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IIT-JEE 2012 / Actual Paper /Questions and Solutions

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AnalysisIIT-JEE

Actual Paper – 12012

IIT-JEE 2012 / Actual Paper /Questions and Solutions

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PART– I : PHYSICS

SECTION I : Single Correct Answer Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

1. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive

surface charge density. The variation of the magnitude of the electric field E(r)ur

and the electric

potential V(r) with the distance r from the centre, is best represented by which graph?

(A) \(B)

(C) (D)

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Sol.

1. D

<= ≥ π ∈

20

0, r RE Q

, r R4 r

≤ π ∈= ≥ π ∈

0

0

Qr R

4 RV

Q, r R

4 r

2. Young’s double slit experiment is carried out by using green, red and blue light, one color at aime.

The fringe widths recorded are G Bβ > β and Bβ , respectively. Then,

(A) G B Rβ > β > β (B) B G Rβ > β > β (C) R B Gβ > β > β (D) R G Bβ > β > β

Sol.

2. D β ∝ λ

λ > λR V

λ > λ > λR G B

β > β > βR G B

β < β < βB G R

3. Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DCvoltage source of potential difference X . A proton is released at rest midway between the two plates.It is found to move at 45º to the vertical JUST after release. Then X is nearly(A) 1× 10–5 V (B) 1× 10–7V (C) 1 × 10–9 V (D) 1 × 10–10 V

Sol.

3. C = =eE eX

gm ml

−= 9X 10

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4. A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive indexn of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the sameradius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the imagedistance will be

(A) –280.0 cm. (B) 40.0 cm. (C) 21.5 cm (D) 13.3 cm

Sol.

4. B = −1

1 1(1.5 1) ,

f R

= − = + = =×2

1 1 1 1 1 7 7(1.2 1) ,

f R f 2R 5R 10R 10 14f = 20 cm

− =−

1 1 1v 40 20

V = 40 cm

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5. A small mass m is attached to a massless string whose other end is fixed at P as shown in thefigure. The mass is undergoing circular motion in the x-y plane with centre at O and constantangular speed ω . If the angular momentum of the system, calculated about O and P are denoted by

OLr

and PLr

respectively, then

(A) OLr

and PLr

do not vary with time.

(B) OLr

varies with time while PLr

remains constant.

(C) OLr

remains constant while PLr

varies with time.

(D) OLr

and PLr

both vary with time.

Sol.

5. C =r uur

2OL mr w = constant

= ×r r uur

PL r mw keep changing in direction

6. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass

= 40 amu) is kept at 300 K in a container. The ratio of the r m s speeds rms

rms

V (helium)V (argon)

is

(A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

Sol.

6. D= = = =rms,He

rms,Ar

3RTV 40M 10 3.16V 43RT

M

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7. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω , asshown in the figure. At time t = 0, a small insect start from O and moves with constant speed v withrespect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The

angular speed of the system remains ω throughout. The magnitude of the torque ( )τ on the system

about O, as a function of time is best represented by which plot?

(A) (B)

(C) (D)

Sol.

7. B = + τ =1 2

2m l dLL m(vt) ,

3 dtτ = ∝2mvt.v t

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8. In the determination of Young’s modulus G

24ML

Yld

=

π by using Searle’s method, a wire of length L

= 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in thelength of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer,respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is100. The contributions to the maximum probable error of the Y measurement(A) due to the errors in the measurements of d and l are the same.(B) due to the error in the measurement of d is twice that due to the error in the measurement of l.(C) due to the error in the measurement of l is twice that due to the error in the measurement of d.(D) due to the error in the measurement of d is four times that due to the error in the measurement

of l.

Sol.

8. A =π 2

4MLgY

ld

∆ ∆ ∆ ∆ ∆= + + +

Y M L d l2

Y M L d l

∆= × =

d 0.52 2 2

d 0.5

∆= =

l 0.52

l 0.25

∆ ∆∴ =

d l2

d l

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9. A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The otherend of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. Theblock is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion

with angular frequency 3πω = rad/s. Simultaneously at t = 0, a small pebble is projected with speed

v from point P at an angle of 45º as shown in the figure, Point P is at a horizontal distance of 10 mfrom O. If the pebble hits the block at t = 1 s, the value of v is (take g = 10 m/s2)

(A) 5 0 m / s (B) 51m/s (C) 5 2 m / s (D) 5 3 m / s

Sol.

9. A°

= =2vsin45

t 110

∴ =V 50

10. Three very large plates of same area are kept parallel and close to each other. They are consideredas ideal black surfaces and have very high thermal conductivity. The first and third plates are maintainedat temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate understeady state condition is

(A)

1265

T2

(B)

1497

T4

(C)

1497

T2

(D) ( )1497 T

Sol.

10. C (2T) t (2T)

+ =4 4 4818T T t

2;

=

1497

t T2

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SECTION II : Multiple Correct Answer(s) Type

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

11. For the resistance network shown in the figure, choose the correct option(s).

(A) The current through PQ is zero (B) I1 = 3A(C) the potential at S is less than that at Q (D) I2 = 2A

Sol.11. A,B,C,D

12 V

I3

I4I5

y (12 – y)

4Ω 4Ω

1Ω 1Ω

2Ωx

IAI4

I1

I – I54

I2 = 2A

I3

(O)

(12 – x)

(12)

− −+ + =

x 12 2x 12 2x.y0

2 2 2

− −

+ + =y 12 2y 12 4y.yx

04 4 4

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5x – 2y = 247y – 4x = 245x – 2y = 7y – 4x9x = 9yx = y

= =×1

12I 3A

6 1218

= × =2

2I 3 2A

3

=QV 8V

=SV 4V

12. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels downthe pipe. When this pulse reaches the other end of the pipe.(A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.(B) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.(C) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.(D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.

Sol.

12.

HP

HP

HP

LP

no phasechange atclosed end

phasechange of at open endπ

D

B

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13. Consider the motion of a positive point charge in a region where there are simultaneously uniform

electric and magnetic fields 0E E j=ur $ and 0B B j=

ur $ . At time t = 0, this charge has velocity vr

in the x-

y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for timet> 0 ?(A) If θ = 0º, the charge moves in a circular path in the x-z plane.(B) If θ = 0º, the charge undergoes helical motion with constant pitch along the y-axis.(C) If θ = 10º, the charge undergoes helical motion with its pitch increasing with time, along the

y-axis.(D) If θ = 90º, the charge undergoes linear but accelerated motion along the y-axis.

Sol.13. C, D

yBE v

z

θ x

=ur $

0E E j

=ur $

0B B j

θ = °10 or θ = ° ⇒0 helix with increasing pitch

(D) θ = ⇒90 correct

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14. A cubical region of side a has its centre at the origin. IT encloses three fixed point charges, –q at0, –a/4, 0), +3q at (0, 0, 0) and –q at (0, +a/4, 0). Choose the correct option(s).

(A) The net electric flux crossing the plane x = a/2 is equal to the net electric flux crossing the planex = –a/2.

(B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing theplane y = –a/2.

(C) The net electric flux crossing the entire region is ε0

q.

(D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing theplane x = +a/2.

Sol.14. A, C, D

=∈0

qtotal flux

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15. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with thehorizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in thefigure. The block remains stationary if (take g = 10 m/s2)

(A) θ = °45 .(B) θ > °45 and a frictional force acts on the block towards P..

(C) θ > °45 and a frictional force acts on the block towards Q.

(D) θ < °45 and a frictional force acts on the block towards Q.

Sol.15. A, C

PQ

Q

1 cos θ

sin + cos

θθ IN

1 sin θ

θ

θIN

(C) θ > °45θ < θ ⇒cos sin needs friction towards Q to keep it in equilibrium.

(A) θ = °45 , θ = θcos sin no need of friction.

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SECTION III : Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0to 9 (both inclusive)

16. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Boththe cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the

magnitude of the magnetic field at the point P is given by µ0N

aJ,12

then the value of N is

Sol.

16. 5

a ××

××P

J→

$ $ $0

1B= µ j×r

2; = µ0

1 ˆB jr2

µ π= µ −

×

20

P 0J. a1

B J.a3a2 2x 42

µ µ

= −0 0aJ aJ2 12

= µ = µP 0 0

5 NB aJ aJ

12 12N = 5

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17. A proton is fired from very far away towards a nucleus with charge Q = 120 e, where is the electroniccharge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units offm) of the proton at its start is: (take the proton mass,

mp = (5/3) × 10–27 kg; h/e = 4.2 × 10–15J.S/C; = ×πε

9

0

19 10 m/F;

4 1 fm = 10–15 m)

Sol.

17. 7−= × 27

P

5m 10 kg

3

−= × 15h4.2 10 J.& / c

e

×=

λ π ∈

2P

20

2m 120e1 e.

4 .r.h h

×6 e

10 h

−λ = × × =15104.2 10 7fm

6

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18. A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of

side a (a<<R) having two turns is placed with its center at =z 3R along the axis of ?? the circular

wire loop, as shown in figure. The plane of the square loop makes an angle of 45° with respect to the

z-axis. If the mutual inductance between the loops is given by µ 2

0p / 2

a,

2 Rthen the value of p is

Sol.

18. 7 φ =1 2Mi

=22 2B a Mi

( )µ

=+

2 20 2

232 2

i R .aMi

2 R 3R

µ=

2072

aM

2 R

⇒ =p 7

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19. An infinitely long solid cylinder of radius R has a uniform volume charge density ρ . It has sphericalcavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitudeof the electric field at the point P, which is at a distance 2R from the axis of cylinder, is given by the

expression ρε0

23 R.

16k The value of k is

Sol.

19. 6

2RP y

x

z

R/2

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... 18IIT-JEE Test Prep

λ= − π ∈ π ∈

P 20 0

QE

2 .2R 4 .4R

( )ρ πλ =

2. R .l

l

ρπ= ρ π =

3 34 R RQ .

3 8 6

ρπ ρ π= − π ∈ × π ∈

22 3

P 2 20 0

R . RB

4 R 6 16 R

= − ∈ 0

PR 11

4 24

=× ∈0

23PR16 6

K = 6

20. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform massdensity and radius 2R, as shown in the figure. The moment of inertia of this lamina about axespassing through O and P is IO and IP, respectively. Both these axes are perpendicular to the plane of

the lamina. The ratio P

O

II to the nearest integer is

\

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Sol.20. 3

2R

2R PO R

3cm

M4 = Cavity

mass

M = full mass

=P

0

I?

I

= − +

2 2 2

0

1 1M MI M.4R .R .R

2 2 4 4 = − =2 2 23 132MR MR MR

8 8

= +2

0 cm3M R

I I .4 9

;

= + +

22

P cm

3M RI I . 4R

4 9

∴ IP – I0 = 3MR2

= + =2 2 2P

13 37I 3MR MR MR

8 8

= ;P

0

I 373

I 13

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PART– II : CHEMISTRY

SECTION I: Single Correct Answer Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

21. The number of optically active products obtained from the complete ozonolysis of the given compoundis

CH – CH = CH – C – CH = CH – C – CH = CH – CH3 3

CH3

H

H

CH3

(A) 0 (B) 1 (C) 2 (D) 4

21. (A)

CH3 CH CH C CH

CH3

H

CH C CH

CH3

H

CH CH3

ozonolysis2CH3 C H

O

+ H C

O

C

CH3

H

C

O

H + H C

O

C

H

CH3

C

O

H

No. one product is optically active.

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... 21IIT-JEE Test Prep

22. The colour of light absorbed by an aqueous solution of CuSO4 is(A) orange-red (B) blue-green (C) yellow (D) violet

22. (A)23. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

(A) HNO3, NO, NH4Cl, N2 (B) HNO3, NO, N2, NH4Cl(C) HNO3, NH4Cl, NO, N2 (D) NO, HNO3, NH4Cl, N2

23. (B)5 2 0 3

23 4HNO,NO, N ,NHC l+ + −

24. As per IUPAC nomenclature, the name of the complex [CO(H2O)4 (NH3)2]Cl3 is(A) Tetraaquadiaminecobalt (III) chloride (B) Tetraaquadiamminecobalt (III) chloride(C) Diaminetetraaquacobalt (III) chloride (D) Diamminetetraaquacobalt (III) chloride

24. (D)

Diamminetetraaoquacobalt (III) chloride

25. The carboxyl functional group (–COOH) is present in(A) picric acid (B) barbituric (C) ascorbic acid (D) aspirin

25. (D)

26. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius]

(A) π

2

2 20

h4 ma (B) π

2

2 20

h16 ma (C) π

2

2 20

h32 ma (D) π

2

2 20

h64 ma

26. (C)

0 2nhmvn =π (r = 4a0)

402

nhmv

a=

π (n = 2)

2 22 2

2 160

44

n hm v

a=

π ; 2

22 2

0

1 12 2 16

hmvma

=

π

2

2 20

.32

hK E

ma=

π

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27. In the allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are)(A) sp and sp3 (B) sp and sp2 (C) only sp2 (D) sp2 and sp3

27. (B)

Structure of allene is H C

H

sp2C C

H

Hsp2sp

28. A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shownbelow. The empirical formula of the compound is

(A) MX (B) MX2 (C) M2X (D) M5X14

28. (B)

No. of M atoms are 2 and X is 4 the formullae is M2X4 i.e. MX2

29. The number of aldol reaction(s) that occurs in the given transformation is

+ →conc.aq.NaOH3CHCHO 4HCHO

OH OH

OHHO(A) 1 (B) 2 (C) 3 (D) 4

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29. (C)

CH2

H

C H

O

Con. aq. NaOH-H2O

CH2C H

OH C H

O

/H2O

Aldol

CH

H

CHO

C

H

H OH

OH¯-H2O

CH

CH2OH

CHOH C H

O

/H2O

Aldol

CH2OH

C

CH2OH

H C H

OOH¯

-H2O

CH2OH

C

CH2OH

C H

OH C H

O

/H2O

Aldol

CH2OH

C

CH2OH

CH2 C H

O

OH

Conc. NaOH

cannizaroreaction

CH2OH

C

CH2OH

CH2 CH2OHHO

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30. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV us. 1/V plot is shown below.The value of the van der Waals constant a (atm. liter2 mol–2) is

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0

30. (C)

( )

( )

2P 0 RT

PV RT

1PV RT

yx

a vvav

av↓↓

+ − =

+ =

= − +

m = -aor Tanθ = -a

or Tanα = a

or22.6 20.1

3 2a− =

or a = 1.5

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... 25IIT-JEE Test Prep

SECTION II : Multiple Correct Answer(s) Type

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

31. Choose the correct reason(s) for the stability of the lyophobic colloidal particles.(A) Preferential adsorption of ions on their surface from the solution(B) Preferential adsorption of solvent on their surface from the solution(C) Attraction between different particles having opposite charges on their surface(D) Potential difference between the fixed layer and the diffused layer of opposite charges around

the colloidal particles.

31. (A, D)

32. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The finalstate Z can be reached by either of the two paths shown in the figure. Which of the followingchoice(s) is (are) correct? [take ∆S as change in entropy and w as work done]

(A) → → →∆ = ∆ + ∆x z x y y zS S S (B) → → →= +x z x y y zw w w

(C) → → →=x y z x yw w (D) → → →∆ = ∆x y z x yS S

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32. (A, C)Enropy is a state function so DS is independent of path.So S S Sx z x y y z→ → →∆ = ∆ + ∆

work done in change is state Y to Z, W 0y z→ = ,since ∆v = 0, (isochoric process)work done is a path dependent function (not state function)So W W W W 0

Wx y z x y y z x y

x y

→ → → → →

= + = +

=

→ work done is not a state function, so option (B) is incorrect→ DS is state function so option (D) is incorrect.

33. Which of the following hydrogen halides react(s) with AgNO3(aq) to give a precipitate that dissolvesin Na2S2O3(aq)?(A) HCl (B) HF (C) HBr (D) HI

33. (A, C, D)

34. Identify the binary mixture(s) that can be separated into individual compounds, by differentialextraction, as shown in the given scheme.

Binary mixture containing

Compound 1 and Compound 2

NaOH(aq)

NaHCO(aq)3

Compound 1

Compound 1

Compound 2

Compound 2

+

+

(A) C6H5OH and C6H5COOH (B) C6H5COOH and C6H5CH2OH(C) C6H5CH2OH and C6H5OH (D) C6H5CH2OH and C6H5CH2COOH

34. (B, D)option B and D differentiated by both NaOH and NaHCO3.

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35. Which of the following molecules, in pure form, is (are) unstable at room temperature?

(A) (B) (C)

O

(D)

O

35. (B, C)

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... 28IIT-JEE Test Prep

SECTION III : Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0to 9 (both inclusive)

36. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons,undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongsin the periodic table?

+ → + α + +63 1 1 129 1 0 1Cu H 6 n 2 H X

36. 836 1 1 129 1 0 1

2 42

Cu H 6 n H X

Where He He+

+ → + α + +

α = =

balancing number of protonsno. of protons in left part of reaction = 29 + 1no. of protons in right part of reaction

= 0 + 2 + 2×1 + x= 4 + x

So 4 + x = 30 ⇒ x = 30 - 4 = 26So atomic number of x = 26So group of atom in periodic table = 8

37. An organic compound undergoes first-order decomposition. The time taken for its decomposition to

1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of ×1/8

1/10

[t ]10

[t ]?

(take log102 = 0.3)

37. 9For first order reaction

18

11 0

18

0

t

0

0

0

0

1

3

R2.303t log

K R

R2.303 2.303t log log8

K R / 8 K

R2.303t log 2.303log10K R /10

2.303 log8t K10 102.303t log10

Klog8a 10 10 log230log2 30 0.3 9

=

= =

= =

× = ×

= × = ×= = × =

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... 29IIT-JEE Test Prep

38. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL–1. The molecular weight of HCl is 36.5 gmol–1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

38. 8Take wt. of HCl stock solution = 100gthen wt. of HCl = 29.2% of 100g

= 29.2 g

So no. of moles of HCl = 29.2 29236.5 365

=

volume of HCl soln =mass 100 10000

density 1.25 125= =

so molarity of HCl stock solution = nv

= 80

298292 125365 1000

10000 1 365 10000125 1000

= × ××

39. The substituents R1 and R2 for nine peptides are listed in the table given below. How many of thesepeptides are positively charged at pH = 7.0?

⊕ Θ

− − − − − − − − − − −3

1 2

H N CH CO NH CH CO NH CH CO NH CH COO| | | |H R R H

Peptide R1 R2

I H H

II H CH3

III CH2COOH H

IV CH2CONH2 (CH2)4NH2

V CH2CONH2 CH2CONH2

VI (CH2)4NH2 (CH2)4NH2

VII CH2COOH CH2CONH2

VIII CH2OH (CH2)4NH2

IX (CH2)4NH2 CH3

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39. 4Either of R1 or R2, or both having amine as a functional group willsufficiently basic to accept H+ ion from neutral H2O, (amide, alco-hol or carboxilic group will not)

40. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in itspyranose form is

2

2

CHO|CH|CHOH|CHOH|CHOH|CHOH

40. 8

HO

CH2OH

OH

H

H

H

OHOH

1

23

4

5

Pyranose form of given compound

CHO

CH2

CHOH

CHOH

CHOH

CH2OH

1

2

3

4

5

6

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given compoundD-form of the compound forms pyranose, so 5 carbon will not formstereoisomer. So stereoisomers is due to carbon 1, 3 and 4.So total number of isomer = 2 × 2 × 2 = 8

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PART– III : MATHEMATICS

SECTION I: Single Correct Answer Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

41. The function [ ] [ ]→f : 0, 3 1,29 , defined by f(x) = 2x3 – 15x2 + 36x + 1, is

(A) one-one and onto. (B) onto but not one-one.(C) one-one but not onto. (D) neither one-one nor onto.

41. b( ) ( )= − + = − −

2 2f (x) 6x 30x 36 6 x 2 x 3

the function is increaing for (0,2) anddecreaing from (2,3) and has glogal maxima at x=2& global minima at x=0the function is onto but not one-one

42. The ellipse + =2 2

1x yE : 19 4

is inscribed in a rectangle R whose sides are parallel to the coordinate

axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricityof the ellipse E2 is

(A) 2

2(B)

32

(C) 12

(D) 34

42. c( )± ±

+ =

=

−∴ = =

2 2

2 2

2

The vartices of rectangle are 3, 2

x yLet equation of ellipse E be 1

16aAs the ellipse passes through the vartices of

the rectangle we get a 12

16 12 1e

16 2

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43. The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1, – 1, 4) with theplane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1, 4) to QR, thenthe length of the line segment PS is

(A) 1

2(B) 2 (C) 2 (D) 2 2

43. a

( )

( ) ( )( )

− − −= =

+

⇒ λ λ + − λ + =

−⇒ = µ + µ µ

x 2 y 3 z 5Equation of line QR is

1 4 1Let Co-ordinates of p be x 2, 4x+3,x+5

As the point lie on the plane 5x-4y-z=1

5( +2)-4 4 3 5 1

2x Let co-ordinates of 5 be 2, 4 +3, +53Direction ration of TS ( )

( ) ( ) ( )

µ µ µ

µ + µ + + µ + =

−⇒ µ =

∴ λ − µ + +

=

2 2 2

will be , 4 +2, +1

as TS + QR We get

1 4 4 2 1 1 0

12

length of line segment ps will be 1 4 1

1

2

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44. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straightline 4x – 5y = 20 to the circle x2 + y2 = 9 is(A) 20(x2 +y2) – 36x + 45y = 0 (B) 20(x2 +y2) + 36x – 45y = 0(C) 36(x2 +y2) – 20x + 45y = 0 (D) 36(x2 +y2) + 20x – 45y = 0

44. a( )

( )

+ = +

1 2

2 21 1

Let Q x ,y be a point on the line 4x-5y=20

Let P h,k be the mid point of chord of contact

of tangents drawn from Q.Equation of chord of contact is

x x yy 9 and also hx+ky=h k

Comparing the co-efficients

( )

( )

= =+ = + +

⇒ − = + +

∴ + − + =

1 12 2

1 1 2 2 2 2

2 2 2 2

2 2

we getx y 9h k h k

9h 9kQ x ,y 0 , is a point on

h k h kthe line 4x-5y=20

9h 9k4 5 20

h k h k

Locus of point `P` is 20 x y 36x 45y 0

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45. Let P = [aij] be a 3 × 3 matrix and let Q = [bij], where bij = 2i+j aij for ≤ ≤1 i, j 3. If the determinant of

P is 2, then the determinant of the matrix Q is(A) 210 (B) 211 (C) 212 (D) 213

45. d

=

= ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ =

11 12 13

21 22 23

31 32 33

2 3 411 12 13

3 4 521 22 23

4 5 631 32 33

11 12 132 3 4

21 22 23

2 2 231 32 33

2 3 4 2 13

a a a

Let P b a a a 2

a a a

2 a 2 a 2 a

Q will be 2 a 2 a 2 a

2 a 2 a 2 a

a a a

2 2 2 2a 2a 2a

2 a 2 a 2 a

2 2 2 2 2 2

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46. The integral ( )

+∫

2

9 /2sec x

dxsecx tanx

equals (for some arbitrary constant K)

(A) ( )

( ) − − + + +

211/2

1 1 1secx tanx K

11 7secx tanx

(B) ( )

( ) − + + +

211/2

1 1 1secx tanx K

11 7secx tanx

(C) ( )

( ) − + + + +

211/2

1 1 1secx tanx K

11 7secx tanx

(D) ( )

( ) + + + +

211/2

1 1 1secx tanx K

11 7secx tanx

46. c

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

+

+ −

+ +

+ +

− −

+ +

− + −

∫ ∫

92

2

2 2

9 92 2

922

92

2

7 112 2

sec x dx

sec x tanx

1 sec secx tanx 1 sec x secx tanx= dx+ dx

2 2sec x tanx secx tanx

1= secx tanx , sec x secx tanx dx

2

1+ secx tanx sec x secx tanx dx

2

secx tanx sec x tanx1=

7 1122 2

( )− = + + +

+

211

2

k

1 1 1 (secx tanx) k

11 7secx tanx

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47. The total number of ways in which 5 balls of different colours can be distributed among 3 persons sothat each person gets at least one ball is(A) 75 (B) 150 (C) 210 (D) 243

47. b

− ⋅ + ⋅

= − + =

5 3 5 3 51 2

Equal to number of noto functions from as setcontaining 5 elements to a set containing3 elements which is equal to

=3 c 2 c 1

243 96 3 150

48. If →∞

+ +− − = +

2

x

x x 1lim ax b 4,then

x 1

(A) a = 1, b = 4 (B) a = 1, b = – 4 (C) a = 2, b = – 3 (D) a = 2, b = 3

48. b( ) ( ) ( )

+ + − + − + =

+

⇒ − =∴ ⇒ ⇒

2

x 1

2

x x 1 ax x 1 b x 1lim 4

x 1

Co efficient of x 0 & co-efficient of x is 41-a=0 a=1 & 1-a-b=4 b=-4

49. Let z be a complex number such that the imaginary part of z is nonzero and z = z2 + z + 1 is real.Then a cannot take the value

(A) – 1 (B) 13 (C)

12

(D) 34

49. d

( ) ( )( ) ( )

( )

+ + + +

= − + + +

−+ = ⇒ =

∴ = −

2

2 2

2

Let Z=x+iy ; where y 0

a= x iy x iy 1

x y 1 i ; 2xy y

1as `a` is Real we get y 2x 1 0 x 23 3

a y which is always len than 4 4

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50.

π= ≠ ∈ =

2x cos ,xLetf(x) x 0, , x IR,0, x 0

then f is(A) differentiable both at x = 0 and at x = 2(B) differentiable at x = 0 but not differentiable at x = 2(C) not differentiable at x = 0 but differentiable at x = 2(D) differentiable neither at x = 0 nor at x = 2

50. b

( )

( )

π≠

= =

ππ π −π′ = + −

π

2

2

the function is continuous at x=0 but not coutinuous at x=2

x cos , x 0f x x

0 , x 0

x cos2

f x 2xcos x sin x x x xcos

xLHD and RHO at x=0 are equal at x=2 are also to opposite signs.It is differentiable at x=0 but not at x=2

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SECTION II : Multiple Correct Answer(s) Type

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

51. Let S be the area of the region enclosed by = = =2xy e , y 0,x 0, and x = 1. Then

(A) ≥ 1Se (B) ≥ − 1S 1

e

(C)

≤ +

1 1S 1

4 e(D)

≤ + −

1 1 1S 1

2 e 2

51. a,b,d− −

− −

− −

− − −

≥ ∈

⇒ ≥

≥ ∀ ∈ ∴ ≥

= +

≤ ⋅ + ≤ + −

∫ ∫

∫ ∫ ∫

∫ ∫

2

2

2 2

2 2 2

x x

1 1x x

0 0

1x x

0

11 12

x x x

10 02

112 1

2

102

e e when x 0,1

e dx e dx

1 1e x 0,1 e dx

e e

e dx e dx e dx

1 1 11 dx e dx 1

2 e 2

52. If y(x) satisfies the differential equation y’ – y tan x sec x and y(0), then

(A) π π =

2y

4 8 2(B)

π π =

2y'

4 18

(C) π π =

2y

3 9 (D) π π π = +

24 2y'

3 3 3 3

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52. a,d

∫ = =

+= ⇒ =

∴ =

+=

tanx dx ln cosx

2

2

2

dyy tanx=2x secx

dx

Integrating factor=e e cosxmultiplying by cosx on both sidescos x dy-y sinx dx = 2x dxintegrationg on both sides we get

ycosx = x cy(0) 0 c 0

xy

cosxdy 2xcosx x sdx 2

inxcos x

53. Tangents are drawn to the hyperbola − =2 2x y 1,

9 4 parallel to the straight line 2x – y = 1. The pointss

of contact of the tangents on the hyperbola are

(A)

9 1,

2 2 2(B)

9 1,

2 2 2

(C) ( )−3 3, 2 2 (D) ( )− −3 3, 2 2

53. a,b( )θ θ

θ

θ

− −∴

Let us consider a point P 3sec ,2tan on the Hyparbola

2Slope of tangent at `P` is which should

3sin1

be equal to 2. which Implies sin =3

9 1 9 1Points of contact will be , & ,

2 2 2 2 2 2

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54. Let [ ]θ ϕ ∉ π, 0 , 2 be such that

θ θ θ − ϕ θ + ϕ − 22cos (1 sin )sin tan cot cos 1

2 2

π − θ > − < θ −3

tan(2 ) 0and 1 sin .2

Then ϕ cannot satisfy

(A) π

< ϕ <02

(B) π π< ϕ < 42 3 (C)

π π< ϕ <4 33 2 (D)

π< ϕ < π

32

2

54. a,c,d

( )

( )

( )

θ θ θ − φ = θ + φ −

⇒ θ − θ φ θ φ

⇒ θ + = θ + φ

−θ θ

π π ⇒ π − < θ < π − θ + φ ∈

π π⇒ π + < + φ < π +

22cos 1 sin sin tan cot cos 12 2

2cos 2cos sin =2sin cos -1

2cos 1 2sin

3given tan <0 and -1<sin <

21

2n 2n & sin ,12 3 2

52k 0 2k

6 6

55. A ship is fitted with three engines E1, E2 and E3. The engines independently of each other with

respective probabilities 1 1 1

, and .2 4 4

For the ship to be operational at least two of its engines must

function. Let X denote the event that the ship is operational and let X1, X2 and X3 denote respectivelythe events that the engines E1, E2 and E3 are functioning. Which of the following is (are) true?

(A) = c

13P X | X

16 (B) P (Exactly two engines of the ship are functioning = 7|X]8

(C) [ ] =25P X | X

16 (D) [ ] =17P X | X

16 1

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55. b,d

( )

( )

( )

( )

= =+ + +

+ += =

+ + +

= + + + = =

= + + + =

c1

2

1

1 1 1x x 12 4 4A) P x / x

1 1 1 1 1 1 1 3 1 1 1 3 8x x x x x x x x2 4 4 2 4 4 2 4 4 2 4 4

1 1 1 1 3 1 1 1 3x x x x x x 72 4 4 2 4 4 2 4 4B) P Exactly two x

1 1 1 1 1 1 1 3 1 1 1 3 8x x x x x x x x2 4 4 2 4 4 2 4 4 2 4 4

1 1 1 1 1 3 5 5C) P X x x x

2 4 2 4 2 4 8 8

1 1 1 3 3 3 1 7D) P X x x x x

4 4 4 4 4 4 4 16

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SECTION III : Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0to 9 (both inclusive)

56. The value of

+ − − −

32

1 1 1 16 log 4 4 4 ...

3 2 3 2 3 2 3 2 is

56. 4

− − =

⇒ = −

⇒ =

∴ + = + = − =

2

3 32 2

1 1Let 4 4 4.... x

3 2 3 21

x 4 x3 2

8x

3 21 4

6 log x 6 log 6 2 493 2

57. Let S be the focus of the parabola y2 = 8x and let PQ be the common chord of the circle x2 + y2 – 2x– 4y = 0 and the given parabola. The area of the triangle PQS is

57. 4=

+ − − =

2

2 2

Solving the equation of parabola y 8x and

the circle x y 2x 4y 0, we get the pointsof contact as P(0,0) and Q(2,4).Fours of the parabola is S(2,0).

1Area of triangle PQS is x 2 x 4 = 4

2

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58. If r ra,b and

rc are unit vectors satisfying − + − + − =

r r r r r r2 2 2a b b c c a 9, then

− +r r r

2a 5b 5c is

58. 3

( )

− + − + − =

⇒ ⋅ + ⋅ + ⋅ = −

⋅ = ⋅ = ⋅ = −

− ∴ + + = + + + ⋅ + ⋅ + ⋅ =

2

2 2 2

a b b c c a 9

3a b b c c a

21

Which is possible only when a b b c c a2

12a 5b 5c 2 5 5 2 2 5 5 5 2 5 3

2

59. Let →f : I R IR be defined as = + −2f(x) x x 1. The total number of points at which f attains either

a local maximum or a local minimum is

59. 5( ) = + − = + − ≥

− + ≤ ≤

− − + − ≤ ≤

− + − ≤ −

2 2

2

2

2

f x x x 1 x x 1 ; x 1

x x 1 ; o x 1

x x 1 ; 1 x 0

x x 1 ; x 1

−− 11 O2 1 12

f(x)

x

the function has local maxima or local minima

at = ± ±1

x 0, and 12

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60. Let p(x) be a real polynomial of least degree which as a local maximum at x = 1 and a localminimum at x = 3. If p(1) = 6 and p(3) = 2 then p’(0) is

60. 9( )

( ) ( ) ( )

( )

( ) ( )( )

∴ = − −

⇒ = − + +

= = ⇒ =

∴ =

1

33

1

P x has local maximum at x=1 and local minimum at x=3

P x a x 1 x 3

xP x a 2x 3x b

3

P 1 b & P 3 2 a 3 and b=2

P 0 9