ANSWERS, HINTS & SOLUTIONS FULL TEST –IISol)-JEE(Main)/15 FIITJEE Ltd., FIITJEE House, 29 -A, Kalu...

AITS-FT-II-PCM(Sol)-JEE(Main)/15

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST –II (Main)

S. No. PHYSICS CHEMISTRY MATHEMATICS 1. C A D 2. A B B 3. A D B 4. D C B 5. B B B 6. C D B 7. C C D 8. B B C 9. B A C 10. B D C 11. A A A 12. C A D 13. B C A 14. B C B 15. C D D 16. B C B 17. B C C 18. B D B 19. D B C 20. C A C 21. B C C 22. C D D 23. A D A 24. A C A 25. B C B 26. B D C 27. C B A 28. A C B 29. D A B 30. B C C

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2

PPhhyyssiiccss PART – I

SECTION - A

1. Q = C.V. , [Q] = T’A’ So, a + b + c + d + e + f + g + h = 2

3. t t

t 0r t r(0) v.dt

,

t t

t 0v t a.dt v 0

t t t 2

2

t 0 0

3tˆ ˆv.dt t 5 i 2 j dt2

= 3 3t tˆ ˆ5t i 2t j3 2

3 3t tˆ ˆr t 5t 10 i 2t 10 j3 2

4.

cA B1 2 3

1 2 3 3 2 3

TT Ta a a

m m m m m m

TA > TB > TC

1A B

2 3

mT T 1

m m

and 1

2 3

m1

m m

So, A BT 2T

TB = TC 2

3

m1

m

and 2

3

m1

m

So, TB < 2TC

Now, 1 2A C

3

m mT T 1

m

And, m1 + m2 < m3 + m3 1 2

3

m mm

< 2

Or TA < 3TC 5. ‘T’ total charge in the system (initially) = zero We know that when two or more charged

conducting plates are peaced parallel to each other, the outermost surfaces get equal charges ‘p’

can be proved using gauss’s law) Where, TP2

; So, P = 0



3

Since the middle plates are not connected total change on than would be conserved the charges on the plates after S1 and S2 are connected are shown in the figure. Again, we utilize the fact that the facing surfaces must carry opposite charges. Finally the potential difference between surface 1 and 8 must be zero. The electric field is non-zero between the conducting plates and zero inside them.

0 0 0

q Q q qd d d 0A A A

d = spacing between the plates A = Area of the plate q = Q/3 6. x = A cos wt

R = sin wt : Q = A cos wt = A sin wt2

7. Electric field at a distance x-on the axis = Electric field due to an infinite sheet – Electric field due to a uniformly charged disc of radius ‘R’

2 2

0 0

x12 2 x r

2 2

0

xE2 x R

2 2

0

exF qE2 x R

2 2

0

du exmudx 2 x R

0

2 1 eRu

m

8. When the point B is about to leave ground the normal reaction to the rigid rod passes through A. Gravity would not cause a torque about centre of mass. Also, the torques must balance just

before B leaves ground.

2F 3mg3 2

F = 9mg4

9. cmV

of the system is non-zero initially and would remain so because extF 0

at all times.



4

10. Initial angular momentum = mvr

Final angular momentum = 2mr w m wr r

2

Text = 0 : L is conserved

Mur = 23 mwr2

Or, 2uwr3

Velocity of mass m first after the string is taut = wr = 2m/s 11. Force on inclined surface in liquid = Pressure at the centroid of the surface × area of surface

ag l sin ab2

12. 21 2

1P u P2

25 3 2110 Pa 10 10 P2

52

1P 10 Pa2

13. P dg2d

P = 5mN, = 1.75×10–3 kg/m, d = 50 mn 14. Potential energy = – 2 × kinetic energy Magnitude of potential energy = 2 × K.E. Clearly, U > K 16. Volume of the gas is increasing (assuming its mass remains constant) and volume of the gas is

directly proportional to T.

nCv (dT) = (dV) km

or V T

The process is isobaric and exchange of heat is given by nCp (dT)

17. 32 1m3 3

Wave speed on the string = 1 (lowest resonant node) ustring = 400 m/s

When a wave passes from one medium to another (the air) of differing wave speed, the frequency remains the same.

Now, airair string

string

vv

= 343 1 m400 3



5

18. Fq = kq2

2 22

2 15 5 105 10

9 12

49 10 4 10 12 N

225 10

= 27.5 N

20. Distance travelled in first two seconds = 15 m Acceleration of the Car at t = 4s is zero Braking force = Ma = 1 Mg×15 m/s2 = 15 KN

21.

1 23 3

2k pcos k psinE ;E

| r | | r |

2

1

E tantanE 2

or 1 tantan2

or 1 tantan2

25.

211 1 2

22 2 12

V / RP H R 2R 2P H R R 1V / R

26. CAB = C/2 ; CCD = 2C ; CEF = C



6

27. A slit would have produced a spherical wave front. A biprism and a glass slab would produce a parallel wave front.

28. 1 2

1 1 1 1 1 11.5 1 ; 1.8 1f 25 50 f 24 30

1 2

1 1 1 8 4 75;F cmF f f 150 75 4

29. hcE 12.79eV

n 1E E E 0.81eV

n 213.6E 0.81eV;n 4n



7

CChheemmiissttrryy PART – II SECTION – A

1. The mass of 1 cc of (C2H5)4Pb is = 1 1.66 = 1.66 g and this is the amount needed per litre.

No. of moles of (C2H5)4Pb needed = 323

66.1 = 0.00514 ml

1 mole of (C2H5)4Pb requires 4 (0.00514) = 0.0206 ml of C2H5Cl Mass of C2H5 Cl = 0.0206 64.5 = 1.33 g 2. The sp3 bound leaving groups can be directly displaced by the nucleophile or the leaving group

can ionise off to form a carbocation (SN1) >C = C – L. However, sp2 bound leaving groups are very difficult to ionise Halogen attached to allylic system are activated towards SN1 due to stability of allyl carbocation.

4. Aromatic stabilisation occurs in rings that have an unbroken loop of p-orbitals. Any cyclic

compound will be stable especially when the ring contains (4n + 2) -electron. 5. The alcohols II & III are 3 but III gives more substituted C = C, I & IV are both 2 but IV can give

a more substituted C = C.

6. 3 2

2 3A X 2A 3X2S 3S

Ksp = [2S]2 [3S]3 = 22. 33. S5

23

25 5551.08 10S 1 10 1 10

108

7. 2

2 4 4H SO 2H SO

2H O H OH At anode: 22OH H O O 2e

2O O O

OH ions are discharged at anode as OH ions have lower discharge potential then 24SO .

8. Meq. of Na2CO3.xH2O in 20 mL = 19.8 101

Meq. of Na2CO3.xH2O in 100 ml = 19.8 10

51

0.7 1000 19.8M 102

M = 141.41 23 2 + 12 + 3 16 + 18x = 141.41 x = 2 9. 3Na2S2O3 + BiCl3 = Na3[Bi(S2O3)3] + 3NaCl Bi2(S2O3)3 + 3H2O = Bi2S3 + 3H2SO4



8

10. N2H5+ + H2O N2H4 + H3O+

Kh = 14 2

2 4 3 w7

2 5 b

[N H ][H O ] k 10 x[N H ] k 9.6 10 0.1

= 1.04 10–8

x = 3.2 10–5

% hydrolysis = 1001.0102.3 5

= 0.032 %

11. NH3 + H+ Cl NH4

+ Cl Solution contains NH4Cl in contact with NH3 (excess) & hence buffer 12. 2.303 log C/C0 = – kt 40% to completion means 60% unreacted 2.303 log 0.6 = – (3.3 10–4)t t = 26 min

13. 9.8 g Cu = 25.63

8.9 = 0.308 mol e–

10000 96500

13 = 0.311 mol e–

If 0.311 mol electrons provided by the current 0.308 mol was used to deposit copper.

The current efficiency is 100311.0308.0

= 99.1%

14. Moles escaped = original – final moles

= 290082.0

105.2290082.0

104

= 0.630 mol

20. )108.3()1026.1(

1026.154

4

= 77% cyclohexene

methyl cyclopentene = (100 – 77) = 23% 21. Let x and y be the molar mass of A & B respectively then

f

f

T 8 1000 80m 1K x 2y 100 x 2y

… (1)

f

f

T 10 1000 100& m 1K 2x y 100 2x y

… (2)

Solving (1) & (2) x = 40; y = 20. 22. Reaction of D-(+)-glucose with methanolic HCl leads to the formation of methyl glucoside

(C, - OH group is methylated) which, being acetal is not hydrolysable by base, so it will not respond Tollen’s reagent.

23. For precise fitting of Zn2+ ions into tetrahedral voids of S2- ions packing r / r ratio must be 0.225.



9

24. 22 21 2

1 1RZn n

For last line of Lyman series of H-specturm; Z = 1, n1 = 1, n2 = . For H line in Balmer series of He+; Z = 2, n1 = 2, n2 = 3. 25. NO+ has lost an antibonding electron where as CO+ has lost bonding electron. 26. Below CMC no micellization takes place. Sodium oleate ionizes almost completely in aqueous

solution.

27. 2 21 1H g Cl g HCl g2 2

… (1)

o o oS S product S reactants

2 2

o o oHCl H Cl

1 1S S S2 2

11187 131 223 10 JK2

28. In 3

4PO , P shows sp3 hybridization. In 3NO , N shows sp2 hybridization’ In 2ICl , I shows sp3 hybridization. 29. OF2 Pale yellow. SF6 & SF4 Gas.



10

MMaatthheemmaattiiccss PART – III SECTION – A

1. Bisectors of the lines in both the cases will remain unchanged 2. 2

1 0D D , n2 4 2

2 1 0 n 0D D D , ..... D D

3. 1 1

0 0

t 1f x cos x t cos x t dt

= 21 2cos x sin x

2

max 2 4 21 4 4f

4. y = x3 – 3x y = –a

2dy 3x 3dx

x = 1 or –1 For three different roots –a (–2, 2)

(–1, 2) 2

y

1

y = –a

(1, –2)

x

5. kx icotn

, k {1, 2, 3, ….., n – 1}

Then coskx n

k1 isinn

x 1x 1

= 2k 2kcos isin

n n

nx 1 cos2k isin2k 1

x 1

(x + 1)n – (x – 1)n = 0 n n 1 n n 3

1 3C x C x ..... 0

Roots of above equation are kicotn

, k {1, ..... n – 1}

22n 1 n 1

k 1 k 1 1 p q n 1

k k picot icot 2 icotn n n

qicotn

=

n3

n1

C0 2

C

n 1 n 2f n

3

6. 15 5c : ,2 2

r1 = 5

23 3c : ,2 2

r2 = 3

c3

c1 c2

p



11

r be radius of third circle and its centre be at c3, then

2

2 2 2 22 3 3 1 2 1 3 3c c pc c c c c pc

2

2 22 22 1 2 1r r r r r r r r

1 2 1 2

21 2

4r r r r 15r8r r

= 8 7. D be the origin and DA a , DB b and DC c a b 41 , b c 36 , c a 7 distance between mid-points of AB and CD

2

22 22 a b c 1d a b c 2a b 2a c 2b c2 4

= 2 22 2 2 21 a b c a b a c b c4

= 137

A C

B

D

a

b

c

8. 2b = a + c 2 sin B = sin A + sin C

B B A C A C2 2sin cos 2sin cos2 2 2 2

B 1sin2 2 2 , B 7cos

2 2 2 7sinB

4

9. Obviously for c (0, 1), f(x) lies above the g(x), also x2 = cx3

x = 0, 1xc

Hence, 1/c

2 33

0

1 2x cx dx312c

1c2

10. sec 40º, sec 80º, sec 160º are the roots of 38 6 1 0

tt

or t3 – 6t2 + 8 = 0 Sum of roots = 6 11. The chord of contact from (h, k) to y2 = 4x is ky = 2(x + h). If it touches the hyperbola

4h2 + k2 = 4 locus is 2 2x y 1

1 4

12. x = sin–1 (a4 + 1) + cos–1 (a4 + 1) – tan–1 (a4 + 1) is defined at a = 0 only

x4

Area A = 21

2 4 4 32



12

13. This can happen, if three lines are real and distinct as well as angle between any two adjacent

sides is 23

f(m) = bm3 + dm2 + cm + a = 0 has three distinct real roots (where m = y/x)

And 2 3 3 11 2

1 2 2 3 1 3

m m m mm m3

1 m m 1 m m 1 m m

3 + m1m2 + m2m3 + m3m1 = 0 and 3bm2 + 2dm + c = 0 has roots , with f() f() < 0 3b + c = 0 14. 2n {x} = 3x + 2[x]

5 xx

2n 3

as 0 {x} < 1

2n 30 x5

{n 2}

It has 5 solution, [x] = 0, 1, 2, 3, 4, only if 2n 34 55

23 n 142

12, 13, 14

15. Let a is first term and d is common difference of the A.P. a + (r – 1)d = 6, a + (s – 1)d = 8, a + (t – 1)d = 12 Solving, we get 2r + t = 3s also f(x) = tx2 + 2rx – 2s f(0) = –2s, f(1) = s f(0) f(1) = –2s2 < 0 hence, exactly one root in (0, 1) 16. (a – b)2 + (b – c)2 + (c – d)2 + (d – a)2 0 2(a2 + b2 + c2 + d2) 2(ab + bc + cd + ad) ab + bc + cd + ad 25 17. f(x) is continuous where 3 sin x + a2 – 10a + 30 = 4 cos x or a2 – 10a + 30 = 4 cos x – 3 sin x LHS 5, RHS 5 (a – 5)2 + 5 = 4 cos x – 3 sin x

a = 5, x = 2n – 1 3tan4

, n I

18. Required ways = 6 4

2 2C C 90 19. z = x + iy

2 2 1x 2 y 1 x y2

20. Let ˆ ˆ ˆa 3i 4 j 5k

ˆ ˆ ˆb tanxi tanyj tanzk

2 22a b a b

tan2 x + tan2 y + tan2 z 8



13

21. P = cos cos 2 cos 3 ….. cos 1004 Q = sin sin 2 ….. sin 1004 Then 21004 PQ = sin 2 sin 4 ….. sin 2008 = (sin 2 ….. sin 1004)[sin (2 – 1003) sin(2 – 1001) ….. sin (2 – )] = (sin 2 sin 4 ….. sin 1004)(–sin 1003) ….. (–sin ) = Q

10041p

2

22. 1 + x2 = t2 x dx = t dt

22 3

tdt dtI 2 1 t c 2 1 1 x c1 tt t

23. If OP OQ for P and Q on the hyperbola then b > a Director circle of the hyperbola is x2 + y2 = a2 – b2, that exist only when a > b 24. Any plane through first line can be written as 3x + y + 2z – 2 + (2x + y + z – 1) = 0

It is parallel to x = y = z if 32

Thus, plane parallel to 2nd line is –y + z – 1 = 0

25. *A A . If AA* = I, then A is unitary

26. Solution is y = m(x + 2) Fixed point p is (–2, 0)

Area = 23

27. 5

5 52 52 10

1 1 11 x x 1 x 1 xxx x

Now, coefficient of x10 in (1 – x5)5 (1– x)–5 is = 14 5 9 5 44 1 4 2 4C C C C C 381

28. Mean = 1007

Variance 2 2 2 2 21 1 3 ..... 2013 10071007

= 338016

d 338016 581.39 29. Let n(A) = n

Then 2

2n n

n n 22 2

n = 3 30. (sin–1 x + sin–1 y)(sin–1 z + sin –1 w) = 2 is possible only when x = y = z = w = 1 or –1 then

1 2

3 2

n n

n n

x y

z w = 0, 2 or –2

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