FIITJEE AITS Solutions 1 Ft 4

7/23/2019 FIITJEE AITS Solutions 1 Ft 4

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

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1

ANSWERS, HINTS & SOLUTIONS FULL TEST– IV

(Paper-1)

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 A

1.

A → p B → r C → s D → q

A → p, r, s, t B → p, r C → q

D → p, r, s

(A) (r)

(B) (s)

(C) (p, r)

(D) (p, q, r, s, t)

2.

A → s B → r C → q

D → q

A → s B → p C → q

D → q

(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

FIITJEE JEE(Advanced)-2015

F I I T J E E S t u d e n t s F r o m A

l l P

r o g r a m s h a v e b a g g e d 3 4 i n T o p 1 0 0 , 6 6 i n T o p 2 0 0 a n d 1 7 4 i n T o p 5 0 0 A l l I n d i a R a n k s . F I I T J E E P e r f o r m a n c e i n J E E ( A

d v a n c e d ) , 2 0 1 4 :

2 5 2 1 F I I T J E E S t u d e n t s f

r o m C

l a s s r o o m /

I n t e g r a t e d S c h o o l P r o

g r a m s & 3 5 7 9 F I I T J E E S t u d e n t s f r o m A

l l P r o g r a m s h a v e q u a l i f i e d i n J E E ( A d v a n c e d ) , 2 0 1 4 .

A

L L I N D I A

T E S T S E R

I E S





2

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.

2

2

3d xKx M

dt

orK

3M .

2.eff

1 1 1

v u F

eff

1 1 1

20 / 3 20 F

or Feff = +10 cm

Noweff m 0

1 1 2F f f

orm

1 1 2

10 f 30

orm

1 1 2 1

f 10 30 30

or f m = 30 cm or R = +60 cm.

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

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

5. Say the beam is incident alongˆ ˆ ˆi j k

I3

and the normal of one of the faces is ˆ ˆN i then its

projection area is2aˆ ˆ AN.I3

. There are three such faces on which the radiation falls.

Total projection area = 23a

Force = 2I. 3a

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 ofincline as well as the cylinder.

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





3

8. Potential difference between plates is same as .

V A – VB = = E.d in both air and liquid.V A = VB and V A’ = 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 = 0Taking about bottom most point:

Mg Rsin θ = m(2Rsinθ/2)2 .

Solving we get,

2

gsin

4Rsin2

R

Mg

M

11. mg – N = mRsinθ

N= m(g - Rsinθ)

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

SECTION - B

1. N A + 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 are4 → 3, 3 → 2 and 2 → 1.

2. Surface tension force is acting downward at both inner and outer surface at contact point on thetube.

F = 2rTcosθ + 2rTcosθ = 4rTcosθ

3. (T A – TB) = T0e –t

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

become half = 40°C =1

(80 )2

. In further one hour temperature difference becomes further half.

T A + TB = 120°C

T A – TB =

2

0

1T 20

2

C

TB = 50°C x = 5





4

4. Vertical component of normal reaction balancesmg and horizontal component of normalreaction balances component of spring forcetowards centre.Nv = mg

NH = F

s cosθ =

kR 4 kR

4 5 5

2 2v HN N N 5 N.

C

P

sF

mg

NVN

HN

5. Ball collide at highest point of projectile Motion,after collision, motion is mirror image of motionthat would have been without collision.Top view of motion is as shown in diagram OQis required distance.

OQ = 2 (OB) sin 30 = 8 m

4 3m

8m

Q

Q'

8 m

8m

B30

O





5

hemistry

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.

N

O O

O

C

O O

O

2sp 2sp

6. 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 =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

8. When a solid mixture of NaCl, K2Cr 2O7 and conc. H2SO4 is heated, the products obtained are

2 2 7 2 4 2 2 4 2Red vapour

4KCl K Cr O 6H SO 2CrO Cl 6KHSO 3H O

2 2 2 4 2Yellow solution

CrO Cl 4NaOH Na CrO 2NaCl 2H O

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) o

eG RTlog Q then G ve then non spontaneous.

(C) PK Q equilibrium

(D) G H T. S

SECTION – C

3.

C C

H

CH3 H

C2H5

4CCl2Br C C

Br H

CH3

H C2H5

Br x x

2 productsOptically active

C C

CH3

H H

C2H5

4CCl2Br C C

Br C2

H5

CH3

H H

Br

2 productsOptically active

total = 2 + 2 = 44. Conductivity of Na2SO4 = 2.6 10

–4

4

2m 2 4

1000 2.6 10Na SO 260 S cm

0.001

2m 4 m 2 4 mSO Na SO 2 Na

= 260 – 2 50 = 160 S cm2 mol

–1





6

Coductivity of CaSO4 solution

= 7 10 –4

– 2.6 10 –4

= 4.4 10 –4

S cm –1

2 2m 4 m m 4CaSO Ca SO

= 120 + 160 = 280 S cm2 mol

–1

Solubility, S =

4

m

1000 K 1000 4.4 10

280

= 1.57 10 –3

cm

2 2sp 4

totalK Ca SO 0.00157 0.00157 0.001

= 4 10 –6

M2

= y 10 –x

M2





7

Mathematics

PART – III

SECTION – A

1. Here,3

tan 2

Thus, locus of P isx

2 + y

2 = 180

Hence, radius = 6 5

AP(h, k)

BO

0

2. Sum = Re(ei(1 + e

i)n)

3. For the given curve, x 3

dy 2

2dx k k = –1

The curve is, x2 + 2x + 4y + 3 = 0

and the tangent is 4x + 2y – 3 = 0

4. Equation of first ellipse is2 2

2 2

x y1

a b

Vertices of second ellipse are (ae, 0)

Equation of second ellipse is2

2 2 ax 2y

2

Equation of tangent to it at (h, k) is2a

xh 2yk2

Homogenising x2 + y

2 – b

2 = 0 with this straight line

22 2 2

2

2x y b xh 2yk 0

a

These are perpendicular, hence4

2 2 2

2

ah 2x a

2b

Locus of (h, k) is x2 + 2y

2 = a

2

5. Given,

1/2 1/3 1/4 1/n

1/2 1/3 1/nx

2 x 3 x 4 x ..... n xlim

2x 3 2x 3 ..... 2x 3

=

1/2 1/3 1/n

1/2 1/3 1/n1/2 1/3 1/nh 0

2 h 3 h ..... n hlim

h 2 3h h 2 3h ..... h 2 3h

[Put1

x

h

]

=

1 1 1 1 1 1

2 3 2 4 2 n

1 1 1 1h 01/2 1/3 1/n2 3 2 n

2 3h 4h ..... nhlim

2 3h h 2 3h ..... h 2 3h

=1/2

2 0 0 ..... 02

2 0 0 .....





8

6. Substitute x2 = t

2 2 2 2t 2 t t t t 2 t1 1

I e 1 t 2t e dt e t e e 2t e dt2 2

= 2t t t1 1

e f t f ' t dt e t e c2 2

= 2 4x 2 x1

e x e c2

7. Let, f(x) = A + Bx + Cx2 + Dx

3 + …..

Now, f(0) = 0 and f (0) = 0

8. Here, cos a1

0

x lnx sinaf ' a dx

lnx

=

1cosa

0

x sinadx =a

tan2

sina

f a da ln 1 cosa1 cosa

9. Here, f (x) = 3x2 – 2x + 100 > 0

f(x) is increasing

10. Here,2 2 2bx cy az b c a

b sinB c sinC a sin Ac a b 2R

k = 2R

11. cot A + cot B + cot C = 2 2 2 2 2 2 2 2 2Rb c a c a b a b c

abc

= 2 2 2

2 2 2

R 1 1 1a b c

abc x y z

k =

12.-14. Let (h, k) be the circumcircle of ABC

Then,4 3p

h2p

;

4 3pk

2q

4

p2h 3

; 4h 12

qk 2h 3

This (p, q) lies on x2 = 4ay

Locus of (h, k) is

21 9 9

y a h2a 8 4

SECTION – B

1. (A) A is idempotent, A

2

= A

3

= A

4

= ….. = A(A + I)

n = n n 2 n n n n n

1 2 n 1 2 nI C A C A ..... C A I C A C A ..... C A

= n n 2 n n1 2 nI C C A ..... C A I 2 1 A

n2 1 127 n = 7(B) (I – A) (I + A + A

2 + ….. + A

7) = (I + A + A

2 + ….. + A

7) – (A + A

2 + ….. A

8)

= I – A8 = I (if A

8 = 0)

(C) A is skew symmetric T n A A ( 1) A

n A I ( 1) 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





9

2. (A) Required event will occur if last digit in all the numbers is 1, 3, 7 or 9. Thus, required

probability =n

n

4

10

(B) Required probability = P (last digit is 1, 2, 3, 4, 6, 7, 8, 9) P (last digit is 1, 3, 7, 9) =n n

n

8 4

10

(C) P (1, 3, 5, 7, 9) P (1, 3, 7, 9) =n n

n

5 4

10

(D) P (required) = P (0, 5) P (5) = n n n n

n n n n

n n

10 8 5 4 10 8 5 4

10 10

SECTION – C

1. Let hyperbola be xy = c2. Then points of intersection is

cct,

t

c2t4 – 2ct

3 – 20t

2 – 4ct + c

2 = 0

If t1, t2, t3 and t4 are its roots

Then, 1

2t

c ; 1 2 2

20t t

c ; 1 2 3

4t t t

c and t1t2t3t4 = 1

= 2, m = 44, n = 56

2. Let, y = 1 sin 2x 1 sin 2x y2 = 2 – 2|cos 2x|

If x 0,4

or3 5

,4 4

or7

, 24

, cos 2x is non-negative

y = 2|sin x| cos x |sin x| except for x in 0,4

and7

, 24

So, it holds for 3 5,4 4

in which 1sinx2

If3

x ,4 4

or

5 7,

4 4

then cos 2x < 0 y = 2|cos x|

So, both inequalities hold

Thus, solution set is7

x ,4 4

– = 6

3. Here, (1 – 2x + 5x2 + 10x

3) n n n 2

0 1 2C C x C x ..... = 1 + a1x + a2x2 + …..

a1 = n – 2; 2

n n 1a 2n 5

2

Since, 21 2a 2a n = 6

4. f(x)(1 – x) –1

= f(x)(1 + x + x2 + …..) = b0 + b1x + b2x

2 + …..

(a0 + a1x + a2x2 + ….. + anx

n + …..)(1 + x + x

2 + ….. + x

n + …..) = b0 + b1x + b2x

2 + …..

Comparing coefficient of xn 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





10

b10 = a0 + a1 + a2 + ….. + a10 = 211

– 1

5. Here, R.H.S. =

2 21

2 2

tan tan 1tan

1 tan tan

=

2 2 2 21

2 2 2 2

sin sin cos costan

cos cos sin sin

=

2 2

1 cos2 cos2 cos2 cos2tan

2cos2 cos2

= 1 cos2 sec 2 cos2 sec 2tan

2

= 2

FIITJEE AITS Solutions 1 Ft 4

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