F040

Physics1. In a com mon base con fig u ra tion

Ie = 1 mA 0.95; α = , the value of basecur rent is(a) 1.95 mA (b) 0.05 mA

(c) 1.05 mA (d) 0.95 mA

2. A Si spec i men is made into p-typesemi con duc tor by dop ing on an av er age onein dium atom per 6 107× sil i con at oms. If thenum ber den sity of at oms in Si be 6 1028× / ,m 3 what is the in dium at oms per cm 3?(a) 1012 (b) 1015

(c) 1018 (d) 1020

3. The min i mum wave length of X-ray emit tedby X-ray tube is 0.4125 Å. The ac cel er at ingvolt age is(a) 30 kV (b) 50 kV

(c) 80 kV (d) 60 kV

4. The ion is ation po ten tial of hy dro gen atom is13.6 V. How much en ergy need to besup plied to ion ise the hy dro gen atom in thefirst ex cited state?(a) 13.6 eV (b) 27.2 eV

(c) 3.4 eV (d) 6.8 eV

5. What is the per cent age er ror in themea sure ment of time pe riod of a pen du lumif max i mum er rors in mea sure ment of l andg are 2% and 4% re spec tively (a) 6% (b) 4%

(c) 3% (d) 5%

6. A body trav el ling along a straight lineone-third of the to tal dis tance with a velocity 4 m/s. The re main ing part of the dis tancewas cov ered with a ve loc ity 2 m/s for half the time and with ve loc ity 6 m/s for the wholetime of motion is(a) 5 m/s (b) 4 m/s (c) 4.5 m/s (d) 3.5 m/s

7. A body of mass 2 kg moves with anac cel er a tion 3 ms−2 . The change inmo men tum in one sec ond is(a)

2

3

1kg ms− (b) 3

2kg ms 1−

(c) 6 kg ms 1− (d) None of these

8. When an axle ro tates in a sleeve, the friction in volved in the pro cess is(a) slid ing (b) roll ing

(c) lim it ing (d) None of these

9. Two bod ies A and B hav ing mass m andre spec tively pas ses same ki netic en ergy.Given that M m> . If pA and pB be theirmo ments, then which of the fol low ingstate ments is true?(a) p pA B= (b) p pA B>(c) p pA B< (d) It cannot be pre dicted

10. A gun of mass M fires a bul let of mass mwith max i mum speed v. Given that m M< .The ki netic en ergy of the gun will be(a)

1

2

2mv (b) 1

2

2Mv

(c) more than 1

2

2mv (d) less than 1

2

2mv

Solved Paper 2013

Indraprastha CETEngineering Entrance Exam

11. If a solid sphere and solid cyl in der of samemass and den sity ro tate about their ownaxis the mo ment of in er tia will be greater for(a) solid sphere (b) solid cylinder

(c) both (a) and (b) (d) equal both

12. If V is the grav i ta tional po ten tial on thesur face of the earth, then what is its value at the cen tre of the earth?(a) 2V (b) 3V (c)

3

2V (d)

2

3V

13. If γ be the ra tio of spe cific heats of a per fectgas, the num ber of de grees of free dom of amol e cule of the gas is(a)

25

2(γ − 1) (b)

3 1

2 1

γγ

−−

(c) 2

1γ −(d)

9

21( )γ −

14. If an L-R cir cuit con nected to a bat tery ofcon stant emf E switch S is closed at time t = 0. If e denotes the in duced emf acrossinductor and i the cur rent in the cir cuit atany time t. Then which of the fol low inggraphs shows the vari a tion of e with i ?

15. Two iden ti cal glass ( / )µ g = 3 2 equiconvexlenses of fo cal length f are kept in con tact.The space be tween the two lenses is filledwith wa ter ( / )µ w = 4 3 . The fo cal length ofthe com bi na tion is(a) f (b) f /2 (c)

4

3

f(d)

3

4

f

16. A slab con sists of two par al lel la yers of twodif fer ent ma te ri als of same thick ness andther mal con duc tiv i ties k1 and k2 . Theequiv a lent ther mal con duc tiv ity of the slabis(a) k k1 2−(b) k k1 2/

(c) 2 1 2

1 2

k k

k k+

(d) k k

k k1 2

1 22

+

17. The re la tion be tween Young’s mod ulus ( )Y ,bulk mod ulus ( )K and mod ulus of elas tic ity ( )η is(a)

1 1 3

Y K= =

η

(b) 3 1 1

3Y K= +

η

(c) 1 3 1

3Y K= +

η

(d) 1 3 1

3η= +

Y K

18. A point par ti cle of mass 0.1 kg is ex e cut ingSHM of am pli tude 0.1 m. When the par ti clepasses through the mean po si tion, its KE is 8 10 3× − J. The equa tion of mo tion of thispar ti cle if its ini tial phase of os cil la tion is 45° is(a) y

t= +

0.1sin4

π4

(b) yt= +

0.1sin2

π4

(c) y t= −

0.1sin 44

π

(d) y t= +

0.1sin 44

π

19. A man weights 60 kg at earth sur face. Atwhat height above the earth’s sur faceweight be come 30 kg?Given radius of earth is 6400 km.(a) 2624 km (b) 3000 km

(c) 2020 km (d) None of these

2 | Indraprastha CET (Engg.) l Solved Paper 2013

l

e

(a)

l

e

(b)

l

e

(c)

l

e

(d)

20. Two bod ies m1 and m2 are at tached to thetwo ends of a string (fig ure). The stringpasses over a pul ley of mass M and ra dius R. If m m1 2> , then the ac cel er a tion of thesys tem is

(a) ( )m m m g

m m m1 2

1 2

− ++ +

(b) ( )m m g

m m1 2

1 2

−+

(c) ( )m m g

m m1 2

1 2

+−

(d) ( )

( / )

m m g

m m m1 2

1 2 2

−+ +

21. A ball falls ver ti cally onto a floor withmo men tum p and then bounces re peat edly,the co ef fi cient of res ti tu tion is e. The to talmo men tum im parted by the ball to the flooris(a) p e( )1 + (b)

1

1 − e

(c) pe

e

1

1

+−

(d) p

e1

1−

22. A ma chine which is 75 per cent ef fi cient uses 12 J of en ergy is lift ing up a 1 kg massthrough a cer tain dis tance. The mass is then al lowed to fall through that dis tance. Whatwill its ve loc ity be at the end of its fall?(a) 24 m/s (b) 32 m / s

(c) 8 m / s (d) 9

23. An un loaded car mov ing with ve loc ity u on africtionless road can be stopped in adis tances s. If pas sen gers add 40 % to itsweight and brak ing force re mains the same,the stop ping dis tance at ve loc i ties is now(a) 1.4 s (b) 1.4 s (c) (1.4)2 s (d)

1

1.4s

24. A hol low charged metal sphere has a ra dius r. If the po ten tial dif fer ence be tween itssur face and a point at dis tance 3r from thecen tre is V , then the electric intensity atdis tance 3r from the centre is(a) V r/6 (b) V r/4

(c) V r/3 (d) V r/2

25. Three charges Q, q and − q are placed at thever ti ces of right an gled iso ce les tri an gle asshown in fig ure. The net elec tro static en ergy of the con fig u ra tion is zero if Q is equal to

(a) −+

q

1 2

(b) −+2

2 2

q

(c) − 2q

(d) + 2

26. A cur rent i is fol low ing in a hex ag o nal coil ofside l (figure). The mag netic in duc tion at the cen tre of the coil will be

(a) 3 0µπ

i

l

(b) µ

π0

3

i

l

(c) µ

π0

3 3

i

l

(d) 3 3 0µ

πi

l

Indraprastha CET (Engg.) l Solved Paper 2013 | 3

+q +qa

Q

O

ii

M

R

T1

m g1

a T2 a

m g2

27. A bat tery of in ter nal re sis tance 4 Ω iscon nected to the net work of re sis tances asshown. In or der that the max i mum powercan be de liv ered to the net work, the value Rin Ω should be

(a) 4

9(b) 2 (c)

8

3(d) 18

28. A full wave rec ti fier cir cuit along with theout put is shown in fig ure. The con tri bu tionsfrom the di ode is (are)

(a) C (b) A C,

(c) B C D, , (d) A B C D, , ,

29. A ra dio ac tive sub stance X de cays intoan other radioac tive sub stance Y . Ini tiallyonly X was pres ent, λx and λy are thedis in te gra tion con stants of X and Y . Nx and N y are the num ber of nu clei of X and Y atany time t. Num ber of Nu clei N y will bemaximum when(a)

N

N N

y

x y

y

x y−=

−λ

λ λ

(b) N

N Nx

x y

x

x y−=

−λ

λ λ

(c) λ λy y k xN N=(d) λ λy x x yN N=

30. An elec tron in hy dro gen atom af terab sorb ing an en ergy pho ton jumps fromen ergy state n1 to n2 . Then it re turns toground state af ter emit ting six dif fer entwavelengths in emis sion spec trum. Theen ergy of emit ted pho tons is ei ther equal toless than the ab sorbed pho tons. Then n1 and n2 are(a) n n2 14 3= =,

(b) n n2 15 3= =,

(c) n n2 14 2= =,

(d) n n2 14 1= =,

31. A ball is dropped ver ti cally from a height dabove the ground and bounces up ver ti callyto a height d/2. Ne glect ing sub se quentmo tion and air re sis tance its ve loc ity vvar ies with height h above the ground as


t

D1

D2

Output

A B C Dt

V

O

R R

RR

6R

R 4R4 Ω

E

dh

(b)

v

dh(c)

v

h(d)

v

d

dh(a)

v

32. Two par ti cles 1 and 2 are al lowed to de scendon two fric tionless chords OP and OQ. Thera tio of the speeds of the par ti cles 1 and 2re spec tively when they reach on thecir cum fer ence is

(a) 1

4(b)

1

2

(c) 1 (d) 1

2 2

33. A body of mass m, hav ing mo men tum p ismov ing on a rough hor i zon tal sur face. If it isstopped in a dis tance x, the co ef fi cient offric tion be tween the body and the surface is

(a) µ = p

mgx2(b) µ = p

mgx

2

2

(c) µ = p

gm x

2

22(d) µ =

2p

g m x2 2 2

34. When a ceiling fan is switched off, itsangular velocity reduces to half its initialvalue after it completes 36 rotations. Thenumber of rotations it will make furtherbefore coming to rest is (Assuming angularretardation to be uniform)(a) 10 (b) 20

(c) 18 (d) 12

35. A uni form metal rod is used as a barpendulum. If the room tem per a ture rises by 10° C and the co ef fi cient of lin ear ex pan sionof the metal of the rod is 2 10 6× − per ° C, thepe riod of the pen du lum will have per cent age increases of (a) − × −2 10 3 (b) − × −1 10 3

(c) 2 10 3× − (d) 1 10 3× −

36. Two iden ti cal springs of con stant k arecon nected in se ries and par al lel as shown infigure. A mass m is sus pended from them.The ra tio of their fre quen cies of ver ti calos cil la tions will be

(a) 2 1: (b) 1 1:

(c) 1 2: (d) 4 1:

37. An as tro naut is ap proach ing the moon. Hesends a ra dio wave of fre quency 5 109× Hzto wards the moon. The fre quency of thera dio echo re ceived by him has a fre quency 9 104× Hz more then that of the realfre quency. The rel a tive ve loc ity of the rocket with re spect to the moon is(a) 5.40 km/s (b) 4.05 km/s

(c) 2.70 km/s (d) 1.35 km/s

38. Ultraviolet light of wave length 300 nm andin ten sity 1.0 W /m2 falls on the sur face of apho to sen si tive ma te rial. If one per cent ofthe in ci dent pho tons pro duce photoelec trons, then the num ber of photoelec trons emit ted from an area of 1.0 cm2 ofthe surface is nearly(a) 19.61 × −1014 1s (b) 4.12 × − −10 13 1s

(c) 1.51 × −1012 1s (d) 2.13 × −1011 1s

39. An X-ray tube op er ated at 50 kV, pro ducesheat at the tar get at the rate of 796 W. If0.5% en ergy of in ci dent elec trons strik ingthe tar get per sec ond will be(a) 1019 (b) 1018

(c) 1017 (d) 1016


m

(a)

m

(b)

K K

60°

Q

O

P

q

40. The masses of two iso topes of chlo rine are34.980 and 36.978. If the ra dius of thecir cu lar path in Bain bridge massspec tro graph cor re spond ing to lighter is5 cm, then the dis tance be tween the spots on pho to graphic plate marked by two iso topeswill be(a) 5.7 cm (b) 0.57 cm

(c) 0.57 mm (d) 0.57 m

41. In the ura nium ra dio ac tive se ries, the ini tial nu cleus is 92 U238 and that the fi nal nu cleusis 82 206Pb . When ura nium nu cleus de cays tolead the num ber of α-particles andβ-par ti cles emitted are(a) 8 6α, β (b) 6 7α, β(c) 6 8α, β (d) 4 3α, β

42. A gas of monoatomic hy dro gen is bom barded with a stream of elec trons that have beenac cel er ated from rest through a po ten tialdif fer ence of 12.75 V. In the emis sionspec trum one cannot ob serve any line of(a) Lyman se ries (b) Balmer se ries

(c) Paschen se ries (d) Pfund se ries

43. The max i mum in ten sity in Young’s dou bleslit ex per i ment is I0 . Dis tance be tween theslits is d = 5 λ, where λ is the wave length ofmono chro matic light used in theex per i ment. What will be the in ten sity oflight infront of one of the slits on a screen ata dis tance D d= 10 ?(a) I0 (b) I0 4/

(c) 3

40I (d) I0 2/

44. A lamp is hang ing at a height of 40 m fromthe cen tre of a ta ble. If its height is in creased by 10 cm, the illuminance on the ta ble willde crease by(a) 10 % (b) 20 % (c) 27 % (d) 36 %

45. Ac cord ing to Maxwell’s equa tion, theve loc ity of light in any me dium is ex pressedas(a)

1

0 0µ ε(b)

1

µ ε(c)

1

µ ε/(d)

µε0

46. Two mag nets of equal mag netic mo ments Meach are placed as shown in fig ure. There sul tant mag netic mo ment is

(a) M (b) 3 M

(c) 2 M (d) M /2

47. The hys ter esis cy cle for the ma te rial ofper ma nent mag net is(a) short and wide (b) tall and nar row

(c) tall and wide (d) short and nar row

48. In the cir cuit shown in fig ure, the value ofre sis tance x, when the po ten tial dif fer encebe tween the points B and D is zero, will be

(a) 9 Ω (b) 8 Ω (c) 6 Ω (d) 4 Ω

49. A mercury drop of ra dius 1 cm is bro ken into 106 drop lets of equal size. The work done is (ρ = × −35 10 2 N / m)(a) 4.35 J× −10 2

(b) 4.35 J× −10 3

(c) 4.35 J× −10 6

(d) 4.35 J× −10 8


60°N

S

S N

A C

B

D

15 Ω

6 Ω8 Ω

3 Ω

4 Ω

4 Ω

4 Ω6 Ω

6 Ω15 Ω

x

50. A space man in train ing is ro tated in a seatat the end of a hor i zon tal rotating arm oflength 5m. If he can withstand ac cel er a tions upto 9 g, then what is the max i mum num berof rev o lu tions per sec ond permissible?

Take g = 10 2m /s .(a) 13.5 rps

(b) 1.35 rps

(c) 0.675 rps

(d) 6.75 rps

Chemistry1. The hy bridi sa tion state of C-atom in

butendioic acid is(a) sp2 (b) sp3

(c) Both (a) and (b) (d) sp

2. The ox i da tion num ber of C-atom in CH Cl2 2and CCl4 are re spec tively(a) − 2 and − 4 (b) 0 and − 4

(c) 0 and 4 (d) 2 and 4

3. Phenolphthalein of pH range ( )8 10− is used in which of the fol low ing type of ti tra tion asa suitable indicator?(a) NH OH4 and HCl (b) NH OH4 and HCOOH

(c) NH OH4 and C H O2 4 2 (d) NaOH and C H O2 4 2

4. Which of the fol low ing spe cies has high estbond en ergy ?(a) O2

2 − (b) O2+ (c) O2

− (d) O2

5. Which of the fol low ing is a weak acid?(a) C H6 6 (b) CH CH3 ≡≡C

(c) CH CH2 == 2 (d) CH C C CH3 3 ≡≡

6. A mix ture con tain ing 60% cetane and 40%iso-oc tane will have(a) cetane num ber 60 (b) cetane num ber 40

(c) octane num ber 40 (d) None of these

7. A B C D→ → →⋅Alc KOH 2Cl

CCl

Ca (OH)2

4

2

Here the compound C will be(a) Lewisite

(b) Westron

(c) Acet y lene tet ra chlo ride

(d) Both (b) and (c)

8. Which of the fol low ing is least hy dro lysed ?(a) BeCl2 (b) MgCl2 (c) CaCl2 (d) BaCl2

9. The vol ume con cen tra tion of a 3% so lu tion of hy dro gen per ox ide would be(a) 9880 (b) 9.88 (c) 22.4 (d) 3

10. The en ergy pro duced re lated to mass de fectof 0.02 amu is(a) 28.2 MeV (b) 931.5 MeV

(c) 18.62 MeV (d) None of these

11. A so lu tion con tains Cl I− −, and SO42 − ions in

it. Which of the fol low ing ion is ca pa ble topre cip i tate all of above when added in thissolution?(a) Pb2 + (b) Ba 2 + (c) Hg2 + (d) Cu2 +

12. The min i mum num ber of car bon at oms inke tones which will show chain isomerism(a) seven (b) four (c) six (d) five

13. In Vic tor Mayer’s method 0.2g of an or ganicsub stance dis placed 56 mL of air at STP, themo lec u lar weight of the compound is(a) 56 (b) 112

(c) 80 (d) 28

14. 146C is a beta-ac tive nu cleus. A sam ple of

144CH gas kept in a closed ves sel shows

in crease in pres sure with time. This is due to the(a) for ma tion of 14

3NH and H2

(b) formation of 143BH and H2

(c) for ma tion of 14C H2 4 and H2

(d) for ma tion of 143

142CH NH, and H2

15. The bond an gle around the cen tral atom ishigh est in(a) SO2 (b) BBr3(c) CS2 (d) SF4


16. For a d elec tron, the or bital an gu larmo men tum is(a) 6

2

h

π(b) 2

2

h

π

(c) 32

h

π(d)

h

2 π

17. A gas eous mix ture of O2 and X con tain ing20 mole% of X, dif fuses through a small holein 234s while pure O2 take 224s to dif fusethrough the same hole. The mo lec u lar massof mix ture is(a) 34.9 (b) 46.6

(c) 32 (d) 44

18. The electronegativity of C,H,O,N and S are2.5, 2.1, 3.5, 3.0 and 2.5 re spec tively. Whichof fol low ing bond is most po lar? (a) O H (b) S H(c) N H (d) C H

19. ZnS can be ex ist ing in the ……… struc tureother than zing blende stucture.(a) bcc

(b) wurtzite

(c) sim ple cu bic

(d) rock salt

20. The re agent (s) for the fol low ing con ver sion,Br

Br H—C C—H → ≡≡? is are

(a) al co holic KOH

(b) alcoholic KOH followed by NaNH2

(c) aque ous KOH followed by NaNH2

(d) Zn CH OH3/

21. Con sider the fol low ing re duc tion and ad visethe best re agent.

(a) HI/Red (b) LiAlH4

(c) NaBH4 (d) Zn - Hg/HCl

22. Which of the re agents is not used in theprep a ra tion of anisole via Williamson’ssynthesis?(a) Na

(b) CH Cl3

(c)

(d)

23. Iden tify A and B based on the fol low ingre ac tion scheme.

CH CH COOH

CH CH COOH

2 2

2 2

KMnO / OH4

←

−

( )(C H O)

( )C H O

Oxime 6 10

1. OH2.

12 18

NH —OH2A

B

−

→

∆


O

OHC

O

O

HOH C2

O

Cl

OH

O

(a)

, O

O

(b)

,

O

CHO

(c)

,

CH

CH

CHO

(d)

,

CHO

24. Which of the fol low ing carboxylic acidun der goes decarboxylation easily?(a) C H CO CH COOH6 5 2

(b) C H CO COOH6 5

(c) C H CH

OH

COOH6 5

(d) C H CH

NH

COOH6 5

2

25. CH NH + CHCl + KOH3 2 3 → Ni tro gencon tain ing com pound + +KCl H O2 . Ni tro gen con tain ing com pound is(a) CH C N3 ≡≡(b) CH NH CH33

(c) CH N C+3 ≡≡

−

(d) CH N C3 ≡≡+ −

26. In the fol low ing re ac tion

The structure of the major product X is

27. Which of the fol low ing monosaccharidesyield an op ti cally in ac tive alditol on NaBH4reduction?

28. The mono mer mel a mine has a chem i calname(a) 2, 4, 6– triamino − 1 3 5, , – triazine

(b) 1, 3, 5,– triamino − 2 4 6, , – triazine

(c) 2, 4-diamino − 1 3 5, , – triazine

(d) 2- amino − 1 3 5, , – triazine

29. For the re ac tion, N O NO2 4 22( ) ( );g g3 the re la tion con nect ing the de gree ofdis so ci a tion ( )α of N O2 4 ( )g with theequi lib rium con stant K p is

(a) α =+

K

pK

p

p

p4

(b) α =+K

K

p

p4

(c) α =+

K

pK

K

p

p4

1 2

p

/

(d) α =+

K

K

p

p4

1 2/

30. If the sol u bil ity of cal cium phos phate (mol. wt = M) in wa ter at 25°C is w g mL/100 , itssol u bil ity prod uct at 25°C is

(a) 1095

w

M

(b) 1075

w

M

(c) 1055

w

M

(d) 1035

w

M


N

O

H

NO2

(a)

N

O

H

(b)

O N2

N

O

H

NO2

(c)

N

O

H

(d)

O N2

CHO

CH OH2

H OH

HO H

HO H

H OH

(a)

CHO

CH OH2

HO H

HO H

HO H

H OH

(b)

CHO

CH OH2

HO H

H OH

HO H

H OH

(c)

CHO

CH OH2

HO H

HO H

H OH

H OH

(d)

N

O

H

Conc. HNO3

Conc. H SO2 4

X

31. Mass of one atom of an el e ment is 6.6 g4 10 23× − . This is equal to(a) 6.6 u4 10 23× − (b) 40.0 u

(c) 1

40u (d) 6.6 u4

32. Sul phide ores of met als are usu allycon cen trated by froth floa ta tion pro cess.Which one of the fol low ing sulphide oresof fers an ex cep tion and is con cen trated bychem i cal leaching?(a) Ar gen tite (b) Ga lena

(c) Cop per py rite (d) Sphalerite

33. Sol diers of Napolean army which at Alpsdur ing freezing win ter suf fered a se ri ousprob lem as re gards to the tin but tons of their uni forms. White me tal lic tin but tons gotcov ered by grey pow der. Thistrans for ma tion is related to(a) an in ter ac tion with ni tro gen of the air at very low

tem per a tures

(b) a change in the partial pressure of oxygen in theair

(c) a change in the crys tal line struc ture of tin

(d) an in ter ac tion with wa ter vapour con tained in thehu mid air

34. Which of the fol low ing is a mixed ox ide?(a) Fe O2 3 (b) PbO2

(c) BaO2 (d) Pb O3 4

35. If the quan tum num bers for the 5th elec tronin car bon atom are 2 1 1 1

2, , , ,+ then for the

6th elec tron, these val ues would be(a) 2 1 0

1

2, , , − (b) 2 0 1

1

2, , , +

(c) 2 1 11

2, , − (d) 2 1 1

1

2, , ,− +

36. For the ho mog e nous re ac tion. 4NH O 4NO H O3 2+ +5 623

the equilibrium constant K c has the units(a) conc,+ 10

(b) conc,+ 1

(c) conc,− 1

(d) It is dimensionless

37. Which of the fol low ing be hav iour is true foran ideal bi nary liq uid solution?(a) Plot of 1/ totalp vs 1/YA (mole frac tion of A in vapour

phase) is lin ear

(b) Plot of 1/ totalp vs 1/YB is linear

(c) Plot of 1/ totalp vs 1/ Y YA B is lin ear

(d) Plot of 1/ totallp vs YA is lin ear

38. A 0.004 M so lu tion of Na SO2 4 is iso tonic with a 0.010 M solution of glu cose at the 25°C.The ap par ent de gree of dis so ci a tion of Na SO2 4 is(a) 25% (b) 50% (c) 75% (d) 85%

39. Cow milk an ex am ple of nat u ral emul sion isstabilised by(a) fat (b) wa ter

(c) ca sein (d) Mg2 + ions

40. The ini tial con cen tra tion of sugar so lu tion is 0.12 M. On do ing fer men ta tion thecon cen tra tion of sugar de creases to 0.06 Min 10 h and to 0.045 M to 15 h. The or der ofthe reaction is(a) 0.5 (b) 1.0 (c) 1.5 (d) 2.0

41. An ath lete is given 100 g of glu cose (C H O6 12 6) of en ergy equiv a lent to 1560 kJ.He uti lises 50% of this gained en ergy in anevent. In or der to avoid stor age of en ergy inthe body what is the weight of wa ter hewould need to per spire ? The enthalpy ofevap o ra tion of wa ter is 44 kJ/mol.(a) 319 g (b) 638 g (c) 14040 g (d) 35.45 g

42. Which of the fol low ing re la tion ship isin cor rect ?(a)

∆ ∆∆H E

n xT

− = con stant (b) ∆ ∆G T S= − Total

(c) q U W= +∆ (d) K e RTG= − °∆ /

43. A mix ture of gases hav ing dif fer entmo lec u lar weights is sep a rated by whichmethod ?(a) Atmolysis

(b) Metathesis

(c) Ostwald and Walker method

(d) Re verse os mo sis


44. Boric acid is poly meric due to(a) its acidic na ture

(b) the presence of hydrogen bonds

(c) its monobasic na ture

(d) its ge om e try

45. The metal ion which does not form colouredcom pound is(a) chro mium (b) iron

(c) zinc (d) man ga nese

46. The type of isomerism pres ent inpentaamine ni tro co balt (III) chlo ride is(a) op ti cal (b) link age

(c) ion is ation (d) poly meri sa tion

47. Which of the fol low ing is known as in vertsoap?(a) Pentaerythritol monostearate

(b) Sodium stearyl sulphate

(c) Trimethylstearyl am mo nium bro mide

(d) Ethoxylated nonyphenol

48. The cell con stant is the(a) re sis tance × con duc tance

(b) resistance × specific conductance

(c) con duc tance × spe cific re sis tance

(d) re sis tance × spe cific re sis tance

49. It has been found ex per i men tally that ifstan dard re duc tion po ten tial of ox i dant– stan dard re duc tion po ten tial of reductant is more than 1.7V then their combination

lead to ex plo sion (though it may bepre vented by kinetic factors).Now go through the following data andanswer the questions.

E ° + =Ag Ag 0.80 V/

E ° − − =ClO ClO 1.23 V4 3/ E ° + + =Fe Fe 0.77 V3 2/ E ° − + =MnO Mn 1.54 V4 3/ E ° − = −N / N2 3.09 V3

E °+ = −Na Na 2.17 V/

E ° = −O H O2 2 V2 1 03/ .Which of the following ioniccombinations may lead to the formationof explosive substance?

(a) So dium ion and azide ion

(b) Silver ion and perchlorate ion

(c) Sil ver ion and azide ion

(d) All of the above

50. For the re ac tion 2A B+ → prod uct ;dou bling the ini tial con cen tra tions of boththe re ac tants in crease the rate by a fac tor of8 and dou bling the concentration of B abovedou bles the rate. The rate law for thereaction is(a) r k A B= [ ] [ ]2

(b) r k A B= [ ] [ ]

(c) r k A B= [ ] [ ]2 2

(d) r k A B= [ ] [ ]2

Mathematics1. Con cen tric cir cles of ra dii 1 2 3, , , ,… 100 cm

are drawn. The in te rior of the small est cir cle is coloured red and the an gu lar re gions arecolo ured al ter nately green and red, so thatno two ad ja cent re gions are of the samecol our. The to tal area of the green re gion insq. cm is equal to(a) 1000 π (b) 5050π(c) 4950π (d) 5151π

2. The value of a for which the qua draticequa tion 3 2 1 3 2 02 2 2x a x a a+ + + − + =( ) ( )po ssesses roots of op po site signs lies in(a) ( , )− ∞ 1

(b) ( )− ∞ 0

(c) ( , )1 2

(d) 3

22,


3. If 2 3 01 2 3z z z− − = , then z z z1 2 3, and arerep re sented by(a) three ver ti ces of a tri an gle

(b) three collinear points

(c) three ver ti ces of a rhom bus

(d) None of the above

4. The term in de pendent of x, in the ex pan sionof 1 1 2

4+ + +

x

x x is

(a) 35 (b) 30

(c) 32 (d) 31

5. The num ber of six-digit num bers whichhave sum of their digits as an odd in te ger, is(a) 45000 (b) 450000

(c) 97000 (d) 970000

6. Con sider the ∆ AOB in the x y, -plane where A B O≡ ≡ ≡( , , ) , ( , , ) , ( , , )1 0 0 0 2 0 0 0 0 . Thenew po si tion of O, when tri an gle is ro tatedabout side AB by 90° can be(a)

4

5

3

5

2

5, ,

(b) −

3

5

2

5

2

5, ,

(c) 4

5

2

5

2

5, ,

(d) 4

5

2

5

1

5, ,

7. Num ber of planes which are at a givenper pen dic u lar dis tance from a given pointand pass ing through a given point is(a) 0 (b) 2 (c) 4 (d) in fi nite

8. If A and B are two in de pend ent events, thenwhich of the fol low ing is not equal to any ofthe re main ing?(a) P A B P A B( ) ( )′ ∩ ′ − ∩(b) P A P B( ) ( )′ + ′ − 1

(c) P B P A( ) ( )− ′(d) P B P A( ) ( )′ −

9. In u nn = 2 cos θ and u u un n1 1− − is equal to(a) un − 2 (b) un + 1

(c) 0 (d) None of these

10. If 12

1< <x , then

cos cos− −+1 1x x x+ −

12

2 is equal to

(a) 24

1cos x− − π(b) 2 1cos− x

(c) π4

(d) 0

11. The num ber of values of θ sat is fy ing 4 3 5cos sinθ θ+ = as well as 3 4 5cos sinθ θ+ = is(a) 1 (b) 2


12. A kite is fly ing with the string in clined at75° to the ho ri zon. If the length of the stringis 25 m, then height of the kite is(a)

25

23 1 2

−( ) (b) 25

43 1 2

+( )

(c) 25

23 1 2

+( ) (d) 25

26 2

+( )

13. The ends of a quad rant of a cir cle have theco or di nates ( , )1 3 and ( , )3 1 . Then, the cen treof such a cir cle is(a) ( , )2 2 (b) ( , )1 1

(c) ( , )4 4 (d) ( , )2 6

14. If the latus rectum of the pa rab ola 2 2 02x ky− + = be 2, then the ver tex is(a) 0

3

4,

(b) 03

2,

(c) 3

40,

(d) ( , )0 0

15. If f : ( , ) ( , )3 4 0 1→ is de fined by f x x x( ) [ ],= −where [ ]x de notes the greatest integerfunction, then ′f x( ) is(a)

1

x x− [ ](b) [ ]x x−

(c) x − 3 (d) x + 3

16. If f x x xx x

( ) cos=−+

−−

−1

1

1 , then ′ −f ( )2 is

(a) 2

5(b)

− 2

5

(c) − 1

5(d) None of these


17. Let f x( ) be an even func tion in R. If f x( ) ismonotonic in creas ing in [ , ]2 6 , then(a) f f( ) ( )3 5> − (b) f f( ) ( )− <2 2

(c) f f( ) ( )− >2 2 (d) f f( ) ( )− <3 5

18. If a n

n x a xe dx−

−∫ =( ) λ, then the value of

a n

n x a xxe dx a n−

−∫ ≠( ) , 2 , is

(a) aλ2

(b) aλ

(c) 2aλ (d) None of these

19. If Ix

xx

dx= ⋅ −

∫1

1 1/

sinπ

π , then I is

equal to(a) 0 (b) π (c) π

π− 1

(d) ππ

+ 1

20. The num ber of sides of the quad ri lat eralwhose joint equa tion is x y x y2 2 2 21+ = + ,and which are touched by the cir cles x y x2 2 2+ = is(a) 4 (b) 3

(c) 2 (d) 1

21. If f x f xf x

( ) ( )( )

+ = + +

2 12

1 4 and f x( ) ,> 0

for all x R∈ , then lim ( )x

f x→ ∞

is

(a) 1 (b) 2 (c) − 2 (d) 0

22. Let f x( ) be a con tin u ous func tion whoserange is [ , . ]2 65 . If h x x f x( ) cos ( )

=+

λ, λ ∈ N

be con tin u ous, where [.] de notes the great est in te ger func tion, then the least value of λ is(a) 6 (b) 7


23.3 22 3 2

++

⋅∫cos

( cos )xx

dx is equal to

(a) sin

cos

x

xc

2 3+

+

(b) 2

2 3

cos

sin

x

xc

+

+

(c) Both (a) and (b)


24. Differential equa tion of the fam ily of cir clestouch ing the line y = 2 at ( , )0 2 is(a) x y

dy

dxy2 22 2 0+ − + − =( ) ( )

(b) x y xdx

dyy2 2 2 2 0+ − − −

=( )

(c) x ydx

dyy y2 22 2 2 0+ − + + −

− =( ) ( )


25. If a b c, , and are non-zero real num bers and az bz c i2 0+ + + = has purely imag i naryroots, then a is equal to (a) bc (b) b c2 (c) − b c2 (d)

1

2

2b c

26. If a b, , and c are three mu tu ally or thogo nalunit vectors, then the tri ple prod uct [ ]a b c a b b c+ + + + is equal to(a) 0 (b) 1 or − 1 (c) 1 (d) 3

27. y x2 4= is a curve and P Q, , and R are threepoints on it, where P Q= =

( , ), ,1 2 1

41 and

the tan gent to the curve at R is par al lel tothe chord PQ of the curve, then co or di natesof R are

(a) 5

8

5

2,

(b)

9

16

3

2,

(c) 5

8

5

2, −

(d)

9

16

3

2,

−

28. A batsman can score 0, 1, 2, 3, 4 or 6 runsfrom a ball. The num ber of dif fer entse quences in which he can score ex actly + 30runs in an over of six balls, is(a) 4 (b) 72 (c) 56 (d) 71

29. If ( ( )) ( ) ( ( )) , ,x f y f x f y x y R= ∀ ∈2 and f f( ) ( )3 5 4+ = , then ′f ( )3 is equal to (a) 1 (b) 3 (c) − 3 (d) 1

30. The num ber of so lu tions for the equa tion

2 1 32

1 2 1 2sin ( ) cos ( )− −− + + − =x x x x π

is(a) 1 (b) 2 (c) 3 (d) iIn fi nite


31. The num ber of so lu tions of the equa tion

−∫ = < <2

0 02

xx dx x|cos | , π, is

(a) 0 (b) 1 (c) 2 (d) 4

32. A per son stand ing on the bank of a riverob serves that the an gle sub tended by a treeon the op po site bank is 60°, when he retires40m from the bank he finds the an gle to 30°.The breadth of river is(a) 40 m (b) 60 m

(c) 20 m (d) 30 m

33. Two circles x y kx2 2 2 0+ − = and x y x y2 2 4 8 16 0+ − − + = touch each otherex ter nally. Then, k is(a) 4 (b) 1

(c) 2 (d) − 4

34. If the line ax by+ = 2 is a nor mal to the cir cle x y x y2 2 4 4 0+ − − = and a tan gent to thecir cle x y2 2 1+ = , then(a) a b= =1

2

1

2,

(b) a b= + = −1 7

2

1 7

2,

(c) a b= =1

4

3

4,

(d) a b= =1 3,

35. The graph of the curve x y xy x y2 2 2 8 8 32 0+ − − − + = falls whollyin the(a) first quad rant (b) sec ond quad rant

(c) third quad rant (d) None of these

36. The num ber of so lu tions of [cos ] |sin |x x+ = 1 in π π≤ <x 3 is(a) 3 (b) 4

(c) 2 (d) 1

37. The slope of the tan gent to the curve

tan y =+ − −+ + −

1 11 1

x xx x

at x = 12 is

(a) 1

3(b) 3

(c) 1 (d) 1

2

38. If the real valued func tion f x x a x( ) ( )= + − +3 23 1 1 be invertible, thenset of pos si ble real val ues of a is(a) ( , ) ( , )− ∞ − ∪ ∞1 1 (b) ( , )− 1 1

(c) [ , ]− 1 1 (d) ( , ] ,− ∞ − ∪ + ∞1 1

39. The value of 0

42

π / sec(sec tan )∫ +

⋅x

x xdx is

(a) 1 2+ (b) − +( )1 2

(c) − 2 (d) None of these

40. The com bined equa tion of straight lines that can be ob tained by re flect ing the lines y x= −[ ]2 in the y-axis is(a) y x x2 2 4 4 0+ + + =

(b) y x x2 2 4 4 0+ − + =

(c) y x x2 2 4 4 0− + − =

(d) y x x2 2 4 4 0− − − =

41. lim ( )x

xx→

+

0

21 , where . de notes the

frac tional part of x, is equal to(a) e2 7− (b) e2 8−

(c) e2 6− (d) None of these

42. f x e xx

x( ) ,

,/

= >≤

− 1 20

0 0, then f x( ) is

(a) dif fer en tia ble at x = 0

(b) continuous but not differentiable at x = 0

(c) dis con tin u ous at x = 0


43.1

12 4 3 4x xdx

( ) /+∫ is equal to

(a) 114

1 4

+

+x

C/

(b) ( ) /x C4 1 41+ +

(c) 114

1 4

−

+`/

xC (d) − +

+114

1 4

xC

/

44. The so lu tion of the differential equation ( ) ( )1 1 02 2 2 2+ + − =x y y dx x y xdy is(a) xy

x

yC= +log (b) xy

y

xC 2log = +

(c) x yy

xC2 2 2= +log (d) None of these


45. Equa tion of chord of con tact of pair oftan gents, drawn to el lipse 4 9 362 2x y+ =from the point (m n, ), where m n m n. = + , m n, be ing non-zero pos i tive in te gers, is(a) 2 9 18x y+ = (b) 2 2 1x y+ =(c) 4 9 18x y+ = (d) None of these

46. The equa tion to the hy per bola of giventrans verse axis whose ver tex bi sects thedis tance be tween the cen tre and fo cus, isgiven by(a) 3 32 2 2x y a− = (b) x y a2 2 23− =

(c) x y a2 2 23− = (d) None of these

47. The plane ax by cz d− − = will con tain theline x a

ay d

bz e

c−

=+

=−3 , pro vided

(a) b d= [ , ]0 3 (b) a d= [ ]2

(c) c d= [ ]3 (d) b d= −[ ]3

48. If z is a com plex num ber ly ing in the firstquad rant such that Re ( ) Im ( )z z+ = 3, thenthe max i mum val ues of [Re ( )] Im ( )z z2 is(a) 1 (b) 2 (c) 3 (d) 4

49. If A xk x

=−

−tan 1 3

2 and

B x kk

=−

−tan 1 23

. Then, A B− is equal to

(a) π2

(b) π3

(c) π6

(d) None of these

50. If in a ∆ABC B, ∠ = 23π, then the

cos cosA C+ lies in(a) [ , ]− 3 3 (b) ( , ]− 3 3

(c) 3

23,

(d) 3

23,

Answers

Physics


1. (b) 2. (b) 3. (a) 4. (c) 5. (c) 6. (b) 7. (c) 8. (a) 9. (c) 10. (d)

11. (b) 12. (c) 13. (c) 14. (a) 15. (d) 16. (c) 17. (b) 18. (d) 19. (a) 20. (d)

21. (c) 22. (c) 23. (a) 24. (a) 25. (b) 26. (a) 27. (b) 28. (b) 29. (d) 30. (c)

31. (c) 32. (b) 33. (c) 34. (d) 35. (d) 36. (c) 37. (c) 38. (c) 39. (c) 40. (b)

41. (a) 42. (d) 43. (d) 44. (d) 45. (b) 46. (a) 47. (c) 48. (b) 49. (a) 50. (c)

Chemistry1. (a) 2. (c) 3. (d) 4. (b) 5. (b) 6. (d) 7. (d) 8. (d) 9. (b) 10. (c)

11. (a) 12. (d) 13. (c) 14. (a) 15. (c) 16. (a) 17. (a) 18. (c) 19. (b) 20. (b)

21. (c) 22. (c) 23. (b) 24. (a) 25. (d) 26. (b) 27. (a) 28. (a) 29. (c) 30. (b)

31. (b) 32. (a) 33. (c) 34. (d) 35. (d) 36. (b) 37. (d) 38. (c) 39. (c) 40. (b)

41. (a) 42. (c) 43. (a) 44. (b) 45. (c) 46. (b) 47. (c) 48. (b) 49. (c) 50. (d)

Mathematics1. (b) 2. (c) 3. (b) 4. (d) 5. (b) 6. (c) 7. (d) 8. (c) 9. (b) 10. (c)

11. (c) 12. (b) 13. (b) 14. (a) 15. (d) 16. (b) 17. (d) 18. (b) 19. (a) 20. (c)

21. (b) 22. (c) 23. (a) 24. (b) 25. (b) 26. (b) 27. (b) 28. (d) 29. (b) 30. (b)

31. (a) 32. (c) 33. (c) 34. (b) 35. (a) 36. (a) 37. (a) 38. (d) 39. (d) 40. (d)

41. (a) 42. (a) 43. (d) 44. (c) 45. (c) 46. (a) 47. (d) 48. (d) 49. (c) 50. (c)

Hints & Solutions

Physics1. We know that, cur rent gain for com mon

emit ter am pli fier is given by

β αα

=−

1

Here α = β0.95 ⇒ = 19

Also we know that β, is given by

β = I

IC

b

Since β = 19 and IC = 1mA

⇒ Ib = 0.05 mA

2. Num ber of Si at oms per cm3

= × × −6 10 1028 6

= ×6 1022

Number of indium atoms per cm3

= ××

=6 10

6 1010

22

715

3. En ergy is given by

Ehc=λ

When an electron is accelerated throughV volts, then we have

eVhc=λ

⇒ Vhc

e=

λSubstituting h = × −6.6 J -s10 34

c e= × = ×− −3 10 16 108 1 19ms C, .

λ = 0.4125 Å = 4125 10 m,× −10

We get

V = × =30 10 303V kV

4. Ion is ation en ergy of nth state = − ⋅En For the

first ex cited state n = 2 and en ergy in that state

is 13.6

2eV 3.4 eV

2=

5. We have Tl

g=

2π

percentage error is T = 1

2(% error in l)

+ 1

22( %)error in g

= + =1

22

1

24 3( %) ( %) %

6. Let to tal dis tance = S

Let the time taken to cover first one thirddistance = t1,

Then ts s

1

3

4 12= =/

Now let t2 be the time for the rest two journeys. Then

2

32 6 82 2 2

st t t= + =

∴ ts s

2

2

24 12= =

∴ Average velocity = Total displacement

Total time

=+

=+

= ×+

=s

t t

s

s s1 22

12 6

12 6

12 64 m/s

7. Change the mo men tum in one sec ond is theforce.

Hence, F ma= = ×2 3

= −6 1kg ms

8. When axle ro tates in a sleeve the fric tionin volved in the pro cess is slid ing.

9.1

2

1

22 2mv MV=

Hence, m v

m V

m

M

2 2

2 2=

That is p

p

m

MA

B

=

1 / 2

Since M m>Therefore p pA B< .

10. From law of conservation of momentum

Vmv

M= ;

Ki netic en ergy of gun

= ×

=

1

2

1

2

22m

mv

Mmv

m

M

Since m M< , hence kinetic energy of the gun

is less than 1

22

mv .

11. Moment of inertia of solid sphere = 2

52mR

Moment of inertia of solid cylinder = MR2

2

Moment of inertia is greater for solid cylinder.

12. We have

Gravitational potential on the earth

VGM

R= −

Gravitational potential at the centre of theearth

VGM

RR r

RGM RC = − − = − −

23

3

23 0

32 2

32( ) ( )

= − =3

2

3

2

GM

RV

13. Since CV nR= ×2

and

CnR

Rn

Rp = + = +

2

2

2

Therefore

γ =C

C

n

n

p

V

= + 2

⇒ n =−

2

1γ

14. VL or e = − = −E V ER iR

i e. .; e i− graph is a straight line with negativeslope and positive intercept.

15. Let R be the ra dius of cur va ture of eachsur face. Then

1 = − +

f R R

( )1.5 11 1

∴ R f=For the water lens

1 4

31

1 1 1

3

2

′= −

− −

= −

f R R g

or1 2

3′= −

f f

Now using 1 1 1 1

1 2 3F f f f= + + we have

1 1 1 1

F f f f= + +

′

= − =2 2

3

4

3f d f

∴ Ff= 3

4

16. Rate of flow of heat through each layer of slabis same

K AT

lK A

T

l1

12

2∆ ∆=

= +K T T A

l

( )∆ ∆1 2

2

or K T k T1 1 2 2∆ ∆=

= +k T T( )∆ ∆1 2

2

= C (say)

∴ ∆ ∆TC

KT

C

K1

22

2

= =,

∆ ∆T TC

k1 2

2+ =

orC

K

C

K

C

K1 2

2+ =

or KK K

K K=

+2 1 2

1 2

17. We know Y K= 3 σ( )1 2−

or σ = 1

21

2−

Y

K

Also Y n= +2 1( σ)

or σ =ηY

21−

∴ 1

21

3 21−

= −Y

K

Y

η

or 13

2− = −Y

K

Y

η

or 33

= +Y Y

Kη

or 3 1 1

3Y K= +

η


∆T1 ∆T2

18. KE at mean po si tion

= = × −1

28 102 2 3m aω

or ω = × ×

−2 8 10 3

2

1 2

ma

/

= × ××

=

−2 8

0.1 (0.1)210

43

1

2

Equation of SHM is

y a t t= + +

sin ( sinω θ) = π0.1 4

4

19. From mg mgR

R h′ =

+

2

2( )

∴ 30 606400

6400

2

2= ×

+( )

( )h

or1

2

6400

6400=

+ h

or 6400 6400 2+ =h

or h = −6400 2 6400 ≈ 2624 km

20. Equa tion of mo tion of the two weights can bewrit ten as

m g T m a1 1 1− = or T m g m a1 1 1= − ..(i)

andT m g m a2 2 2− =or T m a m g2 2 2= + ...(ii)

Subtract Eq. (ii) from Eq. (i), we get

T T m m g m m a1 2 1 2 1 2− = − − +( ) ( ) ... (iii)

As Torque, τ α= I

∴ ( )T T R MRa

R1 2

21

2− =

⋅

or T TMa

1 22

− =

Put it in Eq. (iii)Ma

m m g m m a2

1 2 1 2= − − +( ) ( )

or am m g

m m m= −

+ +( )

( / )1 2

1 2 2

21. When a par ti cle un der goes nor mal col li sionwith a floor or a wall, with co ef fi cient ofres ti tu tion e, the speed af ter col li sion is etimes the speed be fore col li sion. Thereforechange in mo men tum af ter Ist im pact is ep p p e− − = +( ) ( )1 . Af ter the sec ond im pact, change in mo men tum would be e ep ep ep e( ) ( ) ( )− − = +1 and so on.There fore, to tal change in momentum of ball

= momentum imparted to floor

= + + + +… = +−

p e e ep e

e( ) [ ]

( )

( )1 1

1

12

22.1

2

1

22 2mv k x=

vk

mx= ⋅ = ×1

24

= 2 2 m/s

23. Let m be the mass of un loaded car then 1

22mu FS= ... (i)

where, F is retarding force. When 40 % weightis added, new mass

m m m1

40

100= +

= 1.4 m

∴ 1

21

21m u FS=

or 1

22

1× =1.4 mu FS ... (ii)

From Eqs. (i) and (ii) we get

S S1 = 1.4

24. Given that

1

4 30π εq

r

q

rV−

=

or 2

4 3 0

q

rV

×=

π ε

∴ q V

4

3

20π ε= .... (i)

Now Eq

r

q

r r= = ×

4 3 4

1

902

0π ε πε( )

= × =3

2

1

9 6

V

r

V

r


M

R

T2

m g2

a

T1

a

m g1

25. The elec tro static en ergy UP of the sys tem is

given by

UQ

a

q

a

Qq

aP

q= + +

1

4 20

2

π ε

When UP = 0, then

Qq

a

q

a

Qq

a+ + =

2

20

or Q q Q2 2 0+ + =or Q q[ ]2 1 2+ = −

∴ Qq q= −

+= −

+1

2 1

2

2 2

26. See fig ure, ev ery an gle of equilated tri an gle is 60°

Distance OGl= 3

2. The magnetic induction

at O is given by B BAB= 6 ×

∴ Bi

l= ×6

4 3 20µ

π ( / ) (sin sin )30 30° + °

= × °6

4 3 22 300µ

πi

l( / )sin

= 3 0µπ

i

l

27. The given cir cuit is that of a Wheatstonebridge.

The circuit is a balanced one sinceresistance across

resistance across

AC

AD

= resistance across

resistance across

CB

BD

Thus no current will flow across 6 R of the sideCD. The given circuit will now be equivalent to

For maximum power net external resistance

or 2 4R =or R = 2 Ω

28. For half cy cle, di ode is for ward bi ased and forthe rest half it is re verse bi ased. There fore itwill con duct only for one half cy cle. There forechoice (b) is correct.

29. Net rate of for ma tion of y at any time t is

dN

dtN N

yx x y y= −λ λ

Ny is maximum when dN

dt

y = 0

or λ λx x y yN N=30. From n2 4= , six lines are ob tained in emis sion

spec trum.

Now E E4 2− = absorbed

F E4 3− < absorbed

and E E E E4 1 3 1 2 1− − − >, , absorbed

Hence n1 2= and n2 4= .


2R 4R

R 2R

4 Ω E

6R

4 Ω E

3R

4 Ω E

2R

R RRR

6R

R 4R

4ΩE

30°

30°

C

F

A

B D

E

√3 l2

l

ii

G

D

2R 4R

2R

R

R

A B

C

4 ΩE

31. (i) For uni formly ac cel er ated/de cel er atedmo tion

v u gh2 2 2= ±i e. , v h− graph will be parabola (becauseequation is quadratic)

(ii) Initially velocity is downwards negative and then after collision it reverses its direction withlesser magnitude i e. ., velocity is upwardspositive. Graph (a) satisfies both theseconditions.

Note that time t = 0 corresponding to thepoint. On the graph where h d=1 → 2, v increases downwards

At 2 → velocity changes its direction

2 → 3, v decreases upwards

32. OP OQ R R= ° =

=cos ( )60 21

2

∴ h OPR

1 602

= ° =cos (R = Radius)

h R2 2=

v

v

gh

gh

h

h1

2

1

2

1

2

2

2

1

4

1

2= = = =

33. The force of fric tion = µmg

∴ Retardation amg

mg= =µ µ

Now v ax gx2 2 2= = µor m v m gx2 2 22= µ or p m gh2 22= µ

∴ µ = p

gm x

2

22

34. ω ω θ202 2= + a n

ω ω θ π= = ×0

236 2,

∴ ω ω α π0

2

02

42 36 2= + × ×

− =3

41440

2ω π or α ωπ

= −×3

4 14402

II case

O = − ××

×ω ωπ

θ02

02

42

3

4 144

θ ω πω

′ = × ×× ×

02

02

4 144

4 2 3 = 24π

Hence number of rotations it make beforecoming to rest

= =24

212

ππ

35. We know that T l g= 2π ( / )

∴ ∆ ∆Tl

l=

⋅2

2

1 2π

lg

/

or∆ ∆ ∆T

T

l

lt= = ∝

2

1

2

= × = ×−

−( ) ( )2 10 10

21 10

65

or∆T

T× = × ×−100 1 10 1005

= × −1 10 3 %

36. We know nk

m= 1

2π

In series kk k

k k=

+1 2

1 2

= × =k k k

2 2

In parallel k k k k= + =1 2 2

∴ nk

m1

1

2 2=

π

and nk

m2

1

2

2=

π

Hence n n1 2 1 2: :=

37. v c= × = × ××

× ×1

2

1

2

9 10

5 103 10

4

98∆ν

ν

= × =− −2.7 ms 2.7 kms103 1 1

38. En ergy in ci dent over 1 10 4m 1.0 J2 = × −

Energy required to produced photoelectron

= × × =− − −1.0 J10 10 104 2 6

Number of photoelectron ejected = Numberof photons which can produce photoelectron = energy required for producingelectron/energy of photon

= = × ×× × ×

− − −

−10 10 300 10

3 10

6 6 9

34 8hC / λ 6.6 10

= × −1.51 1012 s 1

39. Heat pro duced/sec = −100 0.5 = 99.5 %

= 0.995 VI

or 796 = × ×0.995 V ne ( )Q I ne=

or n =× × × × −

796190.995 (50 10 1.6 103 )

≈ 1017


40. Dis tance be tween the spots on pho to graphicplate for two iso topes

= − = −

2 22 1

2 1

11( )r r

m m

mr

= × − ×2(36.978 34.980)

34.9800.05

= 0.0057 m = 0.57 cm

41. Let number of α-par ti cles emit ted be k andnumber of β-par ti cles emit ted be y Differencein mass number 4 238 206 32x = − =

k = 8

Difference in charge number

2 1 92 82 10x y− = − = 16 10 6− = ⇒ =y y

42. En ergy of elec tron stream = 12.75 eV. The

ground state elec tron hav ing en ergy −13.6 eVwill ac quire this en ergy and have to tal en ergy = − −13.6+ 12.75 = 0.85 eV. This cor re sp-onds to en ergy of elec tron in 4 th or bit. It willra di ate energy.

When electron jumps from 4th orbit. Hence Pfund series cannot be observed.

43. Path dif fer ence ∆xxd

D= ... (i)

Infront of one of the slits

xd=2

, but d = 5 λ

∴ x = 5

2

λ and

D d d= = =10 10 5 50( ) λ

∴ From Eq. (i), 45

2

5

50 4x = =λ λ

λλ

Corresponding phase difference

φ πλ

πλ

λ π= = × =2 2

4 2( ) .∆x

As I I=

0

2

2cos

φ

∴ I I II=

=

=02

0

20

4

1

2 2cos

π

44.E

E

r

r2

1

12

22

2

2

40

50

16

25= = =( )

( )

E E2 1

16

25=

Decrease in intensity

= − × =E E

E1 2

1

100 36 %

45. Ve loc ity of light in a me dium

cr r

= =1 1

0 0µ ε µ ε µ ε

46. At mag netic mo ments are di rected along SN,an gle be tween M and M is θ = ° 120(From given fig ure)

∴ Resultant magnetic moment

= + + °M M MM2 2 2 120cos

= + + −

=M M M M2 2 221

2

47. Per ma nent mag net should have largecoercivity and large retentivity. There fore thehysteresis cy cle of the ma te rial should be talland wide.

48. Here, P = + =15 6 21 Ω, (where P is

re sis tance in AB)

Qx

x=

++8

83 (where Q is resistance in BC)

R = + ×+

=156 6

6 618 Ω

(where R is resistance in AD)

S = + ××

=44 4

4 46 m

(where S is resistance in CD)

AsP

Q

R

S= , So Q

PS

R= = × =21 6

187 Ω

38

87+

+=x

x

On solving x = 8 Ω49. It r is the ra dius of small drop let and R is the

ra dius of big drop, then ac cord ing toquestion,

4

310

4

33 6 3π πR r= ×

or rR

R= =100

0.01

= × − −0.01 m =10 m10 2 4

Work done = S.A

= × × ×− −35 10 10 42 6 4 2[ )π (10− × −4 2π (10 )3 ] = × −4.35 10 J2

50. Here m g mr mr v× = =9 42 2 2ω π

or vg

r= = ×

×

×

9

4

9 10

422

75

2 2π

= 0.675 rps


Chemistry1. The struc tures of two geo met ri cal iso mers of

butenedioic acid

HO — C

O

HC==C

C

O

— OH

H

π σ π σ

σ

π

cis-butendioic acid

(maleic acid)

H

HO — C

O

C==CC

O

— OH

Hσ π

π σ

σ

π

trans-butendioic acid

(fumaric acid)

In the given molecule, each C-atom undergoes sp2 hy bridi sa tion, thus leav ing one 2pz or bital

in the unhybridised state. The sp2 hy brid

orbitals of car bon at oms form sigma bondbe tween them, with hy dro gen and ox y genat oms. The unhybridised 2pzorbitals of car bon and ox y gen at oms un der goes side waysover lap ping with the unhybridised 2pz or bitalof the neigh bour ing atom, thereby form ing a πbond be tween the two atoms.

2. In mol e cule CH Cl2 2, car bon is at tached to two

H-at oms, which are less electronegative thancar bon and two Cl-at oms, which are moreelectronegative than car bon, therefore,

H C

Cl

H

Cl+ −

−

+

1 1

1

1

oxidation number of carbon

x + + + − =2 1 2 1 0( ) ( )

or x = 0

On the other hand, in molecule CCl4 carbon isattached to four Cl- atoms, which all are moreelectronegative than carbon therefore

Cl C

Cl

Cl

Cl− −

−

−

1 1

1

1

Oxidation number of carbon

x + − =4 1 0( ) or x = 4

3. For titration of weak acid like CH COOH3

against a strong base like NaOH so lu tion,only phenolphthalein is a suit able in di ca tor. Itcan be ex plained as

Phenolphthalein is a weak organic acid whichmay be represented as HPh.

HPh H PhColourless Pink( ) ( )

3+ −+ …(i)

Thus, HPh is colourless while Ph ions havepink colour. As the medium changes fromacidic to basic, the equilibrium shifts towardsright because in the basic medium, OH− ionswill combine with the H+ ions to formunionised H O2 molecules, and now thesolution has pink colour.

4. (i) MO con fig u ra tion of O2

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz, * , , * , ,

π π π π2 2 2 22 2 1 1p p p px y x y= =, * *

Bond order of

O2

1

2

1

210 6 2= − = − =( ) ( )N Nb a

(ii) MO con fig u ra tion of O2+ ion

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz, * , , * , ,

π π π π2 2 2 22 2 1 0p p p px y x y, , * , *

Bond order of O2+ ion

= − = − =1

2

1

210 5 2

1

2( ) ( )N Nb a

(iii) MO con fig u ra tion of O2− ion (superoxide

ion)

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz, * , , * , ,

π π π π2 2 2 22 2 2 1p p p px y x y, , * , *

Bond order of O2− ion

= − = − =1

2

1

210 7 1

1

2( ) ( )N Nb a

(iv) MO con fig u ra tion of O22 − ion (per ox ide

ion)

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz, * , , * , ,

π π π π2 2 2 22 2 2 2p p p px y x y, , * , *

Bond order of O22 − ion

= − = − =1

2

1

210 8 1( ) ( )N Nb a


Thus, bond orders of these species are

O O O O2 2 2 222 2

1

21

1

21= = = =+ − −, , ,

As the bond energies or bond dissociationenergies are directly proportional to the bondorders, therefore the bond energies of thesemolecular species are in the order

O O O O2 2 2 22+ − −> > >

5. In ethyne and other ter mi nal al kynes (al kynesin which the tri ple bond is at end of the chain)or 1- al kynes, the hy dro gen at oms at tached to the tri ple bonded car bons, i e. ., acetylenichy dro gens are acidic in nature,

2CH C CH+ 2 NaTerminal alkyne

3 ≡≡ →∆

2CH C C Na + H3+

2Acetylide

≡≡ ↑−

The acidity of alkynes can be explained interms of sp-hybridisation of a triply bondedcarbon. The electrons in an sp-hybrid orbitalare more tightly held by the nucleus thanelectrons in an sp2-or sp3-orbitals, because ithas 50% s-character, thus, sp-hybrid orbital is more electronegative than sp2 and sp3

orbitals. Due to this greater electronegativitythe electrons of C H bonds are displacedmore towards the carbon and hydrogen atom can be removed as a proton ( )H+ by a strongbase. Consequently, alkynes behave asacids.

6. The oc tane num ber of a fuel is de fined as theper cent age of iso-oc tane (2 2 4, , −trimethylpentane) pres ent in a mix ture ofiso-oc tane and n-heptane, in which n-heptane has zero oc tane number and iso-oc tane has100 oc tane number.

CH C

CH

CH

CH C H

CH

CH

2,2,4-trimethylpentane

3

3

3

2

3

3

(isooctane)

CH CH CHheptane

3 2 5 3 −( )

n

Whereas the cetane number of a fuel is thepercentage of cetane (n-hexadecane) in amixture of cetane and α-methyl naphthalenein which cetane has 100 cetane number andα-methylnaphthalene has zero cetanenumber.

7. (i) Re ac tion : A B→⋅AlC KOH

Where, ‘A’ is 1, 2 dihaloethane and ‘B’ isacetylene.

CH Br

CH BrKOH (alc)

2

2

2 + →∆

CH

CHKBr 2H O

Acetylene

2 + +2

(ii) Reaction B C→CCl

Cl

4

2

is an example of addition of chlorine toacetylene in CCl4 solution forming first1 2, -dichloroethene and then1 1 2 2, , , -tetrachloroethane,

HC CH

Cl

HC C

H

Cl

CCl

Cl

1,2 dichloroet

≡≡ → ==

−4

2

trans hene

→

CCl

Cl

H C

Cl

Cl

C

Cl

Cl

H4

2

1, 1, 2, 2, -tetrachloroethane (Westron) (oracetylene tetrachloride)

When westron is passed over heatedCa OH( )2, westrosol is obtained. It is used as asolvent for oils, fats, waxes, as a refrigerantand as an anaesthetic.

CHCl CHClCa OH

dehydrochlorination

2 2

2⋅ →

∆

( )

( )

CHCl CCl HClWestrosol

== +2

8. The chlo rides of al ka line earth met als arewa ter sol u ble and their sol u bil ity de creasesfrom BeCl2 to BaCl2.

BeCl MgCl CaCl BaCl2 2 2 2> > >This is due to the decreasing hydrationenergy of ion because of their increased size.

Be Mg Ca Ba2 2 2 2+ + + +< < <


α–methylnaphthalene

CH3

CH —(CH ) —CH3 2 14 3

n-hexadecane (cetane)

9. 100 mL of H O2 2 so lu tion con tain H O g2 2 3=∴ 1000 mL of H O2 2 solution will contain H O2 2

= × =3

1001000 30 g

Consider the chemical equation

2 22H O H O + O2 2 2→2 34 68× = g 22.4 L at NTP

68 g of H O2 2 gives O2 at NTP 2.4= 2 L

∴ 30 g of H O2 2 will give O2 at NTP

= × =22 4

6830

.9.88 L = 9880 mL

But 30 g of H O2 2 are present in 100 mL of

H O2 2 solution. Hence, 100 mL of H O2 2solution gives O2 at NTP = 9880 mL

∴ 1 mL of H O2 2 solution will give O2 at NTP

= =9880

10009.88 mL

Hence, the volume strength of 3% H O2 2solution = 9.88

10. The bind ing en ergy of a nu cleus can becal cu lated from its mass de fect by us ingEin stein’s equation.

∆ ∆E = mc 2

where, ∆m = × × −0.02 1.66 kg10 27

(Q 1 amu = × −166 10 27. kg)

C = × −2.99 ms8 108 1

∴ ∆E = × × × ×−0.02 1.66 2.99810 1027 8 2( )

= × −2.9 J8 10 12

= 1 08.6 MeV (Q 1 eV = × −1.6 J02 10 19 )

11. (i) All the chlo rides are sol u ble in wa ter,ex cept Ag, Hg, Cu and Pb salts.

(ii) All the iodides except AgI, PbI2, Hg2I2and HgI2, are water soluble.

(iii) All the metal sulphate soluble in water ;exceptionally PbSO4 and BaSO4 arewater insoluble.

Hence, it is clear that Pb2 + ion is capable toprecipitate all the three ions, i e. ., Cl−, I− and

SO42 − ions,

Pb HCl PbCl HWhite ppt.

+ → ↓ +2 2 2

Pb HI PbI Hyellow ppt.

+ → ↓ +2 2 2

Pb H SO PbSO SO H O2 4White ppt.

+ → ↓ + +2 24 2 2

12. Metamerism is a com bined form of chain andpo si tion isomerism, which is ob served only incom pounds hav ing polyvalent func tionalgroup like ether, ketone, ester, 2° and 3°amines.

C

O

O N C

O

O; ; ,

Compounds having same molecular formulabut different number and arrangements ofcarbon atoms in alkyl groups on either side ofthe functional group are called metamers andthe phenomenon is called metamerism.Metamers always include same class ofcompounds.

Metamerism in ketones

Acetone and butanone do not showmetamerism.

CH C

O

CH CH C

O

CH CH3 3 2 3Acetone Butanone

3 ;

Ketones with atleast 5 carbon atoms(pentanone) can show metamerism.

Pentanone ( )C H O5 10

CH C

O

CH CH CH3 2Pentan one

− −2 3

2

;

CH C

O

C H

CH

CH3

ethylbutan one

− − −3

3

3 2R

CH CH C

O

CH CHPentan one

3 2 2 33

− −

Thus, metamers are position as well as chainisomers. Hence, it is more appropriate toconsider the isomers of compounds withpolyvalent groups as metamers, rather thanposition or chain isomers.

13. Mo lec u lar mass

= ×Weight of organic substance

air displaced at STP22400

= × =0.2

5622400 80


14. 146C is a un sta ble nu cleus with n p/ ra tio

higher than 1. Thus it emits a β-par ti cle so asto at tain the sta bil ity. Since, a β-emis sionin creases the atomic num ber by 1 with nochange in atomic mass.

Thus, the new el e ment formed will oc cupy one place to the right of the par ent el e ment in thePeriodic Table.

1416

147 1

0C Nparticle

→ + −−

eβ

Here, 14N is a stable nucleus with n p/ ratio is

equal to one, but as we know that the valencyof nitrogen is 3 and that of carbon is 4.

Thus, nitrogen forms three N H bonds withthree hydrogen atoms and one hydrogenatom remains as 1

1H, which further reacts with

another 11H atom to form H2 gas.

2 2614

4 714

3 2CH NH Hemission

→ +−β

As one molecule of methane gas now breaksdown into two gaseous products, i e. .,ammonia and hydrogen gas molecules, thusthe pressure on the walls of the vesselsincreased.

15. Struc ture of SO2 S atom in SO2 in volves

sp2-hybridisation.

S in excited state

Thus, the bond angle O S O (119.5°) isless than 120° because of lone pair of electron on sulphur.

Struc ture of BBr3 The cen tral atom bo ron has sp2 hy bridi sa tion and coplanar structure.

B in excited state :

Thus, the bond angle in BF3 is 120°.

Struc ture of CS2 The cen tral atom car bonhas sp hy bridi sa tion and linear structure.

C in ex cited state

Thus, the bond angle in CS2 is 180°.

Struc ture of SF4 S atom has sp3d

hy bridi sa tion and trigonal bipyramidal shape.

S in excited state

16. The or bital an gu lar mo men tum for a mov ingelec tron

= +hl l

21

π( )

For an electron moving in d-subshell, ( )l = 2

∴ Orbital angular momentum = +h

22 2 1

π( )

= 62

h

π


3s 3p 3d

Lonepair

These formtwo s

bonds withoxygen

These formtwo p bonds P-P and d-P

with O atoms

2 sp – hybridisation

119.5°O O

Spπ–pπ dπ–pπ

2s 2p


F

B F F

120°

2s 2p

Form p bonds with S atoms

Form σ bonds with S atoms


S C Sπ π

180°

3s 3p 3d

3 sp d – hybridisation

S

F

F

F

F

120º

90º

Structure of SF4

17. Com par i son of times taken for dif fu sion of thesame vol ume of two dif fer ent gases

t

t

M M

M

o x

o

x2

2

2

2

0

0

+ =+

or 234

224 32

02=+M Mx

or M Mo x2

234

22432

2

+ =

×

or M Mo x2+ = 34.92

∴ The molecular mass of mixture is 34.92.

18. Electronegativity of an el e ment is its ten dencyto at tract the shared pair of elec trons to wardsit self in a co va lent bond. Fur ther, if theelectronegativity of two at oms form ing a bond is dif fer ent the shared pair of elec trons isat tracted more to wards the moreelectronegative atom. As a re sult two polesare de vel oped and the mol e cule is said to bepolar. Thus, among the given elements

C (electronegativity) = 2.5

H (electronegativity) = 2.1

O (electronegativity) = 3.5

N (electronegativity) = 3.0

S (electronegativity) = 2.5

The most electronegative and leastelectronegative elements are O and H atomsrespectively. Thus, these form most polarbond with an electronegativity difference of ( )3.5 2.1 1.4− = and with 30% partial ioniccharacters.

19. In fact, zinc sul phide ex ists in two forms,called zinc blende and wurtzite. Both have aclose packed ar range ment of S2 + ions. Inboth, Zn2 + ions oc cupy the tet ra he dral voids.Both have 4 : 4 struc ture. They dif fer only in the fact that zinc blende has ccp or fccar range ment whereas wurtzite has hcpar range ment of S2 + ions. As a re sult unit cellof wurt zite has 6 for mula units of ZnS,whereas zinc blende has 4 formula units.

20. Sim ple alkyl halides are dehydrohalogenatedby us ing strong base such as alc. KOH,whereas vi nyl halides re quire stron gest baselike NH2

− for elim i na tion of hy dro gen halide.

BrBr CH ==CH—Br

Alc KOH

vinyl halide2 →⋅

→ ≡≡NaNH2 H—C C—H

21. Al de hydes and ke tones are re duced to thecor re spond ing al co hols by all the givenre duc ing agents, but NaBH4 is most suit ablere agent for this re ac tion as it re duces onlyal de hy dic group but can not re duce lactonegroup. Thus, lactone group remains intact.

22. Wil liam son’s syn the sis of alkyl aryl ethersin volves the treat ment of an alkyl ha lide withso dium phenoxide. The so dium phenoxideneeded for the pur pose is pre pared by theac tion of so dium on phenol.

23. Ox i da tion of ke tones In the ox i da tion ofke tones by al ka line KMnO4 the rupture ofcar bon bond occurs on ei ther side of the ketogroup and carboxylic acid is formed.

(ii) Ketone gives oxime with hydroxyl amine.

(iii) Aldol condensation In this reaction, twomolecules of ketones condense to formp-hydroxyketones in the presence ofdilute alkali. On heating the productundergoes dehydration leading to theformation of α β, -unsaturated ketone. Thedriving force for this dehydration is theconjugation of C C double bond with C O double bond in the final product.


O

OHC

O

O

HOH C2

ONaBH4

OH

+ 2Na

– +O Na

+ H2

Sodium phenoxide

– +O Na

+ CH – Cl3

+δ –δ S 2N

∆

OCH3

+ NaCl

Anisole

2

O

–KMnO /OH4

CH CH COOH2 2

CH CH COOH2 2

1,6 – hexanedioic acid(Adipic acid)

ONH –OH2 NOH

Oxime

24. The carboxylic acid which can pro duce asta ble car ban ion lib er ates CO2 readily, Forex am ple

On the other hand, the other carboxylic acidsproduce relatively less stable carbanions withno resonance.

25. Pri mary amines when re act with chlo ro form inthe pres ence of al kali, they pro duce foulsmell ing car byl amines (iso cya nides), whichen able us to con firm the pres ence of NH2.

(1° amine) group. This is known ascar byl amine re ac tion (Isocyanide test)

CH NH CHCl KOH3 2 3 + + →3

CH N C KCl+ H O3 2Methylisocyanide(foulsmell)

≡≡ ++

− 3

26. ••NH group is ac ti vat ing and ortho-para

di rect ing whereas

C

O

is de ac ti vat ing andmeta di rect ing.

When the two functional groups present in thebenzene ring direct differently, i e. ., one isortho-para and the other is meta directing,then ortho-para directing group takesprecedence, because ortho-para directinggroup is ring activating group, which makesthe nucleophilic substitution reaction easier.

27. So dium borohydride re duc tion of an aldosemakes the cor re spond ing al co hol.CHO

CH OH

CH OH

CH

→

( ) ( )CHOH CHOH

Glucose

NaBH44

2

2

4

2OHSorbitol

Among the given monosaccharides the onlyone, which is given as option (a) can form anoptically inactive or meso compound. Thereason of its optically inactivity is that it has aplane of symmetry.

The two halves of the molecule rotate theplane of polarised light in opposite directionand hence cancel the effect of each other(internal compensation ) and make themolecule optically inactive.


C H —C—CH —C—O—H → C H —C—CH6 5 2 6 5 2

||O

||O

–OH

∆p-keto acid

|

||O

C H —C CH6 5 2

–O

[Resonance stabilized structures]

——

C H —C—C—O—H → C H —C6 5 6 5

||O

||O

|O

–HO

D

–

–

(unstable)

C H —CH—C—O—H → C H —CH—OH6 5 6 5

–HO

D

–

(unstable)

|OH

||O

C H —CH—C—O—H → C H —CH—NH6 5 6 5 2

–HO

D

–

(unstable)

|NH2

||O

N

O

H

Conc. HNO3

Conc. H SO2 4

X

N

O

H

O N2

p–isomer (major)

HNO2

+

N

O

o–isomer (minor due to steric hindrance)

O + O

–OH

O

H

OH

∆

O

α,β-unsaturated

ketone

CHO

CH OH2

H OHHO H

HO H

H OH

Aldose

NaBH4

CH OH2

CH OH2

H OHHO H

HO HH OH

Aldoitol

Plane of symmetry

28. Struc ture of mel a mine.

29. Dis so ci a tion of N O2 4

N O ( ) 2NO ( )2 4 2g g3

Initial moles a 0

Moles at equilibrium a ( )1 − α 2aαTotal number of moles at equilibrium

= − +a a( )1 2α α= +a ( )1 α

If p is total pressure of the equilibrium mixture,then

pa

ap pN O2 4

= −+

× = −+

×( )

( )

( )

( )

1

1

1

1

αα

αα

pa

ap pNO2

=+

× =+

×2

1

2

1

αα

αα( ) ( )

∴ Equilibrium constant, Kp

pp =

( )NO

N O

2

2 4

2

= +− +

[ /( )]

( ) /( )

2 1

1 1

2α αα αp

p =

− +4

1 1

2αα α

p

( ) ( )

∴ Kp

p =−

4

1

2

2

αα

K pp ( )1 42 2− =α α

or K K pp p− =α α2 24

or K p Kp p= +4 2 2α α

or K p Kp p= +α 2 4( )

or α 2

4=

+K

p K

p

p

On dividing both numerator and denominat or by p

α 2

4=

+

K

p

p

p

K

p

p

p

or α =+

K

pK

p

p

p4

1 2/

30. Ca PO3 4( )2 ionises completely in the solution

as Ca PO 3Ca PO3 42+

43( )2 2→ + −

Solubility of calcium phosphate

[ ( ) ]Ca PO3 4 2 = wg/100mL (given)

∴ Solubility in 1 L solution = 10 w g/ litre and

molarity =

10 w

M per litre.

Ionisation of Ca PO3 4( ).

Ca (PO ) 3Ca PO3 4 22 +

4s s s

3

3

3

2

2+ −

∴ Ksp2 +

43Ca PO= −[ ] [ ]3 2

= ( ) ( )3 23 2s s

= ×27 43 2s s

= 108 5s

=

10810

5w

M

= × × ×

108 10 102 55

.w

M

Solubility product of Ca PO3 4 27

5

10( ) =

w

M

= ×

1.08 1075

w

M

≈

1075

w

M

31. As we know that

1 amu (or 1u) = × −1.66 10 24 g

∴ 6.64 10× − 23g has mass in amu

= ××

−

−6.64 10

1.66 10

23

24

= 40.0 u

32. The froth floa ta tion method is based upon thefact that the sur face of sul phide ores ispref er en tially wet ted by oils while that ofgangue is pref er en tially wet ted by water.

This method is widely used for theconcentration of zinc blende or sphalerite(ZnS), copper pyrites (Fool’s gold-CuFeS2)and galena (PbS).

The process of leaching consists in treatingpowdered ore with a suitable reagent (acid,base or other chemical) which can selectivelydissolve the ore but not the impurities.

The method is applied for extracting silverfrom argentite ( )Ag S2 ore.


N

N N

NH2

NH2

H N2

1

2

3

4

5

6

2, 4, 6-triamino-1, 3, 5-triazine

33. Tin (Sn) ex ist in three al lo tropic forms.

α β −−° °

tin tin(Grey)

15.2 C

(White)

161 C3 3

γ −°

tin Liquid tin(brittle)

232 C

3

Thus, below 15.2°C transition of white to greytin occurs. In cold countries, the conversion of white tin to grey tin is accompanied by anincrease in volume and the latter, being verybrittle easily crumbles down to powder. Thephenomenon is called tin dis ease, tin pastor tin plague. This is the rea son to why whitetin but tons of sol dier’s uni forms get cov eredby grey pow der.

34. The ox ide of an el e ment, show ing functions ofmore than one sim ple ox ides of the same ordif fer ent types, is called mixed ox ide. Redlead (Pb O )3 4 is con sid ered to be a mix ture oflead di ox ide ( )PbO2 and lead ox ide ( )PbO . Red lead shows the prop er ties of both ox idessi mul ta neously with formula (PbO PbO)2 ⋅2 .

PbO HNO2Red lead

3⋅ + →2 4PbO

PbO + 2Pb(NO ) + 2H2(from dioxide)

3 2from lead oxide

2( )

O

Another example of mixed oxides is Fe O3 4(ferrosoferric oxide), which is a combination of Fe O2 3 and FeO and can be written as Fe O FeO2 3 ⋅ .

35. Car bon has six elec trons. The first fourelec trons are filled, two each in 1s and 2sorbitals. The re main ing two elec trons will go to any two of the three 2p-orbitals in ac cor dancewith Hund’s rule.

In general, for a given value of l ( . ., )i e l = 2 , m l= − to + l including 0 (i e. ., m = − 1, 0 1, + ).Thus, the 6th electron of carbon atom canhave the value of m either 0 or − 1 because 5th electrons already have m = 1.

Further, all the singly occupied orbitals willhave parallel spins i e. ., in the same direction,viz. either clockwise or anticlockwise. This isdue to the fact that two electrons with parallelspin (of course in different orbitals) willencounter less interelectronic repulsions inspace than when they have opposite spins.

For C atom

The quantum number for 5th electron

= +2 1 11

2, , ,

The quantum number for 6th electron

= − +2 1 11

2, , ,

36. For a gen eral reaction,

Equilibrium constant ( ) ( )Kcn= −mol L 1 ∆

where, ∆n n n= −products reactants

(only gaseous molecules)

Thus, for homogenous reaction

4 5 6 2NH O 4NO3 2( ) ( ) ( )g g (g) g+ +3 H O

∆n = − =10 9 1

∴ KCn= −( )mol L 1 ∆

= − +( )mol L 1 1 or ( )conc + 1

37. Sup pose the two com po nents form ing anideal so lu tion are A and B and their molefrac tions in the liq uid phase are xA and xBwhile in the va pour phase, these arerep re sented by yA and yB re spec tively.

Thus, total vapour pressure of solution = ptotal

vapour pressure of A in solution p x pA A A= °and vapour pressure of B in solution, p x pB B B= °

where, pA° and pB

° are vapour pressures of

solvents before mixing.

Now yp

p

x p

x p x pA

A A A

A A B B

= =+

°

° °total

=− +

°

° ° °x p

p p x pA A

A B A B( )

or x p y p p x y pA A A A B A A B° ° ° °= − +( )

or x p p p y y pA A B A A A B[ ( ) ]° ° ° °+ − =

or xy p

p p p yA

A B

A B A A

=+ −

°

° ° °( )

But yp

p

x p

pA

A A A= =°

total total

Putting the value of xA , we get

yp

p

y p

p p p yA

A A B

A B A A

= ×+ −

° °

° ° °total ( )

or1

p

p p p y

p pA B A A

A Btotal

= + −° ° °

° °( )

or 1 1

p p

p p

p pB

B A

A Btotal

= + −°

° °

° °


1s 2s 2p

m = +1, –1, 0

38. π ( )Na SO2 4 = icRT = i RT( )0.004

π ( )Glucose = CRT = 0.010RT

As solutions are isotonic

i RT RT( )0.004 0.010=∴ i = 2.5

Now Na SO 2Na + SO2 4mole

+4

11

02

2

0−

−

α α α

3

Total number of particles ( )i = − + +1 2α α α i = +1 2α

or α = −i 1

2

= − =2.50.75 =75%

1

2

39. Cow milk is an ex am ple of nat u ral emulsion inwhich wa ter is dispersion medium and oil orfat is the dis persed phase in the form ofdrop lets. As the two in volved liq uids areotherwise im mis ci ble and re quire thepres ence of a sur face ac tive or emulsifyingagent for sta bil ity, thus in milk na turallyoccuring protein casein act as an emul si fierand stabilzes is oil-in-water emulsion.

40. Fermentation of glu cose.

C H O 2C H OH+ 2CO6 12 6

Zymase

2 5 2→

By hit and trial method, if reaction is of firstorder.

Case I kt

a

a x=

−2.303

log

[where, a a x= −0.12 M, ( )= 0.06 M

and t = 10h]

∴ k = 2.303

10

0.12

0.06log

= ×2.303

10log 2

= × −2.303

100.3010 = 0.0693 h 1

Case II k = 2.303

15

0.12

0.045log

= 2.303

152.6log

=2.303

150.4149 = 0.0697 h× − 1

Case III As in 10h, the con cen tra tion ofre ac tant re mains half of the orig i nal value.

∴ For first order reaction, tk

1 2/ = 0.693

= =0.693

0.069310

As k comes out to be constant and thecalculated value and the given of t1 2/ is same,the given reaction is of first order.

41. En ergy left un uti lised = =1560

2780 kJ

The enthalpy of evaporation of water

= −44 1kJ mol

∴ For losing 44 kJ of energy, water to beevaporated

= =1mole 18 g

Thus, for losing 780 kJ of energy, water to be

evaporated = × =18

44780 319g g

42. (a) For ideal gases, re la tion ship be tween heatof re ac tion at con stant pres sure ( )∆H and thatof at con stant vol ume ( )∆E .

∆ ∆ ∆H E n RTg= =

or∆ ∆

∆H E

n TR

g

− = (constant)

(b) Spontaneity in terms of free energychange ( )∆G Total entropy change for asystem.

∆ ∆ ∆S S STotal system surroundings= + …(i)

For a process, being carried out at a constant T and p, the entropy change and heat lost bysurroundings.

∆ ∆S

H

Tsurroundings = −

∴ ∆ ∆ ∆S S

H

Ttotal = − (Q ∆ ∆S Ssystem = )

Multiplying throughout by T, we get

T S T S H∆ ∆ ∆total = −But we know that

∆ ∆ ∆G H T S= −∴ T S G∆ ∆total = −


1/ptotal

1°pB

slop = °pB – °pA

° °p pA B

yA

Thus, for a process to be spontaneous, ∆S totalmust be positive and ∆G must be negative.

(c) According to first law of thermodynamics

∆U q W= + or q U W= −∆

where, ∆U = internal energy change

q = heat supplied

W = work done

(d) Relationship between standard freeenergy change and equilibrium constant.

At equilibrium,

∆G RT K° = − ln

or ln KG

RT= − °∆

or K e G RT= − °∆ /

43. The pro cess of sep a ra tion of a mix ture ofgases on the ba sis of their dif fer ent rates ofdif fu sion due to dif fer ence in their den si tiesand mo lec u lar weights is called atmolysis . Ithas been ap plied with suc cess for thesep a ra tion of iso topes and other gas eousmix ture as the rates of dif fu sion of gases in amix tures are in versely pro por tional to thesquare roots of their molecular weights.

rM

∝ 1

44. In bo ric ( )H BO3 3 , the cen tral bo ron atom is

sp2-hy brid ised and bo rate in ( )BO33 − has

trigonal pla nar struc ture.

B in exited state 2 2 2s p px y′ ′ ′

(form three B O− − σ-bonds)

Thus, H BO3 3 has layered structure in whichplanar BO3

3 − units are joined by H-bonds.

45. Gen er ally, all the com pounds of zinc are white coloured (for ex am ple ZnO, ZnS, ZnCl , ZnSO2 4 etc. are whitecol oured com pounds). It can be ex plained aszinc can show only + 2 va lancy and Zn2 + ion

has com pletely filled d-subshell ( )3 10d .

As it does’t have any un paired elec tron as well as any va cant d-or bital, hence not ca pa ble ofshow ing d d− tran si tion and forms col our less compounds.

46. Pentaaminenitrocobalt (III) chlo ride ex hib itslink age isomerism in which ei ther N or O-atom of ambidentate ligand NO2

− act as a do nor

giv ing two iso mers. Thus, two dif fer entpentaaminenitrocobalt. (III) chlo ride havebeen pre pared each con tain ing the NO2

group in the complex ion.

47. In cationic de ter gents a ma jor part is cat ionwhich is in volved in their cleansing action.They are called in vert soaps be cause theircleans ing ac tion is due to pres ence of cat ionsrather than an ions. e g. .,trimethylstearylammonium bromide.

CH (CH )

CH

CH

Br3 2 17+

3

3

3

stearyltr

−N CH

imethylammonium bromide



B

O

H

O

B

O

H

OH

O

HO

OH

H

OH

B

O

H

O

B

OBO

O OHH HH

H H

NH3

CO

H N3

H N3

ONO

NH3

NH3

2+

Pentaamine nitritocobalt (III)ion (Red)

NH3

CO

H N3

H N3

NO2

NH3

NH3

2+

Pentaamine nitrocobalt III ion (Yellow)

48. Re sis tance, Rl

a= ρ , where, ρ = spe cific

re sis tance

also ρκ

= 1, where κ = specific conductance

∴ Rl

a= 1

κ

or l

aR( ) .cell constant = κ

∴ Cell constant = resistance × specificconductance.

49. As SRP of oxidant − SRP of re duc tant > 1.7,

then their com bi na tion may lead to explosion.In option (a) both are reductant and in option(b) both are oxidant. Only in option (c) silver isreductant and azide ion is oxidant.

E ° − = −N /N2 3 V309.

E ° + =Ag Ag V/ .0 80

∴ E = − −0 80 309. ( . )

= + 389. V

As the value of E is greater than 1.7 thus thiswill lead to be the explosion.

50. Since dou bling the con cen tra tion of Bdou bles the rate there fore, the or der ofre ac tion w.r.t. [ ]B is one and since on dou blingthe con cen tra tion of A and B rate in creases 8times, there fore, or der of re ac tion w.r.t. A willbe 2.

i e. ., r A B∝ [ ] [ ]2

or r k A B= [ ] [ ]2

Mathematics1. π [( ) ( ) ( )]r r r r r r2

212

42

32

1002

992− + − + … + −

∴ r r r r r r2 1 4 3 100 99 1− = − =… = − == + + + + … +π [ ]r r r r r1 2 3 4 100

= + + + … +π [ ]1 2 3 100r

= 5050 π sq. cm

2. The qua dratic equa tion

3 2 1 3 2 02 2 2x a x a a+ + + − + =( )

will have two roots of opposite signs, if theproduct of the roots is negative

i e. ., if a a2 3 2

30

− + <

⇒ 1 2< <a

3. zz z

21 32

2 1= +

+ So, z2 divides the line segment joining z z1 3,in the ratio 2 1: internally.

Note If z z1 2, and z3 are re lated by

az bz cz1 2 3 0+ + = , where a b c+ + = 0, then

z z1 2, , and z3 will be col lin ear points.

4. Let Ex

x x= + + +

11 2

4

⇒ E xx

x xx

= + + +

1 14

= +

+xx

x1

14

4( )

Coefficient of xn in x

xx

2 4

411

+

+( )

= coefficient of x4 in ( ) ( )1 14 2 4+ +x x

Now, general term in the expansion of

( ) ( )1 14 2 4+ +x x is 4 4C Cr k⋅ ⋅ +xr 2k

r K+ =2 4

r : 0 2 4

k : 2 1 0

∴ Coefficient of x4 in given expression is 4 4 4 4

14 4

0 2 2 4 2 31C C C C C C+ + =

5. For the first place we have nine choices, foreach of the next four to choices. At this stageadd the num bers al ready writ ten and thechoose digit for the unit’s place is 5 ways.


R G1

2

3

4

G

100

6. Equa tion of line AB isx y z− =

−= =1

1 2 0λ

Now, AB OC⊥⇒ 1 1 2 2 0( ) ( ) ( )λ λ+ + − − =⇒ 5 1λ = −

⇒ λ = − 1

5

⇒ c is 4

5

2

50, ,

Now, x y z12

12

122 4+ − + =( ) …(i)

and ( )x y z12

12

121 1− + + = …(ii)

Now, OC CD⊥

⇒ x1

4

5

4

5+

⋅ + y1

2

5

2

5−

⋅

+ − ⋅ =( )z1 0 0 0 …(iii)

From Eqs. (i) and (ii), we get

− + =4 2 01 1y x ⇒ x y1 12=From Eqs. (iii), putting x y1 12=

⇒ 24

51y =

⇒ y1

2

5= ⇒ x1

4

5=

Put these values of x1 and y1 in Eqs. (i), we get

z1

2

5= ±

7. Let plane is ax by cz d+ + + = 0

Plane passes through ( , , )x y z1 1 1

⇒ a x x b y y c z z( ) ( ) ( )− + − + − =1 1 0

Let its distance from ( , , )x y z2 2 2 be P, then| ( ) ( ) ( )|a x x b y y c z z

a b cP2 1 2 1 2 1

2 2 2

− + − + −

+ +=

Now, there are infinite values of a b c, , whichsatisfy this equation.

Hence, there are infinite planes.

8. P A B P A B( ) ( )′ ∩ ′ − ∩

= ′ ′ −P A P B P A P B( ) ( ) ( ). ( )

= − − − ( ) ( ) ( ) ( )1 1P A P B P A P B

= − −1 P A P B( ) ( )

P A P B P A P B( ) ( ) ( ) ( )′ + ′ − = − + − −1 1 1 1

= − −1 P A P B( ) ( )

P B P A P B P A( ) ( ) ( ) ( ( ))− ′ = − −1

= + −P A P B( ) ( ) 1

P B P A P B P A( ) ( ) ( ) ( )′ − = − −1

= − −1 P A P B( ) ( )

9. u u u nn n1 1 2 2− =− cos cosθ θ− −2 1cos ( )n θ

= − + +2 1 1[cos ( ) cos ( ) ]n nθ θ = −2 1cos ( )n θ = +2 1cos ( )n θ = +un 1

10. Ex pres sion = +− −cos cos1 1x

x x⋅ + − ⋅ −

1

21 1

1

2

22

= + −− − −cos cos cosx x1 1 11

2

Q1

2

1

2

1 1< ⇒ >

− −x xcos cos

= =−cos 1 1

2 4

π

11. Add ing, 7 10(cos sin )θ θ+ =

or cos sinθ θ+ = 10

7

But maximum value of cos sinθ θ+ is

210

7= ≠1.41

So, no θ is possible.

12. Let P be the po si tion of the kite and AP be thestring of length 25 m in clined at 75° with thehor i zon tal BP the height of the kite.


B(0,2,0)

D(x , y , z )1 1 1

C(λ+ 1,–2λ,0)

O X

A(1, 0, 0)

Z

x – 11

=y

–2=

z0

= λ

Y

A BX

25m

P (Kite)

75°

In ∆ABP,

sin 75° = BP

AP

⇒ BP = ° = +25 75

25 3 1

2 2. sin

( )

= +25

43 1 2( ) m

13. If ( , )α β is the centre, then

( ) ( ) ( ) ( )α β α β− + − = − + −1 3 3 12 2 2 2 …(i)

and βα

βα

−−

⋅ −−

= −3

1

1

31

⇒ ( ) ( ) ( ) ( )α α β β− − + − − =1 3 1 3 0 …(ii)

From Eq. (i), we get

4 4 0α β− =

⇒ α β=From Eq. (ii), we get

2 1 3 0( )( )α α− − =⇒ α = 1 3,

∴ ( , ) ( , ), ( , )α β = 1 1 33

14. 2 3 02x ky− + =

⇒ xk

yk

2

2

3= −

∴ Latus rectum = = =kk

22 4( )Q

So, the vertex =

=

03

03

4, ,

k

15. Let x k= +3 , 0 1< <k

Then, y f x x x= = − =( ) [ ] 3 3 3+ − = = −k k x

∴ x y= + 3 ( ( ) )Q f y y− = +1 3

∴ f x x′ = +( ) 3

16. Here, f x x

x

( ) cos=−

+

− 12

2

11

11

. At xx

= − <21

12

,

So, f xx

( ) tan= −211

∴ ′ =+

⋅ −

= −+

f x

x

x x( ) 2

1

11

1 2

12

2 2

∴ ′ − = −+ −

= −+

= −f ( )

( )2

2

1 2

2

1 4

2

52

17. As f x( ) is even, f f( ) ( )− =2 2 .

As f x( ) is monotonic increasing in [ , ]2 6 .

f f f( ) ( ) ( )3 5 5< = −Also, f f( ) ( )3 5< ⇒ f f( ) ( )− <3 5

18. Put a x z− = , then

I x e dxa n

n x a x=−

−∫ . ( )

= − ⋅ −− −∫n

a n a z za z e dz( ) ( )( )

∴ I a z e dza n

n a z z= −−

−∫ ( ) ( )

= −−

−−

−∫ ∫a e dx xe dxa n

n x a x

a n

n x a x( ) ( )

∴ 2I a e dx aa n

n x a x= ⋅ ⋅ =−

−∫ ( ) λ

19. Put 1

xz= , then − =1

2xdx dz

∴ I zz

z x dz= ⋅ −

⋅ −∫π

π1 21/sin ( )

= −

∫

−

ππ1 1

21

zz

z z dz. sin

= ⋅ −

= −∫ππ1

1 1

zz

zdz Isin

∴ 2 0I =⇒ I = 0

20. Sides are ( ) ( )x y2 21 1 0− − =⇒ x x y y+ = − = + = − =1 0 1 0 1 0 1 0, , , ,

Clearly, only y ± =1 0 touch the circle.

21. Clearly,

lim ( ) lim ( ) lim ( ) ,x x x

f x f x f x l→ ∞ → ∞ → ∞

+ = + = =2 1

(say)

Then, taking limit, I ll

= +

1

2

4

⇒ 1

2

2l

l=

⇒ l 2 4=

⇒ l f x x= > ∀2 0[ ( ) , ]Q

22. As, 2 65≤ ≤f x( ) .

and − ≤ ≤1 1cos x , we get

1 7 5≤ + ≤cos ( ) .x f x

∴ 1 7 5

λ λ λ≤ + ≤cos ( ) .x f x

Clearly, h xx f x

( )cos ( )= +

λ cannot be

continuous, if 1

λ λ,7.5

contains any integer

So, λ = …8 9, ,


23. Di vide above and be low by sin2 x, then

( )

( cot )

3 2

2 3 2

cosec cosecx cot

cosec

2 x x

x xdx

++∫

Put 2 3cosec x x t+ =cot

∴ ( cot )− ⋅ − =2 3 2cosec x cosecx x dx dt

= − = + =+

+∫dt

t tC

x xC

2

1 1

2 3cosec cot

=+

+sin

( cos )

x

xC

2 3

24. Equa tion of cir cle will be

x y y2 22 2 0+ − + + − =( ) ( )λ λ

Differentiating 2 2 2 0x ydy

dx

dy

dx+ − + =( ) λ

∴ Equation is

x y ydx

dyx y2 22 2 2 2 4 0+ − − − ⋅ + −

=( ) ( )

Solving this equation we get

x y xdx

dyy2 2 2 2 0+ − − − =( )[ ]

25. Let iα, where α is purely real, be one of theimag i nary roots. Then,

− + + + =a ib c iα α2 0

⇒ c a b= + =α α2 1 0,

⇒ α 2 = c

a and α = − 1

b

⇒ 12b

c

a= ⇒ a b c= 2

26. Here [ ]abc = ± 1

[ ]a b c a b b c+ + + + = + + × + ⋅ +( ) ( ) ( )a b c a b b c

= × + ⋅ + ( ) ( )c a b b c

= × + × ⋅ + ( ) ( )c a c b b c

= × ⋅ c a b

= [ ]a b c = ± 1

27. Let R t t= ( , )2 2 . Then, m of the tan gent at R is

dy

dx t t

⋅( , )2 2

i e. ., 1

t.

The slope of PQ = −

−=2 1

11

4

4

3

So,1 4

3t= ∴ t = 3

4

∴ R = ×

=

9

162

3

4

9

16

3

2, ,

28. Possibilities Permutations

6, 6, 6, 6, 6, 0 6

56

!

!=

6, 6, 6, 6, 4, 2 6

430

!

!=

6, 6, 6, 6, 3, 3 6

4 215

!

! !=

6, 6, 6, 4, 4, 4 6

3 320

!

! !=

∴ Required number of sequences = 71

29. f xf y f x f y ( ) ( ) ( )= 2

Put x = 1, f f y k f y( ( ) ( )= 2,

where, k f= ( )1 ⇒ f x k x( ) .= 2

f f k k( ) ( )3 5 3 5 4+ = + =

f x x( ) = 1

22

∴ ′ =f x x( )

⇒ ′ =f x x( )

30. x x x22

11

2

3

40− + = −

+ >

For sin ( )− − +1 2 1x x

⇒ 0 1 12< − + ≤x x

⇒ x x2 0− ≤ …(i)

For cos ( )− −1 2x x

⇒ x x2 0− ≥ …(ii)

From Eqs. (i) and (ii) , we get

x x2 0− =∴ x = 0 1,

31. If 02

< <xπ

,

− −

−

∫ ∫=2 2

2x

x dx x dx|cos | |cos |π

+−∫ π /

|cos |2

xx dx

= − +−

−

−∫ ∫2

2

2

π

π

/

/cos cosdx x dx

x

= − +−−

−[sin ] [sin ]//x x x

22

2π

π

= − + +1 2 1sin sin x

∴−∫ =

20

xx dx|cos |

⇒ 1 2 1− + +sin sin x

∴ 0

0x

x dx∫ =|cos |


⇒ 1 2 1 0− + + =sin sin x

⇒ sin sinx = − < −2 2 1

∴ No solution in 02

,π

32. Let PQ be the tree and AP x= be breadth of

the river, so that ∠ =QAP 60º.

Let ∠ = °QBA 30 , then AB = 40 m PQ x x= ° = + °tan ( ) tan60 40 30

⇒ x [tan tan ]60 30 401

3° − ° = ×

⇒ x 31

3

40

3−

=

⇒ x = 20 m

33. Here, r r d1 2+ =

So, k + + −2 4 162 2

= distance between ( , )k 0 and ( , )2 4

⇒ k k+ = − +2 2 42 2( )

⇒ ( )k k k+ = − +2 4 202 2

⇒ 8 16k =∴ k = 2

34. The cen tre of x y x y2 2 4 4 0+ − − = is ( , )2 2 . It

will be on ax by+ = 2

∴ 2 2 2a b+ =⇒ a b+ = 1

ax by+ = 2 touches x y2 2 1+ =

So, 12

2 2= −

+a b

⇒ 2 2 3 02a a− − =

∴ a =± +

= ±2 4 24

4

1 7

2

∴ b a= −1

= − ±1

1 7

2

= ±1 7

2

35. Here, x y x y y2 28 2 8 32 0− + + − + =( )

∴ x y y= + ± −4 4 1

For real x y, ≥ 1 and then

x y y= + ± −4 4 1

= + − ± >4 1 1 02( )y

Similarly, for real y x, ≥ 1 and then y x= +4

± − >2 1 0x

Hence, the point on the graph will have theircoordinates positive.

36. Clearly, [cos ] , ,x = −0 1 1

If [cos ] ,x = − 1 the equation is

− + =1 1|sin |x

⇒ |sin |x = 2 (absurd)

If [cos ]x = 0, the equation is |sin |x = 1.

This is possible only when x = 3

2

5

2

π π, .

If [cos ]x = 1, then equation is |sin |x = 0.Hence x = 2π

37. tan( )

( )y

x x

x x=

+ − −+ − −

1 1

1 1

2

=+ + − − −1 1 2 1

2

2x x x

x

[ ( ) ]Q a b a b ab− = + −2 2 2 2

=− −1 1 2x

x

∴ yx

x=

− −−tan 121 1

Put x = sin 2θ. Then,

[ sin cos ]1 22 2− =θ θ

y = −−tancos

sin1 1 2

2

θθ

=⋅

−tan

sin

sin cos1

22

2

θθ θ

[ cos sin ]Q 2 1 2 2θ θ= − sin sin cos2 2θ θ θ=

= =−tan tan1 θ θ

∴ dy

dx

dy d

dx d x= = =

−

/

/ cos

θθ θ

1

2 2

1

2 1 2

∴ at xdy

dx= =

⋅ −

1

2

1

2 11

4

,

= 1

3


30° 60°AB P

Q

40m x

38. ′ = + −f x x a( ) ( )3 3 12 2

∴ ′f x( ) is minimum, if x a2 2 1 0+ − > for all

x R∈ and this happens, if a2 1 0− > .

Clearly, f x( ) cannot be maximum because thecoefficient of x2 in ′ =f x( ) 340

Now, a2 1 0− ≥⇒ a ∈ − ∞ − ∪ + ∞( , ] [ , )1 1

39. Let sec tanx x z+ = , then

sex x x x dx dz(tan sec )+ =

∴ Idz

z

z= = −

+ − +

∫1

1 2

3

2

1

1 2

2

= −+

−

1

2

1

2 11

2( )`

= − − −1

22 1 12( )

= − −1

22 2 2( )

= −2 1

40. If we re flect y x= −[ ]2 in y-axis, it will be come

y x x= − − = +| | | |2 2

Since, reflected lines are y x y x= + = − −2 2,their combined equation is,

( ) ( )y x y x z− − + + =2 0

⇒ y x2 22 0− + =( )

⇒ y x x2 2 4 4 0− − − =

41. ( ) ( ) [( ) ]/ / /1 1 12 2 2+ = + − +x x xx x x

Now, lim ( ) /

x

xx e→

+ =0

2 21

⇒ lim [( ) ] [ ]/

x

xx e e→

+ = −0

2 2 21

= −e 2 7

42. ′ = −→ ∞

fe

hh

h

( ) lim/

00

2− 1

= =→

−

→

−

lim lim/

log

log

h

h

h h

hh

e

ee

0

1

0

122

as hh

→ −0

12

, tends to − ∞ at a much faster

rate as compared to − → ∞log h

= =→

− −lim

log

h

hh

e0

12

0

Thus, f x( ) is differentiable at x = 0.

43.dx

x xx

dx

xx

2 44

3 45

4

3 4

11

11

+

=+

∫ ∫/ /

Put 114

4+ =x

t

⇒ − =4

45

3

xdx t dt

⇒ − = − +∫t

tdt t C

3

3

= − +

+114

1 4

xC

/

44. x ydx xdyxdy ydx

y

22

( )+ = −

⇒ ( ) ( )xyx

yd xy d

x

y

= −

⇒ x y x

yk

2 2

2= −

+log

45. Given, m n n( )− =1

n is divisible by n − 1

⇒ n = 2 ⇒ m = 2

Hence ; chord of contact of tangents drawn

from ( , )2 2 to x y2 2

9 41+ = is

⇒ 2

9

2

41

x y+ =

⇒ 4 9 18x y+ =

46. Let equation of hyperbola be x

a

y

b

2

2

2

21− = .

Trans verse axis is of length 2a units. If c is thecen tre, s is the fo cus and a be the ver tex, then cs ae=

Also,cs

a2

= ⇒ aae=2

⇒ e = 2

b a e a a2 2 2 2 21 4 1 3= − = − =( ) ( )

∴ Equation of hyperbola is

x

a

y

a

2

2

2

231− =

⇒ 3 32 2 2x y a− =

47. We must have a b c2 2 2 0− − = and

a bd c2 23 0+ − =⇒ − =b bd2 3

⇒ b b d( )− − =3 0

⇒ b = 0 is neglected

∴ b d= − 3


48. Let z a ib= + where a b> >0 0,

Since, Re ( ) Im ( )z z+ = 0 (given)

⇒ a b+ = 3 …(i)

Now, let E z z= Re ( ) . Im ( )2

⇒ E a b a a= = −2 2 3( ) …(ii)

Now, E is maximum or minimum if dE

da= 0

⇒ 6 3 02a a− =⇒ 3 2 0a a( )− =either a = 0

or a = 2

Again, d E

da

2

20< , where a = 2.

Hence, E will attain its maximum, if a = 2

∴ Maximum value = − =( ) ( )2 3 2 43

49. Here, A Bx

k x− =

−

−tan 1 3

2

− −

−tan 1 2

3

x k

k

= − + + −− + −

−tan( )

12 2

2 2

3 4 2 2

3 2 2

xk xk x k xh

k kx x xk

= + −+ −

−tan( )

12 2

2 2

2 2 2

3 2 2 2

k x xk

k x xk

=

=−tan 1 1

3 6

π

50. cos cosA C+

= +

⋅ −

22 2

cos cosA C A C

Q A B CA C B+ + = + = −

1802

902

,

= ⋅ −

22 2

sin cosB A C

= ⋅ ⋅ −

23

2 2cos

A C

0 60≤ + ≤ < °A C

⇒ 0 60< − < °A C

⇒ cosA C−

2

⇒ 3

21,

⇒ cos cosA C+ lies in 3

23,


F040

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