RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

Medical - UG Time : 3 Hrs. Marks : 720

Date : 03-05-2015

AIPMT : 2015 CODE - E PHYSICS

1.

Surface tension 

force MLT 2  length L

 Surfacetension  MT 2  Energy   ML2T 2 Velocity   LT 1 S  k Ea V b T c

 RHS    LHS  a

b

c

MT 2   ML2T 2   LT 1  T  M L0T 2  M a L2 a b T 2 a bc M a 1 L  2a  b  0  b   2

T   2a  b  c  2  2  2  c  2

C  2

 S   k E 'V 2 T 2 2.

N

From Figure sin 45o 

PS OS W VA = 10 km/h

1 PS  2 100

O 45°

100 km

P

100 100 2 PS    50 2 2 2

E

VB = 10 km/h S

relative velocity between A and B is

PS – shortest distance

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

1 11

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

VBA  VA2  VB2  10 2

t

50 2 10 2

t  5h

3.

a

d d dv d    dt dx dt dx

d   2 n x  2 n  1 dx a    2n  x 2 n  1     x 2 n 

a   2 n 2 x  4 n  1 4. M A  4 kg , M B  2 kg , M C  1 kg M  M A  MB  MC F  M .a. F  (4  2  1) a 14  7 a a2

F  F   4a 14  F   4  2 14  8  F F  6 N

5.

m2 g  T  m2a T   m1 g  m1a m1 m 2 g  m1T  m1 m 2 a Tm2  m1 gm2  m1m2a m1mg 1     T  m1  m2   0 T

m1m2 g 1    m1  m2

22

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

6.

K P  KQ

In 1st case : elongation (X) is same. 1 1 K P X 2 , WQ  K Q X 2 2 2

 WP 

 WP  WQ

In 2nd case : Force of elongation is same.  X1 

F F , X2  KP KQ

WP 

1 1 F2 K1 X 12  2 2 KP

1 1 F2 2 W2  K 2 X 2  2 2 KQ

 WP  WQ

7.

Work done by force  K .E  K .E f  K .Ei W .D  K .Ei  K .E f 30

1

 0.1xdx  2 10 10

2

 K .E f

20

0.1 

x2 2

30 20

500  K .E f

0.1  900  400   500  K .E f 2

0.1

500  500  K .E f 2

25  500  K .E f  475

8.

FV = constant = k m

dv V k dt k

 Vdv   m dt v2 k  t 2 m 

V 

2k t m SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

3 33

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

F m

dv dt

2 k 1 1 2 t m 2

m

mk 12 t 2

 9.

From conservation energy initial total energy 1 1 m1u12  m2u22 2 2

Total final energy 1 1 m2v22  m1v12   2 2 1 1 1 1 m1u12  m2u22  m1v12  m2 v22   2 2 2 2



10.

1 1 1 1 m1u12  m2u22    m1v12  m2 v22 2 2 2 2 d-x B

A x W

N1

N2

force balance

N1  N2  W _________(1) Torque balance about C.O.M. of rod

N1x  N2  d  x  Putting value of N2 from equation (1)

N1x  W  N1  d  x 

 N1x  Wd  Wx  N1d  N1x  N1d  W  d  x   N1 

11.

W  d  x d

By COAM R  mv0 R0  mv '  0   2 

 v '  2v0 44

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

1 1 2 So final K .E .  mv '2  m  2v  2 2  2mv 2

12.

I  I1  I 2  I 3

2 5mr 2 I 2  I3  mr 2  mr 2  3 3 I1 

2 2 mr 3

I  2  5



13.

mr 2 2 2  mr 3 3

12mr 2  4mr 2 3

GMm mv 2  Centripetal force r2 r

v2 

T

GM ____________(1) r

2 r v

T2 

4 2r 2 v2

Putting value of v2 from (1)

4 2 r 2 T   GM     r  2

T2 

4 2 r 3 __________(2) GM

T 2  kr 3 ____________(3) from (2) and (3)

4 2  k  GMK  4 2 GM 14.

R M R

5m 2R 2R 4R 12R

M x  5 M  9R  x  SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

5 55

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

X  45R  5x 6 x  45R 2 x  15R x  7.5 R

15.

Wein’s Displacement Law

mT  Constant P  violet max m Q  Red max m R  Green max m

r  g  v Tr  Tg  Tv TP  TR  TQ

16.

Pressure at the bottom P   gd  103  10  2700 P  27  106 Pa  Fractional compression = compressibility x pressure

 45.4 1011 Pa1  27 106 Pa

17. 18.

 1.2 102 In both cases the temperature difference between the ends of the rool is 100C  rate of heat flow is also 4J/s in the second case. By Bernoullis theorem

1 1 P1   v12  P2   v22 2   2  (Assuming is root width is very small)  inside

outside

Pressure difference P1  P2 

1   v22  v12  2

P1  P2 

1  1.2  40 2  0 2  2



1 0.6  1.2 1600 2

 960 N / m 2 Force acting on the roof is f = 960 x 250  960 

1000 4

66

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

 24  104

19.

 2.4 105 N upwords As initial and final points are same so

U ABC  U AC

A  B is isochoric process dWAB  0 dQAB  dU AB  400 J

B  C is isobaric process dQBC  dU BC  dWBC 100 J  dU BC  6  104  2 103 

100 J  dU BC  12 10 dU BC  100  120  20 J So, U ABC  U AC U AB  U C  dQAC  dWAC 1   400  20  dQAC   2  10 4  2  10 3   2  10 3  4  10 4  2  

380 J  dQAC   40  40  dQAC  380  80  460 J 20.

T2

2

W 1 low temp

T1

Q1  w  Q2

  1

T1 T2

1 T  1 1 10 T2

T1 1 9  1  T2 10 10 SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

7 77

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

Q2 T2  Q1 T1 Q1  w T2  Q1 T1 Q1  10 10  Q1 9

9Q1  90  10Q1 90  Q1

21.

U  nCv dT  5R   n   TB  TA   2  

5nR  PBVB PAVA     2  nR nR 

 5/ 2  PBVB  PAVA   5 / 2  2  103  5  103  4   5 / 2  8  10 3 

 20kJ 22.

Cv 

n R 2

C p  Cv  R 

n RR 2

n  C p    1 R 2 

n  1 R C p  2  n2    Cv n / 2R n

  1 23.

2 n

y1  a sin t  y2  b sin  t   / 2 

y R  y1  y2  a 2  b 2 sin t    SHM with complitude (4) 88

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

24.

V12   2  a 2  x12  _____(1) V22   2  a 2  x22  _____(2)

(1) ––– (2) V12  V22   2  x22  x12 

V12  V22 x22  x12



x22  x12 V12  V22

T  2

25.

Closed organ pipe nc 

V 4l

Open organ pipe n0 

V 2l '

IInd overtone of open organ pipe n  3n0 n 

3V 2l 

nc  n V 3V  4l 2l  l   6l l  6  20 l   120 cm C1=C C2 q'=q

26.

disconnected

V

q  CV  V  q / c

C2  CK U1 

q2 2c

U2 

q2 2CK SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

9 99

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

U

q2  1    1 2C  k 

1 1  U  CV 2   1 2 k 

C2  CK

27.

V '

q CK

V'

V K

E  Ar E  Aa E  Aa

1 q  Aa 4 0 a 2

q  4 0 Aa 3 28.

L=4 , 800

1cm  1mv 100cm  100mv 400cm  400mv  0.4V V  0.4  8 R 

2 8 8 R

16 160   40 0.4 4

R = 32

29.

1 1 1   R RB RC

X

A RA = R

RB = 1.5 R B Q

P

C RC = 3 R

1 1 1   R 1.5 R 3R V

1 2 1  R 3R R  R

VXP  VA  iR VPQ  VB  VC  iR

VA  VB  VC 1100

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

30.

The area cross section of conductor is non uniform so current density will be different but the numbers of flow of electron will be same so current will be constant.

31.

BI 

0 i (k ) 4 R BIII 

0 i (  k ) 4 R

BII 

0 i  (i ) 4R

B  BI  BII  BIII B

0 i ( 2k   i ) 4 R

B

r

32.

0i  (2k   i ) 4 R

e i

X

B i

e T

i

e e  2 n  e    ne 2  2 /   2

B

0i 0 ne  2R 2R x A

33. i

C

BX V

V B

D

P.d. across the AB is

VA  VB  B1  a V 

0i aV 2  x  a / 2 

P.d. across the CD is

VC  VD  B2  a V

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

11 1 1 11

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics



 0i aV 2  x  a / 2 

Net P.d. in the loop is

 VA  VB   VC  VD  

0iaV  1 1     2  x  a / 2 x  a / 2 



0iaV 2



  2a     x  a / 2  x  a / 2  

1

 2 x  a  2 x  a 

34. N P0  Vrms irms

Vrms  irms R

Vrms Vrms R R

Vrms  irms R



2 Vrms 2 P0   Vrms  P0 R R

L

R

N

P  Vrms irms cos  2  Vrms

R z2

 P0 R 

P  P0 35.

R z2

R2 z2

E  h 

hc 

 h E Pi    C   h  E Pr    C    2 E  P light  Pr  Pi  C

1122

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

 2E Momentum transfered to the surface    P light  C

1 1  (1.5  1)   f R

36.

1



f lens

1 f concave

1 f concave

0.5 R

 1 1  (1.7  1)   R R 



0.7  2 1.4  R R

1 0.5 1.4 0.5 0.4 0.4      f eq R R R R 20 f eq   50cm

37.



a

sin  

2D/a

 a

y   D a y

D a

width of centeral maxima is 38.

2 D a

d 1 mm  1  103 m D 1m

  500  109 m

a width of central maxima = width of 10 maxima

2 a

s

2  D   10   a  d  a

2d 10 D SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

13 1 3 13

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

a

2  10 3 10  1

a  2  104 m a  0.2 mm

39.

 A  sin    2    A sin   2  A  sin   A  2  cot  2  A sin   2  A  A  cos   sin   2  2   A  A sin   sin   2 2  A A  sin     sin    2 2  2 2

     40.

hc  hv    eV0 ______(1)  hc    e V0  ________(2) 2 hc  2  2eV0 ________(3) 

 3  1  O    V0   V0e   

hc T

T 

hc 

hc  4eV0 

 4

1144

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Physics

41.



h p

p  h

like  yx  c   3 Ahyperbola 42.

Energy of electron in He  3rd orbit = 13.6 

Z2 eV n2

4 eV 9

E3   13.6 

4  1.6  1019 J 9

  13.6 

In Bohr’s model : E3   K .E3  9.7  10 19 J 

2  9.7  1019  1.46  106 m / sec 31 9.1  10

  43.

1 me  2 2

Radius of the nucleus R = R0 A 1/3 1/ 3

RAl  A1    RTe  A2 

1/ 3

 27     125 

1

RAl  33  3     3/ 5 RTe  53  RTe 

5 RAl 3

5v 44. As it is forward - bias takes + ve value 45.

A 0 0 1 1

B 0 1 0 1

Y1 1 1 0 0

Y2 1 0 1 0

Y 0 0 0 1

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

15 1 5 15

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

CHEMISTRY Q.46 CH 2  CN  2 H N  C C C  N

 6  4 

H O

 4  1 

O=C O-H O Xe

 4  4 

O

O O

 3  4 

N  C C  N

Q.47 Ar, K & Ca belong to the same period. Being Nobel gas radii of Ar is heighest. We also know for isoelectronic species size decreases with increase in positive charge of the ion. Q.48 Sodium pump used for conduction of signals from brain to organ and vice versa. In which Na+ and K+ are exchanged. Q.49 Metal nitrates are usualy not found as nitrates in their ores because they are highly soluble in water. Q.50 Down the group solubility of sulphate of alkaline earth metal decreases because of increase in lattice enrergy gradually. Q.51 due to lanthanoid contraction Zr (Zirconium)and Hf (Hafnium) have almost same atomic radii. (O )

Q.52

1.

( III )

Fe  H 2O   Fe2O3  H 2 Steam (O )

( III )

2.

Moisture Fe  O2   Fe2O3. x H 2O

3.

CuSO4  Fe   FeSO4  Cu

4.

Fe  5 CO   Fe CO 5

(O )

(O )

( II )

(O )

 During formation of Fe  CO 5 oxidation number of Fe doesn’t change.

116 6

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.53 S

Cl O

O

O

O

O

O

O

O O

N

O=C O

O

 2  ClO3 & SO3 are isoelectronic and are pyramidal

Q.54 Order of bond order O2   1s 2 *1s 2 2 s 2 * 2 s 2 2 p z2  2 p x2   2 p 2y 



*

2 p x2   * 2 p1y 

Bond order  

no.of electrons in BMO-no of electrons ABMo 2

10  7 3   1.5 2 2

O2   1s 2 *1s 2 2 s 2 * 2s 2 2 p z2   2 p x2   2 p 2y 



*

2 p1x   * 2 p 0y 

Bond order =

10  5 5   2.5 2 2

O2   1s 2 *1s 2 2s 2 * 2s 2 2 p z2  2 p x2   2 p 2y 



*

2 p1x   * 2 p1y 

Bond order 

10  6 4  2 2 2

Order ofbond order : O2  O2  O2 Q.55 NaHSO3 is used as food preservative as it produces SO2 on decomposition which checks the oxidation of food. NO3

NO2

NO2

NO2

Hybridisation

sp 2

sp 2

sp 2

sp (linear)

Bond angle

1200

 1200  1200 1800

Q.56 Species

so NO2 has maximum bond angle.

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

17 1 7 17

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.57 Magnetic moment =

n  n  2  B.M.

where n = number of unpaired electrons  2.84  n  n  2  B.M.

 2.84 

2

 n  n  2

8  n 2  2n On solving n = 2

only Ni 2   Ar  3d 8 4s 0 has 2 unpaired electrons. 3

Q.58 1) Co  NH 3 6  Cl3  Co  NH 3 6   3Cl  2) Co  NH 3 3 Cl3   Co  NH 3 3 Cl3  1

3) Co  NH 3 4 Cl2  Cl  Co  NH 3 4 Cl2   Cl  2

4) Co  NH 3 5 Cl  Cl2  Co  NH 3 5 Cl   2Cl  so Co  NH 3 3 Cl3  does not ionize so doesn’t give test for chloride ions. Q.59 Co  CN  6 

3

Co3   Ar  3d 6 eg orbital t2g orbital CN– is using strong field ligand so zero unpaired electrons are present. Q.60 By Arrhenius equation k  Ae  Ea / RT

ln k  ln A 

Ea RT

on plotting ln k vs

1 T

lnA lnk

slope = -Ea/R

1/T  By knowing slope activation energy can be determined.

118 8

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.61 For ideal solution Pobs  PRaoult

P  0

H mix  0

S mix   ve

Vmix  0 Q.62 Let the masses of H2 and O2 be x g and 4x g

nH 2 x/2 4   Molar ratio n  4 x / 32  1 O2 Q.63

Edge length a = 361 pm, effective number in unit cell = 4 (given)

 It is a FCC structure  face diagonal = 4r

2 a  4r



r



2  361  127 pm 4

Q.64 for ‘n’ order reaction t1/ 2  C01 n

where C0 is initial concentration of reactant. for first order reaction n = 1, half life is independent of intial concentration Q.65 R  P

K

 P eq  R eq

 1.6 1012 

 P eq  R eq

  P eq   R eq

so mostly the products will be present in the equilibrium mixture. Q.66 In Fuel cell energy of combustion is converted into electrical energy. Q.67 Molality of solution of x = molality of solution of y = 0.2 mol/kg by elevation in boiling point relation Tb  iK b m or Tb  i  Tb of solution of ‘x’ > Tb of solution of ‘y’

 ‘i’ of solution of x > i of solution of ‘y’  solute ‘x’ undergoing dissociation

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

19 1 9 19

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.68 Al2  SO4 3  2 Al 3  3SO42  Volume of Vant Hoff’s factor i = 5

K 4  Fe  CN 6   4 K    Fe  CN 6 

4

i  5

K 2 SO4  2 K   SO42  i  3 K 3  Fe  CN 6   3K    Fe  CN  6 

3

i  4

Al  NO3 3  Al 3  3 NO3  i  4  Q.69 Electronic configuration of Fe 2 is  Ar  3d 6 4s 0  Number of ‘d’ electrons = 6

Ne – 1s 2 2 s 2 2 p 6  6  p electrons  Mg – 1s 2 2s 2 2 p 6 3s 2  6  s electrons  Cl – 1s 2 2 s 2 2 p 6 3s 2 3 p 5 11  p electrons 

 Ar  3d 6 4s 2  6  d electrons  Q.70 Order of bond order O2   1s 2 *1s 2 2 s 2 * 2 s 2 2 p z2  2 p x2   2 p 2y 



*

2 p x2   * 2 p1y 

Bond order = 

no.of electronsin BMO-no of electrons in ABMo 2

10  7 3   1.5 2 2

O2   1s 2 *1s 2 2 s 2 * 2s 2 2 p z2   2 p x2   2 p 2y 



*

2 p1x   * 2 p 0y 

Bond order =

10  5 5   2.5 2 2

O22    1s 2 *1s 2 2 s 2 * 2 s 2 2 pz2  2 p x2   2 p 2y 



*

2 px0   * 2 p y0 

Bond order 

10  4 6  3 2 2

Order ofbond order : O22  O2  O2

220 0

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.71 Angular momentum of electron in ‘d’ orbital  l  l  1

h 2

h      2  

 2  2  1  6

K sp

Solubility

Ag 2CrO4

1.1 1012

S

AgCl

1.8  1010

S  K sp  1.34 10 5

AbBr

5  10 13

S  K sp  0.71 10 6

AgI

8.3  1017

S  K sp  0.9  108

Q.72 Salt

3

K sp 4

 0.65  104

 solubility of Ag 2CrO4 is highest it will precipitate last.

Q.73 Tyndall effect is the scattering of light by sol particles. Which can’t be affected by charge on them. Q.74 we know G  G 0  2.303RT log Q but at equilibrium

G  0 , Q  k G 0  2.303RT log K Q.75

Bithional functions as antiseptic.

Q.76

Under weakly acidic conditions Nitro benzene on electrolytic reduction gives aniline but under strongly acetic condition we get p - aminophenol. NO2

NHOH -

4e 4H+

NH2

2e2H+ weakly H+ (strongly) NH2

OH

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

21 2 1 21

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.77 PN 2  Ptotal  PH 2O  725  25  700mm PVM 700  40  28  RT 760 1000  300  0.0821

WN2 

 0.042 g

% of Nitrogen 

0.042  100 0.25

 16.573%

Q.78 CH 3

CH 3 CH 3  C  Cl

  CH 3  C   Cl 

CH 3

(more stable due to 9 hyper conjugation structures & 3 +I

CH 3

groups) Q.79

Its nucleophilic substitution reaction. It follows SN2 mechanism.

Q.80

Electrophilic addition takes place via more stable carbo cation. HBr C6 H 5  CH  CH  CH 3  

C6 H 5  CH  CH 2  CH 3 Br

Q.81 CH 3 O3   CH 3  Zn , H 2O

OHC CH 2

CH 3

CH 3

C

C

H

CH 2

O

H





Q.82

+

 O + CH 3 Li  + –

Li O  CH 4

There is no   H in the structure I and II Q.84 Allylic and benzylic halides show high reactivity towards SN1 reaction. Q.83

H + + H 3C  C  CH  CH 2  Cl  CH 3  CH  CH  CH 2  H 3C  CH  CH  CH 2  Cl 

222 2

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / Chemistry

Q.85

The enolic form of ethyl acetoacetate has 16 single bonds (= 16 sigma bonds) and 2 double bonds (equal to 2 sigma and 2 pi bonds)

Q.86 In keto - enol tautomerism the keto form should have   hydrogen (I & II) H3C

H3C

CH3

CH3



I)

OH

O

CH3 CH3 

II)

CH3 CH3

O

OH

In IIIrd ketone

H  H  III) 

CH3 CH3

CH3 CH3

O

OH

here  H participates in tautomerism Q.87

The enthalpy of hydrogenation is inversely proportional to stability of alkene III  II  I

Q.88 It is an alternating polyamide copolymer of glycine  H 2 N  CH 2  COOH  and amino caproic acid  H 2 N  CH 2 5 COOH  and is biodegradable

O

O

O

H 2 N  CH 2  C  OH  HN CH 2 5 COOH   H 2 N  CH 2  C  NH  CH 2 5 C 

n

H Nylon -2-nylon-6 Q.89

There are 4 bond’s, hence total  -electrons are 8.

Q.90

For the Idoform test there must be a terminal CH 3 group, similarly for the Tollen’s test CHO group is required. Hence, as per question the structure would be O CH 3  CH 2  C  CH 2  CH 3

SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

23 2 3 23

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / BIOLOGY

BIOLOGY Q.91 Agaricus is saprophytic fungus (Not parasitic) Phytophthora belong to Phycomycetes Mucor belong to Phycomycetes. Q.92 Selaginella is heterosporous producing micro & megaspores. Coralloid roots in Cycas have Anabaena . Q.93 Bryophytes and pteridophytes have an independant gametophyte. In spermatophytes gametophyte is reduced and dependant on sporophyte Q.94 Mannitol is stored food in Phaeophyceae (Brown Algae) Rhodophyceae members store food in the form of Floridean starch. Q.95 Methanogens in cow & buffalo guts are helpful for partial digestion of cellulose (which is a major constituent of grass) Q.96 Sprirogyra shows morphological isogamy, physiological anisogamy where male gamete is motile but not having flagella Polysiphonia belong to Rhodophyte which is characterised by non-motile male gametes (non-flagellated) Q.97 Since cambium is absent, Vascular bundles are closed and secondary growth is not seen. Q.98 Floral formula is of Solanaceae family hence is of Petunia, Brassica is from Brassicacene family whereas Allium belong to Liliaceac & Sesbania belong to Fabaceae Q.99 Monocot stem has open and scattered vascular bundles. Roots in monocots have radial vascular bundles. Q.100 Keel is a petal in vexillary aestivation which is a characteristic of family Fabacae Tomato-Solanaceae.Tulip and Aloe-Lilliacea. Q.101 Guava & Cucumber - epigynous china rose - Hypogynous Q.102 In Opuntia leaves become modified into spines to reduce the rate of transpiration. Q.103 Flattened membranous sacs  Thylakoids Stack of thylakoids  Grana Q.104 The chromosomes in which centromere is situated close to one end are Acrocentric, if it is at the centre metacentric, near to the centre - sub - meta centric, at one end - Telocentric Q.105 Rough ER - Production of proteins. Whereas smooth ER is the site for synthesis of lipids & glycogen. Q.106 Anabaena is prokaryotic blue green alga, showing no true nucleus (nucleoid). Q.107 Polysomes are Polyribosomes, attached to mRNA. Q.108 Transpiration cause water to rise in plants by pulling through xylem components Root pressure causes water to rise in plants by pushing water in xylem elements Q.109 Zinc, Cu, Manganese, selenium & boron are micromolecules. Q.110 Because of greater elongation of cells on shaded side, it makes the plant bend towards light source. Q.111 In a ring girdled plant, since phloem is removed, translocation of food does not take towards roots, so root dies first. Q.112 Sigmoid curve shows 3 major phases like initial lag phase, exponential log phase & steady or stationary phase 224 4

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / BIOLOGY

Q.113 Levitt (1954) proposed Active Potassium Transport theory in which stomatal movements (opening & closing) is because of influx & efflux of K+. Q.114 The hilum is a scar on the seed where funicle was attached. In many of the seeds like mungbean, pigeon pea it is very district. Q.115 In Geitonogamy pollen grains are transferred between two different flowers on the same plant. Eg Maize. Q.116 Honey is made by bees by nectar (regurgitation & evaporation). Q.117 Transmission tissue is characteristic feature of solid style which helps in directing pollen tube. Q.118 Runners - in Cynadon Offsets - in Eicchornia & Pistia Bulbils - in Dioscorea Q.119 Nectar & pollen are floral rewards to the animal pollinators which are used by animals like bees as food. Q.120 Seven pairs of contrasting characters in pea plants , were studied by mendel is his experiments. Q.121 Recombination (crossing over) is the most common mechanism of genetic variation in the population of a sexually reproducing organism Q.122 A technique of micropropagation is somatic embryogenesis (which forms clones Of ex-plant in the form of embryo) Q.123 Translocation is reciprocol of chromosomal segment between non-homologous chromosome. Movement of gene from one linkage group to another is called crossing over. Inversion : - When a piece of chromosome is removal & rejoined in reverse order, its known as inversion. Duplication : In this mutation deleted chromosomal segment is attached to its normal homologus chromosome. Here a gene or many genes are repeated twice or more times in the same chromosome. Q.124 Multiple alleles are present on the same locus of the chromosome Q.125 Genetic engineering approval committee regulates GM research & safety of introducing GM organisms for public services. Q.126 The Bt-toxin is an inactive protoxin which is activated due to alkaline pH in the insect gut. Q.127 Glyphosate is a herbicide and crops are engineered for glyphosate are resistant/tolerant to herbicides (By which main plants can be kept safe while using glyphosate as weedicide in fields). Q.128 Ribosomes  60 to 65% rRNA 30 to 40% Protein Q.129 Polyethylene glycol & sodium nitrate increases fusion of protoplasts & hence can be used as fusogen in somatic hybridization or protoplast fusion. Q.130 The United Nation Climate Change Conference was held in Durban South Africa from 28 November to 11 December 2011 to establish a new treaty to limit carbon emissions. Q.131 Horizontal distribution of different species occupying different levels in a biotic community is known as zonation. Q.132 Tissue culture, seed bank & cryopreservation are ex- situ conservation Sacred groves, National park, wildlife sanctuary is in-situ conservation. Q.133 Newly cooled lava, Bare rock & newly created pond  Primary succession. Q.134 The mass of living material at a trophic level at a particular time is standing crop. The inorganic substances present at a trophic level at a particular time is called as standing state SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

25 2 5 25

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / BIOLOGY

Q.135 The rate of production of organic matter during photosynthesis is gross primary productivity. Q.136 Exoskeleton is responsible for diversification of insects on land. Q.137 Trichinella spiralis is viviparous roundworm. Q.138 In aves skin is non glandular, in mammals pinnae is absent in whale, cyclostamata paired appendages are absent. Q.139 Platypus is an egg laying mammal. Q.140 Erythropoiesis starts in liver in foetus. Q.141 Arthroidal membrane joins terga , sterna and pleura of cockroach body. Q.142 Rough endoplasmic reticulum produces nuclear envelope in telophase. Q.143 Cytochromes which are Iron containing electron acceptors are present on cristae of mitochondria, helpful in ETS. Q.144 The competitive inhibitor increases the Km of the enzyme for the substrate. Q.145 Option (3) suggest correct match. Q.146 After S phase, amount of DNA gets doubled but number of chromosomes remain same Chromosome number Amount of DNA

Ga mete (n) Single

Cell (after S phase) (2n) Four times

Q.147 Brunner’s glands are present in the submucosa of duodenum and secrete mucous. Q.148 Gastric juice of infants contains Pepsinogen, Lipase, rennin Q.149 Rise in CO2 concentration stimulates chemoreceptors present in aorta and carotid artery which stimulates respiratory centre. Q.150 Blood pressure in the mammalian aorta is maximum during systole of the left ventricle, because blood is pumped into aorta under high pressure. Q.151 A fluid connective tissue, constituents of which are blood is plasma plus blood cells ie. RBC, WBC & platelets. (ie. plasma plus blood corpuscles) (55% plasma & 45% corpuscles) Q.152 Removal of proximal convoluted tubule from the nephron results in lack of reabsorption of high threshhold substances from renal tubules, and obligatory reabsorption of water is also affected leading to more diluted urine. Q.153 During muscle contraction actin filament slides over myosin filament, where none of them shorten in length. Q.154 Glenoid canty of scapula articulate with head of humerus. Q.155 Language comprehension is the function of wernicke’s area of parietal lobe of cerebrum. Q.156 Vestibular apparatus (otolith organs ie. utricle & sacculus & semicircular canal) help in maintaining body posture, equilibrium & balancing the body. Thus it will help gymnast to balance its body upside down even in total darkness. Q.157 Epinephrine / Adrenaline is a hormone - a chemical signal that has both endocrine and neural roles. Q.158 Renin does not favour for the formation of large quantities of dilute urine as it activates RAAS ( Renin angiotensin activating system) so it causes reabsorption of sodium which leads to formation of concentrated urine.

226 6

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / BIOLOGY

Q.159 Capacitation refers to process occurring in uterus in which changes occur in sperms which helps to enable them to fertilize - ovum. These include Dilution of inhibiting factors in semen. Removal of cholesterol layer (galea) on acrosome. Increase in permeability of membrane to Ca 2 Q.160 Prolactin has no role in parturition, it helps in lactation process, (lactogenic hormone), development of mammary glands (mammotropin) & maintainance of corpus luteum (leutotropin). Q.161 Chikungunga virus is transmitted through aedes aegypti mosquito and does not get transmitted through semen. Human immunodeficiency virus, hepatitis B virus & Ebola virus are sexually transmitted disease, thus are transmitted through semen. Q.162 Spermatogonia i.e. sperm mother cell is diploid whereas spermatids, first polar body and second polar body are Haploid. Q.163 Hysteroctomy is removal of uterus (Hystera means uterus) Q.164 Encephalitis is not a sexually transmitted disease. Q.165 Formation of abnormal ova i.e., 22 + XX in the mother will lead to birth of human baby with ‘XXX’ genotype. Q.166 Alleles or Allelomorphs are the two (or more) alternative forms of a gene (factors), controlling same characters occuping identical loci (positions) or homologous chromosomes these terms were coined by Bateson. Woman B

Q.167 Man A [ I AI A / I AI O ]

[ I BI B / I I O ]

Progeny I A I O I B I O

I AI O I O I O

A

B

AB

O

Q.168 Regulation of lac operon by repressor is referred to as negative regulation. Q.169 Chargaff’s rule A = T, G = C

C = 17% So G = 17%

A+G =1 T+C

A + G + C + T = 100  A + 17 + 17 + T = 100

Since A = y So A = 33%, T = 33%

 A + T = 100 – 34 = 66

Q.170 Homo habilis has the smallest brain capacity of 650-800cc. Homo erectus has capacity of 900 cc Homo sapiens has capacity of 1350 cc Homo Neaderthalis has capacity of 1400 cc. Q.171 A population will not exist in Hardy - weinberg equilibrium if individuals mate selectively as mutation , nonrandom mating and genetic driff affects equilibrium. SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

27 2 7 27

RAO IIT ACADEMY / Medical - UG / AIPMT - 2015 / Solutions / Code : E / BIOLOGY

Q.172 Vaccines for the following diseases are prepared with following :(a) Tuberculosis - harmones (attenuated) bacteria - mycobacterium Tuberculae (b) Whooping cough - killed bacteria - bordetella partusis bacteria (c) Diphtheria - Inactivated toxin (d) Polio - Harmless virus. Q.175 High value of BOD  water is highly polluted. Because of presence of high amount of organic matter number of organisams increases which increase demand for oxygen. Q.173 HIV that causes AIDS, first starts destroying helper T - cells. Q.174 Active form of Entamoeba histolytica feeds upon erythrocytes , mucosa and submucos of colon. Q.176 Tropical rain forest shows striated vegetation because of which most animals are tree dwellers Q.177 Because of more success for food, population of B shows increase in no. of individuals Q.178 In cryopreservation, genes/gene pools are stored at -196oC under liquid nitrogen which is an advance exsitu conservation of bio diversity. Q.179 Silent spring  Pesticide pollution. The book documented the detrimental effects on the environmentparticularly on birds of the indiscriminate use of pesticides. Q.180 Due to depletion of stratospheric ozone, there is increased risk of skin cancer, damage to cornea, conjunctiva etc. as well as reduces our immune system.

228 8

AIPMT-2015-Solutions-Code-E.pdf

Loading… Page 1. Whoops! There was a problem loading more pages. Retrying... AIPMT-2015-Solutions-Code-E.pdf. AIPMT-2015-Solutions-Code-E.pdf. Open.

2MB Sizes 2 Downloads 144 Views

Recommend Documents

No documents