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PHYSICS TEAM

IIT-JEE 2012 PAPER-II WITH SOLUTION [CODE - 6]

Supported by a pool of 20 Faculty Members

Nitin VijayEx-Sr. Faculty Bansal Classes

B.Tech IT-BHU

Amit Verma Ex-Sr. Faculty Bansal Classes

M.Sc. Gold Medalist

Nirbhay Pandey Ex-Faculty Bansal Classes

Ravi Vijay Sr. Faculty Physics

New

Kumar Sourabh Ex-Sr. Faculty Vibrant Academy

Ex-HOD Career point

New

Bhupendra Sudan Ex-Sr. Faculty Vibrant Academy

Ex-Sr. Faculty Bansal Classes

PART - I [PHYSICS]

SECTION – A

Single Correct

1. Consider a disc rotating in the horizontal planewith a constant angular speed ω about its centreO. The disc has a shaded region on one side ofthe diameter and an unshaded region on the other

side as shown in the figure. When the disc is in

the orientation as shown, two pebbles P and Qare simultaneously projected at an angle towards

R. The velocity of projection is in the y-z planeand is same for both pebbles with respect to the

disc. Assume that (i) they land back on the disc

before the disc has completed 1/8 rotation, (ii)

their range is less than half the disc radius, and

(iii) ω remains constant throughout. Then

Q

R

P

y

x

O

(A) P lands in the shaded region and Q in the

unshaded region(B) P lands in the unshaded region and Q in the

shaded region(C) Both P and Q land in the unshaded region

(D) Both P and Q land in the shaded regionSol. C

Q

R

P

y

x

O

45°

At 45° P & Q both land in unshaded region.P & Q also have velocity Rω but since P lands inpoint of less Radius & Q lands in point with moreradius.

2. A student is performing the experiment of

Resonance Column. The diameter of the column

tube is 4 cm. The frequency of the tuning fork is512 Hz. The air temperature is 38°C in which the

speed of sound is 336 m/s. The zero of the meter

scale coincides with the top end of the

Resonance Column tube. When the first

resonance occurs, the reading of the water levelin the column is

(A) 14.0 cm (B) 15.2 cm

(C) 16.4 cm (D) 17.6 cm

Sol. B

for Ist resonanceλ/4 =x + 0.6r ....(1)

∴ v = fλ336 = 5/2 × λ ∴ λ = 336/5/2⇒ λ/4 16.4 cm16.4 = x + 0.6 × 2x = 16.4 – 1.2

= 15.2 cm3. Two identical discs of same radius R are rotating

about their axes in opposite directions with the

same constant angular speed ω. The discs are inthe same horizontal plane At time t = 0, the

points P and Q are facing each other as shownin the figure. The relative speed between the

two points P and Q is νr.In one time period (T) ofrotation of the discs, νr as a function of time isbest represented by

RP Q

R

(A)

T0 t

(B)

T0 t

(C)

T0 t

(D)

T0 t



Sol. A

V2 – V1 = VRel

∴ ( )θ

= ω

2 1

2V – V 2R sin

2

R R

v v

= 2 Rω sinθ = ω ω2R sin t

4. A loop carrying current I lies in the x-y plane as

shown in the figure. The unit vector K is coming

out of the plane of the paper. The magnetic

moment of the current loop is

ax

y

I

a

(A) 2 ˆa IK (B) 2 ˆ1 a IK

2

π +

(C) 2 ˆ– 1 a IK

2

π +

(D) ( ) 2 ˆ2 1 a IKπ +

Sol. B

A = a2 +

2a44 2

π ×

A = a2 + πa2/2= a2 (1+ π/2)

M = iA = Ia2(1 + π/2)k

5. An infinitely long hollow conducting cylinder withinner radius R/2 and outer radius R carries auniform current density along its length. The

magnitude of the magnetic field B as a function

of the radial distance r from the axis is bestrepresented by

(A)

rR/2 R

B

(B)

rR/2 R

B

(C)

rR/2 R

B

(D)

rR/2 R

B

Sol. D

XX

X

XX

X

X

X

r

R

0 inB.dl i= µ∫

B . 2πr = µ0 × J 2

2 Rr –

4

ππ

;

B =

220J Rr –

2 r 4

µ ππ

π

As r = R/2 ; there fore B = 0

6. A thin uniform cylindrical shell, closed at bothends, is partially filled with water. It is floatingvertically in water in half-submerged state. If ρcis the relative density of the material of the shell

with respect to water then the correct statementis that the shell is(A) more than half-filled if ρc is less than 0.5(B) more than half-filled if ρc is more than 1.0(C) half-filled if ρc is more than 0.5(D) less than half-filled if ρc is less than 0.5.

Sol. AFor ρc = 0.5 exactly it will be half submergedbecause upthrust of half solid portion of containerwill balance its total weight.For ρc < 0.5 more water is required and ρc > 0.5less water is required.

7. In the given circuit a charge of +80 µC is givento the upper plate of the 4µF capacitor. Then inthe steady state, the charge on the upper plateof the 3 µf capacitor is.



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(A) +32µC (B) +40µC (C) +48µC (D) +80µCSol. C

xV 0

Ceq = ( )( )3 2 4 20

93 2 4

+ ×=

+ +

v = q/c = 80

920

× = 36 Volt

Apply charge conservation

(x – 36)4 + 3 x + 2x = 0x = 16.∴ q2 = C2V2 =3 × 16 = 48 µC

8. Two moles of ideal helium gas are in a rubberballoon at 30°C. The balloon is fully expandableand can be assumed to require no energy in its

expansion. The temperature of the gas in theballoon is slowly changed to 35°C. the amountof heat required in raising the temperature isnearly (take R = 8.31 J/mol. K)(A) 62 J (B) 104J (C) 124J (D) 208J

Sol. D

Q = nCp∆T= 2 × 5/2 × R × 5= 208 J

Section B

Paragraph(9 to 10)Paragraph for Question Nos.9 to 10Most materials have the refractive index, n > 1.So, when a light ray from air enters a naturallyoccuring material, then by Snell's law,

1

2

2

1

n

n

sin

sin=

θ

θ, it is understood that the refracted

ray bends towards the normal. But it neveremerges on the same side of the normal as theincident ray. According to electromagnetism, therefractive index of the medium is given by the

relation, ,v

cn rrµε±=

= where c is the speed

of electromagnetic waves in vaccum, v its speed

in the medium, rε and rµ are the relative permit-

tivity and permeability of the medium respec-

tively. In normal materials, both rε and rµ are

positive, implying positive n for the medium.

When both rε and rµ are negative, one must

choose the negative root of n. Such negative

refractive index materials can now be artificiallyprepared and are called meta-materials. Theyexhibit significantly different optical behavior,without violating any physical laws. Since n isnegative, it results in a change in the directionof propagation of the refracted light. However,

similar to normal materials, the frequency of lightremains unchanged upon refraction even in meta-materials.

9. Choose the correct statement.(A) the speed of light in the meta-material is

ncv =

(B) The speed of light in the meta-material is

n

cv =

(C) The speed of light in the meta-material is v= c.(D) The wavelength of the light in the meta-

material ( )mλ is given by mλ = nairλ , where airλ is

the wavelength of the light in air.



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Sol. BBy definitionn = C/Vso V = C/n.

10. For light incident from air on a meta-material,

the appropriate ray diagram is

(A) (B)

(C) (D)

Sol. Cn1sinθ1 = n2 sinθ2sinθ1 = –x sinθ2 (x is R.I)θ2 must be negative.so (C)

Paragraph for Question Nos. 11 to 12The β -decay process, discovered around 1900,is basically the decay of a neutron (n). In thelaboratory, a proton (p) and an electron (e-) areobserved as the decay products of the neutron.Therefore, considering the decay of a neutronas a two-body decay process, it was predictedtheoretically that the kinetic energy of the elec-

tron should be a constant. But experimentally, itwas observed that the electron kinetic energyhas a continuous spectrum. Considering a three-

body decay process, i.e. ,vepn e++→ − around

1930, Pauli explained the observed electron en-

ergy spectrum. Assuming the anti-neutrino ( )ev to

be massless and possessing negligible energy,and the neutron to be at rest, momentum and

energy conservation principles are applied. Fromthis calculation, the maximum kinetic energy ofthe electron is 0.8 x 106 eV. The kinetic energycarried by the proton is only the recoil energy.

11. What is the maximum energy of the anti-neutrino?(A) Zero(B) Much less than 0.8 x 106 eV.(C) Nearly 0.8 x 106 eV.(D) Much larger than 0.8 x 106 eV.

Sol. C

12. If the anti-neutrino had a mass of 3 eV/c2 (wherec is the speed of light) instead of zero mass,what should be the range of the kinetic energy,K, of the electron?

(A) eV10x8.0K0 6≤≤

(B) eV10x8.0KeV0.3 6≤≤

(C) eV10x8.0KeV0.3 6<≤

(D) eV10x8.0K0 6<≤

Sol. D

Paragraph for Question Nos. 13 to 14The general motion of a rigid body can beconsidered to be a combination of (i) a motionof its centre of mass about an axis, and (ii) itsmotion about an instantaneous axis passing

through the centre of mass. These axes neednot be stationary. Consider, for example, a thinuniform disc welded (rigidly fixed) horizontally atits rim to a massless stick, as shown in the figure.When the disc-stick system is rotated about theorigin on a horizontal frictionless plane with

angular speed ω , the motion at any instant canbe taken as a combination of (i) a rotation ofthe centre of mass of the disc about the z-axis,and (ii) a rotation of the disc through aninstantaneous vertical axis passing through itscentre of mass (as is seen from the changed

orientation of points P and Q). Both these motionshave the same angular speed ω in this case.

Now consider two similar systems as shown inthe figure: Case (a) the disc with its face verticaland parallel to x-z plane; case (b) the disc withits face making an angle of 45o with x-y planeand its horizontal diameter parallel to x-axis. Inboth the cases, the disc is welded at point P,

and the systems are rotated with constantangular speed ω about the z-axis.









th

REGISTRAT

ION

OPEN


Case (a) Case (b)13. Which of the following statements about the in-

stantaneous axis (passing through the centre ofmass) is correct ?(A) It is vertical for both the cases (a) and (b).

(B) It is vertical for case (a); and is at 45o tothe x-z plane and lies in the plane of the disc forcase (b).(C) It is horizontal for case (a); and is at 45o tothe x-z plane and is normal to the plane of thedisc for case (b).

(D) It is vertical for case (a); and is at 45o tothe x-z plane and is normal to the plane of thedisc for case (b).

Sol. AConsider case (a)

Q

P

A (out of paper)

(insidepaper)

B

at t = 0

AB

at t = T/4

at t = T/2 at t = 3T/4

Q

P

A(inside)(outside) B

A B

Hence axis is vertical.For case (b)

14. Which of the following statements regarding theangular speed about the instantaneous axis(passing through the centre of mass) is correct

(A) It is ω2 for both the cases.

(B) It is ω for case (a); and 2

ω for case (b).

(C) It is ω for case (a); and ω2 for case (b).

(D) It is ω for both the cases.Sol. D

Given ω is same.

Section (C)

Multiple Correct15. The figure shows a system consisting of (i) a

ring of outer radius 3R rolling clockwise withoutslipping on a horizontal surface with angular

speed ω and (ii) an inner disc of radius 2R rotatinganti-clockwise with angular speed ω/ 2. The ringand disc are separated by frictionless ballbearings. The system is in the x-z plane. Thepoint P on the inner disc is at a distance R fromthe origin, where OP makes an angle of 30o with

the horizontal. Then with respect to thehorizontal surface.

(A) the point O has a linear velocity iR3 ω .

(B) the point P has a linear velocity 11 3 ˆˆR i R k4 4

ω + ω

(C) the point P has a linear velocity kR4

3iR

4

13ω−ω

(D) the point P has a l inear velocity

kR4

1iR

4

33 ω+ω

−

Sol. A,BFor point P



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30°

P

O

Vp = 3 Rω i – Rω/4 i +3 ˆR k4

ω

= 11/4 Rω i + 3 ˆR k4

ω

16. In the given circuit, the AC source has ω = 100rad/s. Considering the inductor and capacitor tobe ideal, the correct choice(s) is (are)

(A) The current through the circuit, I is 0.3 A.

(B) The current thorgh the circuit, I is 0.3 2 A.

(C) The voltage across 100Ω resistor = 10 2 V..

(D) The voltage across 50Ω resistor = 10 V.Sol. A,C

25020V2100

I1

I2

XL = ω L = 50

XC = c

1

ω= 100

2100

20

R

VI1 == at 45° with voltage.

250

20I2 = =

25

2 at 45° with voltage.

I1

I2

45°

45°

Net current

22

21 III ++= = 0.3 A approx.

17. A current carrying infinitely long wire is kept alongthe diameter of a circular wire loop, withouttouching it. The correct statement(s) is(are)

(A) The emf induced in the loop is zero if thecurrent is constant.(B) The emf induced in the loop is finite if thecurrent is constant.(C) The emf induced in the loop is zero if thecurrent decreases at a steady rate.

(D) The emf induced in the loop is finite if thecurrent decreases at a steady rate.

Sol. A,CAs current inside = 0so Magnetic flux = 0 in all cases.

18. Six point charges are kept at the vertices of aregular hexagon of side L. and centtre O, as

shown in the figure. Given that 20 L

q

4

1K

πε= ,

which of the following statement(s) is(are)

correct? L(A) The electric field at O is 6K along OD.

(B) The potential at O is zero.(C) The potential at all points on the line PR issame.(D) The potential at all points on the line ST issame.

Sol. A,B,C



LCan be simplified to2q

2q 60°60°4q = 6q E

2L

)q6(kE = = 2

0

1 6q

4 L×

πε = 6 K

At O there are equidistant +ve & – ve charges.Hence v = 0. At PR also +ve & – ve charges aresame and equidistant, so V = 0 at all points. Butat ST potential is +ve on left of O & – ve onright of O.

19. Two solid cylinders P and Q of same mass andsame radius start rolling down a fixed inclinedplane from the same height at the same time.Cylinder P has most of its mass concentratednear its surface, while Q has most of its massconcentrated near the axis. Which statement(s)

is(are) correct?(A) Both cylinders P and Q reach the ground atthe same time.

(B) Cylinder P has larger linear acceleration thancylinder Q.

(C) Both cylinders reach the ground with sametranslational kinetic energy.(D) Cylinder Q reaches the ground with largerangular speed.

Sol. DIP > IQ

a = 2

gsin

1 I /MR

θ

+

Hence ap < a0tp > tQVp < vQAnd as ω = v/RSo ωP < ωQ

20. Two spherical planets P and Q have the same

uniform density ρ , masses Mp and MQ, and

surface areas A and 4A, respectively. A spherical

planet R also has uniform density ρ and its mass

is (MP + MQ). The escape velocities from theplanets P, Q and R, are Vp, VQ and VR, respectively.Then(A) VQ > VR > VP (B) VR > VQ > VP

(C) VR / VP = 3 (D) 2

1V/V QP =

Sol. B,D

r

P

2r

Q

r0

R

MP = M MQ = 8M MR = 9MRadius of R will be slightly larger than 2r,

2GMV 2gR R

R= = ∝

Hence VR > VQ > VP

Also 2

1

V

V

Q

P = .

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R.R. Dwivedi Sr. Faculty Chemistry

B.Tech. IT-BHUExp. 12 yrs

V.P. Singh Ex-HOD Chemistry Bansal Classes

Exp. 12 yrs

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IT-BHU,Exp. 8 yrs

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Exp. 5 yrs

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CHEMISTRY TEAMSupported by a pool of 28 Faculty Members

PART - II [CHEMISTRY]

SECTION – A

SINGLE CORRECT

21. The compound that undergoes decarboxylation mostreadily under mild condition in

(A)

COOH

CH COOH2

(B)

COOH

O

(C)

COOH

COOH(D)

CH COOH2

O

Sol. B

C

O

O O– CO2

OH

β - keto acid

22. Using the data provided, calculate the multiple bondenergy (kJ mol–1) of a C C bond in C2H2. That energyis (take the bond energy of a C–H bond as 350 kJ mol–1)2C(s) + H2(g) C2H2(g) ∆H = 225 kJ mol

–1

2C(s) 2C(g) ∆H = 1410 kJ mol–1

H2(g) 2H(g) ∆H = 330 kJ mol–1

(A) 1165 (B) 837 (C) 865 (D) 815Sol. D

Given,C2H2(g) 2C(s) + H2(g) ∆H = –225 kJ mol

–1 ...(i)

2C(s) 2C (g) ∆H = 1420 kJ mol–1 ...(ii)H2(g) 2H(g) ∆H = 330 kJ mol–1 .(iii)Adding C2H2 (g) → 2C (g) + 2 H(g) →

∆H = 1525 kJ mol–1

Σ(C C) + 2Σ (C – H) = 1515

⇒ Σ(C C) + 1515 – 2 × 350 = 815

23. The major product H of the given reaction sequenceis

CH –CH –CO–CH3 2 3

CNG

95%H SO2 4H

(A) CH –CH=C–COOH3

CH3

(B) CH –CH=C–CN3

CH3

(C) CH –CH =C–COOH3 2

CH3

OH

(D) CH –CH=C–CO–NH3 2

CH3

Sol. A

CH – CH – C – CH3 2 3 CH – CH – C – CH3 2 3

CH – CH – C – CH3 2 3CH – CH = C 3

OCN

O

CN

CN

H

OH

CH3

95% H SO2 4

COOH

Heat

24. NiCl2P(C2H5)2(C6H5)2 exhibits temperaturedependent magnetic behaviour (paramagnetic /

diamagnetic). The coordination geometries of Ni2+ inthe paramagnetic and diamagnetic states arerespectively.

(A) tetrahedral and tetrahedral(B) square planar and square planar(C) tetrahedral and square planar

(D) square planar and tetrahedralSol. [NiCl2 P(C2H5)2 (C6H5)2]

4s 4p

4s 4p

4sp3

–––––––

–– ––

4dsp2––––

weakfield

strongfield

Diamagnetic

–

3d

3d

25. In the cyanide extraction process of silver from

argentite ore, the oxidizing and reducing agent usedare(A) O2 and CO respectively(B) O2 and Zn dust respectively(C) HNO3 and Zn dust respectively(D) HNO3 and CO respectively

Sol. B

Ag2S + NaCN Na[Ag(CN)2] + Na2S

Na2S + H2O + O2 Na2SO4 + NaOH + S826. The reaction of white phosphorus with aqueous NaOHgives phosphine along with another phosphoruscontaining compound. The reaction type; the oxidationstates of phosphorus phosphine and the other product

are respectively



(A) redox reaction ; –3 and –5(B) redox reaction ; + 3 and +5(C) disproportionation reaction ; –3 and +5(D) disproportionation reaction ; –3 and +3

Sol. P4 + NaOH + H2O PH3 + NaH2PO2 + H2OWrong option, correct option is (–3, +1)

27. The shape of XeO2F2 molecule is(A) trigonal bipyramidal

(B) square planar(C) tetrahedral (D) see-saw

Sol. D

XeO2F2 Xe

O

O

F

F:

28. For a dilute solution containing 2.5 g of a non-voltile non-electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2ºC.

Assuming concentration of solute is much lower thanthe concentration of solvent, the vapour pressure (mmof Hg) of the solution is (take Kb = 0.76 K kg mol

–1)(A) 724 (B) 740 (C) 736 (D) 718

Sol. A

∆Tb = 2 = Kf × molality = 0.76 × m

m = 76200

Also,0 s A

0 B

P P x 200 18 36

P x 76 1000 760

−= = × =

1 – 76036

P

P

0

s =

76036

1P

P

0

s −= ⇒ PS = 724 As, (P0 = 760)

ParagraphParagraph for Question Nos. 29 and 30

Bleaching powder and bleach solution are produced ona large scale and used in several house hold products.The effectiveness of bleach solution is often measuredby iodometry.

29. 25 mL of household bleach solution was mixed with30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the

titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3was used to reach the end point. The molarity of thehousehold bleach solution is(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 MSol. C

CaO Cl2 + 2CH3COOH Ca(CH3COO)2 + H2O + Cl2Cl2 + 2KI 2KCl + I2I2 + 2Na2 S2O3 2NaI + Na2S4O6

mol of Cl2 × 2 = 100025.048 ×

mol of Cl2 = 2100025.048

××

= 1000

M25 × ⇒ M = 0.24

30. Bleaching powder contains a salt of an oxoacid as

one of its components. The anhydride of that oxocacidis

(A) Cl2O (B) Cl2O7 (C) ClO2 (D) Cl2O6

Sol. A

Oxidation state of Chlorine in ClO– = + 1Hence option is Cl2O

(As it has + 1 oxidation state)

Paragraph for Question Nos. 31 and 32The electrochemical cell shown below is a concentrationcell.M | M2+ (saturated solution of a sparingly soluble salt,Mx2) || M

2+ (0.001 mol dm–3) | M The emf of the cell

depends on the difference in concentrations of M2+ ionsat the two electrodes. The emf of the cell at 298 K is0.059 V.

31. The solubility product (Ksp; mol3 dm–9) of MX2 at

298 K based on the information available for the givenconcentration cell is (take 2.303 × R × 298 /F = 0.059 V)

(A) 1 × 10–15 (B) 4 × 10–15

(C) 1 × 10–12 (D) 4 × 10–12

Sol. B

Ecell = 1

2

C

Clog

2059.0

0.059 = 1C001.0

log2059.0

log 1C001.0

= 102⇒ 1C001.0

= 2

C1 = 10–5

Also, MX2 M+2 + 2X–

s sKsp = s × (2s)

2 = 4s3

∴ 4 × (10–5)3 = 4 × 10–15

32. The value of ∆G (kJ mol–1) for the given cell is (take1 F = 96500 C mol–1)

(A) –5.7 (B) 5.7(C) 11.4 (D) –11.4

Sol. D

∆G = –nF Ecell= –2 × 96500 × 0.059 Joule = –11.4 k J mol–1









th

REGISTRAT

ION

OPEN


SECTION – AParagraph

Paragraph for Question Nos. 33 to 34

In the following reaction sequence, the com-

pound J is an intermediate.

I (CH CO) O3 2

CH COONa3 J

(i) H , Pd/C2

(ii) SOCl2(iii) anhyd. AlCl3

K

J (C9H8O2) gives effervescence on treatment

with NaHCO3 and a positive Baeyer's test.

33. The compound I is

(A)

HO

(B)

OH

(C)

CH3O

(D)

HH

H

Sol. A

C CH=CH –C–OH2

HO

(CH CO) O3 2 H Pd/C2

SOCl2

AlCl3(anhydrous)

Intra molecularFriedel craft acylation

Perkin reaction

(Cinnamic acid)

(J)(I) CH –C–ONa3

O

O

(K)

CH CH –C–OH2 2–

O

CH CH –C–2 2– Cl

O

=O

34. The compound K is

(A) (B) O O

(C)

O

(D) O

Sol. C

Multiple Correct35. Which of the given statement(s) about N, O, P

and Q with respect to M is (are) correct?

HO HO HH HOH

Cl CH3

Cl

CH3 Cl CH3

H OH HHO H HO

M N O

OH

HHO

Cl

H

CH3

H

HHO

Cl

HO

CH3

P Q(A) M and N are non-mirror image stereoisomers(B) M and O are identical(C) M and P are enantiomers(D) M and Q are identical

Sol. ABC

HO

HO

H

H

Cl

CH3

R

S

(M)

HO

H

H

OH

CH3

Cl

R

R

(N)

HO

HO

H

H

Cl

CH3

R

S

(O)

H

H

OH

OH

Cl

CH3

S

R

(P)

HO

H

H

OH

Cl

CH3

R

R

(Q)

36. The reversible expansion of an ideal gas underadiabatic and isothermal conditions is shown in

the figure. Which of the following statement(s)is are correct?

P

V

(P ,V T )1 1 1,isothermal

adiabatic

(P ,V ,T )2 2 2

(P ,V ,T )3 2 3

(A) T1 = T2 (B) T3 > T1(C) wisothemal > wadiabtic

(D) ∆Uisothermal > ∆UadiabaticSol. ACD

T1 = T2 (for isothermal)Wiso = area under the curve∴ Wiso > Wadia

Also,

∆Uisothermal = nCv ∆T =0(∆U)Adiabatic = nCv(T3–T1) = negative as T3<T1∴ (∆U)isothermal > (∆U)Adiabatic







37. The given graphs / data I,II,III and IV representgeneral trends observed for different

physisorption and chemisorption processes undermild conditions of temperature and pressure.Which of the following choice (s) about I, II, IIIand IV is (are) correct ?

Amount of gas adsorbed

T

I P constantAmount of gas adsorbed

T

II P constant

Amount of gas adsorbed

T

III

250 K

200 K

Potential Energy

IV

Distance of molecule from the surface

H =150 kJ molads

-1

Eads

0

(A) I is physisorption and II is chemisorption(B) I is physisorption and III is chemisorption(C) IV is chemisorption and II is chemisorption(D) IV is chemisorption and III is chemisorption

Sol. AC

38. For the given aqueous reaction, which of thestatement(s) is (are) true ?

excess KI + K3[Fe(CN)6] dil.H SO2 4 brownish-yellow solution

ZnSO4

white precipitate + brownish-yellowfiltrate

Na S O2 2 3

colourless solution(A) The first reaction is a redox reaction(B) White precipitate is Zn3[Fe(CN)6]2(C) Addition of filtrate to starch solution gives blue colour.(D) White precipitate is solution in NaOH solution.Sol. ACD

KI + K3[Fe(CN)6]dil. H SO2 4 I2 + K4[Fe(CN)6]

(excess)

I2 + KI KI3 (brown -yellow sol)'

Zn2+ + K4[Fe(CN)6] → K2Zn3[Fe(CN)6]2 + K+

(white ppt) orZn2[Fe(CN)6] + K

+

(white ppt)

I2 + Na2S2O3 → NaI + Na2S4O6

K2Zn3[Fe(CN)6]2 + NaOH → K+ + [Zn(OH)4]2–

(white ppt) + [Fe(CN)6]4–

39. With respect to graphite and diamond, which ofthe statement(s) given below is (are) correct?

(A) Graphite is harder than diamond(B) Graphite has higher electrical conductivity than

diamond.(C) Graphite has higher thermal conductivity than

diamond.(D) Graphite has higher C – C bond order than

diamond.Sol. BD

40. With reference to the scheme given, which ofthe given statement(s) about T, U, V and W isare correct?

O

O

H3C

U WV

LiAlH4excess

(CH CO) O3 2

CrO /H3

T

(A) T is soluble in hot aqueous NaOH

(B) U is optically active(C) Molecular formula of W is C10H18O4

(D) V gives effervescence on treatment with aqueousNaHCO3

Sol. ACD O

O

H3C

(U)

(T)

(V)

(W)

LiAlH4

excess(CH CO) O3 2

CrO /H3

CH2–OHCOOH

CH2 3–O– C – CH

OH

COOHH3CH3C

H3C

O

O– C – CH3

O



PART - III [MATHEMATICS]

Prakash Gupta Ex-Sr. Faculty Bansal Classes

B.Tech. IIT-BombayExp. 7 yrs

Arjun Gupta Ex-Sr. Faculty Bansal Classes

B.Tech Exp. 5 yrs

Mahe Alam Sr. Faculty Mathematics

B.Tech, (Hons.), Exp. 5 yrs

Arvind Singh Ex-Sr. Faculty Vibrant AcademyEx-Sr. Faculty Bansal Classes

Exp. 9 yrs

Amit Gupta Ex-Sr. Faculty Bansal Classes

Ex-HOD Career Point Exp. 10 yrs

Suneet VijaySr. Faculty Mathematics

Exp. 4 yrs

New

New

MATHEMATICS TEAMSupported by a pool of 23 Faculty Members

SECTION – A

Single Correct

41. Four fair dice D1, D2, D3 and D4, each having sixfaces numbered 1, 2, 3, 4, 5 and 6, are rolledsimultaneously. The probability that D4 shows anumber appearing on one of D1, D2 and D3 is

(A) 216

91(B)

216

108(C)

216

125(D)

216

127

Sol. A

1D

2D

3D

4D

3A + 2A, D + 3D

6.6.6.6

362026156 ××+××+=

6.6.6.6

3601806 ++=216

91

42. The value of the integral

∫π

π−

−π

+π+

2/

2/

2

x

xnx cos x dx is

(A) 0 (B) 42

2

−π

(C) 42

2

+π

(D) 2

2π

Sol. B

I= dxxcosx

2/

2/eveneven

2∫π

π− ↓↓ + dxxcosx

x

4n

2/

2/even

odd

2∫π

π− ↓↓

−π

+π

= 2 ∫π

+2/

0

2 0dxxcosx

= 2 [ ] 2/

02 xsin2xcosx2xsinx

π−+

)x(sinxdxxcosx 22 =∫– 2x(–cos x)

= 2

×−

π12

4

2

+ 2(– sin x) = 42

2

−π

43. Let a1, a2, a3, ..... be in harmonic progressionwith a1 = 5 and a20 = 25. The least positiveinteger n for which an < 0 is [Class Illustration](A) 22 (B) 23 (C) 24 (D) 25

Sol. D

a1 = 5, a20 = 25

t1 = 5

1, t20 = 25

1 = t1 + 19.d ⇒

25

1=5

1 + 19.d

⇒– 19.25

4 = d ⇒

5

1–

19.25

4)1n( − < 0 ⇒

19.25

4)1n( − >

5

1

⇒ (n – 1) > 4

5.19n > 24.7

44. The equation of a plane passing through the lineof intersection of the planes x + 2y + 3z = 2 and

x – y + z = 3 and at a distance 3

2 from the

point (3, 1, –1) is [Class Illustration]

(A) 5x – 11y + z = 17 (B) 2 x + y = 3 2 – 1

(C) x + y + z = 3 (D) x – 2 y = 1 – 2

Sol. A

(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0(1 + λ)x + (2 – λ) y + (λ + 3)z – (2 + 3λ) = 0

⇒ 222 )3()2()1(

|)32()3(1)2(3.)1(|

+λ+λ−+λ+

λ+−+λ−λ−+λ+ =

3

2

⇒ 1443 2 +λ+λ = 3

2 ⇒

1443

|2|

2 +λ+λ

λ−=

3

2

= 3λ2 + 4λ + 14 ⇒ λ = –7/2(x + 2y + 3z – z) – 7/2 (x – y + z – 3) = 0–5x + 11y – z + 17 = 05x – 11y + z = 17

45. Let PQR be a triangle of area ∆ with a = 2, b =

2

7 and c =

2

5, where a, b and c are the lengths

of the sides of the triangle opposite to the angles

at P, Q and R respectively. Then P2sinPsin2

P2sinPsin2

+

−

equals

(A) ∆4

3(B)

∆4

45(C)

2

4

3

∆(D)

2

4

45

∆



Sol. C

a = 2, b = 2

7, c =

2

5

P2sinPsin2

P2sinPsin2

+

− 5/2

P

Q R2

7/2

Pcos1

Pcos1

+

− = tan2

2

P

=

2

)as(s

−

∆ =

2)cs()bs(

∆

−−

=

2

.2

1.

2

3

∆ =

2

4

3

∆

46. If a and b

are vectors such that |a

+ b|= 29

and )k4j3i2(a ++×

= b)k4j3i2(

×++ then a

possible value of )ba(

+ . )k3j2i7( ++− is

(A) 0 (B) 3 (C) 4 (D) 8

Sol. C

|a + b|= 29

a × (2, 3, 4) – (2, 3, 4) × b

= 0

a × (2, 3, 4) + b

× (2, 3, 4) = 0

)ba(

+ × (2, 3, 4) = 0

ba

+ = λ(2, 3, 4)

| ba

+ | = 29

| ba

+ |2 = 29

λ2 (4 + 9 + 16) = 29λ2 = 1 ⇒ λ = ±1

( ba

+ ) . (–7, 2, 3) = λ(2, 3, 4) . (–7, 2, 3)

= λ[–14 + 6 + 12]= λ[4]= 4, –4

47. If P is a 3 × 3 matrix such that PT = 2P + I,where PT is the transpose of P and I is the 3 × 3identity matrix, then there exists a column matrix

X =

z

y

x

≠

0

0

0

such that

(A) PX =

0

0

0

(B) PX = X

(C) PX = 2X (D) PX = – XSol. D

Take transpose of the given equation (PT)T = (2P + I)T

P = 2PT + I

P = 2(2P + I) + I ; P = 4P + 3I P = – I ; Px = – x

48. Let α(a) and β(a) be the roots of the equation

( ) ( ) ( ) 01a1x1a1x1a1 623 =−++−++−+

where a >–1 Then line +→0aLim

α(a) and

+→0aLim

β(a) are

(A) –2

5 and 1 (B) –

2

1 and –1

(C) –2

7 and 2 (D) –

2

9 and 3

Sol. B

using product, a 0 a 0 a 0lim . lim lim

+ + +→ → →α β = αβ

6

3a 0

1 a 1lim

1 a 1+→

+ −=

+ −

3

3 3a 0

1 a 1 1lim

2( 1 a 1)( 1 a 1)+→

+ −= =

+ − + +

Paragraph Paragraph for Question Nos. 49 to 50

Let an denote the number of all n-digit positive

integers formed by the digits 0, 1 or both such

that no consecutive digits in them are 0. Letbn = the number of such n-digit integers ending

with digit 1 and cn = the number of such n-digit

integers ending with digit 0.

49. Which of the following is correct ?(A) a17 = a16 + a15 (B) c17 ≠ c16 + c15(C) b17 ≠ b16 + c16 (D) a17 = c17 + b16

Sol. A

1 + 1 + 15C1 + 8C7 +

8C8 = 9C8 +

16C1 + 1

50. The value of b6 is

(A) 7 (B) 8 (C) 9 (D) 11

Sol. B

1 + 4C1 + 3 = 8









th

REGISTRAT

ION

OPEN


Paragraph for Question Nos. 51 to 52Let f(x) = (1 – x)2 sin2x + x2 for all x ∈ IR, and

let g(x)= ∫

−

+

−x

1

tn1t

)1t(2 f(t)dt for all x∈(1, ∞)

51. Which of the folloiwng is true ?(A) g is increasing on (1, ∞)(B) g is decreasing on (1, ∞)(C) g is increasing on (1, 2) and decreasing on (2, ∞)(D) g is decreasing on (1, 2) and increasing on (2, ∞)

Sol. B

f(x) = (1-x)2 sin2x + x2

2(x 1)g'(x) nx f(x)

x 1

− = − +

42 nx

x 1

= − − +

(f(x) is positive)

Let 4

h 2 nxx 1

= − −+

h'(x)= 2

1 4

x (x 1)

−+

+

2

2

(x 1)

x(x 1)

− −=

+ (always negative)

at x = 1 ; h = 0 so always negative

⇒ g'(x) always negative

52. Consider the statements :P : There exists some x ∈ IR such that f(x) + 2x = 2(1 + x2)

Q : There exists some x ∈ IR such that 2f(x) + 1 = 2x(1 + x)

Then(A) both P and Q are true

(B) P is true and Q is false

(C) P is false and Q is true(D) both P and Q are false

Sol. C

f(x) = 2(1 + x2) – 2x = 2(1 + x2 –x)

(1 – x)2 sin2x + x2 = 2 + 2x2 – 2x

– (1 – x)2 cos2x = 1, Not possible so P is falsefor Q, (1 – x)2 sin2x + x2 = x + x2 – 1/2

(1 – x)2 sin2x = x – 1/2 obvious x > 1/2(1 – x)2 sin2x +1/2 = x

Let y = (1 – x)2 sin2x – x + 1/2

at x = 1/2 ⇒ y = 21 1

sin4 2

(positive)

at x = 1 ⇒ y = 1

2− (negative) so Q is true

Paragraph for Question Nos. 53 to 54A tangent PT is drawn to the circle x2 + y2 = 4

at the point P ( )1,3 . A straight line L,

perpendicular to PT is a tangent to the circle(x – 3)2 + y2 = 1.

53. A possible equation of L is

(A) x – 3 y = 1 (B) x + 3 y = 1

(C) x – 3 y = –1 (D) x + 3 y = 5

Sol. A

y = (1/ 3 ) (x – 3) ± 3/11 +

54. A common tangent of the two circles is

(A) x = 4 (B) y = 2

(C) x + 3 y = 4 (D) x + 2 2 y = 6

Sol. D

(0,0)

(3,0)

(6,0)

P

(6, 0) satisfies so only option D is correct

Multiple Correct55. Let f : (–1, 1) → IR be such that

f(cos 4θ) = θ− 2sec2

2 for θ ∈

π

4,0 ∪

ππ

2,

4.

Then the value(s) for f

3

1 is (are)

(A) 1 – 2

3(B) 1 +

2

3

(C) 1 – 3

2(D) 1 +

3

2

Sol. A,B2

2

2cos 1 cos2f(cos4 )

cos22cos 1

θ + θθ = =

θθ −

⇒2 1

2cos 2 13

θ − = 2 2 2

cos 2 cos23 3

⇒ θ = ⇒ θ = ±

1 3f 13 2

= ±







56. For every integer n, let an and bn be real numbers.

Let function f : IR → IR be given by

f(x) =

−∈π+

+∈π+

)n2,1n2(xfor,xcosb

]1n2,n2[xfor,xsina

n

n,

for all integers n.

If f is continuous, then which of the folloiwnghold(s) for all n ?

(A) an–1 – bn–1 = 0 (B) an – bn = 1(C) an – bn+1 = 1 (D) an–1 – bn = –1

Sol. B, D

at x = 2n ; LHL = bn + 1 & RHL = an ⇒ an–bn= 1

at x = 2n – 1 ; LHL = an–1 & RHL = bn – 1

⇒ an–bn= –1

at x = 2n+1 ; LHL = an & RHL = bn+1 –1

⇒ an–bn+1 = –1

57. If f(x)= ∫ −−x

0

t dt)3t()2t(e2

for all x ∈ (0, ∞),

then(A) f has a local maximum at x = 2

(B) f is decreasing on (2, 3)

(C) there exists some c ∈ (0, ∞) such that f''(c) = 0(D) f has a local minimum at x = 3

Sol. ABCD

2xf '(x) e (x 2).(x 3)= − −

Increasing in (0,2) ∪ (3,∞ ), Decreasing in (2,3)Maximum at 2 & minimum at 3

& f''(x) = 2 2x 2 xe (2x 3) (x 5x 6) 2x e− + − + ⋅ ⋅ =0

3 22x 5 2x 10x 12x 0⇒ − + − + =

3 22x 10x 14x 5⇒ − + − = 0

so there exists some c ∈(0,∞ ) such that f''(c) = 0

58. If the straight lines 2

1x − =

k

1y + =

2

z and

5

1x +

= 2

1y + =

k

z are coplanar, then the plane(s)

containing these two lines is (are)

[FRM Paragraph 495-497]

(A) y + 2z = –1 (B) y + z = –1(C) y – z = –1 (D) y – 2z = –1

Sol. B, C

2 0 0

2 k 2 0

5 2 k

= ⇒ k = ± 2

use the value of k for finding the equation of

planes

59. Let X and Y be two events such that P(X|Y)=2

1.

P(Y|X) = 3

1 and P(X ∩ Y) =

6

1. Which of the

following is (are) correct ? [Class Illustration]

(A) P(X ∪ Y) = 3

2

(B) X and Y are independent

(C) X and Y are not independent

(D) P(XC ∩ Y) = 3

1

Sol. A, B

P(X ∩ Y) = X

P .P(Y)Y

, 1 1 1

P(Y) P(Y)6 2 3

= × ⇒ =

YP(X Y) P .P(X)

X

∩ =

, 1 1 1

.P(X) P(X)6 3 2

= ⇒ =

P(X Y) P(X) P(Y) P(X Y)∪ = + − ∩ = 1 1 1

2 3 6+ −

(A) 2

P(X Y)3

∪ =

(B) 1 1 1

P(X).P(Y) . P(X Y)2 3 6

= = = ∩ independent

(D) c c 1 1 1

P(X Y) P(X ).P(Y) .2 3 6

∩ = = =

60. If the adjoint of a 3 × 3 matrix P is

311

712

441

,

then the possible value(s) of the determinant of

P is (are) [Class Illustration]

(A) –2 (B) –1 (C) 1 (D) 2

Sol. A, D

|adj(P)| =

1 4 4

2 1 7

1 1 3 = 4

A.adj(A) = |A|.I3 , |A|.|adjA|=|A|n

so |adjA| = |A|n-1 ⇒ |A|

n-1 = 4 & n = 3 ⇒ |A| = ±2

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