SOLUTIONS PAPER-2(PCB) CODE-BA (UPSEE 2017)

PHYSICS Sol.1. (C) ray will be still totally internally reflected at interface. As n2 decreases , i

C

1

 n2   also decreases, so condition i  iC is still satisfied and there  n1 

 sin 

will be still total internal reflection at interface. If angle of incidence is increased then ray will be still totally internally reflected at interface because i  iC . Sol.2. (B) 2eV KEmax  h max    3  1  2eV

Sol.3. (C) 100 cm A

C

B

D

F

F

30

10

20

10

30

AD=30+10+20+30+10=100cm Sol.4. (B) conduction As atoms in the spoon vibrates about their equilibrium positions and transfer energy from one end to other end. This process is conduction. Sol.5. (B) along +y axis dq dE P dq

Consider any two symmetric dipole elements, net contribution due to these elements at point P is along +y axis . Similarly by principle of superposition we can say that net electric field at P is directed along +y axis. Sol.6. (D) 8 kR 2 KEi  PEi  KE f  PE f 1 1 2 2 k  OA  R   KE f  k OB  R  2 2 1 1 2 2  k 5 R  R   KE f  k  R  R  2 2 2  KE f  8kR  0

Sol.7. (B) 9 J 1 1 (3)(6) 2 U  Ceq V 2   3  9J 2 2 (3  6)

Sol.8. (A) experiences a force directed along the radial direction only. Circular motion is a special case in gravitational field. There may be straight line,elliptical paths, but force will be always directed toward the centre of the sphere.

Sol.9. (C) speed is maximum at t=4s. t 3

t  0, 2, 4

t 1

x  A

x0

x  A

v 0

vmax

v 0

mg k

Sol.10. (A) kx

F.B.D. of lower rod

kx

kx

kx

T2  2mg

T1  mg

kx  kx  2mg mg x k

T2  2mg

T1  mg

Sol.11. (C) 9 We consider two possible cases: Case  II

Case  I nC

5 nB

nB

4

nA nC

4 5

nA

nC  nA  4  5  9 nA  nC  1

Sol.12. (A) 0.6 m / s By momentum conservation (10  103 )1000  (10  103 )  400  10v  v  0.6m / s Sol.13. (B) AP  AQ Q P N

O A

Q

P

For complimentary angles range ON is same for P and Q as points O and N are on same horizontal plane. From figure AP  AQ Sol.14. (D) 60 Hz Second overtone n=3  f3  3f0  60Hz Sol.15. (B) 4m / s 2 d2x x  slope (t )  2t  2  4 dt 2

2

Sol.16. (C) 1s In one time constant 63% change occurs in the value of current .The 63% of maximum current is 1.26 A .It is obvious from graph that current 1.26A corresponds to time which is slightly greater than 0.9sec. Hence option having 1s is most appropriate. Sol.17. (C) zero a b

At the position of smaller loop, Magnetic field due to larger loop is parallel to plane of smaller loop. Due to larger loop, magnetic flux linked with smaller loop is zero. Hence mutual induction is zero. 1  Mi2  0  Mi2  M  0

Sol.18. (D) 3 m / s Impulse during t=2 to t=4 is

1 1  3  2  5    4  3  5   0 2 2

Impulse =change in momentum mv f  mvi  0  v f  vi  3  speed  3

Sol.19. (D) 10 8

5

1

      aQ  ˆi ,aP  5ˆi  8 ˆj  arel  aP  aQ  6ˆi  8 ˆj , arel  10

Sol.20. (A) pressure of 85cm of Hg Pgas  Patm  P  76cm of Hg  9cm of Hg  85cm of Hg

Sol.21. (C) work done by gas is negative Volume does not remain constant throughout the process AB . As T 

PV nRT

,temperature

decreases initially as both P and V decreases .By area under curve ,net work done is negative. P

A

     

B

V

2 3

Sol.22. (A) 20 kgm s

 dKE   Pall  F.v  ma.v  100 103   20ˆi  10 ˆj  .10ˆi  20 dt

Sol.23. (C) It remains stationary Maximum possible friction force F frictionMax.   0.5 2 g  10 N Maximum applied force Fapplied Max.  8 1  8 N

F

applied Max.

F

frictionMax.

so the block will remain stationary

Sol.24. (A) 84kPa  V p  B   V

Sol.25. (D)

0.004  6  84 kPa   2100  10  100  2

m1R 

m2 R 2 3

I  I ABC  I AOC   mi R 2  m2

 AC 

2

12

 R 2  mi  m2

2R  12

2

 m1 R 2  m2

R2 3

Sol.26. (A) dark Path difference   13.5  13  12     n  12   







So there will be minima. Sol.27. (A) It will be clockwise As collision is elastic ,so after collision ball moves towards left with speed v .As walls and ground are smooth ,there is no tangential torque on the ball. Only normal forces and mg force pass through the centre of the ball ,so their torques about the centre are zero. Torques on the ball about its centre is zero . By   I  ,angular acceleration is zero hence angular velocity does not change. Sol.28. (A) electron R

mv qB

v ,q and B are same so

Rm

Mass of electron is minimum for given options. Sol.29. (C) 8 Nm 1   iAB sin 90 o  2   2  2  2  8 Nm 2 2 Sol.30. (A) 84 m

Result should have only two significant numbers (same as 12m). Sol.31. (C) 27 C 18V i 6F

6 i 6

3F

i

18  1 .5 A and V2  R2i  6  1.5  9V ; q2   3  9  27C 66 x

Sol.32. (B) 0

t

Initially velocity is constant ,so slope of x-t graph is constant and finite .Finally velocity becomes zero hence slope of x-t graph becomes zero.

Sol.33(C)

20m 3

mg   airVg  26 g  1.3Vg  V  20m 3

Sol.34. (D) 5 3 0V

R

2 E

Since equivalent internal resistance of equivalent cell across the external resistor R is ,Hence power delivered to R will be maximum if R  5  Sol.35. (A) RA 

1 4

l   2R 

2  3  5

2

,RB 

l   R

2



RA RB



1 VA R A I 1    4 VB RB I 4

Sol.36. (A) Al Sol.37. (A) 7 kgm2s1 L  2  2  2 sin 90o  3  3  3 sin 0o  1 1 1sin 90o  8  1  7 2V 2 Sol.38. (A) 3R 2 2 V V V 2 2V 2 P    Req R  R 3R 3R 2 2 Sol.39. (C) A  0, B  1, C  0 A0 B 1 C 0

1

F 1

1

Sol.40. (D) remains same Intensity after passing through polaroid A is

I0

Intensity of unpolarised light 2 2 2 I B  I A cos  IA 



Intensity after passing through polaroid B is Here  is the angle between pass axes of A and B .Here their pass(transmission) axes always remain parallel to each other i.e.

  0 .  IB  IA 

I0 2

 constant.

intensity of transmitted light through polaroid B remains same. Sol.41. (B) 4 days 8000 T1/2

 4000  2000  1000

4days  4days

4days

4days

Hence during the rotation,

Sol.42. (B) 1035 A o 

12420 12420 12420    1035 A o E 4  (16) E 2  E1

Sol.43. (D) 2V In reverse bias ,it is equivalent to open circuit condition. From figure given below 2V

VAB  2V

A B

100

Sol.44. (B) greater than P but less than 16P P  T4 ,P  T4 HereT  273  50 ,T  273  100  T  T  2T 4

P  273  100  P 4   1   2 P  273  50  P

Sol.45. (D) It is not possible It is not possible, because it will violate the second law of thermodynamics (Clausius statement) .If we consider imaginary case in which temperature of sample becomes more than 600K then it will radiated power is more than absorbed power. Hence it will correspond to decreasing temperature situation. So it is not possible to heat the sample to 900K. Sol.46. (C)

k p r2 kpcos 900 r2

r

A

kpcos r2

p

kpcos 0 r2

k p cos  k p  2 r2 r 2 2 b  a  

VA 

Sol.47. (C)

R       B1 .A1  B2 .A 2  Bb 2 cos 0o  Ba 2 cos 180o  B  b 2  a 2    b 2  a 2  t

d 2 2  dt  b  a   i   R R R

Sol.48. (C)

F MR F sin 30O O F cos 30O 30

120O

C

  I  F sin 300 R 

MR 2 F  2 MR

Sol.49. (C) acceleration may be zero at t=2 Let us consider an example in which particle is projected vertically upward from ground at t=0 and it reaches highest point at t=2. Then at that instant(t=2) velocity v=0 but acceleration=−g. Here displacement is non zero for the duration t=0 to t=2. For t>2 the ball again acquires velocity. In this example options(A),(B) and (D) are incorrect. Let us consider another example in which a particle is moving on horizontal plane and it comes to rest permanently at t=1 ,then this is the one of the special case in which acceleration of particle is zero at t=2. Sol.50. (C)  4ˆi  2ˆj m / s v  1 v 0 e   separation    y  vy  2 2 40  vapproach  y vx remains same,Hence v final  4iˆ  2jˆ

CHEMISTRY Sol.51. (A) [Co(CO)4 ] and Ni(CO)4 [Co(CO)4 ] has total electrons =3+2×4+1=12, Ni(CO)4 has total electrons =4+2×4=12 ;Thus isoelectronic Sol.52(B) (i), (ii) & (iv)

XeF6 has distortion and become Pentagonal bipyramid with one lone pair. F

O

F

F

Xe

O

Cl

I

F

O

O



XeF6

O

Cl

Sb

F

F

Cl

O



Cl

Cl

IO 56

Cl

Orthoperiiodate  5  

5  6  7   1  48 Octahedral

Octahedral 7  6  6   5  48 electrons Octahedral

SnCl62   4  6(7)  2  48 6 set of electrons  octahedral

Sol.53. (B) H2S < NH3 < H2O < HF Sol.54. (D) BF3

BF3 can form π bond also in addition to σ bond. As F is more electronegative and octate in B is not complete. Sol.55. (D) H2O H2O will act as Brönsted acid as provided H+ ion. Sol.56. (B) 0.354 gm  NH 4Cl   pOH   log K  log  NH 4Cl   pOH  log  NH 4Cl  pOH  pK b  log b  NH 3   NH 3   NH 3  Kb  14  9.45  log

 NH 4Cl   NH 3  Kb

 log 3  0.47

 log1014  (log109  log3)  log

 NH 4Cl   NH3  Kb

 log

1014  NH4Cl   1014   NH 4Cl   log 109  3  NH3  Kb 109  3  NH 3  Kb

1014 1014 NH K  NH Cl   0.011.85 105   NH 4Cl   0.35  3 b  4  9 9 10  3 10  3 2 Kp Sol.57. (D) 1 2 K p 

2 HI ( g )  H 2 ( g )  I 2 ( g ) At eq.2(1  x ) x x Total moles at equilibrium =2-2x+x+x=2 ,here x= degree of dissociation

 x  x  2    P 2 Kp x2 x 2 2 K p     2   2 Kp  x 2 1 x 1 2 K p 4 1  x  P 2 4 1  x  4 Sol.58. (C) 171 6g 60 A 6% solution of sucrose C22H22O11 conc.=  0.06ml 1  molel 1 100 ml 342 30 For unknown solution conc.= 3 g per 100cc  30 gl 1  mole l 1 m 60 30 30  342 For isotropic solution  m  171 342 m 60 Sol.59. (D) 491 kJ CH 3COOH  2O2  2CO2  2 H 2O 869  2   395   2   285    f H ( CH 3COOH )

Simplification gives 2  434.5  2   395   2   285    S H (CH 3COOH ) 1 or  434.5   395    285    f H  491   f H 2 Sol.60. (B) Sulphur Sol.61. (A) Cathode is Lead dioxide (PbO2) and anode is Lead (Pb) Sol.62. (D) ( ∆Ssystem+ ∆S surrounding ) > 0 Sol.63. (A) PVm=RT At low pressure & high pressure V is very high ,thus a2 and b are negligible, finally reduced Vm

eq. PVm=RT Sol.64. (D)

t 3/4

 A 0 If t1/2 vs. A0 will be

is constant ,it is radioactive decay hence is not of zero order. Thus answer t 3/ 4

 A 0 Sol.65. (C) Hexane Sol.66. (A) 5  1010 gm

Let W be the weight of

Ra 238

in equilibrium as

Th232 t1/2Th 1 N0 / 232 1.4 1010    ; N 0  AvagardoNo W  N0 / 238 7 Ra 238 t1/2 Ra W

238  5  1010 gm 9 232  2  10

Sol.67. (B)

n=4,

l=0,

m=0, s=+

1 2

For option (A) electron is in 3d For option (B) electron is in 4s For option (C) electron is in 4p For option (D) electron is in 5s According to Aufbau principle 4s<3d<4p<5s. Hence Answer will be n=4,l=0, m=0,s=+ Sol.68. (A) 4, 6 Sol.69. (C)

COOH

COOH

Terlhalic acid. Sol.70. (B) cis-1, 3-dimethylcyclohexane 



Two chiral centre of plane of symmetry so it is meso compound. Sol.71. (C) 2 ethyl-3 methyl pentanal OH

Sol.72. (A)  CHCl3  aq. NaOH 

In Reimer-Tieman ,phenol reacts with CHCl3 & aq. KOH/NaOH to give Selicil Aldehyde. Sol.73. (B) B < S < P < F I.P. in period increases  left to right. I.P. in group up to down  decreases. Sol.74. (B)

20 Carbocation will be more stable than 10 Substituted carbocation will more stable than simple. Sol.75. (A) Be3N2 Sol.76. (C) Cannizzaro reaction Sol.77. (C) BN Sol.78. (C) i > iv > ii > iii Acidity ∝ 1 basicty

Sol.79. (C) (i),(ii)



Br

(i)

1,2 hydride shift

Br (ii)



H



1 2

Sol.80. (A) iii < i < iv < ii Acidic strength ∝ electron withdrawing group strength. R  H  I 

1 1 1   I H R

Sol.81. (A) Only –I effect Due to SIR effect –NO2 group goes out of plane of the paper (-I). Sol.82. (C) AgNO3 Sol.83. (A) 6-ethyl-2-methyl-5-propyldecane CH3 (4)

CH2 CH2 CH  CH3 

H

C(6 )

CH3 CH2 CH2CH2

(10 )

(9 )

(8)

(3)

(7)

(5)

CH2 CH3

Sol.84. (B) D-fructose

(1)

(2)

CH2CH2 CH3

C H

CHO

CHO

OH

OH



OH     OH

OH

CH2 OH

base

O



OH     base OH

OH

OH

OH

OH

OH

OH

CH2OH

CH2 OH

CH2OH

D Manose

D Glucose

D Fructose

O Sol.85. (B) Benzoquionone

 NH  Cr O   H SO 4 2 2

2

7

4

O

Benzoquinone

Conjugated diketones Sol.86. (C) 19 ,  H

H

H

C H H

H

19 and 

H H H

H

Sol.87. (B) (i) and (iii)

As a plane of symmetry exist in compound(ii),there is no chirality in it. Hence (i) &(iii) will be optically active. Sol.88. (A) BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3 BCl3 and AlCl3 both are Lewis acids but BCl3 is more electron deficient than AlCl3 so BCl3 s stronger Lewis acid than AlCl3 due to high electron negativity.(E.N. B-2.0,Al 1.5) Sol.89. (C) I, II, IV O

Compound C CH3

cannot be obtained by Friedel Craft acylation, In

,NO2 NO2

NO2

Group has -M or –R effect so it cannot be used in Friedel Craft acylation ,this is deactivating & metadirecting group in ESR. O

O  H3 C

Cl AlCl 3

C

C

Cl

NO2

NO2

Sol.90. (A) 4-butyl-1-ethyl-2-methylcycloheptane Naming according to closest set of locant rule

1 7 6 2 5 4 3

Sol.91. (D) 5-chloro-2-methylcyclohexanol CH3 3 4

1

2

5

OH

6

Cl

Sol.92. (A) -hydroxyaldehyde 



R R CH 2

H or OH R CH 2  CHO   Dilute

CH 2

C

OH

H

R condensation  CHO

R CH 2

 H 2O 

CH

C

CHO

Aldoladdition  ,  unsaturated aldehyde

In aldol addition it is β hydroxyl aldehyde(aldol both alcohol and aldehyde group) But in aldol condensation the product is α,β unsaturated aldehyde Sol.93. (D)

O 2N

OH

NO2

O 2N

O 2N

OH

NO 2

OH

OH

Highest dipole moment Sol.94. (A) 4 Grignard reagent will react on CO bond there are four C-O bond .These will be addition of Grignard reagent at four positions. HO

1mol

O 1mol

COOEt 2mol

total  4mol

Sol.95. (A) KO2 KO2  K   O2

SiO2  Si4  2s2 2 p6

 17 e in valance cellso unpaired electron present &show paramagmetism

2O2  Diamagnetic 2s 2 2 p6

TiO2  Ti4  3d 0

2O2  Diamagnetic 2s 2 2 p6

Ba2 

BaO2 

2O2

XeConfiguration

 Diamagnetic

NeConfiguration

CH 3

Sol.96. (C) Nucleophilicity order  CH 3  NH 2  OH   F

Sol.97. (D) For lead +2, for tin +4

PbO 2 + Pb    2PbO,   ΔG 0 < 0

i.e. ΔG0 is negative so reaction is fisiable i.e. for Pb,+2 Oxidation state is more stable. SnO 2 + Sn     2SnO,   ΔG 0 > 0

i.e. ΔG0 is positive so reaction is nonfisiable i.e. for Sn,+4 Oxidation state is more stable. Correct answer is (D) For lead +2, for tin +4 For lead +2, for tin +4 Oxidation State are more characteristics. Sol.98. (A) 1-Phenyl-2-butane (i) Ph  CH 2  CH  CH  CH 3 Geometrical isomerism. CH 2  CH  CH  CH3

(ii)

No Geometrical isomerism

Ph CH 2  C  CH 2  CH3

(iii)

No Geometrical isomerism

Ph

Ph

(iv)

Ph

C  CH  CH3

No Geometrical isomerism

Sol.99. (C) associate At CMC they associate. Sol.100. (B) CH3 CH3

O

CHO

O

CHO

will be most reactive.

Hence most appropriate option is CH3

O

CHO

BIOLOGY Sol.101: (C) Colloidal suspensions Sol.102: (A) Thermaregulation Sol.103: (C)Oriental cockroach Sol.104: (B) Monera ,Protista ,Fungi, Plantae and Animalia Sol.105: (C) Orchid Sol.106: (D)Drupe Sol.107: (C) Endodermis Sol.108: (B) Nucleoside Sol.109: (B) Lamellae Sol.110: (D) Extra chromosomal DNA segment Sol.111: (B) Foot Sol.112: (C) Cycas rachis Sol.113: (C) Haploid Sparophyte Sol.114: (A) Total stem parasite Sol.115: (B) Guttation Sol.116: (A) RNA molecule Sol.117: (C) Red Sol.118: (C) Physical Process Sol.119: (B) Gibberellin Sol.120: (C) Removal of anthers Sol.121: (C) Opening of flower Sol.122: (B) Gamete Sol.123: (A) Mutation Sol.124: (B) Zygoti intra fallopian transfer Sol.125: (B) Nutrition Sol.126: (A) Cretaceous Sol.127: (A) Colon Sol.128: (D) Carpora allata Sol.129: (D) Liver Sol.130: (D) Spleen Sol.131: (B) Glycoprotein Sol.132: (A) All vertebrates Sol.133: (B) Seed coat colour Sol.134: (D) Cuticle Sol.135: (B) Helper T lymphocytes Sol.136: (D) all of the options Sol.137: (C) Placenta and fully developed foetus Sol.138: (B) Leishmania donovani

Sol.139: (B) Fats Sol.140: (D) no eyes Sol.141: (C) Bombyx Sol.142: (C) G1, S , G2 and M Sol.143: (C) Excretion Sol.144: (D) Cryopresevation Sol.145: (A) Epidermis of root Sol.146: (B) Typoid fever Sol.147: (B) Ventral Nerve cord Sol.148: (D) Fungi Sol.149: (D) Sporophyll Sol.150: (C) Bentham and Hooker