2010 sri chaitanyas_iitjee_paper-ii_solutions

Sri Chaitanya IIT Academy.A.P. Ø1× :: IIT JEE 2010 Paper - II ::

Sri Chaitanya IIT Academy, Plot No 875, Ayyappa Society, Madhapur, Hyderabad - 500081

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Sri Chaitanya IIT Academy

Sri Chaitanya�s Solutions to

IIT � JEE - 2010(PAPER � 2)

Time: 3 Hours Maximum Marks: 237

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose

INSTRUCTIONS

A. General :

1. This question Paper contains 32 pages having 57 questions

2. The question paper CODE is printed on the right hand top corner of this sheet and also on the

back page (page no 32 of this booklet)

3. No additional sheets will be provided for rough work

4. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and

electronic gadgets in any form are not allowed

5. Log and Antilog tables are given in the page numbers 30 and 31 respectively.

6. The answer sheet, a machine - gradable Objective Response Sheet (ORS), is provided

separately.

7. Do not Tamper/multilate the ORS or this booklet

8. Do not break the seals of the question paper booklet before instrucuted to do so by the

invigilators.

B. Filling the bootom half of the ORS :

9. The ORS has CODE printed on its lower and upper Parts.

10. Make sure the CODE on the ORS is the same as that on this booklet. If the Codes do not

match, ask for a change of the Booklet.

11. Write your Registration No., Name and Name of centre and sign with pen in appropriate

boxes. Do not write these anywhere else.

12. Darken the appropriate bubbles below your registration number with HB pencil.

C. Question paper format and marking scheme :

13. The question paper consists of 3 Parts (Chemistry, Mathematics and Physics), and each part

consists of four Sections

14. For each question in Section I, you will be awarded 5 marks if you have darkened only the

bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In

all other cases, minus two (�2) mark will be awarded.

15. For each question in Section II, you will be awarded 3 marks if you darken the bubble corre

sponding to the correct answer and zero mark if no bubble is darkened. No negative marks

will be awarded for incorrect answers in this section

16. For each question in Section III, you will be awarded 3 Marks if you darken only the bubble

corresponding to the correct answer and zero mark if no bubbles are darkened. In all other

cases, minus one(�1) mark will be awarded.

17. For each question in Section IV, you will be awarded 2 marks for each row in whihch you

have darkened the bubble(s) corresponding to the correct answer. Thus, each question in

this section carries a maximum of 8 marks.

There is no negative marks awarded for incorrect answer(s) in this Section





IIT JEE 2010 (PAPER - II, CODE :: 00)

PART - I , CHEMISTRY

PART - I :: CHEMISTRY

SECTION - I (Single Correct Choice Type)

This section contains 6 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY

ONE is correct.

1. The species having pyramidal shape is

A) SO3

B) BrF3

C) SiO3

2� D) OSF2

Ans : D

Sol : O F

S

F

Hybridisation sp3

pyramidal shape due to the presence of 1 lone pair on sulphur

2. The compounds P, Q and S

COOH

HOP

OCH3

H3CQ

C

S

O

O

were separately subjected to nitration using HNO3/H

2SO

4 mixture. The major

product formed in each case respectively, is

A)

COOH

HONO2

OCH3

H3CNO2

C

O2NNO2

O

B)

COOH

HO NO2

OCH3

H3C NO2

C

NO2

O

O





C)

COOH

HONO2

OCH3

H3C NO2

C

NO2

O

O NO2

D)

COOH

HONO2

OCH3

H3C NO2

CO

O

NO2

Ans : C

Sol : E(+) attacks w.r.t more activating group in AES mechanism

3. Assuming that Hund�s rule is violated, the bond order and magnetic nature of thediatomic molecule B

2 is

A) 1 and diamagnetic B) 0 and diamagnetic

C) 1 and paramagnetic D) 0 and paramagnetic

Ans : A

Hint: As per MOED

4. 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%

Ans : D

Sol : .v

V FV

=

= 2

2

2 rLπ

, but r = 2 2

L as per diagram

= 4π

= 0.7854





5. In the reaction T, the structure of the Product T is

A) B)

C) D)

Ans : C

Sol : Which is Hoffmann bromamide reaction followed by benzoylation

6. The complex showing a spin - only magnetic moment of 2.82 B.M. is

A) Ni(CO)4

B) [NiCl4]2� C) Ni(PPh

3)

4D) [Ni(CN)

4]2�

Ans : B

Sol : Cl� is weak ligand

hence Ni2+ configuration in [NiCl4]2� is

[Ar]

∴No.of unpair electrons = 2

( )2n nµ = +

( )2 2 2∴ + = 2.82 BM





SECTION - II (Integer Type)This Section contains a group 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 by the solidand the dashed lines as shown in the graph below. If the work done along the solidline path is w

s and that along the dotted line path is w

d, then the integer closest to

the ratio wd/w

s is

Ans : 2

Sol : Wd = Area under the dotted lines in the PV graph

= ( ) ( ) ( )4 1.5 1 1 2.5 0.75× + × + ×

= 8.875 lit atm

= 900 J

Ws = 2.303 n RT log

2

1

V

V

since. It is isothermal expansion

where T = 24.36 K as per PV = nRT equation

= 2.303 × 1 × 8.314 × 24. 36 log 5.50.5

= 470 J

1.91d

s

W

W∴ =

= 2 (nearest integer)





8. Among the following, the number of elements showing only one non-zero oxida-tion state is

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

Ans : 2

Sol : Na, F

9. Silver (atomic weight = 108 g mol�1) has a density of 10.5 g cm�3. The number ofsilver atoms on a surface of area 10�12 m2 can be expressed in scientific notation

as 10xy × . The value of x is

Ans : 7

Sol : Silver crystallises in CCP arrangement

As per density value, r3 = 243 10 CC−×

r = 81.45 10 cm−×

The area of one silver atom in circle form is ( 2rπ ) = 20 26.6 10 m−×

Hence the number of silver atoms in 12 2 710 1.5 10m− = ×

10. The total number of diprotic acids among the following is

H3PO

4H

2SO

4H

3PO

3H

2CO

3H

2S

2O

7

H3BO

3H

3PO

2H

2CrO

4H

2SO

3

Ans : 6

Sol : H

2SO

4H

3PO

3H

2CO

3H

2S

2O

7

H2CrO

4H

2SO

3

11. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3) (NH

3)] is

Ans : 3

Sol : Conceptual





SECTION - III (Paragraph Type)

This section contains 2 paragraphs. Based upon each of the paragraphs 3 mul-tiple choice questions have to be answered. Each of these questions has fourchoices A), B), C) and D) 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 K2CO

3 to give

compound R, which upon treatment with HCN provides compound S. Onadicification and heating, S gives the product shown below :

12. The compound P and Q respectively are

A) B)

C) D)

Ans : B

13. The compound R is

A) B)

C) D)

Ans : A





14. The compound S is

A) B)

C) D)

Ans : D

Solutions for 12 -14

CH3 -CH - C - H

O

CH3

(P)

+ HCHO

(Q)

K2CO3 CH3 -C - CHO

CH3

CH2OH (R)

HCN←CH3 -C - CH-CN

CH3

CH2OH (S)

OH


The hydrogen like species Li2+ is in a spherically symmetric state S1 with one ra-

dial node. Upon absorbing light the ion undergoes transition to a state S2 The

state S2 has one radial node and its energy is equal to the ground state energy of

the hydrogen atom.

15. The state S1 is

A) 1s B) 2s C) 2p D) 3s

Ans : B

Sol : S1 state has 2s, 2p orbitals, but one radial node is only for 2s





16. Energy of the state S1 in units of the hydrogen atom ground state energy is

A) 0.75 B) 1.50 C) 2.25 D) 4.50

Ans : C

Sol : Ground state H � atoms (E1) S

1 state of Li2+ (E

2)

energy : �13.6 eV 2

2

13.6 32

×− eV

2

1

2.25E

E∴ =

17. The orbital angular momentum quantum number of the state S2 is

A) 0 B) 1 C) 2 D) 3

Ans : B

Sol : S2 state is 3p

For p � subshell, l = 1

SECTION - IV (Matrix 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 question, statement B matches with the statements given in

q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS

18. Column - I Column - II

Ans : A - r,s,t B - t C - p,q D - r





19. All the compounds listed in Column I react with water. Match the result of therespective reactions with the appropriate options listed in column II

Column - I Column - II

A) (CH3)

2 SiCl

2p) Hydrogen halide formation

B) XeF4

q) Redox reaction

C) Cl2

r) Reacts with glass

D) VCl5

s) Polymerization

t) O2 formation

Ans : A - p,s B - p,q,r,t C � p,q D - p

(Note VCl5 does not exist According to J.D.Lee Book)





IIT JEE 2010 (PAPER - II)

PART - II , MATHEMATICS

PART - II :: MATHS



ONE is correct.

20. Let S={1,2,3,4). The total numbe of unordered pairs of disjoint subsets of S isequal to

A) 25 B) 34 C) 42 D) 41

Ans : D

Sol : A B A B∩ = ∅

1) null set any set 24 16

2) single ton single ton2

4 6C = 6

3) Single ton 3 element set 4 4

4) 2 element set 2 element 3 3

5) single ton -2 element 4 3× 12

Total 41

21. If the distance of the point P(1, -2, 1) from the plane α+ − =2 2x y z , where α > 0 ,

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

Ans : A

Sol : Given point is ( )1, 2,1−

given plane 2 2x y z α+ − =

r⊥ distance | 5 |

53

α+= =

( )10 0α α⇒ = >

foot of the r⊥ from (1, -2, 1) to x + 2y - 2z = 10

is 1 2 1 5

1 2 2 3

h k l− + −= = =− ; ( ) 8 4 7

, , , ,3 3 3

h k l− ⇒ =





22. A single which can be green or red with probability 4 15 5

and respectively, is re-

ceived by station A and then transmitted to station B. The probability of each

station receiving the signal correctly is 3

.4

If the single received at station B is

green, then the probability that the original single was green is

A) 35

B) 67

C) 2023

D) 920

Ans : C

Sol : 4 1 3 1

( ) , ( ) , ( ) , ( )5 5 4 4

P G P R P C P C= = = =

given

i) 1 3 1 3

. .5 4 4 80

BAR R G→ → =

ii) 1 1 3 3

. .5 4 4 80

R G G→ → =

iii) 4 1 1 4

, ,5 5 4 80

G R G→ → =

iv) 4 3 3 36

, ,5 4 4 80

G G G→ → =

Req. probability by Baye�s theorem

( ) ( ) 4 36 20

( ) ( ) ( ) ( ) 3 13 4 36 23

i ii iv

i ii iii iv

+ += =+ + + + + +

23. Two adjacent sides of a parallelogram ABCD are given by

= + + = + +JJJG JJJJGˆ ˆˆ ˆ ˆ ˆ2 10 11 2 2AB i j k and AD i j k . The side AD is rotated by an acute angle αin the plane of the parallelogram so that AD becomes AD�. If AD� makes a rightangle with the side AB, then the cosine of the angle by

A) 89

B) 179

C) 19

D) 4 5

9

Ans : B

Sol : 2 20 22 8

cos3 15 9

β − + += =×

2

πα β+ = , 2

πα β= − , 17

cos sin9

α β= =





24. 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 all ∈ −( 1,1)x , and let f-1 be the inverse function of f.

Then ( )−1 ’f (2) is equal to

A) 1 B) 13

C) 12

D) 1e

Ans : B

Sol : ( )1 4( ) ( ) 1xe f x f x x− − = +

10 (0) (0) 1x f f= ⇒ − = ; 1(0) 3f⇒ =

( ) ( )( )1 1 1 1( ) ( ) ( ) 1f f x x f f x f x− − −= ⇒ =

( ) ( )111

10

(0)f f

f−⇒ =

( )11 1(2)

3f −⇒ = .

25. For r = 0, 1, .....10, let ,r r rA B andC denote, respectively, the coefficient of rx in the

expansions of ( ) ( )+ +20 301 1 .x and x Then =

−∑10

10 101

( )r r rr

A B B C A is equal to

A) −10 10B C B) ( )−210 10 10 10A B C A C) 0 D) −10 10C B

Ans : D

Sol : 10 , 20 , 30r r rr C C CA B C= = =

( ) ( )10 10

10 10

10 101 1

. 10 20 .20 30 .10r r rr r r C C C C C

r r

A B B C A= =

− = −∑ ∑

10 10

10 10

1 1

20 10 .20 30 10 .10r r r rC C C C C C

r r= =

= −∑ ∑

( ) ( )10 10 10 10 10 10 10 1020 30 1 30 20 1 20 30C C C C C C C B= − − − = − + = −







26. Let k be a positive real number and let

− −

= − = − − − − −

2 1 2 2 0 2 1

2 1 2 1 2 0 2 .

2 2 1 2 0

k k k k k

A k k and B k k

k k k k If det (adj A) + det (adj B) =

106, then [k] is equal to

[Note:- adj M denotes the adjoint of a square matrix M and [k] denotes thelartgest integer less than or equal to k].

Ans : 4

Sol :

2 1 2 2

| | 2 1 2

2 2 1

k k k

A k k

k k

−

= −

− −

2 1 2 2

2 2 1

2 1 2

k k k

k k

k k

−

= − −

− −

2 1 2 2 2 2 2 1 2 2

2 2 1 2 2 1 0 2 1

0 1 22 1 2 2 1 2

k k k k k k k k

k k k k k

kk k k k

− −= − = − + −

−− −

( )( )2

1 2 2

2 . 1 2 1 1 4 1

1 1 2

k k k k

k

− + − −−

( )2 22 4 8 3 4 1k k k k= + + − +

det |adj A| = 106

2 6 3 3| | 10 | | 10 (2 1) 10A A k3= ⇒ = + =

9[ ] 4

2k k⇒ = ⇒ =





27. Let f be a function defined on R (the set of all real numbers) such that

= − − − −2 3 4’( ) 2010( 2009)( 2010) ( 2011) ( 2012) ,f x x x x x for all ∈ .x R If g is a function

defined on R with values in the interval ∞(0, ) such that = A( ) ( ( )),f x n g x for all

∈ ,x R then the number of points in R at which g has a local maximum is

Ans : 1

Sol : ( ) ( )( ) log ( ) ( ) f xf x g x g x e= ⇒ =

1( ) 0g x = ; ( ) 1. ( ) 0f xe f x =

( )g x has a local maximum at x = 2009.

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

1 2 1 215,27 2 0 2 3, 4,....,11.k k ka a a a a for k− −= − > = − = If 2 2 21 2 11.....

90,11

a a a+ + + = then the

value of 1 2 11....

11

a a a+ + + is equal to

Ans : 0

Sol : Clearly 3

2d < −

2 2 2 21 1 2 1

1

( ) ( 2 ) ....( 10 )90 15

11

a a d a d a dand a

+ + + + + + = =

27 30 27 0 3d d d⇒ + + = ⇒ = − since 3

2d < −

1 2 3 11.....0

11

a a a a+ + + +∴ =

29. Two parallel chords of circle of radius 2 are at a distance 3 1+ apart. If the

chords subtend at the center, angles of 2

,andk k

π π where k > 0, then the value of

[k] is

[Note:- [k] denotes the largest integer less than or equal to k]





Ans : 6

Sol : Total distance 2cos 2cos 3 12k k

π π+ = +

33

kk

π π= ⇒ = .

30. Consider a triangle ABC and let a, b and c denote the lengths of the sides oppo-site to vertices A, B and C respectively. Suppose a = 6, b = 10 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

Ans : 3

Sol : .r s∆ =

236 100 1cos

2.6.10 2

cC

+ − −= =

2 136 60c⇒ − =

2 196 14c c= =

6 10 1415

2S

+ += = ; 15 3 .15 3r R⇒ = ⇒ = ; 2 3R =






Consider the polynomial 2 3( ) 1 2 3 4 .f x x x x= + + + Let s be the sum of all distinct

real roots of f(x) and let t = |s|.

31. The real number s lies in the interval

A) 1

,04

− B)

311,

4 − −

C) 3 1

,4 2

− − D)

10,

4

32. The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies inthe interval

A) 3

,34

B) 21 11

,64 16

C) (9,10) D) 21

0,64

33. 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

,4

t −

C) increasing in ( , )t t− D) decreasing in ( , )t t−

Ans : (31 TO 33) C,A, B

Sol : 3 2 1 2( ) 4 3 2 1, ( ) 12 6 2f x x x x f x x x= + + + = + + has imaginary

roots 1( ) 0f x⇒ > ∴ f(x) is increasing only

( )f x⇒ has onlyh one real root.

31)By verfication 3 1 1 1

0 04 2 2 4

f and f− − − = < = >

Root lies between 3 1 3 1

, ,4 2 4 2

s− − − − ⇒ ∈

32) 1 3

| | ,2 4

t s = ∈

Area between curve and x-axis 4 3 2

0

( )t

f x k t t t= = + + +∫





Using 1 3

,2 4

t ∈

the area 4 3 2 15 252

,16 256

t t t t = + + + ∈

Which lies only in 3

,34

33) 1f is increasing 1 1( ) 24 6 0

4f x x x⇒ = + > ⇒ > −

Increasing in 1

, ,4

t−

decreasing in

1,

9t

− −


Tangents are drawn from the point P(3, 4) to the ellipse 2 2

19 4

x y+ = touching the

ellipse at points A and B.

34. The coordinates of A and B are

A) (3, 0) and (0, 2) B) 8 2 161 9 8

, ,5 15 5 5

and − −

C) 8 2 161

, (0,2)5 15

and

− D)

9 8(3,0) ,

5 5and −

Ans : D

Sol : 3 cos .2sin

19 4

x yθ θ+ =

cos 2sin 1θ θ+ =

24sin 1 cos 24θ θ θ= + −

24 4cos 1 cos 2cosθ θ θ− = + −

3cos cos cos 1

5θ θ θ= − = =

4sin

5θ = ;

9 8,

5 5 −

; (3,0)A





35. 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

Ans : C

Sol : (3, 4), (3,0), ( 9 / 5,8 / 5)P A B −

equation of PB 2x + y = 6

equation of AB 3x + y = 6

solving is these two lines we get 11/5, 8/5

36. The equation of the locus of the point whose distance from the point P and theline AB are 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+ − + + − =

Ans : Equation of AB x + 3y -3 = 0

required locus ( ) ( )2

2 2 ( 3 3)3 4

10

h kh k

+ −− + − =

2 29 6 54 62 241 0x xy y x y− + − + + =SECTION - IV (Matrix 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 question, statement B matches with the statements given in

q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS


A) The set of points z satisfying p) an ellipse with eccentricity 4

5

| | || | | ||z i z z i z− = + is contained

in or equalto

B) The set of points z satisfying q) the set of points z satisfying Im z = 0

| 4 | | 4 | 10z z+ + − =

C) If |w| = 2, then the set of points r) the set of points z satisfying | Im | 1z ≤

1z w

w= − is contained in or equal to

D) If |w| = 1, then the set of points s) the set of points z satisfying | Re | 2z ≤

1z w

w= + is contained in or equal to t) the set of points z satisfying | | 3z ≤





Ans : ; ; , , ; , , ,A q B p C p s t D q r s t→ → → →

Sol : A) locus of z is perpendicular bisector i|z|, -i |z|

that is x-axis, that is y = 0 Im z = 0

B) PA + PB = 10 but AB = 8 < 10. So locus of z is elipse with e = 4/5

C) z = x + iy, 2cos 2sinw iθ θ= +

1 3 5cos sin

2 2w i

wθ θ− = + . Therefore locus of

2 2

19 / 4 25 / 4

x yzis + =

eccentricity is 4/5

3 3Re( ) cos

2 2z θ= ≤

D) cos sinw iθ θ= + . 1

2cosww

θ+ =

So locus of z is imaginary z = 0 | | 2z ≤


A) A line from the origin meets the lines p) -4

82 1 1 3 13

1 2 1 2 1 1

xx y z y zand

−− − + + −= = = =− −

at P and Q

respectively. If length PQ = d, then d2 is

B) The values of x satisfying q) 0

1 1 1 3tan ( 3) tan ( 3) sin

5x x− − − + − − =

are

C) Non-zero vectors ,a b and cG G G

satisfy . 0a b =G G

, r) 4

( ) ( ). 0 2 .b a b c and b c b a− + = + = −G G G G G G G G

If 4 ,a b cµ= +G G G

then the possible values of µ are

D) Let f be the function on [ , ]π π− give by s) 5





f(0) = 9 and f(x) =

9sin

20.

sin2

x

for xx

≠

The value of 2

( )f x dxπ

ππ −∫ is t) 6

Ans : ; , ; ;A s B p r C s D r→ → → →

Sol : A) ( )2, 2 1, 1P s s s+ − + −

82 , 3, 1

3Q t t t + − − +

2 2 1 18 3 123

s s s

t tt

+ − + −= =− − ++

16 83 2 6 4 2

3 3

sst s t st t⇒ − − − − = − + − +

9 7 12 26 0 (1)st s t+ − − = − − −

2st s⇒ = −

9( 2) 7 12 26 0s s t− + − − =

24 3( 2) 11 0s s s− − − =

13

2s or s= = ;

13;3 1

3s t t= = ⇒ = ,

1 3; 3

2 2 2

ts t= = − ⇒ = − ,

5 1 10,0, ; ,0, 2

2 2 3 − − −

2 625 9 625 810

36 4 36d

+= + + =

B) 1 13 ( 3)

tan tan1 ( 3)( 3) 4

x x x

x x− − + − − = + + −

2

64

1 9 4

xx

x= ⇒ = ±

+ −

C) 2 . . 0b b c a c+ − =





2 24 8 . 4b b c c+ + 2 2b a= +

2 2 23 8 . 4 0b b c c a+ + − =

20 ( ) 4( . )b b cµ= + ; ( )2 4 .a a c=

( ) ( ). . 4 . 0a c b c b cµ⇒ − + =

( ) ( ). 4 . 0a c b cµ µ⇒ + − =

( )2 23 8 . 4 4 . 0b b c c a c+ + − =

( ) ( ) ( )( )( )3 . . 8 . . . 4 . 0a c b c b c a c b c a cµ⇒ − + + − − =

( )( )5 . 0b cµ⇒ − =

D) 0

9 9sin sin

2 22sin sin

2 2

x x

dx dxx x

π π

π−

=∫ ∫

0

92sin cos

2 222sin cos

2 2

x x

x x

π

= ∫

0

sin 5 sin 42

sin

x xdx

x

π += ∫

0

sin 52 0

sin

xdx

x

π

= +∫

0

sin 54

sin

xdx

x

π

= ∫ using 2

sin 2sin( 1)

sin 1n n n

nxI dx I n x I

x n −= ⇒ = − +−∫

We get 4. 22

π π= =

2int 2 4given egral π

π∴ × =





IIT JEE 2010 (PAPER - II)

PART - III , PHYSICS

PART - III :: PHYSICS



ONE is correct.

39. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m longuniform string is vibrating in its second harmonic and it resonates with the fun-damental frequency of the pipe. If the tension in the wire is 50 N and the speed ofsound is 320 ms-1, the mass of the string is

A) 5 grams B) 10 grams C) 20 grams D) 40 grams

b: B

Sol; ( )2

2 / 4

T v

m L=

A A

2

2

50 320 320

16 0.5 16 0.8 0.8

T vm

m L m

×⇒ = ⇒ = =× ×A

110

100m kg g⇒ = =

40. A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0onwards it is subjected to a time-dependent force F(t) in the x direction. The forceF(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5seconds is

A)4.50 J B) 7.50 J C) 5.06 J D) 14.06 J

Ans : C





Sol; Area of graph = F dt ma dt=∫ ∫

0

v

mdv mv=∫

= 4.5

( )2

2

mvK

m∴ =

= 4.5 4.5

5.062 2

J× =×

41. A uniformly charged thin spherical shell of radius R carries uniform surface chargedensity of σ per unit area. It is made of two hemisphericla shells, held together bypressing them with force F (sec figure). F is proportional to

A) 2 2

0

1Rσ

ε B) 2

0

1Rσ

ε C) 2

0

1

R

σε D)

2

20

1

R

σε

Ans: A

Sol: Considering equilibrium of hemispherical shell we have

22

02F R

σ πε

= × 2 2

0

RF

σε

∴ ∝

42. A tiny spherical oil drop carrying a net charge q is balanced in still air with a

vertical uniform electric field of strength 581

107

π × Vm|�1|. When the field is

switched off, the drop is observed to fall with terminal velocity 3 12 10 ms− −× . Given

g = 9.8 ms�2 viscosity of the air 5 21.8 10 Ns m− −= × and the density of oil = 900 kg m�3,

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−×

Ans : D





Sol: (Neglect density of air) terminal velocity = 22

9

gV r

ρη

=

2 9

2

vr

g

ηρ

=

5310

7r m−= ×

Eq mg∴ =

6 rv mgπη =

6 rv Eqπη =

198 10q C−⇒ = ×

43. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distancebetween thelens and the mirror is 10 cm. A small object is kept at a distance of 30cm 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

Ans: B

Sol: 111

1 1 1 1 1 130

15 30v cm

v u t v− = ⇒ = − =

2

1 1 1

10 15v− =

2 2

1 16

10 15v v cm⇒ = + ⇒ =

therefore distance of image from mirror is 16 cm and nature is real image there-fore answer is B.





44. A vernier calipers has 1 mm makes on the main scale. It has 20 equal divisions onthe vernier scale which match with 16 main scale divisions. For this vernier cali-pers, the least count is

A) 0.02 mm B) 0.05 mm C) 0.1 mm D) 0.2 mm

Ans: D

Sol: 20 VSD = 16 MSD

1 VSD = 4

5MSD

LC = 1 MSD � 1 VSD

4 11

5 5MSD MSD mm= − =

= 0.2 mm



45. To determine the half life of a radioactive element, a student plots a graph of

( )ln

dN t

dt versus t. Hence ( )dN t

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

number of radioactive nuclei of this element decreases by a factor of p after 4.16years, the value of p is

Ans : 8

Sol:dN

Ndt

λ=

dNn n n N

dtλ= +A A A





( )0 0tdN

n n n N t N N edt

λλ λ −= + − ∴ =A A A

1

2λ∴ =

0tN N e λ−=

00

tNN e

pλ−=

1 tep

λ−=

4.16

2e−

=

2.08 8p e+⇒ = �

46. Image of an object approaching a convex mirror of radius of curvature 20 m along

its optical axis is observed to move from25 50

3 7m to m in 30 seconds. What is the

speed of the object in km per hour ?

Ans: 3

comvex mirror

Sol: 102

Rf cm= + = +

for first position of object

1

25

3v m= +

1 1 1

f u v= +

1

1 1 3

10 25u= +

1 50u m= −

for second position of object





2

50

7v m= +

2

1 1 7

10 50u= +

2 25u = −

speed of object = 1 2u u

t

−

( ) ( )50 25

30

− − −=

5

/6

m s= −

5 18

6 5kmph

−= ×

= � 3 kmph

object is moving towards mirror

therefore answer is 3

47. A large glass salb ( )5 / 3µ = 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 slabfrom a circular area of radius R cm.What is the value of R ?

Ans : 6

Sol: Cθ8 cm

R

tan8c

Rθ = , 1 3 3

sin tan5 4c cθ θ

µ= = ⇒ =

therefore R = 6 cm





48. A diatomic ideal gas is compressed adiabatically to 1

32 of its initial volume. In the

initial temperature of the gas is Ti (in Kelvin) and the final temperature is aT

i, the

value of a is

Ans : 4

Sol:1 1

1 2

2 1

V V

T T

γ γ− −

=

1

1 2

2 1

V Ta

V T

γ −

= =

( )7

1532 a

− =

( )2

532 a=

a = 4

49. At time t = 0, a battery of 10 V is connected across points A and B in the givencircuit. If the capacitors have no charge initially, at what time (in seconds) doesthe voltage across them become 4 V ?

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

Ans : 2

Sol: ( )/0 1 tv v e τ−= − RCτ =

6 61 10 4 10RCτ −= = × × × = 4 sec

( )/0 1 tv v e τ−= − 4 = ( )/ 410 1 te−−

3 5 0.54

tn n− = − = −A A t = 2 sec





SECTION - III (PARAGRAPH TYPE)

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice

questions have to be answered. Each of these questions have four choices A), B), C), andD) out of which ONLY ONE is correct.

Paragraph for Questions 50 to 52

When liquid medicine of density ρ is to put in the eye, it is done with the help ofa dropper. As the bulb on the top of the dropper is pressed, a drop forms at theopening of the dropper. We wish to estimate the size of the drop. We first assumethat the drop formed at the opening is spherical because that requires a minimumincrease in it surface energy. To determine the size, we calculate the net verticalforce due to the surface tension T when the radius of the drop is R. When this forcebecomes smaller than the weight of the drop, the drop gets detached from thedropper.

50. If the radius of the opening of the dropper is r, the vertical force due to the surfacetension on the drop of radius R (assuming r <<R) is

A) 2 rTπ B) 2 RTπ C) 22 r T

R

πD)

22 R T

r

π

Ans : C

Sol: 2 sinF rTπ θ=

where sinr

Rθ =

therefore 22 r T

FR

π=

Tr θ

R θθ

51. If 4 3 3 2 15 10 , 10 , 10 , 0.11 ,r m k gm g ms T Nmρ− − − −= × = = = 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−×

Ans : A

Sol: Upwards force = weight of the drop





34

3F R gπ ρ=

12 43

2

r TR

gρ

=

= 31.4 10−× m

52. 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−×

Ans : B

Sol: surface energy of drop

24 R Tπ=

( )23224 1.4 10 0.11

7−= × × × ×

62.7 10 J−×�

Paragraph for Questions 53 to 55

The key feature of Bohor�s theory of spectrum of hydrogen atom is the quantiza-tion of angular momentum when an electron is revolving around a proton. Wewill extend this to a general rotational motion to find quantized rotational energyof a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr�squantization condition.

53. A diatomic molecule has moment of inertia I. By Bohr�s quantization condition itsrotational 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

Ans : D

Sol: for diatomic molecule

L Iω=

and 21

2E Iω=

for quantization 2

nhL

π=

2 2

28

n hE

Iπ⇒ =





54. 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−×

Ans: B

Sol: for first exeistation

E hf∆ =

( )2

2 22

2 18

hhf

Iπ− =

3446 2

22 11

3 3 2 101.87 10

48 8 10

hI kgm

f

ππ π

π

−−× ×= = = ×

× ×

55. 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) 114.4 10 m−×

Ans : C

Sol: 2I dµ=

Id

µ=

where 461.87 10I −= ×

271 2

1 2

12 16 510

12 16 3

m m

m mµ −×= = × ×

+ +

278010

7−= ×

101.3 10d m−⇒ = ×





SECTION - IV (MATCHING TYPE)This section contins 2 questions. Each question has four statements (A,B,C and D) given in Column I andfive statements (p,q,r,s and t) in Column II. Any given statement in Column I can have correct matchingwith one or more statements(s) given in Column II. For example, if for a given question, statement Bmatches with the statements given in q and r, then for that particular question, against statement B, darken

the bubbles corresponding to q and r in the OMR

56. Two transparent media o refracrive indices 1 3andµ µ 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 1 2,µ µ and 3µ are given. Match them to the ray

diagrams shown in Column II.Column-I Column-II

A) 1 2µ µ< p

B) 1 2µ µ> q)

C) 2 3µ µ= r)

D) 2 3µ µ> s)

t)

Ans : A-pr; B-qst; C-prt; D-qs

Sol: A) 1 2µ µ< light entering from rarer to denser light bends towards the normal

B) 1 2µ µ> light entering from derser to rarer light bends away from normal

C) 2 3µ µ= No bending (No refraction)

D) 2 3µ µ> light entering from denser medium to rarer medium light bends away

from normal





57. You are given many resistances, capacitors and inductors. These are connected to

a variable DC voltage source (the firs 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 V1 and V

2. (indicated in circuits) are related as shown in

Column I. Match the two

Column-I Column-II

A) 00,I V≠ is proportional to I p)

B) 2 10,I V V≠ > q)

C) 1 00,V V V= = r)

D) 20,I V≠ is proportional to I s)

t)





Ans : A-rst; B-qrs; C-pq; D-qrst

Sol: In steady state currents through capacitor will be zero and current through induc-tor will be constant for D.C voltage source current will be non-zero in case of A.Cvoltage source.

for p 1 20,V V V= =

for q 1 2 2 10, 0 , 0,V V I V V= ≠ ≠ >

for r 1 2, , 0V I L V I R Iω= = ≠

31 100 6 10V I π −∴ = × × ×

2 2V I= ×

2 1V V∴ <

for s 1 2,I

V I L VC

ωω

= =

( )( )31 100 6 10V I π −∴ = ×

2 6100 3 10

IV

π −=× ×

2 1V V∴ >

for t 1 2 61000 ,

100 3 10

IV I V

π −= × =× ×

0I ≠

2 1V V>

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