IIT JEE 2012 Paper 1 Answer Key

8/2/2019 IIT JEE 2012 Paper 1 Answer Key

1/25

~ T R I U M P H ~ I ~[~11~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S =Part I: Physics

Section - I (Single Correct Answer Type)This section contains 10multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.

1. Abi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n ofthe first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius ofcurvature R = 14em. For this bi -convex lens, for an obj ect distance of 40 em, the image distance will be

n = l . 5 . . _ , n -12- .J,

R=14 em(A) -280.0 em (B) 40.0 em (C) 21.5 em (D) 13.3em

Sol: [B] The focal length of each part_ ! _ = (1.5-1) (_1 _ _ ! _ ) = _1 cm'1; 14 00 28_1 = (1.2-1) ( _ ! _ _ _ I ) = _1 cm'1 2 00 14 70Hence equivalent focal length1 1 1 1 1 1 -1-=-+-=-+-=-cm1 1; 1 2 28 70 20

: : : : : ? I;= 20 emFor u = --40 em = -21 => v = 21 = 40 cm

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2/25

~ T R I U M P H ~ I ~[~21~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~2. A thin uniform rod, pivoted at 0, is rotating in the horizontal plane with

constant angular speed OJ, as shown in the figure. At time t= 0, a small Zinsect starts from and moves with constant speed v with respect to therod towards the other end. It reaches the end of the rod at t = T andstops. The angular speed of the system remains O J throughout. The mag-nitude of the torque ( I T I ) on the system about 0,as a function of time isbest represented by which plot?

(A) (B)

iIfll F"

(C) ~_L_ __ "o T

(D)

Sol: [B] The angular momentum ofthe system aboutthe rotational axis isL = 4 r u + ( A fX 2 ) m = 4 r u +J \ 1 ( v t t c o

or L = 4 r u +WaidL 2r=-=2MvoXdt

Hence I T l a t, till the particle reaches the end and stops, and then becomes zero3. Three very large plates of same area are kept parallel and close to each other. They are considered as

ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained attemperatures 2T and 3Trespectively. The temperature of the middle (i.e. second) plate under steady statecondition is

1

(A) ( ~ y T 1(D) ( 9 7 F TSol: [C] The net heat received by the middle plate from the hotter plate is equal to the net heat given to the

cooler plate hence

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3/25

~ T R I U M P H ~ I ~[~31~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~or 2T~ = 8IT4 + 16T4

or _ ( 9 7 ) 1 / 4= : : : ? To - - T24. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface

charge density. The variation of the magnitude of the electric field I E ( r ) 1 and the electric potential VCr )with the distance r from the centre, is best represented bywhich graph?

IE(r)1 V(r) IE(r)1 V(r)1 " , ------.,.. ,, . . . . .f . . . . . .(A) " . . (B) . . ., t " "" . _ . . . ., . . . . . .I

0 R r 0 R r

IEfr}1 V(r)I"I ..., . . . . .

(C) . . . . . . . . .I . . . . . .II0 R r

IE 1 ' ) 1 VCr)

(D)

o R rSol: [D] For a spherical shell

E = 0 ( r < R )E = :~ ( r > R)whereas the electric potentialV = kQ (constant) for r < RR

kQV=- for r >RrThe relevant graph is (D).

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4/25

~ T R I U M P H ~ I ~[=41~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ = = = P = H = Y = S = I = C = S =5. Inthe determination of Youngs modulus ( Y = 4::::) by using Searle's method, awire oflengthL = 2m

and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension 1= 0.25 em in the length ofthe wireis observed. Quantities d and I are measured using a screw gauge and a micrometer, respectively. Theyhave the same pitch of 0.5mm. the number of divisions on their circular scale is 100. The contributions tothe maximum probable error of the Y measurement(A) due to the errors in the measurements of d and I are the same(B) due to the error in the measurement of d is twice that due to the error in the measurement of I(C) due to the error in the measurement of I is twice that due to the error in the measurement of d(D) due to the error in the measurement of d is four times that due to the error in the measurement of I

Here both screw gauge and micrometer used have same least count and hence maximum probable error inmeasurement of f anddwill be same i.e.,

~ C lhus, ~fmax 0.25mm

2x& &0.50 mm 0.25

and hence contributions due to the errors in the measurement of d and f are the same6. A small block is connected to one end of a massless spring of un -stretched length 4.9 m. The other end of

the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretchedby 0.2 m and released from rest at t= O.Itthen executes simple harmonic motion with angular frequency

JrOJ = " 3 rad I s . Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angleof 45 as shown inthe figure. PointPis at a horizontal distance of! 0 m from O. Ifthe pebble hits the blockatt = Is, the value ofv is (take g = 10 m I S 2 )

zo L 4 ~ ? > .-+- 10 III ...p x(A).J50 m/s (B) J5i m/s (C).J52 m/s (D).J53 m/s

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5/25

~ T R I U M P H ~ I ~[=51~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~Sol: [A] The block starts its oscillation from its positive exrtreme and hence its positionxblock from 0 as a

function of time t can be written asXblock =4.9 + 0.2 cos 0J tXblock ( t = I s ) = 4.9 + 0.2 cos(; X I ) = 5.0 mHence range of the pebble R = 10 -5 = 5 m

v2sin2B tr:So = 5 ~ v = v 50 m / sg7. Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The

fringe widths recorded are fJ G' fJ Rand f J B ' respectively. Then,

ADSol: [D] Fringe width, fJ =dnd hence fJ o: ASince A r e d > A g r e e n > A b l u eSo f J r e d > f J g re e n > f J b lu e

8. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. Themass is undergoing circular motion in the x - y plane with centre at 0 and constant angular speed O J . If theangular momentum of the system, calculated about 0 and P are denoted by L a and L p respectively, then

(A) L a and L p do not vary with time(B) L a varies with time while L p remains constant(C) L a remains constant while L p varies with time(D) L a and i;both vary with time

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6/25

~ T R I U M P H ~ I ~[~61~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~Sol: [C] The net force acting on the mass m is mco'r acting towards point 0 and hence net torque about

point 0is zero but about point P is non -zero. Hence L o remains constant while Lp varies with time.

9. A mixture of2 moles of helium gas (atomic mass =4 amu) and 1mole of argon gas (atomic mass = 40

amu) is kept at 300 K in a container. The ratio of the rms speed ( V ~ ~ h e l i u m ] J isVrms argon(A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

1Sol: [D] Both gases are at same temperature and thus vrms oc JM where M is molar mass of the gasvrms(Helium) = MArgan = ~ = 3.16vrms(Argan) M Helium 4

10. Two large vertical and parallel metal plates having a separation of 1em are connected to a DC voltagesource of potential difference X.A Proton is released at rest midway between the two plates. Itis foundto move at 45 to the vertical JUST after release. Then X is nearly(A) 1 x 10-5y (D) 1 x 10-IOy

Sol: [C] As the net force is at 45 to the vertical so

1 em

qEK 4 5 0Mg

qE=mg

or Xe-=mgd

or x = mgd = 1.6x 10-27 X lOx 10-2 _ -9e 1.6x 10-19 - 1x lOY

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7/25

~ T R I U M P H ~ I ~[=71~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~Section-II (Multiple Correct Answer Type)

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out ofwhich ONE or MORE THAN ONE is I are correct.

11. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, -q at(0, - a 1 4,0 ), + 3q at (0,0,0) and -q at (0, +a I 4,0). Choose the correct option(s)

z

y

x

(A) The net electric flux crossing the plane x = + a!2 is equal to the net electric flux crossing the planex = - a12.The net electric flux crossing the plane y = +a12 is more than the net electric flux crossing the planey = - a12.

(B)

(C)(D)

qThe net electric flux crossing the entire region is -&0The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the planex = + a/2.

a aSol: [A, B, C, D] The given distribution is symmetric about planes x = " 2 and z = " 2 . Further the distribu-ation is separately symmetric about planes y = " 2 . The total charge enclosed is q so accordingly all

options are correct.

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8/25

~ T R I U M P H ~ I ~[~81~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S =12. For the resistance network shown in the figure, choose the correct option (s)

12 V

(A) The current through PQ is zero (B) 1 1 = 3A(D) 1 2 =2AC) The potential at S is less than that at Q

Sol: [A, B, C, D] Due to the symmtrypotentials Vp = VQ and Vs = VTSo current through PQ is zero.

QT2n~ __~1=2~V1~ ~

I = 12V = 3A _ 12 V _Further 1 ( 6 1 1 12)n and 1 2 - 6n - 2AAcross each resistor potential decreases by 4V in the directin of current so V s < V Q

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9/25

~ T R I U M P H ~ I ~[~91~--------------------------------------~II=T~JE=E~--2~O~12~A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~13. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle () with the

horizontal. A horizontal force of 1N acts on the block through its center of mass as shown in the figure.The block remains stationary if (take g= 10m/s")

Q

P

(A) ()= 45(B) () > 45 and a fricitional force acts on the block towards P.(e) ()> 45 and a frictional force acts on the block towards Q.(D) () < 45 and a fricitional force acts on the block towards Q.

Sol: [A,C]

The resultant of weight (1 N dowanrd) and applied force (IN) isFnet = ( 1 ) (sin(}-cos(})For () > 45 F is down the incline so friction acts towards Q where as for () < 45 F is up the incline, net , netso friction acts towards P.For () = 45 the block remains at rest without friction

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10/25

~ T R I U M P H ~ I ~[1-0~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ = = = P = H = Y = S = I = C = S =14. C onsider the m otio n o f a po sitive po int c harge in a regio n w here there are sim ultaneo us unifo rm elec tric

and m agnetic fields E = Eo) and B = Eo) .A t tim e t = 0 , this c harge has v elo city v in the x -y plane,m aking an angle () w ith the x axis. W hic h o f the fo llo wing o ptio n( s) is( a re) c orrec tfo r tim e t> O .(A) If () = 0 , the c harge mo ves in a c irc ular path in the x-z plane(B) If ( ) = 0 , the c harge undergo es helic al m otio n w ith c onstant pitc h alo ng the y- axis.(C ) I f ( ) = 1 0, the c harge u ndergo es helic al m o tio n w ith its p itc h in creasin g w ith tim e, alo ng the y- ax is.(D ) I f () =90 , the c harge undergo es linear but ac celerated m otio n alo ng the y-axis.

Sol : [C, D] y--E1 1Fo r 0 ~ () < 90, c harge partic le, undergo es helic al m otio n w ith inc reasing pitch . Fo r () = 9 0, c ha rg eundergo es lin ear b ut ac celerated m o tio n alo ng the y -ax is.

15. A perso n blo ws into o pen-end o f a lo ng pipe. As a result, a high-pressure pulse o f air travels down thepipe. W hen this pulse reaches the o ther end o f the pipe(A) a high-pressure pulse starts traveling up the pipe, if the o ther end o f the pipe is o pen.(B ) a lo w-pressure pulse starts traveling up the pipe, if the o ther end o f the pipe is o pen(C) a lo w -pressure pulse starts traveling up the pipe, if the o ther end o f the pipe is c lo sed(D ) a high-pressure pulse starts traveling up the pipe, if the o ther end o f the pipe is c lo sed.

Sol : [B,D ] Pressure w av e undergo es a phase c harge o f J[ rad o n refletio n fro m o pen end and no phase c hangeo cc urs o n reflec tio n fro m c lo sed end. A cco rdingly B , D are c orrec t.

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11/25

~ T R I U M P H ~ I ~[1-1~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~Section - III (Integer Answer Type)

T h is sec tio n c o ntain s 5 questions. T he answer to eac h o f the questio ns is a single-digit integer, ran ging fromo to 9 . (b oth e nc lu siv e)16. A ninfinitely lo ng so lid c ylinder o f radius R has a unifo rm v olum e c harge density p. Ith as a sp heric al c av ity

o f r ad ius Rl2 w ith its c entre o n the ax is o f the c ylinder, as sho w n in the figure. T he m agnitude o f the elec tricfield at the po int P, which is at a distanc e 2R from the axis o f the cylinder, is given by the expressio n23pR1 6k& o . Thevalueofkis

z

p

IIX I It==)Sol. [6] L et us assum e that the spheric al c avity is filled w ith sam e c harge density then

~ ~ ~E = E comple te cylinder - E spherical cavity

4 R3= p1CR2f1 _ 31Cg

2m>of l X 2R 4m> o (4R)2pR [ 1 1] 23pR 23pR= ~ 4 - 96 = 96 Eo = 16k ~ 0

= : : : ? k = 6

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12/25

~ T R I U M P H ~ I ~[1-2~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~17 A c ylind ric al c av ity o f d iam eter a ex ists inside a c ylin der o f d iam eter 2a as sho wn in the figure. B oth the

c ylin der a nd th e c av ity a re in fin ite ly lo n g. A u nifo rm c urre nt d en sity J flo w s alo ng the leng th. If the m agni-Ntude o f the magnetic field at the po int P is giv en by 12 J 1 oaJ, then the value o fN is

Sol: [5] L et us assume that c ylind ric al c av ity is also c arrying c urrent w ith sam e c urrent density~ ~ ~Bnet = Bcomplet ecylind er - Bcavity

or

=).l Ja [ _ ! _ - _1 ]_ 5 ).loJaa 2 12 - 12

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13/25

~ T R I U M P H ~ I ~[1~3~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~18. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and

radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through 0 andI pPis 1 0 and Ip, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio Io

to the nearest integer is

p

Sol: [2] Let us assume that the smaller disk is also filled with same surface mass density. Then as mass ispropotional to area so if

mm bO = m then m SD = 4

I = m(2R)2 _ m [R2 + R2] = mR2 ( 2 _ l ) = 1 3 mR2o 2 4 2 8 8

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14/25

~ T R I U M P H ~ I ~[1~4~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ = = = P = H = Y = S = I = C = S =19. A circular wire loop ofradius Ris placed in the x-yplane centered at the origin O. A square loop of side

a(a < < R) having two turns is placed with its center at z = J 3 R along the axis ofthe circular wire loop,as shown in figure. The plane of the square loop makes an angle of 45 with respect to the z-axis. If the

a2mutual inductance between the loops is given by ~~2 ,then the value of p is2 R

z

t- ._. -J 3 R

x

Sol: [7] The mutual inductanceNs(BeAs cos 45)M= .

IeHere Ns = no ofturns in square coil and Bebe the magnetic field of circular coil at the position of squarecoil.

p=7

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15/25

~ T R I U M P H ~ I ~[1~5~1---------------------------------------~I~IT~J~E~E---20~1~2A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = = = P = H = Y = S = I = C = S = = ~20. A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic

charge. Itmakes a closest approach of 10 fin to the nucleus. The deBroglie wavelength (in units of fin) ofthe proton at its start is (take the proton mass, mp =(SI3)x IO-27 kg; h/e=4.2xIO-15 J.s/e;

1 9--=9xIO m/F I fin=IO-15m)4XEo '

Sol: [7] QG----+.At closest approachf.,.Ue =-f.,.KkQe 2P h--=-=--

A_- fh\:: _h ~v~ eV ~=42 10-15 IOxIo-15 x3. x 27 9

2x5xIO- x9xIO xI20= 4.2 X 10-15 x 10 = 7fm

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16/25

alTRIUMPHI~~[~11~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = YPart II

Section-I : Straight Correct Answer TypeThis section contains 10multiple choice questions. Each question has four choices (A), (B), (C) and (D), outof which ONLY ONE is correct.

21. AsperIUPACnomenclaure, the name of the complex [Co(H20)4(NH3)2]Cl3 is(A) Tetraaquadiaminecobalt (llI) chloride(C) Diaminetetraaquacoblat (Ill) chloride

Ans. (D)Sol. Diamminetetraaquacobalt(llI) chloride

(B) Tetraaquadiamminecobalt (llI) chloride(D)Diamminetetraaquacobalt (llI) chloride

22. Inallene (C3H4), thetype(s) of hybrid isation oft he carbon atoms is(are)(A) sp and Sp3 (B) sp and Sp2 (C) only Sp2

Ans. (B)

23. For one mole of a vander Waals gas when b = 0 and T = 300 K, the PV vs. 1 / V plot is shown below.The value of the van der Waals constant a (atm. liter' mol") is

.:- 24.6~ 23.1a 21.6~ 20.1~'-; : . . .a,

[Graph not to scale]

o

(A) 1.0Ans. (C)

(B) 4.5 (C) 1.5Created with

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17/25

alTRIUMPHI~~[~21~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = YSol. Vander Walls equation for 1mol of gas

(P+~2}V-b)=RT

givenb=O(p+ ~2)V = RT

apV +-= RTVapV=RT-- V

1PV vs V graph is a straight line have RT intercept and -a as slope_ 20.1- 21.6 _ 15Slope - - - .3-2

so a = 1.5

24. The number of optically activeproducts obtained from the complete ozonolysis of the given compound isCH3 H, 1

CH3- CH::;CH - y - CH = CH- 9-CH = CH-CH3:: . .H CH3

(A) 0Ans.(A)

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

Sol. Ozonolysis products areCH3CHO and OHC-CH(CH3)CHO

none is optically active

25. A compound MpXq has cubic close packing (ccp) arrangement ofX. Its unit cell structure is shownbelow. The empircial formula fo the compound is

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18/25

alTRIUMPHI~~[~31~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = YSol. 1particle ofM is located at bodycentre and 4 are on edge centre so total particle ofM is

1-x4+1xl=24X is located at all comer and face centre

1 1-x8+-x6=48 2

26. The number of aldol reaction( s) that occurs inthe given transformation is

CH3CHO+ 4HCHO. c o ne. ag. NaOH) E . [ < > HH~OH

(A) 1ADs. (C)

(B) 2 (C) 3 (D) 4

Sol. CH 3CH0 has 3a - hydrogenatoms. First three steps areAlodol reactions. Last step is cannizaro reaction.

27. The colour oflight abosrbed by an aqeuous solution of CuSO 4 is(A) oragne-red (B) blue-green (C) yellow

ADs. (A)(D)violet

Sol. The complenentary colour of organe-red is blue

28. The carboxyl functional group (-COOH) is present in(A) picric acid (B) barabituric acid (C) ascorbic acid

ADs. (D)(D) aspirin

Sol.

COOH~OCOCH3Aspirin l S : 2 J

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

ADs. (C) Created withnitroPDF' p ro fess i ona ldownload the free trial online at nttrcpdf.eomzprofees onaI


19/25

alTRIUMPHI~~[~41~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = C = H = = E = M = I = S = T = R = = Y ~Sol. according to Bohr's postulate

nhmvr=- 2nsquaring both side

n2h 2m2v2r2=--(2n)21 2 n2h2- mv = - :- - -- ,- -2 8n2m r2

placing r2 = 16a~ for n = 2

30. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?(A) HN03, NO, NH4Cl, N2(C) HN03, NH4Cl, NO, N2

ADs. (B)

(B) HN03, NO, N2, NH4Cl(D) NO, HN03, NH4Cl, N2

Sol. In HN03 oxidation state ofN = + 5NO oxidation state ofN = + 2NH4Cl oxidation state ofN =_ 3N 2oxidation state ofN = 0So order RN03, NO, N 2' NH4Cl

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alTRIUMPHI~~[~51~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = C = H = = E = M = I = S = T = R = = Y ~

Section-IIMultiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question four 4 choices (A), (B), (C) and (D), out ofwhich ONE or MORE are correct.

31. Identify the binaryrnixturef s) that can bse separated into individual compound, by differential extraction,as shown in the given scehem.

BinalYmixture containi~g _jC om pound 1 and Com pound 2 ~

1'--__ N _ a h _ _ c _ O a . . ; ; . , . . . . .a....) , - - + I Compound 1 ] + I Compound 2N aO H (aq) Compound 1 I + I Compound2

(A) C6HsOH and C6HsCOOH(C) C6HsCH20H and C6HsOH

Ans. (B, D)

(B) C6HsCOOH and C6HsCH20H(D) C6HsCH20H and C6HsCH2COOH

Sol. Both these mixtures can be separated by NaOH( aq] as well as NaHC03 (aq)

32. Choose the correct reason(s) for the stability ofthe lyophobic colloidal particles.(A) Preferential adsorption ofions on their surface from the solution(B) Preferential adsorptionof 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 particlesAns. (A, D)Sol. Preferential adsorption of ions and potential difference betwen fixed and diffused layer of oppsite charges.

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

( A l O -c o(C) ()Ans. (B, C)Sol: Both are antiaromatic.

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alTRIUMPHI~~[~61~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = Y34. Which of the following hydrogen halides react( s) with AgN 3 (aq] to give a precipitate that dissolves in

Na2S203(aq) ?W~ ~il l ~~ ~m

ADs. (A, C)Sol. Bothe AgCl and AgBr precipitates dissolve in sodium thisulphate solution.

35. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The fmal stateZ can be reached by either of the two paths shown int he figure. Which of the following choice(s) is( are)correct?[taks ~S as change in entropy and w as workd done]

x y

z

V (liter)

(B) Wx-o z = Wx -oy +Wy-->z(C) w x-->y-->zWx -ey

ADs. (A,C)Sol. Entropy is state function

so ~Sx-->z= ~Sx-->y+ ~Sy-->zand Wx-->y-->zWx -eybecause process y ~ z is constant volume so work done is zero.

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alTRIUMPHI~~[~71~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ i f ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ; = C = H = = E = M = I = S = T = R = = Y ~

Section-IIIInteger Answer Type

This section contains 5 questions. The answer to each is a single-digit integer, ranging from 0 to 9 (bothinclusive).36. The substitutents Rl and R2 for nine peptides are listed int he table given below. How many of these

peptides are positively charged at pH = 7.0?e . eH N-CH-CO-NH CH-CO-NH-CH-CO-NH-CH-COO~ . I I I IH R, R 2 H

Peptide Rj R2I H HII H CH3III CH2COOH HIV CH2CONH2 (CH2) 4NH2V CH2CONH2 CH2CONH2VI (CH2) 4NH2 (CH2) 4NH2VII CH2COOH CH2CONH2VIII CH20H (CH2) 4NH2IX (CH2) 4NH2 CH3

Ans. (4)37. The periodic table consists of 18groups.An isotope of copper, on bombardment with protons, undergoes

a nuclearreaction yielding element X as shown below. Towhich group, element X belongs int he periodictable?~~Cu+;H ~ 6 ~n+a+2 ;H+X

Ans. (8)Sol. ~~Cu+ ;H~6~n+a+2 ;H+X

mass no of reactant side63+ 1= 64

mass no of product side6 x 1 + 4 + 2 x 1 +M,

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alTRIUMPHI~~[~81~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = Yatomic no ~

29 + I = 2 + 2 + ZxZx =26

group no = 838. When the following aldohexose exits inits D-configuration, the total number of stereoisomers inits pyranose

form isCHOICH2ICHOHICHOHICHOHICH20H

ADs. (8)Sol. Three chiral centres 23 = 8

39. 29.2% (w/w)HCI stock solution has a density of 1.25gmL-I. ThemolecularweightofHClis 36.5 g mol ".The volume (mL) os stock solution required to prepare a 200 mL solution ofOA M HCI is

ADs. (8)Sol. 29.2% (w/w) ofHCI solution means

29.2 g ofHCI = 100 g of solution100g ..= 1.25 mL of solution (given d = 1.25g I mL )

Now 200 mL ofOA M HCI is to be preparedmeans = 0.08 mol= 2.92 g ofHCI= 8mL of given solution

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alTRIUMPHI~~[~91~--------------------------------------~II~T~JE~E~-~2~O~12A C ~ ~ ~ ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ I = C = H = E = M = I = S = T = R = Y40. An organic compound undergoes first-order decomposition. The time taken for its decomposition to 118

and 1I 10 of its intial concentration are t 11 8 and t 1110 respectively. What is the value of [[t 118 ]] X 10?t 1110(take lcg., 2 = OJ)

ADs. (9)1/8aoSol. t1l8 x k = -2J03log-- a o1/ r o ,t1l10 x k = -2J03log---"- a o

~=0.9t1110~xl0=9t1l10

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MATHEMATICS41. The locus of the mid-point of the chord of .Ans: [A]42. The total number of ways in which 5 balls 0 .Ans: [B]

43. x:;t:O IR,XEx= O

Ans: [B]44. The function f: [0, 3] ---j- [1,29] , defined .Ans: [B]

45. If lim ( x2 + X + 1 ax - bI4 thenx-e-col x +1 )Ans: [B]

46. Let z be a complex number such that the .Ans: [0]

2 247. The ellipse E, : ~ +L1 is inscribed .9 4Ans: [e]

48. Let P =[aii] be a 3 x 3 matrix and let Q = [ b jj], wherebjj = i+jajj for 1:S;i, js 3 .

Ans: [0]2. f sec x49. The mtegral 9/2 dx equals .(sec x-r tan x)

Ans: [e]50. The point P is the intersection of the striaght line join-

ing the points Q(2, 3, 5) and R(I, -1,4) .Ans: [A]51. Let 8, < p E[O, 2 r c ] be such that .Ans: [A,e, 0]

52. Let S be the area of the region enclosed byy = e _x2 , y = 0, x = 0, and x = 1 .Then

Ans: [A,B, 0]53. A ship is fitted with three engiens Ej' E2and E3 The

engines function independently of each other with re-spective probabilities

Ans: [B,O]2 254. Tangents are drawn to the hyperbola x 9 - Y 4 = 1 , paral-

lel to the straight line 2x - Y=1.The .Ans: [A,B]55. If y(x) satisfies the differential equation

y' - ytanx =2xsec x and yeO)=0, thenAns: [A,0]56. Let f : IR- - -7IR be defined as f ( x ) = l x l + l x 2 - 1 1 . The to-

tal number of points at which .Ans: [5]

57. The value of 6+10gi[!rz 4- 3 ! r z ~ 4 - 3 ! r z ) 4 - 3 ! r z ]Ans: [4]

58. Let p( x) be a real polynomial ofleast degree whichhas a local maximum at x = I and a local minimum ....Ans: [9]59. If - b -a, and c are unit vectors satisfying

1 ii- b 1 2 + 1 b - C1 2 + 1 c - ii 2 = 9 , then .Ans: [3]

60. Let Sbethe focus of the parabola l= 8x andletPQbe the common chord .

Ans: [4]40. An organic compound undergoes first-order ...ADS. (9)

IIT JEE 2012 Paper 1 Answer Key

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Transcript of IIT JEE 2012 Paper 1 Answer Key