XY3

Sol. : Moles of Y2 = 120 260

=

Moles of XY3 forned = 43

Mass of XY3 formed = 4 1003

× 133.3 g=

29. Answer (4)

Hint : Mixing formula

Sol. : Final molarity = 1 2

1 2

V 1 V 0.25(V V )

× + ×+

30. Answer (2)

Hint : % of Br = Mass of Br 100Molecular mass

×

Sol. : Let molecular mass of compound is ‘M’

Mass of bromine in compound = 3 × 80 = 240

7.68% of M = 240

M = 3125

M = 6 × 12 + 3 × 1 + n(104) + 240 = 3125

n = 27

31. Answer (3)

Hint : 1 1 1 1

2 2 2 2

L mv r nL mv r n

= =

1 1 1 2

2 2 2 1

P mv z nP mv z n

= =



Sol. :

12 3

1 1 1 2

22 2 1

2

vf 2 r z n

vf z n2 r

π= =

π

2 21

1 1 2

22 2 12

1mvK.E z n21K.E z nmv2

= =

32. Answer (3)

Hint : Intensity ∝ number of photoelectron.

Sol. : Photoelectric current does not increase with the increase in frequency of incident light.

33. Answer (3)

Hint : Molality = moles of solutekg of solvent

Sol. : 100 ml solution contains 93 g H2SO4

∴ Mass of solution = 1.84 × 100 = 184 g

∴ Mass of solvent = 184 – 93 = 91 g

M = 93 / 9891/ 1000

= 10.42

34. Answer (1)

Hint : P ⇒ One radial node 2s, 3p, 4d

Sol. : Q ⇒ Zero radial node 1s, 2p, 3d, 4f

R ⇒ Two radial node 3s, 4p, 5d

35. Answer (3)

Hint : Radii of maximum probability of e– is

smaller for orbital having higher value of l for

given same value of n.

Sol. : r : 3s > 3p > 3d

36. Answer (2)

Hint : Number of valence e– = group number.

Sol. : Pd → 5th period, 10th group.

37. Answer (3)

Hint : EA increases from left to right

Sol. : EA decreases from top to bottom. Period

2 elements have lower EA values because of

small size. ‘N’ has anomalous EA value

38. Answer (4)

Hint : 22 21 2

1 1RZn n

ν = −

2 21 1 5Rx R2 3 36

= − =

or R = 365

x

Sol. : Use the value of R and get the wave

length and ν

22

1 108 xR(2) 1 3R52

ν = − = =

39. Answer (3)

Hint : λ = hmv

Sol. : h2mK.E

λ =

If K.E is same, 1m

λ ∝

40. Answer (2)

Hint. : For radial node, Ψ2s = 0

Sol. : So, 0

r2 0a

− = , r = 2a0.

41. Answer (1)

Hint : Cathode rays are electrons

Sol. : Anode rays are positively charged ions.

42. Answer (1)

Hint : Due to shielding, r : Ga < Al

Sol. : Correct order is B < Ga < Al < In

43. Answer (2)

Hint : Size increases down the group.

Sol. : For isoelectronic ions, size increases with

the negative charge.



44. Answer (2)

Hint : I → Inert gas, highest IE1 and IE2 and

positive ∆Heg.

II → Alkali metal, low IE1 and high IE2

Sol. : III → Large negative value of ∆Heg.

IV → Less negative value for ∆Heg, so least reactive non-metal.

45. Answer (2)

Hint : Across a period increasing nuclear charge dominates.

Sol.: Down the group, shielding effect dominates the nuclear charge.

46. Answer (04.00)

Hint : 4NaClO

122.5n 1122.5

= =

Sol. : By stoichiometry, 4 moles of Cl2 are required

47. Answer (20.00) Hint : Mass is lost in the form of O2

Sol. : 3 23KClO (s) KCl(s) O (g)2

∆→ +

2

2O

0.384n 1.2 1032

−= = ×

3 2

3KClO O

2n n 8 103

−= × = ×

3KClOW decomposed =8 × 10–3 × 122.5 = 0.980 g

Percentage of decomposition = 0.98 1004.9

×

= 20%

48. Answer (01.50) Hint : The data with least significant figures is

deciding factor.

Sol. : Average value = 25.2 25.25 25.03

+ +

= 25.15 ∴ Least precise data contains 3 significant

figures

Reported answer = 25.2

Significant figures = 3.

49. Answer (14.00)

Hint : 26Fe → 1s2, 2s2, 2p6, 3s2, 3p6, 3d6, 4s2

Sol. : l + m = 0

⇒ l = 0, m = 0 s subshell

⇒ l = 1, m = –1 one orbital of p

⇒ l = 2, m = –2 one orbital of d

50. Answer (10.00)

Hint : λ = 150V

Sol. : E = φ + K.E.

K.E. = 4.5 – 3.0

= 1.5 eV

150 10 Å1.5

λ = =

PART - C (MATHEMATICS)

51. Answer (3)

Hint : Find a, b satisfying a2 + 3b2 = 28, a, b ∈N

& use A ∩ B

Sol.: ( )( )( ){ }1, 3 4, 2 5, 1=A

( )( ){ }4, 2 5, 1 ..........=B

( ) 2⇒ ∩ =A B .

52. Answer (4)

Hint : If n(a) = n ⇒ nP(A) = 2n

Sol.: ( )( ) 22 4= =n P A

( )( )( ) 42 16= =n P P A

( )( )( )( ) 162=n P P P A

( )( )( )( )( ) 1622=n P P P P A .



53. Answer (4)

Hint : Use property of G.I.F.

Sol.: 22 5 3 − < − ≤ x

⇒ − ≤ − <21 5 4x

⇒ ≤ <24 9x

( ] [ )⇒ ∈ ∪–3, – 2 2,3x .

54. Answer (1)

Hint : Use for logax : x > 0, a ≠ 1, a > 0

Sol.: ( )( )1/3 4log log 5 0− >x

( ) ⇒ − <

0

41log 53

x

5 4− <x

9 8< ⇒ =x x .

55. Answer (3)

Hint : Use properties of log function.

Sol.: 123log3log 27

log3 2log2a a= ⇒ =

+

3

31 2log 2

⇒ =+

a

33

3 11 2log 2 3 log 21 2

−+= ⇒ = a

a

Now, 36

3

6log 26log2log 64log2 log3 log 2 1

= =+ +

3 1 3 33 362

3 3 31 12 21

2

− − − = = =− +

− ++

a aaa aa a

a a a

363

aa

− = + .

56. Answer (2)

Hint : Put x = 1x

to find f(x)

Sol.:

( )

( ) ( )

( )

410

4

104 4

13 log (i)

13 1/ log (ii)

8 3log log

f x f xx

f x f xx

f x x x−

− = …

− = …

= +

( ) 4 48 10 3log10 log10− −= +x x xf

12 4= − +x x 8= − x

⇒ ( )10− = −xf x .

57. Answer (2) Hint : as {x} ∈ [0, 1) Sol.: Let {n} = f

( )2 23 32 2 1 14 4

−− − = − − + −f f f f

( )21 14

= − −f .

as 0 1≤ <f

0 1 1f⇒ < − ≤ −

( )20 1 1f⇒ < − ≤

( )21 1 0f⇒ − ≤ − − <

( )23 1 114 4 4

⇒ − ≤ − − <f

Hence range is 10,2

58. Answer (3)

Hint : By ploting graphs of ex and lx|

Sol.:

Only one point of

intersection



59. Answer (4) Hint : Use property of {x} & [x] functions

Sol.: ( ) { } { } { }2 2 2 1 ........ 2 99= + + + +f

0.414 0.414..........100 terms= +

0.414 100 41.4= × =

( )2 41=f .

60. Answer (3)

Hint : For domain 2

2– 7 10 0– 5 6

x xx x

+≥

+

Sol.: ( ) ( )( ) ( )

– 5 – 20

– 3 – 2x xx x

≥

⇒ ( ) ( ] { }– , 3 5, – 2x ∈ ∞ ∪ ∞

Domain = (–∞, 3) ∪ ( ] { }5, – 2∞

61. Answer (3)

Hint : Use concept of {x} & [x] functions.

Sol.: x = 0 or 53

62. Answer (3)

Hint : ex > 0

Sol.: f(x) = e2x + 2ex + 1 + 2

= ( )21 2xe + +

( )0,xe ∈ ∞

∴ ( ) ( )3,f x ∈ ∞

Range = (3, ∞)

63. Answer (2)

Hint : By transformation of graphs.

Sol.: Draw y = x2 then apply graph theory.

64. Answer (2)

Hint : Using concept of {x} in inequality

Sol.: {n} –1 is always negative so the only solution is when {n}({n} –1)({n}+2) is zero. That is only possible when ndI.

65. Answer (1)

Hint : Using AM ≥ GM

Sol.: min of f(n) is 1, when nd[1, 2]

min of g(n) is 2, by Am – Gm inequality

Hence m1 – m2 = –1.

66. Answer (3)

Hint : Here use concept f(x) + f(1 – x) = 1

Sol.: f(n) + f(1 – n) =1

13 3

3 3 3 3

n n

n n

−

−+

+ +

3 3

3 3 3 3·3

n

n n= ++ +

3 3 13 3 3 3

n

n n= + =

+ +

1 2 99..........100 100 100

f f f +

1 99 2 2 501 .............

100 100 100 100 100f f f f f + + + − +

1+1+ …..49 form + 12

f

49 + 0.5 = 49. 5

67. Answer (3)

Hint : Use the concept of domain.

Sol.: 1 ≤ x ≤ 3

⇒ ( )221 log 3 – 2 3x x≤ + ≤

22 3 – 2 8x x⇒ ≤ + ≤

⇒ –5 ≤ x ≤ – 4 or 1 ≤ x ≤ 2

68. Answer (2)

Hint : Use the concept of [x] function.

Sol.: [n]2 + [n] + 1 –3 = 0

y2 + y – 2 = 0

(y + 2) (y – 1) = 0

⇒ y = – 2 and y = 1

⇒ nd[–2, –1)j[1, 2).



69. Answer (2)

Hint : Use the properties of log function.

Sol.: ( ) ( )3log 9 2log log log 1e e ex x x= − +

( ) ( ) ( )⇒ = +2 2log log – log 1e e ex x x

Let loge x y=

2y2 – y –y – 1 = 0 ⇒ (2y + 1)(y – 1) = 0

1 and 12

y y−= =

1log and log 12

x x−= =

But logx cannot be negative, hence only one

solution.

70. Answer (4)

Hint : By inspection check the value of x & y.

Sol.: yx = xy and x = 2y is satisfied only when

y = 2 & x = 4

so x2 + y2 = 20.

71. Answer (05.00)

Hint : {x} ∈ [0, 1)

Sol.: { } 21 10, –xe e

∈

1 17 and 2a b

⇒ = =

72. Answer (02.00)

Hint : Use concept of domain

Sol.: f(x) = ( ) { }2 2log 4 – logx

x x

+

⇒ [x2] > 0 and [x2] ≠ 1

⇒ [x2] ≥ 2

⇒ x∈(–∞, ( ) ( )– ,– 2 2,∞ ∪ ∞

and also 4 – |x| > 0

⇒ |x| < 4

⇒ x∈(–4, 4)

So we have

x∈ ( ) ( )–4, – 2 2, 4∪ …(i)

Now also for { } 0x >

⇒ x I∉

⇒ 2x I∉ …(ii)

So from (i) and (ii)

x = 2, 3

73. Answer (05.50)

Hint : Use f(x) + f(1 – x) = 1 concept.

Sol.: Here φ(–x) = 11 xe+

⇒ φ(x) + φ(–x) = 1

So, S = [φ(5) + φ(–5)] + [f(4) + φ(–4)] + … + φ(0)

= 1 + 1 + 1 …..5 times + 0

11 e+

= 1 1152 2

+ = = 05.50

74. Answer (01.00)

Hint : Take log both sides

Sol.: ( )3/2 3/2 xxx x=

3/2 3log log2

x x x x⇒ =

3/2 32

x x⇒ = on log 0x =

⇒ x = 1

32

x =

94

x⇒ = .

75. Answer (04.00)

Hint : Use property of |x| function.

Sol.: 3 > |n – 1| ⇒ –3 < n – 1 < 3

⇒ nd(–2, 4) – {1}.

∴ x = –1, 0, 2, 3

answers & hints...2020/10/18 · test - 1 (code-a)(answers) all india aakash test series for...

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