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1

SECTION - I

Single Correct Choice Type

This section contains 6 multiple choice questions. Each

question has 4 choices (A), (B), (C) and (D) for its an-

swer, out of which ONLY ONE is correct.

1. In the reaction

3H C C

C2NH

Cl

O

O

2(1) /NaOH Br

(2)

T

the structure of the Product T is

(a)

3H C CO C−

OO

(b) C

NH

O

3CH

(c)

3H C

C

NH

O

(d)

3H C CNH C−

OO

For IIT-JEE / AIEEE / MEDICAL ENTRANCE EXAMS

SOLUTIONS OF IIT-JEE 2010

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PAPER - II

[ CHEMISTRY]

R

JABALPUR : 1525, Wright town

NAGPUR : 5 E.H.C. Road, New Ramdaspeth & 24 Pragati Colony

GWALIOR : 21, Ravi Nagar, Near G.D.A.

1.Sol.(c)

3H C CO

2NH

21. /NaOH Br→

CO

Cl2.

Step I is Hofmann’s bromamide reaction ∴

3H C CO

2NH

21. /NaOH Br→3H C

2NH

Cl C− −

O

NPSR 3H C NH C− −

O

2. Assuming that Hund’s rule is violated, the bond

order and magnetic nature of the diatomic

molecule 2B is

(a) 1 and diamagnetic

(b) 0 and diamagnetic

(c) 1 and paramagnetic

(d) 0 and paramagnetic

2.Sol.(a) 2 * 2 2 * 2 2

2 1 1 2 2 2 yB s s s s pσ σ σ σ π→

B.O.[ ]6 4

12

−=

No unpaired electron is hence diamagnetic

3. The compounds ,P Q and S

HO 3H C

COOH 3OCH

P Q



2

C

O

O

S

were separately subjected to nitration using

3 2 4/HNO H SO mixture. The major product

formed in each case respectively, s

(a) HO 3H C

COOH 3OCH

2NO 2NO

2O N

C

O

O

(b) HO 3H C

COOH 3OCH

2NO

2NO

2NO

C

O

O

(c) HO 3H C

COOH 3OCH

2NO2NO

2NO

C

O

O

(d) HO 3H C

COOH 3OCH

2NO2NO

2NO

C

O

O

3.Sol.(c)

COOH

HO(P)

3

2 4

HNO

H SO

+→COOH

HO

2NO

Because OH− is more powerful group

3OCH

HO(Q)

3

2 4

HNO

H SO

+→3OCH

3H C 2NO

Because 3OCH− is more powerful group

CO

O

3

2 4

HNO

H SO

+→

S

CO

O 2NO

because O− is more powerful group & P-is

unoccupied.

Ans =c

4. The species having pyramidal shape is

(a) 3SO (b) 3BrF

(c) 2

3SiO −(d) 2OSF

4.Sol.(d) 2SOF

2 4 03 3 3S s p d→Ground state

S-

Exited state

S-

O FF



3

O

S

FF

5. The complex showing a spin-only magnetic

moment of 2.82 . .B M is

(a) 4( )Ni CO (b) [ ]24NiCl

−

(c) 3 4( )Ni PPh (d) [ ]24( )Ni CN

−

5.Sol..(b) ( )2n nµ = +

( )2.82 2n n= +

( )7.9 2n n= +

28 2n n= +2 2 8 0n n+ − =

2n = +

( )4

0Ni CO = ; ( )2

4 2Ni Cl−=

( )3 40Ni PPh = ; ( )

2

40Ni CN

− =

2 8 03 4Ni d s+ =

2unpaird

e-

xx x

Cl Cl Cl Cl

sp3

6. The packing efficiency of the two-dimensional

square unit cell shown below is :

L

(a) 39.27% (b) 68.02%

(c) 74.05% (d) 78.54%6. (d) Given diagram is a bcc arrangement

∴volume of two sphere =34

23

rπ×

and 3

4

ar =

3

2

42

3.4

23

r

P Fr

r

π× ×=

×

3

2

2 4 3

3 16 2

r

r r

π× ×=× ×

4

π=

3.14

4=

0.785=78.5%=

SECTION - II

Integer Answer Type

This section contains a group of 5 questions. The answer

to each of the questions is a single digit integer, ranging

from 0 to 9. The correct digit below the question no. in the

ORS is to be bubbled.

7. One mole of an ideal gas is taken from a to b

along two paths denoted y the solid and the

dashed lines as shown in he graph below. If the

work done along the solid line path is sw and

that along the dotted line path is dw , then the

integer closest to the ratio /d sw w is :

4.54.03.53.02.52.01.51.00.50.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.5 5.0 6.0

( .)V lit

P( .)atm

b



4

7.Sol. Since process is the isothermal because

1 1 2 2PV PV=

i.e of point “a” 1 1 4 0.5 2PV = × =

and at 2V lit= , 1P atm=

∴ 2 2 2 1 2PV = × =

2

1

2.30 logs

Vw RT

V

= −

5.52.303 log

0.5PV= − ×

2.303 2log11= − × ........(1)

4.7sw ≈

[ ]4 2.5 1 1 2.5 0.6dw = − × + × + ×

10= −

∴10

24.6

d

s

w

w

−= =−

Ans =2

8. Among the following, the number of elements

showing only one non-zero oxidation state is

, , , , , , , ,O Cl F N P Sn Tl Na Ti

8.Sol. F→Only -1 ∴ three elements

Na→Only+1

Tl→Only +1

Ans= 3

9. Silver (atomic weight 1108 g mol−= ) has a den-

sity of 310.5 g cm− . The number of silver atoms

on a surface of area 12 210 m− can be expressed

in scientific notation as 10xy× . The value of x

is

9 Sol. 31 10.5cm gm=

13 310.5 46 10

108 3rπ= × × ×

8 1010 10r cm m− −∴ = =Volume corresponding to 12 2 1210 10 2m r− −= ×

∴22

6 3

10.53 10

10

gmwt

m

−−= × ×

∴no. of atom=

22 23

6

2 10 10.5 6.023 10

108 10

−

−

× × × ××

71.17 10= ×∴Ans =7

10. Total number of geometrical isomers for the

complex 3 3[ ( )( )( )]RhCl CO PPh NH is

10.Sol. [ ]Mabcd type of complexes exist in 3 isomeric

form

3NH

CO

Rh

3PPh

Cl

3NH

CORh

3PPh

Cl

3NHCO

Rh

3PPhCl

Ans=3

11. The total number of diprotic acids among the

following is :

3 4H PO 2 4H SO 3 3H PO

2 3H CO 2 2 7H S O 3 3H BO

3 2H PO 2 4H CrO 2 3H SO

11.Sol 2 4 3 3 2 3 2 2 7, , ,H SO H PO H CO H S O

2 4 2 3,H CrO H SO

Ans= 6

SECTION - III

Paragraph Type

This section contains 2 paragraphs. Based upon each of

the paragraphs 3 multiple choice questions have to e

answered. Each question has 4 choices (A), (B), (C) and

(D) for its answer, out of which ONLY ONE is correct.

Paragraph for questions 12 to 14

Two aliphatic aldehydes P and Q react in the presence of

aqueous 2 3K CO to give compound R, which upon

treatment with HCN provides compound S. On acidification

and heating, S gives the product shown below :



5

3H C

3H C

OH

O O

12. The compounds P and Q respectively are

(a) 3H C C C

O O

CH

3CH

3andH H C H

(b) 3H C C C

O O

CH

3CH

andH H H

(c) CH C C

O3CH O

2CH3H C3andH H C H

(d) CH C C

O3CH O

2CH3H C andH H H

13. The compound R is

(a) C

O

2CH

CH

OH

3H C

3H C

(b) C

O

CH

CH

OH

3H C

3H C

3H C

(c)

3CH

2CH

O

CH CH

OH

CH3H C

(d)

3CH

CH

O

CH CH

OH

CH3H C

3H C

14. The compound S is

(a)

3CH

2CH

O

CH CH

CN

CH3H C

(b)

2CHCN

C3H C

3H CO

CH

(c)

3CH

2CH

CN

CH CHOH

OH

CH3H C

(d)

2CH

CN

CHOH

OH

C3H C

3H C



6

12 to 14 Sol.

C

3H C

3H C

CH

C N≡

OH2CHOH

S

HCN

Nu−←addition

C

3H C

3H C

C

H

O

2CHOH

R

2 3.Aq K CO←

C H C+ −

H

O

H3CH3H C

C

O

H

P Q

or

H C C− −

H

OH

3CH

3CH

C

O

H

3H C

3H C

C

2CH

CH

OH

C

O

OH

OH

Acidificationfollowed by

S

2H O

∆−→

3H C

3H C

OHO

Desired product

∴Ans ( ) ( ) ( )12 ;13 ;14b a d→ → →

Paragraph for questions 15 to 17

The hydrogen-like species 2Li + is in a spherically

symmetric state 1S with one radial node. Upon absorbing

light the ion undergoes transition to a state 2S . The state

2S has one radial node and its energy is equal to the

ground state energy of the hydrogen atom.

15. The state 1S is

(a) 1s (b) 2s

(c) 2p (d) 3s

16. Energy of the state 1S in units of the hydrogen

atom ground state energy is

(a) 0.75 (b) 1.50

(c) 2.25 (d) 4.50

17. The orbital angular momentum quantum number

of the state 2S is

(a) 0 (b) 1

(c) 2 (d) 3

15 - 17 Solution :

15.(b) Ground state energy of hydrogen atom

s 13.6eV= −

∴2

213.6 13.6

Z

n− = − ×

2

2

313.6

n= − ×

∴ 3n = ∴ 2 3S = hence 1 2S S<

Since 1S is spherically symmetric an one radial

node therefore it will be 2s .

16.(c) Energy of 2s = 1E of H-atom

2

2

Z

n×

1E of H-atom

2

2

3

2×

12.25 E= × of H-atom



7

17.(b) Since state 2S has also one radial node. Hence

it will be 3p .

∴ 1l =

SECTION - IV

Matrix - Match Type

This section contains 2 questions. Each question has

four statements (A,B,C and D) given in Column-I and five

statements (p, q, r, s and t) in Column-II. Any given

statement in Column-I can have correct matching with

ONE OR MORE statement(s) given in Column-II. For

example, if for a given questions, statement B matches

with the statements given in q and r, then for that particular

questions, against statement B, darken the bubbles

coresponding to q and r in the ORS.

18. Match the reactions in Column-I with appropriate

options in Column-II.

COLUMN - I

(A)

2N Cl + OH 2/NaOH H O00 C

N N= OH

(B)

3 3H C C C CH− − −

OH

3CH

OH

3CH

2 4H SO3H C

3CH3CH

3CHC

C

O

(C) C CH

3CH 3CH

O OH41.LiAlH

32.H O+

(D) HS ClBase S

COLUMN - II

(p) Racemic mixture

(q) Addition reaction

(r) Substitution reaction

(s) Coupling reaction

(t) Carbocation intermediate

18.Sol. (A)

OHElectrophilic

substitution←N N=

2N Cl+

+ OH

O & P-directing groupElectrophile

∴ In electrophilic substitution reaction

carbocation i.e. cycloarenium is formed

Ans = ,r s

(B) 3 3CH C C CH− − −

OHOH

3CH 3CH

2 4H SO→

Pinacol- pinacolone rearrangement in which first

we get carbocation

Ans= t

(C) C

O

3CH

4

3

1.

2.

LiAlH

H O+→

First is addition reaction & then we get optically

active compound ∴Racemic mixture

Ans= ,p q∴

(D) ClSH

NGP

EffectHCl→ + S

Ans= r∴ Ans=

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

19. All the compounds listed in Column-I react with

water. Match the result of the respective reactions

with the appropriate options listed in Column-II.

COLUMN - I COLUMN - II

(A) ( )3 22CH SiCl (p) Hydrogen halide

formation

(B) 4XeF (q) Redox reaction

(C) 2Cl (r) Reacts with glass

(D) 5VCl (s) Polymerization

(t) 2O formation



8

19.Sol.

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

3CH

3CH

SiCl

Cl

2H O→3 3CH Si CH− −

OH

OH

2HCl+

HO Si OH− −3CH

3CH

+HO Si OH− − →3CH

3CH

O Si O Si O− − − − − −(( )3CH

3CH 3CH

3CH

Polymersilicone

n

(B)

4 2 3 2

13 6 2 12 1

2XeF H O Xe XeO HF O+ → + + +

(C) 2 2Cl H O HCl HOCl+ → ++1-1

(Redox hydrogen halide)

2 2 3 2

13 2

2Cl H O H O Cl O+ −+ → + +

(Oxygen formation)

(D) ( )I V in five oxidation state on hydrolysis form

5 3

2 4V VO VO+ + −→ ++5 +5

No change in oxidation state when solution of

vanatate ion is concentrated they are polymerised

to form [ ]42 7V O−and higher vanadates i.e.

[ ]610 28V O

− [ ]412 32V O

−

Section - I

Single Correct Choice Type

20. Two adjacent sides of a parallelogram ABCD

are given by ˆˆ ˆ2 10 11AB i j k= + +

and

ˆˆ ˆ2 2AD i j k= − + +

The side AD is rotated by an acute angle α in

the plane of the parallelogram so that ADbecomes AD′ . If AD′ makes a right angle withthe side AB , then the cosine of the angle α is

given by

(a) 8

9(b)

17

9

(c) 1

9(d)

4 5

9

Sol.(b) 090 ADBα = −∠

A B

CDD'

α

( )0cos cos 90 ADBα = −∠

sin ADB= ∠

cosAB AD

ADBAB AD

⋅∠ =

408 / 9

15 3= =

⋅

( )2 17cos sin 1 8 / 9

9ADB∴ α = ∠ = − =

21. Let f be a real-valued function defined on the

interval ( )1, 1− such that

( ) 4

0

2 1

x

xe f x t dt− = + +∫ , for al l

( )1, 1x∈ − , and let 1f − be the inverse function

of f . Then ( )1f −′ (2) is equal to

PAPER - II

[ MATHEMATICS]



9

(a) 1 (b) 1

3

(c) 1

2(d)

1

e

Sol.(b) ( )( )1f f x x− =

( )( ) ( )( )1 1 1f f x f x− − ′′⇒ =

( )( )( )( )

1

1

12

2f

f f

−

−

′⇒ =′

Now, ( )0 2f =

( )( ) ( )1 1 12

0 3f

f

− ′⇒ = =′

22. For 0, 1,....., 10r = , let 2 , rA B and rC denote,

respectively, the coefficient of rx in the

expansions of ( ) ( )10 201 , 1x x+ + and

( )301 x+ . Then ( )10

10 10

1

r r r

r

A B B C A=

−∑ is

equal to

(a) 10 10B C−

(b) ( )2

10 10 10 10A B C A−

(c) 0 (d) 10 10C B−

Sol.(d)10 20 30, ,

r r r r r rA C B C C C= = =

( )10 10r r rA B B C A−∑

( )210 10

10 20 30 10

10 10

1 1

r r r

r r

B C C C C= =

= −∑ ∑ ...(1)

( ) ( )20 30 30 20

10 20 10 101 1C C C C= − − −

30 20

10 10C C= −

23. A signal which can be green or red with probability

4

5 and

1

5 respectivley, is received by station A

and then transmitted to station B . The probability

of each station receiving the signal correctly is

3

4. If the signal received at station B is green

then the probability that the original signal was

green is

(a) 3

5(b)

6

7

(c) 20

23(d)

9

20Sol.(c)

24. Let 1, 2,3, 4S − . The total number of unordered

pairs of disjoint subsets of S is equal to

(a) 25 (b) 34

(c) 42 (d) 41

Sol.(d)

25. If the distance of the point ( )1, 2,1P − from the

plane 2 2x y z α+ − = , where 0α > , is 5 ,

then the foot of the perpendicular from P to the

plane is

(a) 8 4 7, ,

3 3 3

−

(b) 4 4 1, ,

3 3 3

−

(c) 1 2 10, ,

3 3 3

(d) 2 1 5, ,

3 3 2

−

Sol.(a) The equation of r⊥ from P

5P on plane is

1 2 1

1 2 2

x y z− + −= =

−its distance form is

1 2 1

1/ 3 2 / 3 2 / 3

x y zd

− + −= = =

−

A point at distance d can be taken

2 21, 2, 1

3 3 3

d d d− + − +

When 5d = when 5d = −

8 4 7 2 16 13, , , ,

3 3 3 3 3 3

− − −



10

Section - II

Integer Type

26. Let 1 2 3 11, , , ...,a a a a be real numbers satisfying

1 215, 27 2 0,a a= − > and 1 22k k ka a a− −= −

for 3, 4,....,11k = .

If

2 2 2

1 2 11...90

11

a a a+ + += , then the value of

1 2 11...

11

a a a+ + + is equal to

Sol.(0) 1 2 3 11, , ..........a a a a are in A. P..

Let common difference ' 'd=

2 1 3 1 11 1, 2 .........., 10a a d a a d a a d= + = + = +

Given ( )10

2

1

0

90 11r

a rd=

+ = ×∑

10 10 102 2 2

1 1

0 0 0

1 2 990r r r

a d r a d r= = =

⇒ + + =∑ ∑ ∑

2 2

1 1

1 1111 10 11 21 2 10

6 2a d a d

⇒ + × × × + ×

990=2 2

1 135 10 90a d a d⇒ + + =

2225 35 150 90d d⇒ + + =235 150 135 0d d⇒ + + =

27 21 9 27 0d d d⇒ + + + =

( ) ( )7 3 9 3 0d d d⇒ + + + =

93,

7d

−⇒ = −

33

2d d

−< ⇒ = −∵

27. Two parallel chords of a circle of radius 2 are at a

distance 3 1+ apart. If the chords subtend at

the centre, angles of k

π and

2

k

π, where 0k > .

then the value of [ ]k is

[Note : [ ]k denotes the largest integer less than

or equal to k ]

Sol.(3) Angle subtended by a chord at the centre

12cosp

r

− θ =

O

A

B

cos2

p rθ

∴ =

Given 2 1 3 1, 2p p r± = + =

22 cos cos 3 1

2 2k k

π π + = +

2 3 1cos cos

2 2 2 2k k

π π + = +

3k = satisfies

3 1cos cos

6 3 2 2

π π+ = +

28. Let k be a positive real number and let

2 1 2 2

2 1 2

2 2 1

k k k

A k k

k k

−

= − − −

and

0 2 1

1 2 0 2

2 0

k k

B k k

k k

−

= − − −

If det ( ) ( ) 6det 10adj A adj B+ = , then [ ]k is

equal to

[Note : adj M denotes the adjoint of a square

matrix M and [ ]k denotes the largest integer

less than or equal to k ]

Sol.(4) ( ) ( ) 1det det

nadjA A

−=

Where n is order

( ) ( )2det ) detadjA A∴ =

( ) ( )2det det 0adjB B= =

B∵ is skew symmetric matrix of odd order



11

Given, ( ) ( ) 6det det 10adjA adjB+ =

( )2 6det 10A⇒ = .......(1)

Expanding det A , we get

( )3det 2 1A k= +

∴ ( )( )23 62 1 10k + =

9 / 2k⇒ =

[ ] 4k∴ =

29. Consider a triangle ABC and let ,a b and c

denote the lengths of the sides opposite to vertices

,A B and C respectively. Suppose

6, 10a b= = and the area of the triangle is

15 3 . If ACB∠ is obtuse and if r denotes the

radius of the incircle of the triangle, then 2r is

equal to

Sol.(3)

B

CAθ

6

10

Area 1

sin2ab C=

115 3 6 10 sin

2C= × × ×

3

2SinC =

0120C∠ =2 2 2

cos2

a b cC

ab

+ −=∵

2 2 21 6 10

2 2 6 10

c+ −− =

⋅ ⋅

14c =

15 3

15r

s

∆= = 2 3r⇒ =

30. Let f be a function defined on R (the set of all

real numbers) such that

( ) ( )2010 2009f x x′ = −

( ) ( )2 32010 2011x x− − ( )42012x − , for all

x R∈ .

If g is a function defined on R with values in the

interval ( )0, ∞ such that ( ) ( )( )f x n g x= ,

for all x R∈ , then the number of points in R at

which g has a local maximum is

Sol.(1)

Section - III

Paragraph Type

Paragraph -1

Tangents are drawn from the point ( )3, 4P to

the ellipse

2 2

19 4

x y+ = touching the ellipse at

points A and B .

31. The coordinates of A and B are

(a) ( )3,0 and ( )0,2

(b) 8 2 161,

5 15

−

and 9 8,

5 5

−

(c) 8 2 161,

5 15

−

and ( )0, 2

(d) ( )3,0 and 9 8,

5 5

−

Sol.(d)

32. The orthocenter of the triangle PAB is

(a) 8

5,7

(b) 7 25,

5 8

(c) 11 8

,5 5

(d) 8 7

,25 5

Sol.(c)

33. The equation of the locus of the point whose

distances from the point P and the line AB are



12

equal, is

(a) 2 29 6 54 62 241 0x y xy x y+ − − − + =

(b) 2 29 6 54 62 241 0x y xy x y+ + − + − =

(c) 2 29 9 6 54 62 241 0x y xy x y+ − − − − =

(d) 2 2 2 27 31 120 0x y xy x y+ − + + − =Sol.(a)

Paragraph -2

Consider the polynomial

( ) 2 31 2 3 4f x x x x= + + +

Let s be the sum of all distinct real roots of ( )f x

and let t s= .

34. The real number s lies in the interval

(a) 1,0

4

−

(b) 3

11,4

− −

(c) 3 1,

4 2

− −

(d) 1

0,4

Sol.(c)

35. The area bounded by the curve ( )y f x= and

the lines 0, 0x y= = and x t= , lies in the

interval

(a) 3,3

4

(b) 21 11

,64 16

(c) ( )9,10 (d) 21

0,64

Sol.(a)

36. The function ( )f x′ is

(a) increasing in 1

,4

t − −

and decreasing in

1,

4t

−

(b) decreasing in 1

,4

t − −

and increasing in

1,

4t

−

(c) increasing in ( ),t t−

(d) decreasing in ( ),t t−

Sol.(b)

Section- IV

Matrix -Match Type

37. Match the statements in Column-I with those in

Column - II.

Column - I

(A) A l ine f rom the origin meets the l ines

2 1 1

1 2 1

x y z− − += =−

8

3 13

2 1 1

xy z

− + −= =−

at P and Q

respectively. If length PQ d= , then 2d is

(B) The values of x satisfying

( ) ( )1 1tan 3 tan 3x x− −+ − −1 3

sin5

− =

are

(C) None-zero vectors ,a b and c

satisfy

0a b⋅ =

, ( ) ( ) 0b a b c− ⋅ + =

and

2 b c b a+ = −

.

If 4a b cµ= +

, then the possible values of µare

(D) Let f be the function on [ ],π π− given by

( )0 9f = and ( )

9sin

2

sin2

x

f xx

=

for 0x ≠ .

The value of ( )2f x dx

π

ππ −∫ is

Column - II

(p) 4−(q) 0

(r) 4



13

(s) 5

(t) 637.Sol. (A) Let the equation of line be

x y z

a b c= = ......(1)

Given lines are

2 1 1

1 2 1

x y z− − += =−

.....(2)

& 8 / 3 3 1

2 1 1

x y z− + −= =−

......(3)

Equations (1) & (2) are coplaner

2 1 1

0

1 2 1

a b c

−

=

−

3 5 0a b c⇒ + + = ...(4)Equations (1) & (3) are also coplaner

8 / 3 3 1

0

2 1 1

a b c

−

=

−

⇒ 3 5 0a b c+ − = ......(5)

Solving (4) & (5) by cross multiplication

20 20 8

a b c⇒ = =− −

( ) ( ), , 5, 5, 2a b c⇒ = −

Let a point on this line be ( )5 , 5 ,2k k k−If it is on (2)

5 2 5 1 2 11

1 2 1

k k kk

− − − += = ⇒ =

−

( )5, 5,2P∴ −If it is on (3)

5 3 2 13 2

1 1

k kk

− + −= ⇒ =

−

10 10 4, ,

3 3 3Q

− ∴

distance

2 2 2

2 10 10 45 5 2

3 3 3PQ

= − + − + + −

50 46

9 9= + =

( ) ( )A t∴ →

38.(B) ( ) ( )1 1 1 3tan 3 tan 3 sin5

x x− − −+ − − =

( ) ( )( ) ( )

1 13 3 3

tan tan1 3 3 4

x x

x x

− − + − − = + + −

2

2

6 316 4, 4

8 4x x

x⇒ = ⇒ = ⇒ = −

−

( ) ( ) ( ),B p r∴ →

38.(C) 0a b⋅ =

( ) ( ) 0b a b c− ⋅ + =

2 0b a b b c a c⇒ − ⋅ + ⋅ − ⋅ =2 0b b c a c⇒ + ⋅ − ⋅ = ...(1)

Also, ( ) ( )2 222 b c b a+ = −

( )2 2 2 24 2 2b c b c b a a b⇒ + + ⋅ = + − ⋅

2 2 23 4 8 0b c a b c⇒ + − + ⋅ = ......(2)

If 4a b c= µ + then

2 2 2 216 8a b c b c= µ + + µ ⋅24a c b c c⋅ = µ ⋅ +

2 4 0a b b b c⋅ = µ + ⋅ =2 24 0b b c b c c+ ⋅ −µ ⋅ − =

(A) → t (B)→p,r (C)→q (D)→ r

38. Match the statements in Column-I with those in

Column - II.

[Note : Here z takes values in the complex plane

and Im z and Re z denote, respectively, theimaginary part and the real part of z .]

Column - I

(A) the set of points z satisfying

z i z z i z− = + is contained in or equal to

(B) The set of points z satisfying

4 4 10z z+ + − = is contained in or equal to

(C) If 2w = , then the set of points 1

z ww

= − , is



14

PAPER - II

[ PHYSICS]QUESTIONS WITH ONLY ONE CORRECT ANSWERS

39. A tiny spherical oil drop carrying a net charge q

is balanced in still air with a vertical uniform

electric field of strength 5 181

107

Vmπ −× . When

the field is switched off, the drop is observed to

fall with terminal velocity 3 12 10 ms− −× . Given

29.8g ms−= , v iscosity of the air

5 21.8 10 Nsm− −= × and the density of oi l

3900mgm−= , the magnitude of q is :

(a) 191.6 10 C−× (b) 193.2 10 C−×

(c) 194.8 10 C−× (d) 198.0 10 C−×Sol. d)

qE mg= .......(1)

346 .

3rV mg r d gπη π= = .......(2)

6 rV qEπη⇒ =

5 3 581

6 1.8 10 2 10 107

r qπ

π − − −⇒ × × × × × = × ×

22q V r dgη =

3 52 109 2 10 1.8 10 18

102 900 9.8 98

r− −

−× × × ×⇒ = = ×

× ×

109

1079

−= ×

53

107

r m−⇒ = ×

9 6Now E rVπη=

5 53

5

6 1.8 10 3 102 10

81 710

7

qππ

− −−× × ×

= × × ××

186 1.8 3 210

81

−× × ×= ×

contained in or equal to

(D) If 1w = , then the set of points 1

z ww

= + is

contained in or equal to

Column - II

(p) an ellipse with eccentricity 4

5

(q) the set of points z satisfying Im 0z =

(r) the set of points z satisfying Im 1z ≤

(s) the set of points z satisfying Re 2z ≤

(t) the set of points z satisfying 3z ≤

38.Sol.

(A). ( )z i z z i z z i z− = + = − −

' 'z⇒ will lie onr⊥ bisector of line joining

&i z i z−

' 'z⇒ will lie on Real axis.

⇒ 0Im z⇒ =

(B). 4 4 10z z+ + − =

' 'z will lie on an ellipse with

2 10a = & 2 8ae = 4 / 5e⇒ =(C) Let 2 iw e θ=

1 12 2

2 2

ii i

i

ez w e e

w e

− θθ θ

θ= − = − = −

3 5cos sin

2 2i= θ+ θ

3 5Re & Im & 5

2 2z z z≤ ≤ ≤

(D) 1ω = ⇒ω= ω

1z = ω+ = ω+ω

ω

Im 0 & Re 1z z⇒ = ≤

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



15

18 18 194 1.810 0.8 10 8 10

9

− − −×= × = × = ×

40. A biconvex lens of focal length 15 cm is in front

of a plane mirror. The distance between the lens

and the mirror is 10 cm. A small object is kept at

a distance of 30 cm from the lens. The final image

is :

(a) virtual and at a distance of 16 cm from the

mirror

(b) real and at a distance of 16 cm from the mirror

(c) virtual and at a distance of 20 cm from the

mirror

(d) real and at a distance of 20 cm from the mirror

Sol. (b)

30 cm

F = 15 cm

O

10 cm

Lens

1 1 1

30 15v+ =

1 1 1 1

15 30 30v= − =

30v cm=∴ virton object for mirror. Image will be at 20

cm from mirror toward left. i.e. virtnd object for

lens at 10 cm

1 1 1

10 15v− =

1 1 1 3 2

10 15 30v

+= + =

6v cm⇒ =41. A Vernier calipers has 1 mm marks on the main

scale. It has 20 equal divisions on the Vernier

scale which match with 16 main scale divisions.

For this Vernier calipers, the least count is :

(a) 0.02 mm (b) 0.05 mm

(c) 0.1 mm (d) 0.2 mm

Sol. (d)

1 1LC MSD VSD= −

161

20mm mm= −

1 0.8 0.2mm= − =42. A uniformly charged thin spherical shell of radius

R carries uniform surface charge density of σper unit area. It is made of two hemispherical

shells, held together by pressing them with force

F (see figure). F is proportional to

F

F

(a) 2 21

o

Rσε (b)

21

o

Rσε

(c)

21

o R

σε (d)

2

2

1

o R

σε

Sol. (a)

2 2 2 2

1 2

2 2

0 0 0

( 4 )QQ R RF

R R

σ π σα = =∈ ∈ ∈

[check dimensions]

43. A hollow pipe of length 0.8 m is closed at one

end. At its open end a 0.5 m long uniform string

is vibrating in its second harmonic and it

resonates with the fundamental frequency of the

pipe. If the tension in the wire is 50 N and the

speed of sound is 320 1ms− , the mass of the

string is :

(a) 5 grams (b) 10 grams

(c) 20 grams (d) 40 grams

Sol. (b)

320 2 50

4 0.8 2 0.5 µ=

× ×

5050

µ⇒ =

1/

50kg mµ⇒ =

∴ mass of string 1

0.5 10050

gm= × ×



16

10 gm=44. A block of mass 2 kg is free to move along the x-

axis. It is at rest and from t = 0 onwards it is

subjected to a time-dependent force F(t) in the x

direction. The force F(t) varies with t as shown in

the figure. The kinetic energy of the block after

4.5 seconds is :

F t( ) N4

O

s3

s4.5

t

(a) 4.50 J (b) 7.50 J

(c) 5.06 J (d) 14.06 J

Sol. (c)1 1

3 4 1.5 22 2

P∆ = × × − × ×

6 1.5 4.5 /kg m s= − =

4.5 /P kg m s∴ =

2 4.5 4.55.06

2 2 2

PK J

m

×= = =

×Intege type

QUESTIONS WITH INTEGER TYPE

45. At time t = 0, a battery of 10 V is connected across

points A and B in the given circuit. If the capacitors

have no charge initially, at what time (in seconds)

does the voltage across them become 4 V ?

[Take : 5 1.6, 3 1.1n n= = ]

2µF

M2 Ω 2µF

M2 Ω

A

B

Sol. (2)

4 Fπ1mΩ

/

0 (1 )tq q e τ−= −

/

0 (1 )tv v e τ−⇒ = −

/ 44 10 1 te− ⇒ = − / 4

0.6t

e−⇒ =

ln 5 ln34

t⇒ = − 0.5=

2t =46. A diatomic ideal gas is compressed adiabatically to

1/32 of its initial volume. In the initial temperature of

the gas is iT (in Kelvin) and the final temperature is

iaT , the value of a is :

Sol. (4)1 1

1 1 2 2T V T Vγ γ− −=

2

5

1 2

1

32T T

⇒ =

2 14T T⇒ =

47. A large glass slab ( 5 / 3)µ = of thickness 8 cm is

placed over a point source of light on a plane

surface. It is seen that light emerges out of the top

surface of the slab from a circular area of radius R

cm. What is the value of R ?

Sol. (6)

R

8 cm c

c

1 3sin

5c

µ= =

3tan

4c⇒ =

3

8 4

R⇒ =

6R cm⇒ =48. To determine the half life of a radioactive element, a

student plots a graph of ( )dN t

ndt

versus t. Here

( )dN t

dt is the rate of radioactive decay at time t. If

the number of radioactive nuclei of this element



17

decreases by a factor of p after 4.16 years, the value

of p is :

3 2

4

5

6

1

2

3

4

5

Years

6

7

8

( )dN tn

dtl

Sol. (8)

0 0

t tdNN N e N e

dt

λ λλ− −= ⇒ − =

0ln ln ( )dN

N tdt

λ λ⇒ = −

Slope of the graph is

111/ 2

2yrλ λ −− = − ⇒ =

14.16

2.082

0

tNe e e

N

λ − ×− −∴ = = =

2.08 8P e∴ = =49. Image of an object approaching a convex mirror of

radius of curvature 20 m along its optical axis is

observed to move from 25

3 m to

50

7 m in 30

seconds. What is the speed of the object in km per

hour ?

Sol. (3)

For 25

, 103

v m f= = +

1 1 1

v u f+ =

3 1 1

25 10u+ =

1 1 3 25 30 5 1

10 25 10 25 10 25 50u

− −− − = = = −

× ×

1 50u⇒ = −Simillarly,

1 1 7 7 50 70 20 1

10 50 50 10 50 10 50 25u

− −= − − = = = −

× ×

2 25u⇒ = −

Object distance 25m in 30 sec80 speed of object

will be 25 18

3 /30 5

km h× =

QUESTIONS WITH COMPREHENSION TYPE

Paragraph For Questions 50 to 52

The key feature of Bohr’s theory of spectrum of

hydrogen atom is the quantization of angular

momentum when an electron is revolving around

a proton. We will extend this to a general

rotational motion to find quantized rotational

energy of a diatomic molecule assuming it to be

rigid. The rule to be applied is Bohr’s quantization

condition.

50. A diatomic molecule has moment of inertia I .

By Bohr’s quantization condition its rotational

energy in the nth level ( n = 0 is not allowed) is

(a)

2

2 2

1

8

h

n Iπ

(b)

2

2

1

8

h

n Iπ

(c)

2

28

hn

Iπ

(d)

22

28

hn

Iπ

Sol. (d)

2

hI nω

π=

2

nh

Iω

π=

2 2 2 22

2 2 2

1 1

2 2 2 8

n h n hI I

I Iω

π π∴= = =

51. It is found that the excitation frequency from

ground to the first excited state of rotation for the

CO molecule is close to 114

10 Hzπ× . Then the

moment of inertia of CO molecule about its center

of mass is close to (Take 342 10h J sπ −= × )

(a) 46 22.76 10 kg m−× (b) 46 21.87 10 kg m−×

(c) 47 24.67 10 kg m−× (d) 47 21.17 10 kg m−×



18

Sol. (b)

22 2

2(2 1 )

8

hhv

Iπ= −

34

2 11

3 3 2 10

488 10

hI

v

ππ π

π

−× ×⇒ = =

× × ×

45453 10

0.187 1016

−−×

= = ×

46 21.87 10 kgm−= ×46 21.87 10 kgm−= ×

52. In a CO molecule, the distance between C (mass

= 12 a.m.u.) and O (mass = 16 a.m.u.), where 1

a.m.u. 275

103

kg−= × , is close to :

(a) 102.4 10 m−× (b) 101.9 10 m−×

(c) 101.3 10 m−× (d) 104.4 10 m−×Sol. (c)

C.o.M.CO

2r 1r

1

12 3

28 7r r r= =

2

16 7

28 4r r r= =

2 2

0 1 2cm r m r I+ =

2 2

2 27 463 4 516 12 10 1.87 10

7 7 3r − −

× + × × × = ×

192 1.87 49 3 10

5 336r

−× × ×⇒ =

×

19 200.163 10 1.63 10− −= × = ×101.28 10r −⇒ = ×

Paragraph For Questions 53 to 55

When liquid medicine of density ρ is to be put

in the eye, it is done with the help of a dropper.

As the bulb on the top of the dropper is pressed,

a drop forms at the opening of the dropper. We

wish to estimate the size of the drop. We first

assume that the drop formed at the opening is

spherical because that requires a minimum

increase in its surface energy. To determine the

size, we calculate the net vertical force due to

the surface tension T when the radius of the drop

is R. When this force becomes smaller than the

weight of the drop, the drop gets detached from

the dropper.

53. If the radius of the opening of the dropper is r, the

vertical force due to the surface tension on the

drop of radius R (assuming r<<R) is :

(a) 2 rTπ (b) 2 RTπ

(c)

22 r T

R

π(d)

22 R T

r

π

Sol. (c)

θR

r

Fr

Net vertically upward force

222

r r TrT

R R

ππ ⇒ =

54. If 45 10r m−= × , 3 310 kgmρ −= , 210g ms−=10.11T Nm−= , the radius of the drop when it

detaches from the dropper is approximately

(a) 31.4 10 m−× (b) 33.3 10 m−×

(c) 32.0 10 m−× (d) 34.1 10 m−×

Sol. (a)

232 4

.3

r TR g

R

ππ ρ=

2 84

3

3 3 25 10 0.11

2 2 10 10

r TR

gρ

−× × ×= =

× ×

31.42 10R m−⇒ ×



19

55. After the drop detaches, its surface energy is

(a) 61.4 10 J−× (b) 62.7 10 J−×

(c) 65.4 10 J−× (d) 68.1 10 J−×Sol. (b)

2 3 24 4 3.14 (1.42 10 ) 0.11E R Tπ −= = × × × ×

62.7 10 J−= ×

QUESTIONS WITH MATRIX MATCH TYPE

56. You are given may resistances, capacitors and

inductors. These are connected to a variable DC

voltage source (the first two circuits) or an AC

voltage source of 50 Hz frequency (the next three

circuits) in different ways as shown in Column

II. When a current I (steady state for DC or rms

for AC) flows through the circuit, the corresponding

voltage 1V and 2V . (indicated in circuits) are

related as shown in Column I. Match the two

Column I Column II

(A) 10,I V≠ (p) V1 V2 mH6

3µF V

is proportional to I

(B) 2 10,I V V≠ > (q)

V1 V2 mH6

2Ω V

(C) 1 20,V V V= = (r)

V1 V2 mH6

2Ω V

~

(D) 20,I V≠ (s)

V1 V2 mH6

V

~

3 Fµ is proportional to I

(t)

V1 V2

V

~

3 Fµ

k1 Ω

Sol.

(a) → r,s,t

(b) →q,r,s,t

(c) →p,q

(d) →q,r,s,t

57. Two transparent media of refractive indices 1µ

and 3µ have a solid lens shaped transparent

material of refractive index 2µ between them as

shown in figures in Column II. A ray traversing

these media is also shown in the figures. In

Column I different relationships between 1µ , 2µ

and 3µ are given. Match them to the ray

diagrams shown in Column II.

Column I Column II

(A) 1 2µ µ< (p)

µ1

µ2

µ3

(B) 1 2µ µ> (q)

µ1

µ2

µ3

(C) 2 3µ µ= (r)

µ1

µ2

µ3

(D) 2 3µ µ> (s)

µ1

µ2

µ3

(t)

µ1

µ2

µ3

Sol.

(a) →p,r,,

(b) →q,s,t

(c) →p,r,t

(d) →q,s

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