fiitjeesouthdelhi.co.in · FPT-X-(Paper-1)-SOL FIITJEE Ltd., FIITJEE House, 29 -A, Kalu Sarai, Sarvapriya Vihar, New Delhi 16, Ph :46106000, 26515949, Fax 26513942. 2 Chemistry PART

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 16, Ph: 46106000, 26515949, Fax: 26513942.

ANSWERS, HINTS & SOLUTIONS (04.05.2016)

PAPER-1

Q. No. CChheemmiissttrryy MMaatthheemmaattiiccss PPhhyyssiiccss 1. D C D

2. A D D

3. B D A

4. B C C

5. A A A

6. D A B

7. B D B

8. D D C 9. A, B, C A, B, D A, B, D 10. A, C, D C, D A, B, D 11. A, B A, D A, B, C 12. A, B B, C A, B, D 13. A D B

14. B B A

15. C C A

16. C C C

17. A D B

18. D B D

1.

A → (p, t) B → (q, s) C → (q, s) D → (q, r, t)

A (r) B (s) C (p) D (r)

(A) → (r, s) (B) → (p) (C) → (p) (D) → (q, t)

2.

A → (p, q, r, s) B → (p, r, t) C → (p, q, s) D → (p, q, s)

A (r, s) B (p, r, s) C (q, t) D (p)

(A) → (p, q) (B) → (p, r) (C) → (p, t) (D) → (p, s)

FIITJEE PRACTICE TEST – X

FPT-X-(Paper-1)-SOL


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CChheemmiissttrryy PART – I

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Which of the following is correct representation of Heinsberg inderterminacy principle

(A) hp. v4

(B) 1p. x2

(C) hp. x2

(D) hE. t4

Sol. D 2. Given reagents are HCl, NaOH, ZnCl2, Na2CO3, NH4Cl and Zn. Which of these reagent(s) can

intensify hydrolysis of FeCl3 when added to its solution (A) NaOH, Na2CO3 (B) NaOH, NH4Cl, HCl (C) Na2CO3, NaOH + HCl (D) NH4Cl, HCl, ZnCl2 Sol. A

FeCl3(aq) being acidic due to cationic hydrolysis, basic reagents can intensify the hydrolysis. 3. Which of the following statement is incorrect? (A) Oxoanions of sulphur have little tendency to polymerise compared with the phosphates and

silicates. (B) In pyrosulphurous acid (H2S2O5), the oxidation states of both the sulphur atoms are not

same and these are + IV and + II (C) Conc. HNO3 oxidises both sulphur and carbon to H2SO4 and H2CO3 respectively. (D) Most metal oxides are ionic and basic in nature while non-metallic oxides are usually covalent

and acidic in nature. Sol. B 4. Co2+ (aq.) + SCN– (aq.) Complex (X). Ni2+ (aq.) + Dimethylglyoxime Complex (Y). The corrdination number of cobalt and nickel in complexes X and Y are four. The IUPAC names of the complexes (X) and (Y) are respectively : (A) tetrathiocyanato-S-cobalt(II) and bis(dimethylglyoximate) nickel(II). (B) tetrathiocyanato-S-cobaltate (II) and bis(dimethylglyoximato)nickel (II). (C) tetrathiocyanato-S-cobaltate (II) and bis(dimethylglyoximato)nickelate(II). (D) tetrathiocyanato-S-cobaltate(III) and bis(dimethylglyoximato)nickel(II). Sol. B (X) = [Co(SCN)4]2- ; (Y) = [Ni(dmg)2]. 5. Average X –Ybond enthalpy in XYn is 309.2 kJ/mol. The value of standard enthalpy of formation

of XYn(g), X(g) and Y(g) are –1100, 275 and 80 kJ/mol respectively. XYn is (A) XY6 (B) XY4 (C) XY2 (D) XY5

Sol. A

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X(g) + nY(g) XYn(g) 0 0 0 0f f n f fH H XY (g) [ H X (g) n H Y (g)]

= –1100 – [275 + 80n] = –1375 – 80 n average X – Y bond enthalpy = 309.2 kJ/mol XYn = – (309.2) n (-309.2)n = –1375 – 80n n = 6 XYn XY6

6. In the reaction sequence

C

O

NH2P2O5 P

i) CH3MgBr

ii) H 3O+ Q

I2/Ca(OH)2 R yellow ppt

Ca(OH)2/

S

Br2/Ca(OH)2

A

+

A & S respectively are

(A) NH2 & C

O

CH3 (B) CH2 NH2 &

O

(C) C

O

NHI & (D) none

Sol. D

C

O

NH2P2O5 i) 1eq CH3MgBr

ii) H 3O+

(Q)

I2/Ca(OH)2

Ca(OH)2/

(S)

Br2/Ca(OH)2

(A)

(i.e Hoffman's degradation)

NH2 + CaCO3 + H2O

C N

(P)

C CH3

O

C OH

O

+ CHI3(yellow ppt)

(R)

C

O

+ CaCO3

7. Consider the following co-ordination compounds

(i) 4Ni(CO) (ii) -4Co(CO) (iii) 2-

Fe(CO)4

The stretching frequency of M – C bond (which is directly proportional to bond strength) follows the order

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(A) (ii) > (iii) > (i) (B) (iii) > (ii) > (i) (C) (i) > (ii) > (iii) (D) (i) > (iii) > (ii)

Sol. B Due to high charge density on central metal electron transfer through back bonding increases

which increases M-C bond strength & thus its stretching frequency. 8. NH2

2NaNO /HClA C

MW 113B D yellow oily liquid

Which of the following is incorrect for above reaction? (A) D is having molecular formula C7H15N (B) Reagent C can be H2/Ni or LiAlH4 (C) B has a characteristic smell (D) Reagent A is present in acidic medium Sol. D NH2

3CHCl /OH

NC

2 4H /Ni or LiAlH

NHCH3

2NaNO /HCl

NHON CH3

yellow oily liquid

Multiple 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 ONE OR MORE is/are correct. 9. Which of the following will give single product on treatment with KOH? (A) PhCOCOPh (B) HCOCOH

(C) OHC(CH2)5CHO (D) O CHO Sol. A, B, C 10. Given below are four reactions. The reactions in which sum of co-efficient of reactants after

balancing either 3 or 4 are:

(A) 22 2 4

2-3

- +Cl + SeO + H O SeO + Cl + H (B) 4 2 3 2 3FeSO Fe O + SO + SO

(C) 42 3 2- - - -Cl + IO + OH Cl + IO + H O (D) 2 2

- - -Br + OH BrO + Br + H O Sol. A, C, D (A) 2 2

2 3 2 4Cl SeO H O SeO 2Cl 2H

(B) 4 2 3 2 32FeSO Fe O SO SO

(C) 2 3 4 2Cl IO 2OH IO 2Cl H O

(D) 2 2Br 2OH BrO Br H O

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11. An organic compound (A) having the molecular formula C4H6O3 upon heating gives (B) C3H6O.

(B)CH3CHO

OH- (C)

(i) CH3Li

(ii) NH4Cl (aq)(D) (major)

(D)Conc. H2SO4

850C(E)

H2/Ni(F)

Br2/ h(G) (major)

mono bromination

Choose the correct options (A)

(B)

(C)

(D)

Compound (D) can be :

Compound (C) is

OH

O

(G) is optically inactive due to external compensation.

Compound (A) can be OH

COOH

Sol. A, B

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COOH

O O

CH3CHO

OH-

(A) (B) (C)

(i) CH3Li

(ii) NH4Cl (aq)

Conc. H2SO4

850C

(E)(D)

H2 / Ni

Br2 / h

Br

*

(F) (G)(racemic mixture)

O

OH

12. From the following reaction sequence:

OH CH2 C

O

HNH3/ HCN

(A)H3O

+

(B)

Compound B is (A)

B is racemic mixture(B)

OH CH2 CH

NH3

COO-

+

The acidic group present in the compound (B) are - COOH and - OH(C)

None of the above.(D) Sol. A, B

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CH2 C H

O

OH

CH2OH CH NH

+: NH3 CH2OH C

NH3

O-

H

CH2OH CH

NH2

OH

-H2O

CN-

CH2OH CH NH

CN

- H3O+

CH2OH CH NH2

COOH

OH CH2 CH

NH3

COO-

+

Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice question based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15

Metals are extracted from their ores by a wide variety of techniques. The most common ores are oxides. Most metals are obtained by direct treatment of theirs ores with chemical agents, but the extraction of certain others require electrolysis. An example of the former type of process is the extraction of iron from its oxide, described by the following equation: 2 3Fe O + 3C 2Fe + 3CO The relative ease of extraction of a metal from its oxide can be estimated using the Ellingham diagram, as shown in the figure. The diagram plots the free energies of formation of various oxides per mole of consumed oxygen as a function of absolute temperature.

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0oC273 K 673 K 1073 K

Temperature

1473 K 1873 K 2273 K800oC 1200oC400oC 1600oC 2000oC

0-100

-200

-300

-400

-500

-600

-700

-800

-900

-1000

-1100-1200

G/k

J m

ol - 1

of m

etal

oxi

de fo

rmat

ion

C + O2 CO2 2Fe + O 2

2FeO

2C + O2 2CO

2CO + O2 2CO2

13. All of the following reagents could be used to extract lead from its oxide except

(A) Molecular fluorine (B) Molecular boron (C) Molecular Nitrogen (D) Elemental carbon

Sol. A 14. Which of the following best explains why the free energy of formation of Fe2O3 becomes less

negative as the temperature is increased: (A) The free energy of formation is independent of the absolute temperature. (B) Entropy drops as a result of the consumption of oxygen. (C) At low temperature, the free energy of formation becomes less dependent on the enthalpy of

formation. (D) As entropy increases, the free energy of formation increases. Sol. B 15. When ferric oxide is reduced to obtain iron metal, CO2 or CO can be produced. The production of

CO2. (A) is directly proportional to the temperature. (B) is inversely proportional to the temperature. (C) is independent of the temperature. (D) requires cooling of the CO. Sol. C


An ordinary D-(+)-glucose has melting point 1460C and specific rotation 25 oα = +112 .D One another

form of D-(+)- glucose has melting point 1500C and specific rotation 25 oα = +18.7 .D The two forms have significantly different optical rotations but when an aqueous solution of either form is allowed to stand, its rotations varies differently with the other. The specific rotation of one form decreases and rotation of the other increases, until both solutions show the same value e.q + 52.70. This change in rotation towards an equilibrium value is called mutarotation.

FPT-X-(Paper-1)-SOL


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O

HO

HO

OH OH

OHCHO

CH2OH

H OH

H OH

HO H

H OH O

HO

HO

OH

OH

OH

25 oα = +112 .D

25 oα = +18.7 .D

16. Mutarotation is a characteristic feature of (A) epimers (B) Enantiomers (C) Anomers (D) ring chain isomers

Sol. C . & forms are anomers

17. Two significant forms which exhibit mutarotation assigned and forms. For mannose, the

mutarotation can be shown in brief as follows

O

HO

HO

OH

OH HOO

HO

HO OH

OHHO

which form of mannose exists predominantly in the aqueous form?

(A) -form (B) -form (C) open chain form (D) None these

Sol. A form is trans to adjacent OH which assumes greater stability. 18. What percentage of -D(+) glucopyranose is found at equilibrium in the aqueous solution?

(A) 50% (B) 100% (C) 36% (D) 64%

Sol. D Let form be x

form is (100 – x) [18.7 × x] + [(100 – x) × 1120] = 52.70

x = 64%

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SECTION – B

Matrix-Match Type This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following.

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Match Column – I with Column – II:

Column – I Column – II (A) CuSO4 (p) Brown precipitate with K4[Fe(CN)6] (B) NaBr (q) Colourless gas evolved with dil H2SO4 (C) KNO2 (r) Gives yellow turbidity with dil HCl (D) CaS2O3 (s) Gives brown fumes with conc. H2SO4 (t) Gives white ppt. with BaCl2 solution

Sol. A → (p, t), B → (q, s), C → (q, s) D → (q, r, t) (A) CuSO4 + 4 6

K Fe CN 2 6Choclate brown

Cu Fe CN

CuSO4 + BaCl2 4white

BaSO + CuCl2

(B) NaBr +

2 4Conc.

H SO 2Brown fumes

Br + NaHSo4 +

2 2colourless gas

SO H O

(C) 2KNO2 + 2 4H SO 2

2unstableHNO NO + NO2 + H2O + K2SO4

(D) CaS2O3 + HCl CaCl2 + yellow turbidity

S + SO2 (Colorless gas)

CaS2O3 + BaCl2 2 3white

BaS O + CaCl2

2. Match the compounds in column I when treated separately with (i) H3O+ and (ii)

CH3MgBr(excess) followed by hydrolysis with the products (major + minor) obtained in column – II

Column – I Column – II (expected products major + minor)

(A) CH3CN (p) CH3COOH (B) CH3CONH2 (q) CH3COCH3 (C) CH3COOC(CH3)3 (r) NH4

+ (D) CH3COCl (s) (CH3)3COH (t) CH4

Sol. A → (p, q, r, s), B → (p, r, t), C → (p, q, s) D → (p, q, s) CH3MgBr is in excess

(A) CH3CNi) CH3MgBr

ii) H3O+ CH3 C

O

CH3

(q)

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i) CH3MgBr

ii) H3O+CH3 C

O

CH3 (CH3) - C - OH(s)

3H O3 3 4 4

(r)(p)CH CN CH COOH NH OH(i.e.NH )

CH3 C

O

NH2(B) H3O+

CH3COOH + NH4

+

(p) (r)

CH3 C

O

NH2 + CH3MgBr CH3 C

O

NHMgBr + CH4(t)

CH3 C

O

O C(CH3)3(C) H3O+

CH3COOH + (CH3)3C-OH(p) (s)

CH3 C

O

O C(CH3)3

i) CH3MgBr

ii) H3O+ CH3 C

O

CH3(q)

CH3 C

O

CH3

i) CH3MgBr

ii) H3O+

(excess)(CH3)3C-OH

(s)

CH3C

O

Cl(D)H3O

+

CH3COOH + HCl(p)

CH3 C

O

Cli) CH3MgBr

ii) H3O+ CH3 C

O

CH3(q)

+ Mg(OH)Cl

CH3 C

O

CH3

i) CH3MgBr

ii) H3O+

(excess)(CH3)3C-OH

(s)

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MMaatthheemmaattiiccss PART – II

SECTION – A

Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONLY ONE is correct.

1. Let S be the set of all points (x, y) in the coordinate plane such that 0 x2

and 0 y2

,

then the area of the subset of S for which 2 2 3sin x sinxsiny sin y4

is

(A) 2

9 (B)

2

8

(C) 2

6 (D)

2316

Sol. C

2 2 3siny sin y 4 sin y4 1 3sin x sin y cosy

2 2 2

.

sinx cos siny cos y sin y3 3 3

Written S, sinx = sin y3

implies x = y -

3 .

However, the case sinx sin y3

implies x y

3

0,2

,2 2

,02

x y3

0,0

2x y3

x y3

when y6

, and 2x y3

when y6

. Those three lines divide the region S into four sub

regions. Testing the points (0, 0), ,0 , 0,2 2

and ,2 2

shows that the inequality is true only

in the shaded sub region. The area of this sub region is 2 2 2 21 1. 2. .

2 2 3 2 6 6

.

2. The expression 20x 2y contain two terms with the same coefficient, a bk x y and a 1 b 1k x y , then a is

(A) 8 (B) 5 (C) 7 (D) 6 Sol. D Let r20 20 r a b

r 1 rT C x 2y k x y and s20 20 s a 1 b 1

s 1 sT C x 2y kx y for given condition

r – s = 1 and 20 1 s 20 s1 s sC .2 C 2 s 13 then r = 14 a 6 .

3. Let

2 2sin x cos xp 144 144 , then the number of integral values of such p’s can take are (A) 124 (B) 123 (C) 144 (D) 122

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Sol. D t 1 tf t 144 144 , where 2t sin x,0 t 1

t 1 tf ' t ln144 144 144 , we have f’(t) = 0 at 1t2

, f t increasing in 1,12

and decreasing

1in 0,2

and min max1f t f 24 and f t2

at f 0 f 1 145 .

Since f is continuous from, the intermediate mean value theorem then says every integers between 24 and 145 is also attainable.

4. Consider ABC with ABC = 2 ACB and BAC 90 given that perpendicular to AC through

C meets AB at D, then value of BC BCAB BD

is

(A) 1 (B) 1/2 (C) 2 (D) 4 Sol. C Using sine rule in

ABC sin 3xBC

AB sinx

, BCD sin 3x

BC 2BD sin x

2

BC BC sin3x cos3x sin3xcosx sin xcos3x sin2x 2AB BD sin x cos x sin xcos x sinxcosx

A B D

C

x

2x 3x 90-3x

90

5. Let f(x) = 11 x

and let f f f f...... xg x,n

n times , then

nlim g x,n

at x = 1 is

(A) 5 12 (B) 5 1

2

(C) 3 22 (D) 2 3

2

Sol. A

Let 1y ,1 x

we get 1 1 5y y1 y 2

because at 1n f x x or

nn 1n n nlim g x,n f lim f x lim f x f y y

.

6. For f x mx the area of the triangle formed by (0, 0) the first quadrant point P a,f a and the

reflection of that point P about y = x is 1000. If m and a are positive integers, then m is (A) 9 (B) 8 (C) 4 (D) 7 Sol. A

Let, then we have points (0, 0), (a, ma), (ma, a) area 0 0 1

1 a ma 1 10002

ma a 1

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2 2a m 1 2000 25.80

so a = 5, 2m 1 80 m = 9.

7. Let z be a complex number such that 12z 1z

, and arg(z) = , then minimum value of 28sin

is (A) 1 (B) 3 (C) 4 (D) 7 Sol. D 8. In a parabola 2y x , vertex and two of its points are chosen to form a right triangle ABC with C

as a right angle and vertex. If A is on 2nd quadrant and area of triangle is 1, then the sum of B’s x-coordinates and y-coordinates is

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

Sol. D AC BC

2 2b a. 1 ab 1

b a …(1)

a < 0, b > 0

11 AC.BC AC.BC 22

2 4 2 4a a b b 2 using (1)

A

B

C (0,0)

2a,a 2b,b

2 4 22 4 21 1 1b b 2 2 b 2

b b b

221b 2 b 1

b

b 0,b 1 Hence B(1, 1) Answer 1 + 1 = 2.

Multiple 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 ONE OR MORE is/are correct.

9. Let functions f, g : R R defined by 1f(x) = 3x 1 2x 1, g x 3x 5 2x 55

. Then

which relation between f and g is/are true (A) gof = fog (B) 1fog gof

(C) fof gog (D) 1fof gog Sol. A, B, D

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f(x) =

1x5x 2x 2 1x

2

, g(x) =

5x x255x 2 x2

after calculation f g x x gf x

1f g and 1g f

Hence 11 1 1fof g og gog

and 1fog gof . 10. If in a right angled triangle, having integer sides the perimeter of triangle is equal to the area of

triangle then possible area of triangle is/are (A) 12 (B) 18 (C) 24 (D) 30 Sol. C, D

2 2 4 a 2aba b a b b

2 a 4

a 5,b 12,a 6,b 8 . 11. For the real values of , the curve 22 2x y 2y 1 x 2y 3 represents

(A) an ellipse, if 5 (B) an ellipse, if 0 5 (C) an hyperbola, if 5 (D) an hyperbola, if 0 5 Sol. A, D

22 2 2x y 2y 1 x y 1 is the distance between point x,y and 0, 1 . The distance

from (x, y) to the line x 2y 3 0 is x 2y 3

5

.

The ratio of these 2 distances is the constant 5

for ellipse 5 1 5

.

12. Let 1 2 3 na ,a ,a ......a be a permutation of 1,2,3......n for which 1 2 3 n

2

a a a ...... a and

n n n n1 22 2 2

a a a ... a

for n as an even positive integer. Also 1 2 3 n 12

a a a ..... a and

n 1 n 1 n12 2

a a ... a

for n as an odd positive integers. Let the total number of permutation of n

be p(n), if 200 p n 500, then values of n is/are (A) 10 (B) 11 (C) 12 (D) 13

Sol. B, C an/2 and a(n+1)/2 are both 1, Note that the number of ways to choose numbers left of an/2 and

n 1 1 2 3 n n 12 2 2

a a a a ...a ;a

is same as that of right. We started with 9

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84

94

n 9 C 70 rejected

n 10 C 126 rejected

10

511

5

n 11 C 252

n 12 C 462

126n 13 C 924 rejected .

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice question based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.


For all x 1 , let f(x) be the function defined as

1 1x . x x , if x x2 x x

f x1f x , otherwisex

and let g(x) = x 102 (where [.] denote greatest integer function) 13. The function f(x) must satisfies the condition

(A) 1f 4 f 48

(B) 1 1f 4 f 44 8

(C) 1f 4 f 43

(D) 1f 4 f 44

Sol. D

For a fixed positive integer n in range 1n,nn

, f(x) = 21 1n x n nx x

2n 2

for example when n = 4 ,f(x) = 14x 162

for 14 x 44

, 1 1f 4 f 44 2

and 1f 4 08

In general are 1 1f n f nn 2

and 1f n 02n

Thus graph of f on 1n,nn

is like V shape.

14. The number of points at which f(x) is not differentiable in the interval x 4,5 , is (A) 8 (B) 7 (C) 4 (D) 2 Sol. B

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For x 4,5 graph is

n 1n2n

1nn

12

15. The number of points at which graphs of f(x) and g(x) intersects is (A) 8 (B) 16 (C) 72 (D) 90 Sol. C x 10g x 2 ,g 9 1/ 2 . g x positive

and increasing, then g intersects each V of graph of f twice in interval 1 x 9 , Total intersection points

92 1 2 3 .... 8 2.8. 722

.

5 4


In a ABC, D, E and F are points on the side BC such that AD, AE and AF are respectively an altitude, an angle bisector and a median of triangle ABC, if AD = 4, AE = 5 and AF = m 16. The values of m for which BAC is an acute angle is (A) 28 7m 100 (B) 0 7m 100 (C) 35 7m 100 (D) m 5 Sol. C 16–18. We may assume AB < AC OFG and AEG are collinear, O is

the circumcentre, E lies between D and F so m>5

2EF m 16 3

ADE GFE so 24 m 16 3AD.EFFG

DE 3

.

A

B

G

C D E F

O 4 5

3

16. If BAC 90 , F lies between O and G AF AO GO FG

24m m 16 33

1007m 100 m 4 0 m7

so 1005 m7

.

17. The values of m for which BAC is an right angle is (A) m > 5 (B) m = 5 (C) 7m 100 (D) 7m 100 Sol. D

If BAC 90 then F = O and AF = FG, 100m7

.

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18. The values of m for which BAC is an obtuse angle is (A) m > 5 (B) 7m 100 (C) 28 7m 100 (D) m> 4

Sol. B

If BAC 90 then O lies between F & G, AF<FG, 100m7

.

SECTION – B

Matrix – Match Type

This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following:

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Match the following

Column – I Column – II (A) The graphs of y x a b and y x c d intersects

at the points (2,5) and (8,3),the value of a b c d is

(p) 5

(B) The positive numbers x and y satisfies xy = 1 the minimum

value of 4 41 1x 4y

is

(q) 4

(C) For all real values of x, f(x) satisfies f x f 2 x 7 and

f 2 x 2 f x then f(x) + f(-x) must be

(r) 2

(D) If the only real solution to 9 25x y 4x y

is the ordered

pair (a, b), then value of a b is

(s) 1

Sol. (A) (r) (B) (s) (C) (p) (D) (r) (A). a 4 ,b = 7 , c 6 , d = 1.

(B). 2

4 4 2 2 2 21 1 1 1 1 1x 4y x 2y x y

, when 2 2x 2y .

(C). Replace x x f x f 2 x 7 …(1)

f 2 x 2 f x …(2) From equation (1)+(2) f 2 x f 2 x 9 Now f(x) + f(2 – x) = 7 …(3) Adding (1) and (3)

FPT-X-(Paper-1)-SOL


19

f x f 2 x f x f 2 x 14

f x f x 14 9 5 .

(D). 2 2x 9 y 25 4x y

2 2x 9 6x y 25 10y 6 10 4x x y y

2 2x 3 y 5

0x y

, the only solution is 3,5 .

2. Match the following

Column – I Column – II

(A) Let a, b, c be real numbers such that 23b cos sin12 12

,

23c cos sin12 12

, 2a b 2c , then which interval contain a

(p) 1/ 2,1

(B) The modulus equation x 1 a 4 a R can have real solutions for x, then a can lies in

(q) 3,

(C) The equation 4x 4x a 0, where a is real, has no real roots then a can lies in

(r) 1,13

(D) The system of equations 2 2 2x y 1,x y a will have 8

solutions, then 2a can lies in

(s) ,4

(t) 4,

Sol. (A) (r, s) (B) (p, r, s) (C) (q, t) (D) (p)

(A). 23cos cos 2 cos cos12 12

2

2 23 23 23 23a b 2c cos sin 2cos sin12 12 12 12

= 2 223 23cos sin 1

12 12 .

(B). x 1 a 4

Since 0 x 1

x 1 a 4

a 4 x 1

Max. value of a = 4 when x 1 0

Min. value of a = – when x 1 a 4 .

(D). Thus there are 8 solutions

X

Y

1

3

4

2

FPT-X-(Paper-1)-SOL


20

PPhhyyssiiccss PART – III

SECTION – A

Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. 1. In gravity free space two particle A and B of same positive charge q

and mass m are connected by light inextensible threads as shown in the figure. An electric field is set up given by E

=y i (where is a

positive constant). Now the system is rotated by small angle and released. The value of 1/2 so that both strings always remains in a

straight line is (Neglect the coulombian force between the particle)

x

y 1

2

A

B (A) only 1/2 (B) only 1/3 (C) only 3 (D) for any value of 1/2.

Sol. D 2. ABC is an equilateral metal wire frame of side length 0 kept in uniform

magnetic field of strength B. The coefficient of linear expansion of the metal frame is . If the temperature of the wire frame slowly changes according to the equation T = T0 sin t and total resistance of circuit is R, then the charge flown through the wire in first quarter cycle is

(A) 20 0B T3

8 R

(B) 20 0B T3

4 R

(C) 20 0B T3

5 R

(D) 20 0B T3

2 R

B C

A

B

Sol. D

dq = d BdAR R

3. A thin tunnel is dug along the chord of earth at a distance R/2 from the

centre of the earth. A tube fully filled with liquid fixed along the tunnel. One end of the tube is at centre of the tunnel have cross-sectional area A1 whereas other end of the tube is at the surface of the earth of cross section area A2 (A1A2). The initial efflux velocity of the liquid at point P will be

(A) 3 gR4

(B) 3 gR8

(C) 7 gR2

(D) 7 gR4

R/2

P Q

Sol. A

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21

4. An evacuated spherical balloon B of volume V in gravity free space is filled with a gas of n moles so that its volume becomes 3V at temperature T. Now this balloon is kept in a completely evacuated chamber C of volume 8 V. The container is maintained at temperature T. The number of moles of the same gas that should be injected into container so that balloon becomes stress free is

B

(A) 3n (B) 5n (C) 7 n (D) 9n Sol. C 5. A bubble of air is formed in a liquid at large depth and start moving upward. Now mark the correct

statement. It will (A) always has +ve upward acceleration (B) has +ve and then ve acceleration (C) gain terminal velocity (D) not move. Sol. A 6. A hypothetical long conductor made up of atoms contain positronium

atom (an atom contain positron as nucleus). When a strong electric field is applied along x-axis electrons detached from nucleus starts moving opposite to the field and the substance act as conductor. If the cross sectional area of the conductor is A, number density of the electron is n, electronic charge absolute value be e and drift speed of electron in ground frame be vd then current through cross section of the conductor will be

+ + + + + + + + + + + +

E

vd

(A) neAvd (B) 2neAvd (C) neAvd/2 (D) neAvd/4 Sol. B 7. One of the plates of the capacitor of capacitance C is given a charge

4Q. Now this capacitor is connected by a battery of emf 4Q/C and then battery is disconnected. Now this capacitor is connected with uncharged capacitor of same capacitance C through a switches S1 and S2 (as shown in the figure). When switches are closed charge flowing through the switch S1 will be

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

Higher potential charged capacitor

S1 S2

Sol. B 8. Each division on the main scale is 1 mm. Which of the following vernier scales will give vernier

constant equal to 0.01 mm? (A) 9 mm divided into 10 divisions (B) 90 mm divided into 100 divisions (C) 99 mm divided into 100 divisions (D) none of these Sol. C

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22

Multiple Correct Answers Type

This section contains 4 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.

9. A uniform disc of mass m and radius R is kept on frictionless horizontal table. Two particles of mass m are connected to disc by two identical light inextensible threads as shown in figure. The particles are given velocity v0 perpendicular to the length of strings. Then

(A) acceleration of the particle will be 20v

10R

(B) angular acceleration of disc will be 202

2v5R

(C) tension in the string is 20mv

20R

(D) tension in the string is 20mv

10R

2R

2R

v0

v0

Sol. A, B, D

T = 20v

m R2R

and 2TR =

2mR2

10. AB is a boundary divides two media I and II having bulk modulus in the ratio 3 : 4 and density in the ratio 4 : 3. Speed of sound in medium I is 300 m/s. Observer can move with a maximum speed of 300 m/s whereas source is stationary. The original frequency is n.

(A) for hearing maximum frequency observer have to move with speed 300 m/s

(B) for hearing maximum frequency observer have to move at an angle tan1(3/4) with x-axis

S

O 4L 3L

7L

A

B

Medium1 Medium2

x

(Observer)

(Source)

(C) for hearing maximum frequency observer have to move at an angle tan1(4/3) with x-axis (D) maximum frequency heard by the observer is 2n. Sol. A, B, D Consider refraction of sound

11. Mark the correct option. (A) There is no gravitational polarisation (like electrostatic polarisation) (B) The flux of gravitation field can never be +ve (C) There is no gravitational polarisation (like electrostatic polarisation) because the flux of

gravitation field can never be +ve (D) There is no gravitational polarisation (like electrostatic polarisation) not because of the flux of

gravitation field can never be +ve Sol. A, B, C

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12. Mark the correct option. (A) Angular momentum of atom in Bohr model is quantized. (B) Force acting on the electron is central (C) Angular momentum of atom in Bohr model is quantized because force acting on the electron

is central (D) Angular momentum of atom in Bohr model is quantized not because of force acting on the

electron is central. Sol. A, B, D

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice question based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15 Figure shows a vertical cylindrical shell of mass M and radius R whose lower end touches the frictionless horizontal floor. The cylinder can only rotate about its axis freely. Height of the cylinder is R. The inner wall of the cylinder is having helical groove. The number of turns of helical path is large, so that we can neglect jerk when a particle of mass M enters the helical path of cylinder from horizontal plane through a small hole which is made in the cylinder.

v0

13. What should be the minimum speed of the particle on horizontal plane so that it reaches on the

top of the cylinder? (A) 2gR (B) 4gR

(C) 8gR (D) 16gR Sol. B 14. What will be the velocity of the particle when it reaches the horizontal plane again? (A) zero (B) 2gR

(C) 4gR (D) 8gR Sol. A 15. If the particle were released from rest from the top of the helical path then what would be its

speed when it reaches the ground? (A) gR (B) 2gR

(C) 3gR (D) 4gR Sol. A 13-15. Use conservation of mechanical energy and angular momentum

FPT-X-(Paper-1)-SOL


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Figure shows inverted T shaped fixed pipe. Liquid is released from rest from the position shown. Liquid is non-viscous. 16. Velocity of fluid particle in vertical part of tube when level of

liquid in it remains H/2 will be

(A) 3gH4

(B) 3gH5

(C) 3gH8

(D) 5gH8

P

H

H H

H/2

Cross sectional area = A

Cross sectional area = A

Cross sectional area = 2A

Sol. C

17. The initial acceleration of the fluid particle in vertical tube will be (A) g (B) g/2 (C) g/3 (D) g/4

Sol. B

18. Initial gauge pressure at point P will be (A) 3gH/4 (B) 5gH/4 (C) gH/2 (D) gH/4

Sol. D 16-18. Using conservation of mechanical energy and equation of continuity

2x 1{ 2Ax}g H 4AH.v2 2

after differentiation we can get for x = 0, a = g/2

SECTION – B


This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following:

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Match the following Column-I Column-II (A) Current density in 2R is maximum (p)

R R 2R

E

(B) Drift speed is in 2R is minimum (q)

R

2R R

E

FPT-X-(Paper-1)-SOL


25

(C) Potential gradient in 2R is minimum

(r)

2R

R R

E

(D) Power dissipated on 2R is neither

maximum nor minimum

(s)

R

2R R

E

(t) R

2R R

E

Sol. (A) → (r, s) (B) → (p) (C) → (p) (D) → (q, t) 2. Match the following Column-I Column-II (A) Concave mirror (p)

Minimum distance between object and image is zero.

(B) Convex mirror (q)

Minimum distance between real object and real image is zero.

(C) Concave lens

(r)

Minimum distance between virtual object and virtual image is zero.

(D) Convex lens (s)

Minimum distance between its real object and real image is 4f

(t) Minimum distance between virtual object and virtual image is 4f

Sol. (A) → (p, q) (B) → (p, r) (C) → (p, t) (D) → (p, s)


ANSWERS, HINTS & SOLUTIONS (04.05.2016)

PAPER-2

Q. No. CChheemmiissttrryy MMaatthheemmaattiiccss PPhhyyssiiccss 1. B D B

2. B C B

3. B A C

4. D A A

5. A, D A, C, D B, C, D

6. A, B, C A, C A, B, C, D

7. A, B, C, D A, C A, B, C, D

8. A, B A, B, D A, B, C

9. A, B, D A, B A, D

1.

A → (r) B → (r, t) C → (p) D → (q)

(A) (p) (B) (t) (C) (p, q, r, s) (D) (q, r, s, t)

(A) → (p, r, t) (B) → (q, s) (C) → (t) (D) →(p, q, r, s)

2.

A → (r, s) B → (q, t) C → (p) D → (s, t)

(A) (s) (B) (p) (C) (r) (D) (t)

(A) → (r) (B) → (p, t) (C) → (q, s) (D) → (p, r)

1. 0 2 5

2. 3 5 4

3. 4 6 4

4. 1 3 1

5. 3 6 4

6. 6 5 5

7. 8 7 5

8. 9 2 3

FIITJEE PRACTICE TEST – X

FPT-X-(Paper-2)-SOL


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CChheemmiissttrryy PART – I

SECTION – A

Straight Objective Type

This section contains 4 multiple choice questions numbered 1 to 4. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The heat of neutralization of four monobasic acids A, B, C & D against a common base are 12.4,

9.4, 13.7 and 11.2 kcal respectively. Then the order of pH of resulting solutions obtained at 1 th4

neutralization point of acids with a common strong base is: (A) B < D < A < C (B) C < A < D < B (C) B < D< C < A (D) none

Sol. B Ka value for acid heat of neutralization

order of Ka value or acidic character for acids is B < D < A < C Higher the value of Ka, lower will the be pH value and upon adding common base, order of pH values for resulting solution will be C < A < D < B

2. Consider the below given statements about alkaline earth metals. (i) Solubility of sulphates decreases down the group. (ii) Solubility of hydroxides decreases down the group (iii) Thermal stability of carbonates increases down the group (iv) Basic nature of oxides increases down the group. The correct one are. (A) (i), (ii) & (iii) (B) (i), (iii) & (iv)

(C) (i) & (iv) only (D) (ii) only Sol. B 3. A yellow metallic powder is burnt in a stream of fluorine to obtain a colourless gas X which is

thermally stable & chemically inert. Its molecule has octahedral geometry. Another colourless gas Y with same constituent atoms as that of X is obtained when sulphur dichloride is heated with soldium fluoride. Its molecule has trigonal bipyramidal geometry. Gases X & Y are respectively. (A) SF4 and S2F2 (B) SF6 and SF4 (C) NaF and NaCl (D) SF4 and SF6

Sol. B Yellow powder is sulphur

ΔS + 3F SF2 6Δ3SCl + 4NaF SF + S Cl + 4NaCl42 2 2

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4. For the reaction, A k B + C

Following data are obtained Conc. of A(in M) 2 0.5 0.25 0.0625 Time (in min) 0 20 30 50 What will be the rate of reaction (in mol. L-1 min-1), when conc of ‘A’ is 3M. (A) 2.079 x 10-3 (B) 6.75 x 10-1 (C) 7.5 x 10-2 (D) none

Sol. D The half life period of ‘A’ is independent of concentration

Reaction is first order with t1/2 = 10 min.

20.693k 6.93 1010

Rate (R) = k [A] = 6.93 x 10-2 x 3 = 0.2079 mol L-1 min-1

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. Which of the following reactions can occur?

(A)

(B)

(C)

(D)

(i) Fe, HCl

NO2

CH3

(ii) OH-

NH2

CH3

NH3 / H2S

NO2

O2N

CH3

C2H5OH

NH2

O2N

CH3

O

O

O

H+

2

OH

HO

OH

OO

CH

O CH CH CH2

CH3

CH2 hn

OCH CH2

CH2

CHCHCH3

FPT-X-(Paper-2)-SOL


4

Sol. A, D (A)

(B)

(C)

(D)

NO2

NH2

CH3

CH

O CH CH CH2

CH3

CH2

OCH CH2

CH2

CHCHCH3

Reduction of - NO 2 group.

will be the product.

O

O

OHOH

(Phenolphthalein) will be the product.

Claisen rearrangement.

6. One mole of an ideal gas (not necessarily mono atomic) is subjected to the following sequence of steps: (i) It is heated at constant volume from 300 K to 400 K. (ii) It is expanded freely into vacuum to triple its volume. (iii) It is cooled reversibly at constant P to 300 K. (A) U for the overall process = 0. (B) H for the overall process = 0. (C) Total work done is 831.4 joule mol-1. (D) Total heat content is 831.4 joule mol-1.

Sol. A, B, C (i) V = 0 w = 0, q = U, H = Cp (T2 – T1), U = CV (T2 – T1)

(ii) w = 0 , U = 0, H = 0, q = 0 (iii) – w = P V = P (V2 – V1) = R (T2 – T1) = -831.4 w = 831.4 Jmol-1

U = CV T H = CPT q = CVT - PV Total U = 0 Total H = 0 w = + 831.4 joule mol-1 q = - 831.4 joule mol-1

FPT-X-(Paper-2)-SOL


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7. Select the correct statement(s) in the following (A) Carbon has pronounced ability to form P P multiple bonds to itself and to other elements

like O and N. (B) In diamond each carbon atom is linked tetrahedrally to four other carbon atoms by sp3

hybridisation. (C) Graphite has planar hexagonal layers of carbon atoms held together by weak Vander Waal’s

forces (D) Out of CO2, SiO2, SnO2 and PbO2. CO2 , SiO2 are acidic, SnO2 is amphoteric and PbO2 is an

oxidizing agent. Sol. A, B, C, D 8. Which of the following species is/are aromatic?

(A)

(B)

(C)

(D)

Sol. A, B

9. Which of the following belong to structural isomeric pair? (A) N-methylacetamide and N, N-dimethylformamide

(B)

O andO CH3

(C) Epoxyethane and ethylene glycol (D) Benzophenone and biphenyl carbaldehyde Sol. A, B, D

SECTION – B

Matrix-Match Type This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following.

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. There are four bottles which contains Bottle – 1 Lead nitrate Bottle – 2 Hydrochloric acid Bottle – 3 Sodium carbonate

FPT-X-(Paper-2)-SOL


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Bottle – 4 Copper sulphate By mixing samples of the contents of the bottles in paris. Match the pairs in column – I with the observation in column – II.

Column – I Column – II (A) Bottle (1) + Bottle (2) (p) colourless gas evolved (B) Bottle (1) + Bottle (4) (q) no visible reaction (C) Bottle (2) + Bottle (3) (r) White precipitate (D) Bottle (2) + Bottle (4) (s) green blue precipitate

(t) the anion of precipitate is sp3

hybridized Sol. A → (r), B → (r, t), C → (p) D → (q) (A) Bottle (1) + Bottle(2) Pb(NO3)2 + 2HCl PbCl2 + 2HNO3

(white ppt) (B) Bottle (1) + Bottle(4)

Pb(NO3)2 + CuSO4 PbSO4 + Cu(NO3)2 (white ppt)

& 24SO is sp3 hybridised

(C) Bottle (2) + Bottle(3) 2HCl + Na2CO3 2NaCl + H2O + CO2(g)

(colourless gas) (D) Bottle (2) + Bottle (4) HCl + CuSO4 no reaction 2. Match the following:

Column – I (Reaction)

Column – II (Observation in products)

(A) Na2CO3 + NaHCO3, titrated by HCl in a same container using phenolphthalein & methyl orange indicator respectively.

(p) 1st titre value will give the indication of strength of 2nd constituent.

(B) Na2CO3 + NaOH, titrated by HCl in a same container using phenolphthalein & methyl orange indicator respectively.

(q) 2nd titre value will give the indication of strength of 1st constituent.

(C) Na2C2O4 + H2C2O4 titrated with NaOH using phenolphthalein indicator followed by KMnO4 titration in acidic medium

(r) Difference in 1st and 2nd titre value will give the estimation of strength of 1st constituent

(D) H2CO3 + NaHCO3, titrated by NaOH in a same container using methyl orange indicator & phenolphthalein indicator respectively.

(s) 1st titre value will give the estimation of 1st constituent.

(t) Difference in 1st and 2nd titre value will give the estimation of 2nd constituent.

Sol. A → (r, s), B → (q, t), C → (p) D → (s, t) (A) Na2CO3 + NaHCO3 Using phenolphthalein indicator: eq of Na2CO3 will react. 1st titre value Na2 CO3 + HCl NaHCO3 + NaCl 1st titre value = eq of Na2 CO3 Using methyl orange indicator : eq of NaHCO3 produced from Na2CO3 & eq of NaHCO3 present will react : NaHCO3 + HCl NaCl + H2O + CO2

FPT-X-(Paper-2)-SOL


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2nd titre value = eq of Na2CO3 + eq of NaHCO3. A r, s

(B) Na2CO3 + NaOH

Using phenolphthalein indicator : Na2CO3 + HCl NaHCO3 + NaCl NaOH + HCl NaCl + H2O 1st titre value = eq of Na2CO3 + eq of NaOH. Using methyl orange indicator : NaHCO3 + HCl NaCl + H2O + CO2 (produced from Na2CO3) 2nd titre value = eq of Na2CO3 B q, t

(C) Na2C2O4 + H2C2O4 Using NaOH & phenolphthalein indicator : H2C2O4 + NaOH Na2C2O4 + H2O 1st titre value of NaOH = eq of H2C2O4 Using KMnO4 in acidic medium : KMnO4 + Na2C2O4 + H+ CO2 + Mn2+ + H2O + Na+ 2nd titre value will give the eq of total Na2C2O4 & H2C2O4 but in different n-factor value. The estimation of Na2C2O4 can be done by the difference in moles of total Na2C2O4 + H2C2O4 in 2nd titration & moles of H2C2O4 in 1st titration. But difference in 1st & 2nd titre value will not give the estimation of 1st constituent. C p

(D) H2CO3 + NaHCO3

Using NaOH & methyl orange indicator : H2CO3 + NaOH NaHCO3 + H2O 1st titre value will give the estimation of 1st constituent. Using NaOH & phenolphthalein indicator : NaHCO3 + NaOH Na2CO3+ H2O (NaHCO3 present + NaHCO3 produced from H2CO3) Difference in 2nd & 1st titre value will give the estimation of 2nd constituent. D s, t

FPT-X-(Paper-2)-SOL


8

SECTION-C

(Integer Type) 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 darkening of bubbles will look like the as shown.

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

7 7 7 7

8 8 8 8

9 9 9 9

6 6 6 6

X Y Z W

1. The sum of E0 (to the nearest integer) for the following reactions at 298 K is:

Ag(NH3)

+ + e Ag + 2NH32

Ag(CN)2 + e Ag + 2CN-

Given : 0Ag/AgE = 0.8 V

0dK for

23 )NH(Ag = 10-8

0dK for

2)CN(Ag = 10-19.072 Sol. 0 0591.0EE 0

Ag/Ag0 log 0

dK

0Ag/)NH(Ag/NH 233

E = 0.8 – 0.0591 log10-8

= 0.3272 V ... (I) 0

Ag/)CN(Ag/CN 2E = 0.8 – 0.0591 log 10-19.072

= - 0.3272 V ... (II) (I) + (II) = 0 2. For a radioactive decay reaction :

X Y + q (where q is the number of stoichiometric coefficient) Following graph was observed:

If the ratio of the volume of emitted helium gas at different time t1 and t2 be ½. Find the value of q. Sol. 3

Time

t1 t2

Concn

FPT-X-(Paper-2)-SOL


9

X Y + q at t1 a – x x qx at t2 a – y y qy at t1, a – x = qx a = qx + x a = x (q + 1)

x = 1q

a

at t2 , a – y = y

y = 2a

volume of the gas no. of moles of the gas

2/a

1q/ayx

qyqx

1q

2yx

Given: 3q21

1q2

21

yx

3. In the following reactions:

2232 CaClbClOCaaCaOCl

HClcSONaSOHNaCl 4242

22322 SOSOHNaCledHClOSNa What will be the value of (a + b + c) – (d + e) in the balanced chemical reactions?

Sol. 4 2232 CaCl5ClOCaCaOCl6

HCl2SONaSOHNaCl2 4242

22322 SOSOHNaCl2HCl2OSNa a = 1, b = 5, c = 2, d = 2, e = 2 (a + b + c) – (d + e) = (1 + 5 + 2) – (2 + 2) = 4 4. In the following sequence of reactions:

O

O

O H3O+

(A) + (B)(alcohol)(acid)

)D()C()A(

)E()C( h/Br2 mono brominated products. What will be the sum of total number of enantiomeric pairs in A, B, C, D and E.

Sol. 1

Time

t1 t2

Concn

Y

X

FPT-X-(Paper-2)-SOL


10

O

OO

OH

OO

+ OH *

(A) (B)Retention in configuration in both (A) and (B)

OH

OO

*

H3O+

O

+ CO2

In (C) and (D) no chiral carbon present.

O

Br2/ h Br

O

Br

O O

Br

+ +

(C) (D)

(E)

*

*

(C)

Chiral carbon only in the 1st compound formed from (C). This will exist in the enantiomeric pair. Sum of total number of optically active products = 4. But total number of enantiomeric pairs = 1 5. Water insoluble gas is being collected over contaminated water in a container of capacity 4 litre at

300 K. Pressure observed was 18.5 atmosphere. Mole fraction of non-volatile solute particles in contaminated water is 0.3427. How many numbers of moles of insoluble gas must have been collected in the container? (aq. tension of pure water at 300 K is 31.8 mm Hg).

Sol. 3

aq. sB

aq.

P Px

P

s31.8 P31.8

= 0.3427

Ps = 31.8 (1 – 0.3427) = 31.8 × 0.6573 = 20.9 Now we know, Pdry = Pwet – Ps

= 18.5 - 20.8 18.47atm760

and ngas = 18.47 40.082 300

= 3 moles

6. What is the pressure at equilibrium of the following reaction at given temperature when vapour

density of equilibrium mixture is 88.5 [initial pressure is 4 atm]

2 5 23N O g 2NO g O g2

Sol. 6

FPT-X-(Paper-2)-SOL


11

2 5 2

o o o

3N O g 2NO g O g23P 1 2P P2

Now,

eq.

th

VD 5 88.5 51 1 0.2VD 2 59 2

PT = Po 512

= 4 × (1.5) = 6 atm

7. Salt AB2 has fluorite structure with ‘A’ in fcc and ‘B’ in all tetrahedral voids. 25% of ‘B’ is replaced

by ‘A’ and then 25% of available ‘A’ is replaced by ‘B’ to get a theoretically possible compound AxBy. What would be value of ‘x + y’?

Sol. 8 AB2 → (Fluorite structure) A B No. of atoms/unit 4 8

After 1st replacement 84 64

88 64

After 2nd replacement 66 4.54

66 64

= 7.5

(9/2) (15/2) So, formula of theoretical compound is A3B5 So, ‘x + y’ = 3 + 5 = 8 8.

2

2

i HNO conc.ii Ph CH CN; X,

Compound ‘X’ is having no. of pairs of e Sol. 9

2HNO conc.

NO

2PhCH CN

Ph N CCN

Ph

FPT-X-(Paper-2)-SOL


12

MMaatthheemmaattiiccss PART – II

SECTION – A

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. Two points A and B in rectangular coordinates system lie on the curve y = 5x – 1 such that

ÔA i 2

, ÔB j 25

where ˆ î and j are the unit vectors along x and y axes respectively. Then

OB 3OA

is equal to (A) 5 (B) 10 (C) 15 (D) none of these Sol. D ˆ ˆ ˆ ÔA 2i 5 j, OB 3i 25j

2 2OB 3OA 3 6 25 15 9 100 109

. 2. If , ( ) are the abscissae of two points on the curve f(x) = x – x2 in the interval (0, 1), then

the maximum value of the expression ( + ) – (2 + 2) will not exceed (A) 2 (B) 1

(C) 12

(D) 14

Sol. C

f(, ) = ( – 2) + ( – 2) = y1 + y2 < 1 1 14 4 2 .

3. If is the only real root of the equation x3 + Ax2+ Bx + 1 = 0, A < B, then the value of tan–1 +

1 1tan

equals

(A) – 2 (B)

2

(C) 0 (D) Sol. A f(x) = x3 + Ax2+ Bx + 1, f(0) = 1, f(–1) = A – B < 0 lies between – 1 and 0 < 0

1 1tan

= – + cot–1

tan–1 – + cot–1 = 2 – = –

2 .

4. If two roots of the equation (p –1)(x2 + x + 1)2 – (p + 1)(x4 + x2 + 1) = 0 are real and distinct and

f(x) = 1 x1 x

, then f(f(x)) + 1f fx

is equal to

(A) p (B) – p (C) 2p (D) – 2p

FPT-X-(Paper-2)-SOL


13

Sol. A

2

2p 1 x x 1p 1 x x 1

222p x 12 2x

p = x + 1x

As, f(x) = 1 x1 x

f(f(x)) + 1f fx

= x + 1x

f(f(x)) + 1f fx

= p.

Multiple Correct Choice Types

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. Let a, b, c be three real numbers such that 1 a b c 0. If (real or complex) is a root of the

cubic equations x3 + ax2 + bx + c = 0 then (A) = 1 (B) 1

(C) 1 (D) = 1 Sol. A, C, D 3 = a2 b c 4 = a3 b2 c = (1 a) 3 + (a b)2 + (b c) + c if 1 , then 4 3 21 a a b b c c

3 3 3 3 31 a a b b c c 1

Hence the only possibility is 1

Then 1 . 6. Let A, B, C be nn real matrices and product is pair wise commutative also ABC = On, if =

det(A3 + B3 + C3). det (A + B + C) then (A) > 0 (B) < 0 (C) = 0 (D) (-, ) – {0} Sol. A, C 3 3 3 3 3 3A B C A B C 3ABC = 2 2 2A B C A B C AB BC CA

= 2 2A B C A B C A B C [ is complex cube root of unity]

= 2 2A B C A B C A B C

thus det(A3 + B3 + C3).det(A + B + C)

det(A + B + C)2.det(A + B + 2C)det 2A B C

= 22 2det A B C . det A B C 0 .

FPT-X-(Paper-2)-SOL


14

7. Let f: A A, where A is a set of non-negative integers and satisfying the condition. (i) x f(x) = 19 [x/19] 90[f(x)/90], x A (ii) 1900 < f(1990) < 2000 then f(1990) equals, (where [.] denote greatest integer function) (A) 1994 (B) 1900 (C) 1904 (D) 1890 Sol. A, C

Put x = 1990, we get f 1990

90

= f(1990) 14

f 1990

90

= 21 or 22

if f 1990

90

= 21 then f(1990) = 1904

or f 1990

90

= 22 then f(1990) = 1994.

8. If the equation x4 + ax3 + bx2 + cx + d = 0 has four real positive roots then. (A) ac 16d (B) b2 36d (C) ac 16b (D) b2 36c Sol. A, B, D xi a , 1 2x x b , 1 2 3x x x c , 1 2 3 4x x x x d

Using AM HM ac = 4

i 1xi

4

i 1

1x

x1 x2 x3 x4 16d

Using AM GM b = 1/ 63 3 3 31 2 1 2 3 4x x 6 x x x x = 6d.

9. In a triangle ABC if area ( 0) is constant. For the given constant angle C if side AB is minimum

then

(A) a = 2sinC (B) b = 2

sinC

(C) a = 4sinC (D) b = 4

sinc

Sol. A, B c2 = a2 + b2 2abcosC = (a b)2 + 2ab(1 cosC) = 1/2 absinC

hence c2 = (a b)2 + 4 1 cosCsinC

= (a b)2 + 4tanc/2 4tanc/2 is constant 80 for c2 to be minimum a = b then 2ab = 2a2 = 4/sinC

a = b = 2sinC .

FPT-X-(Paper-2)-SOL


15

SECTION – B


This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following:

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Let C1 x2 + y2 = 16 and C2 (x – 8)2 + y2 = 4 be the equations of two circles. Tangents are

drawn from any point P on C1 touching C2 at A and B respectively. If Q be a variable point on C2 and L, M, N are the feet of perpendiculars from Q on PA, AB, PB respectively. Then match the following

Column – I Column – II (A) If ‘d’ is the maximum distance of P from the radical axis of C1 and

C2, then [d] (where [.] denote greatest integer function) (p) 6

(B) If PBPG PB

, then is equal to (where G is the foot of

perpendicular from P on the radical axis of C1 and C2)

(q) 10

(C) If P(2, 2 3 ) and Q(9, 3 ), then QL . QM > (r) 13

(D) If a line through P(2, 2 3 ) intersects C2 at R and Q then PR PQ <

(s) 12

(t) 16 Sol. (A) (p) (B) (t) (C) (p, q, r, s) (D) (q, r, s, t) (A). Equation of radical axis of C1 and C2

is C1 – C2 = 0 x = 194

Maximum distance is obtained when P coincides with H.

(B). Using the property |PA|2 =

2|PG||OO| (C). Using the property |QL||QM| = |QN|2 equation of AB is 3x – 3 y – 22 = 0

|QN| = 1312

, |QN|2 = 16912

.

(D). Using property PT2 = PR . PQ.

P

H(–4, 0)

B

M

Q

L A

N

O O

FPT-X-(Paper-2)-SOL


16

2. Let (x + 1)(x + 2)(x + 3) ... (x + n – 1)(x + n) = A0 + A1x + A2x2 + ... + Anxn. Column – I Column – II

(A) A0 + A1 + A2 + ... + An = (p) (n + 1)! 1 1 11 ...

2 3 n 1

(B) A0 + 2A1 + 3A2 + ... + (n + 1)An = (q) (n + 1)! 1 1 1...

2 3 n 1

(C) nA0 + (n – 1)A1 + (n – 2)A2 + ... + An – 1 = (r) (n + 1)! 1 2 3 n...

2 3 4 n 1

(D) 2A0 + 3A1 + 4A2 + ... + (n + 2)An = (s) (n + 1)!

(t) (n + 1)! 1 1 11 1 ...2 3 n 1

Sol. (A) (s) (B) (p) (C) (r) (D) (t) (A). Put x = 1. (B). Multiply by x, then differentiate and put x = 1.

(C). Replace x by 1x

and then differentiate, put x = 1.

(D). Multiply by x2, then differentiate and put x = 1.

SECTION – C

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. 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 darkening of bubbles will look like the following:

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

7 7 7 7

8 8 8 8

9 9 9 9

6 6 6 6

X Y Z W

1. If the equations x2 3x ai = 0 has integral roots ai N and ai 300 then i1 a

800 =

_________ Sol. 2 D = 9 + 4a = odd number = perfect square (2k + 1)2 = 9 + 4ai, k I (4k2 + 4k + 1) = 9 + 4ai 4(k2 + k) = 4(ai + 2) k(k + 1) 2 = ai Thus ai = 4, 10, 18, 28, 40, …………. So general term tr = r2 + 3r T15 = 270, t16 = 304

FPT-X-(Paper-2)-SOL


17

15 15 15

2i

i 1 i 1 i 1a r 3 r 1600

.

2. If a 9 digit number is formed using the digits 1 to 9 without repetition are if the probability that it

will be divisible by 11 is p/q then q – p2 equals _________ (p and q are co-prime to each other) Sol. 5 x = x1 + x3 + x5 + x7 + x9 Y = x2 + x4 + x6 + x8 (x y) = 11k, k z+ of x y = 11 and x + y = 45 ie x = 28, y = 17 total numbers (3 4! 5!) + (6 + 4! + 5!) of y x = 1 y + x = 45 ie x = 17, y = 28 total numbers 2 4! 5!

Then all such 9 digit numbers are 1!.4!.5! Required probability p 1!.4!.5! 11q 9! 126

3. Number of ways to distribute 5 identical ball among 3 children such that no child goes empty

handed is _________ Sol. 6 Required number of ways = 5 1 4

3 1 2C C 6 .

4. f(x) + 1f 1x

= 1 + x for x R – {0, 1}. The value of 4f(2) is equal to _________

Sol. 3

Let y = 1 – 1x

f(y) + 11fy

= 1 xf f1 xx x 1

= 11 1f f 21xx 1 x

…(1)

put z = 11 x

f(z) + 1f 1z

= 11f f 1x

1 x1 x

…(2)

subtract

f(x) – 1 11f 111 x xx

f(x) = 1 1 1 x2 1 x x

.

Alternate:

Putting 1x 2,2

and – 1 successively

f (2) + f (1/2) = 3 … (1) f (1/2) + f (– 1) = 3/2 … (2) and f (– 1) + f (2) = 0 … (3) Solving, we get f(2) = 3/4.

FPT-X-(Paper-2)-SOL


18

5. If the line 2x y 1 1, 1 always touches a fixed hyperbola 2 2

2 2x y 1a b

, let the

eccentricity of the hyperbola is 1k56

k 1

, then k is equal to_____________.

Sol. 6 Applying the condition of tangency 2 2 2 2c a m b , we get 2 2 2 21 a b 1 or

2 2 2 2a b b 1 0, > 1 which is an identity in , so a2 – b2 = 0 and b2 – 1 = 0

c = 2 k = 6. 6. If , 2 N are the roots of the equation x2 – 11x + = 0, then the number of possible distinct

value of is/are ________ Sol. 5 + 2 = 11 and (2) = (, 2) = (9, 2), (7, 4), (5, 6), (3, 8), (1, 10).

7. The number of integral values of a for which 2

2ax 3x 4a 3x 4x

takes all values for all real values of x,

is _________ Sol. 7

Let y = 2

2ax 3x 4a 3x 4x

a(1 – y)2 + 4(4 + ay)(a + 4y) 0 9 + 16a > 0 and 4(2a2 + 23)2 – 4(a + 16a)2 0 a [1, 7]. 8. The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60. If

the third side is 3, then remaining fourth side is _________ Sol. 2

BD2 = 22 + 52 – 2 2 5 12

= 32 + x2 – 2 x 3 – 12

x2 + 3x – 10 = 0 x = 2.

A

D

C

B 2

5

3

FPT-X-(Paper-2)-SOL


19

PPhhyyssiiccss PART – III

SECTION – A

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. 1. A ring is falling under gravity in a uniform external magnetic field

which is present in space has magnitude B0 and is radially outward from the axis of the ring. Then

(A) The emf generated in the loop will be zero.

g A

B

C

(B) The flux of magnetic field over ring at any instant of time is zero. (C) The emf generated in the loop will be zero because the flux of magnetic field over ring at any

instant of time is zero. (D) The emf generated in the loop will be zero not because of the flux of magnetic field over ring

at any instant of time is zero.

Sol. B

2. A vehicle is moving on a circular path of radius R with constant speed gR . A simple pendulum of length hangs from the ceiling of the vehicle. The time period of oscillations of the pendulum is

(A) 2g

(B) 2

2g

(C) 22g

(D) none of these

Sol. B 3. A tube of radius r and sufficient height is dipped in a liquid of surface tension S and density .

Heat developed upto steady state will be

(A) 2 2S cos

g

(B)

2 2S cos2 g

(C) 2 22 S cos

g

(D)

2 24 S cosg

Sol. C Work done by 2rScos = gain in P.E. + Heat developed 4. In the figure AB is a frictionless fixed rod. The ring R is constrained to

move along AB. P is a particle which is connected with the ring by means of a thread. Mass of the ring and thread are negligible in comparison to that of particle P so that you can take them to be zero. A horizontal constant force F is applied on the particle. The time required to move the ring by distance x is

A B

R

P F

(A) 2mxF

(B) mxF

(C) mx4F

(D) mx

Sol. A

FPT-X-(Paper-2)-SOL


20

Multiple Correct Answers 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. 5. Consider a nuclear reaction A 1 B + C B 2 C A converts into B and C with decay constant 1. B is also an unstable nucleus which further

decay into stable nucleus C with decay constant 2. Mark the correct statement(s).

(A) CdNdt

= (1NA + 2NB) (B) AdNdt

= 1NA

(C) CdNdt

= (1NA + 2NB) (D) BdNdt

= 1NA 2NB

Sol. B, C, D 6. Figure shows an electric circuit contain four resistors of equal

resistances 4. Cells E1, E2, E3 are ideal of unknown emf where as cell E4 has some unknown internal resistance and emf 4V. It is found that current through EA, DH, FB and HE are 7A, 6A, 2A and 5A respectively.

(A) Internal resistance of E4 is 2 (B) Current through DC is 0.5 A. (C) Current through AD is 5.5 A. (D) E2 = 6V.

E3

R

R R

R

A D

C B

F G

H E E1

E4

E2

6A 7A

2A

5A

Sol. A, B, C, D 7. A particle of mass m is suspended from two identical threads each

of length area of cross section A and high Young’s modulus Y so

that elongation in the string is small in comparison to its original length.

45 45

(A) The maximum downfall of the particle is 2mgYA

(B) The down fall of the particle at equilibrium is mgYA

(C) Equilibrium of particle is stable

(D) Maximum elastic potential energy of the system is 2 2m g

2YA

Sol. A, B, C, D

FPT-X-(Paper-2)-SOL


21

8. Six charges of equal magnitude but opposite sign are fixed on an insulating ring at equal separation. P is a general point on the axis of ring at large distance from the centre of ring. PR is a path perpendicular to the axis of the ring and R is at infinite separation from P.

(A) Amount of work done to bring a charge from R to P via RP is zero.

(B) Net dipole moment is zero and electric field is conservative field.

(C) Amount of work done to bring a charge from R to P via RP is zero because net dipole moment is zero and electric field is conservative field.

+q

q q

+q +q

q

P

R

(D) Amount of work done to bring a charge from R to P via RP is zero not because of net dipole moment is zero and electric field is conservative field.

Sol. A, B, C 9. The side of a cube is 8 cm. The volume of the cube with due respect to significant figure is (A) 500 cm3 (B) 512 cm3

(C) 510 cm3 (D) 5 102 cm3

Sol. A, D

SECTION – B


This section contains 2 questions. Each question contains 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; then the correct darkening of bubbles will look like the following:

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Match the following Column-I Column-II (A) All the particles perform SHM with

same amplitude (p)

Plane progressive harmonic wave on string

(B) All the particles perform SHM with different amplitude

(q)

Standing wave on string

(C) Total mechanical energy of a particular particle does not change with time.

(r)

Plane progressive harmonic sound wave

(D) Total mechanical energy of a particle changes with time.

(s)

Vibration of air column

(t) Simple harmonic motion of a block. Sol. (A) → (p, r, t) (B) → (q, s) (C) → (t) (D) → (p, q, r, s)

FPT-X-(Paper-2)-SOL


22

2. AB is a part of a uniform rod having cross sectional area A thermal conductivity k. Length of AB is L. Temperature of rod is defined as T = T0 cos (2/L)x. Assume that surface of the rod is thermally isolated from atmosphere.

x = 0 A B L

Y

x

Match the following Column-I Column-II (A) For small piece of rod at x = 0 (p) Heat flows from left to right (B) For small piece of rod at x = L/4 (q) Heat flows from right to left (C) For small piece of rod at x = 3L/5 (r) There is source of heat (D) For small piece of rod at x = L/8 (s) There is sink of heat (t) There is neither source nor sink of heat Sol. (A) → (r) (B) → (p, t) (C) → (q, s) (D) → (p, r)

SECTION –C

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. 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 darkening of bubbles will look like as shown.

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

7 7 7 7

8 8 8 8

9 9 9 9

6 6 6 6

X Y Z W

1. Find the value of x/4 (in cm) for which all the (paraxial)

rays coming from O emerges from the plane refracting surface along a single path.

f = 10 cm

=2

R = 10 cm

30 cm 60

O

x

Sol. 5 2. Two particles A and B of same mass are connected by means of light

inextenisble string of length and kept on an inclined plane of inclination (sin = 3/5). The coefficient of friction between A and plane is where as no friction act between B and inclined plane. B is projected with some speed to complete circle on inclined plane. The minimum value of for completion

of circle is 21n

, find the value of n.

A

B

Sol. 4

FPT-X-(Paper-2)-SOL


23

3. Find the ratio of radius of second Bohr’s orbit of helium(hydrogen like) and first orbit of hydrogen. Sol. 4 r n2

4. In the figure shown there is no slipping anywhere. Mass of plank

and each sphere is m. The ratio of acceleration of C.M. of bigger and smaller sphere will be

2R R Sol. 1 5. The figure shows a circular ring of radius R = 0.5 m and

mass 0.5 kg. Part ABC and ADC carries uniform charge +C and C. When uniform electric field of 1 N/C i is applied and ring is slightly rotated and released. Find the angular frequency (in rad/sec) of small oscillation.

B

A

C

D

x

y

Sol. 4 6. In a photoelectric experiment when electromagnetic wave given by E = E0 sin t is incident

electron just ejects. When E = E0 sin 2t is incident Kmax = K1 and when E = E0 cos6t is incident Kmax = K2. Find K2/K1.

Sol. 5 7. One end of a uniform string is tied to ceiling. Lower end of the string is vibrated according to the

law x = A sin(10t + ). Find the value of A (in cm), so that maximum acceleration of particle at any point equals acceleration of wave at that point. (consider all physical quantities in SI units in the equation)

Sol. 5 8. The figure shows a cubical container fully filled with a homogeneous

liquid and sealed from all sides. This container is moved horizontally with acceleration g. Find the ratio of force of gauge pressure acting on ADHE to force of gauge pressure on EFGH.

D

E F

G H

A B C

g

Sol. 3

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