Solutions (1)

AIITS-HCT-VIII (Paper-2)-PCM (Sol)-JEE(Advanced)/16

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1

ANSWERS, HINTS & SOLUTIONS

HALF COURSE TEST–VIII (Paper - 2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A C A

2. C C B

3. B A C

4. A A A

5. C A B

6. A D C

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

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

9. B, D A, D A, C, D

10. B D B

11. C C A

12. B C B

13. C C B

14. B D B

1

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

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

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

2.

(A) (p) (B) (q) (C) (s) (D) (p)

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

(A) (p) (B) (p, r) (C) (r) (D) (p, r)

1. 9 3 9

2. 1 3 5

3. 8 7 5

4. 8 7 1

5. 4 8 3

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2

PPhhyyssiiccss PART – I

SECTION – A

1.

2 y H h2 y hx 2gh 2g H h

g g

y = H

Now solve for x.

y 2gh

x

h

2. 2dV D dh.

dt 4 dt

2dh 4 16.5 0.529 cm / sdt 3.14 (6.30)

3. 2 2px qv r

dx dv2px 2qv 0dt dt

pa xq

Particle is executing S.H.M. with angular frequency p.q

4.

22

2 2GMm 3 Gmmr

r 3r

32 rT 2

mG M3

mr

M

m

m

6. Here V 1/r

r distance between Sun and Planet.

7. 2L

00

0

LM xdx

2

L 2

2A

0

MLx dx x2

8. (A) 1 2

1 2

m ma g

m m

2g 2m / s5

21h at2

t = 2 sec.

(B) 2UH 4

2g where U = at = 4.8 m.



3

(C) 6 kg block will be lifted after time

2U2 2.8sec.g

10. (Amp.) 1( ) 3.14ms and 2 where 5Hz 11. Equation of standing wave is

(0.1 ) sin sin(10 )2

y m x t

Putting 0.5x m and t 0.15sec we get 110 2

y m .

12. From the data given if orbital speed v = 2 105 m/s

2

2

mv GMmr r

, where M is mass of the massive object and m is mass of particle

2RvM

G

15 1037

11

6 9.5 10 4 10 10 kg6.67 10

.

13. 37

730

Mass of the object 10 10 50Mass of sun 10

It is a black hole.

14.

11 243

s 22 8

2GM 2 6.67 10 6 10R 10C 3 10

Radius compression factor 3

93

10 106400 10

SECTION – C

1. W = Area of P-V diagram from A to B

= 31 3(1 2) 1 10 1500J2 2

f f i iCvU (P V PV )R

= 5 (3 1000) 7500J2

Q = W + U = 9000 J = 9 Kj 3. The force acts along the tangential direction 4. P = v2

2mdv .v vdt



4

0

0

2v t

v 0

dv dtv m

ln2 .tm

mln2t

4 2t 0.6930.693

t = 8 seconds. 5. After 5 s, speed of detector = 50 – 10 × 5 = 0 and that of source = 0 + 10 × 5 = 50 m/s

and the source has fallen a distance 21 10 (5) 1252

m

and the detector has rises a distance 2150 5 10(5) 1252

m

So, 300 0f ' 130300 50

= 156 Hz. Ground

O v=0

50 m/sS



5

CChheemmiissttrryy PART – II

SECTION – A 3. A s 2B aq 2C s D aq

1 1/ 3 0 01 x 1/ 3 2x 2x x

202

1/ 610B

B = 4 1011

4. rms

3RTVM

5. NO2 at para position will go out of plane of benzene ring due to steric inhibition of resonance. 7. 4 2 3 2 2P 3OH 3H O PH 3H PO 8 2 2 3 2 2S 12NaOH 2Na S O 4Na S 6H O 2 3 23Cl 6NaOH 5NaCl NaClO 3H O 3 3 22B 6NaOH 2Na BO 3H

8. Molarity = 10 d x 1M

M

Volume = 100 + 100 = 200 ml

Mass of H2SO4 = 200 1 98 19.6 g1000

11. Meq of FeSO4 = meq of KMnO4

w 1000 25 0.05 5152 / 1

= 0.95 gm

2 4FeC Ow 10 0.95 9.05 gm

Percentage of FeC2O4 = 90.5% 12. I.E. full filled orbital.

SECTION – B

1. Energy sequence in (p) is due to only Aufbau energy rule. In rest, rules/principles are obey as mentioned in answer.



6

SECTION – C

2.

O

Si

Si

O

O

SiO

O

O

O

OO

3. 3 2

3 3 3 3 3 3BO , H BO , BF , HCHO, CH COOH, CO , C graphite 4. 1 mole CuSO4 required = 5 moles of H2O 0.6 mole of CuSO4 required = 0.6 5 = 3 mole H2O 1 mole of Na2CO310H2O gave 3 mole H2O means resulting formula is Na2CO37H2O. 5. Volume of gas temperature

2 4 2 2N O N 2O

4 1 4 4 2

Total volume = 4(1 ) + 4 + 8 = 4 + 8 = 50% = 0.5 V = 4 + 8 0.5 = 8 L



7

MMaatthheemmaattiiccss PART – III

SECTION – A

1. Clearly, 2 23 sin2 k sin2 04

3k4

2. a 2 5R2sin30º 2sin30º

AH = 2R cosA 3 2 5 .

A

B C

H

30º

a = 2+ 5

3. (27)40 = (3)120 3119 = (4 – 1)119 119 119 119 118 119

0 1 118C 4 C 4 .... C 4 1 = 4k – 1 3120 = 12k – 3 = 12(k – 1) + 9

5. 1 + 4sinA/2 sinB/2 sinC/2 = 74

r = 4R sinA/2 sinB/2 sinC/2

R 4r 3 .

6. 2s(2s – 2a) = bc

2 Asin2 4

0 14

0 < < 4.

7. (A) sin2x – cos2 = – cos2x 1

(B) 6 1 1sinx cosx5 2 3

6 5 sin x 15 6

(C) sin6x + cos6x = 231 sin2x 14

(D) cos2x + sin2x = 1 – sin2x . cos2x = 211 sin2x 14



8

8. 1tan 3,2 2 .

9. Clearly (a – c)2 + (a – 2b)2 = 0 a = c, a = 2b

2 2 2b c acosA

2bc

cosA = ¼

similarly, 7cosB8

10-11. Clearly, an = bn + cn bn = an – 1 cn = an–2.

12-13. cos(A – B) = 4/5

A B 1tan2 3

1 1absinC 6 3 sin90º2 2

= 9

14. 1 6 3 1 92

2 2a a b

sinA 1

6 45sinA

2sinA5

.

SECTION – B

2. (A) sinBcos A2sinC

2 2 2b c a b

2bc 2c

c = a

isosceles. (B) cosA sinC + 2 sinC . cosC = sinB . cosA + 2sinB . cosB

cosA(sinB – sinC – 2sin(B – C)) = 0

A = 90º or b = c (C) Multiplying both the sides by abc, we get b2 + c2 = a2

(D)

2 2

2 2

sin A Bsin A sin Bsin A Bsin A sin B

(sin2A + sin2B) . sin(A – B) = sin2C sin(A – B)

A = B or a2 + b2 = c2.



9

SECTION – C

1.

5nm

n n

5n 2 3 5n 1 dss 2ns s 2 3 n 1 d2

d = 6 a2 = 9.

2. Clearly,

n n 1k k 1 1224

2

k = 25 k – 20 = 5

3. |2z – 6 + 5i| = 2 5iz 32

9i2 z 3 2i2

5.

4. |a + b + c2|

1 i 3 1 i 3a b c2 2

2a b c b 3 c 3i2 2

2 2 21 a b b c c a2

for |a + b + c2| to be minimum any two values are zero and third is minimum magnitude integer i.e. 1

|a + b + c2|min = 1.

5. 1 6 10 sinC 15 32

3sinC2

C = 120º

2 2 2a b ccosC

2ab

21 36 100 c

2 2 6 10

C = 14

and a b c 152

A

B C

c b=10

a=6

15 3r 3s 15

r2 = 3.

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Solutions (1)