iit jee 2010 solution paper 1 english

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Date : 11-04-2010 Duration : 3 Hours Max. Marks : 252

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12. The question paper consists of 3 parts (Chemistry, Mathematics and Physics). Each partconsists of four Sections.

13. For each question is Section�I, you will be awarded 3 marks if you have darkened only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. In allother cases, minus one (�1) will be awarded.

14. For each question is Section�II, you will be awarded 3 marks if you have darken only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. Partialmarks will be answered for partially correct answers. No negative marks will be awarded inthis Section.

15. For each question is Section�III, you will be awarded 3 marks if you have darken only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. In allother cases, minus one (�1) will be awarded.

16. For each question is Section�IV, you will be awarded 3 marks if you darken the bubblecorresponding to the correct answer and zero mark if no bubbles is darkened. No negativemarks will be awarded for in this Section.

PAPER - 1

id32178296 pdfMachine by Broadgun Software - a great PDF writer! - a great PDF creator! - http://www.pdfmachine.com http://www.broadgun.com

RESONANCE Page # 2

Useful Data :

Atomic Numbers : Be 4; N 7; O 8; Al 13 ; Si 14; Cr 24 ; Fe 26; Fe 26; Zn 30; Br 35.

1 amu = 1.66 × 10�27 kg R = 0.082 L-atm K�1 mol�1

h = 6.626 × 10�34 J s NA = 6.022 × 1023

me = 9.1 × 10�31 kg e = 1.6 × 10�19 C

c = 3.0 × 108 m s�1 F = 96500 C mol�1

RH = 2.18 × 10�18 J 40 = 1.11 × 10�10 J�1 C2 m�1

RESONANCE Page # 3

PART- I

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 species which by definition has ZERO standard molar enthalpy of formation at 298 K is :

(A) Br2(g) (B) Cl2(g)

(C) H2O(g) (D) CH4(g)

Ans. (B)

Sol. Hf° (Cl2,g) = 0, As Hf° of elements in their standard state is taken to be zero.

2. The bond energy (in kcal mol�1) of a C�C single bond is approximately :

(A) 1 (B) 10

(C) 100 (D) 1000

Ans. (C)

Sol. EC � C 100 KCal/mole.

3. The correct structure of ethylenediaminetetraacetic acid (EDTA) is :

(A)

(B)

(C)

(D)

Ans. (C)

CHEMISTRY

RESONANCE Page # 4

Sol. Structure of EDTA is :

4. The ionization isomer of [Cr(H2O)4Cl (NO2)]Cl is :

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

(C) [Cr(H2O)4Cl(ONO)]Cl (D) [Cr(H2O)4Cl2(NO2)].H2O

Ans. (B)

Sol. The ionisation isomer for the given compound will be obtained by exchanging ligand with counter ion as :

[Co(H2O)4Cl2](NO2).

5. 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) BrCH2CH2CH2CH2CH3 and CH3C CH (D) BrCH2CH2CH2CH3 and CH3CH2C CH

Ans. (D)

Sol. CH3�CH2�C C�H 2NaNH CH3�CH2�C C Br�2CH�2CH�2CH�3CH

CH3�CH2�C C�CH2�CH2�CH2�CH3

3�octyne

6. The correct statement about the following disaccharide is :

(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

Ans. (A)

Sol. Ring (a) is pyranose with -glycosidic linkage and ring (b) is furanose with -glycosidic linkage.

CHEMISTRY

RESONANCE Page # 5

7. In the reaction OCH3 HBr the products are :

(A) Br OCH3 and H2 (B) Br and CH3Br

(C) Br and CH3OH (D) OH and CH3Br

Ans. (D)

Sol.

8. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows

Arrhenius equation is :

(A) (B)

(C) (D)

Ans. (A)

Sol. k = Ae�Ea/RT

So, variation will be

CHEMISTRY

RESONANCE Page # 6

SECTION - II

Multiple Correct Correct Type


(D) out of which ONE OR MORE may be correct.

9. Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa 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) CH3COOH and CH3COONa

Ans. (C,D)

Sol. (C) HNO3 + CH3COONa mixture can act as buffer solution if volume of HNO3 solution taken is lesser than

volume of CH3COONa solution because of following reaction :

CH3COONa + HNO3 CH3COOH + NaNO3

(D) CH3COOH + CH3COONa - mixture will act as buffer.

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

(A) molar conductivity (B) electromotive force

(C) resistance (D) heat capacity

Ans. (A,B)

Sol. Molar conductivity and electromotive force (emf) are intensive properties as these are size independent.

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

(A) Ca3(PO4)2 (B) Ca (OH)2

(C) Na2CO3 (D) NaOCl

Ans. (B,C)

Sol. (B) HCO3� + OH� H2O + CO3

2� and CaCO3 will get precipitated.

(C) CO32� + Ca2+ CaCO3, CaCO3 (white) precipitated.

CHEMISTRY

RESONANCE Page # 7

12. In the reaction the intermediate (s) is (are) :

(A) (B) (C) (D)

Ans. (A,C)

Sol. NaOH

It is ortho-para directing.

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

X and Y can respectively be :

(A) H and H (B) H and C2H5

(C) C2H5 and H (D) CH3 and CH3

Ans. (B,D)

Sol.

3

323

3

CH|

CH�CH�C�CH|CH

,

3

523

3

CH|

HC�C�CH|CH

CHEMISTRY

RESONANCE Page # 8

SECTION - III

Comprehension Type

This section contains 2 Paragraphs. Based upon the first paragraph 3 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 Question Nos. 14 to 16

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

Ores of copper include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance

(Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore

chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron

and self-reduction.

14. Partial roasting of Chalcopyrite produces :

(A) Cu2S and FeO (B) Cu2O and FeO

(C) CuS and Fe2O2 (D) Cu2O and Fe2O2

Ans. (A)

Sol. 2 CuFeS2 + 4O2 Cu2S + 2FeO + 3SO2

15. Iron is removed from chalcopyrite as :

(A) FeO (B) FeS

(C) Fe2O3 (D) FeSiO3

Ans. (D)

Sol. Iron is removed in the form of slag of FeSiO3

FeO + SiO2 FeSiO3

16. In self-reduction, the reduction species is :

(A) S (B) O2�

(C) S2� (D) SO2

Ans. (C)

Sol. S2� acts as reducing species in self reduction reaction

2Cu2O + Cu2S 6Cu + SO2

CHEMISTRY

RESONANCE Page # 9


The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside.

The resulting 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 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.

17. For the above cell :

(A) Ecell < 0 ; G > 0 (B) Ecell > 0 ; G < 0

(C) Ecell < 0 ; Gº > 0 (D) Ecell > 0 ; Gº < 0

Ans. (B)

Sol. M (s) | M+ (aq, 0.05 M) || M+ (aq, 1 M) | M(s)

Anode : M (s) M+ (aq) + e�

Cathode : M+ (aq) + e� M (s)

_____________________________________

M+ (aq) |c M+ (aq) |a

Ecell = E°cell � 1

0591.0 log

c

a

|)aq(M

|)aq(M

= 0 � 1

0591.0 log

1

05.0

= + ve = 70 mV and hence G = � nFEcell = � ve.

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

potential would be :

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

Ans. (C)

Sol. Ecell = 10591.0

log

10025.0

= 1

0591.0 log

2005.0

= 70 mV + 1

0591.0 log 20 = 140 mV..

CHEMISTRY

RESONANCE Page # 10

SECTION - IV

(Integer Type)

This section contains TEN questions. The asnwer 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. The concentration of R in the reaction R P was measured as a function of time and the following data is

obtained :

18.012.005.00.0.)(mint

10.040.075.00.1)molar](R[

The order of the reaction is :

Ans. 0

Sol. K = t

CC0 = 05.0

75.01 =

05.025.0

= 5

K = 07.0

40.075.0 =

07.035.0

= 5

So, reaction must be of zero order.

20. The number of neutrons emitted when U23592 undergoes controlled nuclear fission to Xe142

54 and Sr9038 is :

Ans. 3

Sol.

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

Ans. 2

Sol. � NH2 and COO are two basic groups.

CHEMISTRY

RESONANCE Page # 11

22. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is :

Ans. 5

Sol. , , , ,

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

O2H,Zn.2

3O.1 Y

heat.2

)aq(NaOH.1

Ans. 1

Sol. O2H/Zn.2

3O.1

NaOH

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

Ans. 5

CHEMISTRY

RESONANCE Page # 12

Sol. , , , &

are soluble in aqueous NaOH.

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

is :

KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2FeCl3 K2CO3 NH4NO3 LiCN

Ans. 3

Sol. Basic solutions will convert red litmus blue.

, their aqueous solution will be basic due to anionic hydrolysis.

26. Based on VSEPR theory, the number of 90 degree F�Br�F angles is BrF5 is :

Ans. 8

Sol.

27. The value of n in the molecular formula BenAl2Si6O18 is :

Ans. 3

Sol. BenAl2Si6O18. The value of n = 3 by charge balancing.

(Be2+)3 (Al3+)2 (Si6O18)12� [2 × n + 2 (+3) + (�12) = 0 n = 3].

28. 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 :

Ans. 3

Sol. Average titre value = 3

0.2525.252.25 =

345.75

= 25.15 = 25.2 mL

number of significant figures will be 3.

CHEMISTRY

RESONANCE Page # 13

PART-II

SECTION - I



(D) out of which ONLY ONE is correct.

29. Let p and q be real numbers such that p 0, p3 q and p3 � q. If and are nonzero complex numbers

satisfying + = � p and 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

Ans. (B)

Sol. Product = 1

Sum =

22

=

2)( 2

Since 3 + 3 = q � p (2 + 2 � ) = q

(( + )2 � 3 ) = � pq

p2 + pq

= 3

Hence sum = )qp(

p3p

qp32

p

3

32

= qp

q2p3

3

so the equation is x2 +

qp

q2p3

3

x + 1 = 0

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

30. Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x) = 2xe +

2x�e ,

g(x) = 2xxe +

2x�e and h(x) = 2x2ex +

2x�e . If a, b and c denote, respectively, the absolute maximum of

f, g and h on [0, 1], then

(A) a = b and c b (B) a = c and a b (C) a b and c d (D) a = b = c

Ans. (D)

MATHEMATICS

RESONANCE Page # 14

Sol. Clearly f(x) = 2xe +

2x�e

f(x) = 2x (2xe �

2x�e ) 0 increasing fmax

= f(1) = e + e1

g(x) = x2xe +

2x�e g(x) = 2xe + 2x2

2xe � 2x2x�e > 0 increasing g

max = g(1) = e +

e1

h(x) = x2 2xe +

2x�e h(x) = 2x2xe + 2x3

2xe � 2x2x�e = 2x

2x2x22x eexe > 0

hmax

= h(1) = e + e1

,

so a = b = c

31. If the angle A, B and C of a triangle are in arithmetic progression and if a, b and c denote the lengths of the

sides opposite to A, B and C respectively, then the value of the expression ca

sin 2C + ac

sin 2A is

(A) 21

(B) 23

(C) 1 (D) 3

Ans. (D)

Sol.ca

sin 2C + ac

sin 2A = R22

(a cos C + c cos A) = Rb

= 2 sin B = 2 sin 60º = 3

32. Equation of the plane containing the straight line 2x

= 3y

= 4z

and perpendicular to the plane containing the

straight line 3x

= 4y

= 2z

and 4x

= 2y

= 3z

is

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

Ans. (C)

Sol. Direction ratio of normal to plane containing the straight line

324

243

k�j�i�

= 8 i� � j� � 10k�

Required plane1018

432

0z0y0x

= 0 � 26x + 52y � 26z = 0 x � 2y + z = 0

MATHEMATICS

RESONANCE Page # 15

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

2 and r

3 are the numbers

obtained on the die, then the probability that 3r2r1r = 0 is

(A) 181

(B) 91

(C) 92

(D) 361

Ans. (C)

Sol. 3r2r1r = 0 ; r1, r

2, r

3 are to be selected form {1, 2, 3, 4, 5, 6}

As we know that 1 + + 2 = 0

form r1, r

2, r

3 , one has remainder 1, other has remainder 2 and third has remainder 0 when divided by 3.

we have to select r1, r

2, r

3 form (1, 4) or (2, 5) or (3, 6) which can be done in 2C

1 × 2C

1 × 2C

1 ways

value of r1, r

2, r

3 can be interchanged in 3! ways.

required probability = 666

!3)CCC( 12

12

12

=

92

34. Let P, Q, R and S be the points on the plane with position vectors � 2 i� � j� , 4 i� , 3 i� + 3 j� and � 3 i� + 2 j�

respectively. The quadrilateral PQRS must 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

Ans. (A)

Sol. PQ = 136 = 37 = RS, PQ PS

PS = 91 = 10 = QR

slope of PQ = 61

, slope of PS = � 3

PQ is not to PS

So it is parallelogram, which is neither a rhombus nor a rectangle

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

z

y

x

=

0

0

1

has

exactly two distinct solutions, is

(A) 0 (B) 29 � 1 (C) 168 (D) 2

Ans. (A)

MATHEMATICS

RESONANCE Page # 16

Sol. a1x + b

1y + c

1z = 1

a2x + b

2y + c

2z = 0

a3x + b

3y + c

3z = 0

No three planes can meet at two distinct points. So number of matrices is 0

36. The value of 0x

lim

x

043 4t

)t1(nt

x

1 dt is

(A) 0 (B) 121

(C) 241

(D) 641

Ans. (B)

Sol.0x

lim

24 x3)4x(

)x1(nx

=

0xlim

41

× 31

= 121

SECTION - II

Multiple Correct ChoiceType


(D) out of which ONE OR MORE may be correct.

37. The value(s) of

1

02

44

x1

)x1(x dx is (are)

(A) �722

(B) 105

2(C) 0 (D)

23

�1571

Ans. (A)

Sol.

1

02

44

x1

)x1(x =

1

02

224

x1

]x2)x1[(x =

1

02

22224

x1

]x4)x1(x4)x1[(x

= dxx1

x4x4)x1(x

1

02

224

=

dx

x1

x4x4xx

2

6546

Now on polynomial division of x6 by 1 + x2 , we obtain

=

dx

x1

1)1xx(4x4xx

224546

=

dx

x1

44x4x5x4x

22456

MATHEMATICS

RESONANCE Page # 17

= 10x

1

0

3567

xtan4x43x4

5x.5

6x4

7x

=

4

3

41

6

4

7

1 � 4

4 =

5

6

12

7

1 �

=

3

71

� = 722

�

38. Let f be a real-valued function defined on the interval (0, ) by f(x) = n x +

x

0

tsin1 dt. Then which of the

following statement(s) is (are) true?

(A) f(x) exists for all x (0, )

(B) f(x) exists for all x (0, ) 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 > 1 such that |f(x)| + |f(x)| for all x (0, )

Ans. (B, C)

Sol. f(x) = n x +

2

0

dttsin1

f(x) = x1

+ xsin1

f(x) = � 2x

1 +

xsin12

xcos

(A) f is not defined for x = � 2

+ 2n , n

so (A) is wrong

(B) f(x) always exist for x > 0

(C) | f| < | f |

Since f > 0 and f > 0

f < f

x1

+ xsin1 < n x +

x

0

dxxsin1

LHS is bounded RHS is Increasing with range

So there exist some beyond which RHS is greater than LHS

(D) | f | + | f| b is wrong as f is M & its range is not bound while is finite

MATHEMATICS

RESONANCE Page # 18

39. Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius

r having AB as its diameter, then the slope of the line joining A and B can be

(A) � r1

(B) r1

(C) r2

(D) � r2

Ans. (C, D)

Sol. A(t1

2, 2t1), B (t2

2 , 2t

2)

centre of circle

21

22

21 tt,

2

tt

| t1 +

t2 | = r , slope of AB =

21 tt2

= ± r2

40. Let ABC be a rectangle such that ACB = 6

and let a, b and c denote the lengths of the sides opposite to

A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 � 1 and c = 2x + 1 is (are)

(A) � 32 (B) 1 + 3 (C) 2 + 3 (D) 4 3

Ans. (A, B)

Sol. cos 6

= )1x)(1xx(2

)1x2()1xx()1x(22

22222

23

= )1x)(1xx(2

)xx)(2x3x()1x(22

2222

23

= )1x)(1xx(2

)1x(x)2x)(1x()1x(22

22

3 = 1xx

)2x(x1x2

2

3 (x2 + x + 1) = 2x2 + 2x � 1

( 3 � 2) x2 + ( 3 � 2) x + ( 3 + 1) = 0

on solving

x2 + x � )533( = 0

x = 3 + 1, � )32(

MATHEMATICS

RESONANCE Page # 19

41. Let z1 and z

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

1 + tz

2 for some real number t with

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

(A) |z � z1| + |z � z

2| = |z

1 � z

2| (B) Arg (z � z

1) = Arg (z � z

2)

(C) 1212

11

zzzz

zzzz

= 0 (D) Arg (z � z

1) = Arg (z

2 � z

1)

Ans. (A, C, D)

Sol. (A) |z � z1| + |z � z

2| = |z

1 � z

2|

AB + BC = AC

(B) Arg (z � z1) � Arg (z � z

2) =

(C)1zz

1zz

1zz

22

11 = 0

1zz

0zzzz

0zzzz

22

2121

11

= 0

2121

11

zzzz

zzzz

= 0

(D) Arg (z � z1) = Arg (z

2 � z

1)

SECTION - III

Comprehension Type

This section contains 2 Paragraphs. Based upon the first paragraph 3 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.

MATHEMATICS

RESONANCE Page # 20


Let p be an odd prime number and Tp be the following set of 2 × 2 matrices :

Tp =

}1�p,.....,2,1,0{c,b,a:

a

b

c

aA

42. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A)

divisible by p is

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

Ans. (D)

Sol. A =

ac

bawhere a, b, c {0,1,2,....p � 1}

Case- I A is symmetric matrix b = c

det(A) = a2 � b2 is divisible by p

(a � b) (a + b) is divisible by p

(a) a � b is divisible by p if a = b, then 'p' cases are possible

(b) a + b is divisible by p if a + b = p, then 2

)1�p( × 2 = (p � 1) cases are possible

Case - II

A is skew symmetric matrix

if a = 0, b + c = 0, then det (A) = b2

b2 can never be divisible by p. So No case is possible

Total number of A is possible = 2p � 1

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

[Note : The trace of 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)

Ans. (C)

Sol. a2 � bc p

a can be chosen in p � 1 ways (a 0)

Let a be 4 & p = 5

so a2 = 16 and hence bc should be chosen such that a2 � bc p

Now b can be chosen in p � 1 ways and c in only 1 (one be is chosen)

MATHEMATICS

RESONANCE Page # 21

Explanation : If b = 1 c = 1

b = 2 c = 3

b = 3 c = 2

b = 4 c = 4

Hence a can be chosen in p � 1 ways

and then b can be chosen in p � 1 ways

c can be chosen in 1 ways

so (p � 1)2

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

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

Ans. (D)

Sol. As = Total cases � (a 0 and |A| is divisible by p) � (a = 0 and |A| is divisible by p)

= p3 � (p �1)2 � (2p � 1) = p3 � p2

� bc p since b & c both we coprime to p

one of them must be zero.

If b = 0 c can be chosen in {0,1,....p � 1}

If c = 0 b can be chosen in {0,1,....p � 1}


The circle x2 + y2 � 8x = 0 and hyperbola 9x2�

4y2

= 1 intersect at the points A and B.

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

(A) 2x � 5 y � 20 = 0 (B) 2x � 5 y + 4 = 0

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

Ans. (B)

Sol. Let equation of tangent to ellipse

1y2

tan�x

3sec

2sec x � 3 tan y = 6

It is also tangent to circle x2 + y2 � 8x = 0

22 tan9sec4

6�sec8= 4

(8sec � 6)2 = 16 (13sec2 � 9)

12sec2 + 8sec � 15 = 0

MATHEMATICS

RESONANCE Page # 22

sec = 65

and 23

� but sec 65

sec = 23

� tan = 25

slope is positive

Equation of tangent = 2x � 5 y + 4 = 0

46. Equation of the circle with AB as 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

Ans. (D)

Sol. x2 + y2 � 8x = 0

9x2

� 4y2

= 1 4x2 � 9y2 = 36

4x2 � 9(8x � x2) = 36

13x2 � 72x � 36 = 0

13x2 � 78x + 6x � 36 = 0

(13x + 6) (x � 6) = 0

x = � 136

and x = 6

But x > 0 x = 6

A(6, 2 ) and B (6, � 2 )

Equation of circle with AB as a diameter x2 + y2 � 12x + 24 = 0

SECTION - IV

(Integer Type)

This section contains TEN questions. The asnwer 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.

MATHEMATICS

RESONANCE Page # 23

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

(y + z) cos 3 = (xyz) sin 3

x sin 3 = y3cos2

+ z

3sin2

(xyz) sin 3 = (y + 2z) cos 3 + y sin 3

have a solution (x0, y0, z0) with y0 z0 0, is

(y + z) cos 3 = (xyz) sin 3

x sin 3 = y3cos2

+ z

3sin2

(xyz) sin 3 = (y + 2z) cos 3 + y sin 3

Ans. 3

Sol. Let xyz = t

t sin3 � y cos 3 � z cos3 = 0 ..........(1)

t sin3 � 2y sin 3 � 2z cos3 = 0 ..........(2)

t sin3 � y (cos 3 + sin3 ) � 2z cos 3 = 0 ..........(3)

y0. z0 0 hence homogeneous equation has non-trivial solution

D =

3cos2�)3sin3(cos�3sin

3cos2�3cos2�3sin

3cos�3cos�3sin

= 0

sin3cos3(sin 3 � cos3) = 0

sin3= 0 or cos3= 0 or tan3 = 1

Case - I sin3= 0

From equation (2)

z = 0 not possible

Case - II cos3 = 0, sin3 0

t. sin3 = 0 t = 0 x = 0

From equation (2)

y = 0 not possible

Case- III tan3= 1

3 = n+ 4

, n I

x. y. z sin3= 0 = 3

n+

12

, n I

x = 0 , sin3 0 = 12

, 125

, 129

Hence 3 solutions

MATHEMATICS

RESONANCE Page # 24

48. Let f be 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

Ans. 9

Sol. Y � y = m (X � x)

y-intercept (x = 0)

y = y � mx

Given that y � mx = x3 x dxdy

� y = � x3

dxdy

� xy

= � x2

Intergrating factor dxx1

�

e = x1

solution y . x1

= dx)x(.x1 2

f(x) = y = � 2x3

+ cx

Given f(1) = 1 c = 23

f(x) = � 2x3

+ 2x3

f(�3) = 9

49. The number of values of in the interval

2,

2� such that

5n

for n = 0, ±1, ± 2 and

tan = cot 5 as well as sin 2 = cos 4is

Ans. 3

Sol. tan = cot 5

cossin

=

sin5cos

cos 6= 0

6 = (2n + 1) 2

= (2n + 1) 12

; n

MATHEMATICS

RESONANCE Page # 25

= 125

�

, �4

, 12

�

, 12

, 4

, 125

.........(1)

sin2 = cos4

sin2 = 1 � 2 sin2 2

2sin22 + sin2� 1 = 0

sin2 = � 1, 21

2= (4m � 1)2

, p + (�1)p 6

q = (4m � 1)4

, 2

p+ (�1)p

12

; m, p I

= �4

, 12

, 125

...........(2)

From (1) & (2)

125

,12

,4

�

Number of solution

50. The maximum value of the expression

22 cos5cossin3sin

1 is

Ans. 2

Sol. f() =

22 cos5cossin3sin

1 =

2)2cos1(5

2sin23

22cos1

1

= 2cos42sin36

2

f()max

= 56

2

= 2

51. If a and b

are vectors in space given by a

= 5

j�2�i� and b

=

14

k�3j�i�2 , then the value of

ba2

. b2�aba

is

Ans. 5

MATHEMATICS

RESONANCE Page # 26

Sol. a =

5

j�2�i� , b

= 14

k�3j�i�2

|a

| = 1 , |b

| = 1

b.a

= 0

)ba2(

. )b2�a()ba(

= )ba2(

. a)b2�a.(b(�b)b2�a.(a

= )ba2(

. a)b.b2�a.b(�b)b.a2�a.a(

= )ba2(

. a)20(�b)0�1(

= )ba2(

. a2b

= )a.b(2b.b)a.a(4)b.a(2

= 0 + 4 + 1 + 0 = 5

52. The line 2x + y = 1 is tangent to the hyperbola 2

2

a

x � 2

2

b

y= 1 . If this line passes through the point

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

Ans. 2

Sol.

x = a/e

y = �2x + 1 e2 = 1 + 2

2

a

b

0 = � ea2

� + 1 = 1 + 2

2

a

)1�a4(

ea

= 21

e2 = 1 + 4 � 2a

1

e = 2a e2 = 5 � 2e

4

MATHEMATICS

RESONANCE Page # 27

c2 = a2 m2 � b2 e4 � 5e2 + 4 = 0

1 = 4a2 � b2 (e2 � 1)(e2 � 4) = 0

1 + b2 � 4a2 = 0 e2 � 1 0 e = 2

53. If the distance between the plane Ax � 2y + z = d and the plane containing the lines

21�x

= 3

2�y =

43�z

and 3

2�x=

43�y

= 5

4�z is 6 , then |d| is

Ans. 6

Sol. Equation of plane is 543

432

3�z2�y1�x

= 0

x � 2y + z = 0 ..........(1)

Ax � 2y + z = d ..........(2)

Compare 1A

= 2�2�

= 11 A = 1

Distance between planes is 6411

d

|d| = 6

54. For any real number, 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

f(x) =

evenis]x[if

,oddis]x[if

x�]x[1

]x[�x

Then the value of

10

10�

2)x(f

10 cos x dx is

Ans. 4

MATHEMATICS

RESONANCE Page # 28

Sol. f(x) =

1n2xn2,}x{1

n2x1n2,}x{

Clearly f(x) is a periodic function with period = 2

Hence f(x) . cos x is also periodic with period = 2

10

2

10

10

dx)xcos()x(f = 2

2

0

dx)xcos()x(f = 2

1

0

dx)xcos(})x{})x{1((

= 22

1

0

dx)xcosx( = � 22 1

02

xcosxsinx

= � 22

2

2 = 4

55. Let be the complex number cos 32

+ i sin 32

. Then the number of distinct complex numbers

z satisfying

z1

1z

1z

2

2

2

= 0 is equal to

Ans. 1

Sol. = cos 32

+ i sin 32

R1 R1 + R2 + R3

z1z

1zz

z2

2

= 0 z

z11

1z1

12

2

= 0

z = 0

2

22

2

�z�10

�1�z0

1

= 0

(z + 2 � )(z + � 2) � (1 � )(1 � 2 ) = 0

z2 = 0

only one solution

MATHEMATICS

RESONANCE Page # 29

56. Let Sk, k = 1, 2,...., 100, denote the sum of the infinite geometric series whose first term is !k1�k

and the common ratio is k1

. Then the value of !100

1002

+

100

1kk

2 S)1k3�k( is

Ans. 3

Sol.

100

2kk

2 S)1k3�k(

for k = 2 |k2 � 3k + 1) Sk| = 1

100

3k!)1�k(11�k

�!)2�k(

1�k

100

3k!)1�k(

1�

!)2�k(1

�!)2�k(

1!)3�k(

1

100

3k!)1�k(

1�

!)3�k(1

S =

!961

�!94

1....

!61

�!4

1!5

1�

!31

!41

�!2

1!3

1�

!11

!21

�11

+

!991

�!97

1!98

1�

!961

!971

�!95

1 = 2 �

!991

�!98

1

E = !100

1002

+ 3 � !98

1 �

!98.991

= !100

1002

+ 3 � !99

100 =

!99.1001002

+ 3 � !99

100= 3

MATHEMATICS

RESONANCE Page # 30

PART-III

SECTION - I


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

four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

57. A block of mass m is on 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 P parallel to the plane. The direction of

force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin � cos) to P2 = mg(sin

+ cos), the frictional force f versus P graph will look like :

(A) (B)

(C) (D)

Ans. (A)

Sol.

P1 = mgsin � mgcos

P2 = mgsin + mgcos

Initially block has tendency to slide down and as tan > , maximum friction mgcos will act in positive

direction. When magnitude P is increased from P1 to P2, friction reverse its direction from positive to negative

and becomes maximum i.e.mgcos in opposite direction.

PHYSICS

RESONANCE Page # 31

58. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required

to take a unit mass from point P on its axis to infinity is :

4R

4R3R

P

(A) 524R7

GM2 (B) 524

R7GM2

(C) R4

GM(D) 12

R5GM2

Ans. (A)

Sol. Wext = U � UP

Wext = 0 �

1.

xGdm

��

Wext = G 222rR16

rdr2

R7

M

= 2R7

GM2

22 rR16

rdr

= 2R7

GM2 z

zdz = 2R7

GM2 [Z]

r3R

4Rdr

P

x

Wext = 2R7

GM2 R4

R3

22 rR16

Wext = 2R7

GM2 R5�R24

Wext = 2R7

GM2 5�24 .

59. Consider a thin square sheet of side L and thickness t, made of a material of resitivity . The resistance

between two opposite faces, shown by the shaded areas in the figure is :

Lt.

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

(C) independent of L (D) independent of t

Ans. (C)

PHYSICS

RESONANCE Page # 32

Sol. R = Al

R = tLL

= t

Independent of L.

60. 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 inter molecular forces become ineffective. So a real gas behaves like

an ideal gas.

61. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with

increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances

R100, R60 and R40, respectively, the relation between these resistance is :

(A) 6040100 R1

R1

R1

(B) R100 = R40 + R60 (C) R100 > R60 > R40 (D) 4060100 R1

R1

R1

Ans. (D)

Sol. 100 = 100

2

'RV

100'R1

= 2V

100

where R�100 is resistance at any temperature corresponds to 100 W

60 = 60

2

'RV

60'R1

= 2V

60

40 = 40

2

'RV

40'R1

= 2V

40

From above equations we can say

100'R1

> 60'R1

> 40'R1

.

So, most appropriate answer is option (D).

PHYSICS

RESONANCE Page # 33

62. To verify Ohm's law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R2,

two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the

experiment is :

(A)

G1

G2

R1RT

R2

V

(B)

G1

G2

R2RT

R1

V

(C)

G1

G2

R2

RT

R1

V

(D)

G1

G2

R2

RT

R1

V

Ans. (C)

Sol. To verify Ohm's law one galvaometer is used as ammeter and other galvanometer is used as voltameter.

Voltameter should have high resistance and ammeter should have low resistance as voltameter is used in

parallel and ammeter in series that is in option (C).

63. An AC voltage source of variable angular frequency and fixed amplitude V connected in series with a

capacitance C and an electric bulb of resistance R (inductance zero). When is increased :

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

(C) total impedence of the circuit is unchanged (D) total impedence of the circuit increases

Ans. (B)

Sol. irms =

222

rms

c

1R

V

when increases, irms increases so the bulb glows brighter

PHYSICS

RESONANCE Page # 34

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

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

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

(A) BL (B)

BL(C)

2BL

(D)

4BL

Ans. (C)

Sol.

2Tsin

2d

= dlB 2T2d

= RdlB T = BR =

2LB

.

SECTION - II

Multiple Correct Choice TypeThis section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D)

out of which ONE OR MORE may be correct.

65. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its

pressure at A is P0. Choose the correct option(s) from the following :

(A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0 n 4

(C) Pressure at C is 4

P0 (D) Temperature at C is 4

T0

Ans. (A, B)

PHYSICS

RESONANCE Page # 35

Sol.

U = 2f

nRT,, where f,n,R are constants. Also temperature T is same at A & B.

UA = UB

Also, WAB = nRT0

i

f

V

V n = nRT0

0

0

V

V4 n = nRT0 n4 = P0V0 n4

So, answers are (A) & (B).

66. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle

of 60º (see figure). If the refractive index of the material 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 AD

(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)

PHYSICS

RESONANCE Page # 36

Sol.

45º

60º

Q

By refraction at face AB :

1.sin60º = 3 . sinr1

So, r1 = 30º

This shows that the refracted ray is parallel to side BC of prism. For side 'CD' angle of incidence will be 45º,

which can be calculated from quadrilateral PBCQ.

By refraction at face CD :

3 sin45º = 1 sinr2

So, sin r2 = 2

3

which is impossible. So, there will be T.I.R. at face CD.

Now, by geometry angle of incidence at AD will be 30º. So, angle of emergence will be 60º.

Hence, angle between incident and emergent beams is 90º

67. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are

shown in the figure. These lines suggest that :

(A) |Q1| > |Q2|

(B) |Q1 | < |Q2|

(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)

PHYSICS

RESONANCE Page # 37

Sol.

Q1

Q2

From the diagram, it can be observed that Q1 is positive, Q2 is negative.

No. of lines on Q1 is greater and number of lines is directly proportional to magnitude of charge.

So, |Q1| > |Q2|

Electric field will be zero to the right of Q2 as it has small magnitude & opposite sign to that of Q1 .

68. A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He

uses a stop watch with the least count of 1sec for this and records 40 seconds for 20 oscillations. For this

observation, which of the following 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. Since, t = nT. So, T = nt

= 2040

or T = 2 sec.

Now, t = n T

andtt=

TT

So,401

= 2T

T = 0.05

Time period, T = g2

So,TT

= gg

21

or gg

= TT

2

So, percentage error in g = gg

× 100

= TT

2

× 100 = � 205.0

2 × 100

= 5%.

PHYSICS

RESONANCE Page # 38

69. A point mass of 1kg 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 ms�1. 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 ms�1

(B) Momentum of 5 kg mass after collision is 4 kg ms�1

(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.

Since collision is elastic, so e = 1

Velocity of approach = velocity of separation

So, u = v + 2 .............(i)

By momentum conservation :

1 × u = 5v � 1 × 2

u = 5v � 2

v + 2 = 5v � 2

So, v = 1 m/s

and u = 3 m/s

Momentum of system = 1 × 3 = 3 kgm/s

Momentum of 5kg after collision = 5 × 1 = 5 kgm/s

So, kinetic energy of centre of mass = 21

(m1 + m2)

2

21

1

mm

um

= 21

(1 + 5)

2

631

= 0.75 J

Total kinetic energy = 21

× 1 × 32

= 4.5 J.

PHYSICS

RESONANCE Page # 39

SECTION�III

Paragraph TypeThis section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and

based upon the second paragraph 2 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.


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 proportional to km

, as 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 particle of mass m moving on the x-axis. Its potential energy is V(x) = x4

( > 0) for |x| near the origin and becomes a constant equal to V0 for |x| X0 (see figure)

70. 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 > V0

Ans. (C)

Sol. When 0 < E < V0 there will be acting a restoring force to perform oscillation because in this case particle will

be in the region |x| x0 .

71. For periodic motion of small amplitude A, the time period T of this particle is proportional to :

(A)

mA (B)

mA1

(C) m

A

(D) mA

1

Ans. (B)

Sol. V = x4

T.E. = 21

m2A2 = A4 (not strictly applicable just for dimension matching it is used)

2 = mA2 2

T

mA1

PHYSICS

RESONANCE Page # 40

72. The acceleration of this particle for |x| > X0 is :

(A) proportional to V0 (B) proportional to 0

0

mX

V

(C) proportional to 0

0

mX

V(D) zero

Ans. (D)

Sol. F = dxdU

as for |x| > x0 V = V0 = constant

dxdU

= 0

F = 0.


Electrical resistance of certain materials, known as

superconductors, changes abruptly from a nonzero value

to zero as their temperature is lowered below a criticalT (B)C

T (0)C

O B

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 TC (B) on B

is shown in the figure.

73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two

different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following

graphs shows the correct variation of R with T in these fields?

(A)

R

O T

B1B2 (B)

R

O T

B1

B2

(C)

R

O T

B1 B2 (D)

R

O T

B1

B2

Ans. (A)

PHYSICS

RESONANCE Page # 41

Sol. As the magnetic field is greater, the critical temperature is lower and as B2 is larger than B1. Graph �A� is

correct.

74. 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 < TC (B) < 100 K

(C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = 10 Tesla, TC (B) = 70 K

Ans. (B)

Sol. For B = 0 TC = 100 k

B = 7.5 T TC = 75 k

For B = 5T 75 < TC < 100

SECTION�IV

Integer TypeThis 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.

75. A piece of ice (heat capacity = 2100 J kg�1 ºC�1 and latent heat = 3.36 × 105 J kg�1) of mass m grams is at

�5 ºC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-

water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat

exchange in the process, the value of m is :

Ans. 8 gm

Sol. S = 2100 J kg�1 ºC�1

L = 3.36 × 105 J kg�1

420 = m S Q + (1) × 10�3 × L

420 = m s(5) + 3.36 × 102

420 � 336 = m(2100) × 5

m = 125

1 × 1000 = 8 gm.

76. 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 constant

speeds much smaller than the speed of sound which is 330 ms�1.

Ans. 7

Sol. Let speed of cars are V1 and V2

frequency received by car f1 =

1vvv

f0

PHYSICS

RESONANCE Page # 42

frequency reflected by car f2 =

v

vv 1

1vvv

f0

f = f2� � f2 =

1

1

2

2

vv

vv

vv

vv f0

f =

)vv)(vv(

)vv)(vv()vv)(vv(

21

2112 f0

f = )vv)(vv(

f )vv(v2

21

012

v

f )vv(2 012

Givenv

f )vv(2 012 = 100

2.1 f0

v2 � v1 = 7.126

Answer in nearest integer is 7.

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

of it to 50cm, the magnification of its image changes from m25 to m50. The ratio 50

25

m

m is :

Ans. 6

Sol. When object distance is 25.

v1

� u1

= f1

v1

� )25(�1

= 201

v = 100 cm.

m25 = uv

= 25�

100 = � 4.

When object distance is 50.

v1

� )50(�1

= 201

u = 3

100 cm

m50 = 50�

3100

= � 32

50

25

m

m =

32

�

4� = 6.

Alternate :

50

25

m

m =

5020f

2520f

=

530

= 6

PHYSICS

RESONANCE Page # 43

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

de-Broglie wavelength are and p respectively. The ratio

p, to the nearest integer, is :

Ans. 3

Sol. P1 = )eV100(m2

P = )eV100(m2

h = )eV100)(m4(2

h

P = 8

The ratio

p, to the nearest integer, is equal to 3.

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

rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2.

If J1 = 2.25 J2 the value of R in is :

Ans. 4

Sol.

i = R2

2

J1 =

2

R22

R

PHYSICS

RESONANCE Page # 44

eq =

11

11

11

=

req = 21

i = R

21

=

1R22

J2 =

2

R212

R

Given J1 = 49

J2

2

R22

R =

49

2

R212

R

R2

2

= R21

3

2 + 4R = 6 + 3R

R = 4.

80. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature 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, what will be the ratio of the rate of total energy radiated by A to that of B ?

Ans. 9

Sol. (m)B = 3(m)A

TA = 3TB

E1 = 4 (6)2 TA4 = 4(6)2 (3TB)4

E2 = 4 (18)2 TB4

2

1

EE

= 9.

81. When two progressive waves y1 = 4 sin (2x � 6t) and y2 = 3 sin

2�t6�x2 are superimposed, the

amplitude of the resultant wave is :

Ans. 5

Sol. Aeq = cosAA2AA 2122

21

Aeq = 2

cos)3)(4(234 22

Aeq = 5.

PHYSICS

RESONANCE Page # 45

82. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its

cross-sectional area is 4.9 × 10�7 m2. If the mass is pulled a little in the vertically downward direction and

released, it performs simple harmonic motion of angular frequency 140 rad s�1. If the Young's modulus of the

material of the wire is n × 109 Nm�2, the value of n is :

Ans. 4

Sol. n = mk

= m

/yA =

myA

1.01

)109.4()10n( 79

= 140 n = 4.

83. A binary star consists of two stars A (mass 2.2 MS) and B ( mass 11 MS) is the same mass of the sun. They

are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the

total angular momentum of hte binary star to the angular momentum of star B about the centre of mass is :

Ans. 6

Sol.A Bc.m.

5d6

d6 11 Ms

2.2 Ms

c.m. about B of momentum Angularc.m. about momentum angular Total

=

6d

6d

)M 11(

6d

6d

)M 11(6d5

6d5

)M2.2(

s

ss

= 6.

84. Gravitational acceleration on the surface of a planet is 116

g, where g is the gravitational acceleration on the

surface of the earth. The average mass density of the planet is 32

times that of the earth. If the escape speed

on the surface of the earth is taken to be 11 kms�1, the escape speed on the surface of the planet in kms�1

will be :

Ans. 3

Sol. g = 2R

GM =

2

3

R

R34

)G(

; g R

g'g =

R'R

'

=

32

R'R

= 116

Given,R

'R =

2263

Ve = R

GM =

R

R34

))()G( 3

Ve R ; Ve = 3 km/hr.

PHYSICS