FIITJEE AITS Solutions 1 Ft 5

AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST V (Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A B B

2. A C A

3. D B C

4. B A D

5. A B C

6. B A C

7. A C B

8. C A D

9. A, C B, C A, C

10. A, D B, C A, D

11. B, C A, B, C, D A, C

12. A, B, D A, B, C C, D

13. C D B

14. C A A

15. B C D

16. D A D

17. A A C

18. C B B

1. 3 2 8

2. 2 4 4

3. 6 0 4

4. 6 1

6

5. 2 4

1

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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15


2

PPhhyyssiiccss PART I SECTION A

1. 1 1 2 2CM1 2

m v m vV 1 m / s

m m

At maximum extension both blocks will move in same direction with VCM now use energy y conservation.

2. 2MF N

2 12

F N = Ma N = Ma1

2

1MN

2 12

Acceleration of hinge O

= 1 1a a2 2

Nx = F/4 a = 5F/4M and = 9F/2M

oF a a i i

2 M

F

a1 1

N N a

3. 0y y sin kxcos t and 9 L4

4. 1 2d KA T Tdt L equation of heat flow through the box.

5. Just after collision.

a b v

eucos

2

22 1

Lm v ucos a b m a12

u sin

v

6. 2

21L constant2 2C

7. Using COM; vcos = u sin

and 22v sine 2tanucos

2tan2 < 1

tan < 1/2 1 1tan2

u

v v

8. Power factor = VRMS RMS cos



3

9. For maxima n pyd

for minima py 2n 12d

10. yy A sin t kx A cos t kxt

11. mediumCV

12. Isothermal PV cons tan t adiabatic rPV constant 13-14. 0h eV 15-16. 2E mc

17-18. 0 0S

C Vf f

C V

SECTION C

1. 3 2km/hr 2. The centre of mass of system will not move horizontally.

3. ReldMF Udt

4. In steady state the current in branch containing capacitor will become zero. 5. F q E V B



4

CChheemmiissttrryy PART II

SECTION A

1. 1 1 3

22 4 3NH OH H N O 2NH 2H 3H O

2.

P S

SP

SP

S

P

S

SS

S

S

Has 3 p - d P S bond Has 3 P = S bonds Has 3 tetrahedral units of P Has 6 P S P linkages. 3. 4 2 3 2KI 2KMnO H O KIO 2KOH MnO (nf = 6) (nf = 3) Moles of KI 6 = 3 0.2 10 10-3 Moles of KI = 10-3 = 1 milli mole = x

22 4 2S C N I H SO HCN I H O

To produce 10-3 moles I ,

Moles of 3SCN 6 10 1

Moles 3

410SCN 1.6 10 moles6

4. O CH3

2

HH O

OH

3CH OH

Tautomerises

O

6 5C H CHOOH

O

CH5C6 H

OH

2 4N H / OH A

CH5C6 H

OH

D 2 enantiomeric pairs possible for (D).



5

5. 2 2CoCl 2KCN Co CN 2KCl

Buff coloured

42 6Co CN 4KCN K Co CN 6. Half cell Reaction Eo L.H.S.

2H 2H 2e ooxE 0.00 V

R.H.S. 2Zn 2e Zn oredE 0.76 V Net 2 o2 cellzn H Zn 2H , E 0.76 V

2

2

H 0.0591K log KnZn

ocell cell0.0591E E log K

n

2

10log H0.05910.46 0.762 0.3

6H 4.60 10 M

2

3 3

6 3

HSO H SO

0.4 M 4.6 10 6.4 10

2 6 33 8

a3

H SO 4.6 10 6.4 10K 7.3 10HSO 0.4

pKa = 7.13 7.

MeOHOMe

OH

O

O

3PClOMe

Cl

O

O

2MeNHOMe

NHMe

O

O

NMe

O

O

C

O

C

O

O

8. Shortest of Lyman series 1 for H-atom H 1 H21

1 1 1R R1

9. (2) 2 23 4B OH NaOH NaBO Na B OH H O can be made to proceed forward by

adding cis-diol to stabilization by chelation. (3) The solution turning milky on passing H2S through group II solution indicates the presence of

oxidizing agent where H2S is oxidised to 8O2. 10. (A) The heat of formation for NO2 is calculated as 45 kJ mol-1. (B) For reaction 2 2 2N 2O 2NO , H 310 kJ / mole. i.e. it is exothermic.

(C) o2 21 1N O NO, H 55 kJ / mol2 2

(D) o2 21NO O NO , H 100 kJ / mol,2

it is exothermic.



6

11. (A) No H-bonding can occur in chlorobenzene, so +I effect increases acidic strength. (B) Due to both steric and polar effect has no ortho effect is observed in case of o-MeOC6H4NH2.

MeOC6H5NH2. p-MeOC6H4NH2 > o-MeOC6H4NH2 > m-MeOC6H4NH2 pKa = 5.24 (pKa = 4.45) (pKa = 4.2)

(C) N > NH2 >

N

N

pKa = 5.25 pKa = 4.63 pKa = 1.3

(D)

NH>

Pyrrolidine

(CH3CH2)2NH

(pKa = 11.27) (pKa = 10.98)

12. (A) is correct, 2sys1

VS nR nV

, V2 > V1.

sysS ve (B) PV = constant, as is high, slope is higher. (D) At boiling point, the process is at equilibrium G 0 . 14. x = 6, y = 3, z = 2 Solution for the Q. No. 13 & 14. 53 2 6 2IO 6OH Cl IO 3H O 2Cl

5 H6 5 6IO H IO

o o

2

100 C 200 C5 6 4 2 54H O2H IO 2HIO I O .

17. V. D. = 28.75

D d 0.6n 1 d

2 4 2N O 2NO

At equilibrium : 1 0.6 1.20.4

2 42

N O

NO

P 0.4P 1.2

If moles of N2O4 escaping out is x, then for NO2 it is (1 x)

x 0.4 46 x 0.1881 x 1.2 92

2 4 2N O escaping out NO escaping outX 0.188, X 0.812

18. 22 4

2

NOP

N O

1.2P 1.6K 2.2

0.4P1.6

KP does not change.



7

SECTION C 1.

As

Cl

Cl

AsCl

ClCl

Cl

Cl

Cl

2.

3 5 5 2

2 2

CrO C H NCH Cl

Collins reagent

B

O

OH

O

O

O

O

H

ArOArO

3. copolymerisation2 2 2nCH CH CH CH nCH CH

CN

2 2 2CH CH CH CH CH CH

CN

n



8

MMaatthheemmaattiiccss PART III

SECTION A 1. If [x3 + x2 + x + 1] = [x3 + x2 + 1] + x, then x is integer log |[x]| = 2 |[x]| has same solutions as log |x| = 2 |x|, x is integer No integral solution 0 2. Each element has 5 choices, as it may be in any one of A1, A2, A3, A4 or may not be in any one The required number of ways = 57 3. (1 + x)n = C0 + C1x + C2x2 ..... Cnxn

n 1 2 3 n 11 2 n

01 x 1 C x C x C x

C x .....n 1 2 3 n 1

n 2 2 3 n 20 1 n1 x C x C x C xx 1 .....

n 1 n 2 n 1 n 1 n 2 2 1 3 2 n 1 n 2

Put x = 2, we get

2 3 n 2n 20 1 nC 2 C 2 C 23 1 2 .....

n 1 n 2 n 1 2 1 2 3 n 1 n 2

2 n 1 n 20 1 nC 2 C 2 C 2 1 1 3.....2 2 3 n 1 n 2 n 1 2 n 1 n 2 2 n 1 n 2

n 2

n3S

2 n 1 n 2

9

78

6

S 3 2 7 8 7S 32 8 9 3

4. It is given that |z| = 1 z = cos + i sin (let), (0, 2)

Now, z z 1z z gives 2 |cos 2| = 1

1 1cos2 ,2 2

5 7 11 2 4 5, , , , , , ,6 6 6 6 3 3 3 3

i.e. 8 values

5. As cot tan = 2 cot 2 tan = cot 2 cot 2

1 1 1tan tan cot cot2 2 4 4 4 4

6. Latus rectum, 4a 2 2

1a2

Vertex = 1 1,2 2

Axis = x y = 0 As origin lies outside the parabola Focus = (1, 1), Directrix = x + y = 0



9

Equation of parabola is 2

2 2x y x 1 y 12

x2 + y2 2xy 4x 4y + 4 = 0 7. Tangent at (0, 0) will be same (3 + sin B)x + (2 cos )y = 0 and 2 cos x + 2cy = 0 are same

22cosc

3 sin

cmax = 1 where sin = 1 and = 0

8. Consider g(x) = xf(x) g is continuous on [0, 1] and differentiable on (0, 1) f(1) = 0 g(0) = 0 = g(1) By Rolles Theorem g(c) = 0 some for c (0, 1) cf(c) + f(c) = 0 9. At x = 2, RHD = 5 + |1 x| = 5 + 1 = 6 whereas LHD = 5 11. Let the coordinates of point A are (ct, c/t) So, the slope of normal at A will be t2. And normal will be parallel to BC. So, t will be 2 c = 2. 12. a must be negative and x2 a = x should have no solution

Now, D < 0 1 + 4a < 0 1a4

0

y=x

13. ABC bPQR a

2b 3 3 3 3ABC a aba 4 4

14. Slope of OP and OQ will be 1amb

and 2amb

respectively

If POQ = than tan = 1 2 1 2

2 2 21 2

1 22

a m m ab m mba b a m m1 m mb

Now, area of sector OPQ of circle x2 + y2 = a2 = 2

2a 1a2 2

Area of sector OAB = 2b 1 1a aba 2 2

where = 1 21

2 21 2

ab m mtan

b a m m



10

15.-16. LCM = 90 and CL = CM

CLM = CML = 4

Let ACB = 2

LAC = 4 and LCB = 3

4

By sine Rule in ABC

a2 + b2 = 2 2 2 34R sin sin4 4

A

D 4

L

4

C

4

M

B

a2 + b2 = 2 2 24R sin cos4 4

= 4R2

Let ADC = ABC = CBM =

4

CAD = 2 4

BAC = CAD BC = CD

17.-18. We want to have an = k if k k 1 k k 1

n2 2

n is an integer, this is equivalent to k k 1 k k 11 1n2 8 2 8

2 21 1k k 2n k k4 4

1 1k 2n k2 2

1k 2n k 12

Hence, n1a 2n2

= 2 =

17. Now, a = 2, b = 3, c = 5 Let A = number of numbers which are divisible by 2 B = number of numbers which are divisible by 3 C = number of numbers which are divisible by 5 Required number = A + B + C A B B C C A + A B C

= 1000 1000 1000 1000 1000 1000 1000 7342 3 5 6 15 10 30

18. a = 2, b = 3, c = 5, d = 7 Hence, the given number is 25 35 53 73 4n + 1 is odd number therefore the factor 2 will not occur in divisor. 3 and 7 are of 4n + 3 form,

odd powers of 3 and 7 will be of 4n + 3 form and even powers will be 4n + 1 form 5 is 4n + 1 form and any power of 5 will be of 4n + 1 form Number of divisors of 4n + 1 type = number of terms in the product (1 + 32 + 34)(1 + 5 + 52 + 53)(1 + 72) + number of terms in the product (3 + 32 + 35)(1 + 5 + 52 + 53)(7 + 73) = 48



11

SECTION C 1. As g(x) is inverse of f(x) Domain of g(x) is range of f(x) a = 3, b = 11 2. Required number of ways = co-efficient of x30 in (x2 + x3 + x4 + x5 + x6 + x7 + x8)5

= co-efficient of x30 in

52 7x 1 x

1 x

= co-efficient of x20 in (1 x7)5(1 x)5 = 24C20 517C13 + 1010C6 = 826 A = 8, B = 2, C = 6 A + B C = 4

3. 2 2

3 3x 1 1 x 1 3 = 3 3x 1 1 x 1 3

For x [0, 2] 1 3x 1 3 f(x) = 2

2

0

f x dx 4

4. Orthocentre is foot of perpendicular drawn from origin on the plane

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