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Code

0 UMULATIVE TEST

-1

(CT-

1)

JEE ADVANCED PATTERN)

TARGET : JEE MAIN+ADVANCED) 2016

COURSE : JP, JF, JR

®

Revision Plan-2 | PAPER-2

Date : 10-04-2016 Time: 3 Hours Maximum Marks : 189

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

GENERAL

1. The sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so.

2. The question paper CODE is printed on the right hand top corner of this sheet.

3. Use the Optical Response Sheet (ORS) provided separately for answering the question.

4. Blank spaces are provided within this booklet for rough work.

5. Write your Name and Roll Number in the space provided on the below cover.

QUESTION PAPER FORMAT AND MARKING SCHEME :

6. The question paper has three parts : Mathematics, Physics and Chemistry. Each part has three sections.

7. Carefully read the instructions given at the beginning of each section.

8. Section 1 contains 7 multiple Choice question with Only one correct option.

Marking scheme: +3 for correct answer, 0 if not attempted and –2 in all other cases.9. Section 2 contains 3 "paragraph" type questions. Each paragraph describes an experiment, a situation or a

problem. Two multiple choice questions will be asked based on this paragraph. Only one option can be correct.

Marking scheme: +3 for correct answer, 0 if not attempted and –1 in all other cases.

10. Section 3 contains 3 "match the following" type question and you will have to match entries in Column I with the entriesin Column-II.

Marking scheme : for each entry in Column I, +2 for correct answer, 0 if not attempted and 0 in all other cases.

OPTICAL RESPONSE SHEET :

11. Darken the appropriate bubbles on the original by applying sufficient pressure.

12. The original is machine-gradable and will be collected by the invigilator at the end of the examination.

13. Don not tamper with or mutilate the ORS.

14. Write your name, roll number and the name of the examination centre and sign with pen in the space provided forthis purpose on the original. Do not write any of these details anywhere else. Darken the appropriate bubbleunder each digit of your roll number.

DARKENING THE BUBBLES ON THE ORS :

15. Use a BLACK BALL POINT to darken the bubbles in the upper sheet.

16. Darken the bubble COMPLETELY.

17. Darken the bubble ONLY if you are sure of the answer.

18. The correct way of darkening a bubble is as shown here :

19. There is NO way to erase or "un-darkened bubble.

20. The marking scheme given at the beginning of each section gives details of how darkened and not darkenedbubbles are evaluated.

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M A T H E M A T I C S

MATHEMATICS

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PART : I MATHEMATICS

SECTION – 1 : (Maximum Marks : 21)

This section contains SEVEN questionsEach question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option iscorrect

For each question, darken the bubble corresponding to the correct option in the ORS

Marking scheme :+3 If only the bubble corresponding to the correct option is darkened0 If none of the bubble is darkened

–2 In all other cases

1. The solution set of inequality

(tan –1

x – 2)(cot –1

x) – (tan –1

x) 12

!" #$% &

' ( + 2 1

2

!" #$% &

' ( >

x –lim) *

–1sec x –2

!+ ,- ./ 0

is (where [.] denotes the

greatest integer function)

(A) (tan1, tan2) (B) ( – cot1, cot2) (C) ( –tan1, tan2) (D) ( –tan1, *)

2. x1, x2, x3, x4 are roots of the equation x4 – x3 sin21 + x2 cos 21 – x cos1 – sin1 = 0 and if4

–1

kk 1

tan x2

3 = 4 then tan4 is equal to (Here 1 6 0,2

!" #% &' (

– 6

!7 89 :; <

)

(A) tan1 (B) sin1 (C) cot1 (D) data insufficient

3. The domain of the function f(x) given by 3x + 3f(x) = min {min (2t3 – 15t2 + 36t – 25), 2 + |sint|} where

2 = t = 4, is(A) ( –*, 1) (B) ( –*, log

3e) (C) (0, log

32) (D) ( –*, log

32)


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4. Let f(x) = x2 + >x + ? cos x, > being an integer and ? a real number. The number of ordered pairs

(>, ?) for which the equations f(x) = 0 and f(f(x)) = 0 have the same (non empty) set of real roots is.

(A) 0 (B) 3 (C) 4 (D) infinite

5. For all n 6 N, let f(n + 1) = ( –1)n + 1 n – 2f(n) and f(1) = f(2016), then2015

r 123f(r) is equal to

(A) 0 (B) –1007

3 (C) 336 (D)

1007

3

6. Solution of the equation2

1 1

2

2x 1 x2sin x tan

1 2x

@ @ " #@% &2% &@' (

is

(A)1 1

,2 2

" #@% &' (

(B)1

,12

" #% &' (

(C) A B1 1

1,1 ,2 2

7 8@@ @ 9 :

; < (D)

1 11,1 ,

2 2

7 8@@ @+ , 9 :/ 0

; <

7. Value of A B A B1 3 1 2 1 2cos 3x 4x cos 1 2x sin 1 x@ @ @@ $ @ $ @ = 1acos x b@ $ then value of a + b is

given1

0 x2

" #C C% &

' (

(A) 2 2! $ (B) 2 2! @ (C) 2 (D) 3


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MATHEMATICS






This section contains THREE paragraphsBased on each paragraph, there will be TWO questions.Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correctFor each question, darken the bubble corresponding to the correct option in the ORS

Marking scheme :

+3 If only the bubble corresponding to the correct option is darkened0 If none of the bubbles is darkened –1 In all other cases

Paragraph for Question Nos. 8 to 9

Let f(x) and g(x) be twice differentiable functions, defined as

f(x) = x2 + xgD(1) + gE(2) and g(x) = f(1) x2 + xfD (x) + fE(x).

8. The value of f(1) + g( –1) is

(A) 0 (B) 1 (C) 2 (D) 3

9. The number of integers in the domain of the function F(x) = –f(x)

– 3 – xg(x)

$

(A) 0 (B) 1 (C) 2 (D) infinite


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MATHEMATICS






Let 4, 1 and F are the roots of the equation x3 + 6x + 3 = 0and A = cos

–1 (sin ((4 + 1) –1 + (1 + F) –1 + (F + 4) –1))

B = cos && (

#%%'

" && (

#%%'

" & (

#%'

" F$1$4@2

sintan 1

C = sec –1 (cosec (1 – 4)(1 – 1)(1 – F)))

10. Find the value of 5A + B – C(A) 1 (B) 21 (C) 0 (D) Can't be calculated

11. Find the range of the function h(x) =1x)1B(x

Bx6x)CA5(34

25

$@$

$@

(A) [0, 3] (B) (0, 3) (C) [0, *) (D) ( –*, *)


Consider f(x) = ax2 + ax + a – 3, g(x) = sinx – 1 and h(x) = k – 1x2$ then

12. The least positive integral value of a, say a1, for which f(x) = g(y) has no real solution for any

x and y, is equal to

(A) 4 (B) 5 (C) 6 (D) 7

13. Value of k for which f(x) = a1x2 + a1x + a1 – 3 has exactly one real point of intersection with

y = h(x) is equal to (where a1 is from above question)

(A) 1 (B) 2 (C)3

2 (D)

3

4


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This section contains THREE questions

Each question contains two columns, Column I and Column II

Column I has four entries (A),(B), (C) and (D)

Column II has four entries (P),(Q), (R) and (S)

Match the entries in Column I with the entries in Column II

One or more entries in Column I may match with one or more entries in Column II

The ORS contains a 4 × 4 matrix whose layout will be similar to the one shown below :

(A)

(B)

(C)

(D)

For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry(A) in Column I matches with entries (P), (Q) and (R), then darken these three bubbles in the

ORS. Similarly, for entries (B), (C) and (D).

Marking scheme :

For each entry in Column I

+2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened

0 If none of the bubbles is darkened

0 In all other cases


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14. Column-I Column-II

(A) Domain of the function (P) [0, 1/2)

f(x) =1

x

+ 2sin –1 x +1

x 1@

is

(B) Find the range of the function (Q) [ –1, 1] – {0}

f(x) =1

[x ]

+ ,- .@ !/ 0

, where [.] represents G.I.F.

(C) Range of the function f(x) ={x}

1 {x}$ (R) G

where {.} represents fraction part function

(D) f(x) = sin –1 [x]; g(x) = 2 – 3x2 then domain (S) { –1, 0, 1}

of the function f(g(x)), where [.] represents G.I.F.


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15. Column-I Column-II

(A) Number of values of 'x' satisfying the equation (P) 4

A B A B1 1 13sin 1 x 2sin x cos x 2@ @ @@ $ 2 @ is

(B) Number of integers in the range of function (Q) 1

A B A B1f x cos cos2x cos x@2 $ is/are

(C) Number of integers in the range of function (R) 5

A B A B1 2f cot 6sin cos 8cos@H 2 H H $ H is

(D) Number of solution of equation (S) 2

A BA BA B1 1sin 4cos cot 2tan x 0@ @ 2 is


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16. Column –I Column –II

(A) If f(x) =

2 34 (x 1) xx 3 24 8 72 4

@ @$ @ @ is defined, then x can be (P) 0

(B) The least positive integral value of ‘x’ satisfying (ex – 2) (Q) 3

sin x4

" #!" #$% &% &

' (' ((x – log22)(sin x – cos x) < 0 is

(C) The number of integral values of m for which f : R ) R; (R) 6

f(x) =3x

3+(m –1)x

2+(m + 5)x + n is bijective is

(D) The number of roots of equation (S) 7

A Bx x 3(x – 1)(x – 3) (x 1)(x 3) – e – e x cosx(x – 2)(x – 4) (x 2)(x 4)

@" #" #$ $ @% &% &$ $' (' (= 0


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PHYSICS

P H Y S I C S


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PART - II : PHYSICS


This section contains SEVEN questions

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is

correctFor each question, darken the bubble corresponding to the correct option in the ORS

Marking scheme :

+3 If only the bubble corresponding to the correct option is darkened

0 If none of the bubble is darkened


17. AB is a uniform rod of mass M and a point mass m is placed at origin as shown in figure. The

direction of force on point mass m due to rod makes an angle ! with positive x-axis, where ! is :

(A) + 45° (B) – 30° (C) + 30° (D) + 15°

(Positive angle means angle measured anti-clockwise from x-direction)


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®



18. Two rays are incident on a spherical concave mirror of radius R = 5 cm and rays are parallel to aline AB which passes through the centre C. The rays are at perpendicular distances 3 cm and4 cm from AB respectively as shown. Find the distance between the points at which these raysintersect the line AB after reflection -

(A)25

cm24

(B)24

cm25

(C) zero (D)1

cm3

19. Consider a uniformly charged solid non conducting sphere of radius R and charge Q. Considerthree spherical regions centered at the centre of the uniformly charged solid non conductingsphere, given below :

Region (1) R1 " r " 0 Region (2) R2 " r " R1 Region (3) $ " r " R2

If electrostatic potential energy stored in each of the three regions is equal then 2

1

R

R is :

(A) 2 (B) 3 (C) 1 (D) 4

20. Two satellites revolve around the ‘Sun’ as shown in the figure. First satellite revolves in a circularorbit of radius R with speed v1. Second satellite revolves in elliptical orbit, for which minimum and

maximum distance from the sun are3

R and

3

R5 respectively. Velocities at these positions are

v2 and v

3 respectively. The correct order of speeds is :

R

SunR/3Orbit(1)

v2

v1

5R/3Orbit(2)

v3

(A) v2 > v

3 > v

1 (B) v

3 < v

2 < v

1 (C) v

2 > v

1 > v

3 (D) v

2 > v

3 = v

1


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21. A circular disc of diameter d lies horizontally inside a metallic hemispherical bowl of radius a. The

disc is just visible to an eye looking over the edge. The bowl is now filled with a liquid of refractive

index % 3 . Now, the whole of the disc is just visible to the eye in the same position. Ifd

a is equal to :

(A) 1 (B) 2 (C)1

2

(D) None of these

22. The electric field at the centre of a uniformly charged hemispherical shell is E0. Now two portions of

the hemisphere are cut from either side and remaining portion is shown in figure. If & = ' =3

(,

then electric field intensity at centre due to remaining portion is

(A)3

E0 (B)6

E0 (C)2

E0 (D) E0

23. A small area is removed from a uniform spherical shell of mass M and radius R. Then the

gravitational field intensity near the hollow portion is :

(A)2

GM

R (B)

2

GM

2R (C)

2

3GM

2R (D) Zero


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®




This section contains THREE paragraphs

Based on each paragraph, there will be TWO questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct


Marking scheme :

+3 If only the bubble corresponding to the correct option is darkened0 If none of the bubbles is darkened


Paragraph for Questions 24 and 25

Consider a hypothetical solar system, which has two identical massive suns each of mass M and

radius r, seperated by a seperation of 2 3 R (centre to centre). (R >>>r). These suns are always

at rest. There is only one planet in this solar system having mass m. This planet is revolving in

circular orbit of radius R such that centre of the orbit lies at the mid point of the line joining the

centres of the sun and plane of the orbit is perpendicular to the line joining the centres of the sun.

Whole situation is shown in the figure.

Answer the following question regarding to this solar system.

24. Speed of the planet is :

(A)GM

8R (B)

GM

4R (C)

GM

2R (D)

GM

3R

25. Average force on the planet in half revolution is :

(A)2

GMm

4 R( (B)

2

GMm

4R (C)

2

GMm

2 R( (D)

2

GMm

8R


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A spherical distribution of charge varies with distance from centre as:

)1 from r = 0 to r = R, and )2 from r = R to r = 2R[Relative permittivity of material of sphere is 1]Then answer the following questions:

)1R)2

2R

26. If the electric field at a point A at a distance R/2 from the surface of the smaller sphere is zero, find

the ratio of )1 and )2.

)2

2R

A3R/2

)1R

(A)8

19* (B)

19

8* (C)

27

8* (D)

8

27*

27. Given the above ratio of )1 and )2 find the magnitude of electric field at a point B at the surface ofthe outer sphere. (figure)

)1)2

2R

BR

(A) 1

0

37 R76

)

+ (B) 1

0

37 R456

)

+ (C) 1

0

37 R228

)

+ (D) 1

0

29 R76

)

+


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Consider a situation in which a negatively charged particle of mass M and charge – q revolves in

elliptical path around a fixed charge Q. the closest and farthest distance of the moving particle

from fixed charge is a and 4a respectively. (k = 1/4(+0)

C

QB

a 4aA

v

,2aF

28. The speed of revolving particle when it reaches point A is

(A)3kQq

10Ma (B)

kQq

10Ma (C)

kQq

2Ma (D)

7kQq

10Ma

29. The rate of change of speed of particle at point C is :

(A) zero (B)24 3

KQqma

(C)23

KQqma

(D) 24KQqma


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®










(A)

(B)

(C)

(D)

For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry

(A) in Column I matches with entries (P), (Q) and (R), then darken these three bubbles in the


Marking scheme :






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30. Consider a solid uniformly charged sphere of radius R1 with a spherical hole of radius R2 displaced

a distance x0 from the centre as shown in the figure. ) is the charge density. If column-- indicates

the electric field intensity E on the x-axis as a function of distance x from centre O, at A, B, C and

D, then match the following.

BR2 AD

O C

R1

x0

Column-#

Column-##

(A) EA (P)0

0

x

3

)

+

(B) EB (Q)

3

2

2

0 0

Rx

3 (x x )

. /)*0 1

+ *2 3

(C) EC (R)

3

2

2

0 0

Rx

3 (x x )

. /)*0 1+ 42 3

(D) ED (S)

3 3

1 2

2 2

0 0

R R

3 x (x x )

. /)*0 1+ *2 3


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31.

Consider the situation shown in column-I a real object is moving towards a fixed optical component

or an optical component is moving towards a fixed object. Match the possible direction and

magnitude of velocity of image as shown in column-II. (All velocities in column-I are equal to v0)

Column-# Column-##

(A)2f

O(P)

(B)R

O(Q)

(C)

O

2f f(R) More than v0

(D)Object isfixed

(S) Less than v0


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32. Consider an arrangement shown in Column--, a value related to arrangement is asked. Choosethe proper value from Column--- and Column--

Column-#

Column-##

(A)

A B

(P) 1

A and B are two metallic plates having charges zero and'3' Q Respectively now plate A is grounded by a wire then

if 1Q is charge flown through wire and 2Q is charge on

right surface of A then1

2

Q

Q is

(B) O A (Q) 2

A is a point on axis of conducting ring having charge ''Q .

Now another point charge of Q125

128 is placed at centre of ring.

If E1 and E2 are Electric fields at A before and after placing point

charge then

1

2

E

E is (radius of ring is 3R and OA = 4R)

(C) (R) 3

A light ray is incident on hemispherical mirror parallel toPrinciple axis. If N1 is total number of reflections whenlight is incident. At angle of 60

0 and N2 is total number of

reflections when light is Incident at 800 then

2

13

N

N is

(D)C2

C1 (S) 4

Spherical cavity of radius2 R is made inside a uniform sphere

of radius R as shown in figure. If maximum difference ofgravitation potential. Between any two points inside cavity

isk

G R2

4!" then

2

k is : ( " is density of material of sphere)


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C H E M I S T R Y

CHEMISTRY


Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected]

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PART III : CHEMISTRY

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,

P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,

Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]


This section contains SEVEN questions

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is

correct


Marking scheme :

+3 If only the bubble corresponding to the correct option is darkened

0 If none of the bubble is darkened


33. 50 mL of H2SO4 (sp. gravity 1.84) is mixed with 150 mL of H2SO4 (sp. gravity 1.2). If there is 20%

reduction in volume after mixing, The specific gravity of the resulting solution is:

(A) 1.9 (B) 1.62 (C) 1.7 (D) none of these


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34. A 10 gm sample of "gas liquor" is boiled with an excess NaOH and resulting NH3 produced is

passed into 60 mL of 0.45 M H2SO4. The residual acid was then neutralised by 10 mL of 0.4 M

NaOH. The % ammonia in the "gas liquor" examined is:

(A) 17 (B) 8.5

(C) 34 (D) none of these

35. A mixture of N2O and CO2 in the mole ratio of a : b has a mean molecular mass equal to “M”. What

would be the mean molecular mass of mixture of the same gases present in the mole ratio of

b : a?

(A)22a 44b

M

! (B)

M

a b!

(C) M (D)a b

M

!

36. 0.5 g of an anhydrous organic acid silver salt leaves on ignition 0.355 g silver. In another

experiment 100 mL of 0.2 M acid solution was neutralised by 400 mL of1

10 M caustic potash

solution. What is the approximate molecular weight of acid?

(A) 60 (B) 90

(C) 120 (D) 80


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37. Industrially TNT (C7H5N3O6, explosive material) is synthesized by reacting toluene (C7H8) with

nitric acid in presence of sulphuric acid. Calculate the maximum weight of C7H5N3O6 which can be

produced by 140.5 gm of a mixture of C7H8 and HNO3.

(A) 140.5 (B) 113.5

(C) 140.52

(D) 140.5 – (3 × 18)

38. The ratio of coefficients of HNO3, Fe(NO3)2 and NH4NO3 in the following redox reaction:

3 3 2 4 3 2Fe HNO Fe(NO ) NH NO H O! " ! ! is respectively-

(A) 10:1:4 (B) 10:4:1

(C) 4:10:1 (D) none of these

39. n-caproic acid, C5H11COOH, found in coconut and palm oil is used in making artificial flavours, has

solubility in water equal to 11.6 g/L. The saturated solution has pH = 3.0. The Ka of acid is

(A) 10 –6

(B) 10 –5

(C) 2 × 10 –5

(D) 2 × 10 –6


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This section contains THREE paragraphs

Based on each paragraph, there will be TWO questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct


Marking scheme :+3 If only the bubble corresponding to the correct option is darkened




A mixture of H2C2O4 (oxalic acid) and NaHC2O4 was dissolved in water and the solution made up

to one litre. 10 mL of the solution required 3 mL of 0.1 N NaOH solution for complete

neutralization. In another experiment, 10 mL of the same solution, in hot dilute H2SO4 medium

required 4 mL of 0.1 N KMnO4 solution for complete reaction.

40. Mass of oxalic acid in orginal mixture is

(A) 1.12 g (B) 1.8 g (C) 0.9 g (D) none of these

41. Instead of KMnO4 if K2Cr2O7 was taken for complete reaction, the volume of 0.2 M K2Cr2O7

needed was:

(A) 1mL (B)2

ml3

(C)1

ml3

(D) none of these


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For the reaction: Ag (aq) Cl (aq) AgCl(s).! #! " Given

Species#

G°f (kJ/mol) at 25°C

Ag+(aq) +77

Cl –(aq) –129

AgCl(s) –109

–

2 –

Ag e Ag; E 0.80 V

Zn 2e Zn; E 0.76V

!

!

! " $ %

! " $ % #

42. The cell representantions of the above reaction and E°cell are:

(A) Ag(s) | AgCl(s) | Cl –(aq) || Ag

+(aq) | Ag(s); E°cell = 0.59 V

(B) Ag(s) | Ag+(aq) | Cl

–(aq) || AgCl(s) | Ag (s); E°cell = 0.59 V

(C) Ag(s) | Cl –(aq) || Ag

+(aq) | Ag; E°cell = 0.59 V

(D) Ag(s) | Ag+

(aq) || Cl –

(aq) || AgCl(s) | Ag (s); E°cell =1.18 V

43. A quantity of 6.5 x 10 –2

g of metallic zinc is added to 100 mL of solution of AgNO 3. The value of

2

2

Znlog

Ag

!

!

& '( )

& '( )

at equilibrium is:

(A) 5.288 (B) 52.88 (C) 26.44 (D) None of these


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The dissociation constant of acetic acid is 2.0 × 10 –5

. An aqueous solution of acetic acid is

prepared by dissolving 0.7 millimole acetic acid in sufficient water to get 10 m3 of solution at 25°C.

44. What is the pH of solution? (log 2 = 0.3, log 7 = 0.85, log 17 = 1.23, 4.49 2.1% )

(A) 7.15 (B) 6.77 (C) 6.85 (D) 7.0

45. What is the concentration of unionized acetic acid in the solution?

(A) 0 (B) 4.9 × 10 –10

M (C) 7 × 10 –8

M (D) 2.45 × 10 –10

M


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(A)

(B)

(C)

(D)

For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry

(A) in Column I matches with entries (P), (Q) and (R), then darken these three bubbles in the


Marking scheme :






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46. H3PO4 is triprotic acid and its dissociation constants are

K1 = 10 –4

K2 = 10 –8

K3 = 10 –12

Column-I Column-II

(Composition of solution) (pH of solution)

(A) 0.1 M H3PO4 (P) 8.0

(B) 0.05 M NaH2PO4 (Q) 2.5

(C) 0.05 M NaH2PO4 + 0.05 M Na3PO4 (R) 6.0

(D) 0.5 M Na2HPO4 (S) 10.0


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47. Cl2 reacts with O2 in either of the following ways depending upon the amount of O2 supplied.

2 2 2

1Cl O Cl O

2! **"

2 2 2Cl 2O 2ClO! **"

Cl2O may also react with O2 as per the following reaction

2 2 2

3Cl O O 2ClO

2! **"

Match composition of final mixture of the product (given in column –II) for initial amount of

reactants (given in column –I).

Column-I Column-II

(Mass of initial reactants) (Mole of final products)

(A) 142 gm of Cl2 is mixed with 64 gm of O2 (P) 0.5 (O2), 1 (ClO2)

(B) 71 gm of Cl2 is mixed with 8 gm of O2 (Q) 4 (Cl2O)

(C) 35.5 gm of Cl2 is mixed with 48 gm of O2 (R) 0.5 (Cl2), 0.5 (Cl2O)

(D) 284 gm of Cl2 is mixed with 64 gm of O2 (S) 2 24 4

(Cl O), (ClO )3 3


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48. Match the following:

Column-I Column-II

(Solution) (Result of electrolysis using inert electrode)

(A) AgNO3(aq) (P) pH of solution increases

(B) CuSO4(aq) (Q) O2 gas evolved at anode

(C) CH3COONa(aq) (R) H2 gas evolved at cathode

(D) CuCl2(aq) (S) Metal deposited at cathode

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