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Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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Website : www.resonance.ac.in | E-mail : ccoonnttaacctt@@rreessoonnaannccee..aacc..iinn | CCIINN :: UU8800330022RRJJ22000077PPLLCC002244002299
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
22-05-2016JEE (Advanced) 2016
®
DATE: 03-10-2021
JEE (ADVANCED) 2021
PAPER 1
Questions & Solutions
Time : 9:00 am – 12:00 pm | Duration : 3 Hrs. | Total Marks : 180
SSUUBBJJEECCTT:: CCHHEEMMIISSTTRRYY
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | TToo KKnnooww mmoorree :: ssmmss RREESSOO aatt 5566667777
Website : www.resonance.ac.in | E-mail : ccoonnttaacctt@@rreessoonnaannccee..aacc..iinn | CCIINN :: UU8800330022RRJJ22000077PPLLCC002244002299
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
PAPER-1 : INSTRUCTIONS TO CANDIDATES
Question Paper-1 has three (03) parts: Physics, Chemistry and Mathematics.
Each part has a total Nineteen (19) questions divided into four (04) sections (Section-1, Section-2, Section-3 & Section-4)
Total number of questions in Question Paper-1 are Fifty Seven (57) and Maximum Marks are One Hundred and Eighty (180).
Type of Questions and Marking Schemes
SECTION-1
• This section contains FOUR (04) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : −1 In all other cases.
SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place
using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal
places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases.
SECTION 3
• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 1
7340010333
PART : CHEMISTRY
SECTION-1
• This section contains FOUR (04) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : −1 In all other cases.
1. The major product formed in the following reaction is
NaNH2
NA liq.NH3
(A)
(B)
(C)
(D)
Ans. (B) Sol. It is a birch reduction. In Birch reduction only internal Alkyne can reduced to trans alkene. While terminal alkyne can’t be reduced to alkene. 2. Among the following, the conformation that corresponds to the most stable conformation of meso-
butane-2,3-diol is
(A)
H Me
H Me
OH
OH
(B)
H OH
MeH
Me
OH
(C)
H
OH
Me
Me
H
OH
(D)
H OH
Me H
Me
OH
Ans. (B) Sol. C & D are not meso compound but A & B are meso compound and out of A and B. B is more stable due
to intramolecular hydrogen bonding.
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 2
7340010333
3. For the given close packed structure of a salt made of cation X and anion Y shown below (ions of
only one face are shown for clarity), the packing fraction is
approximately (packing fraction = 100
efficiencypacking)
(A) 0.74 (B) 0.63 (C) 0.52 (D) 0.48 Ans. (B)
Sol. 2[rx + ry] = 2 a {a = 2ry ; ry = 0.59}
2[rx + ry] = 2 (2ry)
2rx = 2 2 ry – 2ry
rx = ry [ 2 – 1]
rx = 0.414 ry = 0.207 a
Packing fraction = 3
3y
3x
a
r3
4r
3
43
= 0.63
4. The calculated spin only magnetic moments of [Cr(NH3)6]3+ and [CuF6]3– in BM, respectively, are
(Atomic numbers of Cr and Cu are 24 and 29, respectively) (A) 3.87 and 2.84 (B) 4.90 and 1.73 (C) 3.87 and 1.73 (D) 4.90 and 2.84
Ans. (A)
Sol. [Cr(NH3)6]3+ Cr3+ 3d3 unpaired e– = 3
= 3.87 BM [CuF6]3– Cu3+ 3d8 unpaired e– = 2
= 2.87 BM
SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place
using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal
places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases.
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 3
7340010333
Question Stem for Question Nos. 5 to 6 Question Stem
For the following reaction scheme, percentage yields are given along the arrow:
H2O Mg2C3 P
(4.0 g)
T 80%
NaNH2 MeI
Q
red hot iron tube
873 K
40%
Hg2+/H+ 333 K
Ba(OH)2 NaOCl U
(y g)
R (x g) 75%
100%
S heat 80%
(decolourises) Baeyer's reagent)
x g and y g are mass of R and U, respectively.
(Use: Molar mass (in g mol–1) of H, C and O as 1, 12 and 16, respectively) 5. The value of x is ______. Ans. (1.62) 6. The value of y is ______. Ans. (3.2) Sol. (Q.5 & Q.6)
H3O+
Mg2C3 CH3–CCH NaNH2
MeI
– –
Hg2+/H+
–
– – –
– – –
CH3–C–CH3
CH3–CC–CH3
0.1 mol 0.1 × 0.75 = 0.075
O
Ba(OH)2
CH3–C–CH=C–CH3
O
NaOCl
CH3
T 0.05 × 0.08 = 0.04 moles
CHCl3 + HO–C–CH=C
O
CH3
CH3
0.04×0.8 ×100 = 3.2 gm
CH3
CH3 CH3
CH3
CH3 CH3
Red hot tube
0.075 × 0.4 ×1/3 = 1.62
So X = 1.62 ; Y = 3.2
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 4
7340010333
Question Stem for Question Nos. 7 and 8
Question Stem
For the reaction, 𝐗(𝑠) 𝐘(𝑠) + 𝐙(𝑔), the plot of ln op
pzversus
T
104
is given below (in solid line), where 𝑝𝐙 is the pressure (in bar) of the gas Z at temperature T and op = 1 bar.
(Given, R
H
T
1d
)K(lnd o
, where the equilibrium constant, op
pzK and the gas constant, R = 8.314 J K–1
mol–1)
7. The value of standard enthalpy, oH (in kJ mol–1) for the given reaction is ___. Ans. (166.28)
8. The value of oS (in J K–1 mol–1) for the given reaction, at 1000 K is ___.
Ans. (141.34)
Sol. (7 & 8)
Gº =Hº – TSº
–RTlnK = Hº – TSº
lnk =
R
ºS
T
1
R
ºH
y = mx + c
For given reaction K = (P2)
Slope = 21012
)3(710
R
ºH 4
= Hº = (2 × 104)R = 16.628 × 104 J = 166.28 kJ
(ii) y = mx + c
– 3 = –2
T
104
+ c
– 3 = –2 [10] + c
c = 17
But c = R
ºS = 17
Sº = 17 × 8.314 = 141.338 J/K
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 5
7340010333
Question Stem for Question Nos. 9 and 10 Question Stem
The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x C. To this solution A, an
equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The
difference in the boiling points of water in the two solutions A and B is y × 10–2 C.
(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts
dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), 𝐾𝑏 = 0.5 K kg mol–1; Boiling point of pure water
as 100 C.)
9. The value of x is ___.
Ans. (100.1)
10. The value of |y| is ___.
Ans. (2.5)
Sol. (9 & 10)
Initially for solution A [0.1 m AgNO3]
Tb = i Kb m
Tb = 2 × 0.5 × 0.1
T’b – 100 = 0.1
T’b = 100.1°C
x = 100.1°C
(II) 2AgNO3 + BaCl2 Ba(NO3)2 + 2AgCl
0.1V 0.1V
ion milimole
Ag+ 0.1V
NO3– 0.1V
Ba2+ 0.1V
Cl– 0.2V
Tb = i Kb × m
= 0.5
V2
V1.0V1.0V1.0
= 0.5 × 2
3.0
T’b – 100 = 0.075
T’b = 100.075
Differece in Tb = 100.1 – 100.075 = 0.025 = 2.5 × 10–2
y = 2.5
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 6
7340010333
SECTION 3
• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.
11. Given :
OH H
H
CH2OH
CHO
OH H
OH H
HO HNO3
P
[D = +52.7º]
D-Glucose
The compound(s), which on reaction with HNO3 will give the product having degree of rotation,
[]D = –52.7º is (are)
(A)
H
H
CH2OH
CHO
OH
OH
HO
HO
H
H
(B)
H
H
CH2OH
CHO
OH
H
HO
HO
H
HO
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 7
7340010333
(C)
H
OH
CHO
CH2OH
H
H
H
HO
HO
HO
(D)
OH
OH
CH2OH
CHO
H
OH
H
H
HO
H
Ans. (CD) 12. The reaction of Q with PhSNa yields an organic compound (major product) that gives positive Carius
test on treatment with Na2O2 followed by addition of BaCl2. The correct option(s) for Q is (are)
(A) O2N F
NO2
(B)
O2N
I
O2N
(C)
MeS
Br
O2N
(D) O2N
Cl
MeS
Ans. (AD)
13. The correct statement(s) related to colloids is(are)
(A) The process of precipitating colloidal sol by an electrolyte is called peptization.
(B) Colloidal solution freezes at higher temperature than the true solution at the same concentration.
(C) Surfactants form micelle above critical micelle concentration (CMC). CMC depends on temperature.
(D) Micelles are macromolecular colloids
Ans. (C) Sol. Surfactants form micelle above critical micelle concentration (CMC). CMC depends on temperature.
14. An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is(are) (p: pressure, V: volume, T: temperature, H: enthalpy, S: entropy)
(A)
(B)
(C) (D)
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
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PAGE # 8
7340010333
Ans. (ABD)
Sol. (i) For isothermal process T constant and for reversible adiabatic process entropy is constant.
(ii) For isothermal process H = 0
(iii) Slope of adiabatic process is times of isothermal process
So, graph AD is correct.
15. The correct statement(s) related to the metal extraction processes is(are)
(A) A mixture of PbS and PbO undergoes self-reduction to produce Pb and SO2.
(B) In the extraction process of copper from copper pyrites, silica is added to produce copper silicate.
(C) Partial oxidation of sulphide ore of copper by roasting, followed by self-reduction produces blister
copper.
(D) In cyanide process, zinc powder is utilized to precipitate gold from Na[Au(CN)2].
Ans. (ACD)
Sol. (A) PbS + PbO Pb + SO2
(Self reduction of Pb)
(B) In extraction process of Cu SiO2 is added to make FeSiO3
(C) Zn + Na[Au(Cu)2] Au(s) + Na2[Zn(CN)4]
16. A mixture of two salts is used to prepare a solution S, which gives the following results:
Dilute NaOH(aq)
Room temperature
white precipitate(s)
only Room temperature
White precipitate(s)
only
S (aq solution of the salts)
Dilute HCl(aq)
The correct option(s) for the salt mixture is (are)
(A) Pb(NO3)2 and Zn(NO3)2 (B) Pb(NO3)2 and Bi(NO3)3
(C) AgNO3 and Bi(NO3)3 (D) Pb(NO3)2 and Hg(NO3)2
Ans. (ABC)
Sol.
PbCl2
Pb+2
HCl Ag+
White
AgCl White
Zn(OH)2 Zn+2
NaOH Hg+2
White
HgO Red brown
Bi(OH)3 White Bi+3
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 9
7340010333
SECTION 4
• This section contains THREE (03) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on-
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases.
17. The maximum number of possible isomers (including stereoisomers) which may be formed on mono-bromination of 1-methylcyclohex-1-ene using Br2 and UV light is _____ .
Ans. 13
Sol. This is allylic substitution at – carbon
Cl2
UV (553K)
Cl
+
Cl
+
•
Cl
1 2
2
+Cl
2
+
Cl
2
+
2
+Cl
2
18. In the reaction given below, the total number of atoms having sp2 hybridization in the major product P is
______. (1) O3 (excess)
then Zn/H2O
(2) NH2OH (excess) P
Ans. 12 Sol.
NH2 – OH
O O
O O
N – OH HO – N
N – OH N – OH
JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY
Resonance Eduventures Ltd.
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Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 10
7340010333
19. The total number of possible isomers for [Pt(NH3)4Cl2]Br2 is ______ . Ans. (6)
Sol. Total number of isomer of complex [Pt(NH3)4Cl2]Br2
(1) Cis - [Pt(NH3)4Cl2]Br2, (2) Trans - [Pt(NH3)4Cl2]Br2,
NH3
Pt
NH3 Cl
Cl
Cis form
NH3
NH3
NH3
Pt
NH3 NH3
NH3
Cl
Cl
Trans form
(3) Cis - [Pt(NH3)4BrCl]ClBr (4) Trans - [Pt(NH3)4BrCl]ClBr
NH3
Pt
NH3 Cl
Br
Cis form
NH3
NH3
NH3
Pt
NH3 NH3
NH3
Br
Cl
Trans form
(5) Cis - [Pt(NH3)4Br2]Cl2 (6) Trans - [Pt(NH3)4Br2]Cl2
NH3
Pt
NH3 Br
Br
Cis form
NH3
NH3
NH3
Pt
NH3 NH3
NH3
Br
Br
Trans form
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | TToo KKnnooww mmoorree :: ssmmss RREESSOO aatt 5566667777
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TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
SECTION 4
• This section contains THREE (03) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on-
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases.
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