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8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis
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Answers by
IIT-JEE 2010(Division of Aakash Educational Services Ltd.)
PAPER - I : CODES
Q.No. 0 1 2 3 4 5 6 7 8 9
01. D C B D D C A B A D
02. A B C A A B D C D A
03. B B C D B A B C A A
04. B C B A C D C B D D
05. C D D B C A C A B C
06. C A A C B D B D C B
07. D A D C D B A A C B
08. A D A B A C D D B C
09. B,D A,B,C D B,D D A,B,C A,B D A,B,C A,B
10. B,C B,C A,B D B,D B,D B,C B,D A,B D
11. A,B D B,C A,B,C A,B D D A,B B,C A,B,C
12. D B,D A,B,C B,C A,B,C B,C B,D A,B,C D B,C
13. A,B,C A,B B,D A,B B,C A,B A,B,C B,C B,D B,D
14. A B A B A B A B A B
15. D C D C D C D C D C
16. C A C A C A C A C A
17. B D B D B D B D B D
18. C C C C C C C C C C
19. 3 3 0 3 3 0 3 4 3 5
20. 2 2 3 0 3 3 0 3 4 3
21. 0 5 2 3 0 3 3 0 3 4
22. 3 3 5 2 3 0 3 3 0 3
23. 4 4 3 5 2 3 0 3 3 0
24. 3 3 4 3 5 2 3 0 3 3
25. 3 0 3 4 3 5 2 3 0 3
26. 3 3 0 3 4 3 5 2 3 0
27. 0 3 3 0 3 4 3 5 2 3
28. 5 0 3 3 0 3 4 3 5 2
29. A D B C A D A C B C
8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis
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Organic Chemistry Inorganic Chemistry Physical Chemistry Total
Easy 1 4 3 8
Medium 7 4 7 18
Tough 1 0 1 2
Total 9 8 11 28
Markwise 27 24 33 84
XI Marks 33 XII Marks 51
XI syllabus 11 XII syllabus 17
Topic-wise Analysis of CHEMISTRY - IIT-JEE 2010
(Paper-1)
29%
64%
7%
Distribution of Level of Questions in Chemistry
Easy Medium Tough
32%
29%
39%
Topic wise distribution in Chemistry
Organic Chemistry
Inorganic Chemistry
Physical Chemistry
39%
61%
Percentage Portion asked from Syllabus of Class XI
& XII - Chemistry
XI syllabus XII syllabus
32%
29%
39%
Topic wise distribution of marks in Chemistry
Organic Chemistry
Inorganic Chemistry
Physical Chemistry
39%
61%
Percentage Portion asked (Mark-wise) from Syllabus of Class XI & XII -Chemistry
XI Marks XII Marks
8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis
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XII XI XI XI XII XII XII
Calculus Trigonometry Algebra 2D- Geometry Probability 3-D + Vector Algebra Total
Easy 1 2 1 2 0 1 0 7
Medium 2 2 4 1 1 3 2 15
Tough 1 0 1 0 0 0 4 6
Total 4 4 6 3 1 4 6 28
Markwise 12 12 18 9 3 12 18 84
XI Marks 39 XII Marks 45
XI syllabus 13 XII syllabus 15
Topic-wise Analysis of MATHEMATICS - IIT-JEE 2010
(Paper-1)
25%
54%
21%
Distribution of Level of Questions in Mathematics
Easy Medium Tough
14%
14%
22%11%
4%
14%
21%
Topic wise distribution in MathsCalculus
Trigonometry
Algebra
2D- Geometry
Probability
3-D + Vector
Algebra
46%54%
Percentage Portion asked from Syllabus of ClassXI & XII
XI syllabus XII syllabus
14%
14%
22%11%
4%
14%
21%
Topic wise distribution of marks in Maths
Calculus
Trigonometry
Algebra
2D- Geometry
Probability
3-D + Vector
Algebra
46%
54%
Percentage Portion asked (Mark-wise) from Syllabus ofClass XI & XII
XI Marks XII Marks
8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis
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XII XI XII XI XII XI XII XII XI&XII
ElectricityHeat &
Thermodynamics
Modern
PhysicsMechanics Optics
Oscillation &
Waves
Magnetism
and EMI
Experimenta
l physics
Miscellan
eousTotal
Easy 3 2 1 2 1 1 1 0 0 11
Medium 1 2 0 5 1 0 1 2 2 14
Tough 0 0 0 2 0 1 0 0 0 3
Total 4 4 1 9 2 2 2 2 2 28
Markwise 12 12 3 27 6 6 6 6 6 84
XI Marks 51 XII Marks 33
XI syllabus 17 XII syllabus 11
Topic-wise Analysis of PHYSICS - IIT-JEE 2010
(Paper-1)
39%
50%
11%
Distribution of Level of Questions in Physics
Easy Medium Tough
15%
14%
4%32%
7%
7%
7%7%
7%
Topic wise distribution in PhysicsElectricity
Heat & Thermodynamics
Modern Physics
Mechanics
Optics
Oscillation & Waves
Magnetism and EMI
Experimental physics
Miscellaneous
61%
39%
Percentage Portion asked from Syllabus of ClassXI & XII - Physics
XI syllabus XII syllabus
15%
14%
4%32%
7%
7%
7%7%
7%
Topic wise distribution of marks in Physics
Electricity
Heat & Thermodynamics
Modern Physics
Mechanics
Optics
Oscillation & Waves
Magnetism and EMI
Experimental physics
Miscellaneous
61%
39%
Percentage Portion asked (Mark-wise) from Syllabus of ClassXI & XII - Physics
XI Marks XII Marks
8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-1 Answer and Solution With Analysis
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Solutions
forforforforforIIT-JEE 2010
Time : 3 hrs. Max. Marks: 252
Regd. Office: Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472
Instructions :
1. The question paper consists of 3 parts (Chemistry, Mathematics, and Physics). Each part consists
of four Sections.
2. For each question in Section I,you will be awarded 3 marks if you have darkened only the bubble
corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,
minus one (1) mark will be awarded.
3. For each question in Section II,you will be awarded 3 marks if you have darkened only the bubble
corresponding to the correct answer and zero mark if no bubbles are darkened. Partial marks will
be awarded for partially correct answers. No negative marks will be awarded in this Section.
4. For each question in Section III,you will be awarded 3 marks if you have darkened only the bubble
corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,
minus one (1) mark will be awarded.
5. For each question in Section IV,you will be awarded 3 marks if you have darkened only the bubble
corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks
will be awarded for in this Section.
DATE : 11/04/2010
PAPER - 1 (Code - 5)
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PARTI : CHEMISTRY
SECTION - I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.
1. The bond energy (in kcal mol1) of a CC single bond is approximately
(A) 1 (B) 10 (C) 100 (D) 1000
Answer (C)
Hints : Bond energy of CC is considered to be 100 kcal mol1 approximately.
2. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is
(A) Br2(g) (B) Cl
2(g) (C) H
2O(g) (D) CH
4(g)
Answer (B)
Hints : Cl2
is considered as standard state of chlorine and enthalpy of formation in standard state is zero (assumed).
3. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that followsArrhenius equation is
(A) k
T
(B) k
T
(C) k
T
(D)k
T
Answer (A)
Hints :aE
RTk A e=
ln k = ln A aE
RT
4. In the reaction OCH3HBr
the products are
(A) OCH3 and H2Br (B) Br and CH Br3
(C) Br and CH OH3 (D) OH and CH Br3
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Answer (D)
Hints :
O
CH3
HBrO
CH3
H
Br OH + CH Br3
5. The correct statement about the following disaccharide is
CH OH2
H
OH H
H OH
H
OH
OH
OCH CH O2 2
CH OH2O
CH OH2
H
OH
H
H
OH
(a) (b)
(A) Ring (a) is pyranose with -glycosidic link (B) Ring (a) is furanose with -glycosidic link
(C) Ring (b) is furanose with -glycosidic link (D) Ring (b) is pyranose with -glycosidic link
Answer (A)
Hints :
CH OH2
H
OH H
H OH
H
OH
OH
OCH CH O2 2
CH OH2O
CH OH2
H
OH
H
H
OH
(a) (b)
5
4
3 2
1
-linkage
Ring (a) is six membered oxygen containing ring.
Pyranose ring and CH2OH of C5 and OR of C1 are across of one another hence, it is -glycosidic linkage.
6. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne.The bromoalkane and alkyne respectively are
(A) BrCH2CH2CH2CH2CH3 and CH3CH2C CH (B) BrCH2CH2CH3 and CH3CH2CH2C CH
(C) BrCH2CH
2CH
2CH
2CH
3and CH
3C CH (D) BrCH2CH2CH2CH3 and CH3CH2C CH
Answer (D)
Hints :
CH CH C CH3 2 NH2
CH CH C C3 2
CH CH CH2 2 2CH3
Br
CH CH C CCH CH CH CH3 32 2 2 23-octyne
7. The ionization isomer of [Cr(H2O)
4Cl(NO
2)]Cl is
(A) [Cr(H2O)4(O2N)]Cl2 (B) [Cr(H2O)4Cl2] (NO2)(C) [Cr(H
2O)
4Cl(ONO)]Cl (D) [Cr(H
2O)
4Cl
2(NO
2)] H
2O
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Answer (B)
Hints : Cr(H2O)4Cl(NO2)]Cl and [Cr(H2O)4Cl2](NO2) have same molecular formula but they give different ions onionization.
8. The correct structure of ethylenediaminetetraacetic acid (EDTA) is
(A) N CH CH N
HOOC CH2
HOOC CH2
CH2 COOH
CH2 COOH
(B) N CH2 CH2 N
HOOC
HOOC
COOH
COOH
(C) N CH2 CH2 N
CH2 COOH
CH2 COOH
HOOC CH2
HOOC CH2
(D) N CH CH N
H
CH2 COOH
HOOC CH2
HCH2
HOOC
CH2
COOH
Answer (C)
Hints :
N CH2 CH2 N
CH2 COOH
CH2 COOH
HOOC CH2
HOOC CH2
SECTION - II
Multiple Correct Choice Type
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.
9. In the reaction
OH
NaOH(aq)/Br2 the intermediate(s) is/are
(A)
O
Br
Br
(B)
O
BrBr
(C)
Br
O
(D)
Br
O
Answer (A, B, C )
Hints : (A) & (C) is usual product. Formation of (B) is justified as
O
BrBr
O
BrBr Br
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10. In the Newman projection for 2, 2-dimethylbutane
XCH3
HH
H C3
Y
X and Y can respectively be
(A) H and H (B) H and C2H
5(C) C
2H
5and H (D) CH
3and CH
3
Answer (B, D)
Hints :
XCH3
HH
H C3
Y
C
CY
CH3H C3
X
HH
When X = H & Y = C2H
5
C
C C H2 5
CH3H C3
H
HH
2
1
2, 2- dimethyl butane
When X = CH3
and Y = CH3
C
C CH3
CH3H C3
CH3
HH
2
3 1
4
2, 2-dimethyl butane
11. Aqueous solutions of HNO3, KOH, CH
3COOH, and CH
3COONa of identical concentrations are provided. The pair(s)
of solutions which form a buffer upon mixing is(are)
(A) HNO3 and CH3COOH (B) KOH and CH3COONa
(C) HNO3
and CH3COONa (D) CH
3COOH and CH
3COONa
Answer (D)
Hints : Solution of weak acid and its salt with strong base behave like buffer.
CH3COOH CH
3COONa
12. The reagent(s) used for softening the temporary hardness of water is(are)
(A) Ca3(PO
4)2
(B) Ca(OH)2
(C) Na2CO
3(D) NaOCl
Answer (B, C)
Hints : Ca(OH)2
and Na2CO
3is used to remove hardness of water. Ca(OH)
2is used in Clarke's method.
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13. Among the following, the intensive property is (properties are)
(A) Molar conductivity (B) Electromotive force (C) Resistance (D) Heat capacity
Answer (A, B)
Hints : Molar conductivity and electromotive force are intensive properties.
SECTION - III
Paragraph Type
This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon thesecond paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B),(C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 14 to 15
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. Theresulting potential difference across the cell is important in several processes such as transmission of nerve impulses
and maintaining the ion balance. A simple model for such a concentration cell involving a metal M isM(s) | M+(aq; 0.05 molar) || M+ (aq; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential |Ecell
| = 70 mV.
14. For the above cell
(A) Ecell
< 0; G > 0 (B) Ecell > 0; G < 0 (C) Ecell < 0; G > 0 (D) Ecell > 0; G < 0
Answer (B)
Hints : Cell reaction
(1M) (0.05 M)M M+ +
Apply Nernst equation
0.059 0.05E E log
1 1=
( )20.059
E log 5 101
=
( )0.059
E 2 log51
= +
15. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potentialwould be
(A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV
Answer (C)
Hints :1
2
E log0.05
E log0.0025=
21
4
2
E log5 10
E log25 10
=
1E 70(given)
2
70 1.3 1E 2.6 2
= =
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Paragraph for Questions 16 to 18
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of
copper include chalcanthite (CuSO45H
2O), atacamite (Cu
2Cl(OH)
3), cuprite (Cu
2O), copper glance (Cu
2S) and malachite
(Cu2(OH)
2CO
3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS
2). The extraction
of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.
16. Partial roasting of chalcopyrite produces
(A) Cu2S and FeO (B) Cu
2O and FeO (C) CuS and Fe
2O
3(D) Cu
2O and Fe
2O
3
Answer (A)
Hints : Partial roasting of chalcopyrite
2O /
2 2CuFeS Cu S FeO +
17. Iron is removed from chalcopyrite as
(A) FeO (B) FeS (C) Fe2O
3(D) FeSiO
3
Answer (D)
Hints : FeO + SiO2 FeSiO
3
gangue flux slag
18. In self-reduction, the reducing species is
(A) S (B) O2 (C) S2 (D) SO2
Answer (C)
Hints : Cu2S + Cu2O 6Cu + SO2
SECTION - IV
(Integer Type)
This section contains 10 questions. The answer to each question is a Single digit integer ranging from 0 to 9. The correct
digit below the question number in the ORS is to be bubbled.
19. Based on VSEPR theory, the number of 90 degree F Br F angles in BrF5
is
Answer (0)
Hints :
Br
FeqFeqFeq
Feq
Fa
Presence of lone pair will distort all the bond angles and none of the bond angle is exactly equal to 90
FaBrFeq = 84
Feq
BrFeq
= 89
20. The value of n in the molecular formula BenAl
2Si
6O
18is
Answer (3)
Hints : Be3Al
2Si
6O
18
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21. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL.
The number of significant figures in the average titre value is
Answer (3)
Hints : Average titre value =25.2 25.25 25
3
+ +
= 25.15
Siginificant figure is 3.
22. The concentration of R in the reaction R P was measured as a function of time and the following data is
obtained
[R] (molar)
t(min.)
1.0 0.75 0.40 0.10
0.0 0.05 0.12 0.18
The order of the reaction isAnswer (0)
Hints : For zero order reaction
dxk
dt=
from 1 to 0.75 (M) from 0.75 to 0.40 from 0.40 to 0.10
0.255
0.05=
0.355
0.07=
0.305
0.06=
It mean order of this reaction is zero.
23. The number of neutrons emitted when 23592 U undergoes controlled nuclear fission to
142
54 Xe and9038 Sr is
Answer (3)
Hints :
142235 90 15492 38 0U Xe+ Sr +3 n
24. The total number of basic groups in the following form of lysine is
H N3 CH2 CH2 CH2 CH2
CHH N2
C
O
O
Answer (2)
Hints :
H N3 CH2 CH2 CH2 CH2
CH
H N2
C
O
O
Basic site
Basic site (Carboxylate ion will
also behave as proton acceptor)
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25. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H
6is
Answer : (5)
Hints : Index of hydrogen deficiency in the compound C4H
6is 2. Therefore either bicyclic compound or cycloalkenes
are possible as cyclic isomers.
Possible isomers are
, , , ,
26. In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is
3
2
1. O 1. NaOH(aq)
2. Zn,H O 2. Heat Y
Answer : (3)
Hints :
1. O
2. Zn H O
3
2
O
O H
O
O
1. O H
O
O+HO3
+
O
H O2
O
+
OH
O
+
O
On dehydration thisalcohol will give threeproducts
27. Amongst the following, the total number of compounds soluble in aqueous NaOH is
NH C3 CH3 COOH OCH CH2 3 OH
CH OH2
NO2 CH CH2 3 COOH
CH CH2 3
OH
NH C3 CH3
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Answer : (4)
Hints : Carboxylic acids and aromatic alcohols dissolves in aqueous NaOH.
COOH OH COOHOH
NH C3 CH3
, , ,
will be soluble in aqueous NaOH.
28. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
KCN K2SO
4(NH
4)2C
2O
4NaCl Zn(NO
3)2
FeCl3
K2CO
3NH
4NO
3LiCN
Answer : (3)
Hints : In basic medium, red litmus paper turn blue
(1)2CN H O HCN OH
+ +(2) 2
3 2 3CO H O HCO OH + +
(3)2CN H O HCN OH
+ +
PARTII : MATHEMATICSSECTION - I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.
29. Let f, gand hbe real-valued functions defined on the interval [0, 1] by2 2 2 2
( ) , ( )x x x x f x e e g x xe e = + = + and2 22( ) x xh x x e e = + . If a, band cdenote, respectively, the absolute maximum of f, gand hon [0, 1], then
(A) a = band cb (B) a = cand ab (C) aband cb (D) a = b = c
Answer (D)
Hints :
The given functions
2 2
( ) x xf x e e = +2 2
( ) x xg x xe e = +
2 22( ) x xh x x e e = +
are strictly increasing in [0, 1], hence at x= 1, the given functions attain absolute maximum all equal to1
e
e
+
and consequently a = b = c.
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30. Let pand qbe real numbers such that p 0, p3qand p3q. If and are nonzero complex numbers
satisfying + = pand 3 + 3 = q, then a quadratic equation having and as its roots is
(A) (p3 + q)x2 (p3 + 2q)x+ (p3 + q) = 0
(B) (p3 + q)x2 (p3 2q)x+ (p3 + q) = 0
(C) (p3 q)x2 (5p3 2q)x+ (p3 q) = 0
(D) (p3 q)x2 (5p3 + 2q)x+ (p3 q) = 0
Answer (B)
Hints :
We have + = p
3 +3 = q= ( + )3 3( + ) = p3 + 3p()
3
3
p q
p
+ =
The quadratic equation with and as roots is
2 0x x
+ + =
2
2 ( ) 2 1 0x x +
+ =
32
2
3
23
1 0
3
p qp
px xp q
p
+
+ =+
(p3 + q)x2 (p3 2q)x+ (p3 + q) = 0
31. The value of 3 400
1 (1 )lim
4
x
x
t n tdt
x t
+
+
is
(A) 0 (B)1
12(C)
1
24(D)
1
64
Answer (B)
Hints :
We have
40
3 4 30 00
(1 )
41 (1 )lim lim
4
x
x
x x
t n tdt
tt n tdt
x t x
+
++=
+
= 4 20
(1 ) 1Lt
4 3x
x n x
x x
+
+
= 20
(1 ) 1 1Lt
123( 4)x
n x
x x
+ =
+
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32. The number of 3 3 matrices A whose entries are either 0 or 1 and for which the system
1
0
0
x
A y
z
=
has
exactly two distinct solutions, is
(A) 0 (B) 29 1 (C) 168 (D) 2
Answer (A)
Hints :
The equation
1
0
0
x
A y
z
=
has two distinct solutions. It should be noted here that the given equation is linear
equation in 3 variables, which may have no solution, or unique solution or infinitely many solutions.
Hence there does not exist any matrix A such that the given equation has exactly two solutions and consequentlynumber of 3 3 matrices is 0.
33. Let P, Q, R and S be the points on the plane with position vectors 2 , 4 , 3 3 and 3 2i j i i j i j + +respectively. The quadrilateral PQRSmust be a
(A) Parallelogram, which is neither a rhombus nor a rectangle
(B) Square
(C) Rectangle, but not a square
(D) Rhombus, but not a square
Answer (A)
Hints :
Let Obe the origin
then 2OP i j =
4OQ i=
3 3OR i j = +
P
S R
Q
O
3 2OS i j = +
Here we have
6PQ i j = +
3QR i j = +
6RS i j =
and 3PS i j = +
5 4PR i j = +
7 2QS i j = +
35 8 27 0PR QS = + =
Diagonals are not perpendicular
and | | | |, | | | |PQ RS QR PS = =
Hence PQRSis a parallelogram which is neither a rhombus nor a rectangle.
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34. Let be a complex cube root of unity with 1. A fair die is thrown three times. If r1, r
2and r
3are the numbers
obtained on the die, then the probability that 31 2 0rr r + + = is
(A)1
18(B)
1
9(C)
2
9(D)
1
36
Answer (C)
Hints :
is a complex cube root of unity with 1
31 2 0rr r + + = where r1, r2, r3 can be successively selected from the sets {1, 4}, {2, 5} and {3, 6}.
n(S) = 6 6 6 = 216
n(E) = 2C1
2C1
2C1
3C1
2C1
= 6 8
Required probability =6 8 2
216 9
=
35. Equation of the plane containing the straight line2 3 4
x y z= = and perpendicular to the plane containing the straight
lines3 4 2
x y z= = and
4 2 3
x y z= = is
(A) x+ 2y 2z= 0 (B) 3x+ 2y 2z= 0 (C) x 2y+ z= 0 (D) 5x+ 2y 4z= 0
Answer (C)
Hints : The equation of the plane containing the line2 3 4
x y z= = may be written as
ax + by + cz =0 while 2a+ 3b+ 4c= 0 ... (i)
The equation of the plane containing straight lines3 4 2
x y z= = and4 2 3
x y z= = can be put in the form
a1x+ b
1y+ c
1z= 0
where 3a1
+ 4b1
+ 2c1
= 0
4a1
+ 2b1
+ 3c1
= 0
1 1 18 1 10
a b c= =
Then the equation of the plane containing given two lines is
8xy 10c= 0 ... (ii)
Since the plane (i) is perpendicular to the plane (ii),
hence 8ab 10c= 0
Also 2a+ 3b+ 4c= 0
4 30 20 32 24 2
a b c= =
+ +
26 52 26
a b c= =
1 2 1
a b c= =
Substituting for a, b, cin equation (i), the required equation of the plane is
x 2y + z= 0
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36. If the angles A, B and Cof a triangle are in an arithmetic progression and if a, band cdenote the lengths of
the sides opposite to A, Band Crespectively, then the value of the expression sin2 sin2a c
C Ac a
+ is
(A)1
2
(B)3
2
(C) 1 (D) 3
Answer (D)
Hints :
Angles A, B, Care in A.P. hence B= 60 =3
sin2 sin2a c
C Ac a
+ =sin sin
2 cos 2 cosC A
a C c Ac a
+
= 2(sinA cos C+ cosA sinC)
= 2 sin(A + C) = 2 sinB=3
2 32
=
SECTION - II
(Multiple Correct Choice Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.
37. Let ABCbe a triangle such that6
ACB
= and let a, band cdenote the lengths of the sides opposite to A,
Band Crespectively. The value(s) of xfor which a= x2
+ x+ 1, b= x2
1 and c= 2x+ 1 is/are
(A) ( )2 3 + (B) 1 3+ (C) 2 3+ (D) 4 3Answer (B)
Hints :2 2 2
cos2
b a cC
ab
+ =
A B
C
6( ) ( ) ( )
( )( )
2 2 22 2
2 2
1 1 2 13
22 1 1
x x x x
x x x
+ + + +=
+ +
2 2 2 2( 2)( 1) ( 1) ( 1) 3( 1)( 1)x x x x x x x x + + + = + +
2 2 23( 1) 2 1x x x x x + + = + +
2(2 3) (2 3) ( 3 1) 0x x + + =
(2 3)& 1 3x x= + = +
But (2 3)x= +
shows that cis negative, which is not possible in a triangle.
Hence 1 3x= + is only possible value of x.1 3x = +
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38. Let z1
and z2
be two distinct complex numbers and let z = (1 t) z1
+ tz2
for some real number t with0 < t< 1. If Arg (w) denotes the principal argument of a nonzero complex number w, then
(A) |zz1| + |zz
2| = |z
1z
2| (B) Arg (zz
1) = Arg (zz
2)
(C)1 1
2 1 2 1
0z z z z
z z z z
=
(D) Arg (zz1) = Arg (z2 z1)
Answer (A, C, D)
Hints : 1 21
zz z
t t=
we have z= (1 t)z1
+ tz2, 0 < t< 1,
( )2 111
t z t z z
t t
+ =
+
( )2 11z t z t z = +
z1, z, z
2are collinear
so Arg(zz1) = Arg(zz
2) = Arg(z
2z
1)
39. The value(s) of( )
441
2
0
1
1
x xdx
x
+is/are
(A)22
7 (B)
2
105(C) 0 (D)
71 3
15 2
Answer (A)
Hints : ( )
441
2
0
1
1
x xdx
x
+
1 8 7 6 5 4
2
0
4 6 4.1
x x x x x dxx
+ += +
16 5 4 2
2
0
44 5 4 4
1x x x x dx
x
= + + +
1 4 5 44 4
7 6 5 3 4
= + +
22
7=
40. Let A and Bbe two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle ofradius rhaving ABas its diameter, then the slope of the line joining A and Bcan be
(A)1
r (B)
1
r(C)
2
r(D)
2
r
Answer (C, D)
Hints :
Mid - point of A & B
a(t1
+ t2) = r
1 2rt ta
+ =
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Slope of( )
( )1 2
2 21 21 2
2 2 2a t t aAB
t t rt t
= = =
+
CB
y = ax2
4
A ( )21 1, 2at at
( )22 2, 2at at In y2 = 4x
a= 1
slope =2
r
If circle in below x-axis, then
Slope = 2
r
41. Let fbe a real-valued function defined on the interval (0, ) by ( )0
1 sinx
f x lnx t dt = + + . Then which of the
following statement(s) is/are true?
(A) f(x) exists for all ( )0,x
(B) f(x) exists for all ( )0,x and f is continuous on (0, ), but not differentiable on (0, )
(C) There exists > 1 such that |f(x)| < |f(x)| for all ( ),x
(D) There exists > 0 such that |f(x)| + |f(x)| for all ( )0,x
Answer (B, C)
Hints : ( ) 1' 1 sinf x xx= + + , x> 0 but not differentiable in (0, ) as sinxmay be 1 and then f(x) can't exist.
Clearly f(x) is continuous on (0, )
(D) Is not true because f(x) tends to as xtends to
(C) Is true because ( )' 3f x if x> 1
But |f(x)| > 3, if x> e3
So we may take = e3
SECTION - III
(Paragraph Type)
This Section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon thesecond paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B),(C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 42 to 43
The circle x2 + y2 8x= 0 and hyperbola2 2
19 4
x y = intersect at the points A and B.
42. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
(A)2 5 20 0x y =
(B)2 5 4 0x y + =
(C) 3x 4y+ 8 = 0 (D) 4x 3y+ 4 = 0
Answer (B)
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Hints : Equation of tangent to hyperbola having slope mis
29 4y mx m = + ...(1)
Equation of tangent to circle is
( ) 24 16 16y m x m = + + ...(2)
(1) and (2) will be identical for2
5m= satisfy
Equation of common tangent is 2 5 4 0x y + =
43. Equation of the circle with ABas its diameter is
(A) x2 + y2 12x+ 24 = 0 (B) x2 + y2 + 12x+ 24 = 0
(C) x2 + y2 + 24x 12 = 0 (D) x2 + y2 24x 12 = 0
Answer (A)
Hints :
The equation of the hyperbola is
2 2
19 4
x y =
and that of circle is
x2 + y2 8x= 0
For their points of intersection
2 28
19 4
x x x+ =
4x2 + 9x2 72x= 36
13x2 72x 36 = 0
13x2 78x+ 6x 36 = 0
13x(x 6) + 6(x 6) = 0
x= 6,
13
6x=
13
6x= not acceptable
Now, for x= 6, 2 3y=
Required equation is
( ) ( )( )26 2 3 2 3 0x y y + + =
x2 12x+ y2 + 24 = 0
x2 + y2 12x+ 24 = 0
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Paragraph for Questions 44 to 46
Let pbe an odd prime number and Tp
be the following set of 2 2 matrices
: , , {0, 1, 2, ....... 1}Pa b
T A a b c p c a
= =
44. The number of A in Tp
such that A is either symmetric or skew-symmetric or both, and det (A) divisible by pis
(A) (p 1)2 (B) 2(p 1) (C) (p 1)2 + 1 (D) 2p 1
Answer (D)
Hints :
If A is skew-symmetric, |A| = b2
p |A| iff b= 0 in which case A is symmetric.
If A is2 2
, | |
a b
A a bb a
=
| |p A if either pa+ bor pab.
The second case occurs only when a = b which gives p choices. The first case again gives p choices with
a= 0 = bcounted again.
The number of such matrices is 2p 1.
45. The number of A in Tp
such that the trace of A is not divisible by pbut det (A) is divisible by pis
[Note : The trace of a matrix is the sum of its diagonal entries](A) (p 1)(p2 p+ 1) (B) p3 (p 1)2 (C) (p 1)2 (D) (p 1) (p2 2)
Answer (C)
Hints :
Tr(A) = 2awill be divisible by pif and only if a= 0.
We want to find all matricesa b
c a
with a= 1, 2, ....., p 1 for which a2 bc 0 (mod p).
There are exactly p 1 ordered pairs (b, c) for any value of a.
Required number is (p 1)2
46. The number of A in Tp
such that det (A) is not divisible by pis
(A) 2p2 (B) p3 5p (C) p3 3p (D) p3 p2
Answer (D)
Hints :
The number of matrices for which pdoes not divide Tr(A) is (p 1)p2 of these (p 1)2 are such that p |A|. The
number of matrices for which pdivides Tr(A) and pdoes not divide |A| are (p 1)2.
Required number is (p 1)p2 (p 1)2 + (p 1)2 = p3 p2
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SECTION - IV
(Integer Type)
This section contains 10 questions. The answer to each question is a Single digit integer ranging from 0 to 9. The correct
digit below the question number in the ORS is to be bubbled.
47. For any real number x, let [x] denote the largest integer less than or equal to x. Let fbe a real valued functiondefined on the interval [10, 10] by
{ } if [ ] is odd,( )
1 { } if [ ] is even
x xf x
x x
=
Then the value of
102
10
( )cos10
f x x dx
is
Answer (4)
Hints :
[ ] if [ ] is odd,( )
1 + [ ] if [ ] is even
x x xf x
x x x
=
10 9 2 1 0 1 2 9 10
f(x)and cosxare both periodic with period 2 and are both even.
10 10
10 0
( )cos 2 ( )cosf x xdx f x xdx
=
2
0
10 ( )cosf x xdx =
1 1 1
0 0 0
( )cos (1 )cos cosf x xdx x xdx u udu = =
2 2 1
1 1 0
( )cos ( 1)cos cosf x xdx x xdx u udu = =
10 1
10 0
( )cos 20 cosf x xdx u udu
=
2
40=
102
10
( )cos 410
f x xdx
=
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48. Let be the complex number2 2
cos sin .3 3
i
+ Then the number of distinct complex numbers z satisfying
2
2
2
1
1 0
1
z
z
z
+
+ =
+
is equal to
Answer (1)
Hints :
Let
2
2
2
1
1
1
A
=
2
0 0 0
0 0 0 , ( ) 0, | | 0
0 0 0
A Tr A A = = =
A3 = 0
2
2
2
1
1 | | 0
1
z
z A zI
z
+
+ = + =
+
z3
=0 z= 0, the number of zsatisfying the given equation is 1.
49. Let Sk, k= 1, 2,.....100, denote the sum of the infinite geometric series whose first term is
1
!
k
k
and the common
ratio is1
.k
Then the value of2 100
2
1
100( 3 1)
100!k
k
k k S=
+ + is
Answer (4)
Hints :
We have,
1
!1
1k
k
kS
k
=
1
( 1)!k=
Now,
2 1( 3 1) {( 2)( 1) 1}( 1)!
kk k S k k k
+ =
1 1
( 3)! ( 1)!k k=
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100
2
1
1 1| ( 3 1) | 1 1 2
99! 98!k
k
k k S=
+ = + + +
21004
100!=
2 1002
1
100| ( 3 1) | 4
100!k
k
k k S=
+ + =
50. The number of all possible values of , where 0 < < , for which the system of equations
(y + z)cos3 = (xyz)sin3
2cos3 2sin3sin3x
y z
= +
(xyz)sin3 = (y+ 2z)cos3 + ysin3
have a solution (x0, y
0, z
0) with y
0z
0 0, is
Answer (3)
Hints :
Given equation can be written as
cos3 cos3sin3 0x
y z
= ...(i)
2cos3 2sin3sin3 0x
y z
= ...(ii)
2 1sin3 cos3 (cos3 sin3 ) 0x
y z + = ...(iii)
Equations (ii) and (iii) imply
2sin3 = cos3 + sin3
sin3 = cos3
5 9
tan3 1 3 , ,4 4 4
= =
or 5 9, ,12 12 12 =
51. Let fbe a real-valued differentiable function on R(the set of all real numbers) such that f(1) = 1. If the y-intercept
of the tangent at any point P(x, y) on the curve y= f(x) is equal to the cube of the abscissa of P, then the value
of f(3) is equal to
Answer (9)
Hints :
The equation of the tangent at (x, y) to the given curve y= f(x) is
( )dy
Y y X x dx
=
intercept =dy
Y y xdx
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According to the question,
3 dyx y xdx
=
2dy y
xdx x =
which is linear is x
1
. .dx
xI F e
=
=1
x
Required solution is
21 1
.y x dx x x=
2
2
y xC
x
= +
3
2
xy Cx
= +
at x= 1, y= 1
1
12
C
= +
3
2C=
Now,27 3
( 3) ( 3)2 2
f = +
27 99
2
= =
52. The number of values of in the interval ,2 2
such that 5
n
for n= 0, 1, 2 and tan = cot 5 as
well as sin2 = cos4 is
Answer (3)
Hints :
tan = cot5
we have,
tan tan 52
=
52
n = +
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62
n
=
12 6
n =
Also,
cos4 sin2 cos 22
= =
4 2 22
n
=
Taking positive
6 22
n
= +
3 12n = +
Taking negative
2 22
n
=
4n
=
Above values of suggests that there are only 3 common solutions.
53. The maximum value of the expression2 2
1
sin 3sin cos 5cos + + is
Answer (2)
Hints :
We have,
2 2
1( )
sin 3sin cos 5cosf =
+ +
Let g() = sin2 + 3sincos + 5cos2
1 cos2 1 cos2 35 sin22 2 2
+ = + +
33 2cos2 sin2
2= + +
min
9( ) 3 4
4g = +
5 13
2 2= =
maxmin
1( ) 2( )
fg
= =
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54. if anda b
are vectors in space given by 2 2 3
and = ,5 14
i j i j k a b
+ +=
then the value of
(2 ).[( ) ( 2 )]a b a b a b +
is
Answer (5)
Hints :
From the given information, it is clear that
2| | 1, | | 1, . 0
5
i ja a b a b
= = = =
Now, (2 ).[( ) ( 2 )]a b a b a b +
= 2 2(2 ).[ ( . ). 2 . 2( . ). ]a b a b a b a b a b a a + +
= [2 ].[ 2 ]a b b a + +
= 4a2 + b2
= 4.1 + 1 = 5 [ as . 0]a b=
55. The line 2x+ y= 1 is tangent to the hyperbola2 2
2 21.
x y
a b = If this line passes through the point of intersection
of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is
Answer (2)
Hints :
F1 12
, 0
Clearly1
2
a
e= ...(i)
Also y= 2x+ 1 is tangent to hyperbola
1 = 4a2 b2 ...(ii)
( )221
4 1ea
=
2
2
45 e
e=
e4 5e2 + 4 = 0
(e2 4)(e2 1) = 0
e= 2, e= 1
e= 1 gives the conic as parabola. But conic is given as hyperbola, hence
e= 2
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56. If the distance between the plane Ax 2y + z = dand the plane containing the lines1 2 3
2 3 4
x y z = = and
2 3 4
3 4 5
x y z = = is 6 , then |d| is
Answer (6)
Hints :
Equation of plane containing the given lines is
1 2 3
2 3 4 0
3 4 5
x y z
=
(x 1)( 1) (y 2)(2) + (z 3) (1)
x+ 1 + 2y 4 z+ 3 = 0
x+ 2y z= 0 ...(i)
Given plane is
x 2y+ z = d ...(ii)
(i) and (ii) are parallel
Clearly A = 1
Now, distance between plane 61 4 1
d= =
+ +
| d| = 6
PARTIII : PHYSICS
SECTION - I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of whichONLY ONE is correct.
57. A thin uniform annular disc (see figure) of mass Mhas outer radius 4Rand inner radius 3R. The work requiredto take a unit mass from point Pon its axis to infinity is
3R4R
P
4R
(A)
2
(4 2 5)7
GM
R (B)2
(4 2 5)7
GM
R (C) 4
GM
R (D)
2
( 2 1)5
GM
R
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Answer (A)
Hints :
Potential4
2 23 16
R
P
R
GdMV
R x
=
+
=
4
2 2 23
27 16
R
R
M xdx G
R R x
+
Solving
2(4 2 5)
7P
GMV
R
=
P
X
dx
Work done in moving a unit mass from Pto
=
2
0 (4 2 5)7PGM
V V R
=
=2
(4 2 5)7
GM
R
58. A block of mass mis on an inclined plane of angle . The coefficient of friction between the block and the plane
is and tan > . The block is held stationary by applying a force Pparallel to the plane. The direction of force
pointing up the plane is taken to be positive. As P is varied from P1
= mg(sin co s) to
P2
= mg(sin + cos), the frictional force fversus Pgraph will look like
P
(A)P
2
P1
P
f
(B)P
2P
1
P
f
(C)P2
P1
P
f
(D)P
2P
1
P
f
Answer (A)
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Hints :
From FBD of block
N= mgcos
P+ f= mgsin
f= mgsin P
fN
mg
P
As Pvaries from (mgsin cos) to (mgsin + cos), fvaries
from + mgcos to mgcos. This value of friction is always
less than or equal to Nin magnitude.
GraphP1
P2
59. A thin flexible wire of length L is connected to two adjacent fixed points and carries a current Iin the clockwise
direction, as shown in the figure. When the system is put in a uniform magnetic field of strength Bgoing into
the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
(A) IBL (B)IBL
(C)
2
IBL
(D)
4
IBL
Answer (C)
Hints :
From FBD of a small element on wire
2 sin2
dT idlB
=
22
dT iRd B
=
idlB
T T
T= BiR
2BilT =
idlB
T T
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60. An AC voltage source of variable angular frequency and fixed amplitude V0
is connected in series with a
capacitance Cand an electric bulb of resistance R(inductance zero). When is increased
(A) The bulb glows dimmer (B) The bulb glows brighter
(C) Total impedance of the circuit is unchanged (D) Total impedance of the circuit increases
Answer (B)Hints :
As is increased, impedance of circuit decreases.
R
Current increases
Bulb glows brighter.
61. To verify Ohms law, a student is provided with a test resistor RT, a high resistance R
1, a small resistance R
2,
two identical galvanometers G1
and G2, and a variable voltage source V. The correct circuit to carry out the
experiment is
(A)
G2
R2
G1
RT
R1
V
(B)
G2
R1
G1
RT
R2
V
(C)
R1
G1
RT
R2
G2
V
(D)
R2
G1
RT
R1
G2
V
Answer (C)
Hints :
G1
is used as voltmeter by connecting it in series with high resistance and applying it across RT
in parallel G2
is used as ammeter by shunting it with small resistance and connecting in series with RT.
R1
G1
R2
G2
RT
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62. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with theincrease in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistance R
100, R
60and R
40, respectively, the relation between these resistance is
(A)100 40 60
1 1 1
R R R= + (B) R
100= R
40+ R
60
(C) R100
> R60
> R40
(D)100 60 40
1 1 1
R R R> >
Answer (D)
Hints :
2VP
R=
1
PR
100 60 40
1 1 1R R R
> >
Normally100 40 60
1 1 1
R R R= + could have been correct. But small change in resistance due to temperature change
can destroy the equality.
63. A real gas behaves like an ideal gas if its
(A) Pressure and temperature are both high (B) Pressure and temperature are both low
(C) Pressure is high and temperature is low (D) Pressure is low and temperature is high
Answer (D)
Hints :
Real gas behaves like an ideal gas if its pressure is low and temperature is high. This insures large intermolecularseparation and no condensation.
64. Consider a thin square sheet of side L and thickness t, made of a material of resistivity . The resistance betweentwo opposite faces, shown by the shaded areas in the figure is
L
t
(A) Directly proportional to L (B) Directly proportional to t
(C) Independent of L (D) Independent of t
Answer (C)
Hints :
l LR
A tL t
= = =
Resistance is independent of L.
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SECTION - II
Multiple Correct Choice Type
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONEOR MORE may be correct.
65. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He usesa stop watch with the least count of 1 second for this and records 40 seconds for 20 oscillations. For this
observation, which of the following statement(s) is (are) true?
(A) Error Tin measuring T, the time period, is 0.05 seconds
(B) Error Tin measuring T, the time period, is 1 second
(C) Percentage error in the determination of gis 5%
(D) Percentage error in the determination of gis 2.5%
Answer (A, C)
Hints :
Relative error in measurement of time 1 second 1
40 second 40= =
Time period = 2 second
Error in measurement of time period1 1
2 second = 0.05 second40 20
= =
2
1g
T
2 2 1 1
40 20g T
g T = = =
100 5%g
g
=
66. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass
reverses its direction and moves with a speed of 2 ms1. Which of the following statement(s) is (are) correct for
the system of these two masses?
(A) Total momentum of the system is 3 kg ms1
(B) Momentum of 5 kg mass after collision is 4 kg ms1
(C) Kinetic energy of the centre of mass is 0.75 J
(D) Total kinetic energy of the system is 4 J
Answer (A, C)
Hints :
Initial
Final
5 kg1 kg
v2
u O
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By conservation of momentum
1u= 2 + 5v
By coefficient of restitution equation
21
v
u
+=
Solving we get
3 m/su= 1m/sv =
Momentum = 1 3 = 3 kg m/s
Momentum of 5 kg block after collision = 5 1 = 5 kg ms1
Kinetic energy of centre of mass
21 1 3
(1 5)2 6
= +
= 0.75 J
Total kinetic energy21 1 3 4.5 J
2= =
67. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA as shown in the figure. Its pressureat A is P
0. Choose the correct option(s) from the following :
V0
4V0
A
B
V
T0
TC
(A) Internal energies at A and Bare the same
(B) Work done by the gas in process ABis P0V
0 ln4
(C) Pressure at Cis 04
P
(D) Temperature at Cis
0
4
T
Answer (A, B)
Hints :
ABis an isothermal process
Internal energy at A = Internal energy at B.
ApplyingPV
T= constant between A and B
V0
4V0
A
B
V
T0
TC
0 0 0 0
0 0
4
4= =B
B
P V P V P P
T TAs it is not clearly indicated that BCis passing through origin, prediction about point Cis not possible.
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68. A few electric field lines for a system of two charges Q1
and Q2
fixed at two different points on the x-axis areshown in the figure. These lines suggest that
Q1
Q2
(A) |Q1| > |Q
2|
(B) |Q1| < |Q
2|
(C) At a finite distance to the left of Q1
the electric field is zero
(D) At a finite distance to the right of Q2 the electric field is zero
Answer (A, D)
Hints :
Lines are denser around Q1
|Q1| > |Q
2|
Since |Q1| > |Q2|, then electric field will be zero at some distance to right of Q2
69. A ray OPof monochromatic light is incident on the face ABof prism ABCDnear vertex Bat an incident angle
of 60(see figure). If the refractive index of the material of the prism is 3 , which of the following is/are correct?
B
D
135
60
7590
A
CP
O
(A) The ray gets totally internally reflected at face CD
(B) The ray comes out through face AD
(C) The angle between the incident ray and the emergent ray is 90
(D) The angle between the incident ray and the emergent ray is 120
Answer (A, B, C)
Hints :
At P
1 sin60 3 sinr =
r= 30
From geometry, angle of incidence at Qis 45
P
D
60
7590
R
60
Q45
45
30
60
30
At Q
33 sin(45 ) 1
2 = >
TIRtakes place
At R
angle of incidence is 30
By symmetry R= 60
From second diagram, angle of deviation is 90
60
60
i
e
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SECTION - III (Paragraph Type)
This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon thesecond paragraph 3 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C)and (D) out of which ONLY ONE is correct.
Paragraph for Questions 70 to 71
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero
as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that
their critical temperature becomes smaller than TC(0) if they are placed in a magnetic field, i.e., the critical temperature
TC(B) is a function of the magnetic field strength B. The dependence of T
C(B) on Bis shown in the figure.
T BC( )
TC(0)
OB
70. In the graphs below, the resistance Rof a superconductor is shown as a function of its temperature Tfor twodifferent magnetic fields B
1(solid line) and B
2(dashed line). If B
2is larger than B
1, which of the following graphs
shows the correct variation of Rwith Tin these fields?
(A)
T
R
O
B1B2(B)
T
R
O
B1
B2
(C)
T
R
O
B1 B2(D)
T
R
O
B1
B2
Answer (A)
Hints :
As B2
> B1; critical temperature becomes less. Correct answer is (A).
71. A superconductor has TC(0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its T
Cdecreases to
75 K. For this material one can definitely say that when
(A) B= 5 Tesla, TC(B) = 80 K
(B) B= 5 Tesla, 75 K < TC(B) < 100 K
(C) B= 10 Tesla, 75 K < TC(B) < 100 K
(D) B= 10 Tesla, TC(B) = 70 K
Answer (B)
Hints :
As critical temperature decreases continuously, with increase in magnetic field,
For B= 5T; TC
> 75 K but TC
< 100 K
Correct answer is (B).
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Paragraph for Questions 72 to 74
When a particle of mass mmoves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic
motion. The corresponding time period is proportional to ,m
kas can be seen easily using dimensional analysis.
However, the motion of a particle can be periodic even when its potential energy increases on both sides of x= 0 in
a way different from kx2
and its total energy is such that the particle does not escape to infinity. Consider a particleof mass mmoving on the x-axis. Its potential energy is V(x) = x4 ( > 0) for | x | near the origin and becomes aconstant equal to V
0for |x| X
0(see figure).
V0
X0x
V x( )
72. If the total energy of the particle is E, it will perform periodic motion only if(A) E< 0 (B) E> 0 (C) V
0> E> 0 (D) E> V
0
Answer (C)
Hints :
For motion to be periodic, it must reverse its path i.e., KE should become zero for a finite value of x.
Now K + U = E
K = E U
Umax
= V0
Kmin = EV0
If Kmin
> 0 particle will escape
So, EV0
< 0
E< V0
Also E= K + U> 0
Correct answer is (C)
73. For periodic motion of small amplitude A, the time period Tof this particle is proportional to
(A) mA (B) 1 mA (C) A m (D) 1A m
Answer (B)
Hints :
As V= x4
[] = ML2T2
By methods of dimensions
1 m
A
= M0L0T
Correct answer is (B).
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74. The acceleration of this particle for | x| > X0
is
(A) Proportional to V0
(B) Proportional to0
0
V
mX (C) Proportional to0
0
V
mX (D) Zero
Answer (D)
Hints :
For X> X0, potential energy is constant
Force is zero
Correct answer is (D).
SECTION - IV
(Integer Type)
This Section contains TEN questions. The answer to each question is a single-digit integer, ranging from 0 to 9. Thecorrect digit below the question number in the ORS is to be bubbled.
75. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectionalarea is 4.9 107 m2. If the mass is pulled a little in the vertically downward direction and released, it performssimple harmonic motion of angular frequency 140 rad s1. If the Young's modulus of the material of the wire isn 109 Nm2, the value of nis
Answer (4)
Hints :
FLY
Al= , where lis elongation
FL FlAYAY
L
= =
Equivalent spring constant isYA
kL
=
k YA
m mL = =
Y= 4 109 N/m2
n= 4
76. A binary star consists of two stars A (mass 2.2MS) and B(mass 11M
S), where M
Sis the mass of the sun. They
are separated by distance dand are rotating about their centre of mass, which is stationary. The ratio of the totalangular momentum of the binary star to the angular momentum of star Babout the centre of mass is
Answer (6)
Hints :
Total angular momentum =21 2
1 2
m md
m m
+
Angular momentum of m2
= m2r22, where 12
1 2
m dr
m m=
+
Required ratio = 1 2
1
2.2 116
2.2
m m
m
+ += =
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77. Gravitational acceleration on the surface of a planet is6
11g, where gis the gravitational acceleration on the surface
of the earth. The average mass density of the planet is2
3times that of the earth. If the escape speed on the
surface of the earth is taken to be 11 kms1, the escape speed on the surface of the planet in kms1 will be
Answer (3)
Hints :
2
4
3
pp p p
p
GMg G R
R= =
p p p
e e e
g R
g R
=
Also, 2ev gR=
6 3
11 2
p p p p e
e e e e p
v g R g
v g R g
= = =
vp
= 3 km/s
78. A piece of ice (heat capacity = 2100 J kg1 C1 and latent heat = 3.36 105 J kg1) of mass mgrams is at 5C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally, when theice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heatexchange in the process, the value of mis
Answer (8)
Hints : Heat absorbed by melting of 1 gram ice is
Q= mL = 0.001 3.36 105 = 336 J
m C (5) = 420 336
m= 8 g
79. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the
source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the
difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constantspeeds much smaller than the speed of sound which is 330 ms1.
Answer (7)
Hints :
Frequency of sound reflected by the car is 0c
c
v vf f
v v
+=
As vc
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80. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of
it to 50 cm, the magnification of its image changes from m25
to m50
. The ratio 25
50
m
mis
Answer (6)
Hints :
fm
f u=
+
25
204
20 25m = =
50
20 2
20 50 3m
= =
25
50
126
2
m
m= =
81. An -particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their
de-Broglie wavelengths are and p respectively. The ratiop
, to the nearest integer, is
Answer (3)
Hints :
2
h h
mv mqV = =
2p
h
m eV =
2 (4 ) (2 ) 2 2
ph
m e V
= =
2 2 3p
=
82. When two identical batteries of internal resistance 1 each are connected in series across a resistor R, the rate
of heat produced in Ris J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 =2.25 J
2, then the value of Rin is
Answer (4)
Hints :
In case 1,
2
1
2
2J R
R
= +
In case 2,
2
2
2
2 1J R
R
= +
J1 = 2.25 J2
R= 4
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83. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1
and T2, respectively. The
maximum intensity in the emission spectrum of A is at 500 nm and in that of Bis at 1500 nm. Considering themto be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?
Answer (9)
Hints :
As per Weins displacement law
3A B
B A
T
T
= =
As per Stefans law
Also,4 2
4
4 2
(6)(3) 9
(18)
A A A
B B B
P A T
P A T= = =
84. When two progressive waves y1 = 4 sin (2x 6t) and 2 3 sin 2 6 2y x t
= are superimposed, the amplitude
of the resultant wave is
Answer (5)
Hints :
2 24 3 2 4 3 cos 52
A
= + + =