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IIT

- ian’s EDUCATION PVT LTD

ANDHERI /VILE PARLE/ DADAR / CHEMBUR / THANE / CHURCHGATE/NERUL/POWAI

TIME : 3 HRS IIT-JEE 2010

A.I.T.S. # 1 PAPER - II

DATE : 03/01/2010 MAX. MARKS : 240

I N S T R U C T I O N S T O C A N D I D AT E 1.

This booklet is your question paper. Answers have to be marked on the provided OMR sheets .

2.

This question paper contains 57 questions. .

3.

Blank sheets are provided for rough work alongwith the question paper. No additional sheet will be provided for rough work

4.

For each question in Section - III , you will awarded 2 marks for each row in which you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marking for incorrect(s) for this section.

5.

If a student is caught copying , he/ she will not be allowed to write the exam.

6.

On the OMR sheet, write in ink your name, and name of the centre and put your signature in the appropriate boxes.

7.

The question paper consists of 3 parts ( Physics , Chemistry & Maths ) Each part has 4 sections.

8.

Marking Scheme: Physics : Q.1 - Q.4 ( + 3, -1) ; Q.5 - Q.9 ( + 4,-1) ; Q.10 - Q.11 ( + 8, 0) ; Q.12 - Q.19 ( +4 ,-1) Chemistry: Q.20 - Q.23 ( + 3, -1) ; Q.24 - Q.28 ( + 4,-1) ; Q.29 - Q.30 ( + 8, 0) ; Q.31 - Q.38 ( + 4 ,-1) Maths : Q.39 - Q.42 ( + 3, -1) ; Q.43 - Q.47 ( + 4,-1) ; Q.48 - Q.49 ( + 8, 0) ; Q.50 - Q.57 ( +4 ,-1)

Name :_____________________________________________________ Batch : ________________________ _____________________________ Centre : ____________________________________________________

SECTION – I Single Correct Choice Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONLY ONE is correct.

1.

At night approximately 500 photons per second must enter an unaided human eye for an object to be seen. A light bulb emits about 5.00  1018 photons per second uniformly in all directions. The radius of the pupil of the eye is about 4  10-3 meters. What is the maximum distance from which the bulb could be seen? (A) 2.0  104m

2.

(B) 2.0  105m

(C) 20 m

(D) 5.0  103m

An infinitely long solenoid passes through the circuit as shown. The magnetic field of the solenoid, directed into the plane of the page, is weakening which produces a constant emf of magnitude  for a closed loop around the outside of the solenoid. Once equilibrium is established in this circuit, what is the potential difference across the switch S?

(A) 0 3.

(B)

(C)

4  3

(D) cannot be determined

Two particles A and B having equal charges after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of mass of A to that of B is: (A) (R1 / R2)2

4.

1  3

(B) (R1 / R2)1/2

(C) (R2 / R1)3/2

(D) (R1 / R2)

Two spheres having masses m1 and m2 are kept on a on a inclined plane as shown. Both of them are in equilibrium. What is a necessary condition for this? ( 1 is coefficient of friction between m1 and inclined plane and  2 is coefficient of friction between m2 and inclined plane). m2 m1



(A) m1 > m2

(B) m1 < m2

(C) 1   2

(D) 1   2

SECTION – II IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONE OR MORE is/are correct.

5.

An electrostatic field line leaves at angle  from point charge q1 , and terminates at point charge – q2 at an angle  (shown in figure). Then the relationship between  and  is:



 q 2

 q1

(A) q1 sin 2   q 2 sin 2 

(B) q1 tan   q 2 tan 

   q 2 sin 2 2 2

(D) q1 cos   q 2 cos 

(C) q1 sin 2 6.

Two capacitor of 2 F and 3 F are charged to 150 volt and 120 volt respectively. The plates of a capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5 F falls to the free ends of the wire. Then after the system comes in steady state :

150V

(A) Charge on the 1.5 F capacitor will become 180 C (B) Charge on the 2 F capacitor will become 120 C (C) + ve charge flows through A from left to right (D) +ve charge flows through A from right to left 7.

 A proton of charge ‘e’ and mass enters a uniform and constant magnetic field B  Biˆ with  an initial velocity V  V0X ˆi  V0Y ˆj. Which of the following will be correct at any later time ‘t’

 qBt  (A) x – component of velocity of the proton will be Vox cos    m   qBt  (B) y – component velocity will be Voy cos    m   qBt  ˆ (C) z – component of velocity will be Voy sin   along  k diercetion  m  (D) x – component of velocity will remains unchanged.

 

8.

Graphs shows a hypothetical speed distribution for a sample of N gas particle IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

(for V < V0 ;

dN  0) dV

(A) The value of aV0 is 2N (B) The ratio Vavg / V0 is equal to 2/3 . (C) The ratio Vrms / V0 is equal to 1/ 2 (D) Three fourth of the total particle has a speed between 0.5 V0 and V0 9.

A U – shaped conducting frame is fixed in space. A conducting rod CD lies at rest on the smooth frame as shown. The frame is in uniform magnetic field B0 , which is perpendicular to the plane of frame. At time t = 0, the magnitude of magnetic field B begins to change with time t as , B  0 , where k is a positive 1  kt

constant. For no current to be ever induced in frame, the speed with which rod should be pulled starting from time t =0 is ( the rod CD should be moved such that its velocity must be lie in the plane of frame and perpendicular to rod CD )

(A) ak

(B) bk

(C) a ( 1 + kt )

(D) b ( 1 + kt)

SECTION – III IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

Matrix – Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. One or more than one may be correct

10.

For the situation shown in the figure below, match the entries of Column-I with entries of Column-II. r q1

q2 Hollow Neutral conductor

Column – I

Column – II

(A)

In the situation shown

(p)

Distribution of charge on inner surface of conductor is uniform

(B)

If outside charge is not present

(q)

Distribution of charge on inner surface of conductor is nonuniform

(C)

If we displace the outside charge while (r) the inside charge remains at centre

Distribution of charge on outer surface of conductor is uniform

(D)

If the inside charge is displaced by (s) small amount from centre then

Distribution of charge on outer surface of conductor is nonuniform

IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

11.

Column-II gives four situations in which three or four semi infinite current carrying wires are placed in xy-plane as shown. The magnitude and direction of current is shown in each figure. Column-I gives statements regarding the x and y components of magnetic field at a point P whose coordinates are P(0, 0, d). Match the statements in column-I with the corresponding figures in column-II. Column – I Column – II (A)

The x component of magnetic field at (P) point P is zero in

y i/3 x

i

i/3 i/3

(B)

The z component of magnetic field at (Q) point P is zero in

y i/2 45 o 45 o

i

x

i/2

(C)

The magnitude of magnetic field at point (R) P is

i/2

0i in 4d

x i i/2

(D)

The magnitude of magnetic field at point (S)

y

i P is less than 0 in 2d i 45 o

45 o

i/2

i/2

x

IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

SECTION – IV Integer Answer Type This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.

12.

In the figure ammeter (I) reads a current of 10mA , while the volt meter reads a potential difference of 3V. What does ammeter (II) ( in mA ) read? The ammeters are identical , the internal resistance of the battery is negligible. ( Consider all ammeters and voltmeters as non ideal ) ( round off the answer to the nearest integer) 100

4V

A A

II

100

13.

I

V

Three identical large metal plates of area A are small at distances d and 2d from each other. Top metal plate is uncharged, while other metal plates have charges +Q and – Q. Top and bottom metal plates are connected by switch S through a resistor of unknown resistance. What energy is dissipated in the resistor when switch is closed? (x 10 4 J) . Find x. ( Given :

0 A  6F, Q  60C) d

A

2d

S

14.

d

A Q

A Q

A uniform square plate of mass m = 100 gm and side a = 24 cm can rotate about a smooth vertical axis passing through one edge It is initially at rest . A particle of mass m =100 gm is moving horizontally and perpendicular to the plane of the plate with velocity u=70 cm/s. It collides with the plate elastically at the centre of the plate. Find the angular velocity ( in rad/s) of the plate just after collision?

15. In the circuit shown S1 and S2 are switches. S2 remains closed for a long time and S1 open. Now S1 is also closed . It is given that R = 10  , L = 1 mH and   3V . Just after S1 is di closed , find the magnitude of rate of change of current ( in ampere/sec) that is , in the dt inductor L is 1000 x. Find x.

IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

16.

Light is incident at an angle  on one planar end of a transparent cylindrical rod of refractive index  . Determine the least value of  so that the light entering the rod does not emerge from the curved surface of rod irrespective of the value of  . Find 2 2 . 

 

17.

90  

An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0  F, and the resulting LC circuit is set oscillating at its natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found that the maximum value of Q is 200 mC. When Q = 100 mC, the value of |dI/dt| is found to be 10x A/s. Find the value of x.

18.

In Searle’s experiment, which is used to find Young’s Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). The maximum possible error in the values of Young’s modulus is x  1010 N/m 2. Find the value of x. Screw gauge and meter scale are free from error.( Round of to nearest integer)

19.

Water at temperature 20o C flows from a tap T into a heated container C. The container has a heating element ( a resistor R) which is generating heat at the rate P, that may be varied. The rate of water in flow from tap is F = 0.021 litres per minute. The heat generated is sufficient so that the water in the container is boiling and getting converted into steam at a steady rate. What is the minimum power P ( in watt) that must be generated as heat in the steady state in resistor R so that the amount of liquid water in the container neither increases nor decreases with time? ( Neglect other losses of heat, such as conduction from the container to the air and heat capacity of container)For water , specific heat c = 4.2 kJ. Kg-1 , latent heat of vaporization Lvap= 2.3 MJ . kg-1, density  = 1000 kg. m-3). Find P/100 . (Round of to nearest integer)

IIT – ian’s PACE education Pvt. Ltd. : Andheri / Dadar/Chembur/ Thane / Nerul / Powai / Kandivali

PART – II SECTION – I Single 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 ONLY ONE is correct.

20.

Which term does not describe  -D-glucopyranose? (A) diastereomers (B) epimers

the

relationship

(C) enantiomers

between

 -D-glucopyranose

and

(D) anomers

21. o

The products of above reaction are (A) phenol and 2-iodopropane (C) iodobenzene & propene

(B) iodobenzene and 2-propanol (D) p-isopropyl phenol

22.

The IUPAC name of Na2[Fe(CN)5(NO)] is (A) Sodium pentacyanonitrosyl ferrate (III) (B) Sodium pentacyanonitrosyl ferrate (II) (C) Sodium pentacyanonitrosyl iron (III) (D) Sodium pentacyanonitrosyl iron (II)

23.

Consider the cell described by the following reaction. Which of the changes listed will cause an increase in the cell voltage H 2(g)  2AgCl(s)  2Ag (s)  2H  (aq)  2Cl (aq) I. increase in H2(g) pressure II. increase in the amount of AgCl(s) III. increase in pH (A) I & III (B) II & III (C) I & II (D) II

SECTION - II IIT – ian’s PACE education Pvt. Ltd. : Andheri/Dadar/Chembur/Thane/Churchgate/Nerul/Powai

Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has 4 choices (A) , (B),(C) and (D) for its answers, out of which ONE OR MORE is/are correct . 24. C6 H 5 H

C

OH + SOCl 2

Pyridine

A + HCl + SO2

H3C

( ) -  -Phenyl ethanol (optically pure) Which of the following statement(s) are true regarding the above reaction. (A) Product ‘A’ has opposite configuration to alcohol (B) Product ‘A’ has same configuration as alcohol (C) Product ‘A’ is leavo rotatory (D) Product ‘A’ is dextro rotatory 25.

Which of the following statement(s) is true? (A) Lyophobic sol shows greater light scattering effect than lyophilic sol. (B) The stability of lyophobic sol is due to presence of charge on sol particles. (C) A catalyst increases the value of rate constant (D) The absolute entropy of an element is zero at 25oC and 1 atm pressure

26.

Pick out correct statements regarding Na2B4O7.10H2O (A) It exists as Na2[B4O5(OH)4].8H2O (B) It contain two tetrahedrons and two triangles (C) Two boron atoms are sp3 hybridised and two boron atomis are sp2 hybridised (D) All boron atoms are sp 2 hybridised

27.

Which of the following statement(s) are correct regarding brown ring com complex [Fe(NO)(H2O)5]2+ (A) oxidation state of iron is + 1 (B) oxidation state of iron is + 2 (C) magnetic momentum (s ) is 3.9 B.M (D) magnetic momentum (s ) is 5.9 B.M

28.

The ions which are planar are (A) CO32 (B) NO32

(C) [I Cl4 ]

(D) ClO3

SECTION - III Matrix - Match Type IIT – ian’s PACE education Pvt. Ltd. : Andheri/Dadar/Chembur/Thane/Churchgate/Nerul/Powai

This section contains 2 questions. Each question constains 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 .

29.

Column – I (A) (B) (C)

(D) 30.

Column – II

Extracted by electrolytic reduction Metal dissolved in acids & bases Metal used as reducing agent in the extraction of Cr and Mn from their oxides. Metal extracted by self reduction Column – I

(A)

(s) Al Column – II

O

H

(p) Pb (q) Mg (r) Cu

O

C

(p) Fehling reagent

CH3

&

(B)

H C OH & H C CH 3 O

O

(C)

H3C C H + H3C C CH3

(r) FeCl3

O

O

(D)

(q) [Ag(NH3)2]+

OH

HO

CH

CH 3

(s) I2 + NaOH

&

C2 H5

SECTION - IV INTEGER ANSWER TYPE IIT – ian’s PACE education Pvt. Ltd. : Andheri/Dadar/Chembur/Thane/Churchgate/Nerul/Powai

This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6 , 0, 9 and 2 respectively, then the correct darking of bubbles will look like the following:

31.

The number of isomeric disubstituted products formed by Borazole (Inorganic benzene) are _______

32.

The normality of 22.4 volume H2O2 is ________

33.

The total number of  and  bonds present in XeO4 are ________

34.

20 ml 5.6 volume H2O2 completely reduces 20 ml of KMnO4 in acid solution. The molarity of KMnO4 _______ x 10-1

35.

The total number of isomers (structural as well as stereo) possible for dichlorocyclopentane are ____

36.

For

37.

If 36.8% of initial amount of a radioactive substance is left after 43.2 minutes, the time taken for 99.9% disintegration of initial amount is ______ hours.

38.

The potential of the cell 0.1MBCl  Cu 2  Pt, H 2 (1atm) Cu is 0.88 V at 25oC 0.2M BOH (1M) If 20 ml of 0.1M BOH is titrated with 0.1M HCl, the pH at equivalence point is ________

sublimation of iodine crystals I 2(s)  I 2(g) at 25oC and atmospheric pressure H  10 K.cal/mole and S  20 cal/K-mole. The temperature at which solid I2 be in equilibrium with gaseous I2 is _________ x 10 2 K

IIT – ian’s PACE education Pvt. Ltd. : Andheri/Dadar/Chembur/Thane/Churchgate/Nerul/Powai

PART III : MATHEMATICS SECTION – I Single Correct Choice Type This section contains 4 multiple choice questions. Each question has 4 options of which ONLY ONE is correct.

39.

 cos x  x   sin x  dx equals x cos x 

 

(A)  x cos x  C

(B)

x sin x  C

(C) 2 x cos x  C

(D) C  2 x cos x

40.

Chord AB of the circle x 2  y 2  100 passes through the point(7,1) and subtends an angle of 60 at the circumference of the circle. If m1 and m2 are the slopes of two such chords then the value of m1m2, is (A) 1 (B) 1 (C) 7/12 (D) 3

41.

Statement 1 : Consider the cubic equation f  x   x 3  nx  1  0 where n  3, n  N , then

f  x   0 has at least one root in (0,1). because Statement 2 : If g(x) is a differentiable function such that g '  x   0 at x  a and x  b and g  a  g  b   0 then the equation g  x   0 has atleast one root in (a,b).

(A) Statement-1 is true, statment-2 is true and statment-2 is correct explanation for statement-1. (B) Statement-1 is true,Statement-2 is true and Statement-2 is not the correct explanation for Statement-1. (C) Statement-1 is true,Statement-2 is false (D) Statement-1 is false,Statement-2 is true Let g : R  R defined by g  x   e x  , where {x} denotes fractional part function. Statement 1 : g(x) is periodic function because Statement 2 : {x} is periodic function. (A) Statement-1 is true, statment-2 is true and statment-2 is correct explanation for statement-1. (B) Statement-1 is true,Statement-2 is true and Statement-2 is not the correct explanation for Statement-1. (C) Statement-1 is true,Statement-2 is false (D) Statement-1 is false,Statement-2 is true 42.

IIT-ian’s PACE, Andheri, Vile Parle, Dadar, Chembur, Thane, Churchgate :26245357 / 5657 / 26245223 / 09 www.iitianspace.com

SECTION – II Multiple Correct Choice Type This section contains 5 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. 43.

Consider the function f : R  R defined as f  x   x  sin x. Which of the following are correct statements ? (A) The function is strictly increasing at every point on R except at ‘x’ equal to an odd integral multiple of  where the derivative of f(x) is zero and where the function f is not strictly increasing. (B) The function is bounded in every bounded interval but unbounded on whole real line. (C) The graph of the function y = f(x) lies in the first and third quadrants only. (D) The graph of the function y = f(x) cuts the line y = x at infinitely many points.

44.

Let f(x) be a non constant twice diffrentiable function defined on R such that

f 2  x  f 2  x 1

and f '    0  f ' 1 . Then which of the following alternatives are correct? 2 (A) f  4   f  8 . (B) Minimum number of roots of the equation f "  x   0 in (0,4) are 4.  4

(C)





45.

2

f  2  x  sin x dx  0.

(D)

 0

4

4

f  t  5cos  t dt   f  4  t  5cos  t dt. 2

 2k  1  1  cos  2k  1  1  cos  4k  1         then which of Let P  k    1  cos 4k   1  cos 4k 4k 4k 









the following is/are correct?

46.

(A) P  3 

3 16

(B) P  4  

2 2 16

(C) P  5  

3 5 32

(D) P  6  

2 3 32

 x2   for x  0 2 f x  If the function f(x) defined as    is continuous but not derivable at  x n sin 1 for x  0  x

x  0 then n can not be

(A) 1/2

(B) 1

(C) 3/2

(D) 2

IIT-ian’s PACE, Andheri, Vile Parle, Dadar, Chembur, Thane, Churchgate :26245357 / 5657 / 26245223 / 09 www.iitianspace.com

47.

A twice differentiable function f(x) is defined for all real numbers and satisfies the following conditions f  0   2;

f '  0   5

f "  0   3. The function g(x) is defined by

and

g  x   e ax  f  x   x  R, where ‘a’ is any constant. If g '  0   g "  0   0 then ‘a’ can be

equal to (A) 1

(B) 1

(C) 2

(D) 2

SECTION – III Matrix – Match Type This section contains 2 questions. Each question contains statements given in two columns, which are 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. 48.

Column I

Column II

4 . One vertex is taken and folded 3

(A)

A rectangle is of width

(B)

over to touch the opposite edge on the rectangle (length not specified). the minimum length of the crease, is The number of solutions of the equation log x 3  5 x  7,

(C)

As ‘x’ range over the interval  0,   , the function

(P)

1

(Q)

2

(R)

3

(S)

4

(T)

5

f  x   9 x 2  173x  900  9 x 2  77 x  900 range over (0,M).

The possible integral value(s) in the range of f(x) is (D)

2

The possible real number ‘k’ for which  x 2  y 2  z 2   k  x 4  y 4  z 4  holds for  x, y, z  R, is

49.

Column I

Column II

e

ln x  ln  x 1 ........ ln 1

 20 is

(A)

The value of the integer x  x  2  satisfying

(B)

1 1 The number of solution of the equation, tan  4  x  cot  x   x  

(C)

(D)

 x  2 !

(P) 1  , is 2

(where [] denotes greatest integer function and {} denotes fractional part function) If the equation x 2  y 2  2 xy  2 gx  2 fy  4  0 represents a pair of real straight lines, then the possible value of ‘g’ can be

(Q)

2

(R)

3

(S)

4

(T)

5

Let A and B are two points in x-y plane which are d units apart where d  5, 6 . The number of lines that are 2 units away from A and 3 units away from B can be

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SECTION – IV Integer Answer Type This section contains 8 questions. The answer to each of the questions is a integer type ranging . The appropriate bubbles below the respective question numbers in the ORS have to be darkened.

50.

Line l is a tangent to a unit circle S at a point P. Point A and the circle S are on the same side of l , and the distance from A to l is 3. Two tangent form point A intersect line l at the point B and C respectively. Find the value of (PB)(PC).

51.

Consider two lines L1 :

x 7 y 7 z 3 x 1 y  1 z  1     . If a line L whose diand L2 : 3 2 1 2 4 3

rection ratios are  2, 2,1  intersect the line L1 and L2 at A and B, then find the distance AB. 52.

1

Let a  3223  1 and for all n  3, let f  n   nC0  a n1  nC1  a n 2  nC2  a n 3  ....   1

n 1

 n Cn 1  a 0 .

If the value of f  2007   f  2008   9k where k  N , then find k.   1   1 x cot     cot 2 If the value of the definite integral 1  1 x   1  x2  1

1

53.

 

x

 2   dx   



a b



c

where a, b, c  N in their lowest form, then find the value of  a  b  c  . 54.

The sides of a triangle have the combined equation x 2  3 y 2  2 xy  8 y  4  0. The third side, which is variable always passes through the point  5, 1 . If the range of values of the slope of the third line so that the origin is an interior point of the triangle, lies in the interval (a,b) then  

find  a 

1  . b2 

55.

Number of ways in which 5A’s and 6B’s can be arranged in a row which reads the same beckwards and forwards, is

56.

If 2x  y 5  y

57.

Find the value of lim x0

1



1 5

then  x 2  1

d2y dy x  ky , then find the value of ‘k’. 2 dx dx

1  cos5 x cos3 2 x cos3 3x . x2

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Paper 2 ( Solution ) 1.

(B) Photons per area per second at a distance r are 5.00  1018/4 r 2 . Photons per second entering the eye, radius R is then this times R 2 . Set this product equal to 500 per second and solve for r. The result is [B].

2.

(C) 

(O)

(O)

R R

 R S

( iR)

i

 3R

VS    iR   

3.

 4  VS  3 3

(A)

mv 2  qvB r



r

2m(KE) 2mqV mv 2mv 1    qB qB qB q B

R1 M1  R2 M2 4.

M1 R 12  M 2 R 22



(B) (m1 + m2) g sin   2f3 fs

N f2 fs

N f1



N + m1 g sin   f1

m 2g sin   N  f 2

m 2g sin   N  N  m1g sin  m 2  m1 5.

(C) Flux passing through the cone will remain same for both Thus, 1  2

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 q1

or

q1 q (1  cos )  2 (1  cos ) (using solid angle) 2 0 2 0 q1 sin 2

6.

q1

   q 2 sin 2 2 2

(ABC) In the initial state, charge on each capacitor is shown in figure-1 1.5F

360C

2F

3F

300C

figure  1

q

1.5F

300  q

360  q 3F

q

A

2F

figure  2

Let charge q flow anticlockwise in the circuit before it achieves steady state as shown in figure.-2. Applying KVL to figure 2.



(360  q) q (300  q)    0 or 180 C 3 1.5 2

 final charge on 1.5 F capacitor is q = 180 C and final charge on 2 F capacitor is 300- q = 120 C

7.

(B,D)The x – component of velocity being parallel to magnetic field , shall remain unchanged. The component of velocity perpendicular to x – axis will always have magnitude Voy, and at any time t it shall make an angle  = t with y – axis as shown . z

  t

y

 v oy

 y – component of velocity is V oy cos t and z – component of velocity along negative z – direction

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at any time t is Voy sin t . Where   8.

qB m

(a,b,c,d) Area under the curve is equal to number of molecules of the gas sample. Hence

N

1  a  V0  aV0  2N 2 

Vavg 



1 1 V N(V)dV   N0 N

V0

 a

 V   v 0

0

  V  dV 

V 2 2 V0  avg  3 V0 3 V

2 Vrms 



  V02 1 1 0 2 a 2 V N(V)dV  V  V dV    N 0 N 0 2  V0 

Vrms 1  V0 2

Area under the curve from 0.5 V0 to V0 is

3 of total area. 4

9.(A) The magnetic flux must remain constant

  m  B0 ab 

B0 bx 1  kt

 B0

C

b x

D

Where x is as shown

 x = a (1 + kt) or v =

dx  ak dt

10.

(A – p,s ; B – p, r ; C – p, s; D – q, s)

11.

(A – p,q,r ; B – p,q,r,s ; C – r ; D – p,q,r,s) IIT – ian’s PACE Education Pvt. Ltd. ; Andheri/Dadar/Chembur/Thane/Nerul / Powai/ Kandivali

(P) i/3 i/3

i i/3



not is downward



0i 0i / 3  4d 4d



0i ( i) 3d

(Q)

0 i / 2 4 d

 0i / 2 4 d

 0i 4d

i  i 2  0 cos 45   0  8d  4d





0 i  1  1   ( j) 4d  2

(R) 0 i / 2 4d

 0i / 2 4d

 0i 4d

 0i (  j) 4 d

 (S)

 0i

0i / 2 4d

4 2 d

  0i / 2 4d

 0i 4d

 0i 4 2 d 12.

( j) 

 0i 4 d

 0i ( j) 4d

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4  V1 V1  1 V1  0   ...(1) 100 RA 100

1V – 0V = ( 10 mA) RA RA = 100

4  V1 V1  1 V1  0   ( By using equation (1) and (2) 100 10Q 100 V1 = 5/3 V

V1  1  ( current in ammeter (II) RA 13.

(1) u i 

5 / 3 1  6.67mA = 7 100

Q 2d 2 0 A

then VC = VA ( At equilibrium condition)

VAB = VBC

2

Uf 

Q12d  Q  Q1  2d Q2d   0 A  0A 30 A

Q1d Q  Q1 (2d)  0 A 0 A Q1 = 2(Q –Q1) 3Q1 = 2Q

Uf 

Q12 d  Q  Q1  2d Q 2 d    0A  0A 3 0 A 2

u  14. (5)

Q 2d Q 2d Q2d  60      0.1mJ . So x = 1 20 A 30 A 6 0A 6(6) The plate is free to rotate about vertical axis yy’. y

y

v cm

u C

y'



C

m

v m

y'

Let v, vcm and  be the velocity of particle, velocity of centre of mass of plate and IIT – ian’s PACE Education Pvt. Ltd. ; Andheri/Dadar/Chembur/Thane/Nerul / Powai/ Kandivali

angular



velocity of plate just after collision. From conservation of angular momentum about vertical axis passing though O

is

mu

a a ma 2  mv   2 2 3

… (1)

Since the collision is elastic, the equation of coefficient of restitution is e

But

v cm  v 1 u

v cm 

… (2)

a 2

… (3)

Solving equation (1), (2) and (3) we get 

12 u = 5 rad/s 7 a

15.(2) When S2 is closed current in inductor remains, i 



  V1   V1    R 2R 2R S1

2    V1   3   R

2R

v1





i

0

 2R

0



Potential difference (V)   

And

L

dt 2  dt 3

 2R



0

2   Ans. 3 3

di 2 = 2000 ; X = 2  dt 3L

16.(4) The light entering the rod does not emerge from the curved surface of the rod when the angle 90 – r is greater than the critical angle. i.e.,  

1 where C is the critical angle. sin C

Here C = 90 – r

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 

1 1  sin(90  r) cos r

As a limiting case  

1 cos r

… (i)

Applying Snell’s law at A



sin  sin   sin r  sin r 

… (ii)

The smallest angle of incident on the curve surface is when  

 . This can be taken as a 2

limiting case for angle of incidence on plane surface. Form (ii) sin r 

sin  / 2 1   sin r

… (iii)

Form (i) and (ii) sin r = cos r 

r = 45o







2 2  4

1 1   cos 45 1/ 2

This is the least value of the refractive index of rod for light entering the rod and not leaving it form the curved surface. 17. Here Q0 = maximum value of Q = 200  C = 2  10-4 C



1 1   10 4 s 1 Let at t = 0, Q = Q0 then 3 6 LC (2 10 H) (5.0 10 F)

Q(t) = Q0 cos t

I(t) 

dQ  Q0  sin t dt

… (1)

… (2)

L  2.0mH

dI(t)  Q0 2 cos(t) dt

From (1) 100 = 200 cos t

… (3)

 Q  For Q  100C  or 0   2  1 or cos(t)  , From equation (3): 2

C  5.0F

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dt 1  (2.0 10 1 C) (10 4 s 1 ) 2   dt 2

18. (1)

dI  10 4 A / s But dt

Maximum percentage error in Y is given by Y 

 Y   D  X L     2  L  Y max   D  X  0.001   0.001   0.1   2     0.0489  0.05   0.125   110  So maximum percentage = 4.89% W  50N; D  0.05cm;  0.05  10 2 m; L = 110 cm = 110 x 10 -2 m = 2.24 x 10 11 N/m2

 11 109 N / m 2 19.

Y

dI  10 x A / s  x = 7 dt

W L  D 2 X 4

It is given that X = 0.125 cm = 0.125 x 10 -2 m ;

50  4  110  10 2 3.14 0.05 102  0.125  10 2



 



11

Y  0.0489  2.24 10  1.09 1010 N / m 2  x = 11

Rate of water coming from Tap T is 0.021 x 10 -5 m3 / minute

2.110 5 3 2.110 5 3 F m / sec F  m / sec 60 60 same amount of water should be ejected in from of steam. So heat supplied must be

Q  mST  mL or Rate of heat transfer

Q m m  ST  L t t t

P  FST  FL

P=

 m    F    t 

1000  2.1105 [4.2  100  20   2300]kJ / sec 60

P = 922.6 (J/sec. or watt). P/100 = 9.22 = 9

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20. C

21. A

27. A,C

Paper – II Solution 23. A 24. A

22. B

25. A,B,C

26. A,B,C

28. A,B,C

29. A – q,s; B – p,s; C – s ; D – p,r 31. 4

32. 4

30. A-q,s; B-p,q,s; C-p,q; D-r

33. 8

34. 2 N1V1 = N2V2 1  20 = 20  N2 N2 = 1 = 5  M  M = 0.2 = 2  10-1 35. 7 Cl Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

36. 5 At equilibrium G  H  TS = 0 H 10  1000 T   500K  5  102 K S 20 43.2 37. 5 Average life T = 1.44  t1/2 =  0.72 h 60 0.72 1 t1/ 2   h 1.44 2 Time for 99.9% disintegration 10  t1/2 1 = 10   5 h 2 38. 5 0.88 = 0.34 – ( 0.059  pH )

pH =

0.54 9 0.059

5 = pKb + log

0.1 0.2

pKb = 5.3

1 pH = 7  5.3  log 5  102  2

5 = pKb  0.3 = 7

1 5.3  0.7  2 2

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=5

Solution Paper # 2 39. (C)

40.

(A)

41.

(B)

42.

(D)

43.

(B,C,D)

44.

(A,B,C,D)

45.

(B.C)

46.

(C,D)

47.

(A,D)

48.

(A)-R , (B)-Q , (C)-P,Q,R,S,T , (D)- R,S,T

49.

(A)-T,(B)-Q,(C)-Q,R,S,T,(D)-R,S

50.

(3)

51.

(18)

52.

(2187)

53.

(7)

54.

(24)

55.

(10)

56.

(25)

57.

(22)

IIT-JEE 2010

A.I.T.S # 1 (Answers Key)

Date : 03/01/2010

ANSWERS KEY Paper I Que. Ans. Que. Ans.

1 B 11 ABCD

2 C 12 A

3 D 13 B

Que. Ans. Que. Ans.

21 B 31 A,B

22 C 32 A,B

Que. Ans. Que. Ans.

41 A 51 A,C

42 C 52 A,B,C

4 D

14 C

5 D 15 B

6 D 16 B

7 D 17 A

8 B 18 A

9 A,B,C,D 19 A-q; B- r; C-s

23 A 33 B

24 D 34 C

25 B 35 A

26 C 36 A

27 B 37 B

28 D 38 A

43 C 53 A

44 B 54 C

45 A 55 B

46 A 56 A,D

47 A 57 A,D

48 D 58 A,C,D

29 A,B 39 A-p,s; B-p,q,r C-p,q; D-p,s 49 B,C 59 A-s; B-r; C-q; D-p

10 A,C 20 A-p,q; B-r,s; C-p,q 30 A,B,C 40 A-q,r,s; B-p; C-q,r,s; D-q 50 A,B,D 60 A-r; B-t; C-q;D-q

Paper II Que. Ans.

1 B

2 C

3 A

4 B

5 C

6 A,B,C

7 B,D

8 A,B,C,D

9 A

Que. Ans.

11 A-p,q,r; Bp,q,r,s; C-r; D-p,q,r,s 20 C

12 7

13 1

14 5

15 2

16 4

17 7

18 1

19 9

21 A

22 B

23 A

24 A

25 A,B,C

26 A,B,C

27 A,C

28 A,B,C

30

31 4

32 4

33 8

34 2

35 7

36 5

37 5

38 5

39 C

40 A

41 B

42 D

43 B,C,D

44 A,B,C,D

45 B,C

46 C,D

47 A,D

49

50

A-t; B-q; C-q,r,s,t; D-r,s

3

51 18

52 2187

53 7

54 24

55 10

56 25

57 22

Que. Ans. Que. Ans. Que. Ans.

Que. Ans.

A-q,s; Bp,q,s; C-p,q; D-r

10 A-p,s; B-p,r; C-p,s; D-q,s

29 A-q,s; B-p,s; C-s; D-p,r

48 A-r; B-q; C-p,q,r,s,t; Dr,s,t

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