aakash-goiit-iit-jee-2010 paper-1 answer and solution with analysis

8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis

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Answers by

IIT-JEE 2010(Division of Aakash Educational Services Ltd.)

PAPER - I : CODES

Q.No. 0 1 2 3 4 5 6 7 8 9

01. D C B D D C A B A D

02. A B C A A B D C D A

03. B B C D B A B C A A

04. B C B A C D C B D D

05. C D D B C A C A B C

06. C A A C B D B D C B

07. D A D C D B A A C B

08. A D A B A C D D B C

09. B,D A,B,C D B,D D A,B,C A,B D A,B,C A,B

10. B,C B,C A,B D B,D B,D B,C B,D A,B D

11. A,B D B,C A,B,C A,B D D A,B B,C A,B,C

12. D B,D A,B,C B,C A,B,C B,C B,D A,B,C D B,C

13. A,B,C A,B B,D A,B B,C A,B A,B,C B,C B,D B,D

14. A B A B A B A B A B

15. D C D C D C D C D C

16. C A C A C A C A C A

17. B D B D B D B D B D

18. C C C C C C C C C C

19. 3 3 0 3 3 0 3 4 3 5

20. 2 2 3 0 3 3 0 3 4 3

21. 0 5 2 3 0 3 3 0 3 4

22. 3 3 5 2 3 0 3 3 0 3

23. 4 4 3 5 2 3 0 3 3 0

24. 3 3 4 3 5 2 3 0 3 3

25. 3 0 3 4 3 5 2 3 0 3

26. 3 3 0 3 4 3 5 2 3 0

27. 0 3 3 0 3 4 3 5 2 3

28. 5 0 3 3 0 3 4 3 5 2

29. A D B C A D A C B C


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Organic Chemistry Inorganic Chemistry Physical Chemistry Total

Easy 1 4 3 8

Medium 7 4 7 18

Tough 1 0 1 2

Total 9 8 11 28

Markwise 27 24 33 84

XI Marks 33 XII Marks 51

XI syllabus 11 XII syllabus 17

Topic-wise Analysis of CHEMISTRY - IIT-JEE 2010

(Paper-1)

29%

64%

7%

Distribution of Level of Questions in Chemistry

Easy Medium Tough

32%

29%

39%

Topic wise distribution in Chemistry

Organic Chemistry

Inorganic Chemistry

Physical Chemistry

39%

61%

Percentage Portion asked from Syllabus of Class XI

& XII - Chemistry

XI syllabus XII syllabus

32%

29%

39%

Topic wise distribution of marks in Chemistry

Organic Chemistry

Inorganic Chemistry

Physical Chemistry

39%

61%

Percentage Portion asked (Mark-wise) from Syllabus of Class XI & XII -Chemistry

XI Marks XII Marks


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XII XI XI XI XII XII XII

Calculus Trigonometry Algebra 2D- Geometry Probability 3-D + Vector Algebra Total

Easy 1 2 1 2 0 1 0 7

Medium 2 2 4 1 1 3 2 15

Tough 1 0 1 0 0 0 4 6

Total 4 4 6 3 1 4 6 28

Markwise 12 12 18 9 3 12 18 84



Topic-wise Analysis of MATHEMATICS - IIT-JEE 2010

(Paper-1)

25%

54%

21%

Distribution of Level of Questions in Mathematics

Easy Medium Tough

14%

14%

22%11%

4%

14%

21%

Topic wise distribution in MathsCalculus

Trigonometry

Algebra

2D- Geometry

Probability

3-D + Vector

Algebra

46%54%

Percentage Portion asked from Syllabus of ClassXI & XII


14%

14%

22%11%

4%

14%

21%

Topic wise distribution of marks in Maths

Calculus

Trigonometry

Algebra

2D- Geometry

Probability

3-D + Vector

Algebra

46%

54%

Percentage Portion asked (Mark-wise) from Syllabus ofClass XI & XII

XI Marks XII Marks


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XII XI XII XI XII XI XII XII XI&XII

ElectricityHeat &

Thermodynamics

Modern

PhysicsMechanics Optics

Oscillation &

Waves

Magnetism

and EMI

Experimenta

l physics

Miscellan

eousTotal

Easy 3 2 1 2 1 1 1 0 0 11

Medium 1 2 0 5 1 0 1 2 2 14

Tough 0 0 0 2 0 1 0 0 0 3

Total 4 4 1 9 2 2 2 2 2 28

Markwise 12 12 3 27 6 6 6 6 6 84



Topic-wise Analysis of PHYSICS - IIT-JEE 2010

(Paper-1)

39%

50%

11%

Distribution of Level of Questions in Physics

Easy Medium Tough

15%

14%

4%32%

7%

7%

7%7%

7%

Topic wise distribution in PhysicsElectricity

Heat & Thermodynamics

Modern Physics

Mechanics

Optics

Oscillation & Waves

Magnetism and EMI

Experimental physics

Miscellaneous

61%

39%

Percentage Portion asked from Syllabus of ClassXI & XII - Physics


15%

14%

4%32%

7%

7%

7%7%

7%

Topic wise distribution of marks in Physics

Electricity

Heat & Thermodynamics

Modern Physics

Mechanics

Optics

Oscillation & Waves

Magnetism and EMI

Experimental physics

Miscellaneous

61%

39%

Percentage Portion asked (Mark-wise) from Syllabus of ClassXI & XII - Physics

XI Marks XII Marks


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Solutions

forforforforforIIT-JEE 2010

Time : 3 hrs. Max. Marks: 252

Regd. Office: Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472

Instructions :

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

of four Sections.

2. 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.

3. For each question in Section II,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. Partial marks will

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

4. For each question in Section III,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.

5. For each question in Section IV,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. No negative marks

will be awarded for in this Section.

DATE : 11/04/2010

PAPER - 1 (Code - 5)


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AakashIIT-JEE -Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, NewDelhi-75 Ph.: 011-47623417/23 Fax : 47623472

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IIT-JEE 2010 (Paper-1)

PARTI : CHEMISTRY

SECTION - I

Single Correct Choice Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

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

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

Answer (C)

Hints : Bond energy of CC is considered to be 100 kcal mol1 approximately.

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

(A) Br2(g) (B) Cl

2(g) (C) H

2O(g) (D) CH

4(g)

Answer (B)

Hints : Cl2

is considered as standard state of chlorine and enthalpy of formation in standard state is zero (assumed).

3. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that followsArrhenius equation is

(A) k

T

(B) k

T

(C) k

T

(D)k

T

Answer (A)

Hints :aE

RTk A e=

ln k = ln A aE

RT

4. In the reaction OCH3HBr

the products are

(A) OCH3 and H2Br (B) Br and CH Br3

(C) Br and CH OH3 (D) OH and CH Br3


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Answer (D)

Hints :

O

CH3

HBrO

CH3

H

Br OH + CH Br3

5. The correct statement about the following disaccharide is

CH OH2

H

OH H

H OH

H

OH

OH

OCH CH O2 2

CH OH2O

CH OH2

H

OH

H

H

OH

(a) (b)

(A) Ring (a) is pyranose with -glycosidic link (B) Ring (a) is furanose with -glycosidic link

(C) Ring (b) is furanose with -glycosidic link (D) Ring (b) is pyranose with -glycosidic link

Answer (A)

Hints :

CH OH2

H

OH H

H OH

H

OH

OH

OCH CH O2 2

CH OH2O

CH OH2

H

OH

H

H

OH

(a) (b)

5

4

3 2

1

-linkage

Ring (a) is six membered oxygen containing ring.

Pyranose ring and CH2OH of C5 and OR of C1 are across of one another hence, it is -glycosidic linkage.

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

(A) BrCH2CH2CH2CH2CH3 and CH3CH2C CH (B) BrCH2CH2CH3 and CH3CH2CH2C CH

(C) BrCH2CH

2CH

2CH

2CH

3and CH

3C CH (D) BrCH2CH2CH2CH3 and CH3CH2C CH

Answer (D)

Hints :

CH CH C CH3 2 NH2

CH CH C C3 2

CH CH CH2 2 2CH3

Br

CH CH C CCH CH CH CH3 32 2 2 23-octyne

7. The ionization isomer of [Cr(H2O)

4Cl(NO

2)]Cl is

(A) [Cr(H2O)4(O2N)]Cl2 (B) [Cr(H2O)4Cl2] (NO2)(C) [Cr(H

2O)

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

2O)

4Cl

2(NO

2)] H

2O


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Answer (B)

Hints : Cr(H2O)4Cl(NO2)]Cl and [Cr(H2O)4Cl2](NO2) have same molecular formula but they give different ions onionization.

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

(A) N CH CH N

HOOC CH2

HOOC CH2

CH2 COOH

CH2 COOH

(B) N CH2 CH2 N

HOOC

HOOC

COOH

COOH

(C) N CH2 CH2 N

CH2 COOH

CH2 COOH

HOOC CH2

HOOC CH2

(D) N CH CH N

H

CH2 COOH

HOOC CH2

HCH2

HOOC

CH2

COOH

Answer (C)

Hints :

N CH2 CH2 N

CH2 COOH

CH2 COOH

HOOC CH2

HOOC CH2

SECTION - II

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.

9. In the reaction

OH

NaOH(aq)/Br2 the intermediate(s) is/are

(A)

O

Br

Br

(B)

O

BrBr

(C)

Br

O

(D)

Br

O

Answer (A, B, C )

Hints : (A) & (C) is usual product. Formation of (B) is justified as

O

BrBr

O

BrBr Br


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10. In the Newman projection for 2, 2-dimethylbutane

XCH3

HH

H C3

Y

X and Y can respectively be

(A) H and H (B) H and C2H

5(C) C

2H

5and H (D) CH

3and CH

3

Answer (B, D)

Hints :

XCH3

HH

H C3

Y

C

CY

CH3H C3

X

HH

When X = H & Y = C2H

5

C

C C H2 5

CH3H C3

H

HH

2

1

2, 2- dimethyl butane

When X = CH3

and Y = CH3

C

C CH3

CH3H C3

CH3

HH

2

3 1

4

2, 2-dimethyl butane

11. Aqueous solutions of HNO3, KOH, CH

3COOH, and CH

3COONa of identical concentrations are provided. The pair(s)

of solutions which form a buffer upon mixing is(are)

(A) HNO3 and CH3COOH (B) KOH and CH3COONa

(C) HNO3

and CH3COONa (D) CH

3COOH and CH

3COONa

Answer (D)

Hints : Solution of weak acid and its salt with strong base behave like buffer.

CH3COOH CH

3COONa

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

(A) Ca3(PO

4)2

(B) Ca(OH)2

(C) Na2CO

3(D) NaOCl

Answer (B, C)

Hints : Ca(OH)2

and Na2CO

3is used to remove hardness of water. Ca(OH)

2is used in Clarke's method.


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

(A) Molar conductivity (B) Electromotive force (C) Resistance (D) Heat capacity

Answer (A, B)

Hints : Molar conductivity and electromotive force are intensive properties.

SECTION - III

Paragraph Type

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon thesecond 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 concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. Theresulting potential difference across the cell is important in several processes such as transmission of nerve impulses

and maintaining the ion balance. A simple model for such a concentration cell involving a metal M isM(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; G > 0 (B) Ecell > 0; G < 0 (C) Ecell < 0; G > 0 (D) Ecell > 0; G < 0

Answer (B)

Hints : Cell reaction

(1M) (0.05 M)M M+ +

Apply Nernst equation

0.059 0.05E E log

1 1=

( )20.059

E log 5 101

=

( )0.059

E 2 log51

= +

15. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potentialwould be

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

Answer (C)

Hints :1

2

E log0.05

E log0.0025=

21

4

2

E log5 10

E log25 10

=

1E 70(given)

2

70 1.3 1E 2.6 2

= =


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Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of

copper include chalcanthite (CuSO45H

2O), atacamite (Cu

2Cl(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 (CuFeS

2). The extraction

of copper from chalcopyrite involves partial 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

3(D) Cu

2O and Fe

2O

3

Answer (A)

Hints : Partial roasting of chalcopyrite

2O /

2 2CuFeS Cu S FeO +

17. Iron is removed from chalcopyrite as

(A) FeO (B) FeS (C) Fe2O

3(D) FeSiO

3

Answer (D)

Hints : FeO + SiO2 FeSiO

3

gangue flux slag

18. In self-reduction, the reducing species is

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

Answer (C)

Hints : Cu2S + Cu2O 6Cu + SO2

SECTION - IV

(Integer Type)

This section contains 10 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 F Br F angles in BrF5

is

Answer (0)

Hints :

Br

FeqFeqFeq

Feq

Fa

Presence of lone pair will distort all the bond angles and none of the bond angle is exactly equal to 90

FaBrFeq = 84

Feq

BrFeq

= 89

20. The value of n in the molecular formula BenAl

2Si

6O

18is

Answer (3)

Hints : Be3Al

2Si

6O

18


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21. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL.

The number of significant figures in the average titre value is

Answer (3)

Hints : Average titre value =25.2 25.25 25

3

+ +

= 25.15

Siginificant figure is 3.

22. The concentration of R in the reaction R P was measured as a function of time and the following data is

obtained

[R] (molar)

t(min.)

1.0 0.75 0.40 0.10

0.0 0.05 0.12 0.18

The order of the reaction isAnswer (0)

Hints : For zero order reaction

dxk

dt=

from 1 to 0.75 (M) from 0.75 to 0.40 from 0.40 to 0.10

0.255

0.05=

0.355

0.07=

0.305

0.06=

It mean order of this reaction is zero.

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

142

54 Xe and9038 Sr is

Answer (3)

Hints :

142235 90 15492 38 0U Xe+ Sr +3 n

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

H N3 CH2 CH2 CH2 CH2

CHH N2

C

O

O

Answer (2)

Hints :

H N3 CH2 CH2 CH2 CH2

CH

H N2

C

O

O

Basic site

Basic site (Carboxylate ion will

also behave as proton acceptor)


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25. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H

6is

Answer : (5)

Hints : Index of hydrogen deficiency in the compound C4H

6is 2. Therefore either bicyclic compound or cycloalkenes

are possible as cyclic isomers.

Possible isomers are

, , , ,

26. In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is

3

2

1. O 1. NaOH(aq)

2. Zn,H O 2. Heat Y

Answer : (3)

Hints :

1. O

2. Zn H O

3

2

O

O H

O

O

1. O H

O

O+HO3

+

O

H O2

O

+

OH

O

+

O

On dehydration thisalcohol will give threeproducts

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

NH C3 CH3 COOH OCH CH2 3 OH

CH OH2

NO2 CH CH2 3 COOH

CH CH2 3

OH

NH C3 CH3


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Answer : (4)

Hints : Carboxylic acids and aromatic alcohols dissolves in aqueous NaOH.

COOH OH COOHOH

NH C3 CH3

, , ,

will be soluble in aqueous NaOH.

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

KCN K2SO

4(NH

4)2C

2O

4NaCl Zn(NO

3)2

FeCl3

K2CO

3NH

4NO

3LiCN

Answer : (3)

Hints : In basic medium, red litmus paper turn blue

(1)2CN H O HCN OH

+ +(2) 2

3 2 3CO H O HCO OH + +

(3)2CN H O HCN OH

+ +

PARTII : MATHEMATICSSECTION - I


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.

29. Let f, gand hbe real-valued functions defined on the interval [0, 1] by2 2 2 2

( ) , ( )x x x x f x e e g x xe e = + = + and2 22( ) x xh x x e e = + . If a, band cdenote, respectively, the absolute maximum of f, gand hon [0, 1], then

(A) a = band cb (B) a = cand ab (C) aband cb (D) a = b = c

Answer (D)

Hints :

The given functions

2 2

( ) x xf x e e = +2 2

( ) x xg x xe e = +

2 22( ) x xh x x e e = +

are strictly increasing in [0, 1], hence at x= 1, the given functions attain absolute maximum all equal to1

e

e

+

and consequently a = b = c.


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30. Let pand qbe real numbers such that p 0, p3qand p3q. If and are nonzero complex numbers

satisfying + = pand 3 + 3 = q, then a quadratic equation having and as its roots is

(A) (p3 + q)x2 (p3 + 2q)x+ (p3 + q) = 0

(B) (p3 + q)x2 (p3 2q)x+ (p3 + q) = 0

(C) (p3 q)x2 (5p3 2q)x+ (p3 q) = 0

(D) (p3 q)x2 (5p3 + 2q)x+ (p3 q) = 0

Answer (B)

Hints :

We have + = p

3 +3 = q= ( + )3 3( + ) = p3 + 3p()

3

3

p q

p

+ =

The quadratic equation with and as roots is

2 0x x

+ + =

2

2 ( ) 2 1 0x x +

+ =

32

2

3

23

1 0

3

p qp

px xp q

p

+

+ =+

(p3 + q)x2 (p3 2q)x+ (p3 + q) = 0

31. The value of 3 400

1 (1 )lim

4

x

x

t n tdt

x t

+

+

is

(A) 0 (B)1

12(C)

1

24(D)

1

64

Answer (B)

Hints :

We have

40

3 4 30 00

(1 )

41 (1 )lim lim

4

x

x

x x

t n tdt

tt n tdt

x t x

+

++=

+

= 4 20

(1 ) 1Lt

4 3x

x n x

x x

+

+

= 20

(1 ) 1 1Lt

123( 4)x

n x

x x

+ =

+


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

1

0

0

x

A y

z

=

has

exactly two distinct solutions, is

(A) 0 (B) 29 1 (C) 168 (D) 2

Answer (A)

Hints :

The equation

1

0

0

x

A y

z

=

has two distinct solutions. It should be noted here that the given equation is linear

equation in 3 variables, which may have no solution, or unique solution or infinitely many solutions.

Hence there does not exist any matrix A such that the given equation has exactly two solutions and consequentlynumber of 3 3 matrices is 0.

33. Let P, Q, R and S be the points on the plane with position vectors 2 , 4 , 3 3 and 3 2i j i i j i j + +respectively. The quadrilateral PQRSmust be a

(A) Parallelogram, which is neither a rhombus nor a rectangle

(B) Square

(C) Rectangle, but not a square

(D) Rhombus, but not a square

Answer (A)

Hints :

Let Obe the origin

then 2OP i j =

4OQ i=

3 3OR i j = +

P

S R

Q

O

3 2OS i j = +

Here we have

6PQ i j = +

3QR i j = +

6RS i j =

and 3PS i j = +

5 4PR i j = +

7 2QS i j = +

35 8 27 0PR QS = + =

Diagonals are not perpendicular

and | | | |, | | | |PQ RS QR PS = =

Hence PQRSis a parallelogram which is neither a rhombus nor a rectangle.


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34. Let be a complex cube root of unity with 1. A fair die is thrown three times. If r1, r

2and r

3are the numbers

obtained on the die, then the probability that 31 2 0rr r + + = is

(A)1

18(B)

1

9(C)

2

9(D)

1

36

Answer (C)

Hints :

is a complex cube root of unity with 1

31 2 0rr r + + = where r1, r2, r3 can be successively selected from the sets {1, 4}, {2, 5} and {3, 6}.

n(S) = 6 6 6 = 216

n(E) = 2C1

2C1

2C1

3C1

2C1

= 6 8

Required probability =6 8 2

216 9

=

35. Equation of the plane containing the straight line2 3 4

x y z= = and perpendicular to the plane containing the straight

lines3 4 2

x y z= = and

4 2 3

x y z= = is

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

Answer (C)

Hints : The equation of the plane containing the line2 3 4

x y z= = may be written as

ax + by + cz =0 while 2a+ 3b+ 4c= 0 ... (i)

The equation of the plane containing straight lines3 4 2

x y z= = and4 2 3

x y z= = can be put in the form

a1x+ b

1y+ c

1z= 0

where 3a1

+ 4b1

+ 2c1

= 0

4a1

+ 2b1

+ 3c1

= 0

1 1 18 1 10

a b c= =

Then the equation of the plane containing given two lines is

8xy 10c= 0 ... (ii)

Since the plane (i) is perpendicular to the plane (ii),

hence 8ab 10c= 0

Also 2a+ 3b+ 4c= 0

4 30 20 32 24 2

a b c= =

+ +

26 52 26

a b c= =

1 2 1

a b c= =

Substituting for a, b, cin equation (i), the required equation of the plane is

x 2y + z= 0


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36. If the angles A, B and Cof a triangle are in an arithmetic progression and if a, band cdenote the lengths of

the sides opposite to A, Band Crespectively, then the value of the expression sin2 sin2a c

C Ac a

+ is

(A)1

2

(B)3

2

(C) 1 (D) 3

Answer (D)

Hints :

Angles A, B, Care in A.P. hence B= 60 =3

sin2 sin2a c

C Ac a

+ =sin sin

2 cos 2 cosC A

a C c Ac a

+

= 2(sinA cos C+ cosA sinC)

= 2 sin(A + C) = 2 sinB=3

2 32

=

SECTION - II

(Multiple Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.

37. Let ABCbe a triangle such that6

ACB

= and let a, band cdenote the lengths of the sides opposite to A,

Band Crespectively. The value(s) of xfor which a= x2

+ x+ 1, b= x2

1 and c= 2x+ 1 is/are

(A) ( )2 3 + (B) 1 3+ (C) 2 3+ (D) 4 3Answer (B)

Hints :2 2 2

cos2

b a cC

ab

+ =

A B

C

6( ) ( ) ( )

( )( )

2 2 22 2

2 2

1 1 2 13

22 1 1

x x x x

x x x

+ + + +=

+ +

2 2 2 2( 2)( 1) ( 1) ( 1) 3( 1)( 1)x x x x x x x x + + + = + +

2 2 23( 1) 2 1x x x x x + + = + +

2(2 3) (2 3) ( 3 1) 0x x + + =

(2 3)& 1 3x x= + = +

But (2 3)x= +

shows that cis negative, which is not possible in a triangle.

Hence 1 3x= + is only possible value of x.1 3x = +


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38. Let z1

and z2

be two distinct complex numbers and let z = (1 t) z1

+ tz2

for some real number t with0 < t< 1. If Arg (w) denotes the principal argument of a nonzero complex number w, then

(A) |zz1| + |zz

2| = |z

1z

2| (B) Arg (zz

1) = Arg (zz

2)

(C)1 1

2 1 2 1

0z z z z

z z z z

=

(D) Arg (zz1) = Arg (z2 z1)

Answer (A, C, D)

Hints : 1 21

zz z

t t=

we have z= (1 t)z1

+ tz2, 0 < t< 1,

( )2 111

t z t z z

t t

+ =

+

( )2 11z t z t z = +

z1, z, z

2are collinear

so Arg(zz1) = Arg(zz

2) = Arg(z

2z

1)

39. The value(s) of( )

441

2

0

1

1

x xdx

x

+is/are

(A)22

7 (B)

2

105(C) 0 (D)

71 3

15 2

Answer (A)

Hints : ( )

441

2

0

1

1

x xdx

x

+

1 8 7 6 5 4

2

0

4 6 4.1

x x x x x dxx

+ += +

16 5 4 2

2

0

44 5 4 4

1x x x x dx

x

= + + +

1 4 5 44 4

7 6 5 3 4

= + +

22

7=

40. Let A and Bbe two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle ofradius rhaving ABas its diameter, then the slope of the line joining A and Bcan be

(A)1

r (B)

1

r(C)

2

r(D)

2

r

Answer (C, D)

Hints :

Mid - point of A & B

a(t1

+ t2) = r

1 2rt ta

+ =


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Slope of( )

( )1 2

2 21 21 2

2 2 2a t t aAB

t t rt t

= = =

+

CB

y = ax2

4

A ( )21 1, 2at at

( )22 2, 2at at In y2 = 4x

a= 1

slope =2

r

If circle in below x-axis, then

Slope = 2

r

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

1 sinx

f x lnx 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 > 1 such that |f(x)| < |f(x)| for all ( ),x

(D) There exists > 0 such that |f(x)| + |f(x)| for all ( )0,x

Answer (B, C)

Hints : ( ) 1' 1 sinf x xx= + + , x> 0 but not differentiable in (0, ) as sinxmay be 1 and then f(x) can't exist.

Clearly f(x) is continuous on (0, )

(D) Is not true because f(x) tends to as xtends to

(C) Is true because ( )' 3f x if x> 1

But |f(x)| > 3, if x> e3

So we may take = e3

SECTION - III

(Paragraph Type)

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


The circle x2 + y2 8x= 0 and hyperbola2 2

19 4

x y = intersect at the points A and B.

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

(A)2 5 20 0x y =

(B)2 5 4 0x y + =

(C) 3x 4y+ 8 = 0 (D) 4x 3y+ 4 = 0

Answer (B)


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Hints : Equation of tangent to hyperbola having slope mis

29 4y mx m = + ...(1)

Equation of tangent to circle is

( ) 24 16 16y m x m = + + ...(2)

(1) and (2) will be identical for2

5m= satisfy

Equation of common tangent is 2 5 4 0x y + =

43. Equation of the circle with ABas its diameter is

(A) x2 + y2 12x+ 24 = 0 (B) x2 + y2 + 12x+ 24 = 0

(C) x2 + y2 + 24x 12 = 0 (D) x2 + y2 24x 12 = 0

Answer (A)

Hints :

The equation of the hyperbola is

2 2

19 4

x y =

and that of circle is

x2 + y2 8x= 0

For their points of intersection

2 28

19 4

x x x+ =

4x2 + 9x2 72x= 36

13x2 72x 36 = 0

13x2 78x+ 6x 36 = 0

13x(x 6) + 6(x 6) = 0

x= 6,

13

6x=

13

6x= not acceptable

Now, for x= 6, 2 3y=

Required equation is

( ) ( )( )26 2 3 2 3 0x y y + + =

x2 12x+ y2 + 24 = 0

x2 + y2 12x+ 24 = 0


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Let pbe an odd prime number and Tp

be the following set of 2 2 matrices

: , , {0, 1, 2, ....... 1}Pa b

T A a b c p c a

= =

44. The number of A in Tp

such that A is either symmetric or skew-symmetric or both, and det (A) divisible by pis

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

Answer (D)

Hints :

If A is skew-symmetric, |A| = b2

p |A| iff b= 0 in which case A is symmetric.

If A is2 2

, | |

a b

A a bb a

=

| |p A if either pa+ bor pab.

The second case occurs only when a = b which gives p choices. The first case again gives p choices with

a= 0 = bcounted again.

The number of such matrices is 2p 1.


such that the trace of A is not divisible by pbut det (A) is divisible by pis

[Note : The trace of a matrix is the sum of its diagonal entries](A) (p 1)(p2 p+ 1) (B) p3 (p 1)2 (C) (p 1)2 (D) (p 1) (p2 2)

Answer (C)

Hints :

Tr(A) = 2awill be divisible by pif and only if a= 0.

We want to find all matricesa b

c a

with a= 1, 2, ....., p 1 for which a2 bc 0 (mod p).

There are exactly p 1 ordered pairs (b, c) for any value of a.

Required number is (p 1)2


such that det (A) is not divisible by pis

(A) 2p2 (B) p3 5p (C) p3 3p (D) p3 p2

Answer (D)

Hints :

The number of matrices for which pdoes not divide Tr(A) is (p 1)p2 of these (p 1)2 are such that p |A|. The

number of matrices for which pdivides Tr(A) and pdoes not divide |A| are (p 1)2.

Required number is (p 1)p2 (p 1)2 + (p 1)2 = p3 p2


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SECTION - IV

(Integer Type)

This section contains 10 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.

47. For any real number x, let [x] denote the largest integer less than or equal to x. Let fbe a real valued functiondefined on the interval [10, 10] by

{ } if [ ] is odd,( )

1 { } if [ ] is even

x xf x

x x

=

Then the value of

102

10

( )cos10

f x x dx

is

Answer (4)

Hints :

[ ] if [ ] is odd,( )

1 + [ ] if [ ] is even

x x xf x

x x x

=

10 9 2 1 0 1 2 9 10

f(x)and cosxare both periodic with period 2 and are both even.

10 10

10 0

( )cos 2 ( )cosf x xdx f x xdx

=

2

0

10 ( )cosf x xdx =

1 1 1

0 0 0

( )cos (1 )cos cosf x xdx x xdx u udu = =

2 2 1

1 1 0

( )cos ( 1)cos cosf x xdx x xdx u udu = =

10 1

10 0

( )cos 20 cosf x xdx u udu

=

2

40=

102

10

( )cos 410

f x xdx

=


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48. Let be the complex number2 2

cos sin .3 3

i

+ Then the number of distinct complex numbers z satisfying

2

2

2

1

1 0

1

z

z

z

+

+ =

+

is equal to

Answer (1)

Hints :

Let

2

2

2

1

1

1

A

=

2

0 0 0

0 0 0 , ( ) 0, | | 0

0 0 0

A Tr A A = = =

A3 = 0

2

2

2

1

1 | | 0

1

z

z A zI

z

+

+ = + =

+

z3

=0 z= 0, the number of zsatisfying the given equation is 1.

49. Let Sk, k= 1, 2,.....100, denote the sum of the infinite geometric series whose first term is

1

!

k

k

and the common

ratio is1

.k

Then the value of2 100

2

1

100( 3 1)

100!k

k

k k S=

+ + is

Answer (4)

Hints :

We have,

1

!1

1k

k

kS

k

=

1

( 1)!k=

Now,

2 1( 3 1) {( 2)( 1) 1}( 1)!

kk k S k k k

+ =

1 1

( 3)! ( 1)!k k=


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100

2

1

1 1| ( 3 1) | 1 1 2

99! 98!k

k

k k S=

+ = + + +

21004

100!=

2 1002

1

100| ( 3 1) | 4

100!k

k

k k S=

+ + =

50. The number of all possible values of , where 0 < < , for which the system of equations

(y + z)cos3 = (xyz)sin3

2cos3 2sin3sin3x

y z

= +

(xyz)sin3 = (y+ 2z)cos3 + ysin3

have a solution (x0, y

0, z

0) with y

0z

0 0, is

Answer (3)

Hints :

Given equation can be written as

cos3 cos3sin3 0x

y z

= ...(i)

2cos3 2sin3sin3 0x

y z

= ...(ii)

2 1sin3 cos3 (cos3 sin3 ) 0x

y z + = ...(iii)

Equations (ii) and (iii) imply

2sin3 = cos3 + sin3

sin3 = cos3

5 9

tan3 1 3 , ,4 4 4

= =

or 5 9, ,12 12 12 =

51. Let fbe a real-valued differentiable function on R(the set of all real numbers) such that f(1) = 1. If the y-intercept

of the tangent at any point P(x, y) on the curve y= f(x) is equal to the cube of the abscissa of P, then the value

of f(3) is equal to

Answer (9)

Hints :

The equation of the tangent at (x, y) to the given curve y= f(x) is

( )dy

Y y X x dx

=

intercept =dy

Y y xdx


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According to the question,

3 dyx y xdx

=

2dy y

xdx x =

which is linear is x

1

. .dx

xI F e

=

=1

x

Required solution is

21 1

.y x dx x x=

2

2

y xC

x

= +

3

2

xy Cx

= +

at x= 1, y= 1

1

12

C

= +

3

2C=

Now,27 3

( 3) ( 3)2 2

f = +

27 99

2

= =

52. The number of values of in the interval ,2 2

such that 5

n

for n= 0, 1, 2 and tan = cot 5 as

well as sin2 = cos4 is

Answer (3)

Hints :

tan = cot5

we have,

tan tan 52

=

52

n = +


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62

n

=

12 6

n =

Also,

cos4 sin2 cos 22

= =

4 2 22

n

=

Taking positive

6 22

n

= +

3 12n = +

Taking negative

2 22

n

=

4n

=

Above values of suggests that there are only 3 common solutions.

53. The maximum value of the expression2 2

1

sin 3sin cos 5cos + + is

Answer (2)

Hints :

We have,

2 2

1( )

sin 3sin cos 5cosf =

+ +

Let g() = sin2 + 3sincos + 5cos2

1 cos2 1 cos2 35 sin22 2 2

+ = + +

33 2cos2 sin2

2= + +

min

9( ) 3 4

4g = +

5 13

2 2= =

maxmin

1( ) 2( )

fg

= =


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54. if anda b

are vectors in space given by 2 2 3

and = ,5 14

i j i j k a b

+ +=

then the value of

(2 ).[( ) ( 2 )]a b a b a b +

is

Answer (5)

Hints :

From the given information, it is clear that

2| | 1, | | 1, . 0

5

i ja a b a b

= = = =

Now, (2 ).[( ) ( 2 )]a b a b a b +

= 2 2(2 ).[ ( . ). 2 . 2( . ). ]a b a b a b a b a b a a + +

= [2 ].[ 2 ]a b b a + +

= 4a2 + b2

= 4.1 + 1 = 5 [ as . 0]a b=

55. The line 2x+ y= 1 is tangent to the hyperbola2 2

2 21.

x y

a b = If this line passes through the point of intersection

of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

Answer (2)

Hints :

F1 12

, 0

Clearly1

2

a

e= ...(i)

Also y= 2x+ 1 is tangent to hyperbola

1 = 4a2 b2 ...(ii)

( )221

4 1ea

=

2

2

45 e

e=

e4 5e2 + 4 = 0

(e2 4)(e2 1) = 0

e= 2, e= 1

e= 1 gives the conic as parabola. But conic is given as hyperbola, hence

e= 2


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56. If the distance between the plane Ax 2y + z = dand the plane containing the lines1 2 3

2 3 4

x y z = = and

2 3 4

3 4 5

x y z = = is 6 , then |d| is

Answer (6)

Hints :

Equation of plane containing the given lines is

1 2 3

2 3 4 0

3 4 5

x y z

=

(x 1)( 1) (y 2)(2) + (z 3) (1)

x+ 1 + 2y 4 z+ 3 = 0

x+ 2y z= 0 ...(i)

Given plane is

x 2y+ z = d ...(ii)

(i) and (ii) are parallel

Clearly A = 1

Now, distance between plane 61 4 1

d= =

+ +

| d| = 6

PARTIII : PHYSICS

SECTION - I


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of whichONLY ONE is correct.

57. A thin uniform annular disc (see figure) of mass Mhas outer radius 4Rand inner radius 3R. The work requiredto take a unit mass from point Pon its axis to infinity is

3R4R

P

4R

(A)

2

(4 2 5)7

GM

R (B)2

(4 2 5)7

GM

R (C) 4

GM

R (D)

2

( 2 1)5

GM

R


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Answer (A)

Hints :

Potential4

2 23 16

R

P

R

GdMV

R x

=

+

=

4

2 2 23

27 16

R

R

M xdx G

R R x

+

Solving

2(4 2 5)

7P

GMV

R

=

P

X

dx

Work done in moving a unit mass from Pto

=

2

0 (4 2 5)7PGM

V V R

=

=2

(4 2 5)7

GM

R

58. A block of mass mis on an inclined plane of angle . The coefficient of friction between the block and the plane

is and tan > . The block is held stationary by applying a force Pparallel to the plane. The direction of force

pointing up the plane is taken to be positive. As P is varied from P1

= mg(sin co s) to

P2

= mg(sin + cos), the frictional force fversus Pgraph will look like

P

(A)P

2

P1

P

f

(B)P

2P

1

P

f

(C)P2

P1

P

f

(D)P

2P

1

P

f

Answer (A)


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Hints :

From FBD of block

N= mgcos

P+ f= mgsin

f= mgsin P

fN

mg

P

As Pvaries from (mgsin cos) to (mgsin + cos), fvaries

from + mgcos to mgcos. This value of friction is always

less than or equal to Nin magnitude.

GraphP1

P2

59. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current Iin the clockwise

direction, as shown in the figure. When the system is put in a uniform magnetic field of strength Bgoing into

the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

(A) IBL (B)IBL

(C)

2

IBL

(D)

4

IBL

Answer (C)

Hints :

From FBD of a small element on wire

2 sin2

dT idlB

=

22

dT iRd B

=

idlB

T T

T= BiR

2BilT =

idlB

T T


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

is connected in series with a

capacitance Cand an electric bulb of resistance R(inductance zero). When is increased

(A) The bulb glows dimmer (B) The bulb glows brighter

(C) Total impedance of the circuit is unchanged (D) Total impedance of the circuit increases

Answer (B)Hints :

As is increased, impedance of circuit decreases.

R

Current increases

Bulb glows brighter.

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

1, a small resistance R

2,

two identical galvanometers G1

and G2, and a variable voltage source V. The correct circuit to carry out the

experiment is

(A)

G2

R2

G1

RT

R1

V

(B)

G2

R1

G1

RT

R2

V

(C)

R1

G1

RT

R2

G2

V

(D)

R2

G1

RT

R1

G2

V

Answer (C)

Hints :

G1

is used as voltmeter by connecting it in series with high resistance and applying it across RT

in parallel G2

is used as ammeter by shunting it with small resistance and connecting in series with RT.

R1

G1

R2

G2

RT


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

100, R

60and R

40, respectively, the relation between these resistance is

(A)100 40 60

1 1 1

R R R= + (B) R

100= R

40+ R

60

(C) R100

> R60

> R40

(D)100 60 40

1 1 1

R R R> >

Answer (D)

Hints :

2VP

R=

1

PR

100 60 40

1 1 1R R R

> >

Normally100 40 60

1 1 1

R R R= + could have been correct. But small change in resistance due to temperature change

can destroy the equality.

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

Answer (D)

Hints :

Real gas behaves like an ideal gas if its pressure is low and temperature is high. This insures large intermolecularseparation and no condensation.

64. Consider a thin square sheet of side L and thickness t, made of a material of resistivity . The resistance betweentwo opposite faces, shown by the shaded areas in the figure is

L

t

(A) Directly proportional to L (B) Directly proportional to t

(C) Independent of L (D) Independent of t

Answer (C)

Hints :

l LR

A tL t

= = =

Resistance is independent of L.


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SECTION - II

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.

65. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He usesa stop watch with the least count of 1 second for this and records 40 seconds for 20 oscillations. For this

observation, which of the following statement(s) is (are) true?

(A) Error Tin measuring T, the time period, is 0.05 seconds

(B) Error Tin measuring T, the time period, is 1 second

(C) Percentage error in the determination of gis 5%

(D) Percentage error in the determination of gis 2.5%

Answer (A, C)

Hints :

Relative error in measurement of time 1 second 1

40 second 40= =

Time period = 2 second

Error in measurement of time period1 1

2 second = 0.05 second40 20

= =

2

1g

T

2 2 1 1

40 20g T

g T = = =

100 5%g

g

=

66. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass

reverses its direction and moves with a speed of 2 ms1. Which of the following statement(s) is (are) correct for

the system of these two masses?

(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

Answer (A, C)

Hints :

Initial

Final

5 kg1 kg

v2

u O


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By conservation of momentum

1u= 2 + 5v

By coefficient of restitution equation

21

v

u

+=

Solving we get

3 m/su= 1m/sv =

Momentum = 1 3 = 3 kg m/s

Momentum of 5 kg block after collision = 5 1 = 5 kg ms1

Kinetic energy of centre of mass

21 1 3

(1 5)2 6

= +

= 0.75 J

Total kinetic energy21 1 3 4.5 J

2= =

67. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA as shown in the figure. Its pressureat A is P

0. Choose the correct option(s) from the following :

V0

4V0

A

B

V

T0

TC

(A) Internal energies at A and Bare the same

(B) Work done by the gas in process ABis P0V

0 ln4

(C) Pressure at Cis 04

P

(D) Temperature at Cis

0

4

T

Answer (A, B)

Hints :

ABis an isothermal process

Internal energy at A = Internal energy at B.

ApplyingPV

T= constant between A and B

V0

4V0

A

B

V

T0

TC

0 0 0 0

0 0

4

4= =B

B

P V P V P P

T TAs it is not clearly indicated that BCis passing through origin, prediction about point Cis not possible.


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

and Q2

fixed at two different points on the x-axis areshown in the figure. These lines suggest that

Q1

Q2

(A) |Q1| > |Q

2|

(B) |Q1| < |Q

2|

(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

Answer (A, D)

Hints :

Lines are denser around Q1

|Q1| > |Q

2|

Since |Q1| > |Q2|, then electric field will be zero at some distance to right of Q2

69. A ray OPof monochromatic light is incident on the face ABof prism ABCDnear vertex Bat an incident angle

of 60(see figure). If the refractive index of the material of the prism is 3 , which of the following is/are correct?

B

D

135

60

7590

A

CP

O

(A) The ray gets totally internally reflected at face CD

(B) The ray comes out through face AD

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

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

Answer (A, B, C)

Hints :

At P

1 sin60 3 sinr =

r= 30

From geometry, angle of incidence at Qis 45

P

D

60

7590

R

60

Q45

45

30

60

30

At Q

33 sin(45 ) 1

2 = >

TIRtakes place

At R

angle of incidence is 30

By symmetry R= 60

From second diagram, angle of deviation is 90

60

60

i

e


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SECTION - III (Paragraph Type)

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon thesecond paragraph 3 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C)and (D) out of which ONLY ONE is correct.


Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero

as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that

their critical temperature becomes smaller than TC(0) if they are placed in a magnetic field, i.e., the critical temperature

TC(B) is a function of the magnetic field strength B. The dependence of T

C(B) on Bis shown in the figure.

T BC( )

TC(0)

OB

70. In the graphs below, the resistance Rof a superconductor is shown as a function of its temperature Tfor twodifferent magnetic fields B

1(solid line) and B

2(dashed line). If B

2is larger than B

1, which of the following graphs

shows the correct variation of Rwith Tin these fields?

(A)

T

R

O

B1B2(B)

T

R

O

B1

B2

(C)

T

R

O

B1 B2(D)

T

R

O

B1

B2

Answer (A)

Hints :

As B2

> B1; critical temperature becomes less. Correct answer is (A).

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

Cdecreases 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 < TC(B) < 100 K

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

(D) B= 10 Tesla, TC(B) = 70 K

Answer (B)

Hints :

As critical temperature decreases continuously, with increase in magnetic field,

For B= 5T; TC

> 75 K but TC

< 100 K

Correct answer is (B).


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When a particle of mass mmoves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic

motion. The corresponding time period is proportional to ,m

kas can be seen easily using dimensional analysis.

However, the motion of a particle can be periodic even when its potential energy increases on both sides of x= 0 in

a way different from kx2

and its total energy is such that the particle does not escape to infinity. Consider a particleof mass mmoving on the x-axis. Its potential energy is V(x) = x4 ( > 0) for | x | near the origin and becomes aconstant equal to V

0for |x| X

0(see figure).

V0

X0x

V x( )

72. If the total energy of the particle is E, it will perform periodic motion only if(A) E< 0 (B) E> 0 (C) V

0> E> 0 (D) E> V

0

Answer (C)

Hints :

For motion to be periodic, it must reverse its path i.e., KE should become zero for a finite value of x.

Now K + U = E

K = E U

Umax

= V0

Kmin = EV0

If Kmin

> 0 particle will escape

So, EV0

< 0

E< V0

Also E= K + U> 0

Correct answer is (C)

73. For periodic motion of small amplitude A, the time period Tof this particle is proportional to

(A) mA (B) 1 mA (C) A m (D) 1A m

Answer (B)

Hints :

As V= x4

[] = ML2T2

By methods of dimensions

1 m

A

= M0L0T

Correct answer is (B).


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74. The acceleration of this particle for | x| > X0

is

(A) Proportional to V0

(B) Proportional to0

0

V

mX (C) Proportional to0

0

V

mX (D) Zero

Answer (D)

Hints :

For X> X0, potential energy is constant

Force is zero

Correct answer is (D).

SECTION - IV

(Integer Type)

This Section contains TEN questions. The answer to each question is a single-digit integer, ranging from 0 to 9. Thecorrect digit below the question number in the ORS is to be bubbled.

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-sectionalarea is 4.9 107 m2. If the mass is pulled a little in the vertically downward direction and released, it performssimple harmonic motion of angular frequency 140 rad s1. If the Young's modulus of the material of the wire isn 109 Nm2, the value of nis

Answer (4)

Hints :

FLY

Al= , where lis elongation

FL FlAYAY

L

= =

Equivalent spring constant isYA

kL

=

k YA

m mL = =

Y= 4 109 N/m2

n= 4

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

S), where M

Sis the mass of the sun. They

are separated by distance dand are rotating about their centre of mass, which is stationary. The ratio of the totalangular momentum of the binary star to the angular momentum of star Babout the centre of mass is

Answer (6)

Hints :

Total angular momentum =21 2

1 2

m md

m m

+

Angular momentum of m2

= m2r22, where 12

1 2

m dr

m m=

+

Required ratio = 1 2

1

2.2 116

2.2

m m

m

+ += =


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77. Gravitational acceleration on the surface of a planet is6

11g, where gis the gravitational acceleration on the surface

of the earth. The average mass density of the planet is2

3times 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

Answer (3)

Hints :

2

4

3

pp p p

p

GMg G R

R= =

p p p

e e e

g R

g R

=

Also, 2ev gR=

6 3

11 2

p p p p e

e e e e p

v g R g

v g R g

= = =

vp

= 3 km/s

78. A piece of ice (heat capacity = 2100 J kg1 C1 and latent heat = 3.36 105 J kg1) of mass mgrams is at 5C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally, when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of mis

Answer (8)

Hints : Heat absorbed by melting of 1 gram ice is

Q= mL = 0.001 3.36 105 = 336 J

m C (5) = 420 336

m= 8 g

79. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the

source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the

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

Answer (7)

Hints :

Frequency of sound reflected by the car is 0c

c

v vf f

v v

+=

As vc


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80. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of

it to 50 cm, the magnification of its image changes from m25

to m50

. The ratio 25

50

m

mis

Answer (6)

Hints :

fm

f u=

+

25

204

20 25m = =

50

20 2

20 50 3m

= =

25

50

126

2

m

m= =

81. An -particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their

de-Broglie wavelengths are and p respectively. The ratiop

, to the nearest integer, is

Answer (3)

Hints :

2

h h

mv mqV = =

2p

h

m eV =

2 (4 ) (2 ) 2 2

ph

m e V

= =

2 2 3p

=

82. When two identical batteries of internal resistance 1 each are connected in series across a resistor R, the rate

of heat produced in Ris J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 =2.25 J

2, then the value of Rin is

Answer (4)

Hints :

In case 1,

2

1

2

2J R

R

= +

In case 2,

2

2

2

2 1J R

R

= +

J1 = 2.25 J2

R= 4


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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 Bis at 1500 nm. Considering themto be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

Answer (9)

Hints :

As per Weins displacement law

3A B

B A

T

T

= =

As per Stefans law

Also,4 2

4

4 2

(6)(3) 9

(18)

A A A

B B B

P A T

P A T= = =

84. When two progressive waves y1 = 4 sin (2x 6t) and 2 3 sin 2 6 2y x t

= are superimposed, the amplitude

of the resultant wave is

Answer (5)

Hints :

2 24 3 2 4 3 cos 52

A

= + + =

aakash-goiit-iit-jee-2010 paper-1 answer and solution with analysis

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