Solutions to IIT-JEE 2008...Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (6)) 7. Let two non-collinear unit

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124(1)

(Division of Aakash Educational Services Ltd.)

Solutions to IIT-JEE 2008

PAPER - II (Code - 0)Time : 3 hrs. Max. Marks: 243

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-45543147/148 Fax : 011-25084124


Instructions :1. The question paper consists of 3 parts (Part I : Mathematics, Part II : Physics, Part III : Chemistry). Each

part has 4 sections.

2. Section I contains 9 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out ofwhich only one is correct.

3. Section II contains 4 questions. Each question contains STATEMENT-1 and STATEMENT-2.Bubble (A) if both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1Bubble (B) if both the statement are TURE AND STATEMENTS-2 is NOT the correct explanation of

STATEMENT-1.Bubble (C) if STATEMENT-1 is TRUE and STATEMENT-2 is FALSE.Bubble (D) if STATEMENT-1 is FALSE and STATEMENT-2 is TRUE.

4. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 3 choices (A), (B), (C) and (D) out of which only one is correct.

5. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements in thefirst column have to be matched with statements in the second column. The answers to these questionshave to be appropriately bubbled in the ORS as per the instructions given at the beginning of the section.

6. For each question in Section I, you will be awarded 3 marks if you have darkened only the bubblecorresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one(–1) mark will be awarded.

7. For each question in Section II, you will be awarded 3 marks if you darkened only the bubblecorresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one(–1) mark will be awarded.

8. For each question in Section III, you will be awarded 4 marks if you darken only the buble correspondingto the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (–1) markwill be awarded.

9. For each question in Section IV, you will be awarded 6 marks if you darken All the bubbles correspondingONLY to the correct answer or awarded 1 mark each for correct bubbling of answer in any row. Nonegative mark will be awarded for an incorrectly bubbled answer.



SECTION - IStraight Objective Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLYONE is correct.

1. A particle P starts from the point iz 210 += , where 1−=i . It moves first horizontally away from origin by

5 units and then vertically away from origin by 3 units to reach a point z1. From z1 the particle moves 2

units in the direction of the vector ji ˆˆ + and then it moves through an angle 2π

in anticlockwise direction on

a circle with centre at origin, to reach a point z2. The point z2 is given by(A) 6 + 7i (B) –7 + 6i (C) 7 + 6i (D) – 6 + 7i

Answer (D)

Hints :

(6, 2)

Z′2Z2

Imaginary axis

Real axis

(7, 6)

Z 0 (1,

2)

90°

11

3

5

iZ 67)6,7()45sin25,45cos26(2 +==°+°×+=′′

by rotation about (0, 0)

)( 222

2

2

2ππ

′=⇒=′

iieZZe

ZZ

iiiiiZ 76))(67(2

sin2

cos)67(2 +−=+=⎟⎠

⎞⎜⎝

⎛ π+

π+=

2. Let the function ⎟⎠

⎞⎜⎝

⎛ ππ−→∞−∞

2,

2),(:g be given by

2)(tan2)( 1 π−= − ueug . Then, g is

(A) Even and is strictly increasing in ),0( ∞

(B) Odd and is strictly decreasing in ),( ∞−∞

(C) Odd and is strictly increasing in ),( ∞−∞

(D) Neither even nor odd, but is strictly increasing in ),( ∞−∞

Answer (C)

Hints : 2

)(tan2)( 1 π−= − ueug for u ∈ (–∞, ∞)

2)(tan2)( 1 π−=− −− ueug

MATHEMATICSPART-I



2))((cot2 1 π−= − ue

2))(tan

22 1 π

−⎟⎠

⎞⎜⎝

⎛ −π

= − ue

2)(tan2 1 π−−π= − ue

)()(tan22

1 ugeu −=−π

= −

g(–u) = –g(u) ⇒ g(u) is odd function.3. Consider a branch of the hyperbola

0624222 22 =−−−− yxyx

with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbolanearest to the point A, then the area of the triangle ABC is

(A) 321− (B) 1

23− (C) 3

21+ (D) 123+

Answer (B)

Hints : 12

)2(4

)2( 22=

+−

− yx

for A(x, y)

23

421 =+=e

22 =−x ⇒ 22 +=x

y

A Cx

B

For C(x, y)

62 ==− aex

26 +=x

AC = 262226 −=−−+

1222===

abBC

Area = 123)26(1

21)26(

21

−=−××=−××BC .

4. The area of the region between the curves xxy

cossin1+

= and xxy

cossin1−

= bounded by the lines x = 0 and

4π

=x is

(A) ∫−

−+

12

022 1)1(

dttt

t (B) ∫−

−+

12

022 1)1(

4 dttt

t

(C) ∫+

−+

12

022 1)1(

4 dttt

t(D) ∫

+

−+

12

022 1)1(

dttt

t



Answer (B)

Hints : ∫

π

⎟⎟⎠

⎞⎜⎜⎝

⎛ +4

0cos

sin–1–cos

sin1 dxx

xx

x ⎟⎠

⎞⎜⎝

⎛ >−

>+ 0

cossin1

cossin1

xx

xx∵

=

2tan1

2tan–1

2tan1

2tan2

1

–

2tan1

2tan–1

2tan1

2tan2

1

2

2

24

0

2

2

2

x

x

x

x

x

x

x

x

+

+−

+

++

∫π

= ∫

π

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

+4

0 2tan1

2tan–1

–

2tan–1

2tan1

dxx

x

x

x

= ∫

π++4

02

2tan–1

2tan1–

2tan1

x

xx

= ∫

π4

0 2

2tan–1

2tan2

dxx

x

Put 2

tan x = t, dtdxx=

2sec

21 2

= ∫

π

⎟⎠⎞⎜

⎝⎛ +

8tan

022 –11

4

tt

dtt

as 1–28

tan =π

So ( )∫+

1–2

022 –11

4

tt

dtt

5. Consider three points )sin),(cos(),cos),sin(( βα−β=β−α−β−= QP and ))sin(),(cos( θ−βθ+α−β=R ,

where 4

,,0 π<θβα< . Then,

(A) P lies on the line segment RQ (B) Q lies on the line segment PR(C) R lies on the line segment QP (D) P, Q, R are non-collinear

Answer (D)

Hints : For collinear points

1)sin()cos(1sin)cos(1cos)sin(

θ−βθ+α−ββα−ββ−α−β−

=Δ

Clearly Δ ≠ 0 for any value of α, β, θ, hence points are non-collinear.



IInd method : (by observation)

P R Qsinθcosθ

(–sin( – ), –cos )β α β (cos( – ), sin )β α β

In this case ⎟⎠

⎞⎜⎝

⎛θ+θ

βθ−βθθ+θ

α−βθ−α−βθ≡

cossincossinsincos,

cossin)sin(sin)cos(·cosR

))sin(),(cos()cossin

)sin(,cossin

)cos(θ−βθ+α−β=⎟⎟

⎠

⎞⎜⎜⎝

⎛θ+θ

θ−βθ+θθ+α−β

≡R , if sinθ + cosθ = 1

Which is not possible if 0 < θ < 4π

Hence points are non-collinear.6. An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A

consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is(A) 2, 4 or 8 (B) 3, 6 or 9 (C) 4 or 8 (D) 5 or 10

Answer (D)

Hints : P(A) = 52

For independent eventsP (A ∩ B) = P (A) P (B)

P (A ∩ B) ≤ 52

⇒ P (A ∩ B) = 104,

103,

102,

101

(Maximum 4 outcomes may be in A ∩ B)

(i) P (A ∩ B) = 101

⇒ P(A) . P(B) = 101

P (B) = 25

101× =

41

, not possible

(ii) P (A ∩ B) = 102

⇒ 102)(

52

=× BP

⇒ P(B) = 105

, outcomes of B = 5

(iii) P (A ∩ B) = 103

P (A) P (B) = 103

)(52 BP× = 10

3

P (B) = 43

, not possible

(iv) P (A ∩ B) = 104

⇒ P (A) . P (B) = 104

⇒ P (B) = 1, outcomes of B = 10



7. Let two non-collinear unit vectors a and b form an acute angle. A point P moves so that at any time t the

position vector OP (where O is the origin) is given by tbta sinˆcosˆ + . When P is farthest from origin O, let

M be the length of OP and u be the unit vector along OP . Then,

(A) 21

)ˆˆ1(and|ˆˆ|

ˆˆˆ baMbabau ⋅+=

+

+= (B) 2

1

)ˆˆ1(and|ˆˆ|


−

−=

(C) 21

)ˆˆ21(and|ˆˆ|


+

+= (D) 2

1

)ˆˆ21(and|ˆˆ|


−

−=

Answer (A)

Hints : tbtaOP sinˆcosˆ +=

ttbatbbtaaOP cossinˆ·ˆ2sin)ˆ·ˆ(cosˆ·ˆ(|| 22 ++=

ttbaOP cossinˆ·ˆ21|| +=

tbaOP 2sinˆ·ˆ1|| +=

baOP ˆ·ˆ1|| max += at sin 2t = 1 ⇒ 4π

=t

)ˆˆ(21

4at batOP +=⎟

⎠

⎞⎜⎝

⎛ π=

Unit vector along |ˆˆ|

ˆˆ4

atbabatOP

+

+=⎟

⎠

⎞⎜⎝

⎛ π=

8. Let ∫ ∫ ++=

++=

−−

−

dxee

eJdxee

eI xx

x

xx

x

1,

1 2424.

Then, for an arbitrary constant C, the value of J – I equals

(A) Ceeee

xx

xx+⎟

⎟⎠

⎞⎜⎜⎝

⎛

++

+−

11log

21

24

24(B) C

eeee

xx

xx+⎟

⎟⎠

⎞⎜⎜⎝

⎛

+−

++

11log

21

2

2

(C) Ceeee

xx

xx+⎟

⎟⎠

⎞⎜⎜⎝

⎛

++

+−

11log

21

2

2(D) C

eeee

xx

xx+⎟

⎟⎠

⎞⎜⎜⎝

⎛

+−

++

11log

21

24

24

Answer (C)

Hints : dxee

eJ xx

x

∫ ++= 42

3

1

dxee

eeIJ xx

xx

∫ ++=− 42

3

1)–(

duuu

u∫ ++

= 42

2

1)1–(

(u = ex)



∫++

⎟⎠⎞

⎜⎝⎛ −

=2

2

2

11

11

uu

duu

∫⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ −

=

1–1

11

2

2

uu

duu

∫= 1–2tdt

⎟⎠

⎞⎜⎝

⎛ +=u

ut 1

Ctt

++−

=11log

21

Cuuuu

+++

+=

11–log

21

2

2

Ceeee

xx

xx+

++

+−=

11log

21

2

2

9. Let )(log)( xfxg = where f(x) is a twice differentiable positive function on ),0( ∞ such that )()1( xfxxf =+ .Then, for N = 1, 2, 3,....,

=⎟⎠

⎞⎜⎝

⎛′′−⎟⎠

⎞⎜⎝

⎛ +′′21

21 gNg

(A)⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++++− 2)12(

1.....251

9114

N(B)

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++++ 2)12(

1.....251

9114

N

(C)⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++++− 2)12(

1.....251

9114

N(D)

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++++ 2)12(

1.....251

9114

NAnswer (A)

Hints : f(x) = eg(x) ⇒ eg(x + 1) = f(x + 1)

= xf(x) = xeg(x)

g(x + 1) = log x + g(x)

i.e. g(x + 1) – g(x) = log x ...(1)

Replacing x by 21

−x

⎟⎠

⎞⎜⎝

⎛ −−⎟⎠

⎞⎜⎝

⎛ +21

21 xgxg = ⎟

⎠

⎞⎜⎝

⎛ −21log x

= log(2x – 1) – log2

∴ 122

21'

21'

−=⎟

⎠

⎞⎜⎝

⎛ −−⎟⎠

⎞⎜⎝

⎛ +x

xgxg

∴ 2)12(4

21''

21''

−

−=⎟

⎠

⎞⎜⎝

⎛ −−⎟⎠

⎞⎜⎝

⎛ +x

xgxg ...(2)

Substituting, x = 1, 2, 3, ......, N in (2) and adding

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−++++−=⎟

⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛ + 2)12(1....

251

9114

21''

21''

NgNg



SECTION - IIReasoning Type

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

10. Suppose four distinct positive numbers a1, a2, a3, a4 are in G.P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 andb4 = b3 + a4.

STATEMENT-1 : The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P.

and

STATEMENT-2 : The numbers b1, b2, b3, b4 are in H.P.

(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True

Answer (C)

Hints : Let a1 = 1, a2 = 2, a3 = 4, a4 = 8

∴ b1 = 1, b2 = 3, b3 = 7, b4 = 15

Clearly, b1, b2, b3, b4 are not in HP.

Statement-2 is false.

11. Let a, b, c, p, q be real numbers. Suppose α, β are the roots of the equation x2 + 2px + q = 0 and β

α1,

are the roots of the equation ax2 + 2bx + c = 0, where }1,0,1{2 −∉β

STATEMENT-1 : 0))(( 22 ≥−− acbqp

and

STATEMENT-2 : qacpab ≠≠ or

(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True

Answer (B)

Hints : 022 =++ qpxx

)ii()i(2

…=αβ…−=β+α

qp

022 =++ cbxax

)iv(

)iii(21

…=βα

…−

=β

+α

ac

ab



)()( 22 acbqp −− = 2

2

2

2

1

2a

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

βα

−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛β

+α

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛αβ−⎟

⎠

⎞⎜⎝

⎛−

β+α

= 0.116

)( 222

≥⎟⎟⎠

⎞⎜⎜⎝

⎛β

−αβ−α a statement-1 is true

Now pa = )(22

β+α=⎟⎠

⎞⎜⎝

⎛ β+α−

aa

b = ⎟⎟⎠

⎞⎜⎜⎝

⎛β

+α−1

2a

correct},1,0,1{,11 2 −≠β≠β⇒β+α≠β

+α⇒≠ bpa

Similarly

If αβ≠βα

⇒≠ aaqac

⇒ 01≠⎟⎟

⎠

⎞⎜⎜⎝

⎛β

−βα

⇒ }1,0,1{01and,0 −≠β⇒≠β

−β≠α

Statement 2 Is true.Both statement 1 and statement 2 are true, But statement 2 do not explains statement 1

12. Consider

L1 : 2x + 3y + p – 3 = 0

L2 : 2x + 3y + p + 3 = 0

where p is a real number, and C : x2 + y2 + 6x – 10y + 30 = 0.

STATEMENT-1 : If line L1 is a chord of circle C, then line L2 is not always a diameter of circle C.

and

STATEMENT-2 : If line L1 is a diameter of circle C, then line L2 is not a chord of circle C.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Answer (C)

Hints : 430259)5()3( 22 =−+=−++ yx

222 2)5()3( =−++ yx

centre = (3, –5)



If L1 is diameter, then, 03)5(3)3(2 =−+−+ pp = 12

∴ L1 is 2x + 3y + 9 = 0L2 is 2x + 3y + 15 = 0

Distance of centre of circle from L2 equals

)circleofradius(2136

32

15)5(3)3(222

<=+

+−+

∴ L2 is a chord of circle C.Statement (2), false

13. Let a solution y = y(x) of the differential equation

dxyydyxx 1––1– 22 = 0

satisfy y(2) = 32

STATEMENT-1 : y(x) = ⎟⎠

⎞⎜⎝

⎛ π6

–secsec 1– x

and

STATEMENT-2 : y(x) is given by 21–1–321

xxy=

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

Answer (C)

Hints :

∫ ∫−

=−

−

−=

11

1

1

22

2

2

xx

dx

yy

dyxx

yydxdy

sec–1 y = sec–1 x + cx = 2

y = 32

c+= −− 2sec32sec 11

c+π

=π

36

c = 6π

−

y = )6/(secsec 1 π−− x

ω=⎥⎥⎦

⎤

⎢⎢⎣

⎡−= −−

23cos1coscos 11

x

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−+= −

43111

23coscos 2

1

xx

211

21

231

xxy−+=



SECTION - IIILinked Comprehension Type

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

Paragraph for Question Nos. 14 to 16

Consider the function f : (– ∞, ∞) → (– ∞, ∞) defined by

f(x) = 11–

2

2

++

+

axxaxx

, 0 < a < 2.

14. Which of the following is true?

(A) (2 + a)2 f " (1) + (2 – a)2 f " (–1) = 0 (B) (2 – a)2 f " (1) – (2 + a)2 f " (–1) = 0(C) f ' (1) f ' (– 1) = (2 – a)2 (D) f ' (1) f ' (– 1) = – (2 + a)2

Answer (A)

Hints : )(xf = ( )

121

2

2

++

−++

axxaxaxx

= 1

21 2 ++−

axxax

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−⋅++−=′

22

2

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

axxaxaxaaxxxf

)1()1(

)1(2)1(

2222

2

22

2…

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−−=

axxxa

axxaax

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+++−−++=′′

42

2222

)1()2()1()1(2)2()1(2)(

axxaxaxxxxaxxaxf

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+−−++= 32

22

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

axxaxxaxxxa

)2()1(

34 32

3

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

++−=

axxaxxa

23 )2(4

)2()2(4)1(

+=

++

=′′a

aa

aaf

23 )2(4

)2()2(4)1(

−−

=−−

=−′′a

aa

aaf

∴0

44)1()2()1()2( 22

=−=−′′−+′′+ aafafa

15. Which of the following is true?

(A) f(x) is decreasing on (–1, 1) and has a local minimum at x = 1(B) f (x) is increasing on (–1, 1) and has a local maximum at x = 1(C) f (x) is increasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1(D) f (x) is decreasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1



Answer (A)

Hints : When x ∈ (–1,1)

011 22 <−⇒< xx

∴ decreasingis)(0)( xfxf ⇒<′

Also, at x = 1, )20(0)2(

4)1( 2 <<>+

=′′ aa

af ∵

f (x)has a local minimum at x = 1

16. Let

g(x) = ∫ +

′xe

dtttf

021)(

Which of the following is true?

(1) g'(x) is positive on (– ∞, 0) and negative on (0, ∞)(2) g'(x) is negative on (– ∞, 0) and positive on (0, ∞)(3) g'(x) changes sign on both (– ∞, 0) and (0, ∞)(4) g'(x) does not change sign on (– ∞, ∞)

Answer (B)

Hints : g′(x) = x

x

xe

eef .

)(1)(2+

′

⎟⎟⎠

⎞⎜⎜⎝

⎛

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

−= x

x

xx

x

ee

aeeea 222

2

1)1(12

,0)( =′ xg if 0..,012 ==− xeie x

If 0)(1,0 2 <′⇒<< xgex x

Paragraph for Question Nos. 17 to 19Consider the lines :

L1 : 31+x

= 1

2+y =

21+z

L2 : 33–

22

12– zyx

=+

=

17. The unit vector perpendicular to both L1 and L2 is

(A)99

ˆ7ˆ7ˆ– kji ++(B)

35

ˆ5ˆ7–ˆ– kji +(C)

35

ˆ5ˆ7ˆ– kji ++(D)

99

ˆ–ˆ7–ˆ7 kji

Answer (B)

Hints : The equations of given lines in vector form may be written as

)ˆ2ˆˆ3()ˆˆ2ˆ(:1 kjikjirL ++λ+−−−=

and )ˆ3ˆ2ˆ()ˆ3ˆ2ˆ2(:2 kjikjirL ++μ++−=



∴ Vector perpendicular to both L1 and L2 is

kjikji

ˆ5ˆ7ˆ

321213

ˆˆˆ

+−−=

∴ Required unit vector = 222 )5()7()1(

)ˆ5ˆ7ˆ(

+−+−

+−− kji

= )ˆ5ˆ7ˆ(35

1 kji +−−

18. The shortest distance between L1 and L2 is

(A) 0 (B) 317

(C) 3541

(D) 3517

Answer (D)

Hints : Shortest distance between L1 and L2 is

35)ˆ5ˆ7ˆ(}ˆ))1(3(ˆ)22(ˆ))1(2{( kjikji +−−⋅−−+−+−−

units 35

1735

)ˆ5ˆ7ˆ()ˆ4ˆ3(=

+−−⋅+=

kjiki

19. The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal isperpendicular to both the lines L1 and L2 is

(A) 752

(B) 757

(C) 7513

(D) 7523

Answer (C)

Hints : The equation of the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to boththe given lines L1 and L2 may be written as

(x + 1) + 7 (y + 2) –5 (z + 1) = 0

i.e. x + 7y – 5z + 10 = 0

The distance of the point (1, 1, 1) from the plane = units 75

132549110571

=++

+−+

SECTION-IVMatrix-Match Type

This section contains 3 questions. Each question contains statements given in two columns, which have to bematched. Statements in Column I are labelled as A, B, C and D whereas statements in Column II are labelledas p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the followingexample.

If the correct matches are A-q, A-r, B-p, B-s, C-r, C-s and D-q, then the correctly bubbled matrix will look like thefollowing :

A

B

C

D

p

p

p

p

q

q

q

q

r

r

r

r

s

s

s

s

p q r s



20. Consider the lines given by

L1 : x + 3y – 5 = 0

L2 : 3x – ky – 1 = 0

L3 : 5x + 2y – 12 = 0

Match the Statements/Expressions in Column I with the Statements/Expressions in Column II and indicateyour answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I Column II

(A) L1, L2, L3 are concurrent, if (p) k = – 9

(B) One of L1, L2, L3 is parallel to at least one (q) k = 56–

of the other two, if

(C) L1, L2, L3 form a triangle, if (r) k = 65

(D) L1, L2, L3 do not form a triangle, if (s) k = 5

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

Hints.

(A) Solving L1 and L3

1521

25121036 −=

−+=

+−yx

∴ x = 2, y = 1

L1, L2, L3 are concurrent, if (2, 1) lies on L2

∴ 6 – k – 1 = 0 ⇒ k = 5

(B) Either L1 is parallel to L2, or L3 is parallel to L2,

then 253or3

31 k

k−

=−

=

⇒ k = –9 or 56−

=k

(C) L1, L2, L3 form a triangle, if they are not concurrent, or not parallel

∴ 65

56,9,5 =⇒−−≠ kk

(D) L1, L2, L3 do not form a triangle, if

56,9,5 −−=k

21. Match the Statements/Expressions in Coilumn I with the Statements/Expressions in Column II and indicateyour answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I Column II

(A) The minimum value of 2

422

+++

xxx

is (p) 0

(B) Let A and B be 3 × 3 matrices of real number, (q) 1where A is symmetric, B is skew-symmetric, and(A + B) (A – B) = (A – B) (A + B). If (AB)t = (–1)k AB,where (AB)t is the transpose of the matrix AB,then the possible values of k are



(C) Let a = log3 log32. An integer k satisfying (r) 2

221 )3( <<−+− ak , must be less than

(D) If φ=θ cossin , then the possible values of (s) 3

⎟⎠

⎞⎜⎝

⎛ π−φ±θ

π 21

are

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

Hints :

(A) Let 2

422

+++

=x

xxy

⇒ x2 + (2 – y)x + (4 – 2y) = 0

⇒ (2 – y)2 – 4(4 – 2y) ≥ 0

⇒ y2 + 4y – 12 ≥ 0

⇒ y ≤ –6, y ≥ 2

∴ minimum value of y is 2

(B) (A + B) (A – B) = (A – B) (A + B)

⇒ A2 – AB + BA – B2 = A2 + AB – BA – B2

⇒ AB = BA

(AB)t = (–1)k AB ⇒ Bt At = (–1)k AB

⇒ –B⋅A = (–1)k + 1AB (∵Bt = –B, At = A)

⇒ B⋅A = (–1)k + 1AB

⇒ (–1)k + 1 = 1

∴ k + 1 is even, or k is odd

(C) 221 )3( <<−+− ak

⇒ 0 < – k + 3–a < 1

2log32loglog 333 =⇒= aa

⇒ 3log3 2=−a .....(2)

∴ 23log2 <<k .....(3)

and 013log1 2 >⇒>>+ kk .....(4)

by (3) & (4), 0 < k < 2 ⇒ k = 1 (k is an integer)

∴ k < 2

(D) φ=⎟⎠

⎞⎜⎝

⎛ θ−π

⇒φ=θ cos2

coscossin

⇒ Znn ∈φ±π=θ−π ,22

⇒ Znn ∈π−=π

−φ±θ ,22

⇒ Znn ∈−=⎟⎠

⎞⎜⎝

⎛ π−φ±θ

π,2

21



22. Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/Expressionsin Column I with the Statements/Expressions in Column II and indicate your answer by darkening theappropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I Column II

(A) The number of permutations containing the (p) 5!

word ENDEA is

(B) The number of permutations in which the (q) 2 × 5!

letter E occurs in the first and the last positions

is

(C) The number of permutations in which none of the (r) 7 × 5!

letters D, L, N occurs in the last five positions is

(D) The number of permutations in which the letters (s) 21 × 5!

A, E, O occur only in odd positions is

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

Hints :

(A) If ENDEA is fixed word then assume this as a single letter

Total letters = 5, total number of arrangement = 5 !

(B) If E is at first and last places then

total permutation = !521!2!7

×=

(C) If D, L, N is not in last five position

D,L,N,N E,E,E,A,O

total permutation = !52!3!5

!2!4

×=×

(D) Total odd position = 5

permutation of AEEEO are !3!5

Total even position = 4

Permutation of N,N,D,L = !2!4

hence total permutation = !52!2!4

!3!5

×=×



PHYSICSPART- II


This section contains 9 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.

23. Consider a system of three charges ,3q

3q

and 32q

− placed at points A, B and C, respectively, as shown

in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60°

Figure :

x

A

C

B

y

60°O

(A) The electric field at point O is 208 R

q

πε directed along the negative x-axis

(B) The potential energy of the system is zero

(C) The magnitude of the force between the charges at C and B is 20

2

54 R

q

πε

(D) The potential at point O is R

q

012πε

Answer (C)

Hints : Force between C and B is

20

2

20 54)3(

133

24

1R

qR

qqπε

=×⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛πε

24. A radioactive sample S1 having an activity of 5 μ Ci has twice the number of nuclei as another sample S2which has an activity of 10 μ Ci. The half lives of S1 and S2 can be

(A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively

(C) 10 years each (D) 5 years each

Answer (A)

Hints : For S1, activity λ1N1 = 5 μCi

For S2, activity λ2N2 = 10 μCi

Also N1 = 2N2

⇒42

1λ

=λ

⇒ T1 = 4T2



25. A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. Thewavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the waveis shown in figure. The velocity of point P when its displacement is 5 cm isFigure :

P

y

x

(A) sm 503 jπ

(B) sm ˆ503 jπ

− (C) sm ˆ503 iπ

(D) sm ˆ503 iπ

−

Answer (A)

Hints : The wave equation can be written as

j)t4.04sin()m1.0(y π−π= .....(1)

as A = 10 cm, π=λπ

==ω

= 42andcm/s10 kk

v .....(2)

Now jxtdtdy ˆ)44.0cos(4.01.0 −ππ×=

From (1), Put y = 5 cm and t = 0, calculate x.From (2), velocity can be calculated

26. A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respectively(see figure I). The other ends are attached to identical supports M1 and M2 not attached to the walls. Thesprings and supports have negligible mass. There is no friction anywhere. The block B is displaced towardswall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance ytowards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B.

The ratio xy

is

Figure :

M2 S2 S1 M112

I

M2 S2 S1 M112

B

B

x

x

II

(A) 4 (B) 2 (C)21

(D)41

Answer (C)

Hints : Using conservation of energy,

22 421

21 kykx ×=×

⇒ 2y = x



27. A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A isjust sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A,satisfies

Figure :

θL

A V

B

(A)4π

=θ (B)24π

<θ<π

(C)4

32

π<θ<

π(D) π<θ<

π4

3

Answer (D)

Hints : At lowest point glv 5=

By conservation of energy,

⎥⎦

⎤⎢⎣

⎡ =θ−+=2

Here)cos1(21

21 22 vVmglmvmv BB

⇒87cos −

=θ

⇒ ⎟⎠⎞

⎜⎝⎛ −=θ −

87cos 1

28. A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valve isin a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemisphericalsoap bubble as shown in figure. Just after opening the valve,

Figure :

12

(A) Air from end 1 flows towards end 2. No change in the volume of the soap bubbles(B) Air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases(C) No change occurs(D) Air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases

Answer (B)

Hints : The sub-hemispherical bubble will have greater radius of curvature. So pressure inside 1 will be more.

29. A vibrating string of certain length under a tension T resonates with a mode corresponding to the first overtone(third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the stringis slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air tobe 340 m/s, the frequency n of the tuning fork in Hz is(A) 344 (B) 336 (C) 117.3 (D) 109.3

Answer (A)



Hints : Hz 344

100754

3403string =

⎟⎠⎞

⎜⎝⎛×

×=f

Now, fstring ~ n = 4On increasing T, fstring increases and difference between the string frequency and tuning fork decreases. So,

fstring < n⇒ n = 344 Hz

30. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant

K = 2. The level of liquid is 3d

initially. Suppose the liquid level decreases at a constant speed V, the time

constant as a function of time t isFigure :

d3

d

CR

(A)tVd

R35

6 0+

ε(B) 222

0

932

)915(

tVtVdd

RtVd

−−

ε+

(C)tVd

R35

6 0−

ε(D) 222

0

932

)915(

tVtVdd

RtVd

−+

ε−

Answer (A)

Hints : Time constant T = RC

Now,

k

xd

xd

AC

−++

ε=

33

2

0

where x = vt.

31. A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and

IV are n0, ,20n

60n

and ,80n

respectively. The angle of incidence θ for which the beam just misses entering

Region IV isFigure :

Region I Region II Region III Region IV

θ n0

2n0

6n0

8n0

0 0.2 m 0.6 m

(A) ⎟⎠

⎞⎜⎝

⎛−

43sin 1

(B) ⎟⎠

⎞⎜⎝

⎛−

81sin 1

(C) ⎟⎠

⎞⎜⎝

⎛−

41sin 1

(D) ⎟⎠

⎞⎜⎝

⎛−

31sin 1

Answer (B)

Hints : For total internal reflection, at IV, θ depends only on I and IV

⇒ 81sin =

μμ

=θI

IV



SECTION - IIAssertion - Reason Type

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

32. STATEMENT-1 : For an observer looking out through the window of a fast moving train, the nearby objectsappear to move in the opposite direction to the train, while the distant objects appear to be stationary.

and

STATEMENT-2 : If the observer and the object are moving at velocities 1V and 2V respectively with reference

to a laboratory frame, the velocity of the object with respect to the observer is .12 VV −

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Answer (B)

Hints : The angular position of farther objects change less. So, they appear stationary. O and O′ represent initialand final position of an observer in motion

near byobject

farther byobject

O′

O

motion

33. STATEMENT-1 : It is easier to pull a heavy object than to push it on a level ground.andSTATEMENT-2 : The magnitude of frictional force depends on the nature of the two surfaces in contact.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

Answer (B)

Hints : During pull, normal reaction decreases so friction decreases

34. STATEMENT-1 : For practical purposes, the earth is used as a reference at zero potential in electrical circuits.andSTATEMENT-2 : The electrical potential of a sphere of radius R and with charge Q uniformly distributed on

the surface is given by .4 0R

Qπε




Answer (B)

Hints : Earth is assumed to have zero potential, although it does have a negative potential. When potential at

infinity is regarded as zero, the potential of a sphere with uniform charge of the surface is RQ

04πε

35. STATEMENT-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magneticmaterial as a core inside the coil.

andSTATEMENT-2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.


Answer (C)

Hints : By a suitable magnetic material field strength increases. Sensitivity is directly proportional to field as

CNBA

I=

θ. Soft iron can be easily magnetised

SECTION-IIILinked Comprehension Type



The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r) [chargeper unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure.The electric field is only along the radial direction.

Figure :

ρ( )r

d

a Rr

36. The electric field at r = R is

(A) Independent of a (B) Directly proportional to a

(C) Directly proportional to a2 (D) Inversely proportional to a

Answer (A)

Hints : The electric field at R (at the surface) is

E = 20

)(4

1RZe

πε

It is independent of a



37. For a = 0, the value of d (maximum value of ρ as shown in the figure) is

(A) 343

RZeπ

(B) 33RZe

π(C) 33

4RZeπ (D) 33 R

Zeπ

Answer (B)

Hints : When a = 0, ⎟⎠

⎞⎜⎝

⎛ −=ρRrd 1

Now, Ze = ∫ ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−π=π×ρ

R R

drRrrddrr

0 0

322 44

⇒ d = 33RZe

π

38. The electric field within the nucleus is generally observed to be linarly dependent on r. This implies

(A) a = 0 (B) a = 2R

(C) a = R (D) a = 32R

Answer (C)

Hints : When E ∝ r ⇒ charge density is uniform (like the case of a uniformly charged sphere).

So, a = R


A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of springconstant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disksymmetrically on either side at a distance d from its centre. The axle is massless and both the springs andthe axle are in a horizontal plane. The unstretched length of each spring is L. The disk is initially at itsequilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping

with velocity iVV ˆ00 = . The coefficient of friction is μ.

Figure

y

d

V0Rd

2d

x

39. The net external force acting on the disk when its centre of mass is at displacement x with respect to itsequilibrium position is

(A) –kx (B) – 2kx (C) 32– kx

(D) 34– kx



Answer (D)

Hints : F – fr = Ma

fr.R = I α ⇒ 2RIafr =

Solving these two,F

fr

2rIM

Fa+

=

Also, F = –2Kx ⇒

2

2

rIM

Kxa+

−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

−×=

2

22

rIM

KxRIfr

For disc 2

21 MRI =

⇒3

4kxfr −=

40. The centre of mass of the disk undergoes simple harmonic motion with angular frequency ω equal to

(A)Mk

(B)Mk2

(C) Mk

32

(D) Mk

34

Answer (D)

Hints :2

2

rIM

kxa+

−=

with a = –ω 2x

⇒2

2

rIM

k

+=ω m

k34

=

41. The maximum value of V0 for which the disk will roll without slipping is

(A)kMgμ (B)

kMg2

μ (C)kMg 3

μ (D)kMg

25

μ

Answer (C)

Hints : MgRIafr μ≤= 2

and mgR

IVμ≤

ω2

0 ( )ω= ovamax∵

mgRVMR

μ≤ω2

02

2

kMgV

4320 μ≤

kMg 3

μ≤



SECTION - IVMatrix Match Type

This section contains 3 questions. Each question contains statements given in two columns, which have to bematched. Statements in Column I are labelled as (A), (B), (C) and (D), whereas statements in Column II are labelledas p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the followingexample.If the correct matches are A-q, A-r, B-p, B-s, C-r, C-s and D-q, then the correctly bubbled matrix will look like thefollowing :

A

B

C

D

p

p

p

p

q

q

q

q

r

r

r

r

s

s

s

s

p q r s

42. Column I gives a list of possible set of parameters measured in some experiments. The variations of theparameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I withthe graphs given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrixgiven in the ORS.

Column I Column II(A) Potential energy of a simple (p) y

xO

pendulum (y axis) as a functionof displacement (x axis)

(B) Displacement (y axis) as a function (q) y

xO

of time (x-axis) for a one dimensionalmotion at zero or constant accelerationwhen the body is moving along the positivex-direction

(C) Range of a projectile (y axis) as (r) y

xO

a function of its velocity (x axis)when projected at a fixed angle

(D) The square of the time period (y axis) (s) y

xO

of a simple pendulum as a functionof its length (x axis)



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

Hints :

(A) → p,s

potential energy of a simple pendulum varies curvilinearly with displacement and is minimum at a certain value

(B) → q, r, s

q and r correspond to zero acceleration, while S corresponds to accelerated motion p is not possible as velocityis negative corresponding to initial portion of the graph.

(C) → s

22 2sin uR

guR ∝⇒

θ=

(D) → q

lTglT ∝⇒π= 22

43. An optical component and an object S placed along its optic axis are given in Column I. The distance betweenthe object and the component can be varied. The properties of images are given in Column II. Match all theproperties of images from Column II with the appropriate components given in Column I. Indicate your answerby darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I Column II

(A)S

(p) Real Image

(B) S (q) Virtual image

(C) S (r) Magnified image

(D) S (s) Image at infinity

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

Hints : In all cases, u is negative

(A) → p, q, r, s

Magnification for mirror, uf

fm−

=

f is negative. Depending on valve of u, m can be positive or negative and | m | can be > 1 or < 1.



(B) → q

uffm−

=

f is positive 0 < m < 1, for any value of u

(C) → p, q, r, s

uffm+

=

f is positive m can be positive or negative and | m | can be > 1 or < 1 depending on u.

(D)→ p, q, r, s

uffm+

=

f is positive m can be positive or negative and |m| can be > 1 or < 1 depending on u.

44. Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describingthe thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles ofthe 4 × 4 matrix given in the ORS.

Column I Column II(A) An insulated container has two chambers (p) The temperature of the gas decreases

separated by a valve. Chamber I containsan ideal gas and the Chamber II has vacuum.The valve is opened.

II

Vacuum

I

ideal gas

(B) An ideal monoatomic gas expands (q) The temperature of the gas increasesto twice its original volume such or remains constant

that its pressure ,12V

P ∝ where V

is the volume of the gas(C) An ideal monoatomic gas expands to (r) The gas loses heat

twice its original volume such that

its pressure ,13/4V

P ∝ where V is

its volume(D) An ideal monoatomic gas expands such (s) The gas gains heat

that its pressure P and volume V followsthe behaviour shown in the graph

P

V1 2V V1

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



Hints : (A) in free expansion W = ΔU = Q = 0

(B), (C) & (D)

PVn = const. if n < γ Q > 0

n > γ Q < 0

n = γ Q = 0

n = 1 T = constant

n > 1 T decreases

n < 1 T increases

CHEMISTRYPART- III


This section contains 9 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.

45. The correct stability order for the following species is

O⊕

O⊕⊕

⊕(I) (II) (III) (IV)

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

Answer (D)

Hints : I & III are stabilized by resonance, hyperconjugation and + inductive effect.

∴ order (I) > (III) > (II) > (IV)

46. Cellulose upon acetylation with excess acetic anhydride / H2SO4 (catalytic) gives cellulose triacetate whosestructure is

(A) H

– O

AcO

OAc

H

H

OAc

O

H

AcO

OAc

H

H

OAc

O

H

H

O

OH

H

AcO

OAc

H

H

OAc

OH H

H

O –

(B)H

– O

AcO

OH

H

H

OH

O

H

AcO

OH

H

H

OH

O

H

H

O

OH

H

AcO

OH

H

H

OH

OH H

H

O –



(C)H

– O

AcO

OAc

H

H

OAc

O

O

H H

AcO

OAc

H

H

OAc

OH

O

H H

AcO

OAc

H

H

OAc

OHH

H

O –

(D)

H

– O

AcO

H H

OAc

O

OH

H

AcO

H H

OAc

OH

O

H H

AcO

H H

OH

H

H

O –

OAc OAc OAcOAc

Answer (C)

Hints : Cellulose is polymer of β-D glucose. Hence structure (C) is correct.

47. In the following reaction sequence, the correct structures of E, F and G are

O O

*Ph OH

Heat [E]I

NaOH2 [F] + [G]

( ∗ implies 13C labelled carbon)

(A)O

*Ph CH3E =

O

Ph O NaF = G = CHI3*

(B)

O*Ph CH3

E = O

Ph O NaF = G = CHI3

(C)O

*Ph CH3E =

O

Ph O NaF = G = CHI3*

(D)O

*Ph CH3E =

O

Ph O NaF = G = CH I3*

Answer (C)

Hints : Ph

O O

OH∗

Δ

–CO2 PhC

CH3

∗

OI + NaOH 2

PhC

O

ONa + CHI3

∗

GEβ-keto acid

iodoform reaction

48. Among the following, the coloured compound is

(A) CuCl (B) K3[Cu(CN)4] (C) CuF2 (D) [Cu(CH3CN)4]BF4

Answer (C)

Hints : In CuF2, copper is in +2 oxidation state

∴ coloured. In other options copper is in +1 oxidation state therefore colourless.



49. Both [Ni(CO)4] and [Ni(CN)4]2– are diamagnetic. The hybridisations of nickel in these complexes, respectively,

are(A) sp3, sp3 (B) sp3, dsp2 (C) dsp2, sp3 (D) dsp2, dsp2

Answer (B)

Hints : [Ni(CO)4] --------- sp3 hybridization

[Ni(CN)4]–2 --------- dsp2 hybridization

50. The IUPAC name of [Ni(NH3)4][NiCl4] is(A) Tetrachloronickel (II)-tetraamminenickel (II) (B) Tetraamminenickel (II) - tetrachloronickel (II)(C) Tetraamminenickel (II) - tetrachloronickelate (II) (D) Tetrachloronickel(II) - tetraamminenickelate (0)

Answer (C)

Hints : IUPAC name is tetraamminenickel(II)-tetrachloronickelate(II).

51. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The timerequired to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol–1)(A) 9.65 × 104 sec (B) 19.3 × 104 sec (C) 28.95 × 104 sec (D) 38.6 × 104 sec

Answer (B)

Hints : Fit

EW

=

96500t101020.01

EW 3 ××

=×=−

t = 19.3 × 104 s

52. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentrationat ambient conditions is(A) CH3(CH2)15N

+(CH3)3Br– (B) CH3(CH2)11OSO3–Na+

(C) CH3(CH2)6COO–Na+ (D) CH3(CH2)11N+(CH3)3Br–

Answer (B)

Hints : Detergent will form micelles in aqueous solution at the lowest molar concentration at ambient condition.

53. Solubility product constants (Ksp) of salts of types MX, MX2 and M3X at temperature 'T' are 4.0 × 10–8,3.2 × 10–14 and 2.7 × 10–15, respectively. Solubilities (mol dm–3) of the salts at temperature 'T' are in the order(A) MX > MX2 > M3X (B) M3X > MX2 > MX(C) MX2 > M3X > MX (D) MX > M3X > MX2

Answer (D)

Hints : (MX) S2 = 4.0 × 10–8

S = 2.0 × 10–4

(MX2) 4S3 = 3.2 × 10–14

S3 = 0.8 × 10–14

S3 = 8 × 10–15

S = 2 × 10–5

(M3X) 27S4 = 2.7 × 10–15

S = 1 × 10–4

∴ Order isMX > M3X > MX2



SECTION - IIAssertion - Reason Type

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

54. STATEMENT-1 : Aniline on reaciton with NaNO2/HCl at 0°C followed by coupling with β-naphthol gives a darkblue coloured precipitate.andSTATEMENT-2 : The colour of the compound formed in the reaction of aniline with NaNO2/HCl at 0°C followedby coupling with β-naphthol is due to the extended conjugation.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True

Answer (D)Hints : Aniline on reaction with NaNO2/HCl at 0° C followed by coupling with β-naphthol gives a orange red dye.

∴ Statement-1 is false but statement 2 is true.55. STATEMENT-1 : [Fe(H2O)5NO]SO4 is paramagnetic.

andSTATEMENT-2 : The Fe in [Fe(H2O)5NO]SO4 has three unpaired electrons.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True

Answer (A)Hints : [Fe(H2O)5NO]SO4 is paramagnetic in nature. Magnetic moment of complex is 3.9 BM. Hence it has 3

unpaired electrons. In this complex, oxidation number of Fe is +1.56. STATEMENT-1 : The geometrical isomers of the complex [M(NH3)4 Cl2] are optically inactive.

andSTATEMENT-2 : Both geometrical isomers of the complex [M(NH3)4Cl2] possess axis of symmetry.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True

Answer (A)Hints : [M(NH3)4Cl2] show cis-trans geometrical isomers and are not optically active due to presence of axis of

symmetry.57. STATEMENT-1 : There is a natural asymmetry between converting work to heat and converting heat to work.

andSTATEMENT-2 : No process is possible in which the sole result is the absorption of heat from a reservoir andits complete conversion into work.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is True

Answer (B)

Hints : According to second law of thermodynamics, It is impossible to convert energy into 100% work.



SECTION-IIILinked Comprehension Type


Paragraph for Question Nos. 58 to 60A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compoundsJ and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reactionwith KOH gives only M.

M = H C3

OPh

HPh58. Compound H is formed by the reaction of

(A) ||Ph CH3

O + PhMgBr

(B) ||Ph CH3

O + PhCH2MgBr

(C) ||Ph H

O + PhCH2MgBr

(D) ||Ph H

O +

|Ph MgBr

Me

Answer (B)

59. The structure of compound I is

(A)CH3Ph

H Ph

(B)H3C Ph

H Ph

(C)

CH3Ph

H CH Ph2

(D)H3C CH3

HPh

Answer (A)

60. The structures of compounds J, K and L respectively, are

(A) PhCOCH3, PhCH2COCH3 and PhCH2COO–K+

(B) PhCHO, PhCH2CHO and PhCOO–K+

(C) PhCOCH3, PhCH2CHO and CH3COO–K+

(D) PhCHO, PhCOCH3 and PhCOO–K+



Answer (D)

Hints : Solution of Q. 58 to 60

Ph – C – CH3 + PhCH MgBr2

O

Ph – C – CH2 – Ph

OH

CH3 H

Conc. H SO dehydration 2 4

Ph – C = CHPh

CH3

I

orPh

H Ph

CH3

Ozonolysis

Ph – C = O

CH3

KOH

Ph – C – H

KOH

O

Cannizzaro'sreaction

PhCH OH + PhCOOK2

K

J

L

Aldol followedby dehydration

Ph H

H3CO

Ph

M

+

Paragraph for Question Nos. 61 to 63In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonalprism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in betweenthem. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphereon a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheresare then placed over the first layer so that they touch each other and represent the second layer. Each oneof these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with athird layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ‘r’.

61. The number of atoms in this HCP unit cell is(A) 4 (B) 6 (C) 12 (D) 17

Answer (B)

Hints : The number of effective atom in a unit cell

= ( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞⎜

⎝⎛

×+×+×

cell unitprimitive hexagonal

inatomsofno

CentreFacecorner

13212

6112 = 6

62. The volume of this HCP unit cell is

(A) 3r224 (B) 3r216 (C) 3r212 (D)3r

3364

Answer (A)



Hints : Volume of HCP unit cell

= Base area × height

= ⎟⎟⎠

⎞×⎟

⎟⎠

⎞⎜⎜⎝

⎛×

32r4a

436 2

= ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛××

32r4r2

436 2

= 3r22463. The empty space in this HCP unit cell is

(A) 74% (B) 47.6% (C) 32% (D) 26%

Answer (D)

Hints : Packing fraction = HeightareaBase

r346 3

×

π×

=

32r4a

436

r346

2

3

×⎟⎟⎠

⎞⎜⎜⎝

⎛×

π× = 0.74

% of empty space = (1 – 0.74) × 100 = 26%

SECTION - IVMatrix Match Type

This section contains 3 questions. Each question contains statements given in two columns, which have to bematched. Statements in Column I are labelled as (A), (B), (C) and (D), whereas statements in Column II are labelledas p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the followingexample.

If the correct matches are A-q, A-r, B-p, B-s, C-r, C-s and D-q, then the correctly bubbled matrix will look like thefollowing :

A

B

C

D

p

p

p

p

q

q

q

q

r

r

r

r

s

s

s

s

p q r s

64. Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. Indicate youranswer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS

Column I Column II

(A) H N—NH Cl2 3

⊕(p) Sodium fusion extract of the compound gives Prussian

blue colour with FeSO4

(B) HONH l3

⊕

COOH(q) Gives positive FeCI3 test



(C) HO NH Cl3

⊕(r) Gives white precipitate with AgNO3

(D) O N2 NH—NH3Br⊕

NO2

(s) Reacts with aldehydes to form the correspondinghydrazone derivative

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

65. Match the conversions in Column I with the type(s) of reaction(s) given in Column II. Indicate your answerby darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS

Column I Column II(A) PbS → PbO (p) Roasting(B) CaCO3 → CaO (q) Calcination(C) ZnS → Zn (r) Carbon reduction(D) Cu2S → Cu (s) Self reduction

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

66. Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answerby darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS

Column I Column II(A) Oribital angular momentum of the electron (p) Principal quantum number

in a hydrogen-like atomic orbital(B) A hydrogen-like one-electron wave function (q) Azimuthal quantum number

obeying Pauli principle(C) Shape, size and orientation of hydrogen-like (r) Magnetic quantum number

atomic orbitals(D) Probability density of electron at the nucleus (s) Electron spin quantum number

in hydrogen-like atomAnswer A(q); B(s); C(p, q, r), D(p, q, r)

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Solutions to IIT-JEE 2008...Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (6)) 7. Let two non-collinear unit