IIT JEE 2012 Paper 1 With Solutions

8/2/2019 IIT JEE 2012 Paper 1 With Solutions

1/36

IIT-JEE 2012 EXAMINATION

CODE : 9 08 / 04 / 2012

Part I : (PHYSICS)

SECTION I (Single Correct Answer Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which

ONLY ONE is correct.

Q.1 A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of

the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of

curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

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

Ans. [B]

Sol. For the combination( )

RRfeq

)1(11 21 +

=

20. =eqf

Here u = 40, f = 20

40=v

Q.2


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Ans. [B]

Sol. Angular momentum about rotational axis

vx

x = vt L(t) = [I + m(vt)

2]

= tmv2dtdL2t ;

torque = (2mv2)t

Q.3

Ans. [C]

Sol. Under steady state ])T3(T[A]T)T2[(A44

14

14 =

(2T)4

41T =4

1T 34T

4

2 41T = (24

+ 34)T

4 ; 2

41T = (16 + 81)T

4

T1 = T2

974/1


3/36

Q.4

Ans. [D]

Sol.

O Rr

E(r)

r < R ; Er= 0

r > R ; Er 1/r2

r < R , V =R

KQ; r = R , V =

R

KQ; r > R , V =

R

KQ

r

(r)


4/36

Q.5

Ans. [A]

Sol.2d

MLg4Y

l= & % ymax = %M + %L + %l + 2%d

Least count of both instrument, l = d = 3105100

5.0 =

%l = %225.0

105100

3

==

l

l

%11005.0

105100

d

dd%

3

=

=

=

here we see that, contribution ofl, = 2%

contribution of d = 2% d = 2 1 = 2%

hence both terms l and d contribute equally.

Q.6


5/36

Ans. [A]

Sol.

4.9 m y = 0 0.2 m

10 m

45

B A

The block is released from A.

x = 4.9 m + (0.2 m) sin (t +2

)

at t = 15 ; x = 5 m

So range of projectile will be 5 m

Nowg

v =

90sin5

2

v2 = 50

v = 50

Q.7

Ans. [D]

Sol. As,d

D=

R> G > B

So R> G > B


6/36

Q.8

Ans. [C]

Sol. prL 00rrr

=

0Lr

is always directed along the axis & its magnitude is constant.

PLr

P'Lr

P

Q.9


7/36

Ans. [D]

Sol.m

KTvrms

3=

16.3104

40

(Ar)

(He) ===rms

rms

v

v

Q.10

Ans. [C]

Sol.

e

mg

qE

45

qE = mg

E =q

mg

q

mg

101

V2

=

19

312

106.1

1001830101.910V

=

= 1012

9.1 114

= 1000 1012

V = 1 109

V


8/36

SECTION II (Multiple Correct Answer(s) Type)

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which

ONE or MORE arecorrect.

Q.11

Ans. [A, C, D]

Sol. out =00

in qq

=

By symmetry

Ans is (A, C, D)


9/36

Q.12

Ans. [A,B,C,D]

Sol. The circuit can be simplified as

12V

4 4 4

2 2 2

11

P

Q T

S

because of symmetry no current passes through 1 resistor.

12V

4 4 4

2 2 2P

Q T

S

I1 I1 12V

12

6I2

Req =612

612

+

= 4

I1 =4

12= 3A ; I2 = =3

3

22A

Here we have

4VV QS = i.e., QS VV <


10/36

Q.13

Ans. [A, C]

Sol.

1N

1N

1 cos

1 sin

P

Q

If = 45 then cos = sin hence block will be at rest.If plane is rough & > 45 then sin > cos so friction will act up the planeIf plane is rough & < 45 then cos > sin so friction will act down the plane so (A, C) are correct

Q.14


11/36

Ans. [C, D]

Sol.

v

X

Y

E0, B0

* If = 0 or 10

then particle moves in helical path with increasing pitch along Y-axis.

* If = 90 then magnetic force on the particle is zero and particle moves along Y-axis with constantacceleration.

Q.15 A person blows into open end of a long pipe. As a result, a high pressure pulse of air travels down the pipe.

When this pulse reaches the other end of the pipe,

(A) a high pressure pulse starts traveling up the pipe, if the other end of the pipe is open.

(B) a low pressure pulse starts traveling up the pipe, if the other end of the pipe is open

(C) a low pressure pulse starts traveling up the pipe, if the other end of the pipe is closed

(B) a high pressure pulse starts traveling up the pipe, if the other end of the pipe is closed

Ans. [B, D]

Sol.

At rigid end there is no phase difference in pressure wave, when end is open phase difference of come inpressure wave.


12/36

SECTION III (Integer Answer Type)

This section contains 5 questions. The answer to each question is a SINGLE DIGIT INTEGER, ranging from 0 to 9

(both inclusive).

Q.16

Ans. [6]

Sol.

2

1

(1) + (2) = Complete cylinder

E1 + E2 = E

E =)R2(2

R

0

2

=04

R

E2 =3

4

3

2

R

)R4(4

12

0=

0424

R

E1 = E E24

R

24

11 =

4

R

64

23

=

616

R23


13/36

Q.17

Ans. [5]

Sol.

OP

a

a/2

I = J a2

B =

a2

aJ 20

4

a

2a32

J 20

B = 0Ja

12

1

2

1

B = 0Ja 12

5

Q.18


14/36

Ans. [3]

Sol.

O

R

2R

P

Let mass of original disc = m

The mass of disc removed =)R4(

m2

R2 =4

m

So M.O.I of remaining section about axis passing

through "O" I0 =2

)R2(m 2

+ 2

2

R4

m

)2(

R

4

m

2mR2

+8

mR2mR 22

83

2 mR2

813

mR2

MOI of remaining section about "P"

IP =

+ 2

2

)R2(m2

)R2(m

+ 2

2

R54

m

)2(

R

4

m

[2mR2 + 4mR2]

+

4

mR5

8

mR 22

6mR2 8

11mR

2

8

37mR

2

O

P

I

I=

8

37

13

8 3


15/36

Q.19

Ans. [7]

Sol. Assume circular wire loop as primary and square loop as secondary coil

secondary = 2/322

20

)RR3(2iR2

+ a2 cos 45

=2

2a

R82

iR 23

20

M =i

ondarysec=

R22

a2/13

20

M =

R2

a

2

7

20

Q.20


16/36

Ans. [7]

Sol.

10 fme

v

Ui = 0

Uf= 15

2

1010

e120k

Ki =2

mv2

1and Kf= 0

From Energy conservation,

Kf+ Uf= Ki + Ui

2

15

2

mv2

1

1010

e120K=

v = 5.76 107

727

34

1076.5103

5

1067.6

mv

h

==

= 7 fm (approx)


17/36

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.

Q.21 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

Ans. [D]

Sol. Factual

Q.22 In allene (C3H4), the type (s) of hybridization of the carbon atoms is (are)(A) sp and sp

3(B) sp and sp

2(C) only sp

2(D) sp

2and sp

3

Ans. [B]

Sol. CH2=C=CH2

sp2

spsp

2

Q.23 For one mole of a van der Waals gas when b = 0 and T= 300 K, the PVvs. 1/Vplot is shown below. The

value of the van der Waals constant a (atm. liter2 mol2)

20.1

3.00

21.6

24.6

23.1

2.0

PV(liter-atmm

ol1)

V

1(mol liter

1)

[Graph not to scale ]

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

Ans. [C]

Sol.

+

2V

aP V = RT

PV = RT V

a

Slope = a

=1

6.211.20= 1.5

a = 1.5


18/36

Q.24 The number of optically active products obtained from the complete ozonolysis of the given compound is

CH3CH=CHCCH=CHCCH=CHCH3

CH3 H

H CH3 (A) 0 (B) 1 (C) 2 (D) 4

Ans. [A]

Sol. CH3CH=CHCCH=CHCCH=CHCH3O3, Zn, H2O

O O O O OO

CH3

H

H

CH3

2CH3.CHO + OHCCCHO + OHCCCHO

CH3

H

H

CH3

Q.25 A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below.

The empirical formula of the compound is -

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

Ans. [B]

Sol.

8

18

X

+ 6 2

1= X4

4

14

M

+ 1 = M2

Simplest ration = X2M

Q.26 The number of aldol reaction (s) that occurs in the given transformation is -

(A) 1 (B) 2 (C) 3 (D) 4Ans. [C]

Sol. CH3CHO + 3HCHO

nrexaldol3

NaOH.aq

.conc HOCH2CCHO + HCHO

CH2OH

CH2OH

nrex

Cannizaro HOCH2CCH2OH + HCOONa

CH2OH

CH2OH

Q.27 The colour of light absorbed by an aqueous solution of CuSO4 is -

(A) orange-red (B) blue-green (C) yellow (D) violet

Ans. [A]

Sol. Factual.


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Q.28 The carboxyl functional group (COOH) is present in -

(A) picric acid (B) barbituric acid (C) ascorbic acid (D) aspirin

Ans. [D]

Sol.

OCOCH3

COOH

Aspirin

Q.29 The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius]

(A)20

2

2

4

h

ma(B)

20

2

2

16

h

ma(C)

20

2

2

32

h

ma(D)

20

2

2

64

h

ma

Ans. [C]

Sol. mvr =2

nh

mv =r2

nh

m2v

2=

22

22

r4

hn

2

1m

2v

2=

22

22

r8

hn

2

1

m

vm 22= K.E =

20

22

2

22

z

nm8

hn

a

=20

22

2

nm8

h

a

==

1z2n

=20

2

2

32

h

ma

Q.30 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

Ans. [B]

Sol. 3

5

ONH

+

ON

2+

2

0

N ClHN 4

3


20/36

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.

Q.31 Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as

shown in the given scheme.

(A) C6H5OH and C6H5COOH (B) C6H5COOH and C6H5CH2OH

(C) C6H5CH2OH and C6H5OH (D) C6H5CH2OH and C6H5CH2COOHAns. [B,D]

Sol. Factual

Q.32 Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

(A) Preferential adsorption of ions of 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

Ans. [A,D]Sol. Factual.

Q.33 Which of the following molecules, in pure form, is (are) unstable at room temperature ?

(A) (B) (C)

O

(D)

O

Ans. [B,C]

Sol. Antiaromaticity

Q.34 Which of the following hydrogen halides react(s) with AgNO3(aq) to give a precipitate that dissolves in

Na2S2O3(aq) ?

(A) HCl (B) HF (C) HBr (D) HI

Ans. [A,C,D]

Sol. AgBr + 2Na2S2O3 Na3[Ag(S2O3)2] + NaBr

Sodium argentothio sulphate

colourless soluble

Similar reaction observed in AgCl, AgI


21/36

Q.35 For an ideal gas, consider onlyP-Vwork in going from an initial state Xto the final stateZ. The final stateZ

can be reached by either of the two paths shown in the figure. Which of the following choices(s) is (are)

correct ? [take Sas change in entropy and w as work done]

(A) Sxz = Sxy + Sy z (B) Wxz = Wxy + Wy z

(C) Wxyz = Wxy (D) Sxyz = SxyAns. [A,C]

Sol. Sx z = Sx y + Sy zState

Wx y z = Wx y + Wy z 0

SECTION III : INTEGER ANSWER TYPE

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

inclusive)

Q.36 The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides

are positively charged at pH = 7.0 ?

H3NCHCONHCHCONHCHCONHCHCOO

H R1 R2 H

Peptide R1 R2

I H H

II H CH3

III CH2COOH H

IV CH2CONH2 (CH2)4NH2

V CH2CONH2 CH2CONH2

VI (CH2)4NH2 (CH2)4NH2VII CH2 COOH CH2CONH2

VIII CH2 OH (CH2)4NH2

IX (CH2)4 NH2 CH3

Ans. [4]

Sol. IV, VI, VIII, IX

Four amides are positively charged at pH = 7.0


22/36

Q.37 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 belongs in the periodic

table ?

X++++ H2n6HCu1

1

1

0

1

1

63

29 Ans. [8]

Sol. Z = 29 + 1 2 2 = 26

Fe 8th groups.

Q.38 When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its

pyranose form is -

CHO

CH2

CHOHCHOH

CHOH

CH2OH

Ans. [8]

Sol. 8 stereoisomers possible for given compound in pyranose form of D-configuration.

Q.39 29.2% (w/w) HCl stock solution has a density of 1.25 g mL1

. The molecular weight of HCl is 36.5 g mol1

.

The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is.Ans. [8]

Sol.wt.mol

10dx V = 200 0.4

5.36

V1025.12.29 = 200 0.4

V = 8 ml.

Q.40 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]t[]t[

10/1

8/1 10 ? (take log102 = 0.3)

Ans. [9]

Sol. =10t

t

10/1

8/1

10/a

alog

K

303.2

8/a

alog

K

303.2

10 = 9


23/36

Part III : (MATHEMATICS)

SECTION I (Single Correct Answer Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.

Q.41 The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line

4x 5y = 20 to the circle x2

+ y2

= 9 is

(A) 20(x2

+ y2) 36x + 45y = 0 (B) 20(x

2+ y

2) + 36x 45y = 0

(C) 36(x2

+ y2) 20x + 45y = 0 (D) 36(x

2+ y

2) + 20x 45y = 0

Ans. [A]

Sol.

x2 + y2 = 9(h, k)P

5

204,

4x 5y = 20

A

B

Equation of chord AB is T = 0

x +

5

204y = 9 (i)

& hx + ky 9 = h2

+ k2

9 (ii)

Q Equation (i) & (ii) both represent the same line

Soh

=

k

5

204

=22 kh

9

+

=22 kh

h9

+=

)kh(4

)kh(20k4522

22

+

++

36h = 45k + 20(h2

+ k2)

20(x2

+ y2) 36x + 45y = 0

Q.42 The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that

each person gets at least one ball is

(A) 75 (B) 150 (C) 210 (D) 243Ans. [B]

Sol. G1 G2 G3

1 1 3

1 2 2

+

!2!2!2!1

!5

!2!3!1!1

!53!

= 150


24/36

Q.43 Let f(x) =

=

0x,0

0x,x

cosx2, x IR,

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

Ans. [B]

Sol. f(0 + h) =0h

lim

0h

0h

cosh2

= 0

f(0 h) =0h

lim h

0h

cosh2

= 0

Q f(0+) = f(0) = 0 = finite

So f(x) is differentiable at x = 0

f(2 + h) =0h

lim

h

0h2

cos)h2( 2

+

+

=

f(2 h) =0h

lim h

0h2cos)h2(2

=

Qf(2+) f(2) but both are finite so f(x) is not differentiable at x = 2 but continuous at x = 2

Q.44 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.

Ans. [B]

Sol. Given f : [0, 3] [1, 29]f(x) = 2x3 15x2 + 36x + 1

f'(x) = 6x2 30x + 36

= 6(x 2) (x 3)

f'(x) > 0 if x(0, 2)& f '(x) < 0 if x(2, 3) Function is many one & continuousNow f(0) = 1

f(2) = 29

Range = co-domain


25/36

0 2 3

(2,29)

(3,28)

(0,1)

Hence function is onto.

Q.45 If

+++

bax

1x

1xxlim

2

x= 4, then

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

Ans. [B]

Sol. xlim

+++

1x

bbxaxax1xx 22

= 4

xlim

+++

1x

b1)ba1(x)a1(x 2= 4

As limit is finite so 1 a = 0

a = 1

Nowx

lim

+

+

x

11

x

b1)ba1(

= 4

1 a b = 4

as a = 1 b = 4

Q.46 Let z be a complex number such that the imaginary part of z is nonzero and a = z 2 + z + 1 is real. Then a

cannot take the value

(A) 1 (B)3

1(C)

2

1(D)

4

3

Ans. [D]

Sol. As a is real

So a = a

z2 + z + 1 = 2z + z + 1

(z z ) (z + z +1) = 0

As z is imaginary

So z z 0

z + z + 1 = 0


26/36

z + z = 1 z = x + iy

x =2

1

so a = (x + iy)

2

+ (x + iy) + 1

= (x2 + x + 1 y2) + (2x + 1)yi x = 2

1

a =4

3 y2

so a 0

So a 4

3

Q.47 The ellipse E1 :4y

9x 22 + = 1 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 eccentricity of the

ellipse E2 is

(A)2

2(B)

2

3(C)

2

1(D)

4

3

Ans. [C]

Sol.

(0, 4)

(0, 2) (3, 2)

(3, 0)

(0,2)

(3,0)

Let equation of ellipse E2 is

1b

y

a

x2

2

2

2=+

it passes through (0, 4)

so b2 = 16

and also passes through (3, 2)

So 1b

4

a

922

=+ 14

1

a

92

=+ a2 = 12

as a < bso 12 = 16 (1 e

2)

e2 =4

1 e =

2

1


27/36

Q.48 Let P = [aij]be a 3 3 matrix and let Q = [b ij], 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

Ans. [D]

Sol. Let P =

333231

232221

131211

aaa

aaaaaa

So Q =

336

325

314

235

224

213

134

123

112

a2a2a2

a2a2a2

a2a2a2

Now det (Q) = 222

32

4

332

322

312

232221

131211

a2a2a2

a2a2a2

aaa

= 29

2 22

333231

232221

131211

aaa

aaa

aaa

= 212

det (P)

= 212

2 = 213

Ans. [C]

Sol. I = + dx)xtanx(secxsec

2/9

2

sex x + tan x = t

(sec x tan x + sec2 x) dx = dt


28/36

sec x dx =t

dt

also, sec x tan x =t

1

sec x =2

1

+t

1t

So, I =

+

2/11t

dtt

1t

2

1

=

+ dtt

1

t

1

2

12/132/9

= K

t

1

11

2

t

1

7

2

2

12/112/7

+

= K11

1

7

t

t

1 2

2/11+

+

=

+++

11

1)xtanx(sec

7

1

)xtanx(sec

1 22/11

+ K

Ans. [A]

Sol. Equation of line

=

=

=

1

5z

4

3y

1

2x

General points { + 2, 4 + 3, + 5}

Intersection point with plane

5( + 2) 4(4 + 3) ( + 5) = 1

5 + 10 16 12 5 = 1

12 8 = 0

= 12

8=

3

2

Point

+

++

5

3

2,3

3

8,2

3

2


29/36

P

3

13,

3

1,

3

4

T(2, 1, 4)

Dr's( , 4 + 2, + 1)

S( +2, 4 + 3, + 5)Dr's(1, 4, 1)

Now

+ 4 (4 + 2) + ( + 1) = 0

+ 16 + 8 + + 1 = 0

18 = 9

= 2

1

Points

+++

5

2

1,32,2

2

1

2

9,1,

2

3

Distance at PS =

222

2

9

3

131

3

1

2

3

3

4

+

+

PS =36

1

9

4

36

1++ =

36

1161 ++=

36

18=

2

1

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.


30/36

Ans. [A, C, D]

Sol. tan (2 ) > 0

0 < 2 2

1ve ; x <

2

1+ve

x 1 f(x) = x2

x 1 f '(x) = 2x 1 ve

11/2 0 1/2 1

Ans. [4]

Sol. Let x = ...23

14

23

14

23

14

x2

= 4 23

1x

23 x2

+ x 12 2 = 0


35/36

x =26

212.23.411 ++

x =26

171+=

23

8

6 + log3/2

9

4= 6 + log3./2

2

2

3

= 6 2 = 4

Ans. [9]

Sol. P'(1) = 0, P'(3) = 0

P'(x) = K(x 1) (x 3)

= K(x2

4 x + 3) P'(0) = 3K

P(x) =3

Kx

3 2K x

2+ 3K x +

3

K 2K + 3K + = 6, 9K 18 K + 9K + = 2

34 K + = 6,

34 K = 4

K = 3

P'(0) = 9

Ans. [3]

Sol.2|ba|

rr + 2|cb|

rr + 2|ac|

rr = 9

a.b + b.c + c.a = 3/2 ....(1)

Q2|cba|

rrr++ 0 ....(2)

a.b + b.c + c.a 2

3....(3)

Q from (1) & (3)


36/36

so |cba|rrr

++ = 0

cbarrr

++ = 0

cbarrr

=

on squaring

1 = 2 + 2 cos B

cos B = 2

1 B = b

r^ cr

.

Let T = |2 ar

+ 5br

+ 5 cr

|

= |3 br

+ 3 cr

|

= |cb|3rr

+

= Bcos223 +

= 3

Ans. [4]

Sol. (x 1)2

+ (y 2)2

= ( 5 )2

(2, 0)

y

SQ

P (2, 4)

y2

= 8x

x

2+ 8x 2x 4.2 x2 = 0

x2

+ 6x 8 x2 = 0

x3/2

+ 6x1/2

28 = 0

x1/2

= t

t3 + 6t 28 = 0

(t 2 ) (t2

2 t + 4) = 0

t = 2 x = 2

y = 4

P(2, 4) Q(0, 0) S(2, 0)

Area PQS =2

1 2 4 = 4

Documents

IIT JEE 2012 Paper 1 With Solutions