all india internal test series chemistry, mathematics

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CHEMISTRY, MATHEMATICS & PHYSICS SET – A

Time Allotted : 3 Hours Maximum Marks: 300

INSTRUCTIONS

Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results. A. General Instructions

Attempt ALL the questions. Answers have to be marked on the OMR sheets. This question paper contains Three Sections. Section – I is “Chemistry”, Section – II is “Mathematics” and Section – III is “Physics”. Each Section is further divided into three Parts: Part – A, Part – B & Part – C. Rough spaces are provided for rough work inside the question paper. No additional sheets will be

provided for rough work. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices,

in any form, are not allowed. B. Filling of OMR Sheet

1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR

sheet. 2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment

No. and write in ink your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers. C. Marking Scheme For All Three Parts.

(i) PART-A (01 – 04) contains 4 Multiple Choice Questions which have Only One Correct answer. Each

question carries +5 marks for correct answer and –3 marks for wrong answer. PART-A (05 – 10) contains 6 Multiple Choice Questions which have One or More Than One Correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer.

(ii) PART-B (01 – 04) contains 4 Matrix Match Type Question which have statements given in 2 columns. Statements in the first column have to be matched with statements in the second column. There may be One or More Than One Correct choices. Each question carries +8 marks for all correct answer however for each correct row +2 marks will be awarded. There is no negative marking.

(iii) PART-C (01 – 06) contains 6 Numerical Based questions with Single Digit Integer as answer, ranging from 0 to 9 and each question carries +4 marks for correct answer and –1 mark for wrong answer.

Name of Candidate :

Batch ID : Date of Examination : / / 2 0 1

Enrolment Number :

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

You are not allowed to leave the Examination Hall before the end of the test.

CL

AS

S X

II

Ai2

TS-6 080163.1 APT - 3

ALL INDIA INTERNAL TEST SERIES

Ai2TS – 6 ( XII ) | SET – A | APT – 3 |

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942

Website: www.fiitjee.com, Mail : [email protected]

SSEECCTTIIOONN-- II ((CCHHEEMMIISSTTRRYY))

PART – A

Straight Objective Type This section (01 - 04) contains 4 multiple choice questions which have only one correct answers. Each question

carries +5 marks for correct answer and – 3 mark for wrong answer.

1.

NH2

p q rs

CF3CH3 C

CH3

FCH3

F

Planar species of above is/are___ (a) p,q (b) p,q,r (c) p,q,r,s (d) q,s

2. 200 ml of 0.15 M H3AsO4 (given 1 2 3

3 7 12

a a aK 10 K 10 K 10 ); 50 ml of 0.6 M Na2HAsO4 solutions and

100 ml of 0.1M NaOH are mixed. pH of the final solution is (a) 6.3 (b) 7.0 (c) 6.7 (d) can not be found.

3. 2

2 3 2( )H O

Mg C Mg OH A

red hot iron tubeA B

2

3 /H O HgA C

2 4.Conc H SOC D

/(3 ) NaOHC moles E

Then the correct option is.... (a) B and E are same (b) C and D are same (c) B and D are same (d) D and E are same 4. A container having two partition of equal volume with a separator contains excess of CaCO3(s) in one partition

and BaCO3(s) in other partition. First partition contains He and CO2 with total pressure 8 atm and mole fraction of He as 0.50 while second partition contains Ne and CO2 with total pressure of 8 atm and mole fraction of Ne as 0.25. Temperature of the container is 600 K at which following equilibriums are established.

600

3( ) ( ) 2( )

K

s s gCaCO CaO CO

600

3( ) ( ) 2( )

K

s s gBaCO BaO CO

If the separator is removed then total pressure inside the container will be___(neglect volume change due to solid compounds) CaCO3(s) BaCO3(s)

Seperator

He + CO2 Ne + CO2

(a) 8 atm (b) 13 atm (c) 9 atm (d) 7 atm

Straight Objective Type This section (05-10) contains 6 multiple choice questions which have one or more than one correct answers.

Each question carries +4 marks for correct answer and – 1 mark for wrong answer.

5. If 50 litre flask containing one mole of N2 and one mole of PCl5 is connected to a 50 litre flask containing 2

moles of PCl5, both vessel heated to 227ºC. The equilibrium pressure is found to be 2.05 atm. Assuming ideal

behaviour and for the reaction PCl5 PCl3 + Cl2 correct option(s) is/are

(a) degree of dissociation of PCl5 is1

3

(b) KP for the reaction is 0.205 atm

(c) One mole of PCl5 is present in each flask (d) 0.5 moles of all gases are present in each flask.

Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 1

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942

Website: www.fiitjee.com, Mail :[email protected]

6.

C

H

CH3

DC

D

HH

2 3Br /CH OHP (Mixture of compounds)

P May be :

(a) CH3

D H

O H

C H

D

Br

CH3

(b) CH3

H D

H O CH3

Br

D

H

(c) H

CH3 D

H O

C H

D

Br

CH3

(d) CH3

D H

O H

Br

CH3

D

H

7. Ashutosh collected 1 ltr water in a jar. Kaustub mixed 0.2 mol of H2S in it while Ananya mixed 0.1 mol of

NaOH in it. Arav brought a big piece of NiS and put it in the solution. After long time, some FIITJEE students are observing the solution. Correct observation(s) made by them is/are

( Given Ka1 (H2S) = 10-5

Ka2 (H2S) = 10-8

Ksp (NiS) = 10-20

) (a) Manish found that Solubility of NiS was increased in that solution from normal solubility (ie s = 10

-10 M) due to reaction of S

2- ion with H

+ present in the solution.

(b) Nischay found pH of the solution to be 5 (c) Shivang said, “Concentration of [S

2-] is 10

-4 M”

(d) Utkarsh said, “Concentration of Ni2+

is 10-10

M” 8. Which is ordered correctly

(a)

> heat of hydrogenation

(b) >

C heat of hydrogenation

(c) <

CH3 heat of combustion

(d) > heat of hydrogenation

9. Consider following reaction and find correct option(s) regarding product(s)

C8H15Cl3

All possible compounds

2 /Cl h

CH3

CH3

CH3

CH3

ClCl

(a) Number of optically active compound formed is 10 (b) Number of pair of enantiomers is 5 (c) Number of pair of diastereomers formed is 4 (d) Number of optically inactive compounds formed is 2 10. Consider following equilibriums :

p1

K

A(s) B(g) C(g)

p2

K

E(s) B(g) D(g)

p3

K

P(s) B(g) Q(s)

Pressure above the solids A(s) , E(s) and P(s) when kept alone are 4 atm, 8 2 atm and 5 atm respectively at

temperature T(K). When three solids A(s),s E(s), P(s) are kept together at the same temperature T(K) then. (a) Partial pressure of B(g) is 5 atm (b) Partial pressure of C(g) is 4/5 atm (c) Partial pressure of D(g) is 32/5 atm (d) Total pressure is 12 atm.




SECTION – B

Matrix Match Type This section (01 – 04) contains 4 Matrix Match Type Question which have statements given in 2 columns.

Statements in the first column have to be matched with statements in the second column. There may be One or More Than One Correct choices. Each question carries +8 marks for all correct answer however for each correct row +2 marks will be awarded. There is no negative marking.

1. Match the Column-I with Column – II (where = degree of dissociation, Mmix = average molecular weight of

mixture ; P = equilibrium pressure)

COLUMN – I COLUMN – II

(a) (g) (g) (g)A B C H ve (p)

p

p

K

4P K

(b) (g) (g)D 2E Energy (q) Yield of product(s) will be high at low temperature

(c) 5(g) 3(g) 2(g)PCl PCl Cl (r) Yield of product(s) will be high at high temperature

(d) 2 4(g) 2(g)N O 2NO (s) reac tantmix

MM

1

(t) Yield of product(s) will be high at low pressure

2.

Column I (Reactions) Column II (Product(s))

(a) Borax P H2B4O7

(b) 2Borax H O Q H3BO3

(c) Borax NaOH R Used in glass formation

(d) Borax HCl S NaBO2

T B2O3

3. Match the Column-I with Column – II

Column – I (Mixing equal volume for H2S Ka1 = 10-6

Ka2 = 10-9

) Column – II

(a) 1 M H2S + 1 M Na2S (P) pH =7.5

(b) 2 M H2S + 1 M Na2S (Q) pH = 6

(c) 1 M H2S + 2 M Na2S (R) pH = 9

(d) 1 M Na2S + 2 M HCl (S) Buffer

(T) pH = 3.15

4.

COLUMN – I COLUMN – II

(a) cis2-butene Br /CCl2 4 (p) Cyclic intermediate

(b) Trans-2-butene cold

alk.KMnO4 (q) Recemic mixture

(c) Trans-2-butene Br /CS2 2 (r) Meso-isomer

(d) Cis-2-butene HOCl

(s) Enantiomeric pairs

(t) Product/s mixture is/are optically inactive

PART – C

(Integer Type) This section contains 6 questions. The answer to each question is a single-digit integer, ranging from 0 to

9. The correct digit below the question number in the ORS is to be bubbled. Each question carries +4 marks for correct answer and –1 mark for wrong answer.

1. Maximum pH that must be maintained in a saturated H2S (0.1 M) to precipitate CdS but not ZnS is x and

concentration of Cd2+

ion at that pH is y × 10–7 then x + y =....

given [Cd2+] = [Zn2+] = 0.1 M initially ?




Ksp (CdS) = 8 × 10–27

Ksp (ZnS) = 1 × 10–21

Ka1 (H2S) = 1.0 × 10–8 ,Ka2 (H2S) = 1.0 × 10–13

2.

One molecule of silicate mineral containing above anion has x number of atoms of a metal M, y number of atoms of Potassium and z number of atoms of Calcium as cations.

If x2+y

2+2z

2 = 2yz + 2x + 4z – 5 then oxidation state of metal M is___

3.

Cl H

HCl P (Number of Products)

CH3 H

4ICl /CClQ (Number of Products)

CH3 H

4Cold Dil Alk KMnOR (Number of Products)

Value of P + Q + R is ______________

4. The equilibrium composition in 1 litre container for the reaction is -

PCl3 + Cl2 PCl5

0.20 0.08 0.40 moles/litre

0.22 mole of Cl2 is added at same temperature and new equilibrium is established and If sum of new

equilibrium concentration of PCl3, Cl2 and PCl5 is p then value of p is_______

5. 3excessof NH strong heating

2 6 3 low temperatureB H NH P Q(anaromaticcompound)

Number of bond between B and N in cation part of P is x, in anion part of P is y and in compound Q is z then the sum x+y+z =….(count only sigma or coordinate bond)

6. Number of diastereomer pairs that can be grouped from the products formed by mono chlorination of following

compound is...

CH3

CH3

2/Cl h products




SSEECCTTIIOONN-- IIII ((MMAATTHHMMEETTIICCSS))

PART – A

Straight Objective Type This section (01 - 04) contains 4 multiple choice questions which have only one correct answers. Each

question carries +5 marks for correct answer and – 3 mark for wrong answer.

1. The normal chord to the parabola y2 = 8x at the point (8, 8) subtends an angle at the focus, where is :

(a) 4

(b)

3

(c)

2

(d)

6

2. The locus of the point of intersection of tangents to an ellipse at two points, sum of whose eccentric angle is

constant is a part of a/an : (a) ellipse (b) hyperbola (c) parabola (d) straight line 3. Equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one if of its

asymptotes is 3x – 4y – 6 = 0 and the equation of other asymptote is ax + by + c = 0, (where a, b, c N ), then the minimum value of a + b + c is

(a) 10 (b) 24 (c) 22 (d) 28 4. If the slope of one of the lines represented by the equation ax

2 + 2hxy + by

2 = 0 be k times that of the other,

then : (a) 4kh = ab(1 + k) (b) kh = ab (1 + k)

2 (c) 4 kh

2 = ab(1 + k)

2 (d) None of these



5. If y = 3x + 1 is a tangent to the hyperbola 2 2

2

x y1

8p , then which of the following cannot be the sides of a

right angle triangle ?

(a) 4p , 10,3 (b)

4p ,10, 3 (c) 4p ,10,3 (d) 4p , 10,3

6. If 2dyysin(x) cos(x)[sin(x) y ]

dx

and 2

y at x3 2

, then the value of y at x

3

is NOT equal to

(a) 1/ 23 (b)

1/ 43 (c) 1/ 43

(d) 1/ 23

7. If the length of intercepts made on the line passing through origin and x + y = 1 by the circle x

2 + y

2 = x – 3y

are equal, then the above mentioned line can be (a) x + y = 0 (b) x – y = 0 (c) x + 7y = 0 (d) x – 7y = 0 8. If the line y = 2x + a lies between the circles x

2 + y

2 – 2x – 2y + 1 = 0 and x

2 + y

2 – 16 x – 2y + 61 = 0 without

intersecting or touching either circle, then a can be (a) – 2 (b) – 4 (c) – 6 (d) – 8

9. If the line x = P divides the area of the region 3S (x,y) R R; x y x,0 x 1 into two equal parts,

then

(a) 1

0 P2

(b) 1

P 12 (c)

4 22P 4P 1 0 (d) 4 2P 4P 1 0

10. If the curve passing through (0, 1) and satisfying the differential equation dy y

dx (2y n(y) y x)

is y = f(x),

then f(e) is greater than (a) e (b) 1 (c) (d) 2




SECTION – B



1. Match the following :

Column – I Column – II

(a) 4y = 5x is a secant to the curve(s) (P) x2 + y

2 = 1

(b) The curve passing through (1, 0) and satisfying

2 dy[1 y ] xy is /are

dx

(Q) x2 – y

2 = 1

(c) The curve for which the lines x2 = 1 are tangent is/are (R) 2x

2 – y

2 = 2

(d) 50,

4

is an exterior point for (S) 2x

2 + y

2 = 2

2. For the parabola 2

2 2 (3x y 7)(x 2) (y 3)

10


(a) The equation of line about which parabola is symmetrical (P) 3x – y + 7 = 0

(b) The equation of line along which minimum length of focal chord occurs

(Q) 3x – y – 3 = 0

(c) The locus of point of intersection of perpendicular tangents (R) x + 3y – 11 = 0

(d) The locus of foot of perpendicular from focus upon any tangent (S) 3x – y + 2 = 0

3. Match the following


(a) If the circles x2

+ y2 = b

2 and x

2 + y

2 + 2gx + 2fy + ac = 0 cut each

other orthogonally, then (P) ac > 0

(b) If the circles x2 + y

2 + 2ax + b = 0 and x

2 + y

2 + 2cx + b = 0 touch

each other (a c)

(Q) b2 = ac

(c) If in (b), the first circle lies completely inside the second circle, then (R) b > 0

(d) If the chord of contact of the tangents drawn to x2 + y

2 = b

2 from any

point on x2 + y

2 = a

2, touches the circle x

2 + y

2 = c

2 then

(S) b = 0

(T) | b | 0

4. Match the following


(a) The area bounded by 2x y and y x 2 is (P) -12

(b) The sum of the distance of any point on the ellipse 3x2 + 4y

2 = 12

from its directrix is (Q) 9/2

(c) The value of c for which 3x2 – 5xy – 2y

2 + 5x + 11y + c = 0 are the

asymptotes of the hyperbola 3x2 – 5xy – 2y

2 + 5x + 11y – 8 = 0 is

(R) 8

(d) If e is the eccentricity of

2 2

2 2

x y1

a b and is the angle between the

asymptotes then 8ecos2

equals

(S) 386

88




PART – C (Integer Type)

This section contains 6 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. Each question carries +4 marks for correct answer and –1 mark for wrong answer.

1. A curve y = f(x) passes through origin and slope of tangent line at any point (x, y) of the curve is 4

2

x 2xy 1

1 x

,

then the value of greatest integer which is less than or equal to f(-1) is 2. If the equation of the directrix of the parabola which touches x-axis and y-axis at (2, 0) and (0, 3) respectively,

is ax + by + c = 0, (a, b, c are integers) then the minimum value of |a + b + c| is

3. If the differential equation corresponding to the family of curves y = (A + Bx)e3x

is given by 2

2

d y dya by

dxdx ,

then (a b)

5

equals

4. Let A be a point on 2 2(x 2) y

116 12

, B and C be its foci. If the locus of the incentre of ABC is an ellipse with

eccentricity e, then 9e2 is equal to

5. If (x, y) satisfies the equation x2 + y

2 – 6x – 8y + 21 = 0 then the minimum value of 2 21

(x y 1) (x y)2

is

equal to 6. Chords of the hyperbola x

2 – y

2 = a

2 touch the parabola y

2 = 4ax. If the locus of their middle points is the curve

ym(x - a) = x

n, then m +n equals




SSEECCTTIIOONN-- IIIIII ((PPHHYYSSIICCSS))

PART – A

Straight Objective Type This section (01 - 04) contains 4 multiple choice questions which have only one correct answer. Each

question carries +5 marks for correct answer and – 3 mark for wrong answer.

1. A variable resistance is connected across a non ideal cell of constant emf and constant

internal resistance. Init ial value of variable resistance is R. Now the magnitude of variable resistance has been made n times, that is, its new value is nR. Surprisingly the new value of power dissipated in variable resistance is stil l same as its init ial value. The value of internal resistance of cell is

(A) 1

2

nR

(B) R n (C)

21

2

nR

(D) R nR

2. A current I flows through a cylindrical rod of uniform cross-section area A and resistivity . The electric flux

through the shaded cross-section of rod as shown in figure is :

(A) I

ρ (B)ρI (C)

ρI

A (D)

ρA

I

3. In the given potentiometer circuit, the resistance of the potentiometer wire AB is 0R . ‘C’is a cell whose internal

resistance r is separately shown. The zero centered galvanometer ‘G’ does not give zero deflection for any

position of the jockey J. Which of the following cannot be a reason for this?

A B

C G

RD

+ -

J

r

(A) r > 0R

(B) R >> 0R

(C) emf of cell C > emf of cell D

(D) The negative terminal of C is connected to A.

4. A unit positive point charge of mass gm is to be projected inside the tunnel

in sphere of radius 4m (as shown in figure), with a velocity of 5 m/sec. The

tunnel has been made inside a uniformly charged 3( 1c / m ) non

conducting sphere. The minimum velocity with which the point charge should

be projected such that it can reach the opposite end of the tunnel, is equal to

(a) 1.0 × 106 m/sec

(b) 12 × 106 m/sec

(c) zero, because the initial and the final points are at same potential (d) 4.8 × 10

6 m/sec

shaded cross section






5. In the given electric circuit which of the following statements are correct (Assume that the given capacitors are

uncharged before they are connected in the circuit and the batteries of negligible internal resistance)

24V

12V

6V 2 F

1 F

3 F

a b

A) 16a bV V V B) Charge on the 2 F capacitor is 20 C

C) Charge on the 1 F capacitor is 4 C D) Charge on the 3F capacitor is 24 C

6. A ring rolls without slipping on the ground. Its centre C moves with a constant sped u. P is any point on the ring. The speed of P with respect to the ground is v, then

A) 0 v 2u

B) v = u, if CP is horizontal

C) 6v u , if CP makes an angle of 300 with the horizontal

D) 2v u , if CP is horizontal

7. A ladder AB is supported by a smooth vertical wall and rough horizontal floor as shown. A boy starts moving

from A to B slowly. The ladder remains at rest, then pick up the correct statements(s):

a) Magnitude of normal reaction by wall on ladder at point B will increase. b) Magnitude of normal reaction by wall on ladder at point B will decrease. c) Magnitude of normal reaction by floor on ladder at point A will remain unchanged d) Magnitude of friction force by floor on ladder at point A will increase

8. A solid uniform sphere is connected with a moving trolley car by a light spring. The trolley car moves with an acceleration a. If the sphere remains at rest relative to the trolley car, then :

a

(A) Spring force = ma (B) Friction between the sphere and trolley car is equal to zero

(C) Friction between the sphere and trolley car is equal to ma

2

(D) Spring force is equal to ma

2

9. As shown in the figure find the correct statement/s :

(a) i at instant switch is closed is 2A (b) i after long time is 2A

(c) charge on 4 f after long time is 16

c3 (when switch is closed)




(d) charge on 2 f at instant switch is closed is 8 c

10. Initially switch is open. After some time switch is closed. Choose the correct statements (a) the charge flown through the switch is 54 c

(b) Point b is at higher potential than point a before the switch is closed (c) the potential of point b is 6v w.r.t. ground, after s is closed (d) the potential of point b is 12v, after s is closed

SECTION – B



1. A ring of mass m and radius R is placed on a rough inclined plane so that it rolls without slipping. Match the following table.

Column I Column II

(A) Linear acceleration of centre of mass (p) Is directly proportional to m

(B) Angular acceleration (q) Is inversely proportional to m

(C) Rotational kinetic energy at any instant (r) Is directly proportional to R2

(D) Translational kinetic energy at any

instant (s) Is directly proportional to R

3

(t) None of these

2. In each situation of column–I a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface

as shown. At t = 0 (initially) the angular velocity of disc is o and velocity of centre of mass of disc is vo (in

horizontal direction). The relation between vo and o for each situation and also initial sense of rotation is given for each situation in column–I. Then match the statements in column–I with the corresponding results in column–II

Column-I Column-II

(A)

(P) The angular momentum of disc about point A (as shown in figure) remains conserved.

(B)

(Q) The kinetic energy of disc after it starts rolling without slipping is less than its initial kinetic energy

(C)

(R) In the duration disc rolls with slipping, the friction force on disc acts backward.

(D)

(S) Before rolling starts acceleration of the disc remain constant in magnitude and direction

(T) Final angular velocity is independent friction coefficient between disc and the surface




3. A galvanometer has a resistance of 100 and a full scale range of 100 A . It can be used as a higher range

ammeter or a voltmeter by providing a resistance. Match the statements of the two columns :

Column - I Column - II

(a) 25 V range with (P) 200K resistance in series

(b) 20 V range with (Q) 10K resistance in series

(c) 5 mA range with (R) 1 resistance in parallel

(d) 10 mA range with (S) None of these

(T) 2 resistance in parallel

4. Four capacitors with capacitances 1C 2 F ,

2C 4 F , 3C 4 F ,

4C 2 F are connected as shown in

figure and are connected to a 12 V source. Charge flow battery Eq and switch is

Sq .

List I List II

(A) Switch S1 is closed value of Eq (P) 32 C

(B) Switch S1 is closed value of Sq (Q) 4 C

(C) Switch S2 is closed value of Eq (R) 48 C

(D) Switch S1 is closed value of Sq (S) 12 C

PART – C

(Integer Type) This section contains 6 questions. The answer to each question is a single-digit integer, ranging from 0 to

9. The correct digit below the question number in the ORS is to be bubbled. Each question carries +4 marks for correct answer and –1 mark for wrong answer.

1. Two identical parallel plate capacitors are placed side by side with a small gap between them as shown in fig.

V1 and V2 are potential difference across the two capacitors respectively. An electron is projected from the end

of upper plate of the left capacitor with velocity 0v towards right as shown. Charge and mass of electron ‘e’ and

‘m’ respectively. V2 is applied such that the electron just grazes the lower plate of right capacitor horizontally while coming out of it. Assume that there is no gravity and no collision of electron with any plate. What is the

ratio 2

1

V

V? (neglect the time taken by electron from one capacitor to another capacitor.)

C1 C2

C3 C4

S1

S2

E1




2. What amount of heat in micro joule will be generated in the circuit , after the switch is shifted from position 1 to

position 2 ? 0C C 2 F,E 6 V

C

1 2

E

C0C

3.

In the adjacent figure a uniform rod of length 5 3m and mass 2 kg is kept at rest in horizontal position on an

elevated edge. Friction is sufficient to prevent slipping between rod and elevated edge. The value of x (consider

the figure) is chosen such that the rod will have maximum angular acceleration , as soon as it is set free. Find

the maximum angular acceleration of the rod (in rad / s2).

4. A battery of emf E and internal resistance r is connected with external resistance R. Resistance R can be adjusted to any value greater than or equal to zero. Graph is plotted between the current passing through the resistance (I) and potential difference (V) across it. Internal resistance of the battery ( in

) is

I(amp)

V(volt)

10

2

5. Two identical intersecting spheres having the same volume charge density

of magnitude 0 (in S.I. unit) opposite in nature of charge, are shown in

figure. Radius of spheres are 5cm. The distance between the centres is

6cm. Then the magnitude of electric field in the shaded portion is 0E

100, then

E0 will to (In S.I. unit)

6. A disc of radius R and mass M is placed on smooth horizontal

surface as shown in figure. A light rod of length is hanged from

the centre of disc and a small mass m is attached at the end as

shown in figure. Now a velocity V0 is given to mass m. The

maximum height which mass m can attain. 0[v 8g, M 3m]

( Assume rod can rotate < 90)

M

R

m v0



CHEMISTRY, MATHEMATICS & PHYSICS

CLASS XIIth

ANSWER KEY

Chemistry ( Section – I ) Mathematics ( Section – II ) Physics ( Section – III )

1 D C111303 1 C M110904 1 B P121517

2 A C110504 2 D M111029 2 B P121411

3 C C111711 3 B M111123 3 A P121515

4 C C110403 4 C M110730 4 B P120104

5 ABC C110402 5 BC M111120 5 BCD P121420

6 CD C111707 6 ABD M122407 6 AD P111825

7 BC C110501 7 BC M110823 7 ACD P111823

8 ACD C111303 8 BCD M110808 8 AB P111820

9 ABC C111405 9 BC M121003 9 AC P120118

10 D C110403 10 BD M122407 10 ABC P120115

11 QST,PQST,RST,PRST C110403 11 PRS,Q,PQRS,PQR M111003 11 T,T,P,P P111829

12 RST,QRS,RS,RQ C111503 12 R,Q,P,S M110909 12 PQRST,PQRST,PQST,PQRST P111822

13 P,QS,RS,T C110502 13 PQ,ST,R,PQ M11082 13 S,P,T,R P120212

14 PQST,PQST,PRT,PQST C111707 14 Q,R,P,R M111123 14 Q,S,P,R P120216

15 9 C110501 15 1 M1122407 15 1 P120106

16 4 C111608 16 5 M1110909 16 4 P121422

17 9 C111707 17 3 M1122402 17 2 P111815

18 8 C110403 18 6 M111029 18 5 P120205

19 8 C111508 19 4 M111082 19 2 P121406

20 5 C111804 20 5 M110903 20 3 P111822

Ai2

TS – 6 080163.1 SET – A

ALL INDIA INTERNAL TEST SERIES



SOLUTIONS

CHEMISTRY :

1. (D)

CH3 and aniline are planar

2. (A) [Concept Code : C110507/504]

3 4 2 4 2H AsO OH H AsO H O

Initial milimols 30 10 0 After reaction 20 0 10

2

3 4 4 2 4H AsO HAsO 2H AsO

Initial milimols 20 30 10 After reaction 0 10 50

So final pH = 2

10pKa log 6.3

50

3. (C)

A=> propyne B=> mesitylele C => acetone D => mesitylene E=> phorone

4. (C)

Kp (CaCO3) = 4 atm

Kp (BaCO3) = 6 atm

Pressure above (CaCO3+BaCO3) = 6 atm

Total pressure = 6+1+2= 9 atm

5. (ABC)

N2 + PCl5 PCl3 + Cl2

Moles at t = 0 1 3 0 0

Moles 1 (3 – x) x x

at equilibrium

Total moles present at equilibrium = 4 + x

Given total pressure at equilibrium = 2.05

Total volume = 100 Litre

PV = nRT

2.05 × 100 = (4 + x) × 0.082 × 500

x = 1.0

Degree of dissociation '' for PCl5

= presentmolesTotal

ddissociateMoles =

3

1 = 0.3333 = 33.33%

Also KP =

22

(3 – )

x P

x n

=

2 2.05

(3 – ) (4 )

x

x x

=)14)(1–3(

05.2)1( 2

KP= 0.205 atm.




6. (CD)

7. (BC)

5

a1

5

K 10

20.1 100.1

H S HS H

At equilibrium

8

a2

54

K 10 2

0.1 1010HS S H

20

sp

164

K 10 2 2

(S)1010

NiS S Ni

8. (ACD)

9. (ABC)

Compounds formed

Total = 10

Optically active =10

Optically inactive=0

Diastereomer pairs = 4

Enantiomer pairsm= 5

10. (D) [Concept Code : C110402]

When solids A(s), E(s) and P(s) are kept together then

2

1

1 1 2 3

Kp 4atm

(s) (g) (g)

p p p p

A C B

2

2

2 1 2 3

Kp 32atm

(s) (g) (g)

p p p p

E D B

3

1 2 3

Kp 5atm

(s) (s) (g)

p p p

P Q B

1 2 3

4 32p , p , p 1

5 5

But 3p 1 means to establish (s) (g) (s)P B Q , it has to go backward which is not

possible due to absence of Q(s) in the container so p3 = 0

Hence 1 2 1 1 2 2(p p )p 4, (p p )p 32

1 2p p = Partial pressure of B(g) = 6 atm

and the total pressure = 12 atm

(Integer Type)

1. (9)

In order to prevent precipitation of ZnS,

[Zn2+] [S2–] < Ksp (ZnS) = 1 × 10–21

(Ionic product)

or (0.1).[S2–] < 1 × 10–21

or [S2–] < 1 × 10–20

This is the maximum value of [S2–] before ZnS will precipitate. Let [H+] to maintain this [S2–] be x.

Thus for H2S 2H+ + S2–,

Ka = ]SH[

]S[]H[

2

–22

= 1.0

)101(x 20–2




= 1.0 × 10–21

or x = [H+] = 0.1 M. => pH= 1

when [S2–] = 1 × 10–20 M and [Cd2+] [S2–] = Ksp (CdS) = 8 × 10–27

Then [Cd2+

] = 8 × 10–7 M

So, x=1 and y=8 => x+y = 9

No ZnS will precipitate at a concentration of H+ greater than 0.1 M.

2. (4)

x2+y

2+2z

2 = 2yz + 2x + 4z – 5

=> x = 1; z = y = 2

3. (9)

3 + 4 + 2 = 9

4. 8

PCl3 + Cl2 PCl5

0.20 0.08 0.40 moles/litre

Kc = 25

If 0.22 mole of Cl2 is added then at equilibrium

0.20 – x 0.30 – x 0.40 + x

25 = )x–30.0)(x–20.0(

x40.0

or x = 0.1

[PCl3] = 0.2 - 0.1 = 0.1 moles

[Cl2] = 0.3 - 0.1 = 0.2 moles

[PCl5] = 0.4 + 0.1 = 0.5 moles

Sum of conc. = 0.8

5. (8)

P is 2 3 42[BH NH ] [ ]BH

Q is borazine

6. (5)




MATHS 1.

y2 = 4ax. The normal chord at (4a, 4a) subtends a right angle at focus.

2. Tangents at (acos , bsin ) and (acos ,bsin )

meet at

acos bsin2 2

, (h,k)

cos cos2 2

k b b

tan constant y = tan xh a 2 a 2

3. Asymptotes of a rectangular hyperbola are perpendicular

4

(x y 1) (3x 4y 6) 0 slope3

4x + 3y = 17 4. y = mx in ax2 + 2hxy + by2 = 0 2 2x [a (2h)m bm ] 0

1 2m km

1 2

2hm m

b

1 2

am m

b

5. 2 2 2y mx a m b

2 29p 8 1 p 1

6. t = y2, First order linear differential equation in variable t. 7. The length of perpendicular from centre of the circle on both the lines will be equal. 8. Length of perpendicular from centre on the line will be greater than the radius of the circle

9. 1

3 3

0 0

1(x x ) dx (x x ) dx

2

10. First order linear D.E. in dx

dy

Integer :

1. 2

2

dy 2xy (x 1)

dx 1 x

2. The line joining origin and (2,3) will be parallel to the axis of the parabola 4. Sum of focal distances for any point = 8

A : (2 4cos( ), 12 sin( ))

5. 2 2(x y 1)

(x y) k2

1 2 1 2C C r r

6. 2 2hx ky h k becomes a tangent to y2 = 4ax

2 2h k h

y xk k




2 2

2

h(k h )a

k

SOLUTIONS (PHYSICS) 1) (B)

2 2

2 2( ) ( )

R nP R

r R r nR

r R n

2) (B)

The current density J, electric field E at the cross-section are related by

E

J

multiplying both side by A.

EA

JA= or I= Where is electric flux.ρ

EA

JA= or I=ρ

or I

3) (A)

In the position of balance no current exists through the galvanometer, and hence through the cell. Therefore the

magnitude of r does not affect the condition for balance

4. (b)

If point B is being crossed then the particle will reach to other end

B AKE (V V )q

2

B

11V K R

6

2

A

4V K R

3

5. BCD With the help of loop law & junction law

16a bV V V

Hence b, c, d are correct 6. AD 7. ACD 8. AB

w.r.t. trolley, cm

F 0 & 0

9. AC

q q

4 8 04 2

16

q3

at t i 0

at t 0 4 2i 8 0 i 2A

10. ABC




MATRIX MATCHING : 1. A–t; B–t; C–p; D–p From theory 2. (A) P,Q,R, S, T; (B) P,Q,R,S,T; (C) P,Q, S,T; (D) P,Q,R,S,T

3. a-S, b-P, c-T, d-R

V ig(G R)

= (100)(10-6

)(100 + 2 × 105)

20V

g

g

i i G

i S

g g

Gi i i

S

6 6100100(10 ) (100)(10 )

1

3 30.1 10 10(10 )

10mA .

22. A-Q, B-S, C-P, D-R

E 1 eq qq C' C . 1

eqC equation capacitance with S1 closed

8

12 3 4 C3

Sq 12 C

4

E 1 eq eq

S

4q C C , 12 4 32 C

3

q 12 4 48 C

NUMERICAL :

1. (1)

The electron covers a distance 2

d across the first capacitor when if comes out of it.

in second capacitor it has to cover 2

d across it.

V1 = V2

2. 4 3. 2 4. 5

When i = 0 and V 10V

10V ...(i)

When V 0, R = 0 and i = 2A

ir

10

r 52




5. (2) E inside shaded portion is = distance between centres

2

20

0 0

6 10 10E 2 10 N / C

3 3

0

0

E 2E 2

100 100

6. 3

v0

v0

v

mV0 = (M + m) V

V = 0mV

M m

From energy conservation

1

2mV0

2 + Mg = mgh + Mg +

1

2 (M+m)V

2

1

2mV0

2 –

1

2

2 2

0m v

M m= mgh

1

2

2

0mV [M m m]

[M m]

= mgh

h = 1

2

2

0V M

g M m

all india internal test series chemistry, mathematics

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