iit jee 2010 solution paper 1 english
DESCRIPTION
IIT JEE 2010 Solution Paper 1 English by ResonanceTRANSCRIPT
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Date : 11-04-2010 Duration : 3 Hours Max. Marks : 252
INSTRUCTIONSA. General :1. This Question Paper contains 32 pages having 84 questions.
2. The question paper CODE is printed on the right hand top corner of this sheet and also on theback page of this booklet.
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8. The ORS has CODE printed on its lower and upper Parts.
9. Make sure the CODE on the ORS is the same as that on this booklet. If the Codes do notmatch, ask for a change of the Booklet.
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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 partconsists of four Sections.
13. For each question is Section�I, you will be awarded 3 marks if you have darkened only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. In allother cases, minus one (�1) will be awarded.
14. For each question is Section�II, you will be awarded 3 marks if you have darken only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. Partialmarks will be answered for partially correct answers. No negative marks will be awarded inthis Section.
15. For each question is Section�III, you will be awarded 3 marks if you have darken only thebubble corresponding to the correct answer and zero mark if no bubbles are darkened. In allother cases, minus one (�1) will be awarded.
16. For each question is Section�IV, you will be awarded 3 marks if you darken the bubblecorresponding to the correct answer and zero mark if no bubbles is darkened. No negativemarks will be awarded for in this Section.
PAPER - 1
id32178296 pdfMachine by Broadgun Software - a great PDF writer! - a great PDF creator! - http://www.pdfmachine.com http://www.broadgun.com
RESONANCE Page # 2
Useful Data :
Atomic Numbers : Be 4; N 7; O 8; Al 13 ; Si 14; Cr 24 ; Fe 26; Fe 26; Zn 30; Br 35.
1 amu = 1.66 × 10�27 kg R = 0.082 L-atm K�1 mol�1
h = 6.626 × 10�34 J s NA = 6.022 × 1023
me = 9.1 × 10�31 kg e = 1.6 × 10�19 C
c = 3.0 × 108 m s�1 F = 96500 C mol�1
RH = 2.18 × 10�18 J 40 = 1.11 × 10�10 J�1 C2 m�1
RESONANCE Page # 3
PART- I
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 species which by definition has ZERO standard molar enthalpy of formation at 298 K is :
(A) Br2(g) (B) Cl2(g)
(C) H2O(g) (D) CH4(g)
Ans. (B)
Sol. Hf° (Cl2,g) = 0, As Hf° of elements in their standard state is taken to be zero.
2. 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. EC � C 100 KCal/mole.
3. The correct structure of ethylenediaminetetraacetic acid (EDTA) is :
(A)
(B)
(C)
(D)
Ans. (C)
CHEMISTRY
RESONANCE Page # 4
Sol. Structure of EDTA is :
4. 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
Ans. (B)
Sol. The ionisation isomer for the given compound will be obtained by exchanging ligand with counter ion as :
[Co(H2O)4Cl2](NO2).
5. 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
Ans. (D)
Sol. CH3�CH2�C C�H 2NaNH CH3�CH2�C C Br�2CH�2CH�2CH�3CH
CH3�CH2�C C�CH2�CH2�CH2�CH3
3�octyne
6. The correct statement about the following disaccharide is :
(A) Ring (a) is pyranose with -glycosidic link (B) Ring (a) is furanose with -glycosidic link
(C) Ring (b) is furanose with -glycosidic link (D) Ring (b) is pyranose with -glycosidic link
Ans. (A)
Sol. Ring (a) is pyranose with -glycosidic linkage and ring (b) is furanose with -glycosidic linkage.
CHEMISTRY
RESONANCE Page # 5
7. In the reaction OCH3 HBr the products are :
(A) Br OCH3 and H2 (B) Br and CH3Br
(C) Br and CH3OH (D) OH and CH3Br
Ans. (D)
Sol.
8. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows
Arrhenius equation is :
(A) (B)
(C) (D)
Ans. (A)
Sol. k = Ae�Ea/RT
So, variation will be
CHEMISTRY
RESONANCE Page # 6
SECTION - II
Multiple Correct Correct Type
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.
9. 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
Ans. (C,D)
Sol. (C) HNO3 + CH3COONa mixture can act as buffer solution if volume of HNO3 solution taken is lesser than
volume of CH3COONa solution because of following reaction :
CH3COONa + HNO3 CH3COOH + NaNO3
(D) CH3COOH + CH3COONa - mixture will act as buffer.
10. Among the following, the intensive property is (properties are) :
(A) molar conductivity (B) electromotive force
(C) resistance (D) heat capacity
Ans. (A,B)
Sol. Molar conductivity and electromotive force (emf) are intensive properties as these are size independent.
11. 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
Ans. (B,C)
Sol. (B) HCO3� + OH� H2O + CO3
2� and CaCO3 will get precipitated.
(C) CO32� + Ca2+ CaCO3, CaCO3 (white) precipitated.
CHEMISTRY
RESONANCE Page # 7
12. In the reaction the intermediate (s) is (are) :
(A) (B) (C) (D)
Ans. (A,C)
Sol. NaOH
It is ortho-para directing.
13. In the Newman projection for 2, 2-dimethylbutane
X and Y can respectively be :
(A) H and H (B) H and C2H5
(C) C2H5 and H (D) CH3 and CH3
Ans. (B,D)
Sol.
3
323
3
CH|
CH�CH�C�CH|CH
,
3
523
3
CH|
HC�C�CH|CH
CHEMISTRY
RESONANCE Page # 8
SECTION - III
Comprehension Type
This section contains 2 Paragraphs. Based upon the first paragraph 3 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 Question Nos. 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 Fe2O2 (D) Cu2O and Fe2O2
Ans. (A)
Sol. 2 CuFeS2 + 4O2 Cu2S + 2FeO + 3SO2
15. Iron is removed from chalcopyrite as :
(A) FeO (B) FeS
(C) Fe2O3 (D) FeSiO3
Ans. (D)
Sol. Iron is removed in the form of slag of FeSiO3
FeO + SiO2 FeSiO3
16. In self-reduction, the reduction species is :
(A) S (B) O2�
(C) S2� (D) SO2
Ans. (C)
Sol. S2� acts as reducing species in self reduction reaction
2Cu2O + Cu2S 6Cu + SO2
CHEMISTRY
RESONANCE Page # 9
Paragraph for Question Nos. 17 to 18
The concentration of potassium ions inside a biological cell is at least 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 | Ecell | = 70 mV.
17. For the above cell :
(A) Ecell < 0 ; G > 0 (B) Ecell > 0 ; G < 0
(C) Ecell < 0 ; Gº > 0 (D) Ecell > 0 ; Gº < 0
Ans. (B)
Sol. M (s) | M+ (aq, 0.05 M) || M+ (aq, 1 M) | M(s)
Anode : M (s) M+ (aq) + e�
Cathode : M+ (aq) + e� M (s)
_____________________________________
M+ (aq) |c M+ (aq) |a
Ecell = E°cell � 1
0591.0 log
c
a
|)aq(M
|)aq(M
= 0 � 1
0591.0 log
1
05.0
= + ve = 70 mV and hence G = � nFEcell = � ve.
18. 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
Ans. (C)
Sol. Ecell = 10591.0
log
10025.0
= 1
0591.0 log
2005.0
= 70 mV + 1
0591.0 log 20 = 140 mV..
CHEMISTRY
RESONANCE Page # 10
SECTION - IV
(Integer Type)
This section contains TEN questions. The asnwer 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. The concentration of R in the reaction R P was measured as a function of time and the following data is
obtained :
18.012.005.00.0.)(mint
10.040.075.00.1)molar](R[
The order of the reaction is :
Ans. 0
Sol. K = t
CC0 = 05.0
75.01 =
05.025.0
= 5
K = 07.0
40.075.0 =
07.035.0
= 5
So, reaction must be of zero order.
20. The number of neutrons emitted when U23592 undergoes controlled nuclear fission to Xe142
54 and Sr9038 is :
Ans. 3
Sol.
21. The total number of basic groups in the following form of lysine is :
Ans. 2
Sol. � NH2 and COO are two basic groups.
CHEMISTRY
RESONANCE Page # 11
22. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is :
Ans. 5
Sol. , , , ,
23. In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is:
O2H,Zn.2
3O.1 Y
heat.2
)aq(NaOH.1
Ans. 1
Sol. O2H/Zn.2
3O.1
NaOH
24. Amongst the following, the total number of compounds soluble in aqueous NaOH is :
Ans. 5
CHEMISTRY
RESONANCE Page # 12
Sol. , , , &
are soluble in aqueous NaOH.
25. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue
is :
KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2FeCl3 K2CO3 NH4NO3 LiCN
Ans. 3
Sol. Basic solutions will convert red litmus blue.
, their aqueous solution will be basic due to anionic hydrolysis.
26. Based on VSEPR theory, the number of 90 degree F�Br�F angles is BrF5 is :
Ans. 8
Sol.
27. The value of n in the molecular formula BenAl2Si6O18 is :
Ans. 3
Sol. BenAl2Si6O18. The value of n = 3 by charge balancing.
(Be2+)3 (Al3+)2 (Si6O18)12� [2 × n + 2 (+3) + (�12) = 0 n = 3].
28. 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 = 3
0.2525.252.25 =
345.75
= 25.15 = 25.2 mL
number of significant figures will be 3.
CHEMISTRY
RESONANCE Page # 13
PART-II
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.
29. Let p and q be real numbers such that p 0, p3 q and p3 � q. If and are nonzero complex numbers
satisfying + = � p and 3 + 3 = q, then a quadratic equation having
and
as its roots is
(A) (p3 + q) x2 � (p3 + 2q)x + (p3 + q) = 0 (B) (p3 + q) x2 � (p3 � 2q)x + (p3 + q) = 0
(C) (p3 � q) x2 � (5p3 � 2q)x + (p3 � q) = 0 (D) (p3 � q) x2 � (5p3 + 2q)x + (p3 � q) = 0
Ans. (B)
Sol. Product = 1
Sum =
22
=
2)( 2
Since 3 + 3 = q � p (2 + 2 � ) = q
(( + )2 � 3 ) = � pq
p2 + pq
= 3
Hence sum = )qp(
p3p
qp32
p
3
32
= qp
q2p3
3
so the equation is x2 +
qp
q2p3
3
x + 1 = 0
(p3 + q) x2 � (p3 � 2q)x + (p3 + q) = 0
30. Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x) = 2xe +
2x�e ,
g(x) = 2xxe +
2x�e and h(x) = 2x2ex +
2x�e . 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 (B) a = c and a b (C) a b and c d (D) a = b = c
Ans. (D)
MATHEMATICS
RESONANCE Page # 14
Sol. Clearly f(x) = 2xe +
2x�e
f(x) = 2x (2xe �
2x�e ) 0 increasing fmax
= f(1) = e + e1
g(x) = x2xe +
2x�e g(x) = 2xe + 2x2
2xe � 2x2x�e > 0 increasing g
max = g(1) = e +
e1
h(x) = x2 2xe +
2x�e h(x) = 2x2xe + 2x3
2xe � 2x2x�e = 2x
2x2x22x eexe > 0
hmax
= h(1) = e + e1
,
so a = b = c
31. If the angle A, B and C of a triangle are in 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 ca
sin 2C + ac
sin 2A is
(A) 21
(B) 23
(C) 1 (D) 3
Ans. (D)
Sol.ca
sin 2C + ac
sin 2A = R22
(a cos C + c cos A) = Rb
= 2 sin B = 2 sin 60º = 3
32. Equation of the plane containing the straight line 2x
= 3y
= 4z
and perpendicular to the plane containing the
straight line 3x
= 4y
= 2z
and 4x
= 2y
= 3z
is
(A) x + 2y � 2z = 0 (B) 3x + 2y � 2z = 0 (C) x � 2y + z = 0 (D) 5x + 2y � 4z = 0
Ans. (C)
Sol. Direction ratio of normal to plane containing the straight line
324
243
k�j�i�
= 8 i� � j� � 10k�
Required plane1018
432
0z0y0x
= 0 � 26x + 52y � 26z = 0 x � 2y + z = 0
MATHEMATICS
RESONANCE Page # 15
33. Let be a complex cube root of unity with 1. A fair die is thrown three times. If r1, r
2 and r
3 are the numbers
obtained on the die, then the probability that 3r2r1r = 0 is
(A) 181
(B) 91
(C) 92
(D) 361
Ans. (C)
Sol. 3r2r1r = 0 ; r1, r
2, r
3 are to be selected form {1, 2, 3, 4, 5, 6}
As we know that 1 + + 2 = 0
form r1, r
2, r
3 , one has remainder 1, other has remainder 2 and third has remainder 0 when divided by 3.
we have to select r1, r
2, r
3 form (1, 4) or (2, 5) or (3, 6) which can be done in 2C
1 × 2C
1 × 2C
1 ways
value of r1, r
2, r
3 can be interchanged in 3! ways.
required probability = 666
!3)CCC( 12
12
12
=
92
34. Let P, Q, R and S be the points on the plane with position vectors � 2 i� � j� , 4 i� , 3 i� + 3 j� and � 3 i� + 2 j�
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
Ans. (A)
Sol. PQ = 136 = 37 = RS, PQ PS
PS = 91 = 10 = QR
slope of PQ = 61
, slope of PS = � 3
PQ is not to PS
So it is parallelogram, which is neither a rhombus nor a rectangle
35. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A
z
y
x
=
0
0
1
has
exactly two distinct solutions, is
(A) 0 (B) 29 � 1 (C) 168 (D) 2
Ans. (A)
MATHEMATICS
RESONANCE Page # 16
Sol. a1x + b
1y + c
1z = 1
a2x + b
2y + c
2z = 0
a3x + b
3y + c
3z = 0
No three planes can meet at two distinct points. So number of matrices is 0
36. The value of 0x
lim
x
043 4t
)t1(nt
x
1 dt is
(A) 0 (B) 121
(C) 241
(D) 641
Ans. (B)
Sol.0x
lim
24 x3)4x(
)x1(nx
=
0xlim
41
× 31
= 121
SECTION - II
Multiple Correct ChoiceType
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.
37. The value(s) of
1
02
44
x1
)x1(x dx is (are)
(A) �722
(B) 105
2(C) 0 (D)
23
�1571
Ans. (A)
Sol.
1
02
44
x1
)x1(x =
1
02
224
x1
]x2)x1[(x =
1
02
22224
x1
]x4)x1(x4)x1[(x
= dxx1
x4x4)x1(x
1
02
224
=
dx
x1
x4x4xx
2
6546
Now on polynomial division of x6 by 1 + x2 , we obtain
=
dx
x1
1)1xx(4x4xx
224546
=
dx
x1
44x4x5x4x
22456
MATHEMATICS
RESONANCE Page # 17
= 10x
1
0
3567
xtan4x43x4
5x.5
6x4
7x
=
4
3
41
6
4
7
1 � 4
4 =
5
6
12
7
1 �
=
3
71
� = 722
�
38. Let f be a real-valued function defined on the interval (0, ) by f(x) = n x +
x
0
tsin1 dt. Then which of the
following statement(s) is (are) true?
(A) f(x) exists for all x (0, )
(B) f(x) exists for all x (0, ) and f is continuous on (0, ), but not differentiable on (0, )
(C) there exists > 1 such that |f(x)| < |f(x)| for all x (, )
(D) there exists > 1 such that |f(x)| + |f(x)| for all x (0, )
Ans. (B, C)
Sol. f(x) = n x +
2
0
dttsin1
f(x) = x1
+ xsin1
f(x) = � 2x
1 +
xsin12
xcos
(A) f is not defined for x = � 2
+ 2n , n
so (A) is wrong
(B) f(x) always exist for x > 0
(C) | f| < | f |
Since f > 0 and f > 0
f < f
x1
+ xsin1 < n x +
x
0
dxxsin1
LHS is bounded RHS is Increasing with range
So there exist some beyond which RHS is greater than LHS
(D) | f | + | f| b is wrong as f is M & its range is not bound while is finite
MATHEMATICS
RESONANCE Page # 18
39. Let A and B be two distinct points on the parabola y2 = 4x. 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) � r1
(B) r1
(C) r2
(D) � r2
Ans. (C, D)
Sol. A(t1
2, 2t1), B (t2
2 , 2t
2)
centre of circle
21
22
21 tt,
2
tt
| t1 +
t2 | = r , slope of AB =
21 tt2
= ± r2
40. Let ABC be a rectangle such that ACB = 6
and let a, b and c denote the lengths of the sides opposite to
A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 � 1 and c = 2x + 1 is (are)
(A) � 32 (B) 1 + 3 (C) 2 + 3 (D) 4 3
Ans. (A, B)
Sol. cos 6
= )1x)(1xx(2
)1x2()1xx()1x(22
22222
23
= )1x)(1xx(2
)xx)(2x3x()1x(22
2222
23
= )1x)(1xx(2
)1x(x)2x)(1x()1x(22
22
3 = 1xx
)2x(x1x2
2
3 (x2 + x + 1) = 2x2 + 2x � 1
( 3 � 2) x2 + ( 3 � 2) x + ( 3 + 1) = 0
on solving
x2 + x � )533( = 0
x = 3 + 1, � )32(
MATHEMATICS
RESONANCE Page # 19
41. Let z1 and z
2 be two distinct complex numbers and let z = (1 � t) z
1 + tz
2 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| = |z
1 � z
2| (B) Arg (z � z
1) = Arg (z � z
2)
(C) 1212
11
zzzz
zzzz
= 0 (D) Arg (z � z
1) = Arg (z
2 � z
1)
Ans. (A, C, D)
Sol. (A) |z � z1| + |z � z
2| = |z
1 � z
2|
AB + BC = AC
(B) Arg (z � z1) � Arg (z � z
2) =
(C)1zz
1zz
1zz
22
11 = 0
1zz
0zzzz
0zzzz
22
2121
11
= 0
2121
11
zzzz
zzzz
= 0
(D) Arg (z � z1) = Arg (z
2 � z
1)
SECTION - III
Comprehension Type
This section contains 2 Paragraphs. Based upon the first paragraph 3 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.
MATHEMATICS
RESONANCE Page # 20
Paragraph for Question Nos. 42 to 44
Let p be an odd prime number and Tp be the following set of 2 × 2 matrices :
Tp =
}1�p,.....,2,1,0{c,b,a:
a
b
c
aA
42. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A)
divisible by p is
(A) (p � 1)2 (B) 2 (p � 1) (C) (p � 1)2 + 1 (D) 2p � 1
Ans. (D)
Sol. A =
ac
bawhere a, b, c {0,1,2,....p � 1}
Case- I A is symmetric matrix b = c
det(A) = a2 � b2 is divisible by p
(a � b) (a + b) is divisible by p
(a) a � b is divisible by p if a = b, then 'p' cases are possible
(b) a + b is divisible by p if a + b = p, then 2
)1�p( × 2 = (p � 1) cases are possible
Case - II
A is skew symmetric matrix
if a = 0, b + c = 0, then det (A) = b2
b2 can never be divisible by p. So No case is possible
Total number of A is possible = 2p � 1
43. The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is
[Note : The trace of matrix is the sum of its diagonal entries.]
(A) (p � 1)(p2 � p + 1) (B) p3 � (p � 1)2
(C) (p � 1)2 (D) (p � 1)(p2 � 2)
Ans. (C)
Sol. a2 � bc p
a can be chosen in p � 1 ways (a 0)
Let a be 4 & p = 5
so a2 = 16 and hence bc should be chosen such that a2 � bc p
Now b can be chosen in p � 1 ways and c in only 1 (one be is chosen)
MATHEMATICS
RESONANCE Page # 21
Explanation : If b = 1 c = 1
b = 2 c = 3
b = 3 c = 2
b = 4 c = 4
Hence a can be chosen in p � 1 ways
and then b can be chosen in p � 1 ways
c can be chosen in 1 ways
so (p � 1)2
44. The number of A in Tp such that det (A) is not divisible by p is
(A) 2p2 (B) p3 � 5p (C) p3 � 3p (D) p3 � p2
Ans. (D)
Sol. As = Total cases � (a 0 and |A| is divisible by p) � (a = 0 and |A| is divisible by p)
= p3 � (p �1)2 � (2p � 1) = p3 � p2
� bc p since b & c both we coprime to p
one of them must be zero.
If b = 0 c can be chosen in {0,1,....p � 1}
If c = 0 b can be chosen in {0,1,....p � 1}
Paragraph for Question Nos. 45 to 46
The circle x2 + y2 � 8x = 0 and hyperbola 9x2�
4y2
= 1 intersect at the points A and B.
45. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
(A) 2x � 5 y � 20 = 0 (B) 2x � 5 y + 4 = 0
(C) 3x � 4y + 8 = 0 (D) 4x � 3y + 4 = 0
Ans. (B)
Sol. Let equation of tangent to ellipse
1y2
tan�x
3sec
2sec x � 3 tan y = 6
It is also tangent to circle x2 + y2 � 8x = 0
22 tan9sec4
6�sec8= 4
(8sec � 6)2 = 16 (13sec2 � 9)
12sec2 + 8sec � 15 = 0
MATHEMATICS
RESONANCE Page # 22
sec = 65
and 23
� but sec 65
sec = 23
� tan = 25
slope is positive
Equation of tangent = 2x � 5 y + 4 = 0
46. Equation of the circle with AB as its diameter is
(A) x2 + y2 � 12x + 24 = 0 (B) x2 + y2 + 12x + 24 = 0
(C) x2 + y2 + 24x � 12 = 0 (D) x2 + y2 � 24x � 12 = 0
Ans. (D)
Sol. x2 + y2 � 8x = 0
9x2
� 4y2
= 1 4x2 � 9y2 = 36
4x2 � 9(8x � x2) = 36
13x2 � 72x � 36 = 0
13x2 � 78x + 6x � 36 = 0
(13x + 6) (x � 6) = 0
x = � 136
and x = 6
But x > 0 x = 6
A(6, 2 ) and B (6, � 2 )
Equation of circle with AB as a diameter x2 + y2 � 12x + 24 = 0
SECTION - IV
(Integer Type)
This section contains TEN questions. The asnwer 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.
MATHEMATICS
RESONANCE Page # 23
47. The number of all possible values of , where 0 < < , for which the system of equations
(y + z) cos 3 = (xyz) sin 3
x sin 3 = y3cos2
+ z
3sin2
(xyz) sin 3 = (y + 2z) cos 3 + y sin 3
have a solution (x0, y0, z0) with y0 z0 0, is
(y + z) cos 3 = (xyz) sin 3
x sin 3 = y3cos2
+ z
3sin2
(xyz) sin 3 = (y + 2z) cos 3 + y sin 3
Ans. 3
Sol. Let xyz = t
t sin3 � y cos 3 � z cos3 = 0 ..........(1)
t sin3 � 2y sin 3 � 2z cos3 = 0 ..........(2)
t sin3 � y (cos 3 + sin3 ) � 2z cos 3 = 0 ..........(3)
y0. z0 0 hence homogeneous equation has non-trivial solution
D =
3cos2�)3sin3(cos�3sin
3cos2�3cos2�3sin
3cos�3cos�3sin
= 0
sin3cos3(sin 3 � cos3) = 0
sin3= 0 or cos3= 0 or tan3 = 1
Case - I sin3= 0
From equation (2)
z = 0 not possible
Case - II cos3 = 0, sin3 0
t. sin3 = 0 t = 0 x = 0
From equation (2)
y = 0 not possible
Case- III tan3= 1
3 = n+ 4
, n I
x. y. z sin3= 0 = 3
n+
12
, n I
x = 0 , sin3 0 = 12
, 125
, 129
Hence 3 solutions
MATHEMATICS
RESONANCE Page # 24
48. 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
Ans. 9
Sol. Y � y = m (X � x)
y-intercept (x = 0)
y = y � mx
Given that y � mx = x3 x dxdy
� y = � x3
dxdy
� xy
= � x2
Intergrating factor dxx1
�
e = x1
solution y . x1
= dx)x(.x1 2
f(x) = y = � 2x3
+ cx
Given f(1) = 1 c = 23
f(x) = � 2x3
+ 2x3
f(�3) = 9
49. The number of values of in the interval
2,
2� such that
5n
for n = 0, ±1, ± 2 and
tan = cot 5 as well as sin 2 = cos 4is
Ans. 3
Sol. tan = cot 5
cossin
=
sin5cos
cos 6= 0
6 = (2n + 1) 2
= (2n + 1) 12
; n
MATHEMATICS
RESONANCE Page # 25
= 125
�
, �4
, 12
�
, 12
, 4
, 125
.........(1)
sin2 = cos4
sin2 = 1 � 2 sin2 2
2sin22 + sin2� 1 = 0
sin2 = � 1, 21
2= (4m � 1)2
, p + (�1)p 6
q = (4m � 1)4
, 2
p+ (�1)p
12
; m, p I
= �4
, 12
, 125
...........(2)
From (1) & (2)
125
,12
,4
�
Number of solution
50. The maximum value of the expression
22 cos5cossin3sin
1 is
Ans. 2
Sol. f() =
22 cos5cossin3sin
1 =
2)2cos1(5
2sin23
22cos1
1
= 2cos42sin36
2
f()max
= 56
2
= 2
51. If a and b
are vectors in space given by a
= 5
j�2�i� and b
=
14
k�3j�i�2 , then the value of
ba2
. b2�aba
is
Ans. 5
MATHEMATICS
RESONANCE Page # 26
Sol. a =
5
j�2�i� , b
= 14
k�3j�i�2
|a
| = 1 , |b
| = 1
b.a
= 0
)ba2(
. )b2�a()ba(
= )ba2(
. a)b2�a.(b(�b)b2�a.(a
= )ba2(
. a)b.b2�a.b(�b)b.a2�a.a(
= )ba2(
. a)20(�b)0�1(
= )ba2(
. a2b
= )a.b(2b.b)a.a(4)b.a(2
= 0 + 4 + 1 + 0 = 5
52. The line 2x + y = 1 is tangent to the hyperbola 2
2
a
x � 2
2
b
y= 1 . If this line passes through the point
of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is
Ans. 2
Sol.
x = a/e
y = �2x + 1 e2 = 1 + 2
2
a
b
0 = � ea2
� + 1 = 1 + 2
2
a
)1�a4(
ea
= 21
e2 = 1 + 4 � 2a
1
e = 2a e2 = 5 � 2e
4
MATHEMATICS
RESONANCE Page # 27
c2 = a2 m2 � b2 e4 � 5e2 + 4 = 0
1 = 4a2 � b2 (e2 � 1)(e2 � 4) = 0
1 + b2 � 4a2 = 0 e2 � 1 0 e = 2
53. If the distance between the plane Ax � 2y + z = d and the plane containing the lines
21�x
= 3
2�y =
43�z
and 3
2�x=
43�y
= 5
4�z is 6 , then |d| is
Ans. 6
Sol. Equation of plane is 543
432
3�z2�y1�x
= 0
x � 2y + z = 0 ..........(1)
Ax � 2y + z = d ..........(2)
Compare 1A
= 2�2�
= 11 A = 1
Distance between planes is 6411
d
|d| = 6
54. For any real number, 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
f(x) =
evenis]x[if
,oddis]x[if
x�]x[1
]x[�x
Then the value of
10
10�
2)x(f
10 cos x dx is
Ans. 4
MATHEMATICS
RESONANCE Page # 28
Sol. f(x) =
1n2xn2,}x{1
n2x1n2,}x{
Clearly f(x) is a periodic function with period = 2
Hence f(x) . cos x is also periodic with period = 2
10
2
10
10
dx)xcos()x(f = 2
2
0
dx)xcos()x(f = 2
1
0
dx)xcos(})x{})x{1((
= 22
1
0
dx)xcosx( = � 22 1
02
xcosxsinx
= � 22
2
2 = 4
55. Let be the complex number cos 32
+ i sin 32
. Then the number of distinct complex numbers
z satisfying
z1
1z
1z
2
2
2
= 0 is equal to
Ans. 1
Sol. = cos 32
+ i sin 32
R1 R1 + R2 + R3
z1z
1zz
z2
2
= 0 z
z11
1z1
12
2
= 0
z = 0
2
22
2
�z�10
�1�z0
1
= 0
(z + 2 � )(z + � 2) � (1 � )(1 � 2 ) = 0
z2 = 0
only one solution
MATHEMATICS
RESONANCE Page # 29
56. Let Sk, k = 1, 2,...., 100, denote the sum of the infinite geometric series whose first term is !k1�k
and the common ratio is k1
. Then the value of !100
1002
+
100
1kk
2 S)1k3�k( is
Ans. 3
Sol.
100
2kk
2 S)1k3�k(
for k = 2 |k2 � 3k + 1) Sk| = 1
100
3k!)1�k(11�k
�!)2�k(
1�k
100
3k!)1�k(
1�
!)2�k(1
�!)2�k(
1!)3�k(
1
100
3k!)1�k(
1�
!)3�k(1
S =
!961
�!94
1....
!61
�!4
1!5
1�
!31
!41
�!2
1!3
1�
!11
!21
�11
+
!991
�!97
1!98
1�
!961
!971
�!95
1 = 2 �
!991
�!98
1
E = !100
1002
+ 3 � !98
1 �
!98.991
= !100
1002
+ 3 � !99
100 =
!99.1001002
+ 3 � !99
100= 3
MATHEMATICS
RESONANCE Page # 30
PART-III
SECTION - I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
57. A block of mass m is on 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 :
(A) (B)
(C) (D)
Ans. (A)
Sol.
P1 = mgsin � mgcos
P2 = mgsin + mgcos
Initially block has tendency to slide down and as tan > , maximum friction mgcos will act in positive
direction. When magnitude P is increased from P1 to P2, friction reverse its direction from positive to negative
and becomes maximum i.e.mgcos in opposite direction.
PHYSICS
RESONANCE Page # 31
58. 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 :
4R
4R3R
P
(A) 524R7
GM2 (B) 524
R7GM2
(C) R4
GM(D) 12
R5GM2
Ans. (A)
Sol. Wext = U � UP
Wext = 0 �
1.
xGdm
��
Wext = G 222rR16
rdr2
R7
M
= 2R7
GM2
22 rR16
rdr
= 2R7
GM2 z
zdz = 2R7
GM2 [Z]
r3R
4Rdr
P
x
Wext = 2R7
GM2 R4
R3
22 rR16
Wext = 2R7
GM2 R5�R24
Wext = 2R7
GM2 5�24 .
59. Consider a thin square sheet of side L and thickness t, made of a material of resitivity . The resistance
between two opposite faces, shown by the shaded areas in the figure is :
Lt.
(A) directly proportional to L (B) directly proportional to t
(C) independent of L (D) independent of t
Ans. (C)
PHYSICS
RESONANCE Page # 32
Sol. R = Al
R = tLL
= t
Independent of L.
60. 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 inter molecular forces become ineffective. So a real gas behaves like
an ideal gas.
61. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with
increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances
R100, R60 and R40, respectively, the relation between these resistance is :
(A) 6040100 R1
R1
R1
(B) R100 = R40 + R60 (C) R100 > R60 > R40 (D) 4060100 R1
R1
R1
Ans. (D)
Sol. 100 = 100
2
'RV
100'R1
= 2V
100
where R�100 is resistance at any temperature corresponds to 100 W
60 = 60
2
'RV
60'R1
= 2V
60
40 = 40
2
'RV
40'R1
= 2V
40
From above equations we can say
100'R1
> 60'R1
> 40'R1
.
So, most appropriate answer is option (D).
PHYSICS
RESONANCE Page # 33
62. 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 :
(A)
G1
G2
R1RT
R2
V
(B)
G1
G2
R2RT
R1
V
(C)
G1
G2
R2
RT
R1
V
(D)
G1
G2
R2
RT
R1
V
Ans. (C)
Sol. To verify Ohm's law one galvaometer is used as ammeter and other galvanometer is used as voltameter.
Voltameter should have high resistance and ammeter should have low resistance as voltameter is used in
parallel and ammeter in series that is in option (C).
63. An AC voltage source of variable angular frequency and fixed amplitude V 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 impedence of the circuit is unchanged (D) total impedence of the circuit increases
Ans. (B)
Sol. irms =
222
rms
c
1R
V
when increases, irms increases so the bulb glows brighter
PHYSICS
RESONANCE Page # 34
64. A thin flexible wire of length L is connected to two adjacent fixed points carries a current in the clockwise
direction, as shown in the figure. When system is put in a uniform magnetic field of strength B going into the
plane of paper, the wire takes the shape of a circle. The tension in the wire is :
(A) BL (B)
BL(C)
2BL
(D)
4BL
Ans. (C)
Sol.
2Tsin
2d
= dlB 2T2d
= RdlB T = BR =
2LB
.
SECTION - II
Multiple Correct Choice TypeThis 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. 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 4
P0 (D) Temperature at C is 4
T0
Ans. (A, B)
PHYSICS
RESONANCE Page # 35
Sol.
U = 2f
nRT,, where f,n,R are constants. Also temperature T is same at A & B.
UA = UB
Also, WAB = nRT0
i
f
V
V n = nRT0
0
0
V
V4 n = nRT0 n4 = P0V0 n4
So, answers are (A) & (B).
66. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle
of 60º (see figure). If the refractive index of the material 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 AD
(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)
PHYSICS
RESONANCE Page # 36
Sol.
45º
60º
Q
By refraction at face AB :
1.sin60º = 3 . sinr1
So, r1 = 30º
This shows that the refracted ray is parallel to side BC of prism. For side 'CD' angle of incidence will be 45º,
which can be calculated from quadrilateral PBCQ.
By refraction at face CD :
3 sin45º = 1 sinr2
So, sin r2 = 2
3
which is impossible. So, there will be T.I.R. at face CD.
Now, by geometry angle of incidence at AD will be 30º. So, angle of emergence will be 60º.
Hence, angle between incident and emergent beams is 90º
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|
(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
Ans. (A, D)
PHYSICS
RESONANCE Page # 37
Sol.
Q1
Q2
From the diagram, it can be observed that Q1 is positive, Q2 is negative.
No. of lines on Q1 is greater and number of lines is directly proportional to magnitude of charge.
So, |Q1| > |Q2|
Electric field will be zero to the right of Q2 as it has small magnitude & opposite sign to that of Q1 .
68. A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He
uses a stop watch with the least count of 1sec 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%
Ans. (A, C)
Sol. Since, t = nT. So, T = nt
= 2040
or T = 2 sec.
Now, t = n T
andtt=
TT
So,401
= 2T
T = 0.05
Time period, T = g2
So,TT
= gg
21
or gg
= TT
2
So, percentage error in g = gg
× 100
= TT
2
× 100 = � 205.0
2 × 100
= 5%.
PHYSICS
RESONANCE Page # 38
69. A point mass of 1kg 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
Ans. (A, C)
Sol.
Since collision is elastic, so e = 1
Velocity of approach = velocity of separation
So, u = v + 2 .............(i)
By momentum conservation :
1 × u = 5v � 1 × 2
u = 5v � 2
v + 2 = 5v � 2
So, v = 1 m/s
and u = 3 m/s
Momentum of system = 1 × 3 = 3 kgm/s
Momentum of 5kg after collision = 5 × 1 = 5 kgm/s
So, kinetic energy of centre of mass = 21
(m1 + m2)
2
21
1
mm
um
= 21
(1 + 5)
2
631
= 0.75 J
Total kinetic energy = 21
× 1 × 32
= 4.5 J.
PHYSICS
RESONANCE Page # 39
SECTION�III
Paragraph TypeThis 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 Nos. 70 to 72
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 proportional to km
, 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) = x4
( > 0) for |x| near the origin and becomes a constant equal to V0 for |x| X0 (see figure)
70. 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)
Sol. When 0 < E < V0 there will be acting a restoring force to perform oscillation because in this case particle will
be in the region |x| x0 .
71. For periodic motion of small amplitude A, the time period T of this particle is proportional to :
(A)
mA (B)
mA1
(C) m
A
(D) mA
1
Ans. (B)
Sol. V = x4
T.E. = 21
m2A2 = A4 (not strictly applicable just for dimension matching it is used)
2 = mA2 2
T
mA1
PHYSICS
RESONANCE Page # 40
72. The acceleration of this particle for |x| > X0 is :
(A) proportional to V0 (B) proportional to 0
0
mX
V
(C) proportional to 0
0
mX
V(D) zero
Ans. (D)
Sol. F = dxdU
as for |x| > x0 V = V0 = constant
dxdU
= 0
F = 0.
Paragraph for Question Nos. 73 to 74
Electrical resistance of certain materials, known as
superconductors, changes abruptly from a nonzero value
to zero as their temperature is lowered below a criticalT (B)C
T (0)C
O B
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.
73. 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 larger than B1, which of the following
graphs shows the correct variation of R with T in these fields?
(A)
R
O T
B1B2 (B)
R
O T
B1
B2
(C)
R
O T
B1 B2 (D)
R
O T
B1
B2
Ans. (A)
PHYSICS
RESONANCE Page # 41
Sol. As the magnetic field is greater, the critical temperature is lower and as B2 is larger than B1. Graph �A� is
correct.
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
Ans. (B)
Sol. For B = 0 TC = 100 k
B = 7.5 T TC = 75 k
For B = 5T 75 < TC < 100
SECTION�IV
Integer TypeThis Section contains 10 questions. The answer to each question is a single-digit integer, ranging from 0
to 9. The correct digit below the question number in the ORS is to be bubbled.
75. A piece of ice (heat capacity = 2100 J kg�1 ºC�1 and latent heat = 3.36 × 105 J kg�1) of mass m grams is at
�5 ºC 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. 8 gm
Sol. S = 2100 J kg�1 ºC�1
L = 3.36 × 105 J kg�1
420 = m S Q + (1) × 10�3 × L
420 = m s(5) + 3.36 × 102
420 � 336 = m(2100) × 5
m = 125
1 × 1000 = 8 gm.
76. 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.
Ans. 7
Sol. Let speed of cars are V1 and V2
frequency received by car f1 =
1vvv
f0
PHYSICS
RESONANCE Page # 42
frequency reflected by car f2 =
v
vv 1
1vvv
f0
f = f2� � f2 =
1
1
2
2
vv
vv
vv
vv f0
f =
)vv)(vv(
)vv)(vv()vv)(vv(
21
2112 f0
f = )vv)(vv(
f )vv(v2
21
012
v
f )vv(2 012
Givenv
f )vv(2 012 = 100
2.1 f0
v2 � v1 = 7.126
Answer in nearest integer is 7.
77. The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front
of it to 50cm, the magnification of its image changes from m25 to m50. The ratio 50
25
m
m is :
Ans. 6
Sol. When object distance is 25.
v1
� u1
= f1
v1
� )25(�1
= 201
v = 100 cm.
m25 = uv
= 25�
100 = � 4.
When object distance is 50.
v1
� )50(�1
= 201
u = 3
100 cm
m50 = 50�
3100
= � 32
50
25
m
m =
32
�
4� = 6.
Alternate :
50
25
m
m =
5020f
2520f
=
530
= 6
PHYSICS
RESONANCE Page # 43
78. An -particle and a proton are accelerated from rest by a potential difference of 100V. After this, their
de-Broglie wavelength are and p respectively. The ratio
p, to the nearest integer, is :
Ans. 3
Sol. P1 = )eV100(m2
P = )eV100(m2
h = )eV100)(m4(2
h
P = 8
The ratio
p, to the nearest integer, is equal to 3.
79. When two identical batteries of internal resistance 1each 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 the value of R in is :
Ans. 4
Sol.
i = R2
2
J1 =
2
R22
R
PHYSICS
RESONANCE Page # 44
eq =
11
11
11
=
req = 21
i = R
21
=
1R22
J2 =
2
R212
R
Given J1 = 49
J2
2
R22
R =
49
2
R212
R
R2
2
= R21
3
2 + 4R = 6 + 3R
R = 4.
80. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature 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. (m)B = 3(m)A
TA = 3TB
E1 = 4 (6)2 TA4 = 4(6)2 (3TB)4
E2 = 4 (18)2 TB4
2
1
EE
= 9.
81. When two progressive waves y1 = 4 sin (2x � 6t) and y2 = 3 sin
2�t6�x2 are superimposed, the
amplitude of the resultant wave is :
Ans. 5
Sol. Aeq = cosAA2AA 2122
21
Aeq = 2
cos)3)(4(234 22
Aeq = 5.
PHYSICS
RESONANCE Page # 45
82. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its
cross-sectional area 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 :
Ans. 4
Sol. n = mk
= m
/yA =
myA
1.01
)109.4()10n( 79
= 140 n = 4.
83. A binary star consists of two stars A (mass 2.2 MS) and B ( mass 11 MS) is the same 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 hte binary star to the angular momentum of star B about the centre of mass is :
Ans. 6
Sol.A Bc.m.
5d6
d6 11 Ms
2.2 Ms
c.m. about B of momentum Angularc.m. about momentum angular Total
=
6d
6d
)M 11(
6d
6d
)M 11(6d5
6d5
)M2.2(
s
ss
= 6.
84. Gravitational acceleration on the surface of a planet is 116
g, where g is the gravitational acceleration on the
surface of the earth. The average mass density of the planet is 32
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. g = 2R
GM =
2
3
R
R34
)G(
; g R
g'g =
R'R
'
=
32
R'R
= 116
Given,R
'R =
2263
Ve = R
GM =
R
R34
))()G( 3
Ve R ; Ve = 3 km/hr.
PHYSICS