solutions to iit-jee 2009 · 2019. 7. 14. · aakash iit-jee regd. office : aakash tower, plot no....

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 47623456 Fax : 25084124(1)

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Solutions to IIT-JEE 2009

PAPER - II (Code - 0)Time : 3 hrs. Max. Marks: 240

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-25084124


Instructions :

1. The question paper consists of 3 parts (Chemistry, Mathematics and Physics). Each part has 4 sections.

2. Section I contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for itsanswer, out of which only one is correct.

3. Section II contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for itsanswer, out of which one or more is/are correct.

4. Section III contains 2 questions. Each question has four statements (A, B, C and D) given in Column I andfive statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching withone or more statement(s) given in Column II. For example, if for a given question, statement B matches withthe statements given in q and r, then for that particular question, against statement B, darken the bubblescorresponding to q and r in the ORS.

5. Section IV contains 8 questions. The answer to each of the questions is a single-digit integer, rangingfrom 0 to 9. The answer will have to be appropriately bubbled in the ORS as per the instructions given atthe beginning of the section.

6. For each question in Section I, you will be awarded 3 marks if you darken the bubble corresponding tothe correct answer and zero mark if no bubble is darkened. In case of bubbling of incorrect answer,minus one (–1) mark will be awarded.

7. For each question in Section II, you will be awarded 4 marks if you darken the bubble(s) correspondingto the correct choice(s) for the answer, and zero mark if no bubble is darkened. In all other cases, minusone (–1) mark will be awarded.

8. For each question in Section III, you will be awarded 2 marks for each row in which you have darkenedthe bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximumof 8 marks. There is no negative marking for incorrect answer(s) for this section.

9. For each question in Section IV, you will be awarded 4 marks if you darken the bubble corresponding tothe correct answer, and zero mark if no bubble is darkened. In all other cases, minus one (–1) mark willbe awarded.



CHEMISTRYPART - I

SECTION - ISingle Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, outof which ONLY ONE is correct.

1. For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the equation

log k = – (2000) 1T

+ 6.0. The pre-exponential factor A and the activation energy Ea, respectively, are

(A) 1.0 × 106 s–1 and 9.2 kJ mol–1 (B) 6.0 s–1 and 16.6 kJ mol–1

(C) 1.0 × 106 s–1 and 16.6 kJ mol–1 (D) 1.0 × 106 s–1 and 38.3 kJ mol–1

Answer (D)

Hints :

log k = – (2000) 17

+ 6.0; comparing this equation with log k = log A – aE2.303 RT

We get log A = 6 ∴ A = 106 s–1

and – aE

2.303 RT = – 2000

T

Ea = 2000 × 2.303 × 8.314

Ea = 38.3 kJ mol–1

2. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is

(A) 0 (B) 2.84 (C) 4.90 (D) 5.92

Answer (A)

Hints :Cr(CO)6

Cr in G.S. Under the influenceof strong ligand

d sp2 3 hybridization

Diamagnetic ⇒ μ = 0

3. In the following carbocation, H/CH3 that is most likely to migrate to the positively charged carbon is

H C—C —C—C —CH3 32 41 + 5

H

HO H

H

CH3

3

(A) CH3 at C-4 (B) H at C-4 (C) CH3 at C-2 (D) H at C-2



Answer (D)

Hints :

H C—C —C—C —CH3 32 41 + 5

H

HO H

H

CH3

3CH —C—C—C—CH3 3

+

:OH H CH3

H H

More stable resonancestabilized carbocation

4. The correct stability order of the following resonance structures is– – – –

2 2 2 2H C N N H C— N N H C— N N H C— N N(I) (II) (III) (IV)

+ + + += = = ≡ =

(A) (I) > (II) > (IV) > (III) (B) (I) > (III) > (II) > (IV)

(C) (II) > (I) > (III) > (IV) (D) (III) > (I) > (IV) > (II)

Answer (B)

Hints :

Resonating structures having maximum number of covalent bonds are more contributing. Among chargeseparated resonating structures, structures where opposite charge are close enough are more contributing.

SECTION - IIMultiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, outof which ONE OR MORE is/are correct.

5. For the reduction of –3NO ion in an aqueous solution, E0 is +0.96 V. Values of E0 for some metal ions are

given below

V2+ (aq) + 2e– → V E0 = – 1.19 V

Fe3+ (aq) + 3e– → Fe E0 = –0.04 V

Au3+ (aq) + 3e– → Au E0 = +1.40 V

Hg2+ (aq) + 2e– → Hg E0 = +0.86 V

The pair(s) of metals that is(are) oxidized by –3NO in aqueous solution is(are)

(A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V

Answer (A, B, D)

Hints :

Oxidation of V

E0 = 0.96 – (–1.19) = 2.15 V

For Fe,

E0 = 0.96 – (–0.04) = 1.0 V

For Au,

E0 = 0.96 – 1.4 = – 0.044 V (not feasible)

For Hg

E0 = 0.96 – 0.86 = 0.1 V



6. Among the following, the state function(s) is(are)

(A) Internal energy (B) Irreversible expansion work

(C) Reversible expansion work (D) Molar enthalpyAnswer (A, C, D)Hints :

In a reversible expansion work, process is very slow, isothermal and work done in a state function.7. In the reaction

2X + B2H6 → [BH2(X)2]+ [BH4]–

the amine(s) X is(are)(A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) (CH3)3N

Answer (A, B)Hints :

Small bases (NH3 and 1º amine) are able to break the bond of bridged hydrogen in a unsymmetrical manner.8. The nitrogen oxide(s) that contain(s) N-N bond(s) is(are)

(A) N2O (B) N2O3 (C) N2O4 (D) N2O5

Answer (A, B, C)Hints :

N N—O≡⊕ s

, N—NO

..

O

O, N—N

O O

OO

9. The correct statement(s) about the following sugars X and Y is(are)

CH OH2

HO

H

H

OH

OH

H

HO

H

HO

OH

H

H

O H

O

HOH C2

CH OH2

X

CH OH2

HO

H

H

OH

OH

H

HO H

Y

CH OH2

HO

HO

OH

H

H

H

O

HOH

(A) X is a reducing sugar and Y is a non-reducing sugar

(B) X is a non-reducing sugar and Y is a reducing sugar

(C) The glucosidic linkages in X and Y are α and β, respectively

(D) The glucosidic linkages in X and Y are β and α, respectivelyAnswer (B, C)Hints :

No hemiacetal linkage is free in X. Therefore it is Non reducing. While Y having hemiacetal linkage would bereducing sugar.

CH OH2

HO

H

H

OH

OH

H

HO

H

OH

HO

H

H

O HO

HOH C2

CH OH2

α-Glucosidic linkage

HOH2C

HO

H

HO H

β-Glucosidic linkage

CH OH2

HO

OH

HO

H

H

H

O

HOH



SECTION-IIIMatrix-Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched.The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s andt. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. Theappropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the followingexampleIf the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbleswill look like the following.

A

B

C

D

p

p

p

p

q

q

q

q

r

r

r

r

s

s

s

s

p q r s

t

t

t

t

t

10. Match each of the reactions given in Column I with the corresponding product(s) given in Column II.

Column I Column II

(A) Cu + dil HNO3 (p) NO

(B) Cu + conc HNO3 (q) NO2

(C) Zn + dil HNO3 (r) N2O

(D) Zn + conc HNO3 (s) Cu(NO3)2(t) Zn(NO3)2

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

Hints : 3 3 2 2(dil.)

3Cu 8HNO 3Cu(NO ) 2NO 4H O+ ⎯⎯⎯→ + +

3 3 2 2 2(conc.)

Cu 4HNO Cu(NO ) 2NO 2H O+ ⎯⎯⎯→ + +

3 3 2 2 2(dil.)

4Zn 10HNO 4Zn(NO ) N O 5H O+ ⎯⎯⎯→ + +

3 3 2 2 2(conc.)

Zn 4HNO Zn(NO ) 2NO 2H O+ ⎯⎯⎯→ + +

11. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given inColumn II.

Column I Column II

(A)O

Br(p) Nucleophilic substitution

(B) OH (q) Elimination



(C)CHO

OH(r) Nucleophilic addition

(D)Br

NO2

(s) Esterification with acetic anhydride

(t) Dehydrogenation

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

Hints :

(A)O

BrS 2N

E2

O

NCl

O

(B) OH ZnCl /HCl2 Cl

(RCO) O2

O RC

O

Elimination or dehydrogenation

CHO

[Nucleophilic substitution]

(C) HC

OH

O NCl

CR O

O

C R

O

HC

OH

NClHO

HC

O R

O

O

(D)Br

NO2

Nus

(SNAr)Mechanism NO2

Nu



SECTION - IVInteger Answer Type

This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. Theappropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correctanswers to question numbers X, Y, Z and W(say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbleswill look like the following :

0

X Y Z W

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9

12. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to thecombustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for theenthalpy of combustion of the gas in kJ mol–1 is.

Answer : (9)

Hints : q = heat capacity × ΔT

= 1.125 kJ for 3.5 g

for 1 mole = 9 kJ.

13. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probablespeed of gas Y at 60 K. The molecular weight of the gas Y is

Answer : (4)

Hints :( X )rms

3RTUM

=

(Y )mp2RTVM

=

Equating both we get

3R(400) 2R(60)40 X

=

X = 4

14. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10–4. The pH of a 0.01 M solutionof its sodium salt is

Answer : (8)

Hints : pH = apK7 logC

2+ + =

4 27 82 2

−+ + =



15. The total number of α and β particles emitted in the nuclear reaction 238 21492 82U Pb→ is

Answer : 8

Hints : 238 214 4 092 82 2 1U Pb X He y e−⎯⎯⎯→ + +

24 = 4x; x = 6

10 = 2x – y

y = 2

∴ x + y = 8

16. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

Answer : 6

Hints : KOH 22 4fuseMnO MnO −⎯⎯⎯⎯→

Oxidation state = 6.

17. The number of water molecule(s) directly bonded to the metal centre in CuSO4.5H2O is

Answer : 4

Hints : In CuSO4.5H2O are molecule of water in H-bonded and other be molecule are liquid to metal centre.

18. The coordination number of Al in the crystalline state of AlCl3 is

Answer : 4

Hints :Al

Cl

Cl

Cl

ClAl

Cl

Cl

19. The total number of cyclic structural as well as stereoisomers possible for a compound with the molecularformula C5H10 is

Answer : 7

Hints : ,

d l,

, ,

d l, , meso

MATHEMATICSPART - II


This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, outof which ONLY ONE is correct.

20. If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

(A)−2 2(4 1)

6n n c

(B)+2 2(4 1)

3n n c

(C)−2 2(4 1)

3n n c

(D)+2 2(4 1)

6n n c



Answer (C)

Hints : Sn = cn2

Sn – 1 = c(n –1)2

tn = c [n2 – (n – 1)2]

= c(2n – 1)

So now 2

1

n

nn

t=∑

= c2[12 + 32 + ........ + (2n – 1)2]

= c2[12 + 22 + ........ + (2n)2 – (22 + 42 + ........ + (2n)2)]

= 2 2 2 2(2 )(2 1)(4 1) – 2 (1 2 ........ )6

n n nc n+ +⎡ ⎤+ + +⎢ ⎥⎣ ⎦

= 2 (2 )(2 1)(4 1) ( )( 1)(2 1)– 4

6 6n n n n n nc + + + +⎡ ⎤

⎢ ⎥⎣ ⎦

= ( )2 (2 1)(4 1) – 2 ( 1)

3nc n n n n+⎡ ⎤

+ +⎢ ⎥⎣ ⎦

= 2 2 2(2 1) (4 – 2 – 2 )3

nc n n n n+⎡ ⎤+⎢ ⎥⎣ ⎦

= 2 2(2 1) (2 – )3

nc n n+

= 2 (2 1)(2 – 1)

3c n n n+

= 2 2(4 – 1)

3c n n

21. A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with thecoordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQequals

(A) 1 (B) 2

(C) 3 (D) 2

Answer (C)

Hints : The equation of given line is – 2 1 – 21 1 1

x y z+= = (as direction cosines are equal and positive)

Any points on the line may be taken as (r + 2, r – 1, r + 2) which lies on 2x + y + z = 9

∴ 2(r + 2) + r – 1 + r + 2 = 9

4r + 5 = 9

r = 1

Point is (3, 0, 3)

PQ = 2 2 2(3 – 2) (0 1) (3 – 2)+ + +

= 1 1 1 3+ + =



22. The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q. If M is the mid point of the linesegment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points

(A)⎛ ⎞± ±⎜ ⎟⎜ ⎟⎝ ⎠

3 5 2,2 7 (B)

⎛ ⎞± ±⎜ ⎟⎜ ⎟⎝ ⎠

3 5 19,2 4

(C)⎛ ⎞± ±⎜ ⎟⎝ ⎠

12 3,7 (D)

⎛ ⎞± ±⎜ ⎟⎜ ⎟⎝ ⎠

4 32 3,7

Answer (C)

Hints : The given equation of ellipse is 2 2

116 4x y

+ =

Any point on ellipse (4 cosθ, 2 sinθ) may be taken as

The equation of normal at (4 cosθ, 2 sinθ) is given by

4 secθ x – 2y cosecθ = 16 – 4

⇒ 2 secθ x – y cosecθ = 6 …(i)

Q = (3 cosθ, 0)

Now M = (h, k)

3cos 4cos2

h θ + θ= ⇒

2cos7h

θ =

k = sinθ ⇒ sinθ = 1

224 1

49h k+ =

Locus of M is

4x2 + 49y2 = 49

Equation of latus rectum is 2 3x = ±

at 2 3x = ± , y2 = 49 – 48 = 1

17

y = ±

23. The locus of the orthocentre of the triangle formed by the lines

(1 + p)x – py + p(1 + p) = 0,

(1 + q)x – qy + q(1 + q) = 0,

and y = 0, where p ≠ q, is

(A) A hyperbola (B) A parabola

(C) An ellipse (D) A straight line

Answer (D)

Hints : We have

(1 + p)x – py + p(1 + p) = 0

⇒ 1– 1x yp p+ =

+



Also

(1 + q)x – qy + q(1 + q) = 0

⇒ 1– 1x yq q+ =

+

Equation of AM is

x

y

C

pcM

N

(– , 0)q(–

, 0)

p

A B

– ( )1

py x ap

= ++

and equation of BM is – ( )1

qy x pa

= ++

⇒ ( ) ( )1 1

p qx a x pp q

+ = ++ +

⇒ x = pq …(i)

Also

– ( )1

py pq qp

= ++

⇒ y = –pq …(ii)

from (i) and (ii)

x = –y

⇒ x + y = 0

Which is the required locus representing a straight line.


This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, outof which ONE OR MORE is/are correct.

24. If π

−π

=+ π∫sin ,

(1 )sinn x

nxI dxx

n = 0, 1, 2, ...., then

(A) In = In + 2 (B) +=

= π∑10

2 11

10mm

I

(C)=

=∑10

21

0mm

I (D) In = In + 1

Answer (A, B, C)

Hints : We have

–

sin(1 )sinn x

nxI dxx

π

π

=+ π∫

⇒ ––

sin(– )(1 )sin(– )n x

nxI dxx

π

π

=+ π∫

⇒–

sin2sinn

nxI dxx

π

π

= ∫

⇒0

sinsinn

nxI dxx

π

= ∫



Clearly, for n = 0, for n = 2

I0 = 0,0

sin2sinn

xI dxx

π

= ∫

0

2 cos x dxπ

= ∫

02[sin ]x π=

= 0

Clearly, for n = even, In = 0

for n = 1

10

sinsin

xI dxx

π

= = π∫Similarly, I3 = I5 = π

Now clearly, n and n + 2 will be either even or odd

⇒ In = In + 2

25. An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal ofthat of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

(A) Equation of ellipse is x2 + 2y2 = 2 (B) The foci of ellipse are (±1, 0)

(C) Equation of ellipse is x2 + 2y2 = 4 (D) The foci of ellipse are ±( 2, 0)

Answer (A, B)

Hints : We have

2x2 – 2y2 = 1

⇒ 2 2 1–2

x y = …(i)

Eccentricity 11 21

e = + =

Let the equation of the ellipse is 2 2

2 2 1x ya b

+ =

given that 2

211–2

ba

=

⇒2

2 22

1 22

b a ba

= ⇒ =

Hence equation of the ellipse is

2 2

2 2 12x yb b

+ =

⇒ x2 + 2y2 = 2b2 …(ii)

Solving (i) and (ii)

2 22 21 2 4 – 1,

3 6b bx y+

= =



Differentiating (i)

2 – 2 0dyx ydx

=

dy xdx y

= …(iii)

By (ii)

2 4 0dyx ydx

+ =

–2

dy xdx y

= …(iv)

As the curves are orthogonal. Hence

– –12

x xy y

⎛ ⎞⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

x2 = 2y2

⇒2 21 2 4 12.

3 6+ −

=b b

⇒ 1 + 2b2 = 4b2 – 1

2b2 = 2

b2 = 1

Hence equation of the ellipse is

x2 + 2y2 = 2

⇒2 2

12 1

+ =x y

Foci = 12 10 ( 1, 0)2

⎛ ⎞± × = ±⎜ ⎟⎝ ⎠

26. For the function = ≥1( ) cos , 1f x x xx

,

(A) For at least one x in the internal [1, ∞), f(x + 2) – f(x) < 2

(B)→∞

′ =lim ( ) 1x

f x

(C) For all x in the interval [1, ∞), f(x + 2) – f(x) > 2

(D) f ′(x) is strictly decreasing in the interval [1, ∞)

Answer (B, C, D)

Hints :

f(x) = 1cosxx

21 1 1( ) sin cos⎛ ⎞⎛ ⎞ ⎛ ⎞′ = − − +⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠f x x

x xx

1 1 1( ) cos sin⎛ ⎞ ⎛ ⎞′ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

f xx x x



Clearly, lim ( ) 1→∞

=x

f x

Hence (B) is correct.

f′(x) = 1 1 1cos sin⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠x x x

2 2 21 1 1 1 1 1 1sin . cos sin⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞′′ = − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠f

x x x xx x x

2 32 1 1 1sin cos . 0⎛ ⎞ ⎛ ⎞′′ = − − <⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠f

x xx x

Hence f ′(x) is strictly decreasing function.

Hence (D) is true.

27. The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T andN, respectively. The locus of the centroid of the triangle PTN is a parabola whose

(A) Vertex is ⎛ ⎞⎜ ⎟⎝ ⎠

2 , 03a (B) Directrix is x = 0

(C) Latus rectum is 23a

(D) Focus is (a, 0)

Answer (A, D)

Hints :

Let P ≡ (at2, 2at)

Equation of tangent at P is

yt = x + at2

at, y = 0 we get x = –at2

Hence T ≡ (–at2, 0)

Similarly equation of normal at P is

y + xt = 2at + at3

at, y = 0, we get x = 2a + at2

Hence N ≡ (2a + at2, 0)

Let centroid of the triangle PTN is (h, k)

⇒2 2 22

3− + +

=at at a ath

⇒22

3+

=a ath …(i)

2 0 0 23 3+ +

= =at atk …(ii)



By (i) and (ii)

2322

3

⎛ ⎞+ ⎜ ⎟⎝ ⎠=

ka aah

⇒2

293 2 .4

= +kh a aa

293 24

= +kh aa

⇒29 3 2

4= −

k h aa

2 4 (3 2 )9

= −ak h a

2 4 23 3⎛ ⎞= −⎜ ⎟⎝ ⎠

a ak h

Hence locus of (h, k) is

2 4 23 3⎛ ⎞= −⎜ ⎟⎝ ⎠

a ay x

which represents a parabola whose vertex is 2 , 03

⎛ ⎞⎜ ⎟⎝ ⎠

a

Directrix : 23 3

− = −a ax ⇒

3=

ax

Latus rectum : 43a

Focus : 23 3

− =a ax and y = 0

⇒ x = a, y = 0

⇒ Focus : (a, 0)

28. For π

< θ <02

, the solution(s) of =

− π π⎛ ⎞ ⎛ ⎞θ + θ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑6

1

( 1)cosec cosec 4 24 4m

m m is(are)

(A)π4

(B)π6

(C)π

12(D)

π512

Answer (C, D)Hints : We have

cosec cosec4 4m

m mT π − π π⎛ ⎞ ⎛ ⎞= θ + θ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



1

sin sin4 4 4

m m=

π π π⎛ ⎞ ⎛ ⎞θ + − θ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin4 4 41

sin sin sin4 4 4 4

m m

m m

⎡ ⎤⎛ ⎞π π π⎛ ⎞ ⎛ ⎞θ + − θ + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥= ⎢ ⎥π π π π⎛ ⎞ ⎛ ⎞θ + θ + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

sin cos cos sin4 4 4 4 4 42

sin sin4 4 4

m m m m

m m

⎡ ⎤π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞θ + θ + − − θ + θ + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥=π π π⎢ ⎥⎛ ⎞ ⎛ ⎞θ + θ + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 cot cot4 4 4

m m⎡ ⎤π π π⎛ ⎞ ⎛ ⎞= θ + − − θ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

⇒ 1 2 cot( ) cot4

T⎛ ⎞π⎛ ⎞= θ − θ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

222 cot cot

4 4T

⎛ ⎞π π⎛ ⎞ ⎛ ⎞= θ + − θ +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

65 62 cot cot4 4

T⎛ ⎞π π⎛ ⎞ ⎛ ⎞= θ + − θ +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Sum 1 2 3 4 5 6 4 2T T T T T T+ + + + + =

⇒ 62 cot cot 4 24

⎡ ⎤π⎛ ⎞θ − θ + =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

cotθ + tanθ = 4

⇒2 2sin cos 4

sin cosθ + θ

=θ θ

⇒1sin cos4

θ θ =

12sin cos2

θ θ =

1sin22

θ =

⇒52 ,

6 6π π

θ =

⇒5,

12 12π π

θ =



SECTION - IIIMatrix - Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched.The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s andt. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. Theappropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the followingexample :

If the correct matches are A - p, s and t; B - q and r; C - p and q; and D - s and t; then the correct darkening of bubbleswill look like the following.

p q r s t

p

p

p

q

q

q

r

r

r

s

s

s

t

t

t

p q r s tA

B

C

D

29. Match the statements/expressions given in Column I with the values given in Column II.

Column I Column II

(A) Root(s) of the equation 2sin2θ + sin22θ = 2 (p)π6

(B) Points of the discontinuity of the function (q)π4

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦

6 3( ) cosx xf x , where [y] denotes the

largest integer less than or equal to y

(C) Volume of the parallelopiped with its edges (r)π3

represented by the vectors + +ˆ ˆ ˆ ˆ, 2i j i j

and + + πˆ ˆ ˆi j k

(D) Angle between vectors a and b where ,a b (s)π2

and c are unit vectors satisfying

+ + =3 0a b c

(t) π

Answer A(q, s), B(p, r, s, t), C(t), D(r)Hints :

We have(A) 2sin2θ + sin22θ = 2

⇒ 2sin2θ + 4sin2θcos2θ = 2sin2θ + 2sin2θcos2θ = 1sin2θ + 2sin2θ(1 – sin2θ) = 13sin2θ – 2sin4θ = 12sin4θ – 3sin2θ + 1 = 02sin4θ – 2sin2θ – sin2θ + 1 = 02sin2θ(sin2θ – 1) – 1(sin2θ – 1) = 0(sin2θ – 1)(2sin2θ – 1) = 0



⇒1sin 1, sin2

θ = ± θ = ±

, , ,2 2 4 4π π π π

θ = − θ = −

(B) 6 3( ) cosx xf x ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥π π⎣ ⎦ ⎣ ⎦

Point of discontinuity are the points where

6 3 and x xπ π

are integer

11

6 4 5, , , , ,6 6 3 2 6 6

Ix I xπ π π π π π

= ⇒ = = ππ

22

3 2, ,3 3 3

Ix I xπ π π

= ⇒ = = ππ

(C)1 1 01 2 01 1

V = = ππ

(D) 3a b c+ = −

⇒ 1 1 2cos 3+ + θ =

⇒1cos2

θ =

3π

θ =

30. Match the statements/expressions given in Column I with the values given in Column II.

Column I Column II

(A) The number of solutions of the equation (p) 1

xesinx – cosx = 0 in the interval π⎛ ⎞

⎜ ⎟⎝ ⎠0,

2

(B) Value(s) of k for which the planes (q) 2kx + 4y + z = 0, 4x + ky + 2z = 0 and2x + 2y + z = 0 intersect in a straight line

(C) Value(s) of k for which (r) 3|x – 1| + |x – 2| + |x + 1| + |x + 2| = 4khas integer solution(s)

(D) If y′ = y + 1 and y(0) = 1, then value(s) of y(ln 2) (s) 4

(t) 5Answer A(p); B(q, s); C(q, r, s, t), D(r)



Hints :

(A) xesinx – (0) x = 0

xesinx = cosx

(0, 1)

π2 e

π2

y

x

Clearly only one solution exists.

(B) For intersecting in aline given system of equations must have a non-zero solution

i.e.

4 14 2 02 2 1

kk =

⇒ k(k – 4) – 4(4 – 2) + 2(8 – k) = 0

⇒ k2 – 4k – 8 + 16 – 2k = 0

⇒ k2 – 6k + 8 = 0

⇒ k = 4, 2

(C) |x – 1| + |x – 2| + |x + 1| + |x + 2| = 4k

Following cases may arise :

Case (i)

x ≤ –2

–x + 1 – x + 2 – x – 1 – x – 2 = 4k

–4x = 4k ⇒ x = – k ⇒ k = 2, 3, 4, 5,...

Case (ii)

–2 < x ≤ –1

x + 2 – x – 1 – x + 2 – x + 1 = 4k

–2x + 4 = 4k, no integral value of k

Case (iii)

–1 < x ≤ 1

x + 2 + x + 1 – x + 1 – x + 2 = 4k

6 = 4k, no integral value of k



1 < x ≤ 2

x – 1 + x + 2 + x + 1 + x + 2 = 4k

= 2x + 4 = 4k

x = 2(k – 1) ⇒ k = 2, 3

Case (iv)

x > 2

x – 1 + x – 2 + x + 1 + x + 2 = 4k

4x = 4k

x = k ⇒ k = 3, 4, 5,...

(D) 1dy ydx

= +

⇒ 1dy dx

y=

+∫ ∫⇒ ln (y + 1) = x + c

⇒ (y + 1) = kex

⇒ y = k.ex – 1

At x = 0, y = 1

1 = k – 1 ⇒ k = 2

∴ y = 2ex – 1

y(ln 2) = ln 22 1e − = 4 – 1 = 3


This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. Theappropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correctanswers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbleswill look like the following :

0 0 0 0

X Y Z W

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9



31. The maximum value of the function f(x) = 2x3 – 15x2 + 36x – 48 on the set A = {x |x2 + 20 ≤ 9x} is

Answer (7)

Hints :

f(x)= 2x3 – 15x2 + 36x – 48

A = {x : x2 + 20 ≤ 9x}

We know x2 – 9x + 20 ≤ 0

⇒ x ∈ [4, 5]

f ′(x) = 6 (x – 2) (x – 3)

f ′(x) = 0 for x = 2, 3

⇒ f(x) is increasing for x ∈ [4, 5]

∴ In the interval 4 ≤ x ≤ 5, f(x) is maximum at x = 5

f(5) = 2. (5)3 – 15 (5)2 + 36 × 5 – 48 = 7

32. Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations:

3x – y – z = 0

–3x + z = 0

–3x + 2y + z = 0

Then the number of such points for which x2 + y2 + z2 ≤ 100 is

Answer (7)

Hints : We have

3x – y – z = 0 ...(i)

–3x + z = 0 ...(ii)

–3x + 2y + z = 0 ...(iii)

Applying (i) and (ii) we get

⇒ y = 0 ...(iv)

Also 3x = z

Points satisfying x2 + y2 + z2 ≤ 100 with integral coordinates can be

(0, 0, 0), (1, 0, 3), (2, 0, 6), (3, 0, 9), (–1, 0, –3), (–2, 0, –6), (–3, 0, –9)

Hence 7 such points exist

33. Let ABC and ABC ′ be two non-congruent triangles with sides AB = 4, AC = AC ′ = 2 2 and angle B = 30°.The absolute value of the difference between the areas of these triangles is

Answer (4)

Hints :

Required area will be the shaded region in the figure

Area of 1 2 4 42

ACC′Δ = × × =30° 45°

22 24

B

A

C C′

22



34. Let p(x) be a polynomial of degree 4 having extremum at x = 1, 2 and →

⎛ ⎞+ =⎜ ⎟⎝ ⎠20

( )lim 1 2x

p xx

. Then the value of

p(2) is

Answer (0)

Hints :

20

( )lim 1 2x

p xx→

⎛ ⎞+ =⎜ ⎟⎝ ⎠

⇒ 20

( )lim 1 2x

p xx→

= =

⇒ 20

( )lim 2 1 1x

p xx→

= − =

⇒ The coefficient of x2 in p(x) is 1 and the coefficient of x in p(x) is 0 and the constant term also is 0.

Now

Let p(x) = ax4 + bx3 + x2

p′(x) = 4ax3 + 3bx2 + 2x

Since p(x) has extremum at x = 1, 2

p′(1) = 0 ⇒ 4a + 3b + 2 = 0 ...(i)

p′(2) = 0 ⇒ 32a + 12b + 4 = 0

⇒ 8a + 3b + 1 = 0 ...(ii)

Solving (i) and (ii) 14

a = and b = –1

Hence 4

3 2( )4xp x x x= − +

p(2) = 0

35. Let f : R → R be a continuous function which satisfies = ∫0

( ) ( )x

f x f t dt . Then the value of f(ln 5) is

Answer (0)

Hints :0

( ) ( )x

f x f t dt= ∫⇒ f′(x) = f(x)

⇒ [ ( ), ( )dy dyy y f x f xdx dx

′= = =

⇒ 1·dy dxdx

= ∫



⇒ lny = x + ln k

⇒ ln y xk=

⇒ y = kex

⇒ f(x) = kex

Now 0

( ) ( )x

f x f t dt= ∫

⇒0 0

x xx t tke ke dt k e dt= =∫ ∫

⇒0

xx tke k e⎡ ⎤= ⎣ ⎦

⇒ kex = k[ex – e0]

⇒ kex = kex – k

⇒ k = 0

∴ f(x) = 0 ∀x

⇒ f(ln 5) = 0

36. The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let Pbe the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 andC, then the radius of the circle C is

Answer (8)

Hints :

The figure below is according to the given situation

Let ∠SPO2 = θ

1sin3

θ =

Now 2

2sin

POQO

= θ

θ

C1 C2

Q

Sθ

3 O2O1

P

1

C

⇒2

3 13QO

=

⇒ QO2 = 9

QS = QO2 – SO2 = 9 – 1 = 8

Hence the radius of circle C is 8

37. The smallest value of k, for which both the roots of the equation

x2 – 8kx + 16(k2 – k + 1) = 0

are real, distinct and have values at least 4, is

Answer (2)



Hints : x2 – 8kx + 16 (x2 – k + 1) = 0

Roots are real and distinct

Let f(x) = x2 – 8kx + 16(k2 – k + 1)

⇒ D > 0 ⇒ x > 1

Let f(4) > 0 ⇒ k ∈ (–∞, 1] ∪ [2, ∞)

∴ Least value of k can be 2.

38. If the function 3 2( )x

f x x e= + and g(x) = f –1(x), then the value of g′(1) is

Answer (2)

Hints : 3 /2( ) xf x x e= + , g(x) = f–1(x)

f′(x) = 3x2 + /212

xe

⇒ f′(0) = 12

f(g(x)) = x

Differentiating both sides

f′(g(x))g′(x) = 1

⇒1(1)

( (1))g

f g′ =

′

for f(x) = 1 x = 0

∴ f –1(1) = 0

∴1 1(1) 2

1(0)2

gf

′ = = =′

PHYSICSPART - III


This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out which ONLY ONE is correct.

39. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of thepoint P is

k1 k2

P M

(A)1

2

k Ak (B)

2

1

k Ak (C)

1

1 2

k Ak k+ (D)

2

1 2

k Ak k+



Answer (D)

Hint : Internal forces in the springs are same

⇒ k1 x1 = k2 x2

Also x1 + x2 = A

⇒ x1 = 2

1 2

k Ak k+

40. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. Thebead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is atrest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of thenew equilibrium position of the bead, where the bead can stay at rest w.r.t. the wire, from the y-axis is

(A)a

gk (B) 2agk (C)

2agk (D) 4

agk

Answer (B)

Hint : For equilibrium

dydx = tanθ =

ag

θ

y

x⇒ 2kx = ag

⇒ x = 2agk

41. Photoelectric effect experiments are performed using three different metal plates p, q and r having workfunctions φp = 2.0 eV, φq = 2.5 eV and φr = 3.0 eV, respectively. A light beam containing wavelengths of550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I - V graph forthe experiment is [Take hc = 1240 eV nm]

(A)

pqr

I

V

(B) pq

r

I

V

(C) pqr

I

V

(D)

pqr

I

V



Answer (A)

Hint :

0 stopping potential1( ) hcVe

⎛ ⎞= − φ⎜ ⎟⎝ ⎠λ

As, φp < φq < φr

V0p > V0q

> V0r

In the case of metal P all wavelength are less than threshold wavelength so current will be maximum. In thecase of metal q the wavelength 550 nm is more than threshold wavelength so the current will be lessercompared to metal p.

In the case of metal r only the wavelength 350 nm will be able to eject the electron, hence current will beleast.

42. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs ofequal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is freeto oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction andreleased. The frequency of oscillation is

(A)1 2

2k

Mπ(B)

12

kMπ

(C)1 6

2k

Mπ(D)

1 242

kMπ

Answer (C)

Hint : Restoring torque

τ = 22 2l lk⎛ ⎞

− θ ×⎜ ⎟⎝ ⎠

θ

θ

kxklθ2

kx kl = θ2=

2–

2kl θ

⇒ α = Iτ

=

2

2

–2

12

kl

ml

θ

= 6– kmθ

= – ω2θ

⇒ ω = 6km

= 2 fπ




This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer,out of which ONE OR MORE is/are correct

43. Two metallic rings A and B, identical in shape and size but having different resistivities ρA and ρB, are kepton top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoidsin identical manner, the rings A and B jump to heights hA and hB, respectively, with hA > hB. The possiblerelation(s) between their resistivities and their masses mA and mB is(are)

A B

(A) ρA > ρB and mA = mB (B) ρA < ρB and mA = mB

(C) ρA > ρB and mA > mB (D) ρA < ρB and mA < mB

Answer (B, D)

Hint :

(A) ρA > ρB

⇒ iA < iBand mA = mB

⇒ hA < hB

(B) ρA < ρB

⇒ iA > iBand mA = mB

⇒ hA > hB

(C) ρA > ρB

⇒ iA < iBand mA > mB

⇒ hA < hB

(D) rA < rB and mA < mB

⇒ hB > hA

44. A student performed the experiment to measure the speed of sound in air using resonance air-column method.Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorterair-column is the first resonance and that with the longer air-column is the second resonance. Then,

(A) The intensity of the sound heard at the first resonance was more than that at the second resonance

(B) The prongs of the tuning fork were kept in a horizontal plane above the resonance tube

(C) The amplitude of vibration of the ends of the prongs is typically around 1 cm

(D) The length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength ofthe sound in air



Answer (D)

Hint :

The length of air column at the first resonance will be slightly less than4λ

due to end correction.

45. The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circleand CDA is half of an ellipse. Then,

C

A

BD

0

1

2

3P

1 2 3 V

(A) The process during the path A → B is isothermal

(B) Heat flows out of the gas during the path B → C → D

(C) Work done during the path A → B → C is zero

(D) Positive work is done by the gas in the cycle ABCDA

Answer (B, D)

Hint : (A) It is incorrect as isothermal process is graphically

P

V

represented like as shown

(B) For B → C → D, ΔU = – ve and W = – ve

⇒ Q = – ve

(C) W = shaded area = + ve

A

B

C(D) Clockwise P – V cycle

⇒ + ve work

46. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit.Find out the correct statement(s)

(A) The angular momentum of the charge –q is constant

(B) The linear momentum of the charge –q is constant

(C) The angular velocity of the charge –q is constant

(D) The linear speed of the charge –q is constant

Answer (A)

Hint : As τ about Q is zero L = constant about + Q (although this should have been mentioned in the problem).Linear velocity, momentum, speed and angular speed vary.



47. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact,B is the centre of the sphere and C is its topmost point. Then,

A

B

C

(A) ( )2C A B CV V V V− = −

(B) C B B AV V V V− = −

(C) 2C A B CV V V V− = −

(D) 4C A BV V V− =

Answer (B, C)

Hint :

The motion can be considered to be pure rotation about point A velocities will be as shown

V VC B = 2

VA = 0

VB

⇒ C B B AV V V V− = −

and | | 2 | |C A B CV V V V− = −

SECTION-IIIMatrix-Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to bematched. The statements in Column I are labelled as A, B, C and D, while the statements in Column II are labelledp, q, r, s and t. Any given statement in Column I can have correct matching with ONE OE MORE statement(s) inColumn II. The appropriate bubbles corresponding to the answers to these questions have to be darkened asillustrated in the following example:

If the correct matches are A–p, s and t; B–q and r; C–p and q; and D–s and t; then the correct darkening of bubbleswill look like the following.

A

B

C

D

p

p

p

p

q

q

q

q

r

r

r

r

s

s

s

s

p q r s

t

t

t

t

t



48. Column II gives certain systems undergoing a process. Column I suggests changes in some of theparameters related to the system. Match the statements in Column I to the appropriate process(es) fromColumn II.

Column I Column II

(A) The energy of the system is increased (p) System : A capacitor, initially uncharged

Process : It is connected to a battery

(B) Mechanical energy is provided to the system, (q) System : A gas in an adiabatic container fitted with anwhich is converted into energy of random adiabatic pistonmotion of its parts Process : The gas is compressed by pushing the

piston(C) Internal energy of the system is converted (r) System : A gas in a rigid container

into its mechanical energy Process : The gas gets cooled due to colderatmosphere surrounding it

(D) Mass of the system is decreased (s) System : A heavy nucleus, initially at rest

Process : The nucleus fissions into two fragments ofnearly equal masses and some neutronsare emitted

(t) System : A resistive wire loop

Process : The loop is placed in a time varyingmagnetic field perpendicular to its plane

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

Hint.

(p) Work done by the battery is partially stored in the capacitor in the form of its electric potential energy.Thus energy of the capacitor increases.

(q) Internal energy of the gas increases as work done on the gas is converted into internal energy i.e. energyof random motion of its parts.

(r) Internal energy of the gas decreases as the gas gives energy to the surroundings.

(s) Energy of the system decreases as the energy is released in the process.

(t) Energy of the wire loop increases due to the development of induced current.

49. Column I shows four situations of standard Young's double slit arrangement with the screen placed far awayfrom the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 – S2P1 = λ/4 and S1P2 – S2P2 = λ/3,where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive indexμ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phasedifference between the light waves reaching a point P on the screen from the two slits is denoted by δ(P) andthe intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for thatsituation.

Column I Column II

(A) S2

S1

PPP

2

1

0

•

•

•

(p) δ(P0) = 0



(B) (μ – 1)t = λ/4 S2

S1

PPP

2

1

0

•

•

•

(q) δ(P1) = 0

(C) (μ – 1)t = λ/2 S2

S1

PPP

2

1

0

•

•

•

(r) I(P1) = 0

(D) (μ – 1)t = 3λ/4 S2

S1

PPP

2

1

0

•

•

•

(s) I(P0) > I(P1)

(t) I(P2) > I(P1)

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

Hint :

Case (A)

0 1 2

0 0 0

04 3

202 3

4 2

P P P

x

I I I I

λ λΔ

π πδ

Case (B)

0 1 2

0 0 0

04 12

02 6

2 4 (2 3)

P P P

x

I I I I

λ λΔ

π πδ

+



Case (C)

0 1 2

0 0

2 4 6

2 3

0 2 3

P P P

x

I I I

λ λ λΔ

π πδ π

Case (D)

0 1 2

0 0

3 54 2 12

3 52 6

2 0 (2 3)

P P P

x

I I I

λ λ λΔ

π πδ π

−


This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example,if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following:

0

1

2

4

8 88 8

9 9 9 9

0

X Y Z W

0 0

1 1 1

2 2 2

3 3 3 3

4 4 4

5 5 55

6 6 66

7 7 77



50. A metal rod AB of length 10x has its one end A in ice at 0ºC and the other end B in water at 100ºC. If apoint P on the rod is maintained at 400ºC, then it is found that equal amounts of water and ice evaporate andmelt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80cal/g. If the point P is at a distance of lf from the ice end A, find the value of λ.

[Neglect any heat loss to the surrounding.]

Answer (9)

Hint :

0ºC 100ºC10x

A 400ºC B

λx

Ice400 0

.

IdmL

x dtK A

−= ×

λ

Also steam400 10010

.

wdmL

x x dtK A

−= ×

− λ

Ice

Steam

400(10 ) 80 4300 540 27

LL

− λ= = =

λ ×

= 8 4

54 27=

⇒ 9(10 – λ) = λ

⇒ 90 = 10λ

⇒ λ = 9

51. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closedand water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottomis opened. Some water comes out from the orifice and the water level in the vessel becomes steady withheight of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice.

[Take atmospheric pressure = 1.0 × 105 N/m2, density of water = 1000 kg/m3 and g = 10 m/s2. Neglect anyeffect of surface tension.]

Answer ( 6)

Hint : P0A (500 – H) = (P0 – ρgh)A.300

⇒5 4

5(10 10 0.2) 300500

10H − × ×

− = ⇒ H = 206 mm

Fall in height = 6 mm

52. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. Theradii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to makebubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A andB, respectively. [Neglect the effect of gravity.]



Answer (6)

Hint : 24 4 0.048 8

2 10ATPR −

×− = = =

×

⇒ PA = 16 N/m2

Similarly, PB = 12 N/m2

As n ∝PV

⇒2

316 2 4 1 112 3 8 64

A A A

B B B

n P Vn P V

×= = × = × =

×

⇒ 6B

A

nn

=

53. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have massesm, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collisionwith it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same staright line.Find the final speed (in m/s) of the object C.

m 2m mA B C

Answer (4)

Hint :

(2 ) 0 2 93B

m m mvm

− × + ×′ = = 6 m/s

A B mC

m 2m m

9 m/s

∴2 6

3cmv

m×

= = 4 m/s2m m

6 m/s

54. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and

QR = 5x. If the magnitude of the magnetic field at P due to this loop is 0

48I

kx

μ⎛ ⎞⎜ ⎟π⎝ ⎠

, find the value of k.

Answer (7)

Hint :

BPR = 0

BPQ = 0

As 4x × 35

PS=

125xPS =

37°

5x3x

P

Q

R4x

S

∴ 0 {sin53 sin37 }4 12 / 5

IBx

μ= × ° + °

π

0 5 4 34 12 5 5

Ix

μ ⎧ ⎫= × +⎨ ⎬π ⎩ ⎭

0 077

4 12 48I I

x xμ μ⎧ ⎫

= = ⎨ ⎬π× π⎩ ⎭

∴ k = 7



55. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocksof masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the blockof mass 0.36 kg during the first second after the system is released from rest.

Answer (8)

Hint :

asystem = 0.72 0.36

1.08g−

0.361.08

g= ×

= 3g

m/s2

∴ T – 0.36 × g = 0.36 × 3g

10.36 13

T g ⎧ ⎫= +⎨ ⎬⎩ ⎭

40.363

g ⎧ ⎫= ⎨ ⎬⎩ ⎭

= 0.12 × 4 × g

= 0.48 g = 4.8 N

∴ 21 12 3 6

g gS = × × =

∴10 484.86 6

W = × = = 8 J

56. A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = kra, where k

and a are constants and r is the distance from its centre. If the electric field at 1is2 8Rr = times that at

r = R, find the value of a.



Answer (2)

Hint :

0. inq

E ds =ε∫

2 2

0 0

1.4 4r

aE r kx x dxπ = πε ∫

rP

xdx

R

2 2

0 0

1.4 4r

aE r kx x dxπ = πε ∫

3

20 3

ak rEar

+⎡ ⎤= ⎢ ⎥

+ε ⎢ ⎥⎣ ⎦

1

0( 3)

akrEa

+

=ε +

/218R RE E=

⇒1

1

0 0

1 [ ]( 3) 2 8 ( 3)

aak R k R

a a

++⎡ ⎤ =⎢ ⎥ε + ε +⎣ ⎦

⇒ a = 2

57. A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N.The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm)between the successive nodes on the string.

Answer (5)

Hint : l = 20 cm

m = 1 gm

T = 0.5 N

n = 100 Hz

2 3

2

0.51002 20 10 10

20 10

n− −

−

=× ×

×

1 0.5 20 1040n

= × ×

1040n

= ×

⇒ n = 4

∴ .2

n lλ=

⇒20

2 4λ= = 5 cm

solutions to iit-jee 2009 · 2019. 7. 14. · aakash iit-jee regd. office : aakash tower, plot no....

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