2010 sri chaitanyas_iitjee_paper-i_solutions

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

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

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Sri Chaitanya IIT Academy

Sri Chaitanyas Solutions to

IIT JEE - 2010(PAPER 1)

Time: 3 Hours Maximum Marks: 252

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

INSTRUCTIONS

A. General :

1. This question Paper contains 43 pages having 84 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. The answer sheet, a machine - gradable Objective Response Sheet (ORS), is provided

separately.

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

7. 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 :

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

9. 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.

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

boxes. Do not write these anywhere else.

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

C. Question paper format and marking scheme :

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

consists of four sections

13. For each question in Section I, you will be awarded 3 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 one (1) mark will be awarded.

14. For each question in Section II, 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. Partial

marks will be awarded for partially correct answers. No negative marks will be awarded in

this Section.

15. 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.

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

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

will be awarded for in this section.





IIT JEE 2010 (PAPER - I)

PART - I , CHEMISTRY

PART - I : CHEMISTRY

SECTION - I (SINGLE CORRECT CHOICE TYPE)This section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY

ONE is correct

1. The bond energy (in kcal mol1) of a CC single bond is approximately

A) 1 B) 10 C) 100 D) 1000

Ans : C

Sol : C C bond energy is 100 kcal/mol approximately

2. The species which by definition has ZERO standard molar enthalpy of formationat 298K is

A) Br2(g) B) Cl

2(g) C) H

2O(g) D) CH

4(g)

Ans : B

Sol : ( )02 ;fH for Cl g∆ which is in native state = 0

3. Plots showing the variation of the rate constant (k) with temperature (T) are givenbelow. The plot that follows Arrhenius equation is

A) k

T

B) k

T

C) k

T

D) k

T

Ans : A

Sol : k = / ;Ea RTA e −⋅ k increases exponentially with temperature





4. In the reaction OCH3 HBr→ the products are

A) OCH3Br and H2

B) Br and CH3 Br

C) Br and CH3 OH D) OH and CH

3 Br

Ans : D

Sol : O CH3H +

→ O CH3

H

Br-(SN2)

OH + CH3 - Br

5. The correct statement about the following disaccharide is

oH

OH

CH2OH

H (a)

H

OH

OH

H

H

OCH2CH2O

OH

OH

OH

H

H

CH2OH

CH2OH(b)

A) Ring (a) is a pyranose with α − glycosidic link

B) Ring (a) is a furanose with α − glycosidic link

C) Ring (b) is a furanose with α − glycosidic link

D) Ring (b) is a pyranose with α − glycosidic link

Ans : A

Sol : Pyranose is six membered ring and furanose is five membered ring. Ring(a) is

attached with α − linkage





6. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture ofsodium amide and an alkyne. The bromoalkane and alkyne respectively are

A) BrCH2CH

2CH

2CH

2CH

3 and CH

3 CH

2C ≡CH

B) BrCH2CH

2 CH

3 and CH

3CH

2CH

2C ≡CH

C) BrCH2CH

2CH

2CH

2CH

3 and CH

3C ≡ CH

D) BrCH2CH

2CH

2CH

3 and CH

3CH

2C ≡ CH

Ans : D

Sol : R Br + sodamide + Alkyne → CH3 CH

2 C ≡ C CH

2 CH

2 CH

2 CH

3

possiblities

1) CH3 CH

2 Br + CH

3 CH

2 CH

2 CH

2 C ≡ C H

(or)

2) CH3 CH

2 CH

2 CH

2 Br + CH

3 CH

2 C ≡ CH ∴ D is correct

7. The ionization isomer of [Cr(H2O)

4 Cl (NO

2)]Cl is

A) [Cr(H2O)

4 (O

2N)] Cl

2B) [Cr(H

2O)

4 Cl

2](NO

2)

C) [Cr(H2O)

4 Cl(ONO)] Cl D) [Cr(H

2O)

4Cl

2 (NO

2)]. H

2O

Ans : B

Sol : Ions present outside of coordination sphere are Cl and 2NO − . The ionisation

isomer of [Cr(H2O)

4 Cl(NO

2)] Cl is [Cr(H

2O)

4Cl

2] (NO

2)

8. The correct structure of ethylenediaminetetraacetic acid (EDTA) is

A) B)

C) D)

Ans : C

Sol : The structure of EDTA is





SECTION - II (Multiple Correct Choice Type)This section contains 5 multiple choice questions. Each question has four choices A), B), C), D) out of whihc ONE OR

MORE may be correct.

9. In the reaction

OH

( ) 2/NaOH aq Br→ the intermediate(s) is(are)

A)

OBr

Br

B)

O

BrBr

C)

O

Br

D)

O

Br

Ans : A,B,C

Sol : Phenol react with NaOH gives phenoxide ion which is ortho,para directing grouphence Br+ attack at ortho, para positions

10. In the Newman projecttion for 2, 2 dimethylbutane

H3CX

CH3

HHY

X and Y can respectively be

A) H and H B) H and C2H

5

C) C2H

5 and H D) CH

3 and CH

3

Ans : B,D

Sol : Rotation between C2 and C

3

H3CCH3(X)

CH3

HHCH3(Y)

Rotation between C1 and C

2

H3CH(X)

CH3

HHC2H5(Y)





11. Aqueous solutions of HNO3, KOH, CH

3COOH, and CH

3COONa of identical con-

centrations are provided. The pair(s) of solutions which form a buffer upon mix-ing is (are)

A) HNO3 and CH

3COOH B) KOH and CH

3COONa

C) HNO3 and CH

3COONa D) CH

3COOH and CH

3COONa

Ans : D

Sol : Weak acid and its salt with strong base

12. The reagent(s) used for softening the temporary hardness of water is (are)

A) Ca3(PO

4)

2B) Ca(OH)

2C) Na

2CO

3D) NaOCl

Ans : B,C

Sol : Calcium hydroxide and Sodium carbonate are used to softening the temporaryhardness of water

13. Among the following, the intensive property is (properties are)

A) molar conductivity B) electromotive force

C) resistance D) heat capacity

Ans : A,B

Sol : Intensive property is independent of quantity

SECTION - III (Paragraph Type)

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the

second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C)

and D) out of which ONLY ONE is correct

Paragraph for Questions 14 to 15

The concentrationof potassium ions inside a biological cell is at least twenty times

higher than the outside. The resulting potential difference across the cell is impor-

tant in several processes such as tramsmission of nerve impulses and maintaining

the ion balance. A simple model for such a concentration cell involving a metal M is :

M(s) | M+ (aq ; 0.05 molar) ||M+ (aq ; 1 molar) |M(s)

For the above electrolytic cell the magnitude of the cell potential |Ecell

| = 70 mV.

14. For the above cell

A) Ecell

< 0 ; 0G∆ > B) Ecell

> 0 ; 0G∆ < C) Ecell

< 0 ; 0 0G∆ > D) Ecell

> 0 ; 0 0G∆ <

Ans : B

Sol : Cell reaction is spontaneous reaction





15. If the 0.5 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then themagnitude of the cell potential would be

A) 35 mV B) 70 mV C) 140 mV D) 700 mV

Ans : C

Sol : For given cell, Ecell

= 0.0591

1 log ( )

10.05

given Ecell

= 70 mV

hence Ecell

= 0.0591

1 log

10.0025

Ecell

= 0.0591

1 log ( )2

1

0.05

= 140 mV


Copper is the most noble of the first row transition metals and occurs in small

deposits in several countries. Ores of copper include chalcanthite (CuSO4.5H

2O),

atacamite (Cu2Cl(OH)

3), cuprite (Cu

2O), copper glance (Cu

2S) and malachite

(Cu2(OH

2CO

3). However, 80% of the world copper production comes from the ore

chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves par-

tial roasting, removal of iron and self-reduction.

16. Partial roasting of chalcopyrite produces

A) Cu2S and FeO B) Cu

2O and FeO C) CuS and Fe

2O

3D) Cu

2O and Fe

2O

3

Ans : A

Sol : CuFeS2 Partial Roasting→ Cu

2S + FeO + 2SO ↑

17. Iron is removed from chalcopyrite as

A) FeO B) FeS C) Fe2O

3D) FeSiO

3

Ans : D

Sol : FeO + SiO2 → FeSiO

2

18. In self-reduction, the reducing species is

A) S B) O2 C) S2 D) SO2

Ans : C

Sol : 2 2 22 6Cu S Cu O Cu SO+ → +





SECTION - IV (Integer Type)

This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The

correct digit below the question number in the ORS is to be bubbled

19. Based on VSEPR theory, the number of 90 degree FBrF angles in BrF5 is

Ans : 0

Sol : Br

F

F

FF

F

Repulsion b/w lone pair of e and bond pair of electrons. Because of this bondangle less than 900

20. The value of n in the molecular formula BenAl

2Si

6O

18 is

Ans : 3

Sol : Cyclic Silicate ; Si6

1218O −

21. A student performs a titration with different burettes and finds titre values of25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the averagetitre value is

Ans : 3

Sol : Average titre value is 25.15 which is round of as number of significant figures assmall as in the individual titre values

22. The concentration of R in the reaction R →P was measured as a function of timeand the following data is obtained.

[R] (molar) 1.0 0.75 0.40 0.10

t(min) 0.0 0.05 0.12 0.18

Ans : 0

Sol : x

kt

=

I) 0.25

50.05

k = = II) 0.6

50.12

k = = III) 0.9

50.18

k = =

k value remains constant

23. The number of neutrons emitted when 23592 U undergoes controlled nuclear fission

to 142 9054 38Xe and Sr is

Ans : 4

Sol : 235 1 142 90 192 0 54 38 04U n Xe Sr n+ → + +





24. The total nuber of basic groups in the following form of lysine is

H3N - CH2 - CH2 - CH2 - CH2

CH - COH2N

O

Ans : 2

Sol : NH2 and O

25. The total number of cyclic isomers possible for a hydrocarbon with the molecularformula C

4H

6 is

Ans : 5

Sol : CH3

, CH2

,

CH3

, ,

26. In the scheme given below, the total number of intramolecular aldol condensa-tion products formed from Y is

3

2

1. 1. ( )2. , 2.

O NaOH aqZn H O heat

Y→ →

Ans : 1

Sol : Y =

OO

H H H H

H H H H

Y is symmetrical diketone. All Hα − atoms are identical ∴ only one type of con-

densation is possible





27. Amongst the following, the total number of compounds soluble in aqueous NaOH is

NH3C CH3 COOH OCH2CH3

CH2OH

OH

NO2

OH

NH3C CH3

CH2CH3

CH2CH3 COOH

Ans : 4

Sol :

COOH

,

OH

,

OH

NH3C CH3

,

COOH

28. Amongst the following, the total number of compounds whose aqueous solutionturns red litmus paper blue is

KCN K2SO

4(NH

4)

2C

2O

4NaCl Zn(NO

3)

2

FeCl3

K2CO

3NH

4NO

3LiCN

Ans : 2

Sol : 2

. .

H O

S B W AKCN KOH HCN→ + → solution becomes alkaline

22 3 2 3

. .

H O

S B W A

K CO KOH H CO→ + → solution becomes alkaline






PART - II : MATHEMATICS

SECTION - I (SINGLE CORRECT CHOICE TYPE)This section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY

ONE is correct

29. Let f, g and h be real - valued functions defined on the interval [0, 1] by− − −= + = + = +

2 2 2 2 2 22( ) , ( ) ( ) .x x x x x xf x e e g x xe e and h x x e e If a, b and c denote, respec-

tively, the absolute maximum of f, g and h on [0, 1], then

A) = ≠a b andc b B) = ≠a c and a b C) ≠ ≠a b andc b D) = =a b c

Ans : D

Sol : −= +2 2

( ) x xf x e e

−= = + 1(0) 2 (1)f and f e e

−= − ≥ ∀ ∈2 21( ) 2 ( ) 0 [0,1]x xf x x e e x

absolute max. of −= + 1a e e

−= +22

( ) x xg x x e e

−= = + 1(0) 1, (1)g g e e

( ) −= + − > ∀ ∈2 22’( ) 1 2 2 0 [0,1]x xg x x e x e x

absolute max. of −= = = + 1(1)g b g e e

−= +222( ) x xh x x e e

−= = + 1(0) 1, (1)h h e e

−= + − ≥ ∀ ∈2 22 2’( ) 2 [ ] 0 [0,1]x xh x x e x ex e x

absolute max. of −= = = + 1(1)h c h e e

∴ = =a b c





30. Let p and q be real numbers such that ≠ ≠ ≠ −3 30, .p p q and p q If α βand are non-

zero complex numbers satisfying α β α β+ = − + =3 3 ,p and q then a quadratic equa-

tion having α ββ α

and as its roots is

A) + − + + + =3 2 3 3( ) ( 2 ) ( ) 0p q x p q x p q B) + − − + + =3 2 3 3( ) ( 2 ) ( ) 0p q x p q x p q

C) − − − + − =3 2 3 3( ) (5 2 ) ( ) 0p q x p q x p q D) − − + + − =3 2 3 3( ) (5 2 ) ( ) 0p q x p q x p q

Ans : B

Sol :α β α β+ − + =3 3,p q

Sum of the roots α β α ββ α αβ

−+= + = =+

32 2

3

2p qp q

Product of the roots α ββ α

= =. 1

∴ Req. equation. −

− + = +

32

3

21 0

p qx x

p q

⇒ + − − + + =3 3 3 3( ) ( 2 ) ( ) 0p q x p q x p q

31. The value of →

++∫

A

3 4

(1 )1lim

4

x

x OO

t n tdt

x t

A) 0 B) 1

12C)

124

D) 1

64

Ans : B

Sol : →

++∫ 4

030

log(1 )4

x

x

t tdt

tLt

x

32. The number of 3 x 3 matrices A whose entries are either 0 or 1 and for which the

system

=

100

x

A y

z has exactly two distinct solutions, is

A) 0 B) −92 1 C) 168 D) 2





Ans : A

Sol : A system of equations in 3 unknowns cant have two distinct solutions.

33. Let P, Q, R and S be the points on the plane with position vectors

− − + − +ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 , 4 ,3 3 3 2i j i i j and i j respectively. The quadrilateral PQRS must be a

A) parallelogram, which is neither a rhombus nor a rectangel

B) square

C) rectangle, but not a square D) rhombus, but not a square

Ans : A

Sol : − − −( 2, 1,0), (4,0,0), (3, 3,0), ( 3, 2,0)A B C D

by transforming all vertices to − −( 2, 1,0) we get above diagram.

Therefore AB, CD are equal AD, BC are equal. Therefore ABCD is a parallelo-gram but not rhombus or rectangel.

34. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times.

If 1 2 3,r r and r are the numbers obtained on the die, then the probability that

ω ω ω+ + =1 2 3 0r r r is

A) 1

18B)

19

C) 29

D) 1

36

Ans : C





Sol :

ω ω ω+ + =

==

+ +

∴ = =

1 2 3

1 2 3

1 1 1

1 1 1

0

1 2 3.........3! 64 5 6..........3! 6...3 , 3 1,3 2,.......

2 .2 .2 .3! 2( ) .

6 .6 .6 . 9

r r r

r r r

n n n

c c cp A

c c c

35. Equation of the plane containing the straight line = =2 3 4

yx z and perpendicular to

the plane containing the straight lines = = = =3 4 2 4 2 3

y yx z x zand is

A) + − =2 2 0x y z B) + − =3 2 2 0x y z C) − + =2 0x y z D) + − =5 2 4 0x y z

Ans : C

Sol : d.rs of the normal to the plane containing = = = =2&

3 4 2 4 2 3y yx x z

are 8, -1, -10

Now d.rs of the normal of the plane containing = = &2 3 4

yx z above normal are

1, -2, 1. Therefore Req. plane − − − + − =1( 0) 2( 0) 1( 0) 0x y z

− + =2 0x y z

36. If the angles A, B and C of a triangle are in an arithmetic progression and if a, band c denote the lengths of the sides opposite to A, B and C respectively, then the

value of the expression +sin2 sin2a c

C Ac a

is

A) 12

B) 3

2C) 1 D) 3





Ans : D

Sol : +2 sin .cos 2 sin cosa c

c c A Ac a

= +.2 sin cos .2 sin cos2 sin 2 sin

a cc c A A

R c R A

+= = = = =

cos cos 2 sin2.sin60 3

a c c A b R BR R R



37. Let ABC be a triangle such that π∠ =6

ACB and let a, b and c denote the lengths of

the sides opposite to A, B and C respectively. The values (s) of x for which

= + + = − = +2 21, 1 2 1a x x b x and c x is (are)

A) ( )− +2 3 B) +1 3 C) +2 3 D) 4 3

Ans : A,B

Sol : π + + + − − += =

+ + −

2 2 2 2

2 2

( 1) ( 1) (2 1)3cos

6 2 2( 1)( 1)x x x x

x x x

( )+ + − − − = + + + + + + − + − − −4 3 2 2 4 2 3 2 4 2 23 1 1 2 2 2 2 1 4 4 1x x x x x x x x x x x x x x

( )+ − − = + − − +4 3 4 3 23 1 2 2 3 2 1x x x x x x x

( ) ( ) ( ) ( )⇒ − + − − − − + + =4 3 22 3 2 3 3 2 3 1 3 0x x x x

= − + = + = ±1(2 3), 1 3, 1x x x

38. Let 1 2z and z be two distinct complex numbers and let = − +1 2(1 )z t z tz for some real

number t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzerocomplex number w, then

A) − + − = −1 2 1 2| | | || |z z z z z z B) − = −1 2( ) ( )Arg z z Arg z z

C) − −

=− −

11

2 12 1

0z z z z

z z z z D) − = −1 2 1( ) ( )Arg z z Arg z z





Ans : A,B,C,D

Sol : Z lie on the line joining the points z1, z

2

AB + BC = AC

Therefore − + − = −1 2 1 2| | | || |z z z z z z

ABC collinear therfore argument of AB, argument of AC equal and also argu-ment of AC, argument of AB equal. So B, D is correct.

Complex slope of AB = complex slope of BC

− −= = − − −− −

1 2

1 2( )

z z z zc

z z z z

39. The value(s) of −

+∫1 4 4

20

(1 )1

x xdx

x is (are)

A) π−227

B) 2

105C) 0 D)

π−71 315 2

Ans : A

Sol : ( )− + − +2 3 4 41 4 6 4 2x x x x

+ − + − + − + − +

+− + −

− −+

+−

− −

+−

2 8 7 6 5 4 6 5 4 2

8 6

7 6 5

7 5

6 4

6 4

4

4 2

2

2

1) 4 6 4 ( 4 5 4 4

4 5 4

4 45

5 54

4 44

4 44

x x x x x x x x x x

x xx x x

x xx x

x xx

x xx

x





Given integral = −− + − + −

17 6 5 31

0

4 45 4 4 tan

7 6 5 3x x x x

x x

π= − =220

7.

40. Let A and B to two distinct points on the parabola =2 4 .y x If the axis of the pa-

rabola touches a circle of radius r having AB as its diameter, then the slope of theline joining A and B can be

A) − 1r

B) 1r

C) 2r

D) −2r

Ans : C,D

Sol :

= =

1 21 2

2 2, , ,A t B tt t

Mid.point of A,B is center of the circle that is + =1 22( )2

t tr . Because circle touches

x -axis

Slope of − −= =

− − +2 1 2 1

2 22 1 2 1 2 1

2( ) 2( )( )( )

t t t tAB

t t t t t t

= 2r

in first quadrant, in fourth quadrant it is − 2r

41. Let f be a real - valued function defined on the interval ∞(0, ) by

= + +∫A

0

( ) 1 sin .x

f x n x t dt Then which of the following statement(s) is (are) true?

A) "( )f x exists for all ∈ ∞(0, )x

B) ’( )f x exists for all ∈ ∞(0, )x and ’f is continuous on ∞(0, ) ,

but not differentiable on ∞(0, )

C) there exists α > 1 such that <’( ) ( )f x f x for all α∈ ∞( , )x

D) there exists β > 0 such that β+ ≤( ) ’( )f x f x for all ∈ ∞(0, )x





Ans : B, C

Sol : = + +∫0

( ) log 1 sinx

f x x tdt

= + +1 1( ) 1 sin .f x x

x exists ∀ ∈ ∞(0, )x

∴ 1( )f x exists

Hence 1( )f x is continuous

−= ++2

1 1"( ) (cos )

2 1 sinf x x

x x

is not definied for 1 +sin x = 0

"( )f x not exist ∀ ∈ ∞(0, )x

1( )f x is not differential for sinx = - 1

So (B) correct.

= + +∫0

( ) log 1 sin .x

f x x t dt

= + +1 1( ) 1 sinf x x

x

If x > 1 then ∈1(0,1)

x

and + ∈ 1 sin 0, 2x

both ranges are finite

∴ ∈1( ) [0, ]f x k for same k.

For α = = + +∫0

, ( ) log 1 sink

k ke f x e t = + ( )k same position

α α∴ < ∈1( ) ( ) ( , )f x f x for x where α = ke . So option C is correct.

Sum of two bounded below functions is bounded below.

Hence finite β is not possible





SECTION - III (Paragraph Type)This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the




The circle + − =2 2 8 0x y x and hyperbola − =22

19 4

yx intersect at the points

A and B.

42. Equation of a common tangent with positive slope to the circle as well as to thehyperbola is

A) − − =2 5 20 0x y B) − + =2 5 4 0x y C) − + =3 4 8 0x y D) − + =4 3 4 0x y

Ans : B

Sol : Tangent for hyperbola = ± −29 4y mx m

If it touches the circle then

r = d

( )± −=

+

2

2

4 9 44 .

1

m m

m

+ = + − ± −2 2 2 216 16 16 9 4 8 9 4m m m m m

− = ± −2 29 20 8 9 4m m m

( )+ − = −4 2 2 281 400 360 64 9 4m m m m

= −4 2576 256m m

+ − =4 2495 104 400 0m m

= =2 4/5, 2 / 5m m

43. Equation of the circle with AB as its diameter is

A) + − + =2 2 12 24 0x y x B) + + + =2 2 12 24 0x y x

C) + + − =2 2 24 12 0x y x D) + − − =2 2 24 12 0x y x

Ans : A





Sol : + − = − −2 2 8 0 (1)x y x

− = − −2 24 9 36 (2)x y

(1)equation x (9) + 2 ⇒ − − =213 72 36 0x x

−= 66,

13x

∴ = 6if x , = ±2 3y

( ) ( )= = −6,2 3 , 6, 2 3A B

Equation of circle with A, B as ends of diameter

( )( ) ( )( )− − + + − =6 6 2 3 2 3 0x x y y

+ − + = − − −2 2 12 24 0 ( )x y x A


Let p be an odd prime number and pT be the following set of 2 x 2 matrices:

= = ∈ −

: , , 0,1, 2,....., 1p

a bT A a b c p

c a

44. The number of A in pT such that A is either symmetric or skew-symmetric or

both, and det (A) divisible by p is

A) − 2( 1)p B) −2( 1)p C) − +2( 1) 1p D) −2 1p

Ans : D

Sol :

= = ∈ −

; , , 0,1,2..... 1p

a bT A a b c p

c a

case. 1 = − ⇒ = = = 0TA A a b c

case. 2

= ⇒ = = = −

2 2| |T a bA A b c and A a b

c a

∴ −a b is divisible by a + b is divisible by P

a = b ⇒ a - b is divisible P. ∴ there are P ways of choosings





a + b is divisible by P when (a, b) = (1, P - 1) or (2, 1, -2).....(P-1, 1)

∴ there are P - 1 ways of choosing

∴ total number of ways is 2P - 1

45. The number of A in pT such that the trace of A is not divisible by p but det (A) is

divisible by p is

[Note:- The trace of a matrix is the sum of its diagonal entries.]

A) − − +2( 1)( 1)p p p B) − −3 2( 1)p p C) − 2( 1)p D) − −2( 1)( 2)p p

Ans : C

Sol : Trace of A = 2a is divisible by p only is a = 0

∴ suppose ≠ 0a then = −2| |A a bc

Since each of b and c is less than P let b = p - r and c = p - k.

Then = −2| |A a bc

Since each of b and c is less than P. Let b = p -r and c = p - k

Then = − − −2| | ( )( )A a p r p k

now |A| is divisible p is a2 = rk

each of r and k can be selected is p - 1 ways

∴ Ans. (P - 1)2.

46. The number of A in pT such that det (A) is not divisible by p is

A) 22p B) −3 5p p C) −3 3p p D) −3 2p p

Ans : B

Sol :




47. For any real number x, let [x] denote the largest integer less than or equal to x.Let f be a real valued function defined on the interval [-10, 10] by

−= + −

[ ] [ ] ,( )

1 [ ] [ ]x x if x is odd

f xx x if x is even





Then the value of π π

−∫102

10

( )cos10

f x x dx is.

Ans : 4

Sol : ( )−

= − −

[ ]; [ ]( )

1 [ ] ; [ ]x x if x is odd

f xx x if x is even

f is periodic with period 1and πcos x has period 2

π ππ π−

∴ = × ×∫ ∫10 22 2

10 0

( ).cos 10 ( ).cos10 10

f x x f x x

π π π ππ π

= − + − = + = ∫ ∫1 2

2 22 2

0 1

2 2(1 )cos 1cos . 4.x xdx x x dx

48. Let w be the complex number π π+2 2

cos sin .3 3

i Then the number of distinct

complex numbers z satisfying

ω ωω ωω ω

++ =

+

2

2

2

11 0

1

z

z

z is equal to

Ans : 1

Sol : + =

+

2

2 1 01

z w w

z z w

z z w

+ =+

2

2

11 1 01 1

w w

z z w

z w;

+ − − =− + −

2

2 2

2

10 1 00 1

w w

z z w w w

w z w w

( )( ) ( )( ) + − + − − − − = 2 2 21 1 0z z w w z w w w w

( ) ( ) − − − − + = 22 2 2 3( 1 . 0z z w w w w w

( ) − + − − + + − = 2 22 1 1 0z z w w w w





( )− + − =2 2 2 2 0z z w

− = → =2( 2 ) 0 0z z w z

49. Let =, 1,2,.....,100,kS k denot the sum of the infinite geometric series whose first

term is − 1!

kk

and the common ratio is 1

.k

Then the value of

=

+ − +∑2 100

2

1

100( 3 1)

100! kk

k k S is

Ans : 4

Sol :

−−= = =

− −−

11 1

1 1 11k

kk kk

Sk k k

k

= =

− − − − − −+ = + − − ∑ ∑

2 22 2100 100

1 1

( 3 1 ( 1) ( 1) 1100 100100 1 100 1k k

k k k kk k

= =

− − −= + = + − − − − ∑ ∑

2 2100 100

1 1

( 1)( 2) 1100 100 1 1100 1 100 3 1k k

k kk k k

= =

+ − − −

∑ ∑2 100 100

3 1

100 1 1100 3 1k kk k

= − − +2100 1 1

4100 98 99

= − = +100 1 14

99 98 99 ; = − − + =

1 100 11 4 4

98 99 99

50. The numebr of all possible value of θ , where θ π< <0 , for which the system of

equations

θ θθ θθ

θ θ θ

+ =

= +

= + +

( )cos 3 ( )sin 32 cos 3 2 sin 3

sin 3

( )sin 3 ( 2 )cos 3 sin 3

y z xyz

xy z

xyz y z y

have a solution 0 0 0( , , )x y z with ≠0 0 0,y z is





Ans : 8

Sol : θ θ θ+ − =cos 3 cos 3 sin 3 0y z t

θ θ θ+ − =(2 sin 3 ) (2 cos 3 ) sin 3 0y z t

θ θ θ θ+ + − =(cos 3 sin 3 ) (2 cos 3 ) .sin 3 0y z t

θ θ θθ θ θ

θ θ θ θ

−− =

+ −

cos 3 cos 3 sin 32 sin 3 2 cos 3 sin 3 0

cos 3 sin 3 2 cos 3 sin 3

θθ θ

θ θ=

−

cos 3 1 1cos 3 2 sin 3 2 1 0

cos 3 sin 3 0 0

( )θ θ θ− =cos 3 . cos 3 sin 3 0

θ θ θ= = =cos 3 0, tan 3 1 ,sin 3 0j

( )π π πθ θ= + = +3 2 1 ,2 3 12

nn

π π πθ∴ = 5, ,

6 2 6

π π π5 11, ,

2 12 12;

π πθ = 2,

3 3

51. Let f be a real-valued differentiable function on R (the set of all real numbers)such that f(1) = 1. If they y-intercept of the tangent at any point P(x, y) on thecurve y = f(x) is equal to the cube of the abscissa of P, then the value of f(-3) isequal to

Ans : 9

Sol : Tangent at P(x1, y

1) is y - y

1 = m (x - x

1)

⇒ −y intercept = − + = 31 1 1mx y x

−− + = ⇒ =

33. .

dy dy y xx y x

dx dx x

= = − 21.

dyy x is DE

dx x with I.F =

1and so

x





.G S is −= − = +∫

21. .

2x

y x dx cx

( )− += ⇒ = ⇒ =

2 33(1) 1

2 2

x xf c y

− −− = =( 3)( 6)( 3) 9

2f

52. The number of values of θ in the interval π π −

,2 2 scuh that

πθ ≠5

n for

θ θ= ± ± =0, 1, 2 tan cot 5n and as well as θ θ=sin 2 cos 4 is

Ans : 3

Sol : πθ π θ = + −

52

n

( )πθ = +6 2 12

n ; ( ) πθ = +2 112

n

n = 0, πθ =12

; n = 1, πθ =4

n = 2, πθ −=

12; n = 2,

πθ = 512

n = 2, πθ = −4

; n = - 3, πθ −= 5

12

πθ π θ = ± − 4 2 2

2n ;

πθ π θ= + −4 2 22

n

πθ = +6 (4 1)2

n ; πθ = +(4 1)12

n

πθ π θ= − +4 2 22

n ;πθ = −2 (4 1)2

n

( )πθ = −4 14

n n = 0, π12

, π−4

n = 1, π π5 3

,12 4

; n = -1, π π−− 5

,4 4





n = 2, π π3 7

,4 4

Common ans. for both π π π−5

, ,12 12 4

53. The maximum value of the expression θ θ θ θ+ +2 2

1sin 3sin cos 5 cos

is

Ans : 2

Sol : ( )θ θ θ θθ=

− + + + + +

1 21 cos 2 3 1 cos 2 6 3sin 2 4 cos 2sin 2 5

2 2 2

( ) [ ]θ θθ θ

+ + ∈ ⇒ ∈ + + 2 2

6 3sin 2 4cos2 1,11 ,26 3sin 2 4cos2 11

54. If G G

a andb are vectors in space given by − + +

= =G G

ˆˆ ˆ ˆ ˆ2 2 3,

5 14i j i j k

a and b then the value

of ( ) ( ) ( ) + × × − G G G G G G

2 . 2a b a b a b is

Ans : 5

Sol : ,a b are unit vectors and mutually perpendicular. Let × =a b c

( ) ( )( ) ∴ + × − = −

2 1 02 . 2 0 0 1

1 2 0a b c a b abc

= = × = 5 5 1 5. sin , ,abc ce a b c are

unit vertors, mutually perpendicular.

55. The line 2x + y = 1 is tangent to the hyperbola − =22

2 2 1.yx

a b If this line passes

throught the point of intersection of the nearest directrix and the x-axis, thenthe eccentricity of the hyperbola is.

Ans : 2

Sol : Tangent 2x + y = 1 . Cuts x-axis at ⇒ = ⇒ =

1 1,0 2

2 2a

e ae





Tangent = − + ⇒ = − ⇒ = − ⇒ = − = −2 2 2 2 2 2 2 2 22 1 1 .4 4 1 1y x e a m b a b b a e

+ −+ −∴ = = =

22

2 2 2

2 22

1 5 44

4

eea b e

ea ee

So ( )( )− + = ⇒ − − = ⇒ =4 2 2 25 4 0 1 4 0 2.e e e e e

56. If the distance between the plane − + =2Ax y z d and the plane containing the

lines − −− − − −= = − =

2 31 3 2 42 3 4 3 4 5

y yx z x zand is 6 , then | |d is

Ans : 6

Sol : Given lines have direction ratios (2, 3, 4) and (3, 4, 5). So normal of plane hasD.Rs (1, -2, 1). Plane through the point (1, 2, 3) is x - 2y + z = 0. It is parallel toAx - 2y + z = d

⇒ distance = = ⇒ =+ +| |

6 | | 61 4 1

dd






PART - III , PHYSICS

PART - III : PHYSICS

SECTION - I (SINGLE CORRECT CHOICE TYPE)

This section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY

ONE is correct

57. A thin uniform annular disc (se3c figure) of mass M has outer rdius 4R and innerradius 3R. The work P on its axis to infinity is .

A) ( )24 2 5

7

GM

R− B) ( )2

4 2 57

GM

R− − C)

4

GM

RD) ( )2

2 15

GM

R−

Ans : A

Sol : ( )2 24 / 2

Gdmdv

V r=

+

dr

r

p

where 22

7

Mdm rdr

rπ

π= ×

2

2

7

Mrdr

R=

( )( )24

G dmdv

r R+

−∴ =

2 2 2

2

7 16

Mrdrdv

R R r

−=+





∴ Total gravitational potential at pt. p (VP)

dv= ∫4

2 2 23

2

7 16

R

R

M rdr

R R r= −

+∫

2 22

216

7

MGR r

R

− = +

24 2 5

7

MG

R

− = −

∴ requirel work done

PV V∞ −

( )24 2 5

7

GM

R= + −

58. A block of mass m is on an inclined plane of angle θ . The coefficient of friction

between the block and the plane is µ and tanθ µ> . The block is held stationary by

applying a force P parallel to the plane. The direction of force pointing up the

plane is taken to be positive. As P is varied from P 1 = ( )sin cosθ µ θ− to P

2 =

( )sin cosθ µ θ+ , the frictional force f versus P graph will look like.

A) B)

C) D)





ANS : A

Sol : AS P is increased from mg ( )sin cosθ µ θ− to sinmg θ to mg ( )sin cosθ µ θ+ frac-

tional force will change from cosmgµ θ ( upwards parallel to inclined plane) to

zero to cosmgµ θ (downwards parallel toinclined plane). Under the approxima-

tion of linear variation of P the correct option is A

59. A thin fiexible wire of length L is connected to two adjacent fixed points and car-ries a current I in the clockwise direction, as shown in the figure. When the systemis put in a uniform magnetic ficle of strength B going into the plane of the paper,the wire takes the shape of a circle. The tension in the wire is .

A) IBL B) IBL

π C) 2

IBL

π D) 4

IBL

π

Ans: C

Sol :

/ 2dθ/ 2dθ

T T

/ 2dθ

magF

/ 2dθ

Sin component of tension balance the magnetic field 2 sin / 2 magT d Fθ =

( )Td iR d Bθ θ=

T = i RB

2

iBL

π=

60. An AC voltage source of variable angular frequency ω and fixed amplitude V0 is

connected to series with a capacitance C and an electric bulb of resistance R (in-ductance zero). When ω is increased

A) the bulb glows diammer

B) the bulb glows brighter

C) total bulb glows brighter

D) total impedance of the circuit increases





Ans : B

Sol :

~

C R

00

22

1

Vi

Rw c2

=+

( )2

2 000

22 2

1bulb

V RP i R

RCω

= =+

∴ as w is increased ( )0bulbP increases it glows brighter

61. To verify Ohms law, a student is provided with a test resistor RT, a high resis-

tance R1, a small resistance R

2, two identical galvanometers G

1 and G

2, and a vari-

able voltage source V. The correct circuit to carry out the experiment is

A) B)

C) D)

Ans : C

Sol : Theoretical





62. Incandescent bulbs are designed by keeping in mind that the resistance of theirfilament increases with the increase in temperature. If at room temperature, 100W, 60 W and 40 W bulbs have filament resistance R

100, R

60 and R

40, respectively,

the rlation between these resistance is.

A) 100 40 60

1 1 1

R R R= + B) 100 40 60R R R= + C) 100 60 40R R R> > D)

100 60 40

1 1 1

R R R> >

ANS : D

SOL : 2V

pR

=

2 2 2

100 60 40

100 ,60 , 40V V V

R R R⇒ = = =

100 60 40

1 1 1

RR R∴ > >

63. A real gas behaves like an ideal gas if its

A) pressure and temperature are both high

B) pressure and temperature are both low

C) pressure is high and temperature is low

D) pressure is low and temperature is high

Ans: D

Sol : At low pressure and high temperature, the interaction between molcules isapproaces to zero. Therefore real gas is behaves as ideal gas.

64. Consider a thin square sheet of side L and thickness t, made of a material of resis-tivity ρ . The resistance between two oppositive faces, shown by the shaded areasin the figure is .

A) directly proportional to L B) directly proportional to t

C) independent ot L D) independent of t

Ans : C

Sol : .

.

LR

t L t

ρ ρ= = which is independent of L.







65. A student uses a simple pendulum of exactly 1m length ot determine g, the accel-eration due to gravity. He uses a stop watch with the least count of 1 sec for thisand records 40 seconds for 20 oscillations. For this observation, which of the fol-lowing statement(s) is (are) true ?

A) Error T∆ in measuring T, the time period, is 0.05 seconds

B) Error T∆ in measuring T, the time period , is 1 second

C) Percentage error in the determination of g is 5%

D) Percentage error in the determination of g is 2.5%

Ans : A,C

Sol : ( )40 1 20s T± =

10.05 sec

20T∆ = =

2 /T l gπ=

2

2

4g

T

π= A

2100 100

g T

g T

∆ ∆∴ × = ×

0.052 100

2= × ×

= 5%

66. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. Aftertheir collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms1.Which of the following statements(s) (are) correct for the system of these twomasses ?

A) Total momentum of the system is 3 kg ms1

B) Momentum of 5 kg mass after collision is 4 kg ms1

C) Kinetic energy of the centre of mass is 0.75 J

D) Total kinetic energy of the system is 4 J

Ans : A,C

Sol : 1 21 0 2 1 5u v× + = − × + ....(1)

22 1

1

21 2

0

vv u

u

+= ⇒ + =− ....(2)





2 1 2v u⇒ = −

( ) 1 21 2 5u v⇒ = − +

( )1 1 12 5 2 2 5 10u u u= − + − = − + −

14 12u⇒ − = −

u1 = 3 m/s

2 1 3 0v u= − =

Total momentum of the system = ( )1 1m u

1 3 3 /kgm s= × =

total KE of the system 2 2

1 1 2 2

1 1

2 2m u m u= +

19 4.5

2J= × =

KE of com= ( )( )2

1 2

1

2 cmm m v+

1 1 2 2

1 2

1 3 0 1

6 2cm

m u m uV

m m

+ × += = =+

KE of COM

21 1 6 3

6 0.752 2 8 4

J = × = = =

67. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, asshown in the figure. Its pressure at A is P

0. Choice the correct option(s) from the

following.

A) Interval enerhgies at A and B are the same

B) Work done by the gas in process AB is P0V

0 nA 4

C) Pressure at C is 0

4

T

D) Temperature at C is 0

4

T

Ans : A,B,C,D

Sol : (i) A to B temperature is same ∴ A is correct

(ii) A to B isothemal process

∴ W = nRT 2

1

lnV

V





00

0

4ln .

VnRT

V

=

0 0 ln 4P V=

∴ B is correct

(iii) B to C isobaric process

1 1

2 2

V T

V T=

0 0 02

0 2

4

4

V T TT

V T= ⇒ =

∴ temperature at C is 0

4

T

∴ D is correct

(iv) C to A isochoric process

1 1

2 2

P T

P T=

C C

A A

P T

P T=

0

0 04CP T

P T=

× 0

4C

PP =

∴ C is correct

68. A few electric field for a system of two charges Q1 and Q

2 fixed at two different

points on the x-axis are shown in the figure. These lines susggest that

A) 1 2Q Q> B) 1 2Q Q<

C) at a finite distance to the left of Q1 the electric field is zero

D) at a finite distance to the right of Q2 the electric field is zero

Ans : A,D

Sol : Number of lines of force is directly proportional to the magnitude of the charge

so 1 2Q Q>

At null point where E = 0, there will be no lines of force of only the given twocharges are considered, there is a chance of getting E = 0 to the right of Q

2 (Null

point will be closer to the charge having small magnitude)





69. A ray OP of monochromatic light is incident on the face AB of prism ABCD nearvertex B at an incident angle of 60º (sec figure). The refractive index of the mate-

rial of the prism is 3 , which of the following is (are) correct ?

A) The ray gets totally internally reflected at face CD

B) The ray comes out through face AB

C) The angle between the incident ray and the emergent ray is 90º

D) The angle between the incident ray and the emergent ray is 120º.

Ans : A,B,C

Sol : At surface AB

sin

sin

i

rµ =

sin 603

sin r=

P

A

B

C

D

Q’r

’θ

θr

’’r

’’i

60º

60º

60º

75º45º

135ºN

N'

r = 30º

in quadrilateral PBCQ

45ºθ =





60º

δ

as 45ºθ = r = 45º

For medium critical angle is less than 45º.

∴ ray has total internal reflection at surface CD

' 60ºθ =

then r = 30º

i = 60º

angle of deviation = 180 (60 + 30) = 90

SECTION - III (Paragraph Type)

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the




Electrical resistance of certain materials, know as superconductors, changesabruptly from a nonzero value to zero as their temperature is lowered below acritical temperature T

C(0). An interseting property of supercondeuctors is that

their critical temperature becomes smaller than TC(0)if they are placed in a mag-

netic field, i.e., the critical temperature TC(B) is a function of the magnetic field

strength B. The dependence of TC(B) on B is shown in the figure.

70. In the graphs below, the resistance R of a superconductor is shown as a functionof its temperature T for two different magnetic fields B

1(solid line) and B

2(dashed

line). If B2 is large than B

1, which of the following graphs shows the correct varia-

tion of R with T in these fields ?





A) B)

C) D)

Ans : A

Sol : Under magnetic field, the critical temperatures decreases as given in the datatherefore the super conductors become conductors for higher magnetic field atlow temperature.

71. A superconductor has TC(0) = 100 K. When a magnetic field of 7.5 Tesla is applied,

its TC decreases to 75 K. For this material one can definitely say that when.

A) B = 5 Tesla, TC (B) = 80 K B) B = 5 Tesla, 75 K < T

C (B) < 100 K

C) B = 10 Tesla, 75 K < TC (B) < 100 K D) B = 10 Tesla, T

C (B) = 70 K

Ans : B

Sol : When B is increased, TC(B) decreased where solid becomes super conductor. If B

= 7.5 tesla given that TC(0) decreased from 100 K to 75 K. If B is less than 7.5 then

TC(B) stightly increases, it should be in between 75 K and 100 K so if B = 5 Tesla, 75

K < TC(B) < 100 K


When a particle of mass m moves on the x-axis in a potential of the form V(x) =kx2, it performs simple harmonic motion. The corresponding time period is pro-

portional to m

k, as can be seen easily using dimensional analysis. However, the

motion of a particle can be periodic even when its potential energy increases onboth sides of x = 0 in a way different from kx2 and its total energy is such that theparticle does not escape to infinity. Consider a particle of mass m moving on the

x-axis. Its potential energy is V(x) = 4xα ( )0α > for |x| near the origin and be-

comes a constant equal to V0 for 0x X≥ (sec figure).





72. If the total energy of the particle is E, it will perform periodic motion only if

A) E < 0 B) E > 0 C) V0 > E > 0 D) E > V

0

Ans : C

73. For periodic motion of small amplitude A, the time period T of this particle isproportional to

A) m

Aα B)

1 m

A α C) Am

αD)

1

A m

α

Ans : B

74. The acceleration of this particle for 0x x> is

A) proportional to V0

B) proportional to 0

0

V

mX

C) proportional to 0

0

V

mX D) zero

Ans : D

Sol : For x > x0 the force is zero, therefore acceleration is zero.

For 0 0x x x> > − the force is 34F xα= − which act as restoring force, therefore the

particle perform osscilation for V0 > E > 0 .

for small amplititude, curvature variation for curve x4

and x2 is identical. There-

fore m

Tk

∝ where 2 4kA Aα=

1 mT

A α∝








75. A 0.1 kg mass is suspended from a wire of negligible mass . The length of the wire

is 1 m and its cross-sectinla area is 7 24.9 10 m−× . If the mass is pulled a little in the

vertically downward direction and released, it performs simple harmonic motionof angular frequency 140 rad S1. If the Youngs modulus of the material of the

wire is 9 210n Nm−× , the value of n is

Ans : 4

Sol : 2

2

12 , ,

m m FL KLT k m Y

k k Ae Aπ ω

ω= ⇒ = = = =

2YAm

Lω=

⇒ n = 4

76. A binary star consists of two stars A (mass 2.2MS) and B (mass 11 M

S), where M

S is

the mass of the sun. They are separated by distance d and are rotating about theircentre of mass, which is stationary. The ration of the total angular momentum ofthe binary star to the angular momentum of star B about the centre of mass is

Ans : 6

Sol : 1 1 2

2 1

6L M M

L M

+= =

77. Gravitational acceleration on the surface of a planet is 6

11g , where g is the gravi-

tational acceleration on the surface of the earth. The average mass density of the

planet is 2

3 times that of the earth. If the escape speed on the surface of the earth is

taken to be 11 kms1 the escape speed on the surface of the planet in kms1 will be

Ans : 3

Sol : 2

4

3

GMg G R

Rπ ρ= =

22 8

3e

GMV G R

R

π ρ= =

2

e

gV α

ρ 2

1

3V

V=





78. A piece of ice (heat capacity = 2100 J kg1 ºC1 and latent heat = 53.36 10× J kg1) of

mass m grams is at 5ºC at atmospheric pressure. It is given 420 J of the heat sothat the ice starts melting. Finally when the ice-water mixture is in equilibrium, itis found that 1 gm of ice has melted. Assuming there is no other heat exchange inthe process, the value of m is

Ans : 8

Sol : +icem

5ºC−

1Q m C T= ∆

icem

0ºC ( )1m g ice−

2 1Q g L= ×

1g water

Given 1 2420 Q Q= +

MC T igL∆ +

3 52100 5 10 3.36 10m −= × + × ×

3420 3368 10 8

10500m kg gm−−= = × =

79. A stationary source is emitting sound at a fixed frequency f0, which is reflectdd by

two cars approaching the source. The difference between the frequencies of soundreflected from the cars is 1.2% of f

0. What is the difference in the speeds of the cars

(in km per hour) to the nearest integer ? The cars are moving at constant speedsmuch smaller than the speed of sound which is 330 ms1.

Ans : 7

Sol : Let v is the speed of car approaching towards a stationary source emitty wave ofspeed υfrequency of the sound appeared to stationary observer is

0

vf f

v

υυ

+ = −

change of frequency is

( )( )

210

1 2

vvf f

v v

υυυ υ

+ +∆ = − − −

since 1 2v and v υ<<

21 2

0

2 12. 10f v v

f υ−∆ − = = ×

1 2 7.128v v−

nearest integer is 7





80. The focal length of thin biconvex lens is 20 cm. When an object is moved from adistance of 25 cm in front of it to 50 cm, the magnification of its image changes

from 25 50.m to m The ration 25

50

m

m is

Ans : 6

Sol : case-I u = 25 cm f = + 20 cm

v1 = 100 m

1 4m⇒ =

case - II u = 50 cm f = +20 cm

2

100

3V cm=

2 2 / 3m =

1

2

6m

m∴ =

81. An particleα − and a proton are acceleration from rest by a potential difference of

100V. After this, their de Broglie wavelength are pandαλ λ respectively. The ratio

,p

α

λλ to the nearest integer, is

Ans : 3

Sol : 2

h

mqVλ =

p

p p

m q

m qα α

α

λλ

= ×

4 2= ×

2 2 2 1.414= = ×

= 2.828

∴ nearest answer is 3

82. When two identical batteries of interval resistance 1Ω each are connected in se-ries across a resistor R, the rate of heat produece in R is J

1. When the same batteries

are connected in parallel across R, the rate is J2. If J

1 = 2.25 J

2 then the value of R in

Ω is

Ans : 4





Sol : in case -I

R

ε

1i

1Ω 1Ωε

1

2

2i

R

ε=+

2

1

2

2J R

R

ε ∴ = +

case-II

ε 1Ω

1Ω

R

ε

2

2

2 1I

R

ε=+

2

2

2

2 1J R

R

ε ∴ = +

given that J1 = 22.25 J× ∴ R = 4 Ω

83. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1

and T2, respectively. The maximum intensity in the emission spectrum of A is at

500 nm and in that of B is at 1500 nm. Considering them to be black bodies, whatwill be the ratio of the rate of total energy radiated by A to that of B ?

Ans : 9

Sol : 4E e ATσ=

where A is the area of body and e = 1 here maxT cantλ = ∴ ( )( )

2

1 214

2 1 2

/9

/

R RE

E λ λ= =

84. When two progressive waves y1 = 4 sin (2x 6t) and y

2 = 3 2 6

2x t

π − − are super-

imposed, the amplitude of the resultant wave is

Ans : 5

Sol : 2 21 2 1 22 cosRA A A A A φ= + + where 1 24, 3, / 2,A A φ π= = = 2 23 4 5RA = + =

2010 sri chaitanyas_iitjee_paper-i_solutions

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