PHYSICS TEAM - Motion IIT- · PDF filePHYSICS TEAM IIT-JEE 2012 PAPER-I WITH SOLUTION ......

394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671

IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected] 1

PHYSICS TEAM

IIT-JEE 2012 PAPER-I WITH SOLUTION [CODE - 7]

Supported by a pool of 20 Faculty Members

Nitin VijayEx-Sr. Faculty Bansal Classes

B.Tech IT-BHU

Amit Verma Ex-Sr. Faculty Bansal Classes

M.Sc. Gold Medalist

Nirbhay Pandey Ex-Faculty Bansal Classes

Ravi Vijay Sr. Faculty Physics

New

Kumar Sourabh Ex-Sr. Faculty Vibrant Academy

Ex-HOD Career point

New

Bhupendra Sudan Ex-Sr. Faculty Vibrant Academy

Ex-Sr. Faculty Bansal Classes

SECTION – A

Single Correct

1. 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 βG,βR andβB, respectively. Then

(A) βG > βB > βR (B) βB > βG > βR

(C) βR > βB > βG (D) βR > βG > βB

Sol. Fringe width

d

Pλ=β

as (λR > λ

G > λ

b)

so βR > β

G > β

B

2. Two large vertical and parallel metal plates having

a separation of 1 cm are connected to a DC

voltage source of potential difference X. A proton

is released at rest midway between the two plates.

It is found to move at 45° to the vertical JUSTafter release. Then X is nearly

(A) 1 × 10–5 V (B) 1 × 10–7 V(C) 1 × 10–9 V (D) 1 × 10–10 V

Sol. Fx = Fy ⇒ qE = mg

qv

d = mg

45°

Fx

Fy

45°

⇒ V = mgd

q= 10–9 v

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

= 40 amu) is kept at 300 K in a container. The

ratio of the rms speeds ( )( )

rms

rms

v helium

v argon

is

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

Sol.M

RT3vrms =

He Ar

Ar He

v M 4010 3.16

v M 4= = = =

4. 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) isfixed. The system lies on a horizontal frictionless

surface. The block is stretched by 0.2 m and

released from rest at t = 0. It then executessimple harmonic motion with angular frequency

ω = 3

πrad/s .Simultaneously at t = 0, a small

pebble is projected with speed ν from point P atan angle of 45° as shown in the figure. Point P is

at a horizontal distance of 10 m from O. If the

pebble hits the block at t = 1s, the value of ν is(take g = 10 m/s2)

45°

10m PxO

Z

(A) 50 m/s (B) 51 m/s

(C) 52 m/s (D) 53m/s

Sol. T = 1 S = g

sinu2 θ

u = g

2sinθ =

10

12

2× = 5025 =

5. Three very large plates of same area are keptparallel and close to each other. They areconsidered as ideal black surfaces and have veryhigh thermal conductivity. The first and thirdplates are maintained at temperatures 2T and3T respectively. The temperature of the middle(i.e. second) plate under steady state conditionis

(A)

1

465

2

T (B)

1

497

4

T

(C)

1

497

2

T (D) ( )1

497 T

PART - I [PHYSICS]



Sol.

2T T = ?0 3T

for middle flate (in unit time)Heat absorbed/ Area = Heat emitted/Areaσ (2T)4 + σ(3T)4 = σ(T

0)4×2

T0 = T

2

974/1

6. A thin uniform rod, pivoted at O is rotating in thehorizontal plane with constant angular speed ω,as shown in the figure. At time t = 0, small insectstarts from O and moves with constant speed νwith respect to the rod towards the other end.it reaches the end of the rod at t = T and stops.The angular speed of the system remains ω

throughout. The magnitude of the torque ( )τ

on

the system about O, as a function of time isbest represented by which plot?

O

Z

(A)

0 tT

(B)

0 tT

(C)

0 tT

(D)

0 tT

Sol. τ = dt

dL = )I(

dt

dω

τ = dI

dtω = ω )II(

dt

dmrod +

as Irod

= com ⇒ τ = ( )insec t

wdI

dt

= )mr(dt

d 2ω = mω dr

2rdt

= 2m rωv

= 2m(vt)ωv ⇒ τ ∝ t

7. In the determination of young's modulus

4MLgY

d2

=

π by using Searle's method, a wire of

length L= 2 m and diameter d = 0.5 mm is used.For a load M = 2.5 kg, an extension l = 0.25 mmin the length of the wire is observed. Quantitiesd and l are measured using a screw gauge and amicrometer, respectively. The have the samepitch of 0.5 mm. The number of divisions on theircircular scale is 100. The contributions to themaximum probable error of the Y measurement(A) due to the errors in the measurements of dand l are the same(B) due to the error in the measurement of d istwice that due to the error in the measurementof l.(C) due to the error in the measurement of l istwice that due to the error in the measurementof d.(D) due to the error in the measurement of d isfour times that due to the error in themeasurement of l.

Sol. 2d

MgL4Y

π=

d

d2

L

L

g

g

M

M

Y

Y ∆+

∆+

∆+

∆+

∆=

∆

d∆=∆ (same instrment)

so d

d2∆=

∆

hence both contribute same in

error (Hence)8. Consider a thin spherical shell of radius R with its

centre at the origin, carrying uniform positive

surface charge density. The variation of the

magnitude of the electric field ( )E r

and the

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

(A)

( )rE

V(r)

0 R r

(B)

( )rE

V(r)

0 R r

(C)

( )rE

V(r)

0 R r

(D)

( )rE

V(r)

0 R r



Sol. Inside shell E = 0& V = const.and outside shell

E ∝ 2r

1 & v ∝

r

1

9. A bi-convex lens is formed with two thin plano-

convex lenses as shown in the figure. Refractiveindex n of the first lens is 1.5 and that of the

second lens is 1.2. Both the curved surfaces areof 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

n=1.5 n=1.2

R =14cm(A) –280.0 cm (B) 40.0 cm

(C) 21.5 cm (D) 13.3 cmSol. Left lens

1

1 1 1 1(1.5 1) –

f 14 28

= − =

∞

Right lens

2

1 1 1 1(1.2 1)

f 14 70

= − + =

∞

1 2

1 1 1 1 1 1

feq f f 28 70 20= + = + =

feq = 20 cm

1 1 1

v u f− = ⇒

20

1

40

1

v

1=+ ⇒ v = 40 cm

10. A small mass m is attached to a massless string

whose other end is fixed at P as shown in the

figure. The mass is undergoing circular motion inthe x-y plane with centre at O and constant

angular speed ω.If the angular momentum of thesystem, calculated about O and P are denoted

by 0L

and PL

respectively, then.

m

P

O

z

(A) 0L

and PL

do not vary with time

(B) 0L

varies with time while PL

remains constant

(C) 0L

remains constant while PL

varies with time

(D) 0L

and PL

both vary with time.

Sol. L0 remains cons. in magnitude and direction but

LP changes its direction continously hence L

P is

variable

L0

v x

L (varies direction)P

SECTION – BMultiple Correct

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

Sol. At closed end prssure doesn't change phase andat open end the phase is reversed.

12. A small block of mass of 0.1 kg lies on a fixed

inclined plane PQ which makes an angle θ with

the horizontal. A horizontal force of 1 N acts onthe block through its center of mass as shown inthe figure. The block remains stationary if (takeg = 10 m/s2)

(A) θ = 45o

(B) θ > 45o and a frictional force acts on the

block towards P.

(C) θ > 45o and a frictional force acts on the

block towards Q.

(D) θ < 45o and a frictional force acts on the

block towards Q.

* ALL INDIA OPEN TEST For AIEEE Aspirants 2012

* Open Scholarship cum Admission TestMotion

For X to XI & XI to XII Moving Students

upto 90% Scholarship for Motion Classroom Programs

Date : 18 April 2012

Time : 3:00 pm to 6:00 pm

th

REGISTRAT

ION

OPEN

9 States | 30 Cities | 18 April : Are you ready for the Challenge ??? th



Sol.

Q

P

at θ = 45° No friction is requiredat θ > 45° , sin θ > cos 45°hence friction is required towards Qat θ < 45°, cos θ > sin θhence friction acts towards P

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

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

the plane x = -a/2.(B) The net electric flux crossing the plane y =+a/2 is equal to the net electric flux crossingthe plane y = -a/2.(C) The net electric flux crossing the entire region

is 0

q

ε

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

Sol. For the charge distribution

for x = +a/2 and x = -a/2flux is symmetric & same

also flux is same throughy = +a/2 and y = -a/2

for net flux 0

00

in /qqqq3q

∈=∈

−−=

∈=φ

z = +a/2 and x = +a/2 aresimilar so same flux.

14. For the resistance network shown in the figure,

choose the correct option(s).

Ω

Ω

Ω

Ω Ω

ΩΩ

Ω

2I

1I

(A) The current through PQ is zero

(B) l1 = 3A(C) The potential at S is less than that at Q(D) l2 = 2A

Sol. By symmetry the 1Ω resistances don't get any

current so circuit reduces to

126

126

12I1

+

×=

= 3A

15. Consider the motion of a positive point charge in

a region where there are simultaneous uniform

electric and magnetic fields E= E0 j and B

= B0

j . At time t = 0. At time t = 0, this charge has

velocity v in the x - y plane, making and angle

θ with the x-axis. Which of the following option(s)

is(are) correct for time t > 0 ?

( A ) I f θ = 0o, the charge moves in a circular

path in the x-z plane.

(B) If θ = 0o, the charge undergoes helical motion

with constant pitch along the y-axis.

(C) If If θ = 10o, the charge undergoes helical

motion with its pitch increasing with time, along

the y-axis.

(D) If If θ = 90o, the charge undergoes linear

but acclerated motion along the y-axis.Sol. (i) The path of the particle will be helical with

increasing pitch in x-z plane .

:: Corporate Office ::394 - Rajeev Gandhi Nagar, Kota ( Rajasthan)Ph. No. : 93141-87482, 0744-2209671www. motioniitjee.com , [email protected]

:: Other Study Centers ::

Bilaspur, Dhanbad, Digboi (Assam), Gwalior,

Jodhpur, Modasa(Gujrat), Nagpur



B0

YE0

x

(ii) if Q = 0 no effect due to B. but due to E there is a

force along E which Accelerates if along +y direction so

speed will increase in +y direction

SECTION – CInteger Answer Type

16. A proton is fired from very far away towards a

nucleus with charge Q = 120 e, where e is theelectronic charge. It makes a closest approach

of 10 fm to the nucleus. The de Broglie wave-

length (in units of fm) of the proton at its startis: (take the proton mass, mp = (5/3) x 10

-27

kg; h/e = 4.2 x 10-15

J.s/C; o4

1

πε= 9 X 10

9 m/F;

1 fm = 10-15

m)

Sol. KE = PE ⇒r

qkqmv

2

1 212 =

15

92

1010

e120e109

m2

p−×

×××=

& p

h=λ (from debroglie)

solving λ = 7 x 10-15 = 7 fm

17. A lamina is made by removing a small disc ofdiameter 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 O and P is Io and IP, respec-tively. Both these axes are perpendicular to the

plane of the lamina. The ratio o

P

I

I to the nearest

integer is

Sol. 0

M/4

MP

Let σ be the density off disc.

∴ 22 R4

M

)R2(

M

π=

π=σ

Here M → Mass of disc without cavity

∴ Mass of cavity = σ x πR2 = M/4πR2 x πR2 = M/4

Io = MI of disc with non cavity

- MI of cavity (About O)

+

−

= 222

o R4

MR

4

M

2

1)R2(M

2

1I

222

o R4

MMR

8

1

2

MR4I −−= =

8

MR3

8

MR16 22

−

= 8

MR13 2

Ip = M.I. of Disc without

cavity about P - M.I. of 0 2R

5 R

R

cavity(aboutP)

= ( )

+

−

+

2222 R5

4

MR

4

M

2

1)R2(M)R2(M

2

1

8

MR37I

2

P =

38.213

37

I

I

o

P ≈≈=∴

18. An infinitely long solid cylinder of radius R has a

uniform volume charge density ρ . It has a spheri-

cal cavity of radius R/2 with its centre on the

axis of the cylinder, as shown in the figure. The

magnitude of the electric field at the point P,which is at a distance 2R from the axis of the

cylinder, is given by the expression o16

R23

κε

ρ. The

value of κ is







th

REGISTRAT

ION

OPEN




Sol. For cylinder λ = 2M Rπ ρ

=

= πR2ρ

at E = Ecylinder

+ Esphere

λ = ρ × πR2 = ( )

2

2

2k kQ

2R 2R

λ+

= ( )

3

2

2

4 Rk

2k R 3 8–2R 2R

× π × ρπ ρ

= π kRρ 1

1–24

= 0 0

R 23 23R

4 24 16 6

π ρ ρ× =

πε ε ×

so k = 6

19. A cylindrical cavity of diameter a exists inside a

cylinder of diamter 2a as shown in the figure.Both the cylinder and the cavity are infinitely

long. A uniform current density J flows along thelength. If the magnitude of the magnetic field at

the point P is given by 012

Nµ aJ , then the value

of N is

Sol. B = B1( 2a dimeter) – B

2(a diameter)

a2

)aJ( 2

π

π×µ –

2

a32

)4/aJ( 2

×π

π×µ

= aJ.12

5µ = N = 5

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

of side a(a<<R) having two turns is placed with

its center at z = 3 R along the axis of the

circular wire loop, as shown in figure. The plane

of the square loop makes an angle of 45o with

respect to the z-axis. If the mutual inductance

between the loops is given by R2

a2/p

20µ

, then value

of p is

R3

o45

Sol. At the position of square loop

B =

20

2 2 3/2

iR

2(R 2R )

µ

+ =

0i

16R

µ

α = BA cos θ

= 20i 1

2a16R 2

µ× ×

=

20 ia

8 2R

µ

M = i

θ =

20a

8 2R

µ =

20

7/2

a

2 R

µ

P =7

KNOW YOUR RANK & GET DETAILED VIDEO SOLUTIONS OF IIT-JEE 2012 PAPER

@ www.motioniitjee.com



R.R. Dwivedi Sr. Faculty Chemistry

B.Tech. IT-BHUExp. 12 yrs

V.P. Singh Ex-HOD Chemistry Bansal Classes

Exp. 12 yrs

G.K. Gupta Ex-Sr. Faculty Vibrant AcademyEx-Sr. Faculty Bansal Classes

Exp. 12 yrs

Mohd. Asif Ex-Sr. Faculty Vibrant Academy

Ex-Sr Faculty Bansal ClassesExp. 11 yrs

Anurag Garg Ex-Sr. Faculty Bansal Classes

Exp. 8 yrs

Alok Kumar Ex-Sr. Faculty Bansal Classes

IT-BHU,Exp. 8 yrs

Pramod Singh Sr Faculty Chemistry

Exp. 5 yrs

New

New

CHEMISTRY TEAMSupported by a pool of 28 Faculty Members

PART - II [CHEMISTRY]

SECTION – A

Single Correct

21.The colour of light absorbed by an aqueous solutionof CuSO4 is

(A) orange-red (B) blue-green

(C) yellow (D) violet

Sol. A

B

G

V

R

O Y

munshel wheel

⇒ Since CuSO4 appears blue-green it would have

absorbed organe-red.

22.Which ordering of compounds is according to thedecreasing 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

Sol. B

HNO3 O.S. of N = +5

NO O.S. of N = +2

N2 O.S. of N = 0

NH4Cl O.S. of N = –3

Decreasing order

HNO3, NO, N2, NH4Cl

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

(A) 2

0

2

2

ma4

h

π (B) 2

0

2

2

ma16

h

π

(C) 2

0

2

2

ma32

h

π (D) 2

0

2

2

ma64

h

π

Sol. C

mvr = π2

nh ⇒ m2v2r2 = 2

22

4

hn

π

21mv2 =

2 2 2

2 2 2

0

n h Z

4 2 m(r n )

×

π × × = 2

0

2

2

ma32

h

π

24.The number of aldol reaction(s) that occurs in thegiven transformation is

CH3CHO + 4HCHO conc.aq. NaOH

OHOH

OHOH

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

Sol. C

OH–

–H O2CH CHO3 CH – C – H2

H–C–H

O

O

CH – CH – C – H2 2

O–

H OH

O

CH –CH–C–H2

OH

H

O

2 times aldol

CH2–C – C – H

CH2OH

CH2OH

CH2OH

O

OH,HCHO

CH2HO

HO

HO–CH –C–CH –OH2 2

H – C – O

+

Ocannizaro

25.For one mole of a van der Waals gas when b = 0 andT = 300 K, the PV vs. l/V plot is shown below. The valueof the van der Waals constant a

(atm. liter2 mol–2) is

20.121.6

23.1

24.6

pV (lite

r-atm

mol

)–1

0 2.0 3.01v (mol liter )

–1

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

Sol. C

Since b = 0

2

2

anP

v

+

(v) = nRT

PV = nRT – 2an

v

∴ slope = – an2 = – an2 = 15.1

a = 1.5



26.In Allen (C3H4), the type (s) of hybridisation of thecarbon atoms is (are)

(A) sp and sp3 (B) sp and sp2

(C) only sp2 (D) sp2 and sp3

Sol. B

In allene (C3H4)

H C = C = C H 2 2

sp2

spsp2

27.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

M =

X =

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

Sol. B

X = 8 × 81

+ 6 + 21 = 4

M = 4 × 41

+ 1 = 2

M =

X =

∴ formula of unit cell

= M2X4

& expirical formula of compound = MX2

Ans. – (B)

28.The number of optically active product obtained fromthe complete ozonolysis of the given compound is

CH –CH=CH–C–CH=CH–C–CH=CH–CH33

CH3

CH3

H

H

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

Sol. A

CH –CH=CH–C–CH=CH–C–CH=CH–CH33

H

CH3

CH3

H

ozo.

2CH3CH=O + O = CH – C – CH = O

CH3

H

+

O = CH – C – CH = O

CH3

H

All products are optical inactive.

29.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

Sol. D

[Co(H2O)4(NH3)2]Cl3Diamminetetraaquacobalt (III) chloride

30.The carboxy functional group (–COOH) is present in

(A) picric acid (B) barbituric acid

(C) ascorbic acid (D) aspirin

Sol. D

(A)Picric acid

OH

NO2NO2

NO2

(B) Barbituric acid

OO

O

NH NH

(C) Ascorbic acid

O O

OHHO

HOHO

(D) Aspirin

O – C – CH3

O

C – OH

O



SECTION – A

Multiple Correct

31.For an ideal gas, consider only P-V work in goingfrom a initial state X to the final state Z. The final stateZ can be reached by either of the two paths shown inthe figure. Which of the following choice (s) is (are)correct ? [take ∆S as change in entropy and w as workdone]

Z

YX

P(a

tmosp

here

)

V(liter)

(A) x z x y y zS S S

→ → →∆ = ∆ + ∆

(B) x z x y y zw w w

→ → →= +

(C) yxzyxww

→→→=

(D) yxzyxSS

→→→∆=∆

Sol. A,C

Explanation → Entropy is a state function andindependent of path taken.

A l s o , W y–z = 0 (Isochoric)

32.Which of the following molecules in pure from is (are)unstable at room temperature

(A) (B)

(C)

O

(D)

O

Sol. B

It is anti-aromatic compound. It dimerise at roomtemperature.

+dimerise room temperature

33.Identify the binary mixture(s) that can be separatedinto individual compounds, by differential extraction, asshown in the given scheme.

Binary mixture containing Compound 1 and compound 2

Compound 1 Compound 2+

+Compound 1 Compound 2

NaOH(aq)

NaHCO (aq)3

(A) C6H5OH and C6H5COOH

(B) C6H5COOH and C6H5CH2OH

(C) C6H5CH2OH and C6H5OH

(D) C6H5CH2OH and C6H5CH2COOH

Sol. B,D

NaOH NaHCO3

Ph–OH

Ph–COOH

Ph–CH –OH2

Ph–CH –COOH2

+ve

–ve

=

=

34.Choose the correct reason(s) for the stability of thelyophobic colloidal particles.

(A) Preferential adsorption of ions on their surface fromthe solution

(B) Preferential adsorption of solvent on their surfacefrom the solution

(C) Attraction between different particles having oppositecharges on their surface

(D) Potential difference between the fixed layer andthe diffused layer of opposite charges around the colloidalparticles.

Sol. A,D

Option A – Due to like charges on DP. Sol. particles arestabiltsed.

Option D – If magnitude of electrokinetic potential ishigh then stability of sol is more .

35.Which of the following hydrogen halides react(s) withAgNO3(aq) to give a precipitate that dissolves inNa2S2O3(aq)?

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

Sol. ACD

SECTION – C

Integer Answer Type

36.An organic compound undergoes first-orderdecomposition. The time taken for its decomposition of1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively.

What is the value of 10]t[

]t[

10/1

8/1× ? (take log102 = 0.3)

Sol. 0009

t1/8 = 2.303 1

logk 1 / 8

t1/10 = 2.303 1

logk 1 /10

10]t[

]t[

10/1

8/1× = 3 log 2 × 10 = 3 × 0.3 × 10 = 9







th

REGISTRAT

ION

OPEN




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

CHO

CH2

CHOH

CHOH

CHOH

CH OH2

Sol. 0008

C–H

CH2

CHOH

CHOH

H–C–OH

CH OH2

O

CH2OH

HH

OHCH2

OH

H

*

*

*

O

Chiral centre is fix for D-configuration

HO

H

Total number of stereoisomers in pyranose form ofD-configuration.

∴ 23 = 8

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

H N–CH–CO–NH–CH–CO–NH–CH–CO–NH–CH–COO3

R1 R2H H

Peptide

I H HH CH3

CH COOH2 H

CH CONH2 2 (CH )2 4NH2

CH CONH2 2 CH CONH2 2

(CH NH2 2)4 (CH NH2 2)4

(CH NH2 2)4 CH3

CH COOH2 CH CO2 NH2

CH OH2 (CH ) NH2 4 2

II

III

IVV

VI

VIIVIII

IX

R1 R2

Sol. 0004

Peptide no. IV, VI, VIII & IX under goes protonation atpH = 7 because side chain alkyl group are basic.

39.The periodic table consists of 18 groups. An isotopeof copper, on bombardment with protons, undergoes anuclear reaction yielding element X as shown below. Towhich group, element X belongs in the periodic table?

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

Sol. 0008

63 1 1 4 1 A29 1 0 2 1 zCu H 6 n 2 H X+ → + α + +

30 = 2 + 2 + z

z = 26 (Fe)

1 2 3 4 5 6 7 8

K Ca Sc Ti V Cr Mn Fe

Sol. Ans. 8

40.29.2% (w/w) HCl stock solution has a density of1.25 g mL

–1. The molecular weight of HCl is 36.5 g

mol–1. The volume (mL) of stock solution required to

prepare a 200 mL solution of 0.4 M HCl is.

Sol. 0008

Molarity = 1000100

251

536

229××

.

.

. = 10 M

From question,

MV = 200 × 0.4

V = 10

40200 .× = 8







PART - III [MATHEMATICS]

SECTION – A

Single Correct

41. The ellipse E1 : 14

y

9

x 22

=+ is inscribed in a

rectangle R whose sides are parallel to the coordinateaxes. Another ellipse E2 passing through the point(0, 4) circumscribes the rectangle R. The eccentricityof the ellipse E2 is

(A) 2

2(B)

2

3(C)

2

1(D)

4

3

Sol. C

116

y

a

x 2

2

2

=+

116

4

a

92

=+

2

(0,4)

(3,2)

a2 = 12

e2 = 1 – 16

121 − =

4

1

e = 2

1

42. The point P is the intersection of the straightline joining the points Q(2, 3, 5) and R(1, –1, 4) withthe plane 5x – 4y – z = 1. If S is the foot of theperpendicular drawn from the point T(2, 1, 4) to QR,then the length of the line segment PS is

(A) 2

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

Sol. A

Line )Let(1

5z

4

3y

1

2xλ=

−=

−=

−

dso (λ + 2, 4λ + 3, λ + 5)Line on plane 5x – 4y – z = 15λ + 10 – 16λ – 12 – λ – 5 = 1–12λ = 8

λ = – 2/3 so P

3

13,

3

1,

3

4

for foot of perpendiuclar of T(2, 1, 4)(λ 4λ + 2, λ + 1) . (1, 4, 1) = 0

λ + 16λ + 8 + λ + 1 = 0λ = – 9/18 ⇒ λ = – 1/2

So R(3/2, 1, 9/2), distance a = 1/ 2

43. The integral ∫+ 2/9

2

)xtanx(sec

xsec dx equals

(for some arbitrary constant K)

(A) – 2/11)xtanx(sec

1

+

+− 2)xtanx(sec7

1

11

1+K

(B) 2/11)xtanx(sec

1

+

+− 2)xtanx(sec7

1

11

1+K

(C) – 2/11)xtanx(sec

1

+

++ 2)xtanx(sec7

1

11

1+K

(D) 2/11)xtanx(sec

1

+

++ 2)xtanx(sec7

1

11

1+K

Sol. C

tan x = t ⇒ sec2x dx=dt

2 9/2

dt

(t 1 t )+ +∫ put 2t 1 t z+ + =

2

13/2

1 z 1dz

2 z

+= ∫ 7/2 11/2

1 1 1 1c

7 11z z

−= − +

2

11/2

1 1 1(sec x tanx) c

11 7(sec x tanx)

= − + + +

+

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

(A) –1 (B) 3

1(C)

2

1(D)

4

3

Sol. D

Im(z) ≠ 0 a = z2 + z + 1z = α + iβ β ≠ 0a = α2 – β2 + 2iαβ + α + iβ + 1= α2 – β2 + α + 1 + iβ(2α + 1)α = –1/2 = 3/4 – β2

Prakash Gupta Ex-Sr. Faculty Bansal Classes

B.Tech. IIT-BombayExp. 7 yrs

Arjun Gupta Ex-Sr. Faculty Bansal Classes

B.Tech Exp. 5 yrs

Mahe Alam Sr. Faculty Mathematics

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

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

Exp. 9 yrs

Amit Gupta Ex-Sr. Faculty Bansal Classes

Ex-HOD Career Point Exp. 10 yrs

Suneet VijaySr. Faculty Mathematics

Exp. 4 yrs

New

New

MATHEMATICS TEAMSupported by a pool of 23 Faculty Members



45. Let f(x) =

≠

≠π

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

Sol. B

46. The total number of ways in which 5 balls ofdifferent colours can be distributed among 3 persons sothat each person gets at least one ball is

(A) 75 (B) 150 (C) 210 (D) 243

Sol. B

47. If ∞→xlim

−−

+

++bax

1x

1xx2

= 4, then

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

Sol. B

∞→xlim

+

−−−−+

1x

baxbxaxxHx 22

= 4

∞→xlim

+

−+−−+−

1x

b1x)ab1(x)a1( 2

= 4

1 – a = 0 ⇒ a = 11 – b – a = 4b = –4

48. The function f : [0, 3] → [1, 29], defined byf(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

Sol. B

f(X) = 2X3 – 15X2 + 36X + 1f'(X) = 6X2 – 30X + 36= 6 (X2 – 5X + 6)= 6(x – 3) (x – 2)

(0,1)

(0,29)(0,28)

(2,0) (3,0)f(0) = 1

f(2) = 2 × 8 – 15 × 4 + 36 × 2 +1 = 29f(3) = 2 × 27 – 15 × 9 + 36 × 3 + 1 = 28onto but not one one so 'B'.

49. The locus of the mid-point of the chord ofcontact of tangents drawn from points lying on thestraight line 4x – 5y = 20 to the circle x2 + y2 = 9 is

(A) 20(x2 + y2) – 36x + 45y = 0(B) 20(x2 + y2) + 36x – 45y = 0(C) 36(x2 + y2) – 20x + 45y = 0

(D) 36(x2 + y2) + 20x – 45y = 0Sol. A

P(h,k)

−

5

20a4,a x + y = 9

2 2

Equation of chord of contact

ax +

−

5

20a4y – 9 = 0

5ax + 4ay – 20y – 45 = 05ax + (4a – 20) y – 45 = 0 .....(i)equation of chord of mid point

hx + ky = h2 + k2 .....(ii)

22 kh

45

k

20a4

h

a5

+=

−= ⇒ a = 22 kh

h9

+

4a – 20 = 22 kh

k45

+ put the value of a

22 kh

h36

+ 20 = 22 kh

k45

+

36x – 20(x2 + y2) = 45y20(x2 + y2) + 45y – 36x = 0

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

(A) 210 (B) 211 (C) 212 (D) 213

Sol. D

11 12 13

21 22 23

31 32 33

a a a

P a a a

a a a

=

,

11 12 13

21 22 23

31 32 33

4a 8a 16a

Q 8a 16a 32a

16a 32a 64a

=

Q = 11 12 13

2 3 4 121 22 23

31 32 33

a a a

2 .2 .2 .2 a a a

a a a

Q = 122 P ⇒ |Q| = 213



Multiple Correct51. If. y (x) satisfies the differential equation

y' – y tan x = 2x sex x and y (0) = 0, then

(A)

2

y4 8 2

π π =

(B)

2

y'4 18

π π =

(C)

2

y3 9

π π =

(D)

24 2y'

3 3 3 3

π π π = +

Sol. A,D

I.F. = cos x

y . cos x = ∫ dx.xcos.xsecx2

y . cos x = x2 + c, c = 0y = x2 sec x

52. A ship is fitted with three engines E1, E2 and E3.

The engines function independently of each other

with respective probabilities 1

2,

1

4 and

1

4. For

the ship to be operational at least two of its

engines must function. Let X denote the eventthat the ship is operational and let X1, X2 and

X3 denote respectively the events that theengines E1 , E2 and E3 are functioning. Which of

the following is (are) true ?

(A) c1

3P X | X

16 =

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

8=

(C) 2

5P X | X

16= (D) 1

7P X | X

16=

Sol. B,D

1

1P(E )

2= 2

1P(E )

4= 3

1P(E )

4=

1 1 1 1 1 3 1 1 1P(X) . . . . 2 . .

2 4 4 2 4 4 2 4 4

= + +

= 1 3 1 6 1

32 16 32 4

++ = =

(A) 1

1 1 1. .P(E X) 12 4 4

P(x) 1 / 4 8

∩= =

(B) 4/1

32/7 = 7/8

(C)

1 1 1 1 1 3 1 1 1. . . . . .

2 4 4 2 4 4 2 4 41

4

+ +=

1 3 5516 32 32

11 / 4 8

4

+= = =

(D) 1

1

1 1 1 1 1 3. . . . 2

P(X X ) 2 4 4 2 4 4 7 / 32 7

1P(X ) 1 / 2 16

2

+ ∩ = = =

53. Let , [0,2 ]θ ϕ∈ π be such that

22cos (1 sin ) sin tan cot cos 12 2

θ θ θ − ϕ = θ + ϕ −

3tan(2 ) 0and 1 sin

2π − θ > − < θ < −

Then ϕ cannot satisfy

(A) 02

π< ϕ < (B)

4

2 3

π π< ϕ <

(C) 4 3

3 2

π π< ϕ < (D)

32

2

π< ϕ < π

Sol. A,C,D

2cos (1 sin ) 2sin cos 1θ − ϕ = θ φ −

2 cos θ + 1 = 2 sin (θ + φ)

2

1 = sin(θ + φ) – cos θ

tan(2 ) 0π − θ > , tan θ < 0 ⇒ –1 < sin θ < –2

3

270º < θ < 300º

54. Let S be the area of the region enclosed by

2xy e ,y 0,x 0,−= = = and x = 1. Then

(A) 1

se

≥ (B) 1

s 1e

≥ −

(C) 1 1

s 14 e

≤ +

(D)

1 1 1s 1

2 e 2

≤ + −

Sol. A,B,D

(B) x ≥ x2

∫∫ −− ≤1

0

x1

0

x dxedxe2

0

1

12/1

1/eA D

EBC

F

(D) S ≤ area OABC + area AFDE

1 1 1s 1

2 e 2

≤ + −

55. Tangents are drawn to the hyperbola2 2x y

19 4

− = ,

parallel to the straight line 2x y 1− = . The points

of contact of the tangents on the hyperbola are

(A) 9 1

,2 2 2

(B) 9 1

,2 2 2

− −

(C) ( )3 3, 2 2− (D) ( )3 3,2 2−







th

REGISTRAT

ION

OPEN




Sol. A,B

4x1 x –9y1y = 36

Slope ; 1

1

4x2

9y=

11

9yx

2= .............(1)

21 1x y 2

19 4

− =

1y 1 / 2= ± & 1x 9 / 2 2= ±

A & B

SECTION – CInteger Answer Type

56. Let S be the focus of the parabola y2 = 8x and let

PQ be the common chord of the circle

x2 + y

2 – 2x – 4y = 0 and the given parabola. The area

of the triangle PQS is

Sol. 0004

S(2,0)P

Q(2,4)

Q is end point of latus rectum

Area is ( ) 4242

1=×

57. Let p(x) be a real polynomial of least degree which

has a local maximum at x = 1 and a local minimum at x =

3. If p(1) = 6 and p(3) = 2, then p'(0) isSol. 0009

p'(x) = a(x – 1) (x – 3) = ax2 – 4ax + 3a

p(x) = 3

ax3

2ax2 + 3ax + c

3

a – 2a + 3a + c = 6 & 9a – 18a + 9a = 2

a = 3

58. Let f : IR → IR be defined as f(x) = |x|+|x2 – 1|.

The total number of points at which f attains either alocal maximum or a local minimum is

Sol. 0005

f(x) =

−≤−−

≤≤−+−−

≤≤++−

=−+

1x;1xx

0x1;1xx

1x0;1xx

1x;1xx

2

2

2

2

–1 –1/2 1/2 10

59. The value of 6+

−−− ......23

14

23

14

23

14

23

1log

2

3 is

Sol. 0004

3/2

16 log 4

3 2

+

1y 4 y

3 2= −

2 1y 4 y

3 2= −

23 2y 12 2 y= −

23 2y y 12 2 0+ − =

1 17y

6 2

− ±= =

17 1 16 8

6 2 6 2 3 2

−= =

3/2

1 8log

3 2 3 2×

3/2

4log

9

2

3/2

26 log

3

+

= 4

60. If candb,a are unit vectors satisfying

9accbba222

=−+−+−

, then c5b5a2

++ is

Sol. 0003

Given is maximum value of expression & hence

angle b/w any two is 120°

1a.b b.c c.a

2

−= = =

so |2a 5b 5c |+ +

is 3





PHYSICS TEAM - Motion IIT- · PDF filePHYSICS TEAM IIT-JEE 2012 PAPER-I WITH SOLUTION ......

Documents

Transcript of PHYSICS TEAM - Motion IIT- · PDF filePHYSICS TEAM IIT-JEE 2012 PAPER-I WITH SOLUTION ......