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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]
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.
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 2
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
Website: www.fiitjee.com, Mail :[email protected]
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 ?
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 3
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
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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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 4
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
Website: www.fiitjee.com, Mail :[email protected]
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
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 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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 5
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
Website: www.fiitjee.com, Mail :[email protected]
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 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
Column – I Column – II
(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
Column – I Column – II
(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
Column – I Column – II
(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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 6
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
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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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 7
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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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 8
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, SarvapriyaVihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
Website: www.fiitjee.com, Mail :[email protected]
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. 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)
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 9
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(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
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. 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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 10
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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
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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
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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
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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.
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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
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= 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)
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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
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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
Ai2TS – 6 ( XII ) | SET – A | APT – 3 | P a g e | 5
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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
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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