IIT-JEE 2012 - XtraEdge Jan 2011

XtraEdge Test Series # 9

Based on New Pattern

Time : 3 Hours

Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions : Section - I

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer.

• Question 10 to 13 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for wrong answer.

• Question 14 to 19 are passage based single type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer..

Section - II

• Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly marked answer in any row.

XtraEdge for IIT-JEE 71 JANUARY 2011 6. AA′ and BB′ are two parallel axis through a body of

mass 'M' shown in figure. Point 'C' is center of mass of the body. If IAA′ = I0, then IBB′ is equal to –

B′ A′

B A

(A) I0 + Md² (B) I0 + 2Md² (C) I0 – Md² (D) I0 + 3Md²

7. An ice-cube is put into a glass of water. Water in glass will get cooled primarily due to –

(A) Convection (B) Conduction (C) Radiation

(D) Conduction & Radiation

8. Minimum height through which an ice cube must fall so that all ice melt (assume that any loss of mechanical energy get converted into thermal energy and Lf : Latent heat of fusion) is –

(A) g L_f 2 (B)

g L_f

L_f 2

(D) will depend on mass of ice-ball

9. A liquid X is kept in calibrated container 'C'.

Coefficient of cubical expansion of liquid and container is γ_l and γ_c respectively. If level of liquid is at 200 cc mark, then –

(A) volume of container over liquid surface will not change if γ_l = γ_c

(B) level of liquid will remain at 200 cc mark, if γ_l = γ_c

(D) none of these

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.

The following questions given below consist of an

"Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false.

(D) If (A) is false but (R) is true.

10. Assertion (A) : A block moving with constant speed on a horizontal surface along a straight line cannot be in translational equilibrium.

Reason (R) : Translational equilibrium means net force acting on it is zero

11. Assertion (A) : At the highest position of a parabolic trajectory of a projectile, time rate of change of speed is zero.

Reason (R) : Angle between velocity vector and acceleration vector is

π at this position.

12. Assertion (A) : large normal force is required to move apart two glass plates enclosing a thin water film.

Reason (R) : Due to surface tension, water pulls each plate towards each other.

13. Assertion (A) : A perfectly black coloured body must acts as a black body.

Reason (R) : A perfectly black coloured body absorb all light falling on it.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16)

A uniform disc of radius R and mass M has a round cut as shown in figure. The mass of remaining portion has mass 'm'.

C′

B′

AA′

XtraEdge for IIT-JEE 72 JANUARY 2011 14. Moment of inertia of body about BB′ is –

(A) 8 11MR²

(B) 8 15MR²

15 MR² (D)

48 11MR²

15. Moment of inertia of body about CC′ is – (A)

8 MR 15 ²

(B) 8 11MR²

15 MR² (D)

71 27MR²

16. If body is made free to rotate about horizontal axis AA′, angular acceleration of body when centre of body and axis AA′ lie on same horizontal plane – (A)

R g 20

7 (B)

R 15

g 8

28 (D) None of these

Passage # 2 (Ques. 17 to 19)

A cylindrical vessel of radius 'R' filled with water is rotated about its vertical axis with a constant angular velocity 'ω'.

17. Shape of the meniscus will be – (A) circular away from centre (B) parabolic away from centre (C) convex

(D) none of these

18. Difference of height of the liquid level at centre and surface of cylinder is –

(A) g 2 ω2

. R (B) g ω2

R²

. R² (D)

2 3

g ω2

R²

19. If pressure at centre is P0 and density of water is 'ρ' then pressure at distance x from centre is –

(A) P0 + 2

2 ρ

ω r (B) P0 – 2

2 ρ

ω r

2 ρ

ω r (D) P0 + 2 3ω r² ²ρ

This section contains 3 questions (Questions 20 to 22).

Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

A B C D

Q R S

S P

P P Q R

R R Q Q

S S P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

20. A ball tied with a string of length 10 cm is free to rotate in vertical circle. It is given velocity

72 meter/sec in horizontal direction when it is at its lowest position. Then match the following : Column-I Column-II (A) Height at which the (P)

160 27 ball loses circular path

(in meter)

(B) Maximum height (Q) 2 1

reached by ball (in meter) (C) Velocity of ball at the (R)

20 3 time of leaving circular path (in meter/sec)

(D) Minimum velocity of (S) 2 2

ball (in meter/sec)

21. Column I contain different process and column II contains molar heat capacity. Match the following:

Column-I Column-II (A) T = T0 e^αV (P) CV +

V 1

R α + (B) P = P0 e^αV (Q)

1 R

− γ

γ + V P

V T₀ α ) 22. Column-I contains some phenomena of heat transfer

and column-II contains the mechanism of heat transfer.

Column-I Column-II

(A) Heating up water in bucket (P) Convection with immersion rod slightly

immersed in it

(B) Freezing of lakes in colder (Q) Conduction

region

bottom

(D) Heating up atmosphere (S) None of these

XtraEdge for IIT-JEE 73 JANUARY 2011

CHEMISTRY

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. The shape of TeCl4 is -

(A) Linear (B) Square planar (C) Tetrahedral (D) See-Saw

2. Oxidation states of carbon and nitrogen in KCN are, respectively -

(A) – 3, + 2 (B) + 2, – 3 (C) + 1, – 2 (D) zero each

3. How many moles of nitrogen is produced by the oxidation of one mole of hydrazine by 2/3 mole bromate ion

(A) 3

1 (B) 1 (C) 1.5 (D)

4. The IUPAC name of the compound is :

C–CH CH₃ CH3

NH₂

(A) 1-amino-1-phenyl-2-methyl propane (B) 2-methyl-1-phenyl propanamine

5. If C2 in above compound is rotated by 120º angle in anticlockwise direction along C2–C3, Which of the following form will be produced

H H

CH3

(1) CH3

(2) (3)

(4)

(A) Partial eclipsed (B) Perfectly eclipsed (C) Staggered (D)Gauche conformation 6. Which of the following compounds will exhibit

geometrical isomerism ?

(A) 1-phenyl-2-butene (B) 3-phenyl-1-butene (C) 2-phenyl-1-butene (D)1,1-dephenyl-1-propene

7. Which of the following is/are correct regarding π acceptors and π donors ?

(A) π acceptor have empty π orbitals with correct symmetry whereas π donors have filled π orbitals with correct symmetry

(B) π acceptor involve the metal ions with lower oxidation state whereas π donor favour the central metal ion at higher oxidation state

(D) All of the above are correct 8. An isotone of ⁷⁶₃₂Geis :

(A) ⁷⁷₃₂Ge (B) ⁷⁷₃₃As (C) ⁷⁷₃₄Se (D) ⁷⁹₃₄Se

9. The energy of an electron in the first Bohr orbit of H atom is –13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are)

(A) – 3.4 eV (B) – 4.2 eV (C) – 6.8 eV (D) + 6.8 eV

The following questions given below consist of an

"Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false.

(D) If (A) is false but (R) is true.

10. Assertion (A) : For each ten degree rise of temperature the specific rate constant is nearly doubled.

Reason (R) : Energy-wise distribution of molecules in a gas is an experimental function of temperature.

11. Assertion (A) : Dimethyl sulphide is commonly used for the reduction of an ozonide of an alkene to get the carbonyl compounds.

Reason (R) : It reduces the ozonide giveing water soluble dimethyl sulphoxide and excess of it evaporates.

XtraEdge for IIT-JEE 74 JANUARY 2011 12. Assertion (A) : In the titration of Na2CO3 with HCl

using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.

Reason (R) : Two moles of HCl are required for the complete neutralization of one mole of Na2CO3. 13. Assertion (A) : Band gap in germanium is small.

Reason (R) : The energy spread of each germanium atomic energy level is infinitesimally small.

Passage # 1 (Ques. 14 to 16)

Several short-lived radioactive species have been used to determine the age of wood or animal fossils.

One of the most interesting substances is 6C¹⁴ half-life 5760 years) which is used in determining the age of carbon-bearing materials (e.g. wood, animal fossils, etc.) Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays).

7N¹⁴ + 0n¹ → ₆C¹⁴+ 1H¹

Thus carbon-14 is oxidised to CO2 and eventually ingested by plants and animals. The death of plants or animals put an end to the intake of C¹⁴ from the atmosphere. After this the amount of C¹⁴ in the dead tissues starts decreasing due to its disintegration as per the following reaction : (

6C¹⁴→ ₇N¹⁴+ –1β⁰

The C¹⁴isotope enters the biosphere when carbon dioxide is taken up in plant photosynthesis. Plant are eaten by animals, which exhale C¹⁴ as CO2. Eventually, C¹⁴participates in many aspects of the carbon cycle. The C¹⁴lost by radioactive decay is constantly replenished by the production of new isotopes in the atmosphere. In this decay-replenishment process, a dynamic equilibrium is established whereby the ratio of C¹⁴to C¹²remains constant in living matter. But when an individual plant or an animal dies, the C¹⁴isotope in it is no longer replenished, so the ratio decreases as C¹⁴ decays. So, the number of C¹⁴nuclei after time t (after the death of living matter) would be less than in a living matter. The decay constant can be calculated using the following formula,

t1/2 = λ 693 . 0

The intensity of the cosmic rays have remain the same for 30,000 years. But since some years the changes in this are observed due to excessive burning of fossil fuel and nuclear tests.

14. Why do we use the carbon dating to calculate the age of the fossil?

(A) Rate of exchange of carbon between atmosphere and living is slower than decay of C¹⁴

(B) It is not appropriate to use C¹⁴ dating to determine age

(C) Rate of exchange of C¹⁴ between atmosphere and living organism is so fast that an equilibrium is set up between the intake of C¹⁴ by organism and its exponential decay

(D) none of the above

15. What should be the age of the fossil for meaningful determination of its age?

(A) 6 years (B) 6000 years (C) 60,000 years

(D) can be used to calculate any age

16. A nuclear explosion has taken place leading to increase in concentration of C¹⁴in nearby areas. C¹⁴ concentration is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1

and T1 at the respective places then

(A) The age of the fossil will increase at the place where explosion has taken place and T1 – T2 =

λ 1ln

2 1

C C

(B) The age of the fossil will decrease at the place where explosion has taken place and T1 – T2 =

λ 1ln

2 1

C C

2 1

T T =

2 1

C C

Passage # 2 (Ques. 17 to 19)

The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compound of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

17. Argon is used in are welding because of its (A) low reactivity with metal

(B) ability to lower the melting point of metal (C) Flammability

(D) high calorific value

XtraEdge for IIT-JEE 75 JANUARY 2011 18. The structure of XeO3 is

(A) linear (B) planar (C) pyramidal (D) T-shaped 19. XeF4 and XeF6 are expected to be

(A) oxidizing (B) reducing (C) unreactive (D) strongly basic

This section contains 3 questions (Questions 20 to 22).

A B C D

Q R S

S P

P P Q R

R R Q Q

S S P Q R S

20. Match the extraction processes listed in Column-I with metals listed in Column-II

Column-I Column-II

(A) Self-reduction (P) Lead (B) Carbon reduction (Q) Silver (C) Complex formation (R) Copper and displacement by

metal

(D) Decomposition of (S) Boron

iodide

21. Match the column :

Column-I Column-II (A) Efflorescent (P) NaOH (B) Deliquescent (Q) KOH

(D) washing soda (S) Na2CO3. 10H2O

(T) Na2SO4

22. Match the half-reaction (in column I) with equivalent mass (molar mass = M) (in column II)

Column –I Column II (A) Cr2O72– → Cr³⁺ (P) M (B) C2O42– → CO2 (Q) M/2 (C) MnO4– → MnO₂ (R) M/6 (D) HC2O4– → C₂O42– (S) M/3

MATHEMATICS

1. Sn, the sum to n terms of the series (n² – 1²) + 2(n² – 2²) + 3(n² – 3²) + ... is (A)

1 n² (n² – 1) (B) 4

1 n(n + 1)² (C) 0 (D) 2n (n² – 1)

2. The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of a man and a woman, is

(A) 1240 (B) 1840

3. If cos (θ – α) = a and sin (θ – β) = b (0 < θ – α, θ – β

< π/2), then cos² (α – β) + 2ab sin (α – β) is equal to (A) 4a² b² (B) a² – b²

(C) a² + b² (D) – a² b².

4. If A and B are acute positive angles satisfying the eqⁿ. 3 sin² A + 2 sin² B = 1 and 3 sin 2A – 2 sin 2B = 0, then A + 2B is equal to

(A) π/4 (B) π/2

5. In a triangle ABC,

C r r

cos 1

2 1

+ is equal to

(A)

∆ abc

2 (B)

∆ + c

b a

abc (D) ₂

∆ abc

6. If P is a point (x, y) on the line y = – 3x such that P and the point (3, 4) are on the opposite sides of the line 3x – 4y = 8, then

(A) x > 8/15, y < – 8/5 (B) x > 8/5, y < – 8/15 (C) x = 8/15, y = – 8/5 (D) none of these

7. C1 and C2 are circles of unit radius with centres at (0, 0) and (1, 0) respectively. C3 is a circle of unit radius, passes through the centres of the circles C1

and C2 and have its centre above x-axis. Equation of the common tangent to C1 and C3 which does not pass through C2 is

(A) x – 3y + 2 = 0 (B) 3x – y + 2 = 0 (C) 3x – y – 2 = 0 (D) x + 3y + 2 = 0

XtraEdge for IIT-JEE 76 JANUARY 2011 8. If the tangent at a point (a cos θ, b sin θ) on the

ellipse x²/a² + y²/b² = 1 meets the auxillary circle in two points, the chord joining them subtends a right angle at the centre; then the eccentricity of the ellipse is given by

(A) (1 + cos² θ)^–1/2 (B) 1 + sin² θ (C) (1 + sin² θ)^–1/2 (D) 1 + cos² θ

9. If p^th, q^th, r^th term of a G.P. are the positive numbers a, b, c, then the angle between the vectors log a²i + log b²j + log c²k and (q – r)i + (r – p)j + (p – q)k is (A) π/3 (B) π/2

(C) sin^–11/ a²+b²+c² (D) none of these This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.

The following questions given below consist of an

"Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false.

(D) If (A) is false but (R) is true.

10. Assertion (A) : If a and b are integers and roots of x² + ax + b = 0 are rational then these roots must be integers.

Reason (R) : If a and b are integers then roots of x² + ax + b = 0 are necessarily integers.

11. Assertion (A) : The coefficient of the term of independent of x in the expansion of

x x 



 



 +9+6 is

! )!

2 ( 3

n n

n n .

Reason (R) : The coefficient of x^r in the expansion of (1 + x)ⁿ is _



 



 r n .

12. Assertion (A) : If in a triangle ABC

sin² A + sin² B + sin² C = 2, then one of the angles must be 90º.

Reason (R) : In any triangle ABC

cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C

13. Assertion (A) : P is a point (a, b, c). Let A, B, C be the images of P in yz, zx and xy planes respectively then equation of the plane passing through the points A, B and C is

a x +

b y +

c z = 1.

Reason (R) : The image of a point P in a plane is the foot of the perpendicular draw from P on the plane.

Passage # 1 (Ques. 14 to 16)

f(x) = sin {cot^–1(x + 1)} – cos (tan^–1 x) a = cos tan^–1sin cot^–1 x

b = cos (2 cos^–1 x + sin^–1 x)

14. The value of x for which f(x) = 0 is (A) – 1/2 (B) 0

15. If f(x) = 0 then a² is equal to (A) 1/2 (B) 2/3

16. If a² = 26/51, then b² is equal to (A) 1/25 (B) 24/25

Passage # 2 (Ques. 17 to 19) Let a > 0, a ≠ 1 and x > 0.

(i) loga x > 0, a > 1 ⇔ x > 0, a > 1

(ii) loga x > 0, 0 < a < 1 ⇔ 0 < x < 1, 0 < a < 1 (iii) loga x < 0, a > 1 ⇔ 0 < x < 1, a > 1 (iv) loga x < 0, 0 < a < 1 ⇔ x > 1, 0 < a < 1 17. The solution set of x log1/10 (x²+ x + 1) > 0 is (A) (– ∞, – 1) (B) (0, 1)



 



 ,∞ 10

18. The solution set of log3

8 – 12

12 – 5

x + log1/3 x ≤ 0 is

(A) 



 



 3 ,2 12

5 (B)  _^



 2 ,1 12



 3 ,1 12

5 (D) 



 



 2 ,1 12

19. The solution set of

1 3 1

–10 1 log ₂

9 /

1 ≤



 



 +



 





π x x is

XtraEdge for IIT-JEE 77 JANUARY 2011

(A) 



 

 3 ,1

0 ∪ 



 



 3 ,10

3 (B) (π, ∞)



 



 3 ,1

0 ∪ (π, ∞) (D) (3, π) ∪ (π, ∞)

This section contains 3 questions (Questions 20 to 22).

A B C D

Q R S

S P

P P Q R

R R Q Q

S S P Q R S

20. Normals at P, Q, R are drawn to y² = 4x which intersect at (3, 0). Then :

Column-I Column-II

(A) Area of ∆PQR (P) 2

(B) Radius of circumcircle (Q) 2 5 of ∆PQR



 



 ,0 2 5

(D) Circumcentre of ∆PQR (S) 



 



 ,0 3 2

21. Value of

Column-I Column-II (A) 4 α (β⁴ – α⁴) (P) 0 if α + iβ, β ≠ 0

is a root of z⁵ = 1 (B)

2+z + |2 – z|² (Q) 1 if zz = 1

1 (R) 4

+ (z z )–w 2

2 1+

if w = z₁z₂ , |z1| = 3, |z2| = 1

(D) z1 – z2 + z3 – z4 if (S) 10 z1, z2, z3, z4 represent

vertices of a parallelogram

22. Column-I Column-II (A) Centroid of the triangle (P) (1, 6, 5) with vertices A(2, 3, 7),

B(6, 7, 5), C(1, 2, 3)

(B) Mid-point of the line (Q) (3, 4, 5) joining the points

A(7, 9, 11) and B(–5, 3, –1)

x = 3 y =

5 z, at a distance 2 from the origin

(D) Coordinates of the point (S) (4/ 38,6/ 38, dividing the join (5, 5, 0) 10/ 38) and (0, 0, 5) in the ratio 2 : 3.

• William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

• The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period.

(It is actually 24 hours and 37 minutes)

• Space debris travels through space at over 18,000 mph.

• The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system.

• A car travelling at a constant speed of 60 miles per hour would take longer than 48 million years to reach the nearest star (other than our Sun), Proxima Centauri. This is about 685,000 average human lifetimes.

XtraEdge for IIT-JEE 78 JANUARY 2011

PHYSICS

1. Define the term electric dipole moment. Is it a scalar

In document XtraEdge Jan 2011 (Page 73-81)