1. QUESTIONS & SOLUTIONS OF IIT-JEE 2009
Date : 12-04-2009 Duration : 3 Hours Max. Marks : 240
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
INSTRUCTIONS
PAPER - 1
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2. PART- I
CHEMISTRY
SECTION - I
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONLY ONE is correct.
1. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3
sol is :
(A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl
Sol. (C)
Most effective coagulating agent for Sb2S3 is Al2(SO4)3 because of high charge.
2. The term that corrects for the attractive forces present in a real gas in the vander Waals equation is :
an2 an2
(A) nb (B) (C) – (D) –nb
V2 V2
Sol. (B)
an 2
Correction factor for attractive force in to the real gas is given by .
V2
3. The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of
N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure
is :
(A) 4 × 10–4 (B) 4.0 × 10–5 (C) 5.0 × 10–4 (D) 4.0 × 10–6
Sol. (A)
PN = KH × x N2
2
1
x N2 = × 0.8 × 5 = 4 × 10–5 per mole
105
In 10 mole solubility is 4 × 10–4 .
RESONANCE Page # 2

3. 4. The reaction of P4 with X leads selectively to P4O6. The X is :
(A) Dry O2 (B) A mixture of O2 and N2
(C) Moist O2 (D) O2 in the presence of aqueous NaOH
Sol. (B)
In presence of N
2
P4 + 3O2  P4O6. Here nitrogen acts as diluent.
Note :
In dry O2 following reactions may take place.
P4 + 3O2  P4O6.
P4O6 + 2O2  P4O10.
In moist O2 the P4O6 gets hydrolysed forming H3PO3.
P4O6 + 6H2O  4H3PO3.
In presence of NaOH.
P4 + 3OH– + 3H2O  PH3 + 3H2PO2–
5. Among celluose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force
of attraction is weakest is :
(A) Nylon (B) Poly(vinyl chloride) (C) Cellulose (D) Natural Rubber
Sol. (D)
The natural rubber has intermolecular forces which are weak dispersion force (van-der-waal forces of attraction)
and is an example of an elastomer (polymer).
6. The correct acidity order of the following is :
(A) (III) > (IV) > (II) > (I) (B) (IV) > (III) > (I) > (II)
(C) (III) > (II) > (I) > (IV) (D) (II) > (III) > (IV) > (I)
Sol. (A) – H

More stable conjugate base (carboxylate anion, –ve charge always on oxygen atom).
RESONANCE Page # 3

4. – H

More stable conjugate base (carboxylate anion, –ve charge always on oxygen atom). + I and hyperconjugative
effect of – CH3 group decreases stability of benzoate anion.
– H

Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion). –Cl group
exhibits – I effect.
– H

Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion).
7. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic
mass of Fe is :
(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05
Sol. (B) 54Fe  5%
56Fe  90%
57Fe  5%
Av. atomic mass = x1A1 + x2A2 + x3A3
= 54 × 0.05 + 56 × 0.9 + 57 × 0.05 = 55.95
RESONANCE Page # 4

5. 8. The IUPAC name of the following compound is :
(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile
(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile
Sol. (B)
2-Bromo-5-hydroxybenzonitrile (–CN group gets higher priority over –OH and –Br)
SECTION - II
Multiple Correct Answer Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONE OR MORE is/are correct.
9. The correct statement(s) about the compound H3C(HO)HC–CH=CH–CH(OH)CH3 (X) is(are) :
(A) The total number of stereoisomers possible for X is 6.
(B) The total number of diastereomers possible for X is 3.
(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4.
(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2.
Sol. (A, D) (X) has configurations
(1) R Z R
(2) S Z S
(3) R Z S
(4) R E R
(5) S E S
(6) R E S
RESONANCE Page # 5

6. 10. The compound(s) formed upon combustion of sodium metal in excess air is(are) :
(A) Na2O2 (B) Na2O (C) NaO2 (D) NaOH
Sol. (A, B)
combustion
6Na + 2O2 (from air)   2Na2O + Na2O2

11. The correct statement(s) regarding defects in solids is(are) :
(A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion.
(B) Frenkel defect is a dislocation defect.
(C) Trapping of an electron in the lattice leads to the formation of F-center.
(D) Schottky defects have no effect on the physical properties of solids.
Sol. (B, C) Frenkel defect is a dislocation defect.
Trapping of an electron in the lattice leads to the formation of F-center.
12. The compound(s) that exhibit(s) geometrical isomerism is(are) :
(A) [Pt(en)Cl2] (B) [Pt(en)2]Cl2 (C) [Pt(en)2Cl2]Cl2 (D) [Pt(NH3)2Cl2]
Sol. (C, D)
(A) [Pt(en)Cl2] : exist only in one form.
(B) [Pt(en)2]Cl2 : exist only in one form
(en is symmetrical lignad).
(C) [Pt(en)2Cl2]Cl2 :
RESONANCE Page # 6

7. (D) [Pt(NH3)2Cl2] : Pt is in +2 oxidation state having 5d8 configuration. Hence the hybridisation of complex is
dsp2 and geometry is square planar.
SECTION - III
Comprehension Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on
a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE
is correct.
Paragraph for Question Nos. 13 to 15
A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by
dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes
intramolecular aldol reaction to give predominantly S.
1. MeMgBr 1. O 3 1. OH
P      Q     R      S
2. Zn, H2O 2. 

2. H , H2O
 
3. H2SO4 , 
13. The structure of the carbonyl compound P is :
(A) (B) (C) (D)
Ans. (B)
14. The structures of the products Q and R, respectively, are :
(A) , (B) ,
(C) , (D) ,
Ans. (A)
RESONANCE Page # 7

8. 15. The structure of the product S is :
(A) (B)
(C) (D)
Ans. (B)
Sol. (13, 14 & 15)
(1) MeMgBr
2H SO , 
4
( 2 ) H , H2O
    
 
OH O / Zn,H O
3 
2
int ramolcular aldol
  
 
condensation
Paragraph for Question Nos. 16 to 18
p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with
a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment
of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an
intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of
the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the
formation of Z.
RESONANCE Page # 8

9. 16. The compound X is :
(A) NaNO3 (B) NaCl (C) Na2SO4 (D) Na2S
Ans. (D)
17. The compound Y is :
(A) MgCl2 (B) FeCl2 (C) FeCl3 (D) ZnCl2
Ans. (C)
Sol. (16 to 17)
+ S2– (X) + + 6Fe3+ (Y)
 6Fe2+ + NH4+ + 2H+ + (methylene blue).
The compound X is Na2S and Y is FeCl3.
18. The compound Z is :
(A) Mg2[Fe(CN)6] (B) Fe[Fe(CN)6] (C) ) Fe4[Fe(CN)6]3 (D) ) K2Zn3[Fe(CN)6]2
Sol. (B)
Fe3+ + [Fe(CN)6]3–  Fe[Fe(CN)6] (brown colouration) (Z).
SECTION - IV
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to
be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are
labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE
statement(s) in Coloumn-II. The appropriate bubbles corresponding to the answers to these questions have
to be drakened as illustrated in the following example.
If the correct matches are A-p, s and t ; B-q and r; C-p and q; and D-s and t; then the correct darkening of
bubbles will look like the following :
RESONANCE Page # 9

10. 19. Match each of the compounds in Column I with its characteristic reaction(s) in Column II.
Column I Column II
(A) CH3CH2CH2CN (p) Reduction with Pd-C/H2
(B) CH3CH2OCOCH3 (q) Reduction with SnCl2/HCl
(C) CH3–CH=CH–CH2OH (r) Development of foul smell on treatment with
chloroform and alcoholic KOH.
(D) CH3CH2CH2CH2NH2 (s) Reduction with diisobutylaluminium hydride (DIBAL-H)
(t) Alkaline hydrolysis
Ans. (A) – p, q, s, t ; (B) – p, s, t ; (C) – p ; (D) – r
Sol.
Pd  C / H
2
(A) CH3CH2CH2CN     CH3CH2CH2CH2NH2 (p)
SnCl / HCl
 2   CH3CH2CH2CHO
  (q)
DIBAl H
    CH3CH2CH2CHO
 (s)
OH
  CH3CH2CH2COO (t)
H2O
2Pd  C / H
(B) CH3CH2OCOCH3     2CH3—CH2 – OH (p)
DIBAl H
    CH3—CHO
 (s)
OH
  CH —COO (t)
H2O 3
2 Pd  C / H
(C) CH3–CH=CH–CH2OH     CH3CH2CH2CH2OH (p)
3 CHCl
(D) CH3CH2CH2CH2NH2    CH3CH2CH2CH2NC (r)
KOH

RESONANCE Page # 10

11. 20. Match each of the diatomic molecules in Column I with its property/properties in Column II.
Column I Column II
(A) B2 (p) Paramagnetic
(B) N2 (q) Undergoes oxidation
(C) O2– (r) Undergoes reduction
(D) O2 (s) Bond order  2
(t) Mixing of 's' and 'p' orbitals
Ans. (A) - p, r, t ; (B) - s, t ; (C) - p, q, r ; (D) - p, r, s
Sol. (A) B2 1s2 1s 2 2s 2 2s 2 2px1 = 2py1
Bond order = 1 Paramagnetic
(B) N2 1s2 1s 2 2s 2 2s 2 2px2 = 2py2 2pz2
Bond order = 3 Diamagnetic
(C) O2– 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 2 = 2py 1
Bond order = 1.5 Paramagnetic
(D) O2 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 1 = 2py 1
Bond order = 2 Paramagnetic
RESONANCE Page # 11

12. PART-II
MATHEMATICS
SECTION - I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONLY ONE is correct.
15
2m 1
21. Let z = cos  + i sin  . Then the value of  m ( z
m 1
) at  = 2º is
1 1 1 1
(A) (B) (C) (D)
sin 2º 3 sin 2º 2 sin 2º 4 sin 2º
Sol. (D)
Given that z = cos  + i sin  = ei
15 15 15
 Im (z2m–1) = Im (ei)2m–1 =  Im e i(2m–1) = sin+ sin3+ sin5 + .......... + sin29
m 1 m 1 m 1
 
   29   15  2 
sin  sin 
 2   2  sin(15) sin (15)

= = =
1
 2  sin 
sin 
4sin2º
 2 
22. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and
3 only, is
(A) 55 (B) 66 (C) 77 (D) 88
Sol. (C)
There are two possible cases
Case 1 : Five 1’s, one 2’s, one 3’s
7!
Number of numbers = = 42
5!
Case 2 : Four 1’s, three 2’s
7!
Number of numbers = = 35
4! 3!
Total number of numbers = 42 + 35 = 77
RESONANCE Page # 12

13. 23. Let P(3, 2, 6) be a point in space and Q be a point on the line r  ( ˆ  ˆ  2k )  ( 3ˆ  ˆ  5k ) . Then the value
i j i j
 ˆ ˆ
of  for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is
1 1 1 1
(A) (B) – (C) (D) –
4 4 8 8
Sol. (A)
Given OQ = (1–3) ˆ + ( –1) ˆ + (5 +2) k , OP = 3ˆ  2ˆ  6k (where O is origin)
i j ˆ i j ˆ
PQ = (1–3– 3) ˆ + (–1–2) ˆ + (5 +2–6) k
i j ˆ
= (–2 – 3) ˆ + (–3) ˆ + (5 –4) k
i j ˆ
PQ is parallel to the plane x – 4y + 3z = 1


 –2–3 –4 + 12 + 15 – 12 = 0
8 = 2 =
1
 
4
x x
24. Let f be a non-negative function defined on the interval [0, 1]. If 1  ( f ( t ))2 dt =
0
  f ( t) dt, 0  x  1 and
0
f(0) = 0, then
 1 1  1 1  1 1  1 1
(A) f   < and f   > (B) f   > and f   >
2 2 3 3 2 2 3 3
 1 1  1 1  1 1  1 1
(C) f   < and f   < (D) f   > and f   <
 2 2  3 3  2 2  3 3
Sol. (C)
x x
Given 1 – ( f ' ( t ))2 dt =  f (t )dt, 0  x 1
0 0

Apply Leibnitz theorem, we get 1 – ( f ' ( x ))2 = f (x)
 1 – (f(x))2 = f2(x)  (f(x))2 = 1–f2(x)
dy
f ' (x) =  1 – f 2 (x) =  1– y2 , where y = f(x)
dx
 
dy
=  dx
1– y2

Integrating both sides
sin –1 (y) =  x + c
 f (0) = 0  c=0
RESONANCE Page # 13

14.  y = ± sin x
y = sin x = f(x) given f(x)  0 for x  [0, 1]
it is known that sinx < x,  x  R+
sin   <  f  < and sin   < f   <
1 1 1 1 1 1 1 1

 2 2  2 2  3 3  3 3
25. Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices
are the roots of the equation zz 3  zz 3  350 is
(A) 48 (B) 32 (C) 40 (D) 80
Sol. (A)
z z (z2 + z 2 ) = 350
 2(x2 + y2) (x2 – y2 ) = 350
 (x2 + y2) (x2 –y2) = 175
Since x, y  I , the only possible case which gives integral solution, is
x2 + y2 = 25 ......... (1)
x2 – y2 =7 ..........(2)
From (1) and (2) x2 = 16 ; y2 = 9
x=  4; y=  3  Area = 48
26. Tangents drawn from the point P(1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0 touch the circle at the points A
and B. The equation of the circumcircle of the triangle PAB is
(A) x2 + y2 + 4x – 6y + 19 = 0 (B) x2 + y2 – 4x – 10y + 19 = 0
(C) x2 + y2 – 2x + 6y – 29 = 0 (D) x2 + y2 – 6x – 4y + 19 = 0
Sol. (B)
For required circle, P(1, 8) and O(3, 2) will be the end points of its diameter.
 (x – 1) (x – 3) + (y – 8) (y – 2) = 0
 x2 + y2 – 4x – 10y + 19 = 0
RESONANCE Page # 14

15. 27. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse
x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and
the origin O is
31 29 21 27
(A) (B) (C) (D)
10 10 10 10
Sol. (D)
Equation of auxiliary circle is x2 + y2 = 9 (1)
x y
Equation of AM is + =1
3 1
(2)
 12 9 
on solving (1) and (2), we get M   , 
 5 5
1
Now, area of AOM = . OA × MN = square unit
27
2 10
  1
28. If a, b, c and d are unit vectors such that (a  b ) . (c  d) = 1 and a . c  , then
       
2
(A) a, b, c are non-coplanar
  
(B) b, c, d are non-coplanar
  
(C) b, d are non-parallel (D) a, d are parallel and b, c are parallel
     
Sol. (C)
Let angle between a and b be 1 , c and d be 2 and a  b and c  d be  .
       
(a  b ) . (c  d) = 1
   
 sin 1 . sin 2 . cos  = 1
 1 = 90º, 2 = 90º ,  = 0º
a  b, c  d, (a  b) | | (c  d)
       

a  b = k (c  d)
 
so a  b = k(c  d) and
     
(a  b) . c = k(c  d) . c and (a  b) . d = k (c  d) . d
           

[a b c ]  0 and [a b d]  0
 

a, b, c and a, b, d are coplanar vectors so options A and B are incorrect
     

Let b || d
   
 b  d
as (a  b ) . (c  d) = 1 ( a  b ) . (c  b ) = ± 1
       

[a  b c b] = ± 1
   

[c b a  b] = ± 1
   
RESONANCE Page # 15

16. c . [b  (a  b)]  1 c . [a  (b . a) b] = ± 1
        
 
c . a  1 ( a . b  0 )
   

which is a contradiction so option 'C' is correct
Let option 'D' is correct
d = ± a and c = ± b
   

as (a  b) . (c  d)  1
   
(a  b ) . (b  a) = ± 1
   

which is a contradiction so option 'D' is incorrect
Alternatively option 'C' and 'D' may be observed from the given figure
SECTION - II
Multiple Correct Choice Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONE OR MORE is/are correct.
x2
a  a2  x 2 
29. Let L = lim 4 , a > 0. If L is finite, then
x 0 x4
1 1
(A) a = 2 (B) a = 1 (C) L = (D) L =
64 32
Sol. (A, C)
1 
 1 x 2 1  2  1 x 4
 
 x2
a  a . 1  . 2   .  .... 
x2  2 a 2 2 a4  4
a  a2  x 2 
L = lim 4 , a > 0 = lim
 
4 x 0 4
 
x 0 x x
x2 1 x4 x2
 . 3  ...... 
lim
= x 0 2a 8 a 4
4
x
Since L is finite  2a = 4  a=2
1
lim
L = x 0 =
8 . a3
1

64
RESONANCE Page # 16

17. 30. Area of the region bounded by the curve y = ex and lines x = 0 and y = e is
e 1 e
x
(A) e – 1 (B)  n (e  1  y) dy (C) e – e dx (D)  n y dy
1
1 0
Sol. (B, C, D)
1
Shaded area = e –  e x dx  = 1

  
0 
e
Also  n( e  1  y) dy
1
put e+1–y=t  – dy = dt
1 e e
= n t dt =
 n t (–dt) = 
e 1
 n y dy  1
1
sin4 x cos 4 x 1
31. If + = , then
2 3 5
2 sin8 x cos 8 x 1
(A) tan2x = (B) + =
3 8 27 125
1 sin8 x cos 8 x 2
(C) tan2 x = (D) + =
3 8 27 125
Sol. (A, B)
sin4 x cos 4 x 1 sin4 x (1  sin 2 x )2 1
+ = + =
2 3 5 2 3 5

sin4 x 1  sin 4 x  2 sin2 x 1
+ =
2 3 5

6
5 sin4x – 4 sin2x + 2 =
5

 25 sin4x – 20 sin2x + 4 = 0
 (5 sin2x – 2)2 = 0
2 3
sin2x = , cos2x =
5 5

tan2x =
2

3
sin 8 x cos 8 x
and + =
1
8 27 125
RESONANCE Page # 17

18. A
32. In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin2 . If a, b and c
2
denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then
(A) b + c = 4a (B) b + c = 2a
(C) locus of points A is an ellipse (D) locus of point A is a pair of straight lines
Sol. (B, C)
A
cos B + cos C = 4 sin2
2
BC BC A
2 cos cos = 4 sin2
2 2 2

A  BC A
 2 sin cos 2  2 sin 2  = 0
2  
B C B  C A
cos   =0 as sin 0
2   2  2
  – 2 cos 

B C B C
cos + 3 sin sin =0
2 2 2 2
 – cos
B C 1
tan tan =
2 2 3

(s  a)(s  c ) (s  b)(s  a) 1
. =
s(s  b) s(s  c ) 3

sa 1
= 2s = 3a b + c = 2a
s 3
  
 Locus of A is an ellipse
SECTION - III
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on
a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE
is correct.
Paragraph for Question Nos. 33 to 35
Let be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries
are 1 and four of them are 0.
RESONANCE Page # 18

19. 33. The number of matrices in is
(A) 12 (B) 6 (C) 9 (D) 3
Sol. (A)
Case I : All three diagonal elements are 1
No. of matrices = 3C1 = 3
Case II : Two diagonal elements are zero & one element is one
No. of matrices = 3C1 . 3C1 = 9
Total matrices = 3 + 9 = 12
 x  1 
34. The number of matrices A in for which the system of linear equations A  y  = 0 has a unique solution,
   
z  0
   
is
(A) less than 4 (B) at least 4 but less than 7
(C) at least 7 but less than 10 (D) at least 10
Sol. (B)
x 1
A  y  = 0
   
z
  0
 
For unique solution det (A) 0
Case I det(A) = = 1 – a2 – b2 – c2 + 2abc  0
Here a, b, c is selected from 1, 0, 0. (No case is possible)
Case II
(i) det(A) = = 2abc – c2  0
Here a, b, c are selected from 1, 1, 0. (2 cases are possible)
(ii) det(A) = = 2abc – b2  0
Here a, b, c are selected from 1, 1, 0. (2 cases are possible)
(iii) det(A) = = 2abc – a2  0
Here a, b, c are selected from 1, 1, 0. (2 cases are possible)
Hence there are exactly 6 matrices for unique solution . Hence option B is correct
RESONANCE Page # 19

20. x 1 
35. The number of matrices A in for which the system of linear equations A  y  = 0 is inconsistent, is
   
z 
  0
 
(A) 0 (B) more than 2 (C) 2 (D) 1
Sol. (B)
x 1
A  y  = 0
   
z 
  0
 
Case I :
1 a b x 1
y  = 0
a 1 c  a, b, c are selected from 1, 0, 0
   
b c 1 z 

0
 
     
 x + ay + bz = 1
ax + y + cz = 0
bx + cy + z = 0
(i) If a = 1 , b = c = 0
then x + y = 1 Inconsitent system of equation
x +y=0
(ii) If a = 0 = c, b = 1
then x + z = 1
y=0 Inconsitent system of equation
x+z=0
(iii) If c = 1, a = b = 0
then x = 1, z = 0, y = 0
Case II :
1 a b  x 1
y
a 0 c 0
 
(i) =   a, b, c are selected from 1, 1, 0
   
z 

b c 0 0

 
    
 x + ay + bz = 1
ax + cz = 0
bx + cy = 0 Clearly, In all three cases, solutions are possible so system is consistent.
RESONANCE Page # 20

21. 0 a b  x 1
(ii) a 1 c  y  = 0
     
b c 0 z  0

 
    
 ay + bz = 1
ax + y + cz = 0
bx + cy = 0
Clearly , b = 0, a= c= 1 gives
y=1
x+y+z=0 Inconsistent system
y=0
More than 2 matrices are possible. Hence option B is correct
Paragraph for Question Nos. 36 to 38
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.
36. The probability that X = 3 equals
25 25 5 125
(A) (B) (C) (D)
216 36 36 216
Sol. (A)
5 5 1
P(X = 3) = . . =
25
6 6 6 216
37. The probability that X  3 equals
125 25 5 25
(A) (B) (C) (D)
216 36 36 216
Sol. (B)
5 5
P(X  3) = . .1=
25
6 6 36
38. The conditional probability that X  6 given X > 3 equals
125 25 5 25
(A) (B) (C) (D)
216 216 36 36
Sol. (D)
 5  5  1   5  6  1 
1 .   .      .    .....

 6   6   6   6 

P((X  6) / (X > 3)) =
P(( X  3) /( X  6)) . P( X  6) =  

=
25
P( X  3)  5  3 1  5  4 1
  .    .  ...... 
 36
 6  6  6  6
 

RESONANCE Page # 21

22. SECTION - IV
Matrix - match Type
This section contains 2 questions. Each question contains statements given in two columns, which
have to be matched. The statements in Column - I are labelled A, B, C and D, while the statements
in Column - II are labelled p, q, r, s and t. Any given statement in Column - I can have correct
matching with
ONE OR MORE statement(s) in Column - II. The appropriate bubbles corresponding to the answers
to these questions have to be darkened as illustrated in the following example :
If the correct matches are A – p, s and t ; B – q and r ; C – p and q ; and D – s and t; then the correct
darkening of bubbles will look like the following.
39. Match the statements/expressions in Column - I with the open intervals in Column - II
Column - I Column - II
(A) Interval contained in the domain of definition of (p) – , 
  
 2 2
non-zero solutions of the differential
equation (x – 3)2 y + y = 0
(B) Interval containing the value of the integral (q)  0, 
 
 2
5
 ( x – 1) ( x – 2) ( x – 3) ( x – 4) ( x – 5) dx
1
  5 
(C) Interval in which at least one of the points of local (r)  ,
8 4 

maximum of cos2x + sinx lies
(D) Interval in which tan–1 (sinx + cosx) is increasing (s)  0, 
 
 8
(t) (– , )
RESONANCE Page # 22

23. Sol. (A)
(x – 3)2 . y + y = 0
dy y
=–
dx ( x  3)2

dy dx
= 
y ( x  3)2
  
1
n y = + n C
( x  3)

1
 y = Ce x 3 , C  0
x  R – {3} (hence correct options p, q, s)
Aliter
Given differential equation is homogeneous linear differential equation and has x = 3 as a singular
point hence x = 3 cannot be in domain of solution.
5
(B) Let I =  ( x  1) ( x  2) ( x  3) ( x  4) ( x  5) dx
1
Let x – 3 = t  dx = dt
2
 I=  (t  2)(t  1) t( t  1) (t  2) dt ,
2
 Integrand is an odd function
 I=0 (hence correct options are p, t)
Aliter
5
Let =  ( x  1) ( x  2) ( x  3) ( x  4) ( x  5) dx .......(1)
1
b b
Using  f ( x) dx =  f (a  b  x) dx
a a
5
=  (5  x) (4  x) (3  x) (2  x ) (1  x ) dx ........(2)
1
On adding (1) and (2), we get 2 = 0  =0
RESONANCE Page # 23

24. (C) f(x) = cos2x + sin x
f(x) = – 2 cos x sin x + cos x = cos x (1 – 2 sin x) = 0 (hence correct options are p, t)
Sign scheme for first derivative
 5
Points of local maxima are ,
6 6
Clearly option (p, q, r, t) are correct.
Aliter
y = cos2x + sin x
2
5 1
y= –  sin x  

4  2
For y to be maximum.
2
1
 sin x   = 0

 2
1
sin x =
2

x = n + (–1)n , n 

6

Hence option (p, q, r, t) are correct
(D) y = tan–1 (sin x + cos x)
dy cos x  sin x
=
dx 1  (sin x  cos x )2
Clearly by graph cos x > sin x is true for option(s)
Ans. (A)  (p, q, s), (B)  (p, t), (C)  (p, q, r, t), (D)  (s)
RESONANCE Page # 24

25. 40. Match the conics in Column - I with the statements/expressions in Column - II.
Column - I Column - II
(A) Circle (p) The locus of the point (h, k) for which the
line hx + ky = 1 touches the
circle x2 + y2 = 4
(B) Parabola (q) Points z in the complex plane satisfying
|z + 2| – |z – 2| =  3
(C) Ellipse (r) Points of the conic have parametric
 1– t2 
representation x = 3 
,
1 t 2 

 
2t
y=
1 t 2
(D) Hyperbola (s) The eccentricity of the conic lies in the
interval 1  x < 
(t) Points z in the complex plane satisfying
Re(z + 1)2 = |z|2 + 1
1 1
Sol. (p) =2  h2 + k2 = locus is a circle.
h k2 2 4

(q) z  2 – z – 2  3 and 2 –(–2) = 4 > 3  locus is a hyperbola.
 1 t2  2t
(r) x= 3  1 t 2  , y=
1 t 2
 
 
Let tan =t
x= 3 cos 2 y = sin 2
x2
Hence + y2 = 1  locus is an ellipse
3
(s) Eccentricity x = 1  Parabola , 1< x <   Hyperbola
(t) z = x + iy
Re(z + 1)2 = |z|2 + 1
 (x + 1)2 – y2 = x2 + y2 + 1
 2x = 2y2
 x = y2 Hence locus is a parabola
Ans. (A)  (p), (B)  (s, t), (C)  (r), (D)  (q, s)
RESONANCE Page # 25

26. PART-III
PHYSICS
SECTION - I
Single Correct Choice Type
Thus section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONLY ONE is correct.
41. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at
(–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4
to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0), respectively.
Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ±a/2. The electric flux through this
cubical surface is
2C 2C 10C 12C
(A)  (B)  (C)  (D) 
0 0 0 0
Sol. (A) From Gauss's law
Qin (8C / 4 ) – 7C  (6C / 2) 2C
0 = 0 =–  .
0
42. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the
particle at t = 4/3 s is
1
x(cm)
0 t(s)
4 8 12
–1
3 2  2 2 3 2
(A)  cm/s2 (B) cm/s2 (C) cm/s2 (D) –  cm/s2
32 32 32 32
RESONANCE Page # 26

27. Sol. (D)
2
T = 8 second., A = 1 cm, x = A sint. = 1sin t.
8
2
 2   2 
a = – 2x = –   sin  t cm/s2
 8   8 
2
4  2  3 2
At, t = second. a= –  sin =– cm/s2.

3  8  3 32

43. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is
found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the
charges given to the shells, Q1 : Q2 : Q3, is
(A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18
Sol. (B)
The charge distribution on the surfaces of the shells are given. As per the given condition.
Q1 Q1  Q2 Q1  Q2  Q3 Q1 Q2 Q3
= = .
4R 2 4 ( 2R )2 4 (3R )2 1 3 5
  
44. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is
4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is
12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2]
(A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s
Sol. (C)
x v 2  10  ( 20 – 12 .8) 4
x' = n , v' = n = = 16 m/s.
rel rel 1 3

45. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of
ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink
used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0,
0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, –a). The y-
coordinate of the centre of mass of the ink in this drawing is
RESONANCE Page # 27

28. a a a a
(A) (B) (C) (D)
10 8 12 3
Sol. (A)
m1y1  m2 y 2  m3 y 3  m4 y 4  m5 y 5
ycm = m1  m2  m3  m4  m5
6m(0)  m(a )  m(0 )  m( a)  m(– a ) a
ycm = = .
m  m  m  m  6m 10

46. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction
between them is 3 . The inclination  of this inclined plane from the horizontal plane is gradually increased
from 0º. Then
(A) at  = 30º, the block will start sliding down the plane
(B) the block will remain at rest on the plane up to certain  and then it will topple
(C) at  = 60º, the block will start sliding down the plane and continue to do so at higher angles
(D) at  = 60º, the block will start sliding down the plane and on further increasing , it will topple at certain 
Sol. (B)
Angle of repose 0 = tan–1 = tan–1 3 = 60º
5 2
tan = = .  < 45º.
15 / 2 3
Block will topple before it starts slide down.
47. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular
orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the
particles move with constant speeds. After making how many elastic collisions, other than that at A, these
two particles will again reach the point A?
RESONANCE Page # 28

29. v A 2v
(A) 4 (B) 3 (C) 2 (D) 1
Sol. (C)
Since masses of particles are equal, collisons are elastic, so particles will exchange velocities after
each collision. The first collision will be at a point P and second at point Q again and before third
collision the particles will reach at A.
48. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a
perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field
increases with time. 1 and 2 are the currents in the segments ab and cd. Then,
(A) 1 > 2
(B) 1 < 2
(C) 1 is in the direction ba and 2 is in the direction cd
(D) 1 is in the direction ab and 2 is in the direction dc
Sol. (D)
Current 1 = 2 ,
Since magnetic field increases with time
So induced net flux should be outward i.e. current will flow from a to b
RESONANCE Page # 29

30. SECTION - II
Multiple Correct Choice Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out of which ONE OR MORE is/are correct.
49. A student performed the experiment of determination of focal length of a concave mirror by u-v method using
an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the
location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56),
(48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incor-
rectly recorded, is (are)
(A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39)
Sol. (C, D)
1 1 1 1 1 1 |u| |f |
  or |v| = | u |  | f |
v u f |v| |u| |f | 
 
( 42) ( 24 )
For |u| = 42, |f| = 24 ; |v| = = 56 cm so (42, 56) is correct observation
42  24
For |u| = 48 or |u| = 2f or |v| = 2f so (48, 48) is correct observation
For |u| = 66 cm ; |f| = 24 cm
( 66 ) ( 24 )
|v| =  36 cm which is not in the permissible limit
66  24
so (66, 33), is incorrect recorded
For |u| = 78, |f| = 24 cm
(78 ) ( 24 )
|v| =  32 cm which is also not in the permissible limit.
78  24
so (78, 39), is incorrect recorded.
50. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame,
one can surely say that
(A) linear momentum of the system does not change in time
(B) kinetic energy of the system does not changes in time
(C) angular momentum of the system does not change in time
(D) potential energy of the system does not change in time
RESONANCE Page # 30

31. Sol. (A)
(A) Since there in no resultant external force, linear momentum of the system remains constant.
(B) Kinetic energy of the system may change.
(C) Angular momentum of the system may change as in case of couple, net force is zero but torque
is not zero. Hence angular momentum of the system is not constant.
(D) Potential energy may also change.
51. Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure,
respectively. Then
(A) Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(B) Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(C) Cp / Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(D) Cp . Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
Sol. (B, D)
5 3
For monoatomic gas, Cp = R, Cv = R. Cp – Cv = R
2 2
7 5
For diatomic gas Cp = R, Cv = R. Cp – Cv = R
2 2
Cp – Cv is same for both
Cp + Cv = 6R (for diatomic)
Cp + Cv = 4R (for mono)
so (Cp + Cv)dia > (Cp + Cv)mono
Cp 7
Cv

5
= 1.4 (for diatomic)
Cp 5
Cv

3
= 1.66 (for monoatomic)
35 2
(Cp) (Cv) = R = 8.75 R2 (for diatomic)
4
15 2
(Cp) (Cv) = R (for monoatomic)
4
so (Cp . Cv)diatomic > (Cp . Cv)monoatomic
52. For the circuit shown in the figure
[Current Electricity]
(A) the current I through the battery is 7.5 mA
(B) the potential difference across RL is 18 V
(C) ratio of powers dissipated in R1 and R2 is 3
(D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9
RESONANCE Page # 31

32. Sol. (A, D)
 6  1 .5 6 16
Req =  7.5  2  = +2= k

  5 5
V 24 3
= R =
eq 16 2
× 5 mA = × 5 = 7.5 mA
for potential difference across R1  V1 = 7.5 × 2 = 15 V
for potential difference across R2  V2 = 24 – 15 = 9 V
2 2
V1 V2
: (15 )2 92 25
for power : PR1 : PR 2 = = : =
R1 R2 2 6 3
2
PRL 
V2 92
= = 54 mW
RL 1 .5
If R1 and R2 are interchanged
 ( 2) (1.5 )  3
R´ = R1 || R2 =  2  1.5  =
  3 .5
R´
V´L =
R2  R´
× 24 V = 3 V
V ´2
L 32
Now power dissipated in RL is P´L = = = 6 mW
RL 1 .5
SECTION–III
Comprehension Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a
paragraph. Each question has 4 choices (A),(B),(C) and (D) for its answer, out of which ONLY ONE is
correct.
Paragraph for Question Nos. 53 to 55
2
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 1 H , known as
deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is
2 2
1 H 1 H 3 He  n  energy . In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into
2
2
deuteron nuclei and electrons. This collection of 1 H nuclei and electrons is known as plasma. The nuclei
move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place.
Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the
plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from
the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of
the criteria, a reactor is termed successful if Lawson number is greater than 5×1014 s/cm3.
e2
It may be helpful to use the following: Boltzman constant k = 8.6×10–5 eV/K ; = 1.44 × 10–9 eVm.
4 0
RESONANCE Page # 32

33. 53. In the core of nuclear fusion reactor, the gas becomes plasma because of
(A) strong nuclear force acting between the deuterons
(B) Coulomb force acting between the deuterons
(C) Coulomb force acting between deuterons-electrons pairs
(D) the high temperature maintained inside the reactor core
Sol. (D)
the high temperature maintained inside the reactor core
54. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each
other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb
potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T
required for them to reach a separation of 4 × 10–15 m in the range.
(A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K
(C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K
Sol. (A)
From energy conservation
3 1 e2
2  kT  

2  4 0 r
(1.44  10 9 ) 1
T= = 1.39 × 109 K
(8.6  10  5 ) 3  4  10 15
 ×
55. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which
of these is most promising based on Lawson criterion ?
(A) deuteron density = 2.0 × 1012 cm–3, confinement time = 5.0 × 10–3 s
(B) deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s
(C) deuteron density = 4.0 × 1023 cm–3, confinement time = 1.0 × 10–11 s
(D) deuteron density = 1.0 × 1024 cm–3, confinement time = 4.0 × 10–12 s
Sol. (B)
nt0 > 5 × 1014 s/cm3
for deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s
nt0 = 7.2 × 1014 s/cm3
Paragraph for Question Nos. 56 to 58
When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer
dimension, its energy can take only certain specific values. The allowed energies of the particle moving in
such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and
x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to
p2
the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = .
2m
Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1,2,3,......., (n = 1,
called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line x = 0
to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C.
RESONANCE Page # 33

34. 56. The allowed energy for the particle for a particular value of n is proportional to :
(A) a–2 (B) a–3/2 (C) a–1 (D) a2
Sol. (A)
h nh
n =a and p= =

2  2a
p2 n2h2 1
Energy E = = E
2m 8ma2 a2

57. If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state
is closest to :
(A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV
Sol. (B)
n2h2
E=
8 ma 2
(1)2 ( 6.6  10 34 )2
For ground state n = 1 E1 =  30 = 8 meV
8  10  ( 6.6  10  9 )2  1.6  10 19

58. The speed of the particle, that can take discrete values, is proportional to :
(A) n–3/2 (B) n–1 (C) n1/2 (D) n
Sol. (D)
nh
P= = mv
2a
 vn
Section–IV
Matrix – Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to
be matched. the statements in Column–I are labelled A,B,C and D, while the statements in Column-II are
labelled p,q,r,s and t. Any given statement in Column-I can have correct matching with ONE OR MORE
statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have
to be darkened as illustrated in the following examples :
If the correct matches A–p,s and t; B–q and r; C–p and q ; and D–s and t; then the correct darkening of
bubbles will look like the following.
RESONANCE Page # 34

35. p q r s t
A p q r s t
B p q r s t
C p q r s t
D p q r s t
59. Six point charges, each of the same magnitude q, are arranged in different manners as shown in
Column–II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field
and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it
is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let
B be the magnetic field at M and  be the magnetic moment of the system in this condition. Assume each
rotating charge to be equivalent to a steady current.
Column–I Column–II
(A) E = 0 (p) Charges are at the corners of a
+ Q
regular hexagon. M is at the
–
– + centre of the hexagon. PQ is
perpendicular to the plane of
+ the hexagon.
P
–
(B) V  0 (q) P Charges are on a line perpendicular to
PQ at equal intervals. M is the
– + – + – + midpoint between the two innermost
M
charges.
Q
(C) B = 0 (r) + Q Charges are placed on two coplanar
– +
insulating rings at equal intervals. M is
the common centre of the rings. PQ is
M
perpendicular to the plane of rings.
– –
P
+
(D)  0 (s) – + – Charges are placed at the corners of a
M rectangle of sides a and 2a and at the
P Q
mid–points of the longer sides. M is at
– + –
the centre of the rectangular. PQ is
parallel to the longer sides.
RESONANCE Page # 35

36. P
(t) + – Charges are placed on two coplanar,
identical insulating rings at equal
+ + M– – intervals. M is the mid–point between
Q the centres of the rings. PQ is
perpendicular to the line joining the
centres and coplanar to the rings.
Ans. : (A)  (p), (r), (s); (B)  (r), (s); (C)  (p), (q), (t) ; (D)  (r), (s)
Sol. (p) By symmetry E=0,V =0,B=0
+ – Q
– + and µ = NA but effective = 0, So µ = 0
+
P
–
(q) P E 0,V=0
– + – + – + Since effective = 0  B = 0 and µ = 0
M
Q
(r) + Q E = 0 (By Symmetry)
– +
V  0 (since distances are different)
– M –
P B  0 (since Radius is different)
+
µ0
(s) – + – E = 0 (By symmetry)
M
P Q
V  0 (since distances are different)
– + –
B0
µ0
P
(t) + – E  0, V = 0, µ = 0, B = 0 and given each rotating charge
to be equivalent to a steady current so B = 0 so (t  C)
+ + M– – and if it was not given that each rotating charge to be
Q equivalent to steady current then B  0.
RESONANCE Page # 36

37. 60. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is
shown. Column I gives some statements about X and and/or Y. Match these statements to the appropriate
system(s) from Column II.
Column I Column II
(A) The force exerted (p) Block Y of mass M left on a
by X on Y has a fixed inclined plane X, slides
magnitude Mg. on it with a constant velocity.
(B) The gravitational (q) Two ring magnets Y and Z,
potential energy of each of mass M, are kept in
X is continuously frictionless vertical plastic
increasing, stand so that they repel each
other. Y rests on the base X
and Z hangs in air in
equilibrium. P is the topmost
point of the stand on the
common axis of the two rings.
The whole system is in a lift
that is going up with a constant
velocity.
(C) Mechanical energy (r) A pulley Y of mass m0 is fixed
of the system X + Y to a table through a clamp X.
is continuously A block of mass M hangs from
decreasing. a string that goes over the
pulley and is fixed at point P
of the table. The whole system
is kept in a lift that is going
down with a constant velocity.
(D) The torque of the (s) A sphere Y of mass M is put
weight of Y about in a nonviscous liquid X kept
point P is zero. in a container at rest. The
sphere is released and it moves
down in the liquid.
(t) A sphere Y of mass M is falling
with its terminal velocity in a
viscous liquid X kept in a
container.
Ans. : (A)  (p), (t); (B)  (q), (s), (t); (C)  (p), (r), (t); (D)  (q)
RESONANCE Page # 37