FIITJEE AITS Solutions 1 Ft 4

1

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JEE(Advanced)-2015 ANSWERS, HINTS & SOLUTIONS FULL TEST– IV (Paper-1)

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

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

B

A

C

2.

C

C

B

3.

A

D

B

4.

D

A

A

5.

C

A

C

6.

B

A

D

7.

B, D

A, D

B, C

8.

A, B, C, D

A, B, D

B, D

9.

C, D

A, B, C

B, C

10.

A

A

C

11.

C

B

C

12.

C

C

C

13.

B

B

B

14.

A A→p B→r C→s D→q A→s B→r C→q D→q

A A → p, r, s, t B → p, r C→q D → p, r, s A→s B→p C→q D→q

A (A)  (r) (B)  (s) (C)  (p, r) (D)  (p, q, r, s, t) (A)  (r) (B)  (t) (C)  (q) (D)  (s)

1.

3

3

4

2.

5

3

6

3.

5

4

6

4.

5

6

2

5.

8

7

2

1.

2.

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

Physics

PART – I SECTION – A

1.

If the pulley comes down by a displacement x then the block comes down by 3x and the string connecting the block has tension Kx. 3d2 x  Kx  M  2 dt or  

2.

K . 3M

1 1 1   v u Feff



1 1 1   or Feff = +10 cm 20 / 3 20 Feff

Now or

1 1 2   Feff fm f0

1 1 2   10 fm 30

1 1 2 1    fm 10 30 30 or f m = 30 cm or R = +60 cm.

or

3.

Use KCL/KVL or suppose the current due to each battery. (Forms wheatstones by removing 1 battery at a time).

4.

There is a tetrahedral configuration with a total of 6 interactions.

5.

î  ˆj  kˆ ˆ  î then its Say the beam is incident along Î  and the normal of one of the faces is N 3 2 ˆ ˆ  a . There are three such faces on which the radiation falls. projection area is AN.I 3  Total projection area = 3a2 I  Force = . 3a2 . c

6.

Mechanical energy and momentum both have to be conserved. In case of rough incline, the sphere will also have some rotational KE. Thus lesser energy available as translational KE of incline as well as the cylinder.

7.

Speed is characteristic of the medium is same for all observers. Here wavelength is different for different directions and thus frequency. For observer (a, 0), (0, a) and (0, –a) observed frequency is more than emitted taking into account time lag.



8.

Potential difference between plates is same as . VA – VB =  = E.d in both air and liquid. VA = VB and VA’ = VB’ Complete plate is at same potential.

9.

Since coils carry current in opposite sense magnetic field is always zero in solenoid even if current is there. Thus net inductance = 0, it is simply  and R joined in series.

10.

Instantaneous axis has zero. instantaneous acceleration  = 0 Taking  about bottom most point: Mg Rsin θ = m(2Rsinθ/2)2 . Solving we get, gsin    4R sin2    2

11.

M R



Mg

mg – N = mRsinθ N= m(g - Rsinθ)

12-14. Use Doppler’s effect, the speed of sound increases with increase in temperature.

SECTION - B 1.

NA + NB + NC = N0. B will first have more production rate than decay and finally there will be no B.

2.

In steady state all capacitors G have as open circuits and inductors as short circuits.  purely resistive network: Assume, i = aE1 + bE2 We get, a = 2, b = 3.  i = 2E1 + 3E2.

SECTION – C 1.

There is only a single atom, thus maximum number of photons obtained are 4 → 3, 3 → 2 and 2 → 1.

2.

Surface tension force is acting downward at both inner and outer surface at contact point on the tube. F = 2rTcosθ + 2rTcosθ = 4rTcosθ

3.

(TA – TB) = T0e Same fractional change takes same time, where T0 = 80°C is initial temperature difference. Since in fixed one hour temperature loss of A = temperature gain of B, means heat capacities are same, means they will always undergo same loss and same gain. After one hour temperature difference 1 become half = 40°C = (80) . In further one hour temperature difference becomes further half. 2 TA + TB = 120°C

–t

2

 1 TA – TB =    T0  20 C  2  TB = 50°C  x = 5



4.

Vertical component of normal reaction balances mg and horizontal component of normal reaction balances component of spring force towards centre. Nv = mg kR 4 k NH = Fs cosθ =   R 4 5 5

NV NH

 mg Fs

N  N2v  NH2  5 N. 5.

N

C P

Ball collide at highest point of projectile Motion, after collision, motion is mirror image of motion that would have been without collision. Top view of motion is as shown in diagram OQ is required distance. OQ = 2 (OB) sin 30 = 8 m

Q Q'

8m

8m 30

B

8m

O

4 3m



Chemistry

PART – II SECTION – A

1.

Singlet carbene gives syn addition on alkene. syn addition on cis alkene gives meso compound.

4.

H = E + nRT,  H <  E if  n < 0

5.

O

O N

C

O

O

O

sp 2

6.

O sp 2

The mass of 1 cc of (C2H5)4Pb is = 1  1.66 = 1.66g and this is the amount needed per litre. No. of moles of (C2H5)4Pb needed =

1.66 = 0.00514 ml 323

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 8.

When a solid mixture of NaCl, K2Cr2O7 and conc. H2SO4 is heated, the products obtained are 4KCl  K 2Cr2O7  6H2SO 4  2CrO2Cl2  6KHSO4  3H2O Red vapour

CrO2Cl2  4NaOH  Na2CrO 4  2NaCl  2H2O Yellow solution

9.

Claisen condensation can be shown only when at least 2  H are there.

SECTION-B 2.

(A) KP > Q the reaction will proceed in forward direction spontaneously. (B) Go  RT loge Q then G   ve then non spontaneous. (C) K P  Q  equilibrium (D) G  H  T.S

SECTION – C 3.

H

C2 H5 C

H3C

Br H CCl4

C

 Br2 

H

C C C 2H 5 x x Br H3C

H

Optically active

2 products H3C

C2 H5 C

H

C

Br C2H5 CCl4

 Br2 

H

H

C

C

H Br

H3C

Optically active

2 products

total = 2 + 2 = 44. Conductivity of Na2SO4 = 2.6  10–4 1000  2.6  104 m Na2SO4    260 S cm2 0.001 m SO24  m Na2SO4   2 m Na







2

= 260 – 2  50 = 160 S cm mol



–1



Coductivity of CaSO4 solution = 7  10–4 – 2.6  10–4 = 4.4  10–4 S cm–1 m  CaSO 4    m Ca2   m SO24





2





–1

= 120 + 160 = 280 S cm mol 1000  K 1000  4.4  104 Solubility, S =  m 280 –3

= 1.57  10 cm K sp  Ca2  SO42  –6

total

  0.00157  0.00157  0.001

2

= 4  10 M –x 2 = y  10 M



Mathematics

PART – III SECTION – A

1.

3 2 Thus, locus of P is 2 2 x + y = 180 Hence, radius = 6 5

Here, tan  

A P(h, k)   

O

B

0

i

i n

2.

Sum = Re(e (1 + e ) )

3.

For the given curve,

dy dx

 x 3

2  2  k = –1 k

 The curve is, x2 + 2x + 4y + 3 = 0 and the tangent is 4x + 2y – 3 = 0 4.

x2 y 2  1 a2 b2 Vertices of second ellipse are (ae, 0) Equation of first ellipse is

 Equation of second ellipse is x 2  2y 2 

a2 2

 Equation of tangent to it at (h, k) is xh  2yk 

a2 2 2

2  Homogenising x2 + y2 – b2 = 0 with this straight line x 2  y 2  b2  2 xh  2yk   0 a  4 a These are perpendicular, hence h2  2x 2  2  a2 2b  Locus of (h, k) is x 2 + 2y2 = a2



1/2

5.

Given, lim

2x 

1/2

1/n

 .....  n  x 

1/3

 1 1  3  

2  3h 2

= lim

h 0

1/2

 2  3h  =

1/4

 4 x 

1/n

 2x  3    2x  3   .....   2x  3  1/2 1/3 1/n 2 h   3 h   .....  n h  lim 1/2 1/3 1/n h 0 1/2 h  2  3h   h1/3  2  3h   .....  h1/n  2  3h  x 

=

1/3

 3x

 1 1  3  

 h 2

 1 1  4  

 4h 2

1/3

 2  3h 



[Put x 

1 ] h

 1 1  n  

 .....  nh 2

 1 1  n  

 .....  h 2

1/n

 2  3h 

2  0  0  .....  0  2 21/2  0  0  .....



6.

Substitute x2 = t 2 2 2 2 1 1  I   e t 1  t  2t 2 e t dt   et  t e t  et  2t 2 et  dt    2 2 2 4 1 1 1 2 =  et  f  t   f '  t   dt  e t t et  c = e x x 2 ex  c 2 2 2







 

2

Let, f(x) = A + Bx + Cx + Dx + ….. Now, f(0) = 0 and f(0) = 0

8.

Here, f '  a  

1

x cos a ln x   sina  ln x

0

 f a   





3

7.





1

a dx =   xcos a sina dx =  tan   2 0

sina da  ln 1  cosa  1  cos a 2

9.

Here, f(x) = 3x – 2x + 100 > 0  f(x) is increasing

10.

Here,

11.

cot A + cot B + cot C =

bx cy az b2  c 2  a2    b sinB  c sinC  a sin A  c a b 2R  k = 2R

=

R a2  b2  c 2 abc





R b2  c 2  a2  c 2  a2  b2  a2  b2  c 2 abc  1 1 1   2  2  2  y z  x





k= 12.-14. Let (h, k) be the circumcircle of ABC 4  3p 4  3p Then, h  ; k 2p 2q 4 4h  12  p ; q 2h  3 k  2h  3  This (p, q) lies on x2 = 4ay  Locus of (h, k) is

1  9   9  y  a  h   2a  8   4

2

SECTION – B 1.

(A) A is idempotent, A2 = A3 = A4 = ….. = A n (A + I) = I  n C1A  nC2 A 2  .....  nCn A n  I  nC1A  n C2 A  .....  nCn A = I   n C1  nC 2 A 2  .....  n Cn  A  I   2n  1 A  2n  1  127  n = 7 2 7 2 7 2 8 (B) (I – A) (I + A + A + ….. + A ) = (I + A + A + ….. + A ) – (A + A + ….. A ) 8 8 = I – A = I (if A = 0) (C) A is skew symmetric  A  A T  ( 1)n A  A I  (1)n   0 as n is odd, hence |A| = O  A is singular (D) A is symmetric then A–1 is also symmetric for matrix of any order



2.

(A) Required event will occur if last digit in all the numbers is 1, 3, 7 or 9. Thus, required 4n probability = 10n 8n  4n (B) Required probability = P (last digit is 1, 2, 3, 4, 6, 7, 8, 9)  P (last digit is 1, 3, 7, 9) = 10n 5n  4n (C) P (1, 3, 5, 7, 9)  P (1, 3, 7, 9) = 10n 10n  8n  5n  4n 10n  8n  5n  4n (D) P (required) = P (0, 5)  P (5) =  10n 10n



 



SECTION – C 1.

2.

 c 2 Let hyperbola be xy = c . Then points of intersection is  ct,  t   c2t4 – 2ct3 – 20t2 – 4ct + c2 = 0 If t1, t2, t3 and t4 are its roots 2 20 4 Then,  t1  ;  t1t 2   2 ;  t1t 2 t 3  and t1t2t3t4 = 1 c c c   = 2, m = 44, n = 56 Let, y =

2

1  sin 2x  1  sin 2x  y = 2 – 2|cos 2x|

   3 5   7  If x  0,  or  ,  or  , 2  , cos 2x is non-negative  4 4 4  4     7   y = 2|sin x|  cos x  |sin x| except for x in 0,  and  , 2   4  4  1  3 5  So, it holds for  ,  in which sin x  4 4 2     3   5 7  If x   ,  or  4 , 4  then cos 2x < 0  y = 2|cos x| 4 4    So, both inequalities hold   7  Thus, solution set is x   ,  4 4  –=6 3.

Here, (1 – 2x + 5x2 + 10x3)  a1 = n – 2; a2 

n  n  1

2 Since, a12  2a2  n = 6 4.

–1



n



2

C0  n C1x  n C2 x2  ..... = 1 + a1x + a2x + …..

 2n  5

2

2

f(x)(1 – x) = f(x)(1 + x + x + …..) = b0 + b1x + b2x + …..  (a0 + a1x + a2x2 + ….. + anxn + …..)(1 + x + x 2 + ….. + xn + …..) = b0 + b1x + b2x2 + ….. n Comparing coefficient of x on both sides, a0 + a1 + a2 + ….. + an – 1 + an = bn Comparing coefficient of xn – 1 on both sides, a0 + a1 + a2 + ….. + an – 1 = bn – 1  bn – bn – 1 = an Also, b0 = a0 and b1 = a0 + a1  b0 = 1; a1 = 2



5.

 b10 = a0 + a1 + a2 + ….. + a10 = 211 – 1  tan2      tan2       1 Here, R.H.S. = tan1   2 2 1  tan      tan        sin2      sin2       cos2      cos 2       = tan1   2 2 2 2  cos      cos       sin      sin         cos 2  cos2 2   cos2  cos2 2  = tan   2cos 2 cos2    cos2 sec 2  cos 2 sec 2  = tan1   2   =2 1 


FIITJEE AITS Solutions 1 Ft 4

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