SOL (4)

AITS-CRT-III-(Paper-2)-PCM (Sol)-JEE(Advanced)/16

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1

ANSWERS, HINTS & SOLUTIONS

CONCEPT RECAPITULATION TEST–III PAPER-2

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A, B A, B, D B, C

2. A, B, C, D

A, B, C, D A, C

3. B, C A, D B, C

4. A, B, C A, B, C, D B, C

5. A, B, C, D A, B, D A, D

6. A, B C, D C, D

7. A, B, C A, C, D A, D

8. A, B, C B, D B, C, D

9. A A C

10. D D C

11. A D B

12. A C B

13. D C A

14. A D C

15. A A B

16. D A C

17. B B D

18. D C C

19. A B D

20. C D B

ALL

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2

PPhhyyssiiccss PART – I

SECTION – A 1. AB A B

ˆ ˆv v v v i 2v j

2vtan 2v

OP

gives the direction of

Rmin = 2ddsin5

0dT

5 v

N

O x

Q

yQ P

(d,0) A B

Rmin

2. 2l mlJ

4 12 cm

JVm

A cmlV V 03

, upper cml JV V2 2m

lower cml 5 JV V2 2 m

3. F.B.D. of ball in frame of container

y4 Vg Vga 3g

V

2L 1 3gt2 2 Lt

3g X

Y 4Vg

Vg

4Va Va

ax =

21L 3at2

2Lt3a

Ball to collide at point Q, 2L Lt3a 3g

a = 2g

4. 3

2VT

constant , 2 VdV dT3 T

, 2RTPV

, W PdV 400 R

VU nC T 900R , Q nC T and Q U W 1300R

Q 13C Rn T 6



3

5. 0x x 90 =

Also, PQ = 2Rcos 90 2Rsin 02mvsin

qB

| v |

constant v = v0

For 2 rotation it takes 2 m timeqB

For 2 2 it takes 2mqB

P

Q

v

v0

C R x

x

6. The consecutive prism produce equal and opposite deviation. Hence if number of prism is even

deviation is zero and if odd, deviation is . 7. When t = 3s block just about to move and

acceleration of block given by t 2a1

t >

3

v 10

0 3

dv t 2 dt

6

2920502

2

10

3

2

ttv

5.315.130 m/s

B A F=t

fk= 1N fs= 1N

fk= 1N fs= 2N

8. As insect starts moving the tension in the string becomes more than Mg and so net externa force

(2T 2Mg) is not zero and the centre of mass of the system accelerates. When insect moves with constant velocity, net force on the system (insect + rod + counter mass) is zero. Hence acceleration of the system (insect + rod + counter mass) is zero.

9. Torque acts parallel to velocity, so change in momentum is parallel to v .

10. dLl mg mglsindt

11. = instantaneous flux through loop = 1 2BA BA Sense of induced emf is opposite in each loop

Hence, the net emf induced 2 21 2

dB dBA A 10 a bdt dt

= 0.942 V

12. 6 –6Q CV 5 10 0.942 4.71 10 C

13. The limiting value of frictional force for planck is = 1 (8 2)(10)5

= 20 N

14. The maximum value of friction force for block = 1 2 105 4N



4

Common acceleration (A) = 40 208 2

= 2 m/s2

For block, ma = mg = 4N

15. v =

m2m 52 gm / 2

16.

m2m ydy g;dt m /

t

1/20 0

dy g dty2

t 2 ( 3 2)g

17. s Lvv g g2

, Ls 2

Lmg v ' g , Lmg Ax g … (i) when temperature is raised by T, Lmg A ' x ' g … (ii) From (i) and (ii) L LA A ' '

LL s

LA A 1 2 T

1 T

for fraction inside the liquid to be same

s Ls

L L s

1 T1 3 T

18. Mass of original sphere = M Radius of original sphere = R Mass of cavity = M/8 Radius of cavity = R/2 [M V r3] Mass of substituted (introduced) sphere = 2M [1/8th volume 16 time denser] Radius of substituted (introduced) sphere = R/2 At any point net potential [applying superposition] = V0 (potential due to original sphere) – VC (potential due to cavity) + VS (potential due to

substituted sphere) 20. Total number of transitions from nth state to ground state is nC2 = 6 n = 4 ( for level C) Since when electrons excited from nB

th to nCth (= 4) it took 0.28 eV of energy, so when it return to

nBth level it released 0.28 eV of energy, but for (nB –1)th level it released more than 0.28 eV and for

(nB + 1)th level it released less than 0.28 eV of energy. So for satisfying these three conditions the only value possible for nB (among 1, 2 and 3) is 2. Hence all transition are from atoms of B level. Maximum no. of different transition in which atoms of level A can participate is zero.

Now, 22 2

1 10.28 13.6Z2 4

Z = 3



5

CChheemmiissttrryy PART – II

SECTION – A 1.

O2N F

NO2

is more reactive than

O2N I

NO2

towards nucleophilic

substitution as intermediate carbanion is more stablized in

O2N F

NO2 2.

CCl3 C

O

H OH CCl3 C

O

H

OH

3CCl HCOOH

3CHCl HCOO 3 2CCl CH OH : chloral hydrate

Cl

Cl O

H

Cl O

H

H

Stable 3.

O O

dil OHInt ramolecular Aldol reaction

O

CH3

OH

O

O

3CH MgCl OH

O

CH3tautomerisation O

O

CH3acid catalysed aldol reaction

O

CH3

OH



6

4. CH3 CH2 C OH

O

2SOCl CH3 CH2 C Cl

O

2 4H Pd /BaSO CH3 CH2 CHO

CH3 CH2 C OH

O

3CH OH/H

CH3 CH2 C OCH3

Oo

DIBAL H78

CH3 CH2 CHO

CH3 CH2 C OH

O

2Ca OH3 2 2CH CH COO Ca 2HCOO Ca

dry distillation CH3 CH2 CHO

CH3 CH2 COOH 4

oLiAlH Cu

3 2 2 3 2300 CCH CH CH OH CH CH CHO

5.

C ON C C

CON

sp3

sp2

The lone pair in sp hybridized orbital on carbon interacts with * orbital of the carbonyl group. 6. 4 4 2 2 2 22NH ClO N X Cl 2O 4H O N2 is least reactive among N2, Cl2 and O2 4 2 7 2 2 3 22NH Cr O N Cr O 4H O

4 2 2 2NH NO N 2H O 7. On addition of HCl it reacts with OH and equilibrium shifts to forward direction. On addition of

NH4Cl, moles of 4NH in solution increases. On dilution dissociation of weak electrolyte 4NH OH increases.

8. At T < Tc : vapour T > Tc : gas At Boyle’s temperature in low pressure region a real gas behaves like an ideal gas. 9.

2 2 2 21H O H O O2

20 molx20 x after 150 min2

so 121

2

ln2 1 20ln t 50 mint 150 2.5

10. In 150 min 17.5 mol of H2O2 are decomposed. So heat released = 50 17.5 kJ = 875 kJ 11. A is I ion.

2 2Hg excess KI2 4

scarlet red ppt. colourless so lub le complexI HgI HgI

12. 3 2BiCl H O

3X orange turbidityblack ppt.I BiI BiOI



7

2H O3 white turbidityX

BiCl BiOCI 2H 2Cl

13. room temp.

2 6 3 2 6 3B H NH excess B H .2NH ; P

high temp.2 6 3 xB H NH excess BN ; Q

high temp.2 6 3 3 3 6B H NH B N H Borazine ; R

1 : 2

N

B

B

N

N+

BH

H

H

H

H

H

(R) inorganic benzene 14.

B B

H

H

H

H

H

Hsp3 sp3

In B3N3H6, ‘B’ is sp2 hybridised B2H6.2NH3 is ionic and comprises

2 333

3 4spsp

H N BH NH & BH ion

15.

CH CH red hotCu tube 3

2 4

conc. HNOH SO

NO2

Sn / HCl

NH2

2NaNO / HCl

N2Cl

A B C D E

C C.CuCH3

C CPh CH3

F

3



8

2 4H / Pd / BaSOC CPh CH3

G

C C

H

CH3Ph

H

4dil KMnOOH

Ph

H

OHH

CH3

K

3Na/NH ( )C CPh CH3

H

C C

CH3

HPh

H

4dil KMnOOH

Ph

H

HOH

CH3

L

(G) and (H) are geometrical isomers (diastereomers) 16.

CHOPh

G

C C

H

CH3Ph

H

3

2

OZn / H O CH3CHO

J I

Ph CHO gives Cannizaro reaction other statements are incorrect. 17.

C OHH

CHO

C HCH3

C OHH

C OHH

CH2OH

C HOH

CHO

C HOH

C OHH

C OHH

CH2OH

C O

CH2OH

C HOH

C OHH

C OHH

CH2OH

D-glucose D-mannose D-fructose

18. 2 3Cl H O NH CN (order of increasing field strength) Stronger is the ligand, more will be crystal field splitting hence more energy is absorbed. As E 19. NH3 is basic in nature so it can be dried by CaO.

20. 2

2 5Fe H O NO :

Fe is in +1 O.S.

B

O

OB

B

O

O

BOH OH

OH

OH

O

sp3

sp2

2

5Fe CN NO

is diamagnetic.

5PCl s exist as 4 6PCl & PCl ions



9

MMaatthheemmaattiiccss PART – III

SECTION – A

1. The given equation implies that sin (B + C) sin (B C) = sin (A + B) sin (A B) sin2 B sin2 C = sin2 A sin2 B 2 sin2 B = sin2 A + sin2 C

1 cos 2B = 1 1(1 cos 2A)2 2

(1 cos 2C) 2 cos 2B = cos 2A + cos 2C.

Moreover, sin A sin (B C) = sin C sin (A B) sin(B C) sin(A B)sinC sinB sin A sinB

cot C cot B = cot B cot A 2 cot B = cot A + cot C. 2. We have

34

= 1 (1 ) (1 ) (1 ) = + + ( + + ) + ,

12

= + (1 ) + ( ) + (1 )

= + + 2,

25

= (1 ) + (1 ) + (1 ) = + + 3

= 1 2 5 4 12 5 10 10

+ + = 1 1 72 5 10

+ + = 3 7 1 3 3 274 10 10 4 5 20

2 + 2 + 2 = ( + + )2 2 ( + + ) =227 7 1

20 5

.

3. Let the direction ratios of the two lines be (0, m1, n1) and (l2, m2, n2) so that 2m1 n1 = 0, l2 + 2m2

n2 = 0 and m1m2 + n1n2 = 0

1 1 1l m n0 1 2 2 2 2l m n

5 2 1

. Hence equations of the required line are

y 2x 1 z 15 2 1

= . For = 1 and = 1, we get (B) and (C).

4. In the interval [0, 1], f(x) = 3x2 2x + 1 f(x) = 6x 2 0 for x 1/3. Obviously f(1) = f(1+) = f(1) and |x 1| is not differentiable at x = 1. From x = 1/3 to x = 3, f(x) is

increasing. 5. Let the focus be 0,

SM SP , 1SM5

P lies on the line 2x y 0 at a distance 15

units from S.

2x y 1

P

S

M

0,



10

1x 0 y

1 2 55 5

2 11x,y ,5 5

or 1 2 1x,y ,

5 5

1 7 2x ,y5 5

or 1 3 2x ,y

5 5

5y 25x 17

5y 25x 13

35x 7 5y 2 3x y 1 0 7x y 1 0

6. Radius of director circle 2 2a b 2

2 22 2 b 4

2b 4 3e2

0,2

2,0

Focus lies on the line x 0 y 21 12 2

at a distance ae units from centre.

x 0 y 2 2 31 12 2

x,y 6,2 6 or 6,2 6

7. 5 2x cos log xdx

5 2 5logx 1 coslogxx cos dx x dx2 2

5 51 1x dx x coslogxdx2 2

6

6tx 1 e cos tdt,12 2

where t logx

6tI e cos tdt

6t 6te ecos t sin t dt6 6

6t 6t 6te e ecos t sin t cos t dt6 36 36

6te 1I 6cos t sin t I

36 36

6t36 eI 6cos t sin t

37 36

The given integral = 6 6x x 6coslogx sinlogx c

12 74



11

If you replace c by 1C12

we get option ((D)

8. Slope of normal dxtandy

The given equation becomes at a general point x,y is

2 2dx dyy x x y 1dy dx

2

2 2dy dyyx y .x y 1dx dx

2

2 2dy dyy x y 1 yx 0dx dx

'2 2 ' ' 2yy xy y xy yx 0

' ' 'yy y xy x y xy 0

dy xydx

or dy xdx y

2xlogy c

2 (or) 2 2x y c

2x2y ke (or) 2 2logy x logk

2 2logky x 9.-10. EF = BF – BE

= a c a2 c b

BF AB

EC AC

(angular bisector theorem)

=

a b c2 c b

DBC DAC A / 2, BCD DAB A / 2 (angles in the same segment)

BD = DC = 2RsinA/2 (ABC, BDC have the same circum circle)

A/2

A

B C

D

A/2

A/2 A/2

A/2

E F

In le BFD, apply cosine rule DF2 = BD2 + BF2 – 2BDBF cosA/2 = (2RsinA/2)2 + (RsinA)2 – 2 2RsinA/2 RsinAcosA/2 = (2RsinA/2)2 + R2 sin2A – 2R2sin2A = R2 2 2 24sin A / 2 4sin A / 2 cos A / 2 DF = 2R sin2A/2 = a/2 tan A/2 ...(1) AD = 2Rsin(A/2 + B) (ABC, ABD have the same circumcircle)

= R 2sin A / 2 B cos A / 2

cos A / 2

= R sin A B sinBcos A / 2

= b c2cos A / 2

AD2 = 2 2 2

2b c b c

24cos A / 2

(given)



12

2 2

2b c 2cb2cos A / 2 = b2 + c2

cosA = 2 22bc

c b …(2)

Using (1) and (2)

DF = a 1 cosA2 1 cos A

= a b c2 b c

11.-12. xe f x dx = x xe f x e f x dx

(Applying integration by parts on the integral on R.H.S) x x x xe f x dx e f x e f x e f x dx

(Repeatedly apply integration by parts on the integral in R.H.S)

x xe f x dx e f x f x f x f x f x ... + n

xnn

d f xlim e dxdx given to be’0’

2 3x x

2 3df x d f x d f xe f x dx e f x ...

dx dx dx

…(1)

Consider f(x) – 2 3 4 5

2 3 4 5df x d f x d f x d f x d f x.

dx dx dx dx dx = x …(2)

Differentiating both sides 6n times (where n can be any natural number)

6n 2 3 4 5 6n

6n 2 3 4 5 6nd df x d f x d f x d f x d f x df x x

dxdx dx dx dx dx dx

= 0

6n 6n 1 6n 3 6n 3 6n 4 6n 5

6n 6n 1 6n 2 6n 3 6n 4 6n 5d d d f x d d f x d f xf x f x

dx dx dx dx dx dx

= 0 n N …(3)

Using (2), (3)

f(x) – 2

2df x d f x ...

dx dx

= f(x) – 2 3 4 5

2 3 4 5df x d f x d f x d f x d f x.

dx dx dx dx dx +

6n 6n 1 6n 3 6n 3 6n 4 6n 5

6n 6n 1 6n 2 6n 3 6n 4 6n 5d d d f x d d f x d f xf x f xdx dx dx dx dx dx

= x + 0

f(x) – 2

2df x d f x ...

dx dx = x

Using (1) xe f x dx = xex

Differentiating w.r.t x exf(x) = xex + ex f(x) = x + 1 f(x – 1) = x = sinx has only one root x = 0

1 1

x x

0 0

e f x dx e dx e 1

f(x) is clearly increasing, one–one function. 13. Writing sin2 = x, we get 2 sin cos d = dx, and hence the given integral is equal to



13

/ 2

2m 1 2n 1

0

1 cos sin2

(2 sin cos ) d

= 1 12m 1 2n 1

m 1/ 22 2 n 1/ 22 2

0 0

1 1 1 1 1cos sin dx 1 x x dx m , n2 2 2 2 2

.

14. Writing x zn , we get

n 1n

n k 1k 1

0 0

x1 x dx 1 z nz ndzn

=

1nk k 1 k

0

n 1 z z dz n (k, n 1) .

15. Slope of f(x) at (1, 9) slope of f–1(x) at (9, 1) = 1

9 m = 1 m = 19

.

16. f(x) is an odd function, so is its inverse 17. (P) 48600 = 3 2 52 5 3 2,2,2,5,5,3,3,3,3,3 We can select one or two 2’s. 5 and 3 cannot be selected together Total number of ways = 2 3 6 2 2 5 = 16 (Q) f x is increasing for x 1

f 1 is maximum if 3

2log k 1 f 1

3

2log k 1 4

3

2log k 5

2 5k 3 k 243 Greatest integer value of k 15 (R) 5C3 [34 – 3C1 24 + 3C2] = 10[81 – 48 + 3] = 360.

(S) 22OP 5 5 4 = 121

OP 11

5 5

10,5

2P

O

18. (P) I =

/ 4 |tanx| 2

sinx/ 4

e sec x dx1 e e 1

=

1e 1

/ 4tanx 2

0

e sec x dx

= 1

(Q) 1

2

0

2x dx = 1/ 2 1

2 2 2 2

0 1/ 2

2x 2x dx 2x 2x dx

= 1/ 2 1

2 2

0 1/ 2

2x dx 2x 1 dx = 3 23 2



14

12 3 2

3 23 2 2

= 2

(R) f(x) > 0 x R

–3 < x 01 x

x 3 01 x

and x 01 x

– 3 x 04 (a, b] 3 ,0

4

4(b – a) = 3 (S) (x – y + 2) + (x + y – 2) = 0 passes through the focus of the parabola x2 = 4y – 4. Hence

shortest intercept on this line by parabola will be latus rectum, whose slope is 0. 19. (P) We know tan 3x – tan x – tan 2x = tan 3x tan x tan 2x

/ 3

0

tan3x tan x tan2x dx

= / 3

0

tan3x tan x tan2x dx

= – / 3 / 3

/ 30

00

1 1ln | cos3x | ln | cos x | ln | cos2x |3 2

= – 13

ln1 + ln1/2 + 1/2ln1/2

= 3/2 ln1/2

(Q) Let I = 1

20

ln x 1 dxx 1

(put x = tan)

I = / 4

0

ln 1 tan d

I = / 4

0

ln 1 tan d4

2I = / 4

0

ln 1 tan tan tan tan d4 4

2I = / 4

0

ln2d

I = 8 ln2.

(R) Let I = 1

5 50

dx

3 x 1 x

Lt f(x) = 5 5

1

3 x 1 x

Clearly f(x) is decreasing f(1) I f(0)

1 3 1I2 2

I [0, 2]



15

(S) [x2 – x] =

1 0 x 1

1 50 1 x2

1 51 x 22

2 2 x 2.3

2.3

2

0

x x dx = –1(1) + 0 1 5 12

+ 1 3 5

2

+ 2(0.3)

= –1 + 3 5 0.62 2 = 1 5 0.6

2

= 2.2 52

20. (P) Sides of the excentral triangle 4RcosA/2, 4RcosB/2, 4RcosC/2

Angles of the excentral triangle A B C, ,2 2 2 2 2 2

Area of excental triangle = 1 B C A4Rcos 4Rcos sin2 2 2 2 2

= 8R2 cosA/2 cosB/2 cosC/2 Area of le ABC = 2R2 sinA sinB sinC

2R2 sinA sinB sinC = 21 A B C8R cos cos cos4 2 2 2

sinA/2 sinB/2 sinC/2 1/8 We know sinA/2 sinB/2 sinC/2 1/8 and equality occurs at A = B = C /3 (Equilateral angles) (Q) Let A B C sinA + sinB + sinC 1 We know in a triangle b + c > a sinB + sinC > sinA

12

> sinA

A B C A3

A 3 A is obtuse

(R) r1 2 3

1 1 1r r r

= 1 Put r1 = s a , r2 =

s b , r3 =

s c

r = /s (valid for all triangles) (S) Let r1 max{r2, r3} max{r1, r2, r3} = r1 r2 + r3 + r = r1 4R(sinB/2 cosA/2 cosC/2 + sinC/2 cosA/2 cosB/2) = 4R(sinA/2)(cosB/2 cosC/2 – sinB/2 sinC/2) tanA/2 = 1 A = /2 le is right angled

SOL (4)

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