IIT JEE 2011 PAPER-2 NARYANA
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IIT – JEE (2011) PAPER II QUESTION & SOLUTIONS
CODE 0
PART I : CHEMISTRY PAPER – II SECTION – I(TOTAL MARKS: 24) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1.
Key. : Sol. :
2.
Key. : Sol. :
Among the following complexes (K – P), K3[Fe(CN)6] (K), [Co(NH3)6]Cl3 (L), Na3[Co(oxalate)3] (M), [Ni(H2O)3]Cl2 (N), K2[Pt(CN)4] (O) and [Zn(H2O)6](NO3)2 (P) the diamagnetic complexes are (A) K, L, M, N (B) K, M, O, P (C) L, M, O, P (D) L, M, N, O (C) K3Fe(CN)6 is paramagnetic [Ni(H2O)6]Cl2 is paramagnetic too. Consider the following cell reaction : 2Fe (s) + O2(g) + 4H+ (aq) 2Fe2+ (aq) + 2H2O (l) E° = 1.67 V At [Fe2+] = 10-3M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is (A) 1.47 V (B) 1.77 V (C) 1.87 V (D) 1.57 V (D) 2Fe(s) O 2 g 4H 2Fe 2 aq 2H 2 O E° = 1.67 V 2
Ecell = Ecell
Fe2 0.0591 log 4 n O2 H 2
103 0.0591 log = 1.67 – 4 4 101 103
0.0591 106 log 13 4 10 0.0591 = 1.67 – 7 = 1.566 1.57 Volt. 4 = 1.67 –
3.
The major product of the following reaction is RCH 2 OH H anhydrous
Key. : Sol. :
O (A) a hemiacetal (C) an ether (B)
(B) an acetal (D) an ester
RCH 2 OH
H
O
O
O
OCH2 R
Acetal NARAYANA IIT ACADEMY
3
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
4.
Key.: Sol. :
5.
Key. : Sol. :
6.
Passing H2S gas into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates (A) CuS and HgS (B) MnS and CuS (C) MnS and NiS (D) NiS and HgS (A) Cu2+ & Hg2+ ions belong to Group – II of salt analysis & their sulphides are very less soluble. Hence CuS and HgS will be precipitated out even in acidic medium by passing H2S gas. Oxidation states of the metal in the minerals haematite and magnetite, respectively are (A) II, III and haematite and III in magnetite (B) II, III in haematite and II in magnetite (C) II in haematite and II, III in magnetite (D) III in haematite and II, III in magnetite (C) Formula Oxidation state of Fe Haematite Fe2O3 +3 Magnetile Fe3O4 (FeO + Fe2O3) +2, +3 Amongst the compounds given, the one that would from a brilliant colored dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of -naphthol is N(CH3 )2
NHCH3
(A)
(B) NH2
CH2 NH2
(C)
(D) H3C
Key. : Sol. :
(C) Azo dye test is given by Aromatic 1° amine NH2 NaNO2 HCl 0 5 C
H3C
OH
NCl
N
H3C
N
N OH
napthol
H3C Red azo dye
7.
Key: Sol.:
The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol-1) is (A) – 2.3 × 10-2 (B) – 5.7 × 10-2 -2 (C) – 5.7 × 10 (D) – 1.2 × 10-2 (A) Tf = i × kf × m W 1000 = i kf B M B WA = 4 1.86
0.1 1000 329 100
= 0.0226 = 2.26 × 10-2 2.3 × 10-2 Freezing point = – 2.3 × 10-2. 8.
The following carbohydrate is
OH
H
H HO HO
H H
4
O OH OH
H
NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Key. : Sol.
(A) a ketohexose (C) an -furanose (B) OH H
(B) an aldohexose (D) an - pyransose
O
H HO HO
H
OH OH
H is an aldohexose.
H
SECTION – II (TOTAL MARKS: 16) (Multiple Correct Answer(s) Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
9.
The equilibrium
Cu Cu|| 2Cu| In aqueous medium at 25°C shifts towards the left in the presence of
Key. : Sol. :
(A) NO3
(B) Cl
(C) SCN (B, C, D) The equilibrium constant for the reaction
(D) CN
Cu Cu 2 is high enough to keep it largely towards right. The Cu+ will exist in water with 2Cu stability when either present as complexes or as insoluble salt CuCl, CuCn and CuSCN are insoluble cuprous salt. 10.
The correct functional group X and the reagent / reaction condition Y in the following scheme are X CH 2 4 X condensation polymer O i Y
ii
O
C
(CH 2) 4
HO
C OH
heat
Key. : Sol. :
(A) X = COOCH3, Y = H2 / Ni / heat (C) X = CONH2, Y = Br2 / NaOH (A, B, C, D) O O (A)
H3C (B)
H2 N (C)
H2 N (D)
O
C (CH2 )4
O
O
C (CH2 )4
C
O
O
C (CH2 )4
C
NC (CH 2 ) 4 CN
C
(B) X = CONH2, Y = H2/ Ni / heat (D) X = CN, Y = H2 / Ni/ heat
H 2 / Ni OCH3 CH 3 OH HO CH 2 CH 2 4 CH 2 OH
NH2
H 2 / Ni NH 2 CH 2 6 NH 2
NH2
Br2 / NaOH
NH 2 CH 2 4 NH 2
H 2 / Ni NH 2 CH 2 6 NH 2
Diamines give polyamides with dicarboxylic acids Similarly di-ols give polyester with dicarboxylic acid.
NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
11.
For the first order reaction 2N2O2 (g) 4NO2 (g) + O2 (g) (A) the concentration of the reactant decreases exponentially with time. (B) the half-life of the reaction decreases with increasing temperature. (C) the half-life of the reaction depends on the initial concentration of the reactant. (D) the reaction proceeds to 99.6% completion in eight half-life diration.
Key. :
(A, B, D)
Sol. :
First order reaction :
d N 2 O5 dt
K. N 2 O5
n2 K With temperature ‘K’ increases and therefore t 1/2 decreases At the completion of 8 half lives Half life period : t1/2 =
Remaining % =
1 1 100 100 = 0.39 28 256
% reacted = 100 – 0.39 = 99.61 %. 12.
Reduction of the metal centre in aqueous permagnate ion involves (A) 3 electrons in neutral medium
(B) 5 electrons in neutral medium
(C) 3 electrons in alkaline medium
(D) 5 electrons in acidic medium
Key. :
(A, C, D)
Sol. :
8H 5e MnO4 Mn 2 4H 2 O 2H 2 O 3e MnO4 MnO2 4OH 2H 2 O 3e MnO4 MnO2 4OH
SECTION – III (TOTAL MARKS: 24) (Integer Answer Type)
This Section contains 6 questions. The answer to each question is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to darkened in the ORS.
13. Key. : Sol. :
6
The number of hexagonal faces that are present in a truncated octahedron is 6 The truncated octahedron is an Archimedean solid with 14 faces out of which 8 faces are hexagonal and rest 6 faces are square shaped.
NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
H3C 14.
The total number of contributing structures showing hyperconjugation (involving C – H bonds) for the following carbocation is
Key. : Sol. :
6 H3 C
+
C
CH2 CH3
CH2 CH3
H = 6 contributing hyper conjugating structure = 6 15. Key. : Sol. :
Among the following, the number of compounds than can react with PCl5 to give POCl3 is O2, CO2, SO2, H2O, H2SO4, P4O10 5 CO2 + PCl5 POCl3 + COCl2 PCl5 + H2O POCl3 + 2 HCl (if equi molar amounts are used, the reaction is more gentle and yields POCl3) PCl5 + SO2 POCl3 + SOCl2 PCl5 + P4O10 10 POCl3 PCl5 + H2SO4 POCl3 + HSO2Cl + HCl chlorosulphonic acid 2PCl5 + H2SO4 SO2 Cl2 + 2POCl3 + 2HCl sulphuryl chloride
16.
Key. : Sol. :
In 1 L saturated solution of AgCl [Ksp (AgCl) = 1.6 × 10-10], 0.1 mole of CuCl [Ksp (CuCl) = 1.0 × 10-6] is added. The resultant concentration of Ag+ in the solution is 1.6 × 10-x. The value of “x” is 7
Ag+ + Cl AgCl(s) x x Cu+ + Cl CuCl(s) y y Net Cl in solution = (x + y) ksp AgCl = 1.6 10-10 = [Ag+] [Cl] = x (x + y) x(x + y) = 1.6 10-10 ksp of CuCl = 1.0 10-6 = [Cu+][Cl] 1.0 10-6 = y (x + y) from equation (i) and (ii)
. . . . .(i) . . . . (ii)
x 1.6 104 y y
104 x 1.6
putting the value of y in eq. (i)
104 x. x x 1.6 1010 1.6 NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
x2
104 1.6 1010 1.6
x2 = (1.6)2 10-14 x = 1.6 10-7 17. Key. : Sol. :
The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to 6 2AgNO3 + [Cr(H2O)5Cl] Cl2 2AgCl + [Cr(H2O)5Cl](NO3)2 milli equivalent . of AgNO3 reacted = milli equivalent of [Cr(H2O)5Cl]Cl2 reacted thus (M n V) AgNO3 (M n V)[Cr(H2O)5 Cl]Cl2
18.
0.1 1 V = 0.01 2 30 V=6
The maximum number of isomers (including stereoisomers) that are possible on mono-chlorination of the following compound, is CH3
C CH3 CH2 Key. : Sol.
H
CH2 CH3
8 H:3C
CH2
CH
CH2
CH3
+ Cl2
Monochlorination
CH2 Cl
CH3
*
CH2
CH
CH2
CH3
CH3 (1 chiral) (A)
Cl H3C
*
*
CH Cl
CH
CH2
CH3 (2 chiral carbon)
CH3
H3C
CH2
CH2
CH
CH2
CH3
CH3 (no chiral carbon)
(B) H3C
C
(C) CH2 CH3
CH2 Cl (no chiral carbon) (D)
Thus ‘A’ has two structures, ‘B’ has 4, ‘C’ has 1, and ‘D’ has one therefore total product will be 8
SECTION – IV (TOTAL MARKS: 16) (Matrix-Match Type)
This section contains 2 questions. Each question four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.
19. 8
Match the reactions in Column I with appropriate types of steps / reactive intermediate involved in these reactions as given in Column II NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Column I (A)
O
H3C O
O
(p)
Column II Nucleophilic substitution
(q)
Electrophilic substitution
(r)
Dehydration
(s)
Nucleophilic
(t)
Carbanion
aq. NaOH
O
(B)
O CH2 CH2 CH2 Cl CH3 MgI
CH3
O
(C)
18
O 18
CH2 CH2 CH2 OH H 2 SO 4
CH2 CH2 CH2
(D)
C (CH3 )2 OH
H 2SO4
CH3
H3C
Key: Sol.
(A –r,s,t); (B–p,s,t); (C–r,s); (D–q,r) H2 C
:
O
OH
O
-H2O
NuC
Addition
O
O
O
O CH2 CH2 CH2 Cl
Nu- addition CH3 MgI (source of carbanion)
CH2 CH2 CH2
Cl
CH3
SNi O
CH3
NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II +
H
O C
H
HO C
CH2
+O
CH2
C
CH2
IMPE
+ 18 O
18
: O:
H
18
O
H
H
+
18
C
O
CH2
H2O
18
H+,
CH2 CH2 CH2 C(CH3 )2
O
CH2 CH2 CH2 C(CH3 )2
H2SO4
+
OH
OH
H -H2O
EAS +
H3C
20.
(p) (q) (r) (s) (t)
Column II Phase transition Allotropic change H is positive S is positive S is negative
(A –p,r,s); (B–r,s); (C–t); (D–q,r,t) (A) CO2(s) + Q CO2 (g)
(B) (C)
(D)
10
H3C
Match the transformations in Column I with appropriate options in column II Column I (A) CO2 (s) CO2 (g) (B) CaCO3 (s) CaO (s) + CO2 (g) (C) 2H H2 (g) (D) P(white, solid) P(red, solid)
Key: Sol.:
CH3
H = +ve , phase transition and S = +ve CaCO3(s) + Q CaO (s) + CO2(g) H = +ve, S = +ve 2H0 H2 (g) two particle give one gaseous particle S = -ve Pwhite , solid Pred, solid H = +ve, S = +ve, allotropic change
NARAYANA IIT ACADEMY
CH3
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
PART II : PHYSICS SECTION – I(TOTAL MARKS: 24) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
21.
Key: Sol.:
22.
Key: Sol.:
A satellite is moving with a constant speed V in a circular orbit about the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is 1 (A) (B) mV2 mV 2 2 3 (C) (D) 2mV2. mV 2 2 (B) At the time of ejection ETotal = 0 GMm Total Kinetic Energy = ETotal – P.E. = r = mV2 A point mass is subjected to two simultaneous sinusoidal displacements in x–direction, x1 (t) A sin t and
2 x 2 (t) A sin t . Adding a third sinusoidal displacement x 3 (t) Bsin(t ) brings the mass to a 3 complete rest. The value of B and are 3 4 2A, (A) (B) A, 4 3 5 (C) (D) A, . 3A, 6 3 (B) x1+x2 x2 x1 + x2 = Asin(t + /3) 4 x1 + x2 + x3 = 0 x3 = A sin t 2/3 3 4/3 x1
x3 23.
Key: Sol.:
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is (A) 0.9% (B) 2.4% (C) 3.1% (D) 4.2%. (C) 0.5mm 1 Least count of screw gauge = = mm. 50 100 1 Diameter D = 2.5 + 20 = 2.7 mm 100 m Now, density 3 4 D 3 2
NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Sol.:
25.
A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the X–Y plane and a steady current I flows through the wire. The Z–component of the magnetic field at the centre of the spiral is 0 NI 0 NI b ba (A) (B) n n 2(b a) a 2(b a) b a NI NI b ba (C) 0 n (D) 0 n . 2b a 2b ba (A) N dN = dr ba N 0 Idr b N 0 I b ba B= ln a 2r 2b a a
(A)
Intensity
100% T
(B)
R
Intensity
(C)
90º
Sol.:
12
X
R
90º
90º
100% T (D)
R
0
Key:
b
T
0
100%
26.
a
I
100%
0
Key: Sol.:
Y
A light ray traveling in glass medium is incident on glass air interface at an angle of incidence . The reflected (R) and transmitted (T) intensities, both as function of , are plotted. The correct sketch is
Intensity
Key:
m D 100 3 100 = 3.1%. m D
90º (C) For 0 c there will be little reflection For c there will be no transmission.
Intensity
24.
% error in density =
T R
0
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg traveling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity of the bullet is (A) 250 m/s (B) 250 2 m/s (C) 400 m/s (D) 500 m/s. (D)
25 = 1s 10 COM v 0.01 0.2 20 0.01 100 v = 500 m/s t
NARAYANA IIT ACADEMY
V m/s
0 20
100
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
27.
Which of the field patterns given below is valid for electric field as well as for magnetic field ?
(A)
(B)
(C)
(D)
Key: Sol.:
(C) In case of non-conservative electric field, lines as shown in (C) are possible. Also, in case of toroid, magnetic field lines are circular
28.
A wooden block performs SHM on a frictionless surface with frequency, E v0. The block carries a charge +Q on its surface. If now a uniform electrical +Q field E is switched on as shown, then the SHM of the block will be (A) of the same frequency and with shifted mean position (B) of the same frequency and with the same mean position (C) of changed frequency and with shifted mean position (D) of changed frequency and with the same mean position. (A) An additional constant force will not change the frequency but will shift the mean position.
Key: Sol.:
SECTION – II (TOTAL MARKS: 16) (Multiple Correct Answer(s) Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
29.
A series R–C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the resistor and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true ? (A) I AR I RB (B) I AR I BR
Key:
(C) VCA VCB (B, C)
Sol.:
Z R2
30.
Key:
(D) VCA VCB .
1 c zA > zB I AR I BR and v CA v CB 2 2
A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision (A) the ring has pure rotation about its stationary CM (B) the ring comes to a complete stop (C) friction between the ring and the ground is to the left (D) there is no friction between the ring and the ground. (C) NARAYANA IIT ACADEMY
10m/s 20 m/s 0.75 m
1m/s
13
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Sol.:
Assuming friction to be absent at point of contact (to check the tendency of sliding), we find that under the impulsive forces shown in figure, the point of contact moves to the right hence friction will act towards left.
F dt Jx= 2 Ns
Jy= 1 Ns 31.
Key: Sol.:
Which of the following statement(s) is/are correct ? (A) if the electric field due to a point charge varies as r–2/5 instead of r–2, then the Gauss law will still be valid (B) the Gauss law can be used to calculate the field distribution around an electric dipole (C) if the electric field between two point charges is zero somewhere, then the sign of the two charges is the same (D) the work done by the external force in moving a unit positive charge from point A at potential VA to point B at the potential VB (VB – VA). (C, D) Q E ds f r enc If E r–2.5 then 0 sphere (A) is incorrect. As field of an electric dipole is not symmetric,
E ds
is incalculable
(B) is incorrect. In case of similar charges, null point is between the two charges. (C) is correct. Wext = q(VB – VA) (D) is incorrect. 32.
Key: Sol.:
Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if (A) dA < dF (B) dB > dF (C) dA > dF (D) dA + dB = 2 dF. (A, B, D) For equilibrium of system V(dA + dB)g = 2V dF g
A B
A
B
SECTION – III (TOTAL MARKS: 24) (Integer Answer Type)
This Section contains 6 questions. The answer to each question is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to darkened in the ORS.
14
NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
S
4 ) in a tank is 18 cm deep. Oil of 3
33.
Water (with refractive index
Key: Sol.:
7 refractive index lies on water making a convex surface of radius of 4 curvature R = 6 cm as shown. Consider oil to act as a thin lens. An object S is placed 24 cm above water surface. The location of its image is at x cm above the bottom of the tank. Then x is 2 Refraction at air-oil boundary
=1.0 R=6 cm
=4/3
7 / 4 1 7/4 1 V1 24 6
7 1 1 1 3 1 4V1 24 8 24 12 7 12 V1 21cm 4 Refraction at oil-water boundary
4/3 7/4 4 1 0 V 21 3V 12 4 1 4 12 V 16cm 3V 12 3 x = 2 cm 34.
Key:
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free– space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the shphere is A × 10Z (where 1 < A < 10). The value of Z is 7
Sol.:
E
hc 1242 6.21eV 200 1 ne V 4 0 r
When emission will stop then
E eV
n 1.6 1019
2
9 109
1.51 1.6 1019 102 1.5 102 1.51 15.1 n 108 107 10 1.6 9 10 1.6 9 16 9 z= 7 35.
Key:
A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60º to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s2 is 5
Sol.:
u y 10sin 60 5 3 m/s
t
2u y g
25 3 3s 10
NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
1 Sx u x t a x t 2 2 1 1.15 5 t a t 2 2 3 1.15 5 3 a 2 3a 5 1.73 1.15 8.65 1.15 2 3a 7.5 2 15 a 5 m/s2 3 36.
Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is
Key:
5
Sol.:
I
A
6V
1
3V
2
B
63 1A 3 +
I 6V
1
A
B +
3V
2
VA 3 2 1 VB VAB 5 V 37.
Key: Sol.:
A block of mass 0.18 kg is attached to a spring of force–constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un–stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = N/10. Then N is 4 Work done by friction
mgx 0.1 0.18 10 0.06 108 10 4
1 2 1 1 2 mv 108 104 kx 2 2 0.06 2 2 2
= 36×10–4
0.09v 2 144 104 12 4 v 101 3 10 N=4 38. Key: 16
A series R–C combination is connected to an AC voltage of angular frequency = 500 radian/s. If the impedance of the R–C circuit is R 1.25 , the time constant (in millisecond) of the circuit is 4 NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Sol.:
Z R 2 XC2 R R
5 4
C
~ 5 R 2 XC2 R 2 4 R 2 XC 2 R 2 C R 2 2 RC R R 2 4 103 4ms 500
SECTION – IV (TOTAL MARKS: 16) (Matrix-Match Type)
This section contains 2 questions. Each question four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.
39.
Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency whose wavelength is denoted as 1. Match each system with statements given in Column II describing the nature and wavelength of the standing waves. Column I Pipe closed at one end
(p)
Column II Longitudinal waves
(q)
Transverse waves
(r)
f = L
(s)
f = 2L
(t)
f = 4L
(A) (B)
0 Pipe open at both ends
(C)
0 L Stretched wire clamped at both ends
(D)
L
0 L Stretched wire clamped at both ends and at mid–point
0
L L/2
Key:
(A –p,t); (B–p,s); (C–q,s); (D–q,r)
Sol.:
(A)
1 L 1 4L 4 Longitudinal wave
A – p, t NARAYANA IIT ACADEMY
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IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
(B)
1 L 1 2L 2 Longitudinal wave
(C)
B – p, s
1 L 1 2L 2 Transverse wave
(D)
C- q, s
1 1 L 1 L 2 2 Transverse wave
40.
D- q, r
One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P–V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.
P 3P
B
1P
A
C 0 1V 3V
(A) (B) (C) (D) Key: Sol.:
Column I Process A – B Process B – C Process C – D Process D – A
(p) (q) (r) (s) (t)
(A –p,r,t); (B–p,r); (C–q,s); (D–r,t) (A) Volume decreases Temperature decreases (B) Pressure decreases Temperature decreases (C) Volume increases Temperature increases (D)
D 9V
Column II Internal energy decreases Internal energy increases Heat is lost Heat is gained Work is done on the gas.
A- p, r, t B- p, r
PV P 9V 9PV TD nR nR nR 3P 3V 9PV TA nR nR
C- q, s
D – r, t
PART I : MATHEMATICS SECTION – I(TOTAL MARKS: 24) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
41.
Key: Sol.: 18
A value of b for which the equations x2 + bx - 1 = 0 x2 + x + b = 0 have one root in common is (A) 2 (C) i 5 (B) x2 + bx - 1 = 0 ... (i) x2 + x + b = 0 ... (ii)
(B) - i 3 (D) 2
NARAYANA IIT ACADEMY
V
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
(i) - (ii) gives x =
b 1 which is the common root b 1
Putting x in (ii) b 1 b 1 +b=0 b 1 b 1 b3 + 3b = 0 b(b2 + 3) = 0 b = 0, b2 = -3 b= 3i 2
b=42.
Key: Sol.:
43.
Key: Sol.:
44.
Key:
3 i.
The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through the point 3 5 (A) ,0 (B) , 2 2 2 3 5 (C) , (D) (-4, 0) 2 2 (D) Equation of family of circle touching y-axis at (0, 2) will be (x - 0)2 + (y - 2)2 + x = 0 Since this passes through, (–1, 0) 1+4-=0 =5 Equation of circle is x2 + y2 + 5x - 4y + 4 = 0 Also passes through (-4, 0). Let f(x) = x2 and g(x) = sinx for all xR. Then the set of all x satisfying (f o g o g o f) (x) = (g o g o f) (x), where (f o g) (x) = f(g(x)), is (A) n , n {0, 1, 2, ... } (B) n , n {1, 2, ... } (C) 2n , n {..., -2, -1, 0, 1, 2, ...} (D) 2n, n {..., -2, -1, 0, 1, 2, ....} 2 (A) f(x) = x2, g(x) = sinx g(f(x)) = g(x2) = sinx2 g(g(f(x)) = sin(sin(x2)) f.(g.(g.(f(x)))) = (sin (sin (x2)))2 f. (g.(g(f(x)))) = g.(g.(f(x)) (sin (sin (x2)))2 = sin (sin (x2)) (sin (sin (x2))) [sin.sin (x2) - 1] = 0 sin (sin (x2)) = 0 [or, sin (sin(x2)) 1) sin(x2) ] 2 sin (x2) = 0 x = n {n = 0, 1, 2, ... }
1 Let 1 be a cube root of unity and S be the set of all non-singular matrices of the form 2 where each of a, b, and c is either or 2. Then the number of distinct matrices in the set S is (A) 2 (B) 6 (C) 4 (D) 8 (A)
NARAYANA IIT ACADEMY
a b 1 c , 1
19
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Sol.:
45.
1 a b Let A = 1 c 2 1 |A| = (1 - a) (1 - c) For A to be non-singular matrix, none of a and c should be 2. So, a = c = While b can take value or 2 So, the number of distinct matrices in the set S is 2. x 2 y2 Let P (6, 3) be a point on the hyperbola 2 2 1 . If the normal at the point P intersects the x-axis at a b (9, 0), then the eccentricity of the hyperbola is 5 2
(A)
Key: Sol.:
(C) 2 (B) Let P be (a sec, btan) a sec = 6, b tan = 3 The equation of normal at P is ax cos + bycot = a2 + b2 a 2 b2 a 2 b2 9 Put y = 0, x = =9 a a cos a. 6 1+
46.
b2 3 e= a2 2
(B)
3 2
(D)
3
3 2
Let f : [-1, 2] [0, ) be a continuous function such that f(x) = f(1-x) for all x [-1, 2]. Let R1 = 2
xf (x) dx, and R2 be the area of the region bounded by y = f(x), x = -1, x = 2, and the x-axis. Then
1
Key:
(A) R1 = 2R2 (C) 2R1 = R2 (C)
Sol.:
R1 =
(B) R1 = 3R2 (D) 3R1 = R2
2
2
xf (x) dx , R2 =
f (x) dx
1
1
2
R1 =
(1 x) f (1 x)dx
(replace x by (-1 + 2 - x))
1 2
=
(1 x) f (x)dx
1
=
2
2
1
1
f (x) dx xf (x) dx
R1 = R2 - R1 2R1 = R2. 1
47.
Key: Sol.:
If lim 1 x ln (1 b 2 ) x = 2bsin2, b > 0 and (-, ], then the value of is x 0 (A) (B) 4 3 (C) (D) 2 6 (D) lim[1 x ln (1 b 2 )]1/ x = 2bsin2 (b > 0) x 0
1 lim .[x ln (1 b2 )]
e x 0 x 20
= 2bsin2 NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
1 + b2 = 2b sin2 1 1 sin2 = (b + ) 1 2 b sin2 = 1 (when b = 1) = 48.
Key: Sol.:
2
Let (x, y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is (A) x2 = y (B) y2 = 2x (C) y2 = x (D) x2 = 2y (C) x y Let P be (h, k). So, h = and k = 4 4 As, y2 = 4x 16k2 = 16h k2 = h So, locus of P is y2 = x
SECTION – II (TOTAL MARKS: 16) (Multiple Correct Answer(s) Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
49.
Let E and F be two independent events. The probability that exactly one of them occurs is probability of none of them occurring is
11 and the 25
2 . If P(T) denotes the probability of occurrence of the event T, 25
then
Key:
4 , P(F) = 5 2 (C) P(E) = . P(F) = 5 (A, D)
Sol.:
P(E) + P(F) - 2P(EF) =
(A) P(E) =
3 5 1 5
1 2 , P(F) = 5 5 3 4 (D) P(E) = , P(F) 5 5
(B) P(E) =
11 25
2 25 7 12 12 P(E) + P(F) = , P(E F) = P(E) . P(F) = 5 25 25 3 4 4 3 P(E) = , P(F) = or P(E) = , P(F) = 5 5 5 5
1 - [P(E) + P(F) - P(E F)] =
50.
Let f : (0, 1) R be defined by f(x) =
(B) f f -1 on (0, 1) and f (b) =
(A) f is not invertible on (0, 1) (C) f = f -1 on (0, 1) and f (b) = Key: Sol.: 51.
bx , where b is a constant such that 0 < b < 1. Then 1 b x
1 f (0)
1 f (0)
(D) f -1 is differentiable on (0, 1)
(A) As range of f(x) R f(x) is not invertible. Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by (A) y - x + 3 = 0 (B) y + 3x - 33 = 0 (C) y + x - 15 = 0 (D) y - 2x + 12 = 0 NARAYANA IIT ACADEMY
21
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Key: Sol.:
52.
(A, B, D) Equation of normal is y + tx = 2t + t3 As it passes through (9, 6) 6 + 9t = t3 + 2t or, t3 - 7t - 6 = 0 or, (t + 1) (t2 - t - 6) = 0 or, (t + 1) (t - 3) (t + 2) = 0 t = -1, -2, 3 equation of normals are y - x + 3 = 0, y - 2x + 12 = 0 y + 3x - 33 = 0
x x 2 , 2 cos x, x 0 If f(x) = 2 0 x 1 x 1, ln x, x 1, then (A) f(x) is continuous at x = 2 (C) f(x) is differentiable at x = 1
Key: Sol.:
(B) f(x) is not differentiable at x = 0 (D) f(x) is differentiable at x = -
3 2
(A, B, C, D) f f (x) 0 lim f (x) lim 2 x x 2
2
f(x) is continuous at x = -/2 f(x) is also continuous at x = 0, 1 x 1 2 sin x x 0 f (x) = 2 1 0 x 1 1 x 1 x f(x) is not diff. at x = 0 f(x) is diff. at x = 1 f(x) is diff. at x = -3/2
SECTION – III (TOTAL MARKS: 24) (Integer Answer Type)
This Section contains 6 questions. The answer to each question is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to darkened in the ORS.
53.
Key:
22
0 1 1 1 1 0 Let M be a 3 × 3 matrix satisfying M 1 2 , M 1 1 , and M 1 0 . Then the sum of the 0 3 0 1 1 12 diagonal entries of M is (9)
NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
Sol.:
a M b c d = 1 e=2 f=3 a d b e c f
d l 0 1 e m 1 2 f n 33 0 33 3
l 1 1 m 1 1 n 0 1
a–d=1a=0 b–e=1b=3 c–f=1c=2 a d l 1 0 b e m 1 0 c f n 1 12 a + d + l = 0 l= 1 b + e + m = 0 m = 5 c + f + n = 12 n = 7 0 1 1 M 3 2 5 2 3 7 Sum of diagonal entries = 9 54.
Key:
Sol.:
The straight line 2x - 3y = 1 divides the circular region x2 + y2 6 into two parts. If 3 5 3 1 1 1 1 S 2, , , , , , , , then the number of point(s) in S lying inside the smaller part is 4 2 4 4 4 8 4 (2) x2 + y2 6 L 2x – 3y – 1 > 0 S x2 + y2 – 6 < 0 3 1 1 Clearly 2, and , satisfies above two inequality 4 4 4
Sol.:
Let = ei/3 , and a, b, c, x, y, z be non-zero complex numbers such that a+b+c=x a + b + c2 = y | x |2 | y | 2 | z | 2 a + b2 + c = z . Then the value of is | a |2 | b |2 | C |2 Value is not fixed.
56.
Let y(x) + y(x) g(x) = g(x) g(x), y(0) = 0, xR, where f (x) denotes
55.
Key: Sol.:
df (x) and g(x) is a given nondx constant differentiable function on R with g(0) = g(2) = 0. Then the value of y(2) is (0) g ( x ) dx I.F. = e eg ( x )
y(x) eg(x) =
e
g(x)
.g(x) g (x)dx = eg(x) (g(x) - 1) + c NARAYANA IIT ACADEMY
23
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
57. Key: Sol.:
58. Key: Sol.:
y(x) = (g(x) - 1) + ce-g(x) Put x = 0, 0 = (-1) + c c = 1 y(2) = (-1) + 1.e-0 y(2) = 0. The number of distinct real roots of x4 - 4x3 + 12x2 + x - 1 = 0 is (2) Let f(x) = x4 – 4x3 + 12x2+ x – 1 f (x) = 4x3 – 12x2 + 24x + 1 O f(x) = 12x2 – 24x + 24 Hence f(x) > 0 x R f (x) is strictly increasing function. So, f(x) will have only one point of extrema. f(0) = 1 Hence f(x) = 0 has two distinct real roots. ˆ b ˆi ˆj and c = ˆi 2jˆ 3kˆ be three given vectors. If r is a vector such that r b c b Let a = ˆi k, and r.a 0 , then the value of r.b is (9) r b cb a (r b) a (c b) (a.b)r (a.r)b a (c b) r 0 a (c b) r a (c b)
ˆi ˆj kˆ ˆ 2) 3iˆ 3jˆ 3kˆ c b 1 2 3 ˆi(0 3) ˆj(3 0) k(1 1 1 0 ˆi ˆj kˆ ˆ 0) r a (c b) 1 0 1 ˆi(0 3) ˆj(3 3) k(3 3 3 3 r 3iˆ 6jˆ 3kˆ ˆ ˆi ˆj) = 3 + 6 = 9 r.b (3iˆ 6jˆ 3k).(
SECTION – IV (TOTAL MARKS: 16) (Matrix-Match Type)
This section contains 2 questions. Each question four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.
59.
Match the statement given in Column I with the intervals/union of intervals given in Column II (A)
(B)
24
Column I
2iz The set {Re : z is a complex 2 1 z number, |z| = 1, z 1} The domain of the function f(x) = sin-1 8(3) x 2 is 2(x 1) 1 3
(p)
(-, -1) (1, )
(q)
(-, 0) (0, )
Column II
NARAYANA IIT ACADEMY
IIT-JEE2011-Code-0-Questions and Solutions-Paper-I and II
(C)
1 tan 1 1 tan , then If f() = tan 1 tan 1 } is 2 3/2 If f(x) = x (3x - 10), x 0, then f(x) is increasing in
(r)
(-, -1] [1, )
(s)
(-, 0] [2, )
(t)
(-, 0] [2, )
the set {f() : 0 < (D)
Key: Sol.:
(A–s), (B–t), (C–r), (D–r) (A) z = ei 2iei 2iei Re Re i2 1 e 1 cos 2 i sin 2
1 2iei (2iei )i Re Re Re = (, 1] [1, ) 2 sin 2sin (sin i cos )i 2sin i2sin cos (B) 3x = t 8t 1 1 9 t2 t 2 8t 9 0 t2 9 (t 9)(t 1) 0 (t 3)(t 3) t
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