IIT JEE 2010 Solution

Transcription

®
SOLUTIONS TO IIT-JEE 2010
Paper-I (Code: 8)
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
INSTRUCTIONS
A. General:
1. This Question Paper contains 28 pages having 84 questions. The question paper CODE is
printed on the right hand top corner of this sheet and also on the back page of this
booklet.
2. No additional sheets will be provided for rough work.
3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and
electronic gadgets in any form are not allowed.
4. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided
separately.
5. Write your Registration No., Name and Name of centre and sign with pen in appropriate
boxes. Do not write these anywhere else.
B. Question paper format and Marking scheme:
6. The question paper consists of 3 parts (Chemistry, Mathematics and Physics). Each part
consists of four Sections.
7. 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.
8. For each question in Section II, you will be awarded 3 marks if you darken only the
bubble(s) 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.
9. For each question in Section III, you will be awarded 3 marks if you darken 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.
10. For each question in Section IV, you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. No negative
marks will be awarded in this Section.
®
IITJEE 2010 SOLUTIONS
2
CHEMISTRY: Paper-I (Code: 8)
PART – I
Useful Data
Atomic numbers: Be = 4; C = 6; N = 7; O = 8; Al = 13; Si = 14; Cr = 24; Fe = 26; Zn = 30; Br = 35.
1 amu = 1.66 × 10–27 kg
R = 0.082 L atm K–1 mol–1
–34
h = 6.626 × 10 Js
NA = 6.022 × 1023
me = 9.1 × 10–31 kg
e = 1.6 × 10–19 C
8
–1
c = 3.0 × 10 m s
F = 96500 C mol–1
RH = 2.18 × 10–18 J
4o = 1.11 × 10–10 J–1 C2 m–1
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 which
ONLY ONE is correct.
Note: Questions with (*) mark are from syllabus of class XI.
1.
Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius
equation is
(B) k
(A) k
(C) k
T
T
(D)
T
Sol.:
As per Arrhenius equation (k  AeEa / RT ), the rate constant increases exponentially with temperature.
Correct choice: (A)
2.
In the reaction
HBr
OCH3
(A) Br
Br and CH3OH
H+
OCH3
Sol.:
T
the products are
OCH3 and H2
(C)
k
+
Br–
O–CH3
(B)
Br and CH3Br
(D)
OH and CH3Br
OH + CH3Br
H
Correct choice: (D)
3.
The correct statement about the following disaccharide is
CH2OH
H
HOH2C
O
H (a)
OH
H
H
OCH2CH2O
HO
H
Sol.:
OH
H
O
H
(b)
HO
CH2OH
OH
H
(A) Ring (a) is pyranose with -glycosidic link.
(C) Ring (b) is furanose with -glycosidic link.
Correct choice: (A)
(B) Ring (a) is furanose with -glycosidic link.
(D) Ring (b) is pyranose with -glycosidic link.
*4.
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) BrCH2CH2CH2CH2CH3 and CH3CCH
(D) BrCH2CH2CH2CH3 and CH3CH2CCH
Sol.:
CH3CH2CCH
NaNH2
–NH3
CH3CH2CC– Na+
CH3CH2CH2CH2Br
(SN2)
CH3CH2CCCH2CH2CH2CH3 + NaBr
3-Octyne
Correct choice: (D)
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5.
Sol.:
6.
3
The ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is
(A) [Cr(H2O)4(O2N)]Cl2
(B) [Cr(H2O)4Cl2](NO2)
(C) [Cr(H2O)4Cl(ONO)]Cl
(D) [Cr(H2O)4Cl2(NO2)].H2O
Ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is [Cr(H2O)4Cl2]NO2.
Correct choice: (B)
The correct structure of ethylenediaminetetraacetic acid (EDTA) is
CH2–COOH
COOH
HOOC–H2C
HOOC
N–CH2–CH2–N
N–CH=CH–N
(A)
(B)
HOOC
HOOC–H2C
COOH
CH2–COOH
(C)
HOOC–H2C
COOH
CH2
H
HOOC–H2C
N–CH–CH–N
(D)
H
CH2–COOH
CH2
CH2–COOH
N–CH2–CH2–N
CH2–COOH
HOOC–H2C
HOOC
Sol.:
The correct structure of EDTA is
HOOC–H2C
N–CH2–CH2–N
HOOC–H2C
CH2–COOH
CH2–COOH
Correct choice: (C)
*7.
Sol.:
*8.
Sol.:
The bond energy (in kcal mol–1) of a C–C single bond is approximately
(A) 1
(B) 10
(C) 100
(D) 1000
348
C–C bond energy = 348 kJ/mol
kcal/mol = 82.85 kcal/mol  100 kcal/mol.
4 .2
Correct choice: (C)
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is
(A) Br2 (g)
(B) Cl2 (g)
(C) H2O (g)
The species in its elemental form has zero standard molar enthalpy of formation at 298 K.
(D) CH4 (g)
Br2(l)  Br2(g) ; H of  0
C(s) + 2H2(g)  CH4(g) ; H of  0
H2(g) + ½O2(g)  H2O(g) ; H of  0
Correct choice: (B)
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
ONE OR MORE is/are correct.
9.
In the reaction
OH
NaOH(aq)/Br2
the intermediate(s) is(are)
O
O
O
O
Br
(A)
(B)
OH
O
NaOH
(I)
O
Br
Br2
–Br–
–H2O
Br
O
O
(D)
Br
Br
Br
Br
Sol.:
(C)
Br
Br2
Br2
–Br–
–Br–
Br
Br
Br
Br
(II)
(III)
(IV)
Product of reaction of phenol with NaOH/Br2 is sodium salt of 2,4,6-tribromophenol. Hence, species (I), (II), (III) are formed
as intermediate.
Correct choice: (A), (C)
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*10.
Sol.:
*11.
Sol.:
4
Among the following, the intensive property is (properties are)
(A) molar conductivity
(B) electromotive force
(C) resistance
(D) heat capacity
Mass independent properties are intensive properties. Resistance and heat capacity are extensive properties.
Correct choice: (A), (B)
The reagent(s) used for softening the temporary hardness of water is(are)
(A) Ca3(PO4)2
(B) Ca(OH)2
(C) Na2CO3
(D) NaOCl
Temporary hardness is due to bicarbonates of calcium and magnesium. Temporary hardness can be removed by Clark’s
process, which involves the addition of slaked lime, Ca(OH)2. Washing soda (Na2CO3) removes both the temporary and
permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates.
Ca(HCO3)2 + Ca(OH)2  2CaCO3 + 2H2O
Ca(HCO3)2 + Na2CO3  CaCO3 + 2NaHCO3
2OCl– + 2H2O
2HOCl + 2OH–
Ca(HCO3)2 + 2OH–  CaCO3 + CO 32 + 2H2O
Correct choice: (B), (C), (D)
*12.
Sol.:
Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa 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) CH3COOH and CH3COONa
Any solution of a weak acid and its salt with strong base acts as an acidic buffer solution.
If volume of HNO3 solution added is less as compared to that of CH3COONa solution, it results in the formation of an acidic
buffer solution.
CH3COONa + HNO3 
 CH3COOH + NaNO3
Excess
limiting
reagent
MV
MV
–
–
M(V – V)
0
MV
MV
(V < V)
Correct choice: (C), (D)
*13.
In the Newman projection for 2,2-dimethylbutane
X
CH3
H3C
H
H
Sol.:
Y
X and Y can respectively be
(A) H and H
(B) H and C2H5
Structural formula of 2, 2-dimethylbutane is
CH3
1
2
3
(C) C2H5 and H
(D) CH3 and CH3
4
CH3–C–CH2–CH3
CH3
(I) Newman projection using C1–C2 bond
H
2
H3 C
CH3
1
H
H
C2H5
(II) Newman projection using C3–C2 bond
CH3
2
CH3
CH3
3
H
H
CH3
Correct choice: (B), (D)
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5
SECTION  III
Paragraph Type
This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second
paragraph 2 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 Question 14 to 16
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper
include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite
(Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of
copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.
14.
Partial roasting of chalcopyrite produces
(A) Cu2S and FeO
(B) Cu2O and FeO
(C) CuS and Fe2O3
Sol.:
2CuFeS2 + 4O2  Cu2S + 3SO2 + 2FeO
Partial roasting of FeS and Cu2S produces FeO and Cu2O respectively.
2FeS + 3O2  2FeO + 2SO2
2Cu2S + 3O2  2Cu2O + 2SO2
Correct choice: (B)
15.
Iron is removed from chalcopyrite as
(A) FeO
(B) FeS
FeO + SiO2  FeSiO3
Correct choice: (D)
Sol.:
16.
Sol.:
(D) Cu2O and Fe2O3
14001450 ºC
In self-reduction, the reducing species is
(A) S
(B) O2–
(C) Fe2O3
(D) FeSiO3
(C) S2–
(D) SO2
Cu2S + 2Cu2O  6Cu + SO2
The reducing species is the one which gets oxidized. So, it is S2– ion getting oxidized to S4+.
Correct choice: (C)
Paragraph for Question 17 to 18
The concentration of potassium ions inside a biological cell is atleast twenty times higher than the outside. The resulting
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 is
M(s) | M+(aq; 0.05 molar) || M+(aq; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential |E cell| = 70 mV.
17.
Sol.:
For the above cell
(A) Ecell < 0; G  0
(B) E cell  0; G  0
(C) E cell  0; Gº  0
(D) E cell  0; Gº  0
(A): M(s) 
 M A (aq)  e 
(C): M C (aq)  e  
 M(s)
MC (aq) 
 MA (aq)
E cell  E ocell 
RT [M A ] RT [M C ]
ln  
ln 
F
F
[M C ]
[M A ]
Ecell = 0.059 log
(E ocell  0)
[M C ]
0.05
0.07V = 0.059 log
(M C )
= 0.059 log 20 [M C ] = 0.059 (log 20 + log [M C ] )
0.05
Since [M C ] is greater than 0.05 M, so Ecell is +ve.
Ecell > 0 & G  0.
Correct choice: (B)
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6
If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be
(A) 35 mV
(B) 70 mV
(C) 140 mV
(D) 700 mV
[M C ]
2



Sol.: Ecell = 0.059 log
= 0.059 log 400 [M C ] = 0.059 log (20) [M C ] = 0.059(2 log 20 + log [M C ] )
0.0025
 0.07 
= 0.059  2 
 = 0.14 V = 140 mV.
0.59 

Correct choice: (C)
SECTIONIV
Integer Answer Type
This section contains TEN questions. The answer to each questions is a single digit integer ranging from 0 to 9.
The correct digit below the question number in the ORS is to be bubbled.
18.
19.
In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is
1. O3
2. Zn, H2O
Y
1. NaOH(aq)
2. heat
OH
O
1. O3
2. Zn, H2O
Sol.:
NaOH(aq)
heat
–H2O
O
O
O
(Y)
The number of intramolecular aldol condensation products (, -unsaturated carbonyl compound) formed from (Y) is 1.
 The answer is 1.
20.
Amongst the following, the total number of compounds soluble in aqueous NaOH is
H3C
CH3
N
COOH
OCH2CH3
OH
CH2OH
OH
NO2
Sol.:
*21.
Sol.:
CH2CH3
CH2CH3
COOH
H3C N CH3
All carboxylic acids and phenols are soluble in aqueous NaOH. Four compounds are soluble in aqueous NaOH.
Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
KCN
K2SO4
(NH4)2C2O4
NaCl
Zn(NO3)2
FeCl3
K2CO3
NH4NO3
LiCN
KCN, K2CO3 and LiCN are the salts of weak acid and strong base. So, their aqueous solutions turns red litmus paper blue.
(NH4)2C2O4 is a salt of weak base (NH4OH ; Kb = 1.8 × 10–5) and weak acid (HOOC.COOH ; K a1 = 5.4 × 10–2 and
K a 2 = 5.2 × 10–5 ). As a result, the extent of hydrolysis of NH 4 will be more than that of C 2 O 24 and hence the resulting
solution will be acidic.
The answer is 3.
*22.
Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is
Sol.:
According to VSEPR theory, number of electron pairs around central atom (Br) are 6.
N
75
=
= 6. (Five are bond pairs and one is lone pair )
2
2
Its geometry is octahedral but due to lone pair–bond pair repulsion, the
four fluorine atoms at corner are forced towards the upper fluorine atom
thus reducing F–Br–F angle from 90° to 84.8°.

F
F
F
Br
F
..
F
The answer is 0.
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7
*23.
Sol.:
The value of n in the molecular formula BenAl2Si6O18 is
It is the formula of Beryl (which has BeO, Al2O3 and SiO2).
The formula of Beryl should be
3BeO. Al2O3.6SiO2
Or
Total cationic charge = Total anionic charge
2n + 6 + 24 = 36
n=3
*24.
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
The least significant figure in titre values is 3.
25.2  25.25  25 75.4
Average titre value =

 25.1
3
3
The number of significant figures in average titre value will also be 3.
Sol.:
25.
Sol.:
26.
Sol.:
The concentration of R in the reaction R  P was measured as a function of time and the following data is obtained:
[R] (molar)
1.0
0.75
t(min.)
0.0
0.05
The order of reaction is
The integrated form of a zero-order reaction is
[A0] – [At] = k0t
1.0 – 0.75 = k0 × 0.05, k0 = 5
1.0 – 0.4 = k0 × 0.12, k0 = 5
235
92 U
The number of neutrons emitted when
92 U
235
 0 n  54 Xe
1
142
 38Sr
90
0.40
0.12
0.10
0.18
undergoes controlled nuclear fission to
142
54 Xe
and
90
38 Sr
is
 y 0n
1
235 + 1 = 142 + 90 + y
y = 4. The number of neutrons emitted are (4 – 1) = 3.
*27.
The total number of basic groups in the following form of lysine is
H3N–CH2– CH2–CH2–CH2
O
CH—C
O
H2N
Sol.:
The basic groups in the given form of lysine is NH2 and CO 2 . The answer is 2.
*28.
The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is
Sol.:
The number of cyclic isomers for a hydrocarbon with molecular formula C 4H6 is 5.
The structures are
CH3
,
,
,
CH3
and
CH2
.
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8
MATHEMATICS: Paper-I (Code: 8)
PART – II
SECTION – I
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) for its
answer, out of which ONLY ONE is correct.
29.
The value of lim
x0
1
x
3
x

0
t ln(1  t )
t4  4
(A) 0
dt is
(B)
x
Sol.:
lim
1
x0 x 3
x

0
t ln1  t 
4
t 4

dt  lim
x4  4
x 0
3x 2
 lim

t ln1  t 
0
t4 4
x0
x ln1  x 
 lim
x
ln1  x 

1
12
x 0 3x x 4  4
(C)
1
24
(D)
1
64
dt
3
0

 form  Applying L. Hospital Rule
0



ln1  x 
1
1
1
lim
lim

4
3 x0
x
12
x0 x  4
Correct choice: (B)
30.
 x  1 
   
The number of 3 3 matrices A whose entries are either 0 or 1 and for which the system A  y   0 has exactly two distinct
 z  0
solutions, is
(B) 29  1
(A) 0
Sol.:
(C) 168
(D) 2
For any matrices A it form three planes. But three planes can not intersect exactly at two distinct points.
Correct choice: (A)
31.
Let P, Q, R and S be the points on the plane with position vectors 2iˆ  ˆj, 4iˆ, 3iˆ  3 ˆj and  3iˆ  2 ˆj respectively. The
quadrilateral PQRS must 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
Sol.:
If O is origin OP  2iˆ  ˆj; OQ  4iˆ ; OR  3iˆ  3 ˆj; OS  3iˆ  2 ˆj
QS  OS  OQ  7iˆ  2 ˆj
PR  5iˆ  4 ˆj
QS  49  4  53 ; PR  25  16  41
PQ  6iˆ  ˆj  37 ; QR   iˆ  3 ˆj  10
RS   6iˆ  ˆj  37 ; SP  iˆ  3 ˆj  10
Opposite sides are parallel and equal but diagonals are not equal.
Correct choice: (A)
®
32.
9
Let  be a complex cube root of unity with   1 . A fair die is thrown three times. If r1 , r2 and r3 are the numbers obtained
on the die, then the probability that r1  r2  r3 = 0 is
(A)
Sol.:
1
18
1
9
(B)
(C)
2
9
(D)
1
36
nS   6 3  216
E is the event that r1  r2  r3  0 , is possible if one among r1 , r2 , r3 is of the form 3n ; another one is of the form
3n  1 and the remaining one is of the form 3n  2 , where n  Z .
For 3n , two possible cases are 3, 6
For 3n  1 , two possible cases are 1, 4
For 3n  2 , two possible cases are 2, 5
nE   2  2  2  3 !  48
PE  
nE  48 2


nS  216 9
Correct choice: (C)
33.
Equation of the plane containing the straight line
x y z
 
and perpendicular to the plane containing the straight lines
2 3 4
x y z
x y z
  and   is
3 4 2
4 2 3
Sol.:
(A) x  2 y  2 z  0
(B) 3x  2 y  2 z  0
Plane P2 contains the line L 
x y z
 
2 3 4
(C) x  2 y  z  0
(D) 5x  2 y  4 z  0
Plane P1 contains the lines L1 and L 2 , where
L1 
x y z
x y z
  and L2   
3 4 2
4 2 3
Let direction ratio of normal to the plane P1 is , ,   , then
3  4  2  0 and 4  2  3  0






8  1  10
Let direction ratio of normal of required plane is (a, b, c), then 8a  b  10c  0 ; 2a  3b  4c  0

a
b
c
a
b
c





26  52 26
1 2 1
 Required plane is x  2 y  z  0
Correct choice: (C)
Alternative:
   
    
n  a  b  c  a .c b  a .b c

 
 
 

 26 3iˆ  4 ˆj  2kˆ  26 4iˆ  2 ˆj  3kˆ  26iˆ  52 ˆj  26kˆ
 direction ratio of normal of required plane is 1,  2, 1
Equation is x  2 y  z  0
*34.
If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides
opposite to A, B and C respectively, then the value of the expression
(A)
1
2
(B)
3
2
a
c
sin 2C  sin 2 A is
c
a
(C) 1
(D)
3
®
Sol.:
10
Since A, B, C are in A.P.

2B  A  C
A  B  C  180
 3B  180  B  60
a
c
sin 2C  sin 2 A
c
a

sin A
sin C
.2 sin C cos C 
2 sin A cos A  2cos C sin A  sin C cos A  2 sinA  C   2 sin 2B  2 sin120
sin C
sin A
 2
3
 3
2
Correct choice: (D)
35.
Let f, g and h be real-valued functions defined on the interval [0, 1] by f ( x)  e x  e  x , g ( x)  xe x  e  x
2
2 x2
h( x)  x e
e
 x2
f x   e x  e  x
2
2
g x   xe x  e  x
2
2
hx   x 2 e x  e  x
2
2
2
and
. If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then
(A) a  b and c  b
Sol.:
2
(B) a  c and a  b

2 
 2
f x   2 x e x  e  x 



2
2 
2

g x   2 x xe x  e  x   e x


2
(C) a  b and c  b
(D) a  b  c
2
2 
 2
 h x   2 x 3 e x  2 x e x  e  x 


for x  [0, 1]
f x , g x  and h x   0

1
1
1
f x , g x  and hx  will be increasing functions and their maximum values are a  e  , b  e  , c  e 
e
e
e
at x  1
 abc
Correct choice: (D)
Alternative:
f x   e x 
1
2
e
x2
2
as x  [0, 1]  e x  [1, e]
1
1
f t   t  , t  [1, e] ; max f t   e 
e
t
 max f x   e 
1
e
 hx  g x  f x
 g x   e 
1
1
1
but at x  1 , g x   e  and similarly at x  1 , hx   e 
e
e
e
So, a  b  c
*36.
Let p and q be real numbers such that p  0, p 3  q and p 3  q . If  and  are nonzero complex numbers satisfying
     p and  3  3  q , then a quadratic equation having


and
as its roots is


(A) ( p 3  q) x 2  ( p 3  2q) x  ( p 3  q)  0
(B) ( p 3  q) x 2  ( p 3  2q) x  ( p 3  q)  0
(C) ( p 3  q) x 2  (5 p 3  2q) x  ( p 3  q)  0
(D) ( p 3  q) x 2  (5 p 3  2q) x  ( p 3  q)  0
®
Sol.:
   p
…(i)
 3  3  q
…(ii)
Sum of roots S 
   2   2   2  2
 

 


Product of roots P = 1
11
…(iii)


From (i) and (ii)  q   p  2   2  


2q  p 3 
p2 
q   p   2  3   
3p
 S

q  p3
3p
q  p3
3p
3 p 3  2q  2 p 3
q  p3

p 3  2q
q  p3
 Required quadratic equation is x 2  Sx  P  0
q  p 3 x 2  p 3  2qx  q  p 3   0
Correct choice: (B)
SECTION – II
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) for its answer, out of
which ONE OR MORE may be correct.
*37.
Let ABC be a triangle such that ACB 

and let a, b and c denote the lengths of the sides opposite to A, B and C
6
respectively. The value(s) of x for which a  x 2  x  1, b  x 2  1 and c  2 x  1 is (are)

(A)  2  3
Sol.:

2 


(D) 4 3
2
x 2  x  1  x 2  1  2 x  12


6
2 x 2  x 1 x 2 1
 

2
3 x 2  3x  2x 2  x  x 2  1

2
2x 2  x  1x 2  1
3 x 2  x  1  x  2x  x 2  1

cos
(C) 2  3
(B) 1  3
x



 

 2  3 x 2  2  3 x 1  3  0
 2  3 2  42  3 1  3   2  3 , 1  3 
22  3 
 2 3 
But x   2  3
 (as a, b, c  0
 x  1)
Correct choice: (B)
x
38.
Let f be a real-valued function defined on the interval (0, ) by f ( x)  ln x 

1  sint dt . Then which of the following
0
statement(s) is (are) true?
(A) f ' ' ( x) exists for all x  (0, )
(B) f ' ( x) exists for all x  (0, ) and f ' x  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 x  (0, )
®
Sol.:
f x  
12
1
 1  sin x
x
Clearly f x  exists and continuous for all x  0  .
Also f x  is not differentiable at point where sin x  1 .
f x   f x   ln x 
1
 g x  , where g x  
x
x

1  sin t dt  1  sin x
0
x

Now
1  sin t dt is always continuous, increasing and has range 0,  .
0
So g x   0 for some k  0,  as
Also ln x 
1  sin x  2 for x  0, 
1
 0 for x  k , 
x
y = lnx
 f x  f x  0 for all x  k  k , 

y = 1/x
f x   f x  for all x  ,  , where   k  k 
k
1
Correct choice: (B), (C)
*39.
Let A and B be two distinct points on the parabola y 2  4 x . If the axis of the parabola touches a circle of radius r having AB
as its diameter, then the slope of the line joining A and B can be
(A) 
Sol.:
1
r
(B)
1
r
(C)
2
r
(D) 
Let mid-point of AB is h,  r 
2
r
B
y2 = 4x
xh yr
 Equation of AB is


cos  sin 
A
So A and B can be  cos   h,  sin   r 
O
C(h, ±r)
x
But these will satisfy y 2  4 x
 sin   r 2  4 cos   h


 2 sin2   2 r sin   2 cos   r 2  4h  0
Sum of roots = 0  AC  CB 
 r sin   2 cos   tan   
2
r
Correct choice: (C), (D)
Alternative:
Slope of AB 
y
2
t1  t 2
B
t22 , 2t2 
t12 , 2t1
y-coordinate of mid-point of AB  t1  t2  r
A
r
x
2
2
 slope of AB can be
or 
r
r
1
40.
The value(s) of

0
22

(A)
7
x 4 (1  x) 4
1 x2
dx is (are)
(B)
2
105
(C) 0
(D)
71 3

15 2
®
1
Sol.:
I


x 4 1  x 2  2x
1 x
0

1 1 4
  4
5 7 6
2
1
2 dx  1 x 4 1  x 2 dx  1 4x 5 dx  1


0
0
 4
 x 2 1
  x
0
13
4x 6
 1  x 2 dx
0

1 1 2 4 4
22
 dx       4   


5 7 3 5 3
7
x 1 
1
2
Correct choice: (A)
*41.
Let
Z1
and z 2 be two distinct complex numbers and let z  (1  t ) z1  tz2 for some real number t with 0  t  1 . If Arg(w)
denotes the principal argument of a nonzero complex number w, then
(A) | z  z1 |  | z  z 2 |  | z1  z 2 |
(B) Arg( z  z1 )  Arg( z  z 2 )
(C)
Sol.:
z  z1
z  z1
z 2  z1
z 2  z1
0
(D) Arg( z  z1 )  Arg ( z 2  z1 )
Az1 , Bz , Cz 2  are collinear
AB  BC  AC

z  z1  z  z 2  z1  z 2
Argz  z1    Argz  z 2 
Argz  z1   Argz 2  z1 
Equation of line A, B, C is
z  z1
z  z1
z 2  z1
z 2  z1
0
Correct choice: (A), (C), (D)
SECTION  III
Comprehension Type
paragraph 2 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out
Paragraph for Questions 42 to 44
Let p be an odd prime number and T p be the following set of 2 × 2 matrices :


a b 


Tp  A  
 : a, b, c {0, 1, 2, ....., p  1}
c
a






42.
The number of A in T p such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is
(B) 2( p  1)
(A) ( p  1) 2
Sol.:
(C) ( p  1) 2  1
(D) 2 p  1
For A to be symmetric b  c
   a 2  b 2 should be multiple of p

(a  b) (a  b) should be multiple of p
 Either (a  b) or (a  b) should be multiple of p
a  b can be multiple of p in ( p  1) ways (excluding a  b  0 )
(a  b) can be multiple of p in p ways (including a  b  0 )

Total number of ways = 2p – 1
Correct choice: (D)
®
43.
14
The number of A in T p such that the trace of A is not divisible by p but det (A) is divisible by p is
[Note: The trace of a matrix is the sum of its diagonal entries.]
(A) ( p  1)( p 2  p  1)
Sol.:
(B) p 3  ( p  1) 2
(C) ( p  1) 2
(D) ( p  1)( p 2  2)
As tr A is not a multiple of p
 a0
  a 2  bc will be a multiple of p
if remainder of
bc
a2
and
is same.
p
p
Let when a 2 divided by p remainder is r
when bc is divided by p for different values of c  {0, 1,... p  1} we get distinct remainder  {0, 1,... p  1}
there exist only one value of b for which remainder of bc divided by p gives r

a and c can be taken in ( p  1) ways each and b can be choose in only one way.

Total ways  ( p  1)2  1
Correct choice: (C)
44.
The number of A in T p such that det (A) is not divisible by p is
(B) p 3  5 p
(A) 2 p 2
Sol.:
The total number of A in T p  p 3
(C) p 3  3 p
(D) p 3  p 2
…(i)
Now the A for which det  A is divisible by p
a0
Case-I:
bc should be divisible by p

either b  0 or c  0 or both

total ways  2 p  1
Case-II:
a0
Total ways  ( p  1)2
[as in previous question]
 p  12  2 p  1  p 2

Total number of ways =

Required ways = p3  p 2
…(ii)
Correct choice: (D)
The circle x 2  y 2  8x  0 and hyperbola
*45.
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
(A) 2 x  5 y  20  0
Sol.:
x2 y2

 1 intersect at the points A and B.
9
4
(B) 2 x  5 y  4  0
(C) 3x  4 y  8  0
(D) 4 x  3 y  4  0
Let 41  cos, 4 sin be a point on the circle.
Equation of tangent to the circle is x cos   y sin   41  cos  = 0
 y
x cos  41  cos 

sin 
sin 
…(i)
Condition of tangency for hyperbola is c 2  a 2 m 2  b 2
®

161  cos 2
2
sin 
9 cos2 
sin2 
 4  3 cos2   32 cos   20  0
2
and 10 (rejected)
3
 cos   
 sin  

5
3
 Slope of tangent from (i) is m = 
 y

15
23
cos 
2


sin 
5 3
5
 1 2 3 
; y  2 x  4
x  4
 5 3
5
5
5


2
5 y  2x  4  2x  5 y  4  0
Correct choice: (B)
*46.
Sol.:
Equation of the circle with AB as its diameter is
(A) x 2  y 2  12x  24  0
(B) x 2  y 2  12x  24  0
(C) x 2  y 2  24x  12  0
(D) x 2  y 2  24x  12  0
y 2  8x  x 2

x 2 8x  x 2

1
9
4
 4 x 2  72x  9 x 2  36  13x 2  72x  36  0  13x 2  78x  6 x  36  0
 13xx  6  6( x  6)  0  x  
6
, x6
13
Clearly 6, 0 is the centre.
 if x  6 , y 2  48  36  y 2  12  y  2 3


 End points of diameter are 6, 2 3 , 6,  2 3

 r  2 3 and C 6, 0

x  62  y 2  12
 x 2  12x  36  y 2  12 
x 2  y 2  12x  24  0
Correct choice: (A)
SECTION  IV
Integer Answer Type
This section contains TEN questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The
correct digit below the question numbers in the ORS is to be bubbled.
47.
Sol.:

   


 2iˆ  ˆj  3kˆ

    
 iˆ  2 ˆj

If a and b are vectors in space given by a 
and b 
, then the value of 2a  b . a  b  a  2b is
5
14




a  1, b  1 and a . b  0
2a  b .[ a 2  2a . b b   a . b  2 b 2 a]  1  0 2a . b  b 2   0  2 2 a 2  a . b   1  4  5




Ans. 5
®
*48.
Sol.:
The line 2 x  y  1 is tangent to the hyperbola
x2
y2
 1 . If this line passes through the point of intersection of the
a
b2
nearest directrix and the x-axis, then the eccentricity of the hyperbola is
y  2 x  1
…(i)
y  mx  a 2 m 2  b 2
…(ii)
2

16
(i) and (ii) identical  m  2 and 4a 2  b 2  1
Satisfying 2 x  y  1 by
 4a 2  e 2
2a
1
e
…(iv)


…(iii)
Also b 2  a 2 e 2  1  b 2  a 2 e 2  a 2

…(v)

From (iii) and (iv)  b 2  e 2  1


 e 2 1  a 2 e 2 1  a  1
 e  2a  e  2
Ans. 2
49.
If the distance between the plane Ax  2 y  z  d
and the plane containing the lines
x 1 y  2 z  3
and


2
3
4
x2 y 3 z 4
is 6 , then | d | is


3
4
5
Sol.:
Let d.c. of normal to the plane is l , m, n :
2l  3m  4n  0
3l  4m  5n  0
l
m
n
 
1 2 1
D.R. of normal to the plane is 1,  2, 1
Equation of plane is x  2 y  z  d 
Satisfying by 1, 2, 3  d   0
Required plane is x  2 y  z  0
…(i)
Given plane is Ax  2 y  z  d
…(ii)
From (i) and (ii) we get  A  1 because planes are parallel.
Now,
d 0
 6  d 6
12  2 2  12
Ans. 6
50.
For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the
if [x] is odd,
 x  [ x]
2
interval [–10, 10] by f ( x)  
. Then the value of
10
1  [ x]  x if [x] is even
10
 f ( x) cos x dx is
10
®
Sol.:
17
Period of f x   2
Graph of f(x)
Period of cos x  2
So, period of f x cos x  2
10
5.2
10
5.2
–5
–3
–2
–1
0
1
2
3
5
 f x cos x dx   f x cos x dx
2

 10 f x  cos x dx
0
2
2
1

1





 10 f x  cos x dx  f x  cos x dx   10 1  x cos x dx  xcos x dx 




1
1
0

0



1

2

 x 1 cos x dx
0
1
 10 1  x  cos x dx  10

1
2
2
1





2
2
 10 cos xdx  x cos x dx  x cos x dx  cos x dx   10 0 

 0 
2
2






0
1
1
0






40
2
2
So,
10
10
 f x cos x dx  4
10
Ans. 4
51.
Let  be the complex number cos
z 1

 z  2
2
2
1
= 0 is equal to
z
1
z 1
Sol.:

Given determinant is
1
1
z
By C1  C1  C2  C3 
0
2
z

z
z  2
1
z
1
z

0 z  2  
0
2
z


2

2
z
2
2
 i sin
. Then the number of distinct complex numbers z satisfying
3
3
1 
0
2
1  2
z    2


 
2
 0  z[ z 2  2    3]  0  z z 2  0  z  0
So number of solution is 1.
Ans. 1
®
*52.
18
Let Sk , k  1, 2,..., 100, denote the sum of the infinite geometric series whose first term is
Then the value of
1002

100!
 k

100
2
k 1
1
and the common ratio is .
k
k!
 3k  1 S k is
k 1
k  1
1
k !   S 
for k  1

k
1 

k  1 !
Rk 
k 
Ak 
Sol.:
The value of


1002 
100 !
 k 2  3k  1S k
1002  100
100 !
k 1
 S1   1 S 2 
 k 2  3k  1S k 
100
k 3
1002
100 !
100
 0 1
 k  1

  k  2 !  k 1 ! 
k
k 3
2

 99 100 
100
3   3
4 
 2
 100
  
 
 0 1 2 
3
 0  1   
   ........  

100
!
99 !
100 !
1
!
2
!
2
!
3
!
98
!
99
!


 




1002
Ans. 3
53.
The number of all possible values of , where 0     , for which the system of equations ( y  z) cos3  ( xyz) sin3 ;
2 cos3 2 sin3
; ( xyz) sin3  ( y  2z) cos3  y sin3 have a solution ( x0 , y0 , z0 ) with y0 z0  0 , is
x sin3 

y
z
Sol.:
x sin3 
1
1
cos3  cos3  0
y
z
x sin3 
1
1
2 cos3  2 sin3  0
y
z
x sin3 
1
1
2 cos3  cos3  sin3  0
y
z
sin3  cos3
 cos3
 2 sin3
0
The system of equations has solution, so sin3  2 cos3
sin3  2 cos3  cos3  sin3
1  cos3
 cos3
 2 sin3
0
 sin3 1  2 cos3
0
0
 cos3  sin3
 sin3sin3  cos3cos3  0
 sin3  0, cos3  0 , tan 3  1


1
But sin3  0  for sin3  0,  0 which is not possible
y


cos3  0 (  for cos3  0,
1
 0 , not possible)
z
tan 3  1 (  for 3 solutions  
 5 3
,
,
)
12 12 4
So number of solution = 3.
Ans. 3
®
54.
Sol.:
19
Let f be 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
dy 

According to given condition:  y  x   x 3
dx 

 xdy  ydx 
   xdx 
 

x2


But f 1  1  c 

y
x2
 y

c
d     x dx 
x
2
x

x3 3 x
3
 The curve is y  
 f 3  9

2
2
2
Ans. 9
*55.
Sol.:
n
  
The number of values of  in the interval   ,  such that  
for n  0, 1,  2 and tan   cot 5 as well as
5
 2 2
sin2  cos 4 is
tan   cot 5  cos 6  0
 6  n 

2
…(i)




 


  
6  3  ,  2  ,    , ,   , 2  as     ,   6   3, 3
2
2
2 2
2
2
 2 2


Also sin 2  cos 4




cos 4  cos  2   4  2m    2 , m  Z
2


2

 6  2m 


or 2  2m 
2
2
…(ii)
From (i) and (ii)  The number of common solution = 3
Ans. 3
Alternative:
sin 2  cos 4
…(i)
 2 sin2 2  sin 2  1  0 

sin2  12 sin2  1  0 
sin 2  1 or sin 2 
1
2
  
    ,   2   , 
 2 2


5 
,
or  
4
12 12
All the three values are satisfying the equation tan   cot 5
Ans. 3
*56.
The maximum value of the expression
Sol.:
Maximum value of the expression

1
sin   3sin cos   5 cos2 
2
is
1
2
sin   3 sin  cos   5 cos 2 
1
1
1


2
5
 3

 3 sin 2  4 cos 2 
3



1

sin
2


2
cos
2


1
3





2
2
 2
 min

 min
Ans. 2
®
20
PHYSICS: Paper-I (Code: 8)
PART – III
SECTION – I
This Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
57.
(A) IBL
Sol.:





A thin flexible wire of length L is connected to two adjacent
fixed points and carries a current I in the clockwise direction, as
shown in the figure. When the system is put in a uniform
magnetic field of strength B going into the plane of the paper, the
wire takes the shape of a circle. The tension in the wire is
(B)

T 2  dR  ILB dR
IBL

T
(C)
IBL
2










(D)
IBL
4










ILB
2
Correct choice: (C)
58.
An AC voltage source of variable angular frequency  and fixed amplitude V0 is connected in series with a capacitance C
and 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
2
Sol.:
 1 
| Z | X C2  R 2  
  R2
 C 
When angular frequency of the source increases the impedance of the circuit decreases, therefore the current in the circuit
increases, hence the power through the resistance increases.
Correct choice: (B)
59.
To verify Ohm’s law, a student is provided with a test resistor RT , a high resistance R1 , a small resistance R2 , two identical
galvanometers G1 and G2 , and a variable voltage source V. The correct circuit to carry out the experiment is
G1
G1
R2
RT
(A)
R1
G2
R1
RT
(B)
V
R1
R2
G1
RT
G2
R2
V
Sol.:
R2
V
G1
(C)
G2
(D)
RT
G2
R1
V
A galvanometer with a high resistance in series behaves as a voltmeter and a low resistance connected in parallel to the
galvanometer behaves as an ammeter.
Correct choice: (C)
®
60.
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in
temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60, R40 respectively, the
relation between these resistances is
(A)
Sol.:
21
1
1
1


R100 R40 R60
R
(B) R100  R40  R60
(C) R100  R60  R40
(D)
1
1
1


R100 R60 R40
V2
, where R is the resistance at working temperature and not at room temperature
P
1
1
1


R100 R60 R40
Correct choice: (D)
*61.
Sol.:
A real gas behaves like an ideal gas if its
(A) pressure and temperature are both high
(C) pressure is high and temperature is low
(B) pressure and temperature are both low
(D) pressure is low and temperature is high
Real gases behave like an ideal gas at very low densities.
 TR
As P  
 M0

P
  

T

Correct choice: (D)
62.
Consider a thin square sheet of side L and thickness t, made of a material
of resistivity . The resistance between two opposite faces, shown by the
shaded areas in the figure is
(A) directly proportional to L (B) directly proportional to t
(C) independent of L
(D) independent of t
t
L
Sol.:
R
L
L  t 


t
Correct choice: (C)
*63.
Sol.:
A thin uniform annular disc (see figure) of mass M has outer radius 4R
and inner radius 3R. The work required to take a unit mass from point P
on its axis to infinity is

(A)
2GM
4 2 5
7R
(C)
GM
4R

(B) 
(D)

P

4R
2GM
4 2 5
7R
2GM
5R
 2  1
3R
4R
Potential due to disc on its axis =  2G R 2  x 2  x 


(R is radius of disc, x is the distance from the centre of the disc on its axis)
By superposition principle potential due to annular disc at point P, VP  2G R12  x 2  R22  x 2 


W  V  VP
As R1  4R, R2  3R , x  4R and  

Work done 


2GM
4 2 5
7R
M
4 R   3R 
2
2
 V  0
Correct choice: (A)
®
*64.
A block of mass m is 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 P parallel to the plane.
The direction of force pointing up the plane is taken to be positive. As P is varied from
P1  mg(sin   cos ) to P2  mg(sin   cos ) , the frictional force f versus P graph will look like
f

(B)
P2
P1
P
f
(C)
P
f
(A)
P1
P2
P1
P2
P
f
(D)
P1
P2
Sol.:
22
P
P
(i) when P  mg sin
P
P  f  mg sin 
f
tan > 
mgsin
f  mg sin   P

(ii) when P = mg sin , f = 0
(iii) when P  mg sin 
P
tan > 
f  mg sin   P
mgsin f
f

P2
P1
P
Correct choice: (A)
SECTION – II
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE
OR 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 uses a stop watch
with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following
statement(s) is (are) true?
(A) Error  T in measuring T, the time period, is 0.05 seconds
(B) Error  T in measuring T, the time period, is 1 second
(C) Percentage error in the determination of g is 5%
(D) Percentage error in the determination of g is 2.5%
®
Sol.:
t = nT;

T
T 
t
n
t
n

T 
t
l
4 2 n 2 l
 g
;
 2
n
g
t2
Correct choices: (A), (C)
66.
23
1
 0.05 sec.
20
g
2t
100 
100 = 5%
g
t
A ray OP of monochromatic light is incident on the face AB of prism ABCD
near vertex B at 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
O
600
(A) The ray gets totally internally reflected at face CD
C
1350
P
(B) The ray comes out through face AD
(C) The angle between the incident ray and the emergent ray is 90°
900
(D) The angle between the incident ray and the emergent ray is 120°
750
D
A
Sol.:
B
sin 600
 3  r = 300
sin r
Let ic be the critical angle, then, sin i c 

600
600
300
1
1350
C
450
450
3
300 < ic < 450
Also, the angle between incident and emergent ray is 90 0
300
750
Correct choices: (A), (B), (C)
A
67.
A few electric field lines for a system of two charges Q1 and Q2 fixed at two
different points on the x-axis are shown in the figure. These lines suggest that
(A) | Q1 |  | Q2 |
D
600
Q2
Q1
(B) | Q1 |  | Q2 |
(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
Sol.:
Number of field lines  magnitude of charge and
E
KQ
r2
Correct choices: (A), (D)
*68.
One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as
shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the
following
(A) Internal energies at A and B are the same
(B) Work done by the gas in process AB is P0V0 n 4
(C) Pressure at C is
Sol.:
V0
P0
4
(D) Temperature at C is
V
B
4V0
C
T0
4
A
T0
T
Process A  B is isothermal, so WAB = P0V0 ln 4 and U AB = 0
Process B  C is unknown (as its not given whether the line BC is directed towards origin or not)
Correct choices: (A), (B)
®
24
*69.
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 ms–1. 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 ms–1
(B) Momentum of 5 kg mass after collision is 4 kg ms–1
(C) Kinetic energy of the centre of mass is 0.75 J
(D) Total kinetic energy of the system is 4 J
Sol.:
Applying conservation of linear momentum
… (i)
1 V  5  0  1 2  5V2
1kg
As collision is elastic,
V2  2  V

V
5kg
2m/s
1kg
5kg
V2
After collision
Before collision
… (ii)
V2  1 ms 1, V  3 ms 1
Total momentum of the system = 3 kg m/s
Vcm 

1 V  0 1
 m/s
1 5
2
K . E cm 
1
1  5Vcm2  0.75J
2
Correct choices: (A), (C)
SECTION  III
Paragraph Type
paragraph 2 multiple choice questions have be answered. Each of these questions has four choices (A), (B), (C) and (D) out
Paragraph for Question Nos. 70 to 72
When a particle of mass m moves on the x-axis in a potential of the form V ( x)  kx 2 ,
it performs simple harmonic motion. The corresponding time period is proportional to
V(x)
m
, as can be seen easily using dimensional analysis. However, the motion of the
k
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 particle of mass m moving on the x-axis. Its potential energy is
V ( x)   x 4 (  0) for | x | near the origin and becomes a constant equal to V0 for
V0
X0
| x |  X 0 (see figure).
*70.
Sol.:
If the total energy of the particle is E, it will perform periodic motion only if
(A) E < 0
(B) E > 0
(C) V0 > E > 0
E  V0 , E  0
Correct choice: (C)
*71.
For periodic motion of small amplitude A, the time period T of this particle is proportional to
m

Using dimensional analysis
Correct choice: (B)
(A) A
Sol.:
*72.
1
A
m

(C) A

m
(D)
1
A

m
The acceleration of this particle for | x |  X 0 is
(A) proportional to V0
Sol.:
(B)
(D) E > V0
(B) proportional to
V0
mX 0
(C) proportional to
V0
mX 0
(D) zero
dU
0
dx
Correct choice: (D)
U  V0 and F = 
x
®
25
TC(B)
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
TC (B) on B is shown in the figure.
TC(0)
B
O
73.
In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different
magnetic field B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correct
variation of R with T in these fields?
R
R
B2
(A)
(B)
B1
B1
B2
T
O
T
O
R
R
B1
(C)
(D)
B1
T
O
Sol.:
B2
B2
O
T
Higher the magnetic field, lower is the temperature at which resistance abruptly becomes zero.
Correct choice: (A)
74.
A superconductor has TC (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases 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
Sol.:
As the magnetic field decreases, TC increases
Correct choice: (B)
SECTION  IV
Integer Type
This section contains TEN 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.
75.
When two identical batteries of internal resistance 1  each are connected in series across a resistor R, the rate of heat
produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 then the value of
R in  is
®
Sol.:
26


In first case
2
I
R  2r 
r
r
2
 2 
Rate of heat produced in R, J1  I 2 R  
 R
 R  2r 
In second case

I '
r
R
2



Rate of heat produced in R, J 2  

 R 

As
J 1  2.25J 2
I
R

r

r
2


 R
r 

2 
I'
R

R  4r = 4
Answer is 4
*76.
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 B is at 1500 nm. Considering them to be blackbodies, what
will be the ratio of the rate of total energy radiated by A to that of B?
Sol.:
For a black body, energy radiated per unit time = AT 4
E A  RA 


E B  R B 
2
 TA

T
 B




(where R is radius of the body)
4
… (i)
According to Wein’s displacement law,  mT  constant
2

4
E A  6   1500
  
 =9
E B  18   500 
Answer is 9
*77.


When two progressive waves y1  4 sin(2 x  6t ) and y 2  3 sin 2 x  6t   are superimposed, the amplitude of the
2

resultant wave is
Sol.:
 
4 2  3 2  2  4  3 cos    25  5
 2
A  A12  A22  2 A1 A2 cos  =
Answer is 5
*78.
A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional areas is
4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic
motion of angular frequency 140 rad s–1. If the Young’s modulus of the material of the wire is n × 109 Nm–2, the value of n is
Sol.:

K

m
YA
Lm

1402  YA  n 10

n=4
 4.9 107
1 0.1
9
Lm
Answer is 4
®
*79.
27
A binary star consists of two stars A (mass 2.2MS) and B (mass 11MS), where MS is the mass of the sun. They are separated
by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the
binary star to the angular momentum of star B about the centre of mass is
Sol.:
CM
2.2MS
11MS
d/6
5d/6
L1  I 1 and L2  I 2 
L1  I1

L2  I 2
 11M S  d / 62  2.2M S 5d / 62

6
2

11M S  d / 6

Answers is 6
*80.
Gravitational acceleration on the surface of a planet is
earth. The average mass density of the planet is
6
g , where g is the gravitational acceleration on the surface of the
11
2
times that of the earth. If the escape speed on the surface of the earth is
3
taken to be 11 kms-1, the escape speed on the surface of the planet in kms–1 will be
Sol.:
VeP

VeE
M P RE
 P RP2


M E RP
 E RE2
As
g

RP g P  E

RE g E  P

VeP
P  E 



VeE
 E   P 

VeP = 3 km/s
… (i)
GM
4 R3
g
G
 R
2

3 R2
R
… (ii)
2
 gP

g
 E
2

 


3  6 

2  11 
2
Answer is 3
*81.
A piece of ice (heat capacity = 2100 J kg–1 °C–1 and latent heat = 3.36 × 105 J kg–1) of mass m grams is at –5°C at
atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in
equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is
Ans.: Heat required for melting ice =
1
 3.36 105  336 J
100
Heat used for raising temperature of ice from (–50C to 00C) = Q  420 336 = 84 J
Q  msT  84  m  2100 5
m = 0.008 kg = 8 gm
Answer is 8
*82.
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 constant speeds much smaller than the speed of sound
which is 330 ms–1.
®
Sol.:
28
 V  V0
Apparent frequency received by car is f1  f 0 
 V



… (i)
 V
The frequency of the reflected wave is f 1 '  f 1 
 V  V0

 V  V0 
 = f0 


 V V 
0 


… (ii)
Differentiating f 1 ' with respect to V0
660 f 0
2f
df1 ' 330  V0   330  V0 

f0 
 0
2
2
dV0
330 V0 
330 V0  330
dV0 
165df1 '
 7 km/h
f0
Answer is 7
83.
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
Sol.:
m25
is
m5 0
1
1
1


V1  25 20
V1  100cm
Similarly
1
1
1 52
3




V2 20 50 100 100
V2 
100
cm
3
100
m 25
 25 = 6
100
m50
3  50
Answer is 6
84.
An -particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie
wavelengths are  and p respectively. The ratio
Sol.:
p

, to the nearest integer, is
The momentum of a charged particle accelerated by potential difference V, P  2qmV
de-Broglie wavelength,  

p


h
P
2  2e  4m  V
2 2 3
2e  m  V
Answer is 3
®
29
®
Paper-II (Code: 8)
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
A. General:
1. This Question Paper contains 24 pages having 57 questions. The question paper CODE is
printed on the right hand top corner of this sheet and also on the back page of this
booklet..
2. No additional sheets will be provided for rough work.
3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and
electronic gadgets in any form are not allowed.
4. Log and Antilog tables are given in the paper.
5. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided
separately.
6. Write your Registration No., Name, Name of the centre and sign with pen in the
appropriate boxes. Do not write these anywhere else.
B. Question paper format and Marking scheme:
7. The question paper consists of 3 parts (Chemistry, Mathematics and Physics), and each
Part consists of four Sections.
8. For each question in Section I, you will be awarded 5 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 two (–2) mark will be awarded.
9. For each question in Section II, you will be awarded 3 marks if you darken only the
bubble corresponding to the correct answer and zero mark if no bubble is darkened. No
negative marks will be awarded for incorrect answers in this Section.
10. For each question in Section III, you will be awarded 3 marks if you darken 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.
11. For each question in Section IV, you will be awarded 2 marks for each row in which you
have darkened the bubble(s) corresponding to the correct answer. Thus, each question in
this section carries a maximum of 8 marks.
®
30
CHEMISTRY: Paper-II (Code: 8)
PART – I
SECTION – I
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
Note: Questions with (*) mark are from syllabus of class XI.
O
1.
In the reaction H3C–
(1) NaOH/Br2
–C
NH2
O
(2)
T
–C
Cl
the structure of the Product T is
O
O
H3C–
–C
(A)
O–C
(B)
–NH
C–
O
–CH3
O
(C) H3C–
(D)
–NH
–C
H3C–
O
NH–C
C–
O
O
O
Sol.:
H3C–
(1) NaOH/Br2
–C
O
–C
Me–
NH2
Cl
–NH2
–HCl
Me–
–NH–C–
Correct choice: (C)
*2.
Sol.:
Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B 2 is
(A) 1 and diamagnetic
(B) 0 and diamagnetic
(C) 1 and paramagnetic
(D) 0 and paramagnetic
Molecular orbital configuration of B2 as per the condition will be
1s2, *1s2, 2s2, *2s2, 2Py2
Bond order of B2 =
64
= 1.
2
B2 will be diamagnetic.
Correct choice: (A)
3.
The compounds P, Q and S
COOH
OCH3
O
C
O
HO
H3C
P
Q
S
were separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case respectively, is
O
COOH
OCH3
C
O
(A)
O2N
HO
H3C
NO2
NO2
O
COOH
OCH3
C
O
(B)
HO
NO2
H3C
NO2
NO2
®
COOH
O
C
OCH3
(C)
HO
H3C
NO2
COOH
O
C
OCH3
H3C
NO2
CO2H
NO2
O
NO2
(D)
HO
31
O
NO2
NO2
CO2H
HNO3/H2SO4
Sol.:
NO2
OH
OH
(P)
OCH3
HNO3/H2SO4
CH3
OCH3
NO2
CH3
(Q)
O
O
HNO3/H2SO4
O
O
NO2
(S)
Correct choice: (C)
*4.
The species having pyramidal shape is
(A) SO3
Sol.:
(B) BrF3
(C) SiO32 
(D) OSF2
N
62
=
= 4.
2
2
It has 1 lone pair.
..
OSF2:
O
S
F
(Shape is trigonal pyramidal)
F
The shapes of SO3, BrF3 and SiO32 are triangular planar, T-shaped and triangular planar respectively.
Correct choice: (D)
5.
Sol.:
The complex showing a spin-only magnetic moment of 2.82 B.M. is
(A) Ni(CO)4
(B) [NiCl4]2–
(C) Ni(PPh3)4
(D) [Ni(CN)4]2–
Ni(CO)4 and Ni(PPh3)4 have Ni in zero oxidation state, while CO and Ph 3P are strong field ligands. Thus, both these
molecules have sp3 hybridization and are diamagnetic. [Ni(CN)4]2– has electronic configuration of Ni2+ as 3d8 and
hybridization is dsp2. It is diamagnetic. [NiCl4]2– has sp3 hybridization with 2 unpaired electrons giving spin only magnetic
moment as 8 or 2.82 B.M.
Correct choice: (B)
6.
The packing efficiency of the two-dimensional square unit cell shown below is
L
(A) 39.27%
(B) 68.02%
(C) 74.05%
(D) 78.54%
®
Sol.:
Packing efficiency =
=
Area occupiedby effective circles
Area of square
2r 2
2
L
=
32
 100 =
2r 2
(2 2 r ) 2
100

 100 = 78.54%.
4
Correct choice: (D)
SECTIONII
Integer Type
This section contains a group of 5 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.
The correct digit below the question no. in the ORS is to be bubbled.
*7.
One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the
graphs below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the integer closest
to the ratio wd/ws is
4.5
4.0 a
3.5
3.0
P 25
(atm)
2.0
1.5
1.0
0.5
b
0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
V (lit.)
Sol.:
3
5

 2 5

wd =   4     11      =   6  1  
2
3

 3 2

wd = 
26
L atm
3
5 .5
= –2.303 PV log 11
1/ 2
ws = –4.606 × 1.04 = –4.8 L atm
26
wd
3 = 1.80 ~ 2.0
=
 4. 8
ws
ws = –2.303 RT log
*8.
Among the following, the number of elements showing only one non-zero oxidation state is
Sol.:
O, Cl, F, N, P, Sn, Tl, Na, Ti
Fluorine generally shows O and –1 oxidation states while sodium shows 0 and +1 oxidation state.
9.
Sol.:
Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2
can be expressed in scientific notation as y × 10x. The value of x is
Silver has cubic close packed structure. It has fcc unit cell of edge length a cm.
4 108
10.5 =
6.0231023 (a ) 3
1/ 3
4 108


–8
a= 
 = 4.087 × 10 cm.
23
 6.02310 10.5 
(i) Assuming that surface area given is the area of any extended face of a FCC unit cell.
Area of square face = (4.087 × 10–8)2 cm2
2
Number of Ag atoms per unit area = 2
a
Surface area = 10–12 m2 = 10–8 cm2.

®
Number of Ag atoms in the given surface area =
2 10 8
(4.08710 8 ) 2
33
= 1.197 × 107.
 x = 7.
(ii) Assuming that surface area given is the area corresponding to closest packed layer.
2 a = 4r, r =
2 a 4.087108  2
=
= 1.44 × 10–8
4
4
In a closest packed layer, the arrangement appears like
6
3
6 3
6 3
(2 × 1.44 × 10–8)2 =
( 2r ) 2 =
 4  2.07 1016
4
4
4
= 21.5 × 10–16 cm2
Number of silver atoms on surface =
3  10 8
21.5 10 16
= 0.139 × 108 = 1.39 × 107.
x = 7.
10.
Total number of geometrical isomers for the complex [RhCl(CO)(PPh 3)(NH3)] is
Cl
Sol.:
PPh3
Rh
Cl
PPh3 Cl
Rh
Rh
CO
NH3
NH3
CO NH3
The number of geometrical isomers is 3.
The answer is 3.
*11.
Sol.:
CO
PPh3
The total number of diprotic acids among the following is
H3PO4
H2SO4
H3PO3
H2CO3
H3BO3
H3PO2
H2CrO4
H2SO3
H2S2O7
Diprotic acids are H2SO4, H3PO3, H2CO3, H2S2O7, H2CrO4 and H2SO3.
The answer is 6.
SECTION  III
Paragraph Type
This Section contains 2 paragraphs. Based upon each of the paragraphs 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 12 to 14
Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with
HCN provides compound S. On acidification and heating, S gives the product shown below.
H3C OH
H3C
O O
®
12.
The compounds P and Q respectively are
CH3
CH3
CH
(A)
C
H3C
H and H3C
H
(C)
CH
C
CH3
O
H and H3C
H
H
(D)
C
O
CH2
H3C
H
C
O
O
CH3–CH–CHO + OH–
CH3
H and
C
H3C
O
CH2
H3C
CH
(B)
C
O
Sol.:
34
CH
C
CH3
O
H and
H
H
C
O
CH3–C––CHO + H2O
CH3
(P)
CH2O–
CH2OH
H2O
–
CH3–C –CHO + CH2 = O → CH3–C–CHO –OH–
(Q)
CH3
CH3
CH2OH
CH2OH
CH3–C–CH=O + HCN → CH3–C–CH–CN
CH3
H3C OH
(R)
CH3–C–CHO
CH3
CH2OH
H+/H2O

CH3–C–CH–COOH
Intramolecular esterification
H3C OH
OH
H3C
H3C
O O
(S)
Correct choice: (B)
13.
The compound R is
O
O
(A)
H3C
H3C
C
C
H3C
CH2
Sol.:
Correct choice: (A)
14.
The compound S is
CH3
CH
(C) H3C
CH
O
C
H
CH
CH
O
O
(B)
H
H3C
C
CN
CH2
CN
CN
CH
OH
(D)
H3C
CH
OH
C
H3C
OH
H
C
H3C
CH
OH
H3C
OH
CN
CH2
Sol.:
OH
CH
C
CH3
CH
(D) H3C
H
CH
CH2
H
CH3
C
CH
CH2
(A) H3C
H3C
O
CH
C
C
H3C
OH
CH3
(C) H3C
(B)
H
CH2
OH
Correct choice: (D)
®
35
Paragraph for questions 15 to 17
The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion
undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the
hydrogen atom.
*15.
Sol.:
The state S1 is
(A) 1s
(B) 2s
(C) 2p
(D) 3s
The spherically symmetric state S1 of Li2+ with one radial node is 2s. Upon absorbing light, the ion gets excited to state S 2,
which also has one radial node. The energy of electron in S 2 is same as that of H-atom in its ground state
 En 
Z2
E1 where E1 is the energy of H-atom in the ground state
n2
(3) 2 E1
for Li2+
n2
E n  E1  n  3

 State S2 of Li2+ having one radial node is 3p.
Orbital angular momentum quantum number of 3p is 1.
Energy of state S1 
(3) 2
E1  2.25E1
(2) 2
Correct choice: (B)
*16.
Sol.:
*17.
Sol.:
Energy of the state S1 in units of the hydrogen atom ground state energy is
(A) 0.75
(B) 1.50
(C) 2.25
Correct choice: (C)
(D) 4.50
The orbital angular momentum quantum number of the state S 2 is
(A) 0
(B) 1
(C) 2
Correct choice: (B)
(D) 3
SECTION  IV
(Matrix Type)
This section contains 2 questions. Each question contains has 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.
18.
Match the reactions is Column I with appropriate options in Column II.
Column I
(A)
–N2Cl +
–OH
Column II
NaOH/H2O
0ºC
–N=N–
O
OH OH
(B) H3C–C–C–CH3
H3C CH3
H2SO4
C
H3C
(C)
–C
CH3
(D) HS–
Sol.:
–Cl
1. LiAlH4
2. H3O+
Base
(p)
Racemic mixture
(q)
Addition reaction
(r)
Substitution reaction
(s)
Coupling reaction
(t)
Carbocation intermediate
CH3
C
CH3
CH3
OH
–CH
CH3
O
–OH
S
(A) – (r), (s) ; (B) – (t) ; (C) – (p), (q) ; (D) – (r)
®
19.
All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate
options listed in Column II.
Column I
Sol.:
36
Column II
(A) (CH3)2SiCl2
(p)
Hydrogen halide formation
(B) XeF4
(q)
Redox reaction
(C) Cl2
(r)
Reacts with glass
(D) VCl5
(s)
Polymerization
(t)
O2 formation
(A) – (p), (s)
Polymn.
(CH3)2SiCl2 + 2H2O 
 [(CH3)2Si–O]n.
 (CH3)2Si(OH)2 
2 HCl
(B) – (p), (q), (r), (t)
XeF4 + H2O  Xe + XeO3 + H2F2 + O2
SiO2 + 4HF  SiF4 + 2H2O
glass
SiF4 + 2HF  H2[SiF6]
Soluble hexathorosilicic(IV) acid
(C) – (p), (q)
Cl2 + H2O  HCl + HOCl
(D) – (p), (q)
VCl5 + 7H2O  [V(H2O)6]3+ + 3Cl– + HCl + HOCl
®
37
MATHEMATICS: Paper-II (Code: 8)
PART – II
SECTION – I
This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out of which ONLY ONE is correct.
x
20.
Let f be a real-valued function defined on the interval (–1, 1) such that e
x
f ( x)  2 

t 4  1 dt, for all x  (1, 1), and
0
let f
1
be the inverse function of f . Then ( f
(A) 1
Sol.:
(B)
e  x f ( x)  2 
1
)' (2) is equal to
1
3
(C)
1
2
(D)
1
e
x

t 4  1 dt
…(i)
0
Putting x  0
 f (0)  2  f
1
(2)  0  f ( f
1
( x))  x  f ' ( f
1
( x)) . ( f
1
)' ( x)  1  ( f
1
)' (2) 
1
f ' (0)
Differentiate equation (i) with respect to x
e  x f ' ( x)  e  x f ( x)  x 4  1
Putting x  0

f ' (0)  2  1  f ' (0)  3 
(f
1
)' (2) 
1
3
Correct choice: (B)
4
1
and
respectively, is received by station A and then transmitted to
5
5
3
station B. The probability of each station receiving the signal correctly is . If the signal received at station B is green, then
4
the probability that the original signal was green is
3
6
20
9
(A)
(B)
(C)
(D)
23
20
5
7
21.
A signal which can be green or red with probability
Sol.:
RG 
Received signal is green
OG 
Original signal was green
If the event XYZ denotes original single sent to A was X, received at A was Y and received at B is Z
O
P G
 RG
 P(OG  RG )
P(GRG)  P(GGG)



P
(
R
)
P(GRG)  P(GGG)  P( RRG)  P( RGG)
G

4 1 1 4 3 3
40
    
20
5
4
4
5
4
4

 80 
46 23
4 1 1 4 3 3 1 3 1 1 1 3
          
80
5 4 4 5 4 4 5 4 4 5 4 4
Correct choice: (C)
®
22.
If the distance of the point P(1,  2, 1) from the plane x  2 y  2 z   , where   0 , is 5, then the foot of the perpendicular
from P to the plane is
7
8 4
(A)  , ,  
3
3 3
Sol.:
38
1 4  2  
 4 4 1
(B)  ,  , 
 3 3 3
 1 2 10 
(C)  , , 
3 3 3 
2 1 5
(D)  ,  , 
3 3 2
5
1 4  4

  5  15

  10
(    0)
P
 Plane is x  2 y  2 z 10  0
F
Let the foot of perpendicular be F, equation of line PF is
x 1 y  2 z 1


1
2
2
So the point F can be taken as
(  1, 2  2, 1  2)
It will satisfy equation of plane.
   1  4  4  2  4 10  0   
5
3
7
8 4
 Point F is  , ,  
3
3 3
Correct choice: (A)
*23.
Let S  {1, 2, 3, 4} . The total number of unordered pairs of disjoint subsets of S is equal to
(A) 25
Sol.:
(B) 34
(C) 42
(D) 41
Given set S  1, 2, 3, 4
For disjoint subsets A and B
aS
a A
a A
a A
aB
aB
aB
Then total number of disjoint subsets will be
34  1
 1 = 41
2
Correct choice: (D)
*24.
For r  0, 1,.... 10 let Ar , Br and C r denote, respectively, the coefficient of x r in the expansions of (1  x)10 , (1  x) 20 and
10
(1  x) 30 . Then
 A (B
r
10 Br
 C10 Ar ) is equal to
r 1
2
(B) A10 ( B10
 C10 A10 )
(A) B10  C10


10
Sol.:
2
 10 20
 C r C10 20 C r  30 C10 10 C r  =


r 1

=
20
C10

30

C10  1  30C10

20

C10  1 =
30
10
20
C10

(D) C10  B10
(C) 0

10
20
Cr
C r  30C10
10
r 1
10
Cr

2
r 1
C10  20C10  C10  B10
Correct choice: (D)
25.
Two adjacent sides of a parallelogram ABCD are given by AB  2iˆ  10 ˆj  11kˆ and AD  iˆ  2 ˆj  2kˆ . The side AD is
rotated by an acute angle  in the plane of the parallelogram so that AD becomes AD' . If AD' makes a right angle with the
side AB , then the cosine of the angle  is given by
(A)
8
9
(B)
17
9
(C)
1
9
(D)
4 5
9
®
Sol.:
D'
AB . AD | AB | | AD | cos(90  )

40  15 3 sin 
sin  
D
39
C

8
17
; cos  
9
9
90–
A
B
Correct choice: (B)
SECTION – II
(Integer Type)
This section contains 5 questions. The answer to each question is a single–digit integer, ranging from 0 to 9. The correct
digit below the question no. in the ORS is to be bubbled.
26.
 2k  1 2 k

Let k be a positive real number and let A   2 k
1
 2 k
2
k

 0
2 k


 2k  and B  1  2k
 k
 1 


2k  1
k 

0
2 k .
2 k
0 

If det (adj A)  det (adj B)  106 , then [k] is equal to
[Note: adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k]
Sol.:
2k  1
det A 
0
2 k
2 k
2k  1  2k  1
2 k
2k  1
 2k  1 0
2 k
1
2k
4 k
0
2k  1

2 k
 1  2k  12k  12  8k  2k  13

1
detB  0 [  B is skew symmetric matrix of order 3]

2k  16  106
 2k  1  10  k 
9
 [k] = 4
2
Ans. 4
27.
Let f be a function defined on R (the set on all real numbers) such that
f ' ( x)  2010( x  2009)( x  2010) 2 ( x  2011) 3 ( x  2012) 4 , for all x  R . If g is a function defined on R with values in the
interval (0, ) such that f ( x)  ln(g ( x)) , for all x  R , then the number of points in R at which g has a local maximum is
Sol.:
f x  
1
. g x 
g x 
 g x   f x  g x 
As g x   0 , So g x  will change sign as
–
+
2009
+
2011
 one point of maxima is x  2009
Ans. 1
*28.
Let a1 , a2 , a3 ,.... a11 be real numbers satisfying a1  15 , 27  2a2  0 and ak  2ak 1  ak 2 for k  3, 4, ..., 11 .
If
2
a  a2  ...  a11
a12  a22  ...  a11
 90 , then the value of 1
is equal to
11
11
®
Sol.:
40
a1  15
As a k  a k 1  a k 1  a k 2
So a1 , a 2 , a3 , ...... are in A.P.
Also 27  2a 2  0
 a 2  a1  
3
3
 d 
2
2
2
11
a12  a 22  ...... a11
1
15  r  1d 2  90
 90 
11
11

r 1
 225 150d  35d 2  90  7d 2  30d  27  0 
But d  
Now
9
7
d  37d  9  0
 d  3,
9
7
3

 d   
2

a1  a 2  ......  a11 1
1
 30  10d   30  30  0
11
2
2
Ans. 0
29.
Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively.
Suppose a  6 , b  10 and the area of the triangle is 15 3 . If ACB is obtuse and if r denotes the radius of the incircle of
the triangle, then r 2 is equal to
Sol.:
  15 3
1
6 10 sin C  15 3
2
 sin C 
cos C 
3
2
 C
3
2
36  100  c 2
120

C is obtuse
 c  14
Hence semi perimeter s  15
 r  3  r2  3
Ans. 3
30.
Two parallel chords of a circle of radius 2 are at a distance
3  1 apart. If the chords subtend at the center, angles of
2
, where k  0 , then the value of [k] is [Note : [k] denotes the largest integer less than or equal to k]
k
Sol.:
p1  p 2  3  1
C
D


2 cos  2 cos
 3 1
k
2k
/k
2
p1
O


cos  cos

k
2k
 k 3
3 1
2
2
A
/2k p2
B
Ans. 3

and
k
®
41
SECTION  III
(Paragraph Type)
This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered.
Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Consider the polynomial f ( x)  1  2 x  3x 2  4 x 3 . Let s be the sum of all distinct real roots of f (x) and let t | s | .
31.
The real number s lies in the interval
 1 
(A)   , 0 
 4 
Sol.:
3

(B)   11,  
4

 1
(D)  0, 
 4
1
 3
(C)   ,  
4
2

Given function f ( x)  1  2 x  3x 2  4 x 3 is monotonically increasing
as f ' ( x)  2[6 x 2  3x  1]  0  x  R  12[( x  1 / 4) 2  5 / 48]

f (x) will have only one real root
 3  1
Since f    . f     0
 4  2
1
 3
 s lies in   ,  
2
 4
Correct choice: (C)
32.
The area bounded by the curve y  f (x) and the lines x  0 , y  0 and x  t , lies in the interval
3 
(A)  , 3 
4 
Sol.:
 21 11 
(B)  ,

 64 16 
 21 
(D)  0,

 64 
(C) (9, 10)
1 3
As t | s |  t   , 
2 4
Hence required area will be the shaded region shown in
the figure.
–3/4
–1/2
O
t
t
Area

A(t )  (1  2 x  3x 2  4 x 3 ) dx  t  t 2  t 3  t 4 which is increasing
0

  1   3 
Area   A , A  
  2   4 
 15 525 
 ,

 16 256 
Correct choice: (A)
33.
Sol.:
The function f ' ( x) is
1

 1 
(A) increasing in   t ,   and deceasing in   , t 
4

 4 
1

 1 
(B) decreasing in   t ,   and increasing in   , t 
4

 4 
(C) increasing in (t , t )
(D) decreasing in (t , t )
1

 1

 f x   24 x    f x  increases  x    ,  
4

 4

1

f x  decreases  x    ,  
4

Correct choice: (B)
®
42
Tangents are drawn from the point P(3, 4) to the ellipse
*34.
Sol.:
x2 y2

 1 touching the ellipse at points A and B.
9
4
The coordinates of A and B are
(A) (3, 0) and (0, 2)
 8 2 161 
 and   9 , 8 
(B)   ,
 5
15 
 5 5

 8 2 161 
 and (0, 2)
(C)   ,
 5
15 

 9 8
(D) (3, 0) and   , 
 5 5
Let  y  mx  9m 2  4 be the tangent passing through point 3, 4 to the given ellipse.

4  3m2  9m 2  4
 one tangent is y 
 m
1
and indeterminate.
2
1
5
 9 8
x  at   ,  and the other is x  3 at 3, 0
2
2
 5 5
Correct choice: (D)
*35.
The orthocenter of the triangle PAB is
 8
(A)  5, 
 7
Sol.:
 7 25 
(B)  ,

5 8 
 11 8 
(C)  , 
 5 5
 8 7
(D)  , 
 25 5 
For othercentre of PAB .
Altitude through B that is y 
8
5
Correct choice: (C)
*36.
Sol.:
The equation of the locus of the point whose distances from the point P and the line AB are equal, is
(A) 9 x 2  y 2  6 xy  54x  62y  241  0
(B) x 2  9 y 2  6 xy  54x  62y  241 0
(C) 9 x 2  9 y 2  6 xy  54x  62y  241 0
(D) x 2  y 2  2 xy  27x  31y  120  0
Equation of AB is x  3 y  3


Required locus is 10 x  32   y  42  x  3 y  32
9 x 2  y 2  54x  62y  6 xy  241  0
Correct choice: (A)
SECTION  IV
(Matrix Type)
This section contains 2 questions. Each question has 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
®
*37.
43
Match the statements in Column –I with those in Column-II.
[Note : Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and the real part
of z].
Column I
(A)
Column II
The set of points z satisfying z  i | z |  z  i | z | is
an ellipse with eccentricity
(q)
the set of points z satisfying Im z = 0
1
is
w
(r)
the set of points z satisfying | Im z|  1
1
is
w
(s)
the set of points z satisfying | Re z|  2
(t)
the set of points z satisfying | z |  3
contained in or equal to
(B)
The set of points z satisfying z  4  z  4  10 is
(C)
If | w | 2 , then the set of points
4
5
(p)
z  w
(D)
If | w | 1 , then the set of points z  w 
Sol.:
(A) – (q), (r); (B) – (p); (C) – (p), (s), (t); (D) – (q), (r), (s), (t)
(A)
z is purely real

Im z = 0
(B)
z is on ellipse
(C)
z
x2 y2

1
25 9
3
5
cos   i sin 
2
2
Also represents ellipse
(D)
38.

3
5
| z |
2
2
4x 2 4 y 2
4

 1 , whose eccentricity =
9
25
5
z  2 cos 
Match the statements in Column –I with the values in Column-II.
Column I
(A)
A line from the origin meets the lines
Column II
x  2 y 1 z 1


and
1
2
1
8
3  y  3  z  1 at P and Q respectively. If length PQ  d , then
2
1
1
x
(p)
–4
d 2 is
(B)
3
The values of x satisfying tan 1 ( x  3)  tan 1 ( x  3)  sin1   are
5
(q)
0
(C)
   

 
 
Non-zero vectors a , b and c satisfy a . b  0 , (b  a ) . (b  c )  0 and

 
 


2 | b  c |  | b  a | . If a  b  4c , then the possible values of  are
(r)
4
(s)
5
(t)
6
(D)
Let f be the function on [–, ] given by
2
 9x 
 x
f ( x)  sin  / sin  for x  0 . The value of

 2 
2

f (0)  9
 f ( x) dx is
and

®
Sol.:
(A)
44
8


Let P    2,  2  1,   1 and Q   2  ,    3,   1
3


Now
2
2  1   1


8    3  1
2 
3
 3  4 
P
O
7
26

3
3
…(i)
     2
…(ii)
5
14
      4 
3
3
…(iii)
From (i) and (ii) 
16
44
  4 
3
3
…(iv)
8
20
  4 
3
3
…(v)
From (ii) and (iii) 
Q
1
3
On solving (iv) and (v)    3,  
 10 10 4 
Hence P  5,  5, 2 and Q   ,
, 
3
3
 3
 d2 
25 25 4

 =6
9
9 9
 (t) is correct.
(B)
tan 1 x  3  tan 1 x  3  sin1
3
5


 6 
6
3
3
  tan 1
  tan 1
 tan 1 
 tan 1 
 8  x 2  8  x  4


2
2
4
4
 1 x  9 
 x 8 
 (p) and (r) are correct.
(C)
b  a. b  c  0




 

  a  b   a  b 
  a
0
b 2  b.c  a.c  0  b 2  b . 
 4   4 

 






a2
 0  4  b 2  a 2
 b2  b2 
4
4
 
 
Also 2 b  c  b  a

 a  b
 
 2 b
 b a
4


4  2 b 2  a 2  4b 2  a 2 

Diving (i) and (ii) 

4
4  
2
4

4  b  a


…(i)
 
 2 b a
4  2  4b 2  3a 2
…(ii)
1
 12  3  16   2  8  4  5   2    0
3
  5 as it doesnot satisfy equation(i) and ii
 (q) is correct.
®

(D)
I  2

sin 9 x 2
dx
x
sin
0
2
 9 9 x 

 
cos 9 x 2
2 
 2
I 2
dx  2
dx
cos x 2
 x
sin



0
0
 2 2
 sin


 9x x 
2
2

 
sin 5 x
2 cos x sin 5 x
 sin 5 x 
 2 2
2I  2  2 
dx  4 
 4 2
dx
8
x
x
sin x
2 sin x cos x
 sin x 
0 2 sin cos
0
0
0
2
2
 sin




2
2
2
 2

1  cos 4 x  dx
sin 4 x 

 sin6 x
2
I  4
dx 
dx   4 3  4 sin 2 x dx  4   3  4
sin 2 x 
2
 sin 2 x


0
0
0
 0


2


2
I  4 1 dx  8
0
I



cos 4 x dx  4  8  2
2
0


4
 cos 4x dx
0
4
2
I1 
2 4

4
 2
 (r) are correct.
45
®
46
PHYSICS: Paper-II (Code: 8)
PART – III
SECTION – I
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
39.
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A
small object is kept at a distance 30 cm from the lens. The final image is
(A) virtual and at a distance of 16 cm from the mirror
(B) real and at a distance of 16 cm from the mirror
(C) virtual and at a distance of 20 cm from the mirror
(D) real and at a distance of 20 cm from the mirror
Sol.:
First image will be formed by lens, Using lens formula
1 1 1
1 1
1
 
 

 v  30 cm
v u
f
v 30 15
This image will act as virtual object for mirror and second image will be
formed at 20 cm in front of mirror. This image will act as virtual object for
lens and final image will be formed at
1 1 1
1 1
1
   

 v  6 cm from lens.
v u
f
v 10 15
This image is real and at a distance of 16 cm in front of mirror.
Correct choice: (B)
40.
O
30cm
A uniformly charged thin spherical shell of radius R carries uniform surface
charge density  per unit area. It is made of two hemispherical shells, held
together by pressing them with force F (see figure). F is proportional to
(A)
Sol.:
f =15cm
1 2 2
 R
0
F
(B)
1 2
 R
0
(C)
F
F
1 2
0 R
(D)
1 2
0 R 2
2
R 2
2 0
2 R2
0
Correct choice: (A)
F
*41.
10cm
Fe
F
A block of mass 2 kg is free to move along the x-axis. It is at rest and from
t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction.
The force F(t) varies with t as shown in the figure. The kinetic energy of the
block after 4.5 seconds is
(A) 4.50 J
(B) 7.50 J
(C) 5.06 J
(D) 14.06 J
F(t)
4N
4.5s
O
Sol.:
3s
Area under F-t graph = Change in momentum of body
1
1
4
p   4  3   1.5   1.5 = 4.5 kg m s–1

2
2
3
p2
k
 5.06 J

2m
Correct choice: (C)
t
®
*42.
Sol.:
47
A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second
harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of
sound is 320 ms–1, the mass of the string is
(A) 5 grams
(B) 10 grams
(C) 20 grams
(D) 40 grams
VS
2V

4L p
2l
l
(VS = velocity of sound)
320
1 T

4  0.8 0.5 


 = 0.02 kg m–1

m =  l = 10 grams
0.8m
Correct choice: (B)
*43.
A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16
main scale divisions. For this Vernier calipers, the least count is
(A) 0.02 mm
(B) 0.05 mm
(C) 0.1 mm
(D) 0.2 mm
Sol.:
1 VSD 
16
MSD
20
Least count = 1 MSD – 1 VSD = 1 mm – 0.8 mm = 0.2 mm
Correct choice: (D)
*44.
A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength
81
105 Vm 1 . When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10 –3 m s–1. Given g =
7
9.8 m s–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m–3, the magnitude of q is
(A) 1.6 × 10–19 C
(B) 3.2 × 10–19 C
(C) 4.8 × 10–19 C
(D) 8.0 × 10–19 C
Sol.:
For equilibrium, qE = mg
q
4r 3g
3E
… (i)
Also,
mg = 6rv


4 3
r g  6 r v
3
(after field is switched off)
… (ii)
From equation (i) and (ii) q = 8.0 × 10–19 C
Correct choice: (D)
SECTION – II
Integer Type
This section contains 5 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.
45.
To determine the half life of a radioactive element, a student plots a
dN (t )
dN (t )
graph of n
versus t. Here
is the rate of radioactive
dt
dt
decay at time t. If the number of radioactive nuclei of this element
decreases by a factor of p after 4.16 years, the value of p is
6
5
ln
dN t 
4
dt
3
2
1
2
3
4
5
6
7
8
Years
®
Sol.:
48
d N t 
d N t 
 N 0 e t ,
ln
 lnN 0    t
dt
dt
1
1
Comparing with y  mx  C , we get   (from the graph,   year1  |slope of graph|)
2
2
N t   N 0 e t ,
 N

N
 0
t
t
  1  T1 / 2
 
 2

1  1  ln 2
 
p 2

4.16
1  1  20.693


 
p 2
Answer is 8
p=8
46.
Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from
25
50
m to
m in 30 seconds. What is the speed of the object in km per hour?
3
7
Sol.:
By mirror formula
R=20m
2 1 1
 
R v u
V
t =30s
Initially at t = 0
t=0
50/7m I
25/3m
2
1
1
 u1  50 m


20 25 / 3 u1
I
Finally at t = 30s
2
1
1
 u 2  25m


20 50 / 7 u 2
velocity of object 

| u 2  u1 |
 3 kmh 1
t
Answer is 3
47.
Sol.:
A large glass slab ( = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light
emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?
 sinic  1
sinic 
R
1 3

 5
R
3
 tan ic 
8 cm
4
=5/3

ic = 37°

R = 6 cm
ic
ic
8cm
Source
Answer is 6
48.
At time t = 0, a battery of 10 V is connected across points A and B
in the given circuit. If the capacitors have no charge initially, at
what time (in seconds) does the voltage across them become 4 V?
(Take: ln 5 = 1.6, ln 3 = 1.1)
2M
A
B
2M
Sol.:
2F
2F
In the given figure, Req = 1 M, Ceq = 4 F
Time constant  = ReqCeq = 4 sec
t

 
Potential across the equivalent capacitor at any time t is V  V0 1  e  






4  10 1  e t / 4
t
 ln5  ln3  t = 2 s
4
Answer is 2
®
49
1
of its initial volume. If the initial temperature of the gas is Ti
32
(in Kelvin) and the final temperature is aTi, the value of a is
*49.
A diatomic ideal gas is compressed adiabatically to
Sol.:
For adiabatic process
TV –1 = constant
TiVi –1 = TfVf –1
V
 T f  Ti  i
 Vf






 1
 4 Ti
[ = 7/5 for a diatomic gas]
a=4
Answer is 4
SECTION  III
Paragraph Type
This section contains 2 paragraphs. Based upon the 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.
When liquid medicine of density  is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the
dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that
the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the
size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force
becomes smaller than the weight of the drop, the drop gets detached from the dropper.
*50.
If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming
r << R) is
(A) 2 rT
Sol.:
(B) 2 RT
(C)
2r 2T
R
(D)
2R 2T
r
Force due to surface tension = T 2r cos 
= T 2r
r 2r 2T

R
R
Correct choice: (C)

r

R
*51.
Sol.:
If r = 5 × 10–4 m,  = 103 kgm–3, g = 10 ms–2, T = 0.11 Nm–1, the radius of the drop when it detaches from the dropper is
approximately
(A) 1.4 × 10–3 m
(B) 3.3 × 10–3 m
(C) 2.0 × 10–3 m
(D) 4.1 × 10–3 m
4
mg    R 3 g
3
r2
4
3 2 T
2 T    R 3 g
r
 R4


R
3
2 g
3
T 
  4.1251012
Substituting the values, R 4   r 2
 g 
2
4 log10 R  12  log10 4.125
4 log10 R  12  0.6154
log R  2.8462
log10 R  3  0.1538
R = 1.425  10–3 m
Correct choice: (A)
®
*52.
Sol.:
After the drop detaches, its surface energy is
(A) 1.4 × 10–6 J
(B) 2.7 × 10–6 J
(C) 5.4 × 10–6 J
50
(D) 8.1 × 10–6 J
US = T × 4R2 = 0.11 × 4 × R2 = 2.7 × 10–6 J
Correct choice: (B)
The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is
revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a
diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.
53.
A diatomic molecule has moment of inertial I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is
not allowed) is
(A)
Sol.:
1  h 2
n 2  8 2 I
L  I 
nh
2




(B)

K
1  h 2
n  8 2 I




 h2
(C) n  2
 8 I




 h2
(D) n 2  2
 8 I




 h2 
L2
n2h2
 2
 n2  2 
 8 I 
2 I 4 2 I


Correct choice: (D)
54.
It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to
4
 1011 Hz . Then the moment of inertia of CO molecule about its center of mass is close to (Take h = 2 × 10–34 J s)

(A) 2.76 × 10–46 kg m2
(B) 1.87 × 10–46 kg m2
(C) 4.67 × 10–47 kg m2
(D) 1.17 × 10–47 kg m2
Sol.:
E  hv
K  K 2  K 1 
h2
8  I
2
2
 12

E  K

hv 

I
3h 2
8 2 I
3h
8 2 v
 1.87 1046 kgm2
Correct choice: (B)
55.
In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a. m. u), where 1 a. m. u. 
close to
(A) 2.4 × 10–10 m
Sol.:
(B) 1.9 × 10–10 m
(C) 1.3 × 10–10 m
5
 1027 kg , is
3
(D) 4.4 × 10–11 m
I  d 2 , where  is reduced mass
 m  mo
I   c
 mc  mo

 2
d


d  1.3 1010 m
Correct choice: (C)
®
51
SECTION  IV
(Matrix Type)
This section contains 2 questions. Each question has 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.
56.
Two transparent media of refractive indices 1 and 3 have a solid lens shaped transparent material of refractive index 2
between them as shown in figure in Column II. A ray traversing these media is also shown in the figures. In Column I
different relationships between 1, 2 and 3 are given. Match them to the ray diagrams shown in Column II.
Column I
(A)
(B)
Column II
1   2
(p)
3
(q)
3
2
1
1   2
(C)  2   3
2
3
1
1
(r)
2
(D)
 2  3
(s)
3
2
1
(t)
3
Ans.:
2
1
(A) – (p), (r) ; (B) – (q), (s), (t) ; (C) – (p), (r), (t) ; (D) – (q), (s)
®
57.
52
You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two
circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II.
When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2.
(indicated in circuits) are related as shown in Column I. Match the two
Column I
Column II
V1
(A)
I  0, V1 is proportional to I
V2
6mH
3F
(p)
V
(B)
I  0, V2  V1
(q)
V1
V2
6mH
2
V
(C) V1  0, V2  V
(r)
V1
V2
6mH
2
~
V
V1
(D)
I  0, V2 is proportional to I
(s)
V2
6mH
3F
~
(t)
V2
1k
3F
~
Sol.:
V
V1
V
(A) – (r), (s), (t) ; (B) – (q), (r), (s), (t) ; (C) – (p), (q) ; (D) – (q), (r), (s), (t)
®
BREAK UP 1 (LEVEL OF DIFFICULTY)
53
®
54
®
55
®
BREAK UP 2 (XI-XII)
BREAK UP 2 ( XI-XII)
56
®
57
®
BREAK UP 2 (XI-XII)
58
®
BREAK UP 3 (TOPICWISE/PARTWISE)
59
®
60
®
61

IIT JEE 2010 Solution

Transcription

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