Sol Aits Jeea Ft IV Paper 2

AITS-FT-IV-(Paper-2)-PCM(S)-JEE(Advanced)/13

JEE(Advanced)-2013

FIITJEE

ANSWERS, HINTS & SOLUTIONS FULL TEST –IV (Paper-2)

ALL INDIA TEST SERIES

From Long Term Classroom Programs and Medium / Short Classroom Program 4 in Top 10, 10 in Top 20, 43 in Top 100, 75 in Top 200, 159 in Top 500 Ranks & 3542 t o t a l s e l e c t i o n s i n I I T - J E E 2 0 1 2

1

Q. No. 1.

PHYSICS C

CHEMISTRY C

MATHEMATICS A

2.

A

B

C

3.

D

B

A

4.

B

C

C

5.

A

D

C

6.

C

C

B

7.

A

A

B

8.

A

C

D

9.

D

C

C

10.

C

A

B

11.

A

C

B

12.

C

B

B

13.

A

B

D

14.

B

D

B

15.

A, B, C

A,C

A, B

16.

A, B, C

A,C

A, B, C, D

17.

A, D

B,D

A, B, C

18.

A, B, D

B,D

C, D

19.

A, C, D

A, B, D

A, C

20.

C, D

B, C, D

A, C

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com


2

Physics

PART – I SECTION – A

1.

A is at rest instantaneously (for no slipping at A) 1 V  V B – V A = B  0  (2R) 2 2 R  1 V  B  0( VC – VA = 2 R  1 V B  0 ⇒ VB – VC = 2 R

2.

3 R) 2 BV0R  2  [R ] = 2 

force = (P – P0) A = Ma by P1V1 = P2V2 P0hA = P(h – x) A P0h ⇒P= h–x  Ph  A =a So force =  0 – P0  h – x   M a=

P0 Ax P Ax ⇒ 0 ; (for small x) M(h – x) Mh

T = 2π

3.

Mh P0 A

τrestoring = (mg sinθ L + (mg sin θ

L 2

 L  L  KL L sin θ  +  K sin θ  +   4 4  2 2 2  3mg 5KL  restoring =  2 + 16  θ (θ : small)   3mg 5kL2 + 1 2 16 frequency of oscillation) = 2π Ι mL2 4mL2 = where Ι = mL2 + 3 3

4.

From law of radioactive decay. Activity µ Number of nuclear of that nucleus present in the sample. Activity does not depend upon the rate of the formation of nucleus. In given reaction B decays into E and C. Total activity of B = rate of decay of B into E + rate of decay of B into C = l5N2 + l3N2

5.

The EMF will be induced in the loop, only when there is a change in the flux through the loop. When the loop was completely inside the field region, for some time (say t1) there is no change in the flux, hence that time would not be included in the calculation.


3

The time for which EMF was induced =


b b 2b + = v v v

6.

That wire will break first whose stress reaches the breaking stress (for that material) first. F F = Stress = where F is the tension in the wire. F is same in both the wires. A πr 2 1 Thus stress ∝ 2 for the wire of smaller radius, the stress is more. r Thus the wire with smaller radius will break first. If rA < rB, then wire A will break first. If rA > rB, then wire B will break first. If rA = rB, either of them could break.

8.

As the magnetic field is uniform and the particle is projected in a direction perpendicular to the field, it will describe a circular path. The particle will not hit the yz-plane if the radius of the circle is mv = d. smaller than d. For the maximum value of v, the radius is just equal to d. Thus Bq Bqd ∴ v= m

9.

F = ηA

u − u0 y

F y + u0 ηA 1 y + 1 = 100 y + 1 ⇒ u = −2 10 × 1

⇒ u =

10.

The velocity of upper plate is same as that of the layer in contact with the plate i.e. layer at y = 2 cm. u = 100 (0.02) + 1 = 3 m/s

11.

T2 ∝ r3 (Kepler's law)



TA2 TB2

12.

=

rA3 rB3

=

r3 3

(4r)

=

4

T 1 1 Get A = 64 TB 8

For geostationary satellite, time period = 1 planet day (by def.) Let T = 1 planet day T0 = 1 planet year 4 π2 3 4 π2 3  m  rG = Now T2 = r Gm Gm  M 

=

4 π2 3 r = T03 or T = T0 GM

13-14. ω =

k k , ω′ = = M M+m

2500 = 10 rad/sec 25

M M+m x1 = A sin φ1

ω′ = ω

1 2 1 kx1 = Mω2 sin2 φ1 2 2 Let v and v′ be the velocity of system just before collision and just after collision, so using COLM Mv MAω cos θ v′ = = (M + m ) (M + m )

Potential energy stored in the spring at (t = t1) =

1 2 1( kx1 + M + m ) v ′2 2 2 2 2  M + msin θ  1 1 2  M + m sin θ  = MA 2 ω2   = kA   =8J 2  M+m  2  M+m 

Total energy after collision = PE + KE =

 M + m sin2 θ  1( 1 M + m ) A ′2 ω′2 = MA 2 ω2    M+m  2 2

15.

Imax (a1 + a2 )2 = Imin (a1 − a2 )2 a1 I = 1 = a2 I2

4 2 = 9 3

Imax = 25 Imin If an identical paper is pasted across second slit shifted C.B.F. will back in central point. tD shift = ( µ − 1) d nλ D fringe width = d shift (µ − 1)t ∴ n= = = 15 . fringe width λ ∴

16.

The change in magnetic flux is zero, hence the current in the ring will be zero.

17.

Both (A) and (D) depend on g. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

5

18.


n(n − 1) =6 ⇒ n=4 2 If the initial state were n = 3, in the emission spectrum, no wavelengths shorter than λ0 would have occurred. This is possible if initial state were n = 2 ∴ (A), (B) and (D)

n=4 3 2

1 19.

From COE (A) is correct. The force outside the earth varies as inverse square of the distance. ∴ motion is not simple harmonic. However, from symmetry of motion, the motion will be periodic. ∴ (C) is correct. From COE 1 GMm  3GMm  mv 2 = − − − 2 2R 2R   2GM R ∴ (A), (C) and (D) are correct. ⇒

20.

v=

The current through R1 is constant.


6


Chemistry

PART – II

SECTION – A

pKa1 + pKa2 4 + 9 = = 6.5 2 2

1.

pH at isoelectric point =

2.

Net reaction does not involve any aqueous species.

5.

Cr ( SCN)2 (NH3 )4  , shows geometrical (or cis-trans) and linkage isomerism.

2+

O

6. H3C

C

O CH2

O

C

(i) EtO−

CH2 CH3  O→

O

(ii) H

H3C

H3C

O

O

C

CH C

O

CH2 O

C

CH

C

H3C

C

CH2 CH3

O

CH2 CH3

CH

C

O

CH2 CH3

CH2 OH H3C

O

C

H

O

O

O

C

CH C

CH3 O

O ←  H3C (Michael)

C

CH 2

∆, −OH2

O C

C

O

CH2 CH3

CH2

EtO− OH

O C

O

CH3

CH2 CH3

O C

O

Et

+

Li / ∆   →

O C

O

O

CH2 CH3

+ H2O + CO2 + Et − OH

O

7.

∆ FeSO4  → Fe2O3 + SO3 + SO2

8.

Energy of e− in H-atom = - E Energy of e− in Li+2 = - 9E 1 Energy supplied by photon = IE + mv 2 2 IE = E ∴ Ep = 9E + v=

1 mv 2 2

2 (Ep − 9E ) m

9.

In reverse osmosis water moves from more concentrated to less concentrated solution

10.

π = iCRT, (i=2)


7 11.


Let conc. of both reactants at equilibrium is x M

PbEDTA 2 −  H+     Keq =  2 x

2

x = 10−3 12.

Chelation increases stability of complexes OH

14. +

aq.K 2 CO3 H /H2 O HCN P + Q  → R  → S  → heat

O

O ←  −H2 O

+

H /H2 O ←  heat

O

O

O

OH

OH

OH

CN OH

OH HO

(S) HCN O

O

O +

H

aq. K 2 CO3 →

H

(P )

H

H OH

(Q)

(R )

15. Cl

KCN  → Cl

2  → Cl

H Pt

Cl

CN

NH2

H O+

CF CO H

HIO4 3 3 3  → epoxide → diol  → 2CH3CHO

16.

Trans-2-butene

17.

He and H2 are not hydrogen like particles

18. 19.

Radius of cation is smaller than parent atom and that of anion is more than parent atom O (A) O

H 2C I

CH

C O

→

O CH2

I

I2 will attack the double bond to form a cyclic halonium ion. It then undergoes on intramolecular reaction. Formation of five membered ring is favoured because (i) attack by a nucleophile on a halonium ion normally takes place at the more substituted site. (ii) Formation of five membered ring is more probable.



(B)

O

O

C

+

Br

 →

O H3C

Br

O CH2

O

(C)

C Option(C)gives

O C

(D)

8

Option(D)gives

O

C

O

1, 4 and 1,5-dicarboxylic acids give anhydrides whereas 1,6 and 1, 7-diacids give ketones when heated with P2O5.


9

Mathematics


PART – III SECTION – A

1.

Normal to y2 = 4ax with slope m is ..... (1) y = mx – 2am − am3 (1) touches x2 − y2 = a2 if (−2am − am3)2 = a2 (m2 − 1) ..... (2) i. e. if m6 + 4m4 + 3m2 + 1 = 0 (2) is a sixth degree equation in m with no odd power terms and whose coefficients are all positive. ⇒ (2) has no real root ∴ |S| = 0

2.

Given x + y + z = 5 → (1) and x2 + y2 + z2 = 9 → (2) 1 2 We have yz = ( y + z ) − y 2 + z2   2 1 2 = ( 5 − x ) − 9 − x 2  (using) (1) & (2))  2 = x2 − 5x + 8 ⇒ y, z must be the roots of the equation

( (

(

)

)

)

u2 − ( 5 − x ) u + x 2 − 5x + 8 = 0 → ( 3 )

As y, z are real, we must have discriminant of (3) ≥ 0 ⇒ (5 − x)2 – 4 (x2 – 5x + 8) ≥ 0 ⇒ 3x2 – 10x + 7 ≤ 0 ⇒ (3x – 7) (x – 1) ≤ 0  7 ⇒ x ∈ 1,   3 ⇒ Number of integer values that x can take is 2 3.

2 sin2x + sin2 2x = 2 ⇒ −2(1 − sin2 x) + 4 sin2 x cos2 x = 0 ⇒ cos2 x.cos 2x = 0 → (1) sin 2x + cos 2x = tan x ⇒ 2 sin x cos2 x + cos 2x cos x = sin x ⇒ sin x cos 2x + cos 2x cos x = 0 ⇒ cos 2x (sin x + cos x) = 0 → (2) Common roots of (1) and (2) are given by cos 2x = 0 ( 2n − 1) π ,n ∈ N ⇒ x= 4 nπ In [0, 4π] we have x = , where n = 1, 3, 5,……,15 → (3) 4 Now, − 2 sin2 x + sin x ≤ 0 ⇒ 2 sin2 x − sin x ≥ 0 ⇒ (sin x) (2 sin x − 1) ≥ 0 1 ⇒ sin x ∈[−1, 0] or ≤ sin x ≤ 1 → (4) 2 5π 7π 13π 15π Among the 8 values given by (3), only 4 values, i.e., x = , , , satisfy (4) 4 4 4 4


10


4.

1 z Take z = eiθ (∵ |z| = 1). We have w = eiθ + e−iθ = 2cos θ We know that w is real and | w | ≤ 2

We have w = z +

(

x2 − 1

5.

)

2

The scalar triple product of the three given vectors is 2x − 1

2x + 1

3x 2 + 2

x2 + 4

−3

1

2

2

2

= x − 1 2x − 1 2x + 1

x

2

x +4

2

(

)

2

2

3x + 2

(

1

)

x +1

0

3x + 2

(

2

0

2

= x − 1 2x − 1 2x + 1

5x 2

2

x +4

2

)

3x + 2

2

x

2

x2 + 1

(using C3 → C3 + C2 + C1)

5x + 7

0

= x 2 − 1 2x 2 − 1 −2x 2 + 3 2

1

2

)

−3 x 2 − 1

2

2

(

2

(

2 x2 − 1

−5x

)(

5x 2

2

(using C2 → C2 − 2C1)

2

5x + 7

)

= x 2 − 1 15x 4 + x 2 + 21

Vector are non–coplanar ⇒ Scalar triple product ≠ 0 ⇒ x2 – 1 ≠ 0 (∵ 15x4 + x2 + 21 ≠ 0) ⇒x ≠ ± 1 6.

(

We have f ( x ) = a − xn

)

1/ n

so that

   n  =  a −  a − x    ⇒ g ( x ) = f° f° .....f ( x ) = x

(

 f° f ( x ) = f  a − xn 

)

(

1/ n

)

1/ n

n

  

1/ n

   

( (

))

1/ n

= a − a − xn

2m times

∴

∫

dx

(g ( x ))

n

(1+ (g ( x )) )

n 1/ n

=

∫x

dx n

  1  t −1/ n+1  1 =−  1 + x −n +c = n 1 1− n  − + 1   n 

(

7.

(1 + x ) n

)

1−

1 n

1/ n

=

∫x

dx n +1

(1 + x ) −n

1/ n

=−

dt

∫ nt

1/ n

(using t = 1 + x−n)

+c

Given y = f ( x ) ,z = g ( x ) we have

d2 y d  f ′  dx dy dy / dx f ′ ( x ) g′f ′′ − f ′g′′ 1 = . = and = =  ′ 2 dz dz / dx g′ ( x ) dx g dz dz   ( g′)2 g′ 8.

f ( x ) is continuous at x = 3

( )

( )

⇒ f 3− = f ( 3 ) = f 3+

( )=

Now f 3

−

lim−

x →3

a | x2 − x − 6 | 6+x−x

2

= lim− x →3

(

−a x 2 − x − 6 6+x−x

2

)=a


11

( )

f 3 + = lim+ x →3


x − [x] x−3 = lim+ =1 x →3 x − 3 x−3

x    x 2 + tdt   x     x tdt   0  3 f ( 3 ) = b lim  3  = 9b   form  = b lim  x →3 x →3  1    x − 3   0         1 ⇒ a = 9b = 1 ⇒ b = and a = 1 9

∫

∫

n

9.

Vn =

∑

n

∑ r = n (n + 1)

Sr = 2

r =1

∴ lim

n→∞

10.

Vn n

2

r =1

= lim

n→∞

n +1 =1 n

(i) Vn = n (n + 1) ⇒ V1 < V2 < V3 < ..... ⇒ Statement − 2 is true n

(ii) Qn =

1 1 = S S 4 r =1 r r +1

∑

n

1 1 = r r 1 4 + ) r =1 (

∑

n

∑  r − r + 1 = 4 1 − n + 1 1

1

1

1

r =1

⇒ Q1 < Q2 < Q3 < ………. ⇒ Statement −1 is false 11.

We have  2a− x , x<0 f (x) =  −x x a + a , x ≥ 0 Clearly, f(x) is continuous in R.

 a2x − 1  log a For x > 0, f ′ ( x ) = −a− x loge a + a x loge a = ( a x – a −x) loge a =   a x  e > 0 for x > 0   ⇒ f ( x ) is monotonic increasing for x > 0 For x < 0, f ( x ) = 2a− x ⇒ f ′ ( x ) = −2a− x loge a < 0

⇒ f ( x ) is monotonic decreasing for x < 0 0

12.

Required area =

∫ 2a

1

−x

dx +

−1

=

∫ (a

−x

)

0

0

−x  −2a  dx =    loge a 

−1

1

x −x  −a + a  +   loge a 

0

2

3a − 2a − 1  1 1   −1   −2 − ( −2a )  + + a  − 0 = aloge a loge a  loge a  a   1

15.

+a

x

1

2 2

The equation of the line is b 3 y + a 3 x + a 3 b 3 = 0 1

This can be written as y = −

a3 1

b3 Consider the parabola y2 = 4ax

x

2 1 − a3 b3

1

⇒ m=

−a 3 1

2

1

, c = −a 3 b 3

b3


12


a = m

a

=−

1

−a 3

1 3 ab 1

2

1

= −a 3 b 3 = c

a3

1

b3 ∴ Condition for tangency for the parabola y2 = 4ax is satisfied ∴ The given line touches y2 = 4ax Similarly the line touches x2 = 4by

16.

1

( x + iy ) 5

= a + ib

∴ x + iy = ( a + ib)5 = a5 + 5a4 (ib) + 10a3 (− b2) + 10a2 (−ib3) + 5ab4 + ib5 Equating the real and imaginary parts x = a5 – 10a3b2 + 5ab4 ⇒ y = 5a4b – 10a2b3 + b5 x y 4 2 2 4 4 2 2 4 4 4 4 4 p = − = (a − 10a b + 5b ) − (5a – 10a b + b ) = −4a + 4b = −4 (a − b ) a b = −4(a2 – b2) (a2 + b2) = −4(a + b)( a − b) ( a +ib) (a −ib) ∴ a − b, a + b, a + ib and a – ib are all factors of p. 17.

Since 0 < θ <

π ,0 < sin2 θ, cos2 θ, tan2 θ < 1 4

∞

( A ) ∑ sin2k θ = k =0

∞

(B)

∑ cos

2k

θ=

k =0

∞

(C)

∑

tan2k θ =

k =0

1 1 − sin2 θ

= s ec 2 θ

1 2

1 − cos θ

1 1 − tan2 θ

(D) cot θ > 1 in 0 < θ <

= cosec 2 θ

=

cos2 θ cos2 θ − sin2 θ

π . Hence, 4

19.

3y = 9 − 3x ⇒ 9 − 3x > 0 ⇒ x < 2 Similarly, y < 2

20.

Area ( T ) =

∞

∑ cot

2k

=

cos2 θ cos 2θ

θ does not exist.

k =0

c.c 2 c 3 = 2 2 c

c3 Area (R ) = − x 2 dx 2

∫ 0

3

3

3

c c c − = 2 3 6 Area ( T ) c3 6 lim+ = lim+ . 3 = 3 c →0 Area (R ) R →0 2 c =


Sol Aits Jeea Ft IV Paper 2

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