2010 sri chaitanyas_iitjee_paper-i_solutions

Sri Chaitanya IIT Academy.A.P. Ø1× :: IIT JEE 2010 Paper - I ::
Sri Chaitanya “IIT” Academy, Plot No 875, Ayyappa Society, Madhapur, Hyderabad - 500081
(: 040-64566177, Website : www.srichaitanya.net
Sri Chaitanya IIT Academy
Sri Chaitanya’s Solutions to
IIT – JEE - 2010
(PAPER – 1)
Time: 3 Hours Maximum Marks: 252
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose
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11. Darken the appropriate bubbles below your registration number with HB pencil.
C. Question paper format and marking scheme :
12. The question paper consists of 3 Parts (Chemistry, Mathematics and Physics). Each part
consists of four sections
13. 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.
14. 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 bubbles are darkened. Partial
marks will be awarded for partially correct answers. No negative marks will be awarded in
this Section.
15. 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.
16. For each question in Section IV, you will be awarded 3 marks if you darken the bubble corre
sponding to the correct answer and zero mark if no bubbles is darkened. No negative marks
will be awarded for in this section.

IIT JEE 2010 (PAPER - I)
PART - I , CHEMISTRY
PART - I : CHEMISTRY
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
1. The bond energy (in kcal mol–1
) of a C–C single bond is approximately
A) 1 B) 10 C) 100 D) 1000
Ans : C
Sol : C – C bond energy is 100 kcal/mol approximately
2. The species which by definition has ZERO standard molar enthalpy of formation
at 298K is
A) Br2
(g) B) Cl2
(g) C) H2
O(g) D) CH4
(g)
Ans : B
Sol : ( )0
2 ;fH for Cl g∆ which is in native state = 0
3. Plots showing the variation of the rate constant (k) with temperature (T) are given
below. The plot that follows Arrhenius equation is
A) k
T
B) k
T
C) k
T
D) k
T
Ans : A
Sol : k = /
;Ea RT
A e−
⋅ k increases exponentially with temperature

4. In the reaction OCH3 HBr
→ the products are
A) OCH3Br and H2
B) Br and CH3
Br
C) Br and CH3
OH D) OH and CH3
Br
Ans : D
Sol : O CH3
H +
→ O CH3
H
Br-
(SN2
)
OH + CH3 - Br
5. The correct statement about the following disaccharide is
o
H
OH
CH2OH
H (a)
H
OH
OH
H
H
OCH2CH2O
O
H
OH
OH
H
H
CH2OH
CH2OH
(b)
A) Ring (a) is a pyranose with α − glycosidic link
B) Ring (a) is a furanose with α − glycosidic link
C) Ring (b) is a furanose with α − glycosidic link
D) Ring (b) is a pyranose with α − glycosidic link
Ans : A
Sol : Pyranose is six membered ring and furanose is five membered ring. Ring(a) is
attached with α − 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) BrCH2
CH2
CH2
CH2
CH3
and CH3
CH2
C ≡CH
B) BrCH2
CH2
CH3
and CH3
CH2
CH2
C ≡CH
C) BrCH2
CH2
CH2
CH2
CH3
and CH3
C ≡ CH
D) BrCH2
CH2
CH2
CH3
and CH3
CH2
C ≡ CH
Ans : D
Sol : R – Br + sodamide + Alkyne → CH3
– CH2
– C ≡ C – CH2
– CH2
– CH2
– CH3
possiblities
1) CH3
– CH2
– Br + CH3
– CH2
– CH2
– CH2
– C ≡C – H
(or)
2) CH3
– CH2
– CH2
– CH2
– Br + CH3
– CH2
– C ≡CH ∴ D is correct
7. The ionization isomer of [Cr(H2
O)4
Cl (NO2
)]Cl is
A) [Cr(H2
O)4
(O2
N)] Cl2
B) [Cr(H2
O)4
Cl2
](NO2
)
C) [Cr(H2
O)4
Cl(ONO)] Cl D) [Cr(H2
O)4
Cl2
(NO2
)]. H2
O
Ans : B
Sol : Ions present outside of coordination sphere are Cl–
and 2NO−
. The ionisation
isomer of [Cr(H2
O)4
Cl(NO2
)] Cl is [Cr(H2
O)4
Cl2
] (NO2
)
8. The correct structure of ethylenediaminetetraacetic acid (EDTA) is
A) B)
C) D)
Ans : C
Sol : The structure of EDTA is

SECTION - II (Multiple Correct Choice Type)
This section contains 5 multiple choice questions. Each question has four choices A), B), C), D) out of whihc ONE OR
MORE may be correct.
9. In the reaction
OH
( ) 2/NaOH aq Br
→ the intermediate(s) is(are)
A)
O
Br
Br
B)
O
BrBr
C)
O
Br
D)
O
Br
Ans : A,B,C
Sol : Phenol react with NaOH gives phenoxide ion which is ortho,para directing group
hence Br+
attack at ortho, para positions
10. In the Newman projecttion for 2, 2 – dimethylbutane
H3C
X
CH3
HH
Y
X and Y can respectively be
A) H and H B) H and C2
H5
C) C2
H5
and H D) CH3
and CH3
Ans : B,D
Sol : Rotation between C2
and C3
H3C
CH3(X)
CH3
HH
CH3(Y)
Rotation between C1
and C2
H3C
H(X)
CH3
HH
C2H5(Y)

11. Aqueous solutions of HNO3
, KOH, CH3
COOH, and CH3
COONa of identical con-
centrations are provided. The pair(s) of solutions which form a buffer upon mix-
ing is (are)
A) HNO3
and CH3
COOH B) KOH and CH3
COONa
C) HNO3
and CH3
COONa D) CH3
COOH and CH3
COONa
Ans : D
Sol : Weak acid and its salt with strong base
12. The reagent(s) used for softening the temporary hardness of water is (are)
A) Ca3
(PO4
)2
B) Ca(OH)2
C) Na2
CO3
D) NaOCl
Ans : B,C
Sol : Calcium hydroxide and Sodium carbonate are used to softening the temporary
hardness of water
13. Among the following, the intensive property is (properties are)
A) molar conductivity B) electromotive force
C) resistance D) heat capacity
Ans : A,B
Sol : Intensive property is independent of quantity
SECTION - III (Paragraph Type)
This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the
second 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 concentrationof potassium ions inside a biological cell is at least twenty times
higher than the outside. The resulting potential difference across the cell is impor-
tant in several processes such as tramsmission of nerve impulses and maintaining
theionbalance.AsimplemodelforsuchaconcentrationcellinvolvingametalMis:
M(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 ; 0G∆ > B) Ecell
> 0 ; 0G∆ < C) Ecell
< 0 ; 0
0G∆ > D) Ecell
> 0 ; 0
0G∆ <
Ans : B
Sol : Cell reaction is spontaneous reaction

15. If the 0.5 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
Ans : C
Sol : For given cell, Ecell
=
0.0591
1
log ( )
1
0.05
given Ecell
= 70 mV
hence Ecell
=
0.0591
1
log
1
0.0025
Ecell
=
0.0591
1
log ( )
2
1
0.05
= 140 mV
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
.5H2
O),
atacamite (Cu2
Cl(OH)3
), cuprite (Cu2
O), copper glance (Cu2
S) and malachite
(Cu2
(OH2
CO3
). However, 80% of the world copper production comes from the ore
chalcopyrite (CuFeS2
). The extraction of copper from chalcopyrite involves par-
tial roasting, removal of iron and self-reduction.
16. Partial roasting of chalcopyrite produces
A) Cu2
S and FeO B) Cu2
O and FeO C) CuS and Fe2
O3
D) Cu2
O and Fe2
O3
Ans : A
Sol : CuFeS2
Partial Roasting
→ Cu2
S + FeO + 2SO ↑
17. Iron is removed from chalcopyrite as
A) FeO B) FeS C) Fe2
O3
D) FeSiO3
Ans : D
Sol : FeO + SiO2 → FeSiO2
18. In self-reduction, the reducing species is
A) S B) O2–
C) S2–
D) SO2
Ans : C
Sol : 2 2 22 6Cu S Cu O Cu SO+ → +

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
19. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5
is
Ans : 0
Sol : Br
F
F
FF
F
Repulsion b/w lone pair of e–
and bond pair of electrons. Because of this bond
angle less than 900
20. The value of n in the molecular formula Ben
Al2
Si6
O18
is
Ans : 3
Sol : Cyclic Silicate ; Si6
12
18O−
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
Ans : 3
Sol : Average titre value is 25.15 which is round of as number of significant figures as
small as in the individual titre values
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) 1.0 0.75 0.40 0.10
t(min) 0.0 0.05 0.12 0.18
Ans : 0
Sol :
x
k
t
=
I)
0.25
5
0.05
k = = II)
0.6
5
0.12
k = = III)
0.9
5
0.18
k = =
k value remains constant
23. The number of neutrons emitted when 235
92 U undergoes controlled nuclear fission
to 142 90
54 38Xe and Sr is
Ans : 4
Sol : 235 1 142 90 1
92 0 54 38 04U n Xe Sr n+ → + +

24. The total nuber of basic groups in the following form of lysine is
H3N - CH2 - CH2 - CH2 - CH2
CH - C
OH2N
O
Ans : 2
Sol : NH2
and O–
25. The total number of cyclic isomers possible for a hydrocarbon with the molecular
formula C4
H6
is
Ans : 5
Sol :
CH3
,
CH2
,
CH3
, ,
26. In the scheme given below, the total number of intramolecular aldol condensa-
tion products formed from ‘Y’ is
3
2
1. 1. ( )
2. , 2.
O NaOH aq
Zn H O heat
Y→ →
Ans : 1
Sol :
Y =
O
O
H H H H
H H H H
Y is symmetrical diketone. All Hα − atoms are identical ∴ only one type of con-
densation is possible

27. Amongstthefollowing,thetotalnumberofcompoundssolubleinaqueousNaOHis
NH3C CH3 COOH OCH2CH3
CH2OH
OH
NO2
OH
N
H3C CH3
CH2CH3
CH2CH3 COOH
Ans : 4
Sol :
COOH
,
OH
,
OH
N
H3C CH3
,
COOH
28. Amongst the following, the total number of compounds whose aqueous solution
turns red litmus paper blue is
KCN K2
SO4
(NH4
)2
C2
O4
NaCl Zn(NO3
)2
FeCl3
K2
CO3
NH4
NO3
LiCN
Ans : 2
Sol : }2
. .
H O
S B W A
KCN KOH HCN→ + → solution becomes alkaline
}2
2 3 2 3
. .
H O
S B W A
K CO KOH H CO→ + → solution becomes alkaline

PART - II : MATHEMATICS
ONE is correct
29. Let f, g and h be real - valued functions defined on the interval [0, 1] by
− − −
= + = + = +
2 2 2 2 2 2
2
( ) , ( ) ( ) .x x x x x x
f x e e g x xe e and h x x e e If a, b and c denote, respec-
tively, the absolute maximum of f, g and h on [0, 1], then
A) = ≠a b andc b B) = ≠a c and a b C) ≠ ≠a b andc b D) = =a b c
Ans : D
Sol : −
= +
2 2
( ) x x
f x e e
−
= = + 1
(0) 2 (1)f and f e e
−
= − ≥ ∀ ∈
2 2
1
( ) 2 ( ) 0 [0,1]x x
f x x e e x
absolute max. of −
= + 1
a e e
−
= +
22
( ) x x
g x x e e
−
= = + 1
(0) 1, (1)g g e e
( ) −
= + − > ∀ ∈
2 2
2
’( ) 1 2 2 0 [0,1]x x
g x x e x e x
= = = + 1
(1)g b g e e
−
= +
222
( ) x x
h x x e e
−
= = + 1
(0) 1, (1)h h e e
−
= + − ≥ ∀ ∈
2 2
2 2
’( ) 2 [ ] 0 [0,1]x x
h x x e x ex e x
= = = + 1
(1)h c h e e
∴ = =a b c

30. Let p and q be real numbers such that ≠ ≠ ≠ −3 3
0, .p p q and p q If α βand are non-
zero complex numbers satisfying α β α β+ = − + =3 3
,p and q then a quadratic equa-
tion having
α β
β α
and as its roots is
A) + − + + + =3 2 3 3
( ) ( 2 ) ( ) 0p q x p q x p q B) + − − + + =3 2 3 3
( ) ( 2 ) ( ) 0p q x p q x p q
C) − − − + − =3 2 3 3
( ) (5 2 ) ( ) 0p q x p q x p q D) − − + + − =3 2 3 3
( ) (5 2 ) ( ) 0p q x p q x p q
Ans : B
Sol :α β α β+ − + =3 3
,p q
Sum of the roots
α β α β
β α αβ
−+
= + = =
+
32 2
3
2p q
p q
Product of the roots
α β
β α
= =. 1
∴ Req. equation.
 −
− + = 
+ 
3
2
3
2
1 0
p q
x x
p q
⇒ + − − + + =3 3 3 3
( ) ( 2 ) ( ) 0p q x p q x p q
31. The value of →
+
+∫
A
3 4
(1 )1
lim
4
x
x O
O
t n t
dt
x t
A) 0 B)
1
12
C)
1
24
D)
1
64
Ans : B
Sol :
→
+
+∫ 4
0
30
log(1 )
4
x
x
t t
dt
t
Lt
x
32. The number of 3 x 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) −9
2 1 C) 168 D) 2

Ans : A
Sol : A system of equations in 3 unknowns can’t have two distinct solutions.
33. Let P, Q, R and S be the points on the plane with position vectors
− − + − +ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ,4 ,3 3 3 2i j i i j and i j respectively. The quadrilateral PQRS must be a
A) parallelogram, which is neither a rhombus nor a rectangel
B) square
C) rectangle, but not a square D) rhombus, but not a square
Ans : A
Sol : − − −( 2, 1,0), (4,0,0), (3,3,0), ( 3,2,0)A B C D
by transforming all vertices to − −( 2, 1,0) we get above diagram.
Therefore AB, CD are equal AD, BC are equal. Therefore ABCD is a parallelo-
gram but not rhombus or rectangel.
34. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times.
If 1 2 3,r r andr are the numbers obtained on the die, then the probability that
ω ω ω+ + =1 2 3
0r r r is
A)
1
18
B)
1
9
C)
2
9
D)
1
36
Ans : C

Sol :
ω ω ω+ + =
=
=
+ +
∴ = =
1 2 3
1 2 3
1 1 1
1 1 1
0
1 2 3.........3! 6
4 5 6..........3! 6
.
.
.
3 , 3 1,3 2,.......
2 .2 .2 .3! 2
( ) .
6 .6 .6 . 9
r r r
r r r
n n n
c c c
p A
c c c
35. Equation of the plane containing the straight line = =
2 3 4
yx z
and perpendicular to
the plane containing the straight lines = = = =
3 4 2 4 2 3
y yx z x z
and is
A) + − =2 2 0x y z B) + − =3 2 2 0x y z C) − + =2 0x y z D) + − =5 2 4 0x y z
Ans : C
Sol : d.r’s of the normal to the plane containing = = = =
2
&
3 4 2 4 2 3
y yx x z
are 8, -1, -10
Now d.r’s of the normal of the plane containing = = &
2 3 4
yx z
above normal are
1, -2, 1. Therefore Req. plane − − − + − =1( 0) 2( 0) 1( 0) 0x y z
− + =2 0x y z
36. 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 +sin2 sin2
a c
C A
c a
is
A)
1
2
B)
3
2
C) 1 D) 3

Ans : D
Sol : +2sin .cos 2sin cos
a c
c c A A
c a
= +.2sin cos .2sin cos
2 sin 2 sin
a c
c c A A
R c R A
+
= = = = =
cos cos 2 sin
2.sin60 3
a c c A b R B
R R R
37. Let ABC be a triangle such that
π
∠ =
6
ACB and let a, b and c denote the lengths of
the sides opposite to A, B and C respectively. The values (s) of x for which
= + + = − = +2 2
1, 1 2 1a x x b x andc x is (are)
A) ( )− +2 3 B) +1 3 C) +2 3 D) 4 3
Ans : A,B
Sol :
π + + + − − +
= =
+ + −
2 2 2 2
2 2
( 1) ( 1) (2 1)3
cos
6 2 2( 1)( 1)
x x x x
x x x
( )+ + − − − = + + + + + + − + − − −4 3 2 2 4 2 3 2 4 2 2
3 1 1 2 2 2 2 1 4 4 1x x x x x x x x x x x x x x
( )+ − − = + − − +4 3 4 3 2
3 1 2 2 3 2 1x x x x x x x
( ) ( ) ( ) ( )⇒ − + − − − − + + =4 3 2
2 3 2 3 3 2 3 1 3 0x x x x
= − + = + = ±1(2 3), 1 3, 1x x x
38. Let 1 2z and z be two distinct complex numbers and let = − +1 2(1 )z t z tz for some real
number t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzero
complex number w, then
A) − + − = −1 2 1 2| | | || |z z z z z z B) − = −1 2( ) ( )Arg z z Arg z z
C)
− −
=
− −
11
2 12 1
0
z z z z
z z z z
D) − = −1 2 1( ) ( )Arg z z Arg z z

Ans : A,B,C,D
Sol : Z lie on the line joining the points z1
, z2
AB + BC = AC
Therefore − + − = −1 2 1 2| | | || |z z z z z z
ABC collinear therfore argument of AB, argument of AC equal and also argu-
ment of AC, argument of AB equal. So B, D is correct.
Complex slope of AB = complex slope of BC
− −
= = − − −
− −
1 2
1 2
( )
z z z z
c
z z z z
39. The value(s) of
−
+∫
1 4 4
2
0
(1 )
1
x x
dx
x is (are)
A) π−
22
7
B)
2
105
C) 0 D)
π
−
71 3
15 2
Ans : A
Sol : ( )− + − +2 3 4 4
1 4 6 4 2x x x x
+ − + − + − + − +
+
− + −
− −
+
+
−
− −
+
−
2 8 7 6 5 4 6 5 4 2
8 6
7 6 5
7 5
6 4
6 4
4
4 2
2
2
1) 4 6 4 ( 4 5 4 4
4 5 4
4 4
5
5 5
4
4 4
4
4 4
4
x x x x x x x x x x
x x
x x x
x x
x x
x x
x
x x
x
x

Given integral =
−
− + − + −
17 6 5 3
1
0
4 4
5 4 4tan
7 6 5 3
x x x x
x x
π= − =
22
0
7
.
40. Let A and B to two distinct points on the parabola =2
4 .y x If the axis of the pa-
rabola touches a circle of radius r having AB as its diameter, then the slope of the
line joining A and B can be
A) −
1
r
B)
1
r
C)
2
r
D)
−2
r
Ans : C,D
Sol :
   
= =   
   
1 2
1 2
2 2
, , ,A t B t
t t
Mid.point of A,B is center of the circle that is
+
=1 22( )
2
t t
r . Because circle touches
x -axis
Slope of
− −
= =
− − +
2 1 2 1
2 2
2 1 2 1 2 1
2( ) 2( )
( )( )
t t t t
AB
t t t t t t
=
2
r
in first quadrant, in fourth quadrant it is −
2
r
41. Let f be a real - valued function defined on the interval ∞(0, ) by
= + +∫A
0
( ) 1 sin .
x
f x n x 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 α > 1 such that <’( ) ( )f x f x for all α∈ ∞( , )x
D) there exists β > 0 such that β+ ≤( ) ’( )f x f x for all ∈ ∞(0, )x

Ans : B, C
Sol : = + +∫0
( ) log 1 sin
x
f x x tdt
= + +1 1
( ) 1 sin .f x x
x
exists ∀ ∈ ∞(0, )x
∴ 1
( )f x exists
Hence 1
( )f x is continuous
−
= +
+2
1 1
"( ) (cos )
2 1 sin
f x x
x x
is not definied for 1 +sin x = 0
"( )f x not exist ∀ ∈ ∞(0, )x
1
( )f x is not differential for sinx = - 1
So (B) correct.
= + +∫0
( ) log 1 sin .
x
f x x t dt
= + +1 1
( ) 1 sinf x x
x
If x > 1 then ∈
1
(0,1)
x
and  + ∈ 1 sin 0, 2x
both ranges are finite
∴ ∈1
( ) [0, ]f x k for same k.
For α = = + +∫0
, ( ) log 1 sin
k
k k
e f x e t = +( )k same position
α α∴ < ∈1
( ) ( ) ( , )f x f x for x where α = k
e . So option C is correct.
Sum of two bounded below functions is bounded below.
Hence finite β is not possible

The circle + − =2 2
8 0x y x and hyperbola − =
22
1
9 4
yx
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) − + =3 4 8 0x y D) − + =4 3 4 0x y
Ans : B
Sol : Tangent for hyperbola = ± −2
9 4y mx m
If it touches the circle then
r = d
( )± −
=
+
2
2
4 9 4
4 .
1
m m
m
+ = + − ± −2 2 2 2
16 16 16 9 4 8 9 4m m m m m
− = ± −2 2
9 20 8 9 4m m m
( )+ − = −4 2 2 2
81 400 360 64 9 4m m m m
= −4 2
576 256m m
+ − =4 2
495 104 400 0m m
= =2
4/5, 2/ 5m m
43. Equation of the circle with AB as its diameter is
A) + − + =2 2
12 24 0x y x B) + + + =2 2
12 24 0x y x
C) + + − =2 2
24 12 0x y x D) + − − =2 2
24 12 0x y x
Ans : A

Sol : + − = − −2 2
8 0 (1)x y x
− = − −2 2
4 9 36 (2)x y
(1)equation x (9) + 2 ⇒ − − =2
13 72 36 0x x
−
=
6
6,
13
x
∴ = 6if x , = ±2 3y
( ) ( )= = −6,2 3 , 6, 2 3A B
Equation of circle with A, B as ends of diameter
( )( ) ( )( )− − + + − =6 6 2 3 2 3 0x x y y
+ − + = − − −2 2
12 24 0 ( )x y x A
Let p be an odd prime number and pT be the following set of 2 x 2 matrices:
  
= = ∈ −  
  
: , , {0,1,2,....., 1}p
a b
T A a b c p
c a
44. The number of A in pT such that A is either symmetric or skew-symmetric or
both, and det (A) divisible by p is
A) − 2
( 1)p B) −2( 1)p C) − +2
( 1) 1p D) −2 1p
Ans : D
Sol :
  
= = ∈ −  
 
; , , 0,1,2..... 1}p
a b
T A a b c p
c a
case. 1 = − ⇒ = = = 0T
A A a b c
case. 2
 
= ⇒ = = = − 
 
2 2
| |T
a b
A A b c and A a b
c a
∴ −a b is divisible by a + b is divisible by P
a = b ⇒ a - b is divisible P. ∴ there are P ways of choosings

a + b is divisible by P when (a, b) = (1, P - 1) or (2, 1, -2).....(P-1, 1)
∴ there are P - 1 ways of choosing
∴ total number of ways is 2P - 1
45. The number of A in pT 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) − − +2
( 1)( 1)p p p B) − −3 2
( 1)p p C) − 2
( 1)p D) − −2
( 1)( 2)p p
Ans : C
Sol : Trace of A = 2a is divisible by p only is a = 0
∴ suppose ≠ 0a then = −2
| |A a bc
Since each of b and c is less than P let b = p - r and c = p - k.
Then = −2
| |A a bc
Since each of b and c is less than P. Let b = p -r and c = p - k
Then = − − −2
| | ( )( )A a p r p k
now |A| is divisible p is a2
= rk
each of r and k can be selected is p - 1 ways
∴ Ans. (P - 1)2
.
46. The number of A in pT such that det (A) is not divisible by p is
A) 2
2p B) −3
5p p C) −3
3p p D) −3 2
p p
Ans : B
Sol :
47. 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 interval [-10, 10] by
−
= 
+ −
[ ] [ ] ,
( )
1 [ ] [ ]
x x if x isodd
f x
x x if x is even

Then the value of
π
π
−
∫
102
10
( )cos
10
f x x dx is.
Ans : 4
Sol : ( )
−
= 
− −
[ ]; [ ]
( )
1 [ ] ; [ ]
x x if x isodd
f x
x x if x iseven
f is periodic with period 1and πcos x has period 2
π π
π π
−
∴ = × ×∫ ∫
10 22 2
10 0
( ).cos 10 ( ).cos
10 10
f x x f x x
π π π π
π π
   
= − + − = + =     
∫ ∫
1 2
2 2
2 2
0 1
2 2
(1 )cos 1cos . 4.x xdx x x dx
48. Let w be the complex number
π π
+
2 2
cos sin .
3 3
i Then the number of distinct
complex numbers z satisfying
ω ω
ω ω
ω ω
+
+ =
+
2
2
2
1
1 0
1
z
z
z
is equal to
Ans : 1
Sol :
+ =
+
2
2
1 0
1
z w w
z z w
z z w
+ =
+
2
2
1
1 1 0
1 1
w w
z z w
z w
;
+ − − =
− + −
2
2 2
2
1
0 1 0
0 1
w w
z z w w w
w z w w
( )( ) ( )( ) + − + − − − − = 
2 2 2
1 1 0z z w w z w w w w
( ) ( ) − − − − + =
 
22 2 2 3
( 1 . 0z z w w w w w
( ) − + − − + + − = 
2 2
2 1 1 0z z w w w w

( )− + − =2
2 2 2 0z z w
− = → =2
( 2 ) 0 0z z w z
49. Let =, 1,2,.....,100,kS k denot the sum of the infinite geometric series whose first
term is
−1
!
k
k
and the common ratio is
1
.
k
Then the value of
=
+ − +∑
2 100
2
1
100
( 3 1)
100!
k
k
k k S is
Ans : 4
Sol :
−
−
= = =
− −−
1
1 1
1 1 11
k
k
k kk
S
k k k
k
= =
 − − − − − −
+ = +  − − 
∑ ∑
2 22 2100 100
1 1
( 3 1 ( 1) ( 1) 1100 100
100 1 100 1k k
k k k k
k k
= =
 − − −
= + = + − 
− − − 
∑ ∑
2 2100 100
1 1
( 1)( 2) 1100 100 1 1
100 1 100 3 1k k
k k
k k k
= =
 
+ − 
− − 
∑ ∑
2 100 100
3 1
100 1 1
100 3 1k kk k
= − − +
2
100 1 1
4
100 98 99
= − = +
100 1 1
4
99 98 99 ;
 
= − − + = 
 
1 100 1
1 4 4
98 99 99
50. The numebr of all possible value of θ , where θ π< <0 , for which the system of
equations
θ θ
θ θ
θ
θ θ θ
+ =
= +
= + +
( )cos3 ( )sin 3
2 cos3 2sin 3
sin 3
( )sin 3 ( 2 )cos3 sin 3
y z xyz
x
y z
xyz y z y
have a solution 0 0 0( , , )x y z with ≠0 0 0,y z is

Ans : 8
Sol : θ θ θ+ − =cos3 cos3 sin 3 0y z t
θ θ θ+ − =(2sin 3 ) (2cos3 ) sin 3 0y z t
θ θ θ θ+ + − =(cos3 sin 3 ) (2 cos3 ) .sin 3 0y z t
θ θ θ
θ θ θ
θ θ θ θ
−
− =
+ −
cos3 cos3 sin 3
2sin 3 2 cos3 sin 3 0
cos3 sin 3 2 cos3 sin 3
θ
θ θ
θ θ
=
−
cos3 1 1
cos3 2sin 3 2 1 0
cos3 sin 3 0 0
( )θ θ θ− =cos3 . cos3 sin 3 0
θ θ θ= = =cos3 0,tan3 1 ,sin 3 0j
( )
π π π
θ θ= + = +3 2 1 ,
2 3 12
n
n
π π π
θ∴ =
5
, ,
6 2 6
π π π5 11
, ,
2 12 12
;
π π
θ =
2
,
3 3
51. Let f be a real-valued differentiable function on R (the set of all real numbers)
such that f(1) = 1. If they 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
Ans : 9
Sol : Tangent at P(x1
, y1
) is y - y1
= m (x - x1
)
⇒ −y intercept = − + = 3
1 1 1mx y x
−
− + = ⇒ =
3
3
. .
dy dy y x
x y x
dx dx x
= = − 21
.
dy
y x is DE
dx x
with I.F =
1
andso
x

.G S is
−
= − = +∫
2
1
. .
2
x
y x dx c
x
( )− +
= ⇒ = ⇒ =
2
33
(1) 1
2 2
x x
f c y
− −
− = =
( 3)( 6)
( 3) 9
2
f
52. The number of values of θ in the interval
π π 
− 
 
,
2 2
scuh that
π
θ ≠
5
n
for
θ θ= ± ± =0, 1, 2 tan cot5n and as well as θ θ=sin 2 cos4 is
Ans : 3
Sol :
π
θ π θ 
= + − 
 
5
2
n
( )
π
θ = +6 2 1
2
n ; ( )
π
θ = +2 1
12
n
n = 0,
π
θ =
12
; n = 1,
π
θ =
4
n = 2,
π
θ
−
=
12
; n = 2,
π
θ =
5
12
n = 2,
π
θ = −
4
; n = - 3,
π
θ
−
=
5
12
π
θ π θ 
= ± − 
 
4 2 2
2
n ;
π
θ π θ= + −4 2 2
2
n
π
θ = +6 (4 1)
2
n ;
π
θ = +(4 1)
12
n
π
θ π θ= − +4 2 2
2
n ;
π
θ = −2 (4 1)
2
n
( )
π
θ = −4 1
4
n ’ n = 0,
π
12
,
π
−
4
n = 1,
π π5 3
,
12 4
; n = -1,
π π−
−
5
,
4 4

n = 2,
π π3 7
,
4 4
Common ans. for both
π π π
−
5
, ,
12 12 4
53. The maximum value of the expression
θ θ θ θ+ +2 2
1
sin 3sin cos 5cos
is
Ans : 2
Sol : ( )θ θ θ θθ
=
− + + + + +  
 
1 2
1 cos2 3 1 cos2 6 3sin2 4cos2
sin2 5
2 2 2
( ) [ ]θ θ
θ θ
 
+ + ∈ ⇒ ∈ + +  
2 2
6 3sin2 4cos2 1,11 ,2
6 3sin2 4cos2 11
54. If
G G
aandb are vectors in space given by
− + +
= =
G G ˆˆ ˆ ˆ ˆ2 2 3
,
5 14
i j i j k
a and b then the value
of ( ) ( ) ( ) + × × −
 
G G G G G G
2 . 2a b a b a b is
Ans : 5
Sol : ,a b are unit vectors and mutually perpendicular. Let × =a b c
( ) ( )( )  ∴ + × − =  
−
2 1 0
2 . 2 0 0 1
1 2 0
a b c a b abc
 = = × = 5 5 1 5. sin , ,abc ce a b c are
unit vertors, mutually perpendicular.
55. The line 2x + y = 1 is tangent to the hyperbola − =
22
2 2
1.
yx
a b
If this line passes
throught the point of intersection of the nearest directrix and the x-axis, then
the eccentricity of the hyperbola is.
Ans : 2
Sol : Tangent 2x + y = 1 . Cuts x-axis at
 
⇒ = ⇒ = 
 
1 1
,0 2
2 2
a
e a
e

Tangent = − + ⇒ = − ⇒ = − ⇒ = − = −2 2 2 2 2 2 2 2 2
2 1 1 .4 4 1 1y x e a m b a b b a e
+ −
+ −
∴ = = =
 
 
 
2
2
2 2 2
2 22
1
5 44
4
e
e
a b e
e
a ee
So ( )( )− + = ⇒ − − = ⇒ =4 2 2 2
5 4 0 1 4 0 2.e e e e e
56. If the distance between the plane − + =2Ax y z d and the plane containing the
lines
− −− − − −
= = − =
2 31 3 2 4
2 3 4 3 4 5
y yx z x z
and is 6, then| |d is
Ans : 6
Sol : Given lines have direction ratios (2, 3, 4) and (3, 4, 5). So normal of plane has
D.Rs (1, -2, 1). Plane through the point (1, 2, 3) is x - 2y + z = 0. It is parallel to
Ax - 2y + z = d
⇒ distance = = ⇒ =
+ +
| |
6 | | 6
1 4 1
d
d

PART - III , PHYSICS
PART - III : PHYSICS
ONE is correct
57. A thin uniform annular disc (se3c figure) of mass M has outer rdius 4R and inner
radius 3R. The work P on its axis to infinity is .
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
−
Ans : A
Sol : ( )
2 2
4/ 2
Gdm
dv
V r
=
+
dr
r
p
where 2
2
7
M
dm rdr
r
π
π
= ×
2
2
7
Mrdr
R
=
( )
( )
2
4
G dm
dv
r R+
−
∴ =
2 2 2
2
7 16
Mrdr
dv
R R r
−
=
+

∴ Total gravitational potential at pt. p (VP
)
dv= ∫
4
2 2 2
3
2
7 16
R
R
M rdr
R R r
= −
+
∫
2 2
2
2
16
7
MG
R r
R
−  = +
 
2
4 2 5
7
MG
R
−  = − 
∴ requirel work done
PV V∞ −
( )2
4 2 5
7
GM
R
= + −
58. A block of mass m is on an inclined plane of angle θ . The coefficient of friction
between the block and the plane is µ and tanθ µ> . 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 P 1
= ( )sin cosθ µ θ− to P2
=
( )sin cosθ µ θ+ , the frictional force f versus P graph will look like.
A) B)
C) D)

ANS : A
Sol : AS P is increased from mg ( )sin cosθ µ θ− to sinmg θ to mg ( )sin cosθ µ θ+ frac-
tional force will change from cosmgµ θ ( upwards parallel to inclined plane) to
zero to cosmgµ θ (downwards parallel toinclined plane). Under the approxima-
tion of linear variation of P the correct option is A
59. A thin fiexible wire of length L is connected to two adjacent fixed points and car-
ries a current I in the clockwise direction, as shown in the figure. When the system
is put in a uniform magnetic ficle of strength B going into the plane of the paper,
the wire takes the shape of a circle. The tension in the wire is .
A) IBL B)
IBL
π
C)
2
IBL
π
D)
4
IBL
π
Ans: C
Sol :
/ 2dθ/ 2dθ
T T
/ 2dθ
magF
/ 2dθ
Sin component of tension balance the magnetic field 2 sin / 2 magT d Fθ =
( )Td iR d Bθ θ=
T = i RB
2
iBL
π
=
60. An AC voltage source of variable angular frequency ω and fixed amplitude V0
is
connected to series with a capacitance C and an electric bulb of resistance R (in-
ductance zero). When ω is increased
A) the bulb glows diammer
B) the bulb glows brighter
C) total bulb glows brighter
D) total impedance of the circuit increases

Ans : B
Sol :
~
C R
0
0
2
2
1
V
i
R
w c2
=
+
( )
2
2 0
00
2
2 2
1bulb
V R
P i R
R
Cω
= =
+
∴ as w is increased ( )0bulbP increases it glows brighter
61. To verify Ohm’s law, a student is provided with a test resistor RT
, a high resis-
tance R1
, a small resistance R2
, two identical galvanometers G1
and G2
, and a vari-
able voltage source V. The correct circuit to carry out the experiment is
A) B)
C) D)
Ans : C
Sol : Theoretical

62. 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 resistance R100
, R60
and R40
, respectively,
the rlation between these resistance is.
A)
100 40 60
1 1 1
R R R
= + B) 100 40 60R R R= + C) 100 60 40R R R> > D)
100 60 40
1 1 1
R R R
> >
ANS : D
SOL :
2
V
p
R
=
2 2 2
100 60 40
100 ,60 ,40
V V V
R R R
⇒ = = =
100 60 40
1 1 1
RR R
∴ > >
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
Ans: D
Sol : At low pressure and high temperature, the interaction between molcules is
approaces to zero. Therefore real gas is behaves as ideal gas.
64. Consider a thin square sheet of side L and thickness t, made of a material of resis-
tivity ρ . The resistance between two oppositive faces, shown by the shaded areas
in the figure is .
A) directly proportional to L B) directly proportional to t
C) independent ot L D) independent of t
Ans : C
Sol :
.
.
L
R
t L t
ρ ρ
= = which is independent of L.

65. A student uses a simple pendulum of exactly 1m length ot determine g, the accel-
eration 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 fol-
lowing 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%
Ans : A,C
Sol : ( )40 1 20s T± =
1
0.05 sec
20
T∆ = =
2 /T l gπ=
2
2
4
g
T
π
=
A
2
100 100
g T
g T
∆ ∆
∴ × = ×
0.05
2 100
2
= × ×
= 5%
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 ms–1
.
Which of the following statements(s) (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
Ans : A,C
Sol : 1 21 0 2 1 5u v× + = − × + ....(1)
2
2 1
1
2
1 2
0
v
v u
u
+
= ⇒ + =
− ....(2)

2 1 2v u⇒ = −
( ) 1 21 2 5u v⇒ = − +
( )1 1 12 5 2 2 5 10u u u= − + − = − + −
14 12u⇒ − = −
u1
= 3 m/s
2 1 3 0v u= − =
Total momentum of the system = ( )1 1m u
1 3 3 /kgm s= × =
total KE of the system
2 2
1 1 2 2
1 1
2 2
m u m u= +
1
9 4.5
2
J= × =
KE of com= ( )( )
2
1 2
1
2
cmm m v+
1 1 2 2
1 2
1 3 0 1
6 2
cm
m u m u
V
m m
+ × +
= = =
+
KE of COM
2
1 1 6 3
6 0.75
2 2 8 4
J
 
= × = = = 
 
67. 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
. Choice the correct option(s) from the
following.
A) Interval enerhgies at A and B are the same
B) Work done by the gas in process AB is P0
V0 nA 4
C) Pressure at C is 0
4
T
D) Temperature at C is 0
4
T
Ans : A,B,C,D
Sol : (i) A to B temperature is same ∴ A is correct
(ii) A to B isothemal process
∴ W = nRT
2
1
ln
V
V

0
0
0
4
ln.
V
nRT
V
 
=  
 
0 0 ln 4P V=
∴ B is correct
(iii) B to C isobaric process
1 1
2 2
V T
V T
= 0 0 0
2
0 2
4
4
V T T
T
V T
= ⇒ =
∴ temperature at C is 0
4
T
∴ D is correct
(iv) C to A isochoric process
1 1
2 2
P T
P T
= C C
A A
P T
P T
= 0
0 04
CP T
P T
=
×
0
4
C
P
P =
∴ C is correct
68. A few electric field 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 susggest that
A) 1 2Q Q> B) 1 2Q Q<
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
Ans : A,D
Sol : Number of lines of force is directly proportional to the magnitude of the charge
so 1 2Q Q>
At null point where E = 0, there will be no lines of force of only the given two
charges are considered, there is a chance of getting E = 0 to the right of Q2
(Null
point will be closer to the charge having small magnitude)

69. A ray OP of monochromatic light is incident on the face AB of prism ABCD near
vertex B at an incident angle of 60º (sec figure). The refractive index of the mate-
rial of the prism is 3 , which of the following is (are) correct ?
A) The ray gets totally internally reflected at face CD
B) The ray comes out through face AB
C) The angle between the incident ray and the emergent ray is 90º
D) The angle between the incident ray and the emergent ray is 120º.
Ans : A,B,C
Sol : At surface AB
sin
sin
i
r
µ =
sin 60
3
sin r
=
P
A
B
C
D
Q’r
’θ
θ
r
’’r
’’i
60º
60º
60º
75º
45º
135ºN
N'
r = 30º
in quadrilateral PBCQ
45ºθ =

60º
δ
as 45ºθ = r’ = 45º
For medium critical angle is less than 45º.
∴ ray has total internal reflection at surface CD
' 60ºθ =
then r’’ = 30º
i’’ = 60º
angle of deviation = 180 – (60 + 30) = 90
Electrical resistance of certain materials, know as superconductors, changes
abruptly from a nonzero value to zero as their temperature is lowered below a
critical temperature TC
(0). An interseting property of supercondeuctors is that
their critical temperature becomes smaller than TC
(0)if they are placed in a mag-
netic 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.
70. In the graphs below, the resistance R of a superconductor is shown as a function
of its temperature T for two different magnetic fields B1
(solid line) and B2
(dashed
line). If B2
is large than B1
, which of the following graphs shows the correct varia-
tion of R with T in these fields ?

A) B)
C) D)
Ans : A
Sol : Under magnetic field, the critical temperatures decreases as given in the data
therefore the super conductors become conductors for higher magnetic field at
low temperature.
71. 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
Ans : B
Sol : When ‘B’ is increased, TC
(B) decreased where solid becomes super conductor. If B
= 7.5 tesla given that ‘TC
(0) decreased from 100 K to 75 K. If ‘B’ is less than 7.5 then
TC
(B) stightly increases, it should be in between 75 K and 100 K so if B = 5 Tesla, 75
K < TC
(B) < 100 K
When a particle of mass m moves on the x-axis in a potential of the form V(x) =
kx2
, it performs simple harmonic motion. The corresponding time period is pro-
portional to
m
k
, as 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 particle of mass m moving on the
x-axis. Its potential energy is V(x) = 4
xα ( )0α > for |x| near the origin and be-
comes a constant equal to V0
for 0x X≥ (sec figure).

72. 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 D) E > V0
Ans : C
73. For periodic motion of small amplitude A, the time period T of this particle is
proportional to
A)
m
A
α
B)
1 m
A α
C) A
m
α
D)
1
A m
α
Ans : B
74. The acceleration of this particle for 0x x> is
A) proportional to V0
B) proportional to
0
0
V
mX
C) proportional to
0
0
V
mX D) zero
Ans : D
Sol : For x > x0
the force is zero, therefore acceleration is zero.
For 0 0x x x> > − the force is 3
4F xα= − which act as restoring force, therefore the
particle perform osscilation for V0
> E > 0 .
for small amplititude, curvature variation for curve x4
and x2
is identical. There-
fore
m
T
k
∝ where 2 4
kA Aα=
1 m
T
A α
∝

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-sectinla area is 7 2
4.9 10 m−
× . 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 9 2
10n Nm−
× , the value of n is
Ans : 4
Sol :
2
2
1
2 , ,
m m FL KL
T k m Y
k k Ae A
π ω
ω
= ⇒ = = = =
2YA
m
L
ω=
⇒ n = 4
76. A binary star consists of two stars A (mass 2.2MS
) and B (mass 11 MS
), 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 ration of the total angular momentum of
the binary star to the angular momentum of star B about the centre of mass is
Ans : 6
Sol :
1 1 2
2 1
6
L M M
L M
+
= =
77. Gravitational acceleration on the surface of a planet is
6
11
g , where g is the gravi-
tational acceleration on the surface of the earth. The average mass density of the
planet is
2
3
times that of the earth. If the escape speed on the surface of the earth is
taken to be 11 kms–1
the escape speed on the surface of the planet in kms–1
will be
Ans : 3
Sol : 2
4
3
GM
g G R
R
π ρ= =
22 8
3
e
GM
V G R
R
π
ρ= =
2
e
g
V α
ρ
2
1
3
V
V
=

78. A piece of ice (heat capacity = 2100 J kg–1
ºC–1
and latent heat = 5
3.36 10× J kg–1
) of
mass m grams is at –5ºC at atmospheric pressure. It is given 420 J of the 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 : 8
Sol : +icem
5º C−
1Q m C T= ∆
icem
0º C ( )1m g ice−
2 1Q g L= ×
1g water
Given 1 2420 Q Q= +
MC T igL∆ +
3 5
2100 5 10 3.36 10m −
= × + × ×
3420 336
8 10 8
10500
m kg gm−−
= = × =
79. A stationary source is emitting sound at a fixed frequency f0
, which is reflectdd 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
.
Ans : 7
Sol : Let v is the speed of car approaching towards a stationary source emitty wave of
speed υ
frequency of the sound appeared to stationary observer is
0
v
f f
v
υ
υ
+ 
=  
− 
change of frequency is
( )
( )
21
0
1 2
vv
f f
v v
υυ
υ υ
 + +
∆ = −  
− −  
since 1 2v and v υ<<
21 2
0
2 12. 10
f v v
f υ
−∆ − 
= = × 
 
1 2 7.128v v−
nearest integer is 7

80. The focal length of 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 25 50.m tom The ration
25
50
m
m is
Ans : 6
Sol : case-I u = –25 cm f = + 20 cm
v1
= 100 m
1 4m⇒ =
case - II u = – 50 cm f = +20 cm
2
100
3
V cm=
2 2/3m =
1
2
6
m
m
∴ =
81. An particleα − and a proton are acceleration from rest by a potential difference of
100V. After this, their de Broglie wavelength are pandαλ λ respectively. The ratio
,
p
α
λ
λ
to the nearest integer, is
Ans : 3
Sol : 2
h
mqV
λ =
p
p p
m q
m q
α α
α
λ
λ
= ×
4 2= ×
2 2 2 1.414= = ×
= 2.828
∴ nearest answer is 3
82. When two identical batteries of interval resistance 1Ω each are connected in se-
ries across a resistor R, the rate of heat produece 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
Ω is
Ans : 4

Sol : in case -I
R
ε
1i
1Ω 1Ωε
1
2
2
i
R
ε
=
+
2
1
2
2
J R
R
ε 
∴ =  
+ 
case-II
ε 1Ω
1Ω
R
ε
2
2
2 1
I
R
ε
=
+
2
2
2
2 1
J R
R
ε 
∴ =  
+ 
given that J1
= 22.25 J× ∴ R = 4 Ω
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 B is at 1500 nm. Considering them to be black bodies, what
will be the ratio of the rate of total energy radiated by A to that of B ?
Ans : 9
Sol : 4
E e ATσ=
where A is the area of body and e = 1 here maxT cantλ = ∴
( )
( )
2
1 21
4
2 1 2
/
9
/
R RE
E λ λ
= =
84. When two progressive waves y1
= 4 sin (2x – 6t) and y2
= 3 2 6
2
x t
π 
− − 
 
are super-
imposed, the amplitude of the resultant wave is
Ans : 5
Sol : 2 2
1 2 1 22 cosRA A A A A φ= + + where 1 24, 3, / 2,A A φ π= = = 2 2
3 4 5RA = + =

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