PAPER-1 (09.05.2017)€¦ · Integer Answer Type This section contains 8 questions. The answer to...

FPT-V-(Paper-1)-SOL

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ANSWERS, HINTS & SOLUTIONS

PAPER-1 (09.05.2017)

Q. No. CChheemmiissttrryy MMaatthheemmaattiiccss PPhhyyssiiccss

1. A D B

2. C B A

3. C B B

4. C B B

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

6. A C, D B, C

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

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

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

1.

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

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

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

2.

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

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

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

1. 2 6 6

2. 1 4 4

3. 3 4 2

4. 6 9 4

5. 3 5 2

6. 4 8 2

7. 9 2 2

8. 3 6 6

FIITJEE PRACTICE TEST – V

FPT-V-(Paper-1)-SOL


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CChheemmiissttrryy PART – I

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. Read the following statements carefully (I) The number of maximum (peaks) in radial distribution is n – (II) Angular moment of 1s, 2s, 3s electron are equal (III) 3d10 configuration is more stable than 3d5 configuration (IV) If electron has zero magnetic number, then it must be present in s-orbital Choose the correct statement: (A) I, II, III (B) II, III, IV (C) I, III, IV (D) I, IV Sol. A 2. 5 mole of equimolar mixture of ferric oxalate and ferrous oxalate will require x mole of KMnO4. In

acidic medium for complete oxidation, x is: (A) 4.0 (B) 4.5 (C) 8.0 (D) 0.9 Sol. C

4KMnO 2 32 4 2 2 4 23 nf 5FeC O Fe C O Mn Fe CO

nf 3 nf 6

1 KMnO4 moles for FeC2O4 oxidation = 32.5 1.55

KMnO4 moles for Fe2(C2O4) = 62.5 3.05

Total moles of KMnO4 = 4.5 3. Which of the following statement is incorrect regarding sulphur compounds (A) H2S2O8 and H2SO5 both have one O – O linkage (B) In SF6 all S – F bonds are equivalent (C) SCl2(OCH3)2 and SF2(OCH3)2 – OCH3 groups in both case occupy the equatorial position (D) In H2S2O3 oxidation number of sulphur is +4 and 0 Sol. C 4. On passing H2S gas into the aqueous solution of an inorganic salt gives a black precipitate which

dissolves in aquaregia. The solution is evaporated to dryness and residue is extracted with dilute HCl. When hot nitrite solution is added along with the excess of acetic acid into this solution, a yellow coloured precipitate is obtained. Hence the inorganic salt may be:

(A) NiCl2 (B) Hg(NO3)2 (C) CoCl2 (D) MnCl2

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Sol. C

2 2CoCl H S CoS 2HClBlack

2 3 2 3KNO CH COOH HNO CH COOK 2 2 2 2

CoCl 2KNO Co NO 2KCl

2 2 2 22 3Co NO 2HNO Co NO NO H O

2 2 3 23 6

Co NO 3KNO K Co NO

Yellow ppt.

Hence (C) is the correct answer.

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. Consider the following cell

(Zn)

(ZnSO4)(CuSO4)

(Cu)

ZnSO4 KNO3

1 M 1 M 1 M

CuSO4

Porous Bone Porous Bone

(I) (II) (III)

If volume of solution in each compartment is 1 L and 2

oZn |ZnE 0.76V

2oCu|CuE 0.34V

Then choose the correct statement regarding the above cell after passage of 0.1 F charge (A) Moles of Cu2+ reduced in compartment (III) is 0.025 (B) Concentration of 3NO in second compartment 0.95 M

(C) In second compartment Cu2+, K+ and 3NO ions present

(D) Concentration of 24SO in compartment (III) is 0.975 M

Sol. A, B, D

Moles of Zn2+ = 0.12

= 0.05

+ve charge in 1st compartment 3NO will migrates from second compartment to 1st compartment

(Zn2+) in 1st compartment = 0.0512

=1.025 M

Compartment of 3NO in second compartment = 1 – 0.05 = 0.95 M

In 3rd compartment moles of Cu2+ reduced = 0.05 0.0252

4SO reducing in 3rd compartment 0.0251 0.975M2

FPT-V-(Paper-1)-SOL


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6. Which of the given statement is/are correct for the given reaction knowing the fact that higher the number of hyperconjugative structures higher the stability of carbocation

3 3 3 33 3 3 3CD C CH CH CH C CD CH

(A) The equilibrium constant should be higher than one for the above reaction (B) The equilibrium constant should be lower than one for the above reaction (C) (CD3)3C+ will be more stable than (CH3)3C+ due to higher electron releasing nature of

deuterium than hydrogen (D) (CD3)3C+ will be more stable than (CH3)3C+ due to presence of stronger C – D bond than

C – H bond Sol. A 7. Which parallelograms in the figure below are unit cells?

1

2

3

4

(A) 1 (B) 2 (C) 3 (D) 4 Sol. A, D 8. Find out which of the following statement(s) is(are) correct? (A) The percentage of ortho hydrogen decreases at lower temperature while that of para form

increases. (B) A solution of chromic acid in sulphuric acid is oxidized by hydrogen peroxide to give a yellow

peroxide of chromium CrO5. (C) Chlorine oxidizes hypo to sodium sulphate while Iodine is declourized due to the formation of

tetrathionate (D) white phosphorous when treated with concentrated NaOH solution in inert atmosphere of

CO2. phosphene is evolved which is inflammable due to the presence of P2H4.

Sol. A, C, D

9. Which of the following is/are correct? (A) Lime water is used for the test of carbonate radical (B) Addition of Na2CO3 to an aqueous solution of Ba2+ and Mg2+ will ppt both (C) Ammonium thiocyanate forms coloured ppt with Co2+ (D) Ammonium sulphate can be used in place of ammonium chloride in the 3rd group Sol. A, B, C

FPT-V-(Paper-1)-SOL


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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) CF2 = CF2 + HF → (p) Free radical addition (B) PhCH = CH2 + Br2 → (q) No reaction (C) HBr

(r) Nucleophilic addition

(D) (NC)2C = CHCN + NH3 → (s) Electrophilic addition (t) Nucleophilic substitution

Sol. A → (r); B → (s); C → (s); D → (r) 2. Match the following:

Column I Column II

(A) Bi3+ → (BiO)+ (p) Heat (B) [AlO2]– → Al(OH)3 (q) Hydrolysis (C) 4 6

4 2 7SiO Si O (r) Acidification

(D) 2-4 7B O → 3

B OH (s) Dilution by water

(t) Decomposition Sol. A → (q, s); B→ (q); C → (r); D → (q, r) . A → (q), (s) Bi3+ + H2O → (BiO)+ + 2H+ B→ (q) NaAlO2 + H2O → Al(OH)3 + NaOH C → (r)

44SiO + 2H+ → 6

2 7 2Si O + H O

D → (q), (r)

Na2B4O7 HCl H BO3 3

Na2B4O7 2H O H3BO3

FPT-V-(Paper-1)-SOL


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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. An mono alkene C6H12 (A) undergo oxidative ozonolysis gives two acids (B) and (C). Neither (B)

nor (C) contains more than four carbon. How many possible pairs can be there for (B) and (C). Sol. 2

and

CH3

2. A 150 ml solution of I2 is divided into two unequal parts. I part reacts with hypo solution in acidic

medium. 15 ml of 0.4M hypo was consumed. II part was added with 100 ml of 0.3M NaOH solution. Residual base required 10 ml of 0.3M H2SO4 solution for complete neutralization. Calculate the value of “10 times” the initial concentration of I2?

Sol. 1 For the first part, millimole of Na2S2O3 consumed = 15 × 0.4 = 6 2 2 2 3 2 4 6I + 2Na S O 2NaI+Na S O millimoles of I2 consumed = 3 millimoles (i) For the second part, the reaction is 3I +6NaOH 5NaI+NaIO +3H O2 3 2 reacted with I2

millimoles of NaOH = 100×0.3 - 10×0.3×2 =30 - 6=24millimoles .

Millimoles of I2 consumed 242

= 12 millimoles. (ii)

Total millimoles of I2 consumed in reaction = (i) + (ii) = 3+ 12 = 15 millimoles

215molarity of I = =0.1M

150 " 10 times initial molarity of 2I " =0.1×10 =1

Ans = 1 3. A 19.6 g of a given gaseous sample contains 2.8 g of molecules, whose density is 0.75 g/litre,

11.2 gm of molecules, whose density is 3g/litre and 5.6 gm of molecules, whose density is 1.5 g/litre. All the density measurements are made at N.T.P. Calculate the total number of molecules (N) present in the given sample. Report your answer in “10-23 N”.

Assume Avogadro’s number as 6 × 1023 Sol. 3 Total weight = 19.6 gm (a) 2.8 g of molecules, density = 0.75 g /litre

Volume = mass 2.8=density 0.75

1 mole 22.4 litresatN.T.P

FPT-V-(Paper-1)-SOL


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2.8 litres0.75

at N.T.P

Moles = 2.8 1×0.75 22.4

Molecules = 23 232.8 6× ×10 =1×100.75 22.4

(b) 11.2 g of molecules, density = 3 g /litre molecules = 1 × 1023 (c) 5.6 g of molecules, density = 1.5 g/ litre molecules = 1 × 1023 Total no of molecules = 3 × 1023 = 10-23 N = 3 4. How many of the following terms are directly associated with Pb(lead)? Pattinson’s process, Sindoor, Calcium plumbate, Cinnabar, Solder, Bronze, Wolframite, Liquation,

Self reduction. Sol. 6 Cinnabar → HgS, Wolframite →FeWO4, Solder → 50% Pb and 50% Sn, Sindoor → Pb3O4,

Pattinson’s process-desilverisation of lead, Bronze does not contain Lead, Calcium plumbate → CaPbO2

5. A glow-worm of mass 5.0g emits red-light (650 nm) with power of 0.10 W entirely in the backward

direction. What velocity (v) in m/sec, will it have accelerated to after 10 years, if released in free space (and assumed to live). Report the answer as (v/7)

Sol. 3 According to the law of conservation of momentum, momentum lost by photons = momentum gained by glow-worm Momentum lost by photons = no of photons x momentum lost by one photon

= Totalenergy h h h× Debroglie' equation λ = .p =energy of 1photon λ p λ

power × time h= ×hc λλ

λ h×power×time×

hc λ=

8

0.1×365×24×60×60×10= 13×10

Momentum gained by glow-worm = 5 × v inm / sec 21000

(1) = (2) Hence = 21 m/sec

Ans. v = 37

.

6. How many of the following cations belongs to group IIA, III, IV and V only 2 2 3 2 3 2 3 2 2

2Hg , Pb , Sb , Mg , As , Cu , Cr , Zn , Sn Sol. 4 2 2 3 2Pb , Cu , Cr , Zn

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7. In the monochlorination of 3- methylpentane, Let x be the number of pairs of isomers, which exist as enantiomers, y be the number of pairs of isomers, which exist as diastereomers, z be the number of isomers, which are achiral. Calculate the value of (x + y + z).

Sol. 9

|

3 2 2 3|

CH3

HCH - CH - C - CH -CH 3- methyl pentane monochlorination, different products are

|

2 2 2 3|

CH3

*

H

ClH - CH - C - CH -CH chiral, one chiral carbon atom pair of enantiomers

3| |

3 2 3| |Cl H

* *

CHH

CH -C - C - CH - CH 2 chiral carbon atoms 4 isomers. 2 pairs of enantiomers

4 pairs of diastereomers.

2

|

3 2 2 3|H

CH Cl

CH - CH - C - CH - CH achiral molecule

3

|

3 2 2 3|Cl

CH

CH - CH - C - CH - CH achiral molecule

X = number of pairs of enantiomers = 3 Y = number of pairs of diastereomers = 4 Z = number of isomers which are achiral = 2 Total x + y + z = 9 8. To 250 ml of water, x g of acetic acid is added. If 11.5% of acetic acid is dissociated, the

depression in freezing point comes out 0.416. What will be the value of x if Kf (water) = 1.86 K kg-1 mol-1 and density of water is 0.997 g ml-1.

Sol. 3

i 10.115n 1

So, i = 1.115 f fT i m K

0.416 = x 1000 1.861.11560 249.25

x = 0.416 60 249.25 3 g1.115 1000 1.86

FPT-V-(Paper-1)-SOL


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MMaatthheemmaattiiccss PART – II

SECTION – A


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. Let f: R R – {3} be a function given by f(x + 10) =

f x 5f x 3

, then 20

10

f x dx is equal to

(A) 110

100

f x dx (B) 120

110

f x dx

(C) 130

120

f x dx (D) 140

130

f x dx

Sol. D If f(x) = 1, then f(x + 10) = 2 For f(x) = 2, f(x + 10) = 3 which is not possible So f(x) 1, 2

Now f(x + 20) =

f x 10 5 2f x 5f x 10 3 f x 3

f(x + 30) =

3f x 5f x 1

, f(x + 40) =

3f x 10 5 f xf x 10 1

f(x) is periodic with period 40. 2. From any point on the line (t + 2)(x + y) = 1, t – 2, tangents are drawn to the ellipse 4x2 + 16y2 =

1. It is given that chord of contact passes through a fixed point. Then the range of ‘t’ for which the fixed point always lies inside the ellipse is

(A) 1, 25

(B) 2, 25

(C) 3, 25

(D) 4 , 25

Sol. B

Any point on (t + 2)(x + y) = 1 is 1,t 2

Equation of chord of contact is 4x + 16y 1t 2

= 1

(4x – 16y) + 16 y 1t 2

= 0

the chord of contact passes through the intersection of x – 4y = 0 and 16 y 1t 2

= 0

the point of intersection is t 2 t 2,4 16

It lies inside the ellipse if 2 2t 2 t 24 16 1 0

4 16

5t2 – 12t + 4 < 0 2 t 25 .

FPT-V-(Paper-1)-SOL


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3. Three boxes are labelled as X, Y, Z and each box contains 5 balls numbered 1, 2, 3, 4, 5. The balls in each box are well mixed. One ball is chosen at random from each of the three boxes X, Y, Z respectively. If , , are the numbers on the balls chosen, then the probability that = + is equal to

(A) 125

(B) 225

(C) 425

(D) 15

Sol. B n(S) = 5 5 5 = 125 n(A) = 514, 541, 523, 532, 413, 431, 422, 312, 321, 211

Required probability = 10 2125 25

.

4. A variable chord of the hyperbola 2 2x y 1

4 8 , subtends a right angle at the centre of the

hyperbola. If this chord touches a fixed circle concentric with the hyperbola, then the equation of the circle is

(A) x2 + y2 = 4 (B) x2 + y2 = 8 (C) x2 + y2 = 16 (D) x2 + y2 = 64 Sol. B Let the variable chord be x cos + y sin = p. Let this chord intersect the hyperbola in A and B

The joint equation of OA and OB is 22 2x y xcos ysin

4 8 p

2 2

2 22 2 2

1 cos 1 sin 2sin cosx y 04 8p p p

2 2

22 2

1 cos 1 sin 0 p 84 8p p

.

The variable line touches the fixed circle, thus perpendicular distance of (0, 0) = Radius equation of the circle is x2 + y2 = 8.

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. The function f(x) = [x](x2 – 25)n(x4 + 3x3 + 7x2 + 9x + 12)m, m, n N attains a local minima at x =

5, then (where [.] denotes greatest integer function) (A) n = 2, m = 3 (B) n = 100, m = 200 (C) n = 27, m = 20 (D) n = 64, m = 189 Sol. A, B, D f(x) = [x](x2 – 25)n(x2 + 3)m(x2 + 3x + 4)m f(5) = 0, f(5–) = 4(negative)n (+)(+) but for a local minima at x = 5, f(5–) > 0 n must be even, m can be any number.

FPT-V-(Paper-1)-SOL


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6. The sum of two non-negative numbers is 2a. If P be the probability that the product of these numbers is not less than m times their greatest possible product, then

(A) P = 23

, m = 45

(B) P = 13

, m = 35

(C) P = 12

, m = 34

(D) P = 34

, m = 716

Sol. C, D x + y = 2a, let z = xy = x(2a – x) = a2 – (x – a)2 z is max if x = a, thus zmax = a2 now P(xy ma2) = P(x(2a – x) ma2) – x2 + 2ax – ma2 0 x2 – 2ax + ma2 0

for m = 34

, P = 12

and for m = 716

, P = 34

.

7. The sequence {xn} is defined by xk + 1 = 2k kx x and x1 = 1

2. Let S =

1 2 m

1 1 1...x 1 x 1 x 1

(where [.] denotes the greatest integer function) (A) S = 0, if m = 40 (B) S = 1 if m = 100 (C) S = 0 m N (D) S = 1 m N Sol. B, D

k 1 k k k k

1 1 1 1x x x 1 x x 1

1 2 m 1 m 1

1 1 1 1 1...x 1 x 1 x 1 x x

also 0 < m 1

1x

< 1

1 2 m

1 1 1...x 1 x 1 x 1

= 1 m N.

8. If f(x) = 5 4 2

3x x kxx x5 4 2

be an increasing function x R, then k2 can be

(A) 3 (B) 5 (C) 7 (D) 13 Sol. A, B, C f(x) = x4 + x3 + 3x2 + 2kx + 1

= 2 22

2 2x 1 k kxx 3 x 1 02 4 2

for this to hold 3 – 21 k 0

4

k2 11.

9. If m

nx 0

1 cos xlim sinx

exists, where m, n N, then

(A) m N, n = 1 (B) m N, n = 2 (C) m N, n = 3 (D) m N, n N

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Sol. A, B

m m 2

n n nx 0 x 0 x 0

1 cos x (1 cos x) sin x / 2lim sin sin lim sin lim 2 mx x x

m N and n = 1 or 2.

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. If I(m) = 2

0

ln 1 2mcos x m dx

, then match the following

Column – I Column – II (A) I(1) = (p) I(–1)

(B)

I 9I 3

= (q) 2

(C) I(81) = (r) 0

(D)

I 25I 5

= (s) 4I(3)

(t) I(3) Sol. (A) (p, r) (B) (q) (C) (s) (D) (q)

I(m) = 2

0

ln 1 2mcos x m dx

I(–m) = 2

0

ln 1 2mcosx m dx

= 2

0

ln 1 2mcos x m dx

= 2

0

ln 1 2mcos x m dx I m

I(m) = I(–m) ...(1)

Also I(m) + I(–m) = 2 4

0

ln 1 2m cos2x m dx

Put 2x = t I(m) + I(–m) = I(m2) ...(2)

2I(m) = I(m2) =

2I m 2I m

.

FPT-V-(Paper-1)-SOL


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2. A rational number is given in the form pq

; p, q I+; pq

(0, 1) and p and q are coprime to each

other. Column – I Column – II

(A) pq = 10!, then number of ordered pairs (p, q) are less than

(p) 128

(B) pq = 20!, then number of ordered pairs (p, q) are equal to

(q) 6

(C) pq = 30!, then number of ordered pairs (p, q) are greater than

(r) 621

(D) pq = 40!, then number of ordered pairs (p, q) are greater than

(s) 2050

(t) 500

Sol. (A) (p, r, s, t) (B) (p) (C) (p, q, t) (D) (p, q, r, t) Any prime factor that occurs in the numerator cannot occur in the denominator. There are 4 prime

factors of 10!. For each of these prime factors i.e. 2, 3, 5, 7. One must decide only whether it occurs in numerator or denominator. Thus 4 decisions can be made in 24 ways. They can be grouped into 8 pairs of reciprocals each containing one factor less than 1. (All other cases are similar to this one)

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. x = 2, 4 are the solutions of the equation |mx + | + |mx + | = c where m > 0 and , , c are non-

zero constants, then m

is equal to __________

Sol. 6 y = |mx + | and y = – |mx + | + c

intersect in two points whose abcsissae are 2 and 4.

AEBD is a parallelogram. The diagonal bisect each other

2 4m m

m

= – 6.

B(4, y2)

D(–/m, c)

A(2, y1)

(–/m, 0) (–/m, 0)E

C

y

x

FPT-V-(Paper-1)-SOL


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2. The sides of a triangle are in the ratio 3 : 5 and the third side is 16, if the largest possible area of

the triangle is k, then k30

is __________

Sol. 4

S = 16 3x 5x2

= 8 + 4x,

Now 3x + 5x > 16 x > 2 3x + 16 > 5x x < 8 x (2, 8) 5x + 16 > 3x x > – 8 Now A2(x) = (8 + 4x)(4x – 18)(x + 8)(8 – x) = (16x2 – 64)(64 – x2) Let x2 = t t (4, 64), f(t) = 16(t – 4)(64 – t) = 16(64t – t2 – 256 + 4t) f(t) = – 16(t2 – 68t + 256), f(t) = 2t – 68 = 0 t = 34 f(t) = – 32 < 0 Maxima occurs at t = 34 F(34) = 16(30)(30) largest possible area = 120.

3. The equation z4 + (5z – 1)4 = 0 has two pairs of complex roots 1, 1 and 2, 2. Each pair i, i

are complex conjugates. Then the value of i i

1 113 = __________

Sol. 4

4 4

4 4 1 1z z 5 0 5 1 cos isinz z

15z

= {cos(2n + 1) + isin(2n + 1)}1/4, n = 0, 1, 2, 3

1 5 cos 2n 1 isin 2n 1z 4 4

1

1 5 cos isinz 4 4

2

1 3 35 cos isinz 4 4

3


4


consequently 1 4 2 3

1 1 1 1, and ,z z z z

are complex conjugates.

Let 1 1 4 1

1 1 1 1andz z

1 4

1 1 i 1 1 i5 , 5z z2 2 2 2

2

1 4

1 1 15z z 22

similarly 2

2 3

1 1 15z z 22

2 2

i i

1 1 1 1 15 5 522 22 2

.

FPT-V-(Paper-1)-SOL


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4. If a, b, c are distinct positive integers such that ab + bc + ca 107, then the minimum value of 16

(a3 + b3 + c3 – 3abc) is __________

Sol. 9 Without loss of the generality, let a > b > c, then a – b 1, b – c 1, a – c 2

a2 + b2 + c2 – ab – bc – ca = 12

[(a – b)2 + (b – c)2 + (c – a)2] 3

now a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) 3(a + b + c) Now (a + b + c)2 = (a2 + b2 + c2 – ab – bc – ca) + 3(ab + bc + ca) 3 + 3 107 a + b + c 18 a3 + b3 + c3 – 3abc 3 18 = 54. 5. Let P(sinA, cotB) be the point on the unit circle centred at origin and tanC, cotC are real numbers

where A, B, C R and the minimum value of the expression (tanC – sinA)2 + (cotC – cotB)2 = p – 2 q, then (p + q) is equal to __________

Sol. 5 Let Q(tanC, cotC), then ˆ ˆPQ tanC sinA i cotC cotB j

2 2 2PQ tanC sinA cotC cotB

PQ OQ OP

2 2 2PQ tan C cot C 1

= 222 2tan C cot C 2 1

min(PQ) = 22 1 3 2 2 . 6. If a, b, c be the sides of a triangle ABC with right angle at C. The medians AD and BE have

slopes 1 and 2 respectively. Then ab = kc2 where 36k equals __________ Sol. 8 Let be the acute angle between AD and BE.

Then tan = 2 1 11 2 3

.

Now + 180 – + + 90 = 360 = + – 90 tan = – tan( + )

– 3 = 2 2

2b 2atan tan 2b 2aa b

2b 2a1 tan tan 3ab1a b

9ab = 2c2.

–

B D C

A

E c

a/2 a/2

FPT-V-(Paper-1)-SOL


16

7. If p is a positive integer and ‘f’ be a function defined for positive numbers and attains only positive values such that f(xf(y)) = xpy4, then p = __________

Sol. 2

f(xf(y)) = xpy4, put x = 1

f y

f(1) =

p 44

p1 yy

f y f y

, for y = 1, f(1) = p1

f 1 f(1) = 1

so f(y) = y4/p …(1) hence f(xy4/p) = xpy4 put y = zp/4 f(xz) = xp zp f(x) = xp …(2)

from (1) and (2) 4 pp p = 2.

8. The area enclosed by the curve max(|x – 1|, |y|) = 4 is A, then 3A32

is equal to __________

Sol. 6 Required area = 8 8 = 64.

x = 5 x = –3

y = 4 |x – 1| < |y| |x – 1| > |y|

FPT-V-(Paper-1)-SOL


17

PPhhyyssiiccss PART – III

SECTION – A


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. An ideal gas has molar heat capacity Cv at constant volume. The gas undergo the

process 0T T 1 V where , T0 is constant and V is volume. Then molar heat capacity of the gas is :

(A) vRC 1 VV

(B) vRC 1 VV

(C) vRC 1 V

2 V

(D) v

RC 1 V2 V

Sol. B

Q = U + W

C = CV + P dVn dT

C = CV R (1 V)V

2. A capacitor and a coil in series are connected to a 6 volt ac source. By varying the frequency of the source, maximum current of 600 mA is observed. If the same coil is now connected to a cell of emf 6 volt and internal resistance of 2 ohm, the current through it in steady state will be

(A) 0.5 A (B) 0.6 A (C) 1.0 A (D) 2.0 A

Sol. A

in a resonance XC = XL and Ie = VR

R = 10

in case of d. c. source i = 610 2

3. Ratio of the de Broglie wavelength of molecules of helium and hydrogen at temperature 270 C and 3270C respectively is (A) 2 (B) 1 (C) 3 (D) 4

Sol. B

4. A converging lens of focal length f1 is placed coaxially and in front of a convex mirror of focal

length f2. Their separation is d. A parallel beam of light incident on lens returns as a parallel

beam from arrangement

(A) The beam diameters of incident & reflected beams must be different

(B) d = f1 2|f2|

(C) d = f1 |f2|

(D) if entire arrangement is immersed in water, the conditions will remain unaltered.

Sol. B

FPT-V-(Paper-1)-SOL


18

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. 5. A point source S of light is fixed on a vertical wall and plane

mirror starts moving with a velocity v towards the wall. The adjoining figure shows the top view of the situation. Now choose the correct option(s)

(A) Velocity of light spot formed on the wall is v i2

(B) Light spot is at rest (C) size of the light spot would increase (D) size of the light spot would remains same.

S

wall

y

x

v

d

Sol. B, D 6. A non conducting ring of radius r has charge Q. A magnetic field perpendicular to the ring

changes at rate dB/dt. Choose the correct statement(s). (A) Electric field at any point on the periphery of ring is zero.

(B) Electric field at any point on the periphery of ring is r dB2 dt

(C) Torque experienced by ring due to induced electric field is 2r dB Q2 dt

(D) Torque experienced by ring due to induced electric field is 2r dB Q4 dt

Sol. B, C

7. In the network shown in the figure. Choose the correct statement(s)

(A) Potential difference between points HB is 5V (B) Potential difference between points DB is 1V (C) Potential difference between points DE is 0.5V (D) Potential difference between points DG is 2V

A B

D C

4V

H G

F E

4V

6V 6

2 3 1

1

Sol. A, B, C, D Current in the branch DH is 0 8. Bi212 decays to Th208 by -emission in 34% of the disintegration and to Ra212 by emission in

66% of disintegration. If the total half value period is 60.5 min, choose the correct statement(s) (A) Decay constant for -emission () is 0.65 104/sec (B) Decay constant for -emission () is 2.5 104/sec (C) Decay constant for -emission () is 1.26 104/sec (D) Decay constant for -emission () is 2.5 104/sec

Sol. A, C = +

= 1/2

0.693T

= 66100

, 34100

FPT-V-(Paper-1)-SOL


19

9. The figure shows, a graph of current in a discharging circuit of a

capacitor through a resistor of resistance 10 . Choose the correct option(s)

(A) The initial potential difference across the capacitor is 100 Volts

(B) The capacitance of the capacitor is 1 F10ln2

(C) The total heat produced in the circuit is 1000 Jln2

(D) The total heat produced in the circuit is 500 Jln2

2.5A

10A

2s t

i

Sol. A, B, 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. Resistance of connecting wire is negligible and L is any inductor shown in figure. Switch is closed at t =0. Then match the column(Neglect mutual inductance)

Column I Column II (A)

k

L=3mH

R=1Ω

E=1volt

L=3mH

L=3mH L=3mH

(p) Current through battery at t=3milli sec

is e 1 amp.e

(B)

k

L=18mH 6Ω

E=3volt

6Ω L=18mH

(q) At t= 3milli sec, Current through

inductor L is 1 e 1 amp2 e

(C)

k

L=30mH

10Ω

E=10volt

10Ω

(r) At t= 3milli sec, Current through

inductor L is e 1 ampe

FPT-V-(Paper-1)-SOL


20

(D)

k

L=3mH

E=2volt

2Ω

2Ω

2Ω

2Ω

(s) Just after closing the switch current through inductor L is 0

(t) Current through the battery after very long time is 1 amp.

Sol. (A) (p, q, s, t), (B) (p, q, s, t), (C) (r, s), (D) (s, t) 2. Spring and strings are massless. Initially block of mass m=1kg is in equilibrium position. It is

displaced through small distance x and released. Match the columns. Column I Column II

(A)

m X(small)

21 /2

k N m

11 /

16k N m

11 /

16k N m

(p) Motion of mass m is periodic

(B)

m

K=1 N/m

x

Entire system is in gravity free space. Liquid is non viscous & neglect resistance of the water.

(q) Motion of mass m is SHM

(C)

m

K=1 N/m

X small

+Q=1columb E=2 V/m

(r) Time period T=2sec

FPT-V-(Paper-1)-SOL


21

(D)

m

9 /16

k N m

areaA

Smooth surface

x

Initially spring and string shown is in their natural length. Young modulus of the string

is Y and it is given that 9 YA16 3L

. Block

is displaced through distance x and released. A is the cross-sectional area of string.

(s) In equilibrium spring will be in its natural length(i.e =0)

(t) Maximum displacement from equilibrium position on either side will be equal in magnitude

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

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. A particle is thrown upward with speed 10 3 m/sec. It strike the

inclined surface as shown in the figure. Collision of particle & inclined surface is perfectly inelastic. What will be maximum height attained by the particle from the ground. (g=10m/sec2)

10 3

O

3m

450

Sol. 6 2. A particle is thrown from the origin, at an angle (0<<90) such that it just crosses a wall of

height 9m. Wall is along the plane x = 12m. Speed of projection is n 30 & particle strike the ground at x = 48 m, value of n is (g = 10m/sec2)

Sol. 4

FPT-V-(Paper-1)-SOL


22

3. Side of the regular hexagon is 2 meter. Magnetic field at point O due to the network shown in the figure is

703B 10 tesla. Then value of B0 is

Sol. 2

i=6A

R

2R

O

3R

3R

5R

4R

4. A parallel beam of light ( = 5000 Å) is incident at an angle = 30 to the slits plane as shown in YDSE experiment Intensity due to each slit at any point on screen is I0. The distance between slits is 1 mm. The intensity at point O on the screen is KI0. The value of K is

Sol. 4

d O

S1

S2

screen 3 m

5. Some plane wavefronts are shown in figure. The value of refractive index of medium is 2K . The value of K is

Sol. 2

vacuum

A

A

A

B

B

B

medium 1 m

2 m

6. A radioactive element P disintegrates into Q which successive disintegrates into R as shown 2P Q R At t = 0 number of nuclei of P, Q and R are N0, 0, 0 respectively. At time t number of nuclei of P,

Q and R are N1, N2 and N3 respectively. The ratio of N1 to N2 when N2 is maximum is K. the value of K is

Sol. 2

21 2

dNN 2 N 0

dt

1

2

N2

N

7. A parallel plate capacitor is connected to a battery of emf V volts as shown. Now a slab of dielectric constant k = 2 is inserted between the plates of capacitor without disconnecting battery. The electric field between the plates of

capacitor after inserting the slab is E = PV2d

. The value of P is

Sol. 2

k = 2

d

V 8. Four identical metal plates are arranged as shown Plates 1

and 4 are connected by a connecting wire. A battery of emf V volts is connected between plates 2 and 3. The electric

field between plates 3 and 4 is 2VKd

. The value of K is

1 2 3

4

d d

2d V

Sol. 6

ANSWERS, HINTS & SOLUTIONS

PAPER-2 (09.05.2017)

Q. No. CChheemmiissttrryy MMaatthheemmaattiiccss PPhhyyssiiccss

1. A D C

2. D B A

3. A A B

4. B C B

5. B D C

6. C C B

7. B D C

8. C C A

9. D B A

10. B C B

11. D C A

12. B D B

1.

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

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

(A) (p) (B) (q) (C) (t) (D) (t)

2.

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

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

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

1. 6 2 2

2. 4 4 6

3. 4 2 0

4. 7 2 2

5. 5 0 8

FIITJEE PRACTICE TEST – VI

FPT-VI-(Paper-2)-SOL


2

CChheemmiissttrryy PART – I

SECTION – A

This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. 1. One mole of an alkene (P) on oxidative ozonolysis gives 4 moles of an aldehyde (Q) and another

compound (R) C6H6O6. Compound (Q) can undergo resin formation with phenol while (R) on heating gives an acidic gas (S) and a compound (T). Which of the following cannot be true?

(A)

'R' can be

OHHOOC

O COOH

(B) ‘T’ can be a diketone

(C) ‘P’ cannot have more than 10 carbon atoms in its formula

(D) ‘R’ can exhibit hydrogen bonding

Sol. A

(P) 2. Two equivalent of the Wittig reagent (CH3)2C = CH – CH = P(C6H5)3 were allowed to react with

C4H4O2 compound. The chief product was 2,11-dimethyldodeca-2,4,6,8,10-pentaene. What is the name of C4H4O2 compound used in this reaction?

(A) 2-butyne-1,4 diol (B) 1,2-cyclobutanedione (C) 1,3-butadien-2,3 diol (D) 2-butenedial Sol. D 3. Which of the following statements is correct? (A) 2

4SO ion possess greater flocculation value than [Fe(CN)6]3– ions, for ferric hydroxide sol.

(B) ||

3 2 14

OCH CH C O Na have higher Critical Miscelle Concentration (CMC)

than ||

3 2 10

OCH CH C O Na

(C) Peptisation works on the principle of relative absorption of common ions from electrolyte (D) All are correct Sol. A

(A) Flocculation value 1 Hardy Schulze rulecharge on counterion

(B) CMC 1length of surfactant chain

(C) Peptisation is relative adsorption of common ions from electrolyte Hence (A) is the correct answer.



3

4. An inorganic salt (A) on heating produces a colourless and odourless gas (X) which is neutral to litmus. Gas (X) when passed into another vessel containing limestone and excess of coke at 1273 K temperature produces a well known Fertilizers “NITROLIM” (CaCN2 + C). Hence salt (A) may be:

(A) NH4NO3 (B) Ba(N3)2 (C) LiNO3 (D) Pb(NO3)2 Sol. B

3 22

Ba N Ba 3N

A x

1273 K3 2CaCO CaO CO

gas x2 2

Nitrolim

CaO C CaC CaCN C

Hence (B) is the correct answer. 5. The electrolytic oxidation of N2O4 was carried out in a solution buffered by NaHCO3 and Na2CO3.

The reaction involved was: 2 4 2 3N O 2H O 2NO 4H 2e 100 ml of 1M solution of N2O4 buffered initially at pOH = 3, was oxidized, with reaction above

going to completion. The total ionic concentration initially, i.e. [NaHCO3 + Na2CO3] was 0.4 M. The pH of the solution after oxidation will be: (For H2CO3.

1 2

6 11a aK 10 and K 10 ) :

(A) 5.7 (B) 6.0 (C) 6.3 (D) 11.0 Sol. B Moles of H+ ions produced during critical oxidation = 0.4 Also, in the buffer, initially pH =

2

2a 3 3pK CO HCO 0.2 M

Now, after electrolysis:

23 3CO H HCO

initially 0.20 0.20moles of H ions added 0.4moles left 0 0.2 0.4

Now,

3 2 3HCO H H CO

0.4 0.2 00.2 0 0.2

So, finally mixture contains equal moles of H2CO3 and 3HCO Hence,

1apH pK 6 6. According to IUPAC nomenclature system, which of the following is correct name of the complex

compound given? Na3[Co(NO2) (CN)2 (NO2) (O2) (O2)]; = 4.899 B.M. Where, NO2 and O2 written separately are different ligands. (A) Sodiumdicyanodinitrodiperoxocobaltate (III) (B) Sodiumdicyanodinitroniumdisuperoxocobaltate (III) (C) Sodiumdicyanonitronitrogendioxideperoxosuperoxocobaltate(III) (D) Sodiumdicyanonitronitroniumdioxygenperoxocobaltate(II)



4

Sol. C = 4.899 B.M. 24 n n 2

n = 4 (no. of unpaired electron) So, Co → 1s2 2s2 2p6 3s2 3p6 3d7 4s2

3e73d 24s 43d 04s

So, Co must be in ‘+3’ oxidation state. ‘O2’ can have charge =0(dioxygen), - 1(super oxide) or, - 2(peroxide) Similarly NO2 can have charge 0(Nitrogendioxide), +1(Nitronium) or, - 1(nitro) So, [Na3[Co(NO2) (CN)2 (NO2) (O2) (O2)] 3 × 1 + 3 + a – 2 + b + c + d = 0 a + b + c + d = - 4 Now, a and b must belong to –1, 0, +1 and c & d must belong to – 2, – 1 , 0 So, on hit – n – trial, a = - 1 or, 0 b = 0 or, - 1 c = - 2 or, -1 d = - 1 or, - 2 So, nomenclature is Sodiumdicyanonitronitrogendioxideperoxosuperoxocobaltate(III)

Comprehension Type This Section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions 7 to 9

Given the following information:

BrO3 (aq) HOBr

1. 44V

Br2 (liq.) Br (aq)?1. 49V 1. 58 V

+ - 0

- - 0

Ag aq +e Ag s E = 0.8V

AgBr s + e Ag s +Br aq E = 0.07V

7. Calculate Ksp value of AgBr(s) at 25oC; given 2.303RT = 0.06F

(A) 10 – 11.67 (B) 10 – 12.167 (C) 10 – 13.167 (D) 10 – 14.167 Sol. B oAgBr s Ag aq. Br aq. E 0.73

ospG 2.303 RTlogK

o

spnE FlogK 12.167

2.303RT



5

8. A galvanic cell using standard hydrogen electrode as anode is constructed in which overall reaction is.

- +2 2 2 3Br +H g +2H O 2Br aq +2H O aq silver ions are added until AgBr precipitates

at the cathode and [Ag+] reaches 0.068M. The cell voltage is then measured to be 1.73 V. Calculate E0 of Galvanic cell? Given data:

2.303RT = 0.06F

11.167 12 13.167 14

12.167 13 14.167 15

10 6.8 10 10 6.8 10

10 6.8 10 10 6.8 10

(A) 0.89 V (B) 1.13V (C) 1.07 V (D) 0.83 V Sol. C Ksp = 10 – 12.167 = 6.8 × 10 -13 = [Ag+][Br – ] [Ag+] = 0.068 M

So, [Br – ] 13

116.8 10 100.068

Ecell = 2 2o 0.06E log Br H

n

[H+] = 1 as SHE is used

Eo = 220.061.73 log102

Eo = 1.07 V 9. Estimate the solubility of bromine in the form of Br2(liq.) in water at 25oC?

(A) 53075-

RTe (B) 53075

RTe

(C) 212300

RTe (D) 106150

RTe Sol. D Use o

eq.G RTlnK

2eq. 2K S for Br 2Br aq.

Paragraph for questions 10 to 12

A white solid (A) reacts with dilute H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified dichromate yields a green solution and a slightly coloured precipitate (D). The substance (D), when burnt in air, gives a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate turns blue with this colourless liquid. The addition of aq. NH3 or NaOH solution to (C) produces a precipitate that dissolves in an excess of both the reagents to form a clear solution. Now, answer the following questions: 10. Excess of precipitate (D) is added to hot NaOH solution, then the main product that will be formed

in the reaction is: (A) Na2S (B) Na2S5 (C) Na2S4O6 (D) SO2



6

Sol. B (A) = ZnS (B) = H2S (C) = ZnSO4 (D) = S8 (E) = SO2 So, 8 2 2 2 3 2S NaOH Na S Na S O H O 2 2 5Na S 3S Na S 11. When NH4OH is added to the solution (C), first a white precipitate is obtained which dissolves in

excess of reagent due to the formation of (X). Hence, compound (X) is: (A) Na2AlO2 (B) (NH4)2 [Zn(OH)4] (C) Zn (OH)2 (D) [Zn(NH3)4]SO4 Sol. D

4 4 4 42 2

ZnSO g 2NH OH Zn OH NH SO

white ppt.

3 32 4 2Zn OH 4NH aq Zn NH OH

Hence, (D) is the correct answer. 12. If 3 moles of gas (E) are mixed with 4 moles of gas (B), then the no. of moles of (D) formed will

be: (assume 100% efficiency of the reaction) (A) 6 (B) 0.75 (C) 4 (D) 7.5 Sol. B

2 2 8 23SO 2H S S 2H O8

Hence, (B) is the correct answer.



7

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:


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)

O

O

O

O

4

3

i LiAlDii H O

(p)

H2C CD2

OH OHWill give

(B)

O

O2D O/D

(q) C6H5 – NHOH

(C) NO2

Electrolytic r e duction inbasic medium

(r) C6H5 – NH – NH – C6H5

(D) NO2

Electrolytic r eduction instrongly acidic medium

(s) Product will give formaldehyde as one of the product when treated with HIO4

(t)

H2C CH2

OD ODWill give

Sol. A → (p, s); B → (s, t); C → (r); D → (q) 2. Match Column – I with Column – II:

Column – I Column - II (A) Compound(s) soluble only in aquaregia (p) NiS (B) Compound(s) soluble in excess of KI solution (q) HgS (C) Compound(s) give white ppt. in hot aq.

Solution of NH3 (r) HgI2

(D) Compound(s) soluble in excess of KCN solution

(s) PbI2

(t) FeCl3 Sol. A → (p, q); B → (r, s); C → (r, s); D → (t)



8

SECTION-C

Integer Answer Type

This section contains 5 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. Find the sum of the double bond equivalents in the following three molecules. 4 6 2 6 12 2 8 11 3 2C H O , C H N , C H O Br N

Sol. 6 2. Two aqueous solutions as shown, are put in an evacuated

chamber. When equilibrium is attained, it is found that one solution contains 0.01% of CH3COOH and other 0.014 % of urea by weight. If degree of dissociation is ‘’ then what is the value of 10?

CH3COOH in water

Urea in water

Evacuated chamber

Sol. 4 At equilibrium relative lowering of vapour pressure of both the solution is same

Then, 1 0.01 0.014 1 1.4

60 60

= 0.4 10 = 4

3. How many of the following compound can produce a buffer solution when HCl is added (more than 1 Mole but less than 2 Mole) in their 1 L, 1 M solution.

Na2CO3, Na3PO4, Na2HPO4, Borax, P4O10, SO2Cl2, NaHCO3, (NH4)2SO4

Sol. 4

4. How many of the given species certainly not contains N = N linkage (one one between N-atoms) in their structures?

N2, N2O, HN3, N2F2, N2H2, N2H4, N4H4, N2O3, N2O4, N2O5

Sol. 7 5. A conversion of 2,5-dimethylfuran into 2,5-dimethylpyrrole may be accomplished in following

steps: Step Step Number

Hydrolysis of furan with aq. Acid 1 Heating the hydrolysis product with dil. NaOH 2 Heating the hydrolysis product with excess HI 3 Heating the hydrolysis product with excess of (NH4)2CO3 4

What will be the addition of number assigned to the steps required for this conversion? Sol. 5



9

MMaatthheemmaattiiccss PART – II

SECTION – A


This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONLY ONE is correct. 1. Let a function f: x y is defined where x = {0, 1, 2, 3, …, 9}, y = {0, 1, 2, …, 100} and f(5) = 5,

then the probability that the function of type f: x B where B Y is of bijective in nature is

(A) 1019 100

r 1r 1

10!

r . C

(B) 101

9101

10 101r

r 1

C .9!

r . C

(C)100

9101

10 101r

r 1

C .9!

r . C

(D) 100

9101

9 100r 1

r 1

C .9!

r . C

Sol. D Favourable cases for bijection is 100

9C 9! For total cases ‘B’ is set of one element, no. of function –1 ‘B’ is set of two element, no. of function –29 ‘B’ is set of three element, no. of function –39

…… …… ‘B’ = Y No. of function –1019

Total cases = 100 9 100 9 100 91 2 1001 C .2 C .3 ..... C .101 =

1019 100

r 1r 1

r . C

Required probability = 100

9101

9 100r 1

r 1

C .9!

r . C

2. The area of a square whose edges are parallel to the coordinate axes and vertices lie on the

curve 2 2x y 1

9 16 is

(A) 1327

sq. units (B) 5767

sq. units

(C) 1447

sq. units (D) none of these

Sol. B Symmetry of one vertex of square is (, ), then other vertices must be ( , ), ( , ), (, )

and 2 2

2 2 1a b

=



10

= 2 2

ab

b a Required area =

2 2

2 24a bb a

exist only if b > a.

3. If ( - 2)x2 + y2 = 4 represents a rectangular hyperbola then the value of is equal to (A) 1 (B) 2 (C) 3 (D) 4 Sol. A 4. Let 1 2 3

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆa i j k,b [x]i {sin(x)} j,c c i c j c k

be three non zero vectors, such that | c | 1

,

angle between a

and b

is4 , c

is perpendicular to both a

andb

, (where [x] = Greatest Integer

function and {x} = Fractional part of x), If 2f(x) a.(b c) ,

then which of the following is correct

for f(x) (A) There are three points of discontinuity for x (0, ]

(B) Lagrange’s mean value theorem is applicable for x ,2

(C) x2

is the point of local minima

(D) xlim f(x) 9

Sol. C

2 2 23[abc] [x] {sinx}2

2 23f(x) [x] {sinx}2

Graphically we can see ’C’ is correct and B, C, D are incorrect. 5. If the perimeter of the triangle formed by foot of altitudes of the triangle ABC is equal to four times

the circum radius of ABC, then ABC is (A) isosceles triangle (B) equilateral triangle (C) right angled triangle (D) none of these Sol. D AC = bcosC, BC = acosC AB = ccosC Similarly AC = bcos B and BC = acos A Now 4R = acos A + bcos B + c cos C sin A.sin B.sin C = 1 This is only possible when A = B = C = /2 so triangle is not possible.

B C

A

B

A

C

6. Let two square matrices A[aij] and B[bij] of order 3 × 3 are such that aij = aji and bij = –bji then (A) An, Bn are symmetric matrix where n = odd natural number (B) A–1 is symmetric and B–1 is skew symmetric (C) If aij = bij for i < j then C = A + B is necessarily a upper triangular matrix. (D) Let aij = bij for i j and C= A + B, |C| is necessarily equal to |A|.

Sol. C



11

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. 7 to 9 Let the lines represented by the equation x2y2 x2 y2 + 1 = 0 formed a square ABCD. A point ‘P’ is taken on the same plane of square such that atleast two sides of all the triangles PAB, PBC, PCD and PDA are equal. 7. The number of possible position of the point ‘P’ (A) 1 (B) 5 (C) 17 (D) 9 Sol. D Clearly the possible point will be the intersection of the

circle with centre vertex and radius as side of the square and the line which is perpendicular bisector of the sides. From this 8 different position of P can be find and one point will be the centre of the square.

P

P

y =0

D (1,1)

0, 3 1

60°

x = 0

A (1,1) B (1,1)

(0,0)

60°

C(1,1)

8. For all the possible position of point ‘P’. How many of the given triangles have all the three sides

equal (A) 0 (B) 4 (C) 8 (D) 5 Sol. C Clearly for all the different position of ‘P’ except origin there will be one equilateral triangle. 9. One of the possible position of the ‘P’ such that atleast one of the given triangle is equilateral is

given by (A) (0, 0) (B) 3 1,0

(C) 0, 3 2 (D) 3 1, 3 1

Sol. B Different position of P except origin will be 0, 3 1 , 0, 3 1 3 1,0 , 3 1,0 ,

0, 3 1 , 0, 3 1 , 3 1,0 and 3 1,0 .



12

Paragraph for Question Nos. 10 to 12 For x, y R the functions are defined here under f1(x, y) = |x| + |y|, f2(x, y) = min(x + y, x – y) f3(x, y) = [x] + [y] (where [.] denotes the greatest integer function). 10. Which of the following is always true (A) f1(x, y) < f2(x, y) x, y R (B) f2(x, y) = f3(x, y) x, y Z+ (C) f1(x, y) = – f3(x, y) x, y Z– (D) none of these Sol. C

11. If h1(x) = f1(sin{x}, sinx) x R (where {.} represents fractional part function). h2(x) = f2(x, sinx) 0 < x < 1 h3(x) = f3(x, –x) x R, then (A) h1(x) is non-periodic function (B) h2: (0, 1) (0, 1), then h2 is bijective (C) h3(x) is periodic function (D) none of these Sol. C

12. Which of the following can never be true (A) f1(x, y) > f3(x, y) x, y R (B) f2(x, y) = x + y x, y < 0 (C) f3(x, y) = f3(–x, –y) x, y R (D) f1(x, y) = f3(x, y) x, y R – Z Sol. 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 statements/expression in column 1 with the values given in column II Column – I Column – II

(A) Let f(x) be a invertible function such that f(2) = 4, f(7) = 14.

If 7

2

f(x)dx 69 then 14

1

4

f (x)dx lies in the subset of

interval

(p) [–30, 30]

(B) If / 2

{x}0

sinx dx A[e ]

(where [x] = greatest integer function

and {x} = fraction part of x) then ‘[A]’ lies in the subset of interval

(q) [0, 5)

(C) The number of the positive integral coordinate(s) on the rectangular hyperbola 5x + 5y –2xy –5 = 0 is a subset of the interval

(r) [–1, 1]

(D) If n

r2n 1

r 0 r

CA,

C

then ‘A’ lies in the subset of interval (s) (0, 20]



13

Sol. (A) (p) (B) (p, q, r) (C) (p, q, s) (D) (p, q, s) (A). 21

(B). / 2

{x}0

sinx dx 0[e ]

(C). Only four points (4, 5), (3,10), (10,3) and (5, 4) are possible

(D). n n n

r r 1 r2n 2n 2n 1

r r 1 r

C C C12C C C

n

r2n 1

r 0 r

C2

C

2 Match the statements/expression in column 1 with the values given in column II

Column – I Column – II

(A) The value of 5

x[x 1/ 2]

5

x.e dx

lies in the subset of interval

(where [.] = greatest integer function]

(p) (0, 5)

(B) Number of divisor(s) of the number 6480 which are neither divisible by 10 nor by 24 lies in the subset of interval

(q) (0, 27)

(C) If the equation x4 + ax3 + bx2 – 9x + 5 = 0 has four positive real root then minimum value of ‘–a’ lies in the subset of interval

(r) [–10, 10]

(D) If the quadratic equation ax2 + (a – 1)x + 1 is positive for exactly two integral values of x then ‘a’ lies in the subset of interval

(s) [0, 12]

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

(A). 5

x[x 1/ 2]

5

x.e dx

= 0 (odd function)

(B). 22 (C). Let the roots of x4+ax3+bx2+9x+5=0 be ,,, + + + = –a b 9 5

By A.M. H.M. we get – 80a9

(D). 3a2



14

SECTION – C

Integer Answer Type This section contains 5 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 area bounded by y = f(x), y-axis and line 2y = (x + 1) where

f(x) = sin-1x + cos-1x + 1 11tan tan xx

is k , then the value of k is ___________

Sol. 2

Here f(x) = sin-1x + cos-1x + 1 11tan tan xx

Domain x[-1, 0) (0, 1]

For x[-1, 0), f(x) = 02 2

x(0, 1], f(x) = 2 2

So, required area = 12 12 2

=

2 .

(1, )

x

y

(0, /2)

(-1, 0) (1, 0)

2. If the equation |z – z1|2 + | z – z2|2 = k represents the equation of a circle, where z1 2+ 3i, z2 4

+ 3i are the extremities of a diameter, then the value of k is ___________ Sol. 4

k = |z– z1|2 + | z– z2|2 = | z1– z2|2 k = 4

3. If the shortest distance between the lines x 1 y 1 z 11 1 1

and x 2 y 3 z 41 1 1

is k ,

then k is equal to _______ Sol. 2



15

Since the given lines are parallel. From figure,

BC = (2 1) 1 (3 1) 1 (4 1) 13 3 3

= 1 2 3 2 33

.

AB = 1 4 9 14 Shortest Distance = AC = 14 12 2 .

A(1,1,1)

B(2,3,4)

x 2 y 3 z 4

1 1 1

x 1 y 1 z 1

1 1 1

C

4. The value of 1

1

x[1 sin x] 1

dx is _________, ([.] denotes the greatest integer)

Sol. 2

1 0 1

1 1 0

x[1 sin x] 1 dx x[1 sin x] 1 dx x[1 sin x] 1 dx

Now -1 < x < 0 [1 + sin x] = 0 0 < x < 1 [1 + sin x] = 1 [x[1 + sin x] + 1] = 1

so 1

1

x[1 sin x] 1 dx 2

5. Let f(x) = P0(x) + P1(x)ex + P2(x)e2x + … + Pn(x)enx, where n be an integer 1 and P0(x), P1(x), .…,

Pn(x) are polynomials. If f(x) = 0 for any arbitrary large number x, then the value of Pn – 1(–2) is equal to ___________

Sol. 0

nnxx

f(x)lim P (x) 0e

Similarly 0 1 n 1P (x) P (x) .......,P (x) 0 .



16

PPhhyyssiiccss PART – III

SECTION – A


This section contains 6 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 the above figure a dielectric is released inside

the capacitor at t = 0 from the end of the capacitor. Then which of the following current(i) Vs time(t) graph is correct in one complete oscillation. If clockwise current is taken as positive and anticlockwise as negative. (a > c). Neglect electrical and mechanical resistance.

V

a b c b

k

a c

(A) i

t

(B) i

t

(C) i

t

(D) i

t

Sol. C 2. Three capacitor of capacitance C each are connected in series as

shown in the figure. Initially switch S is open. Now capacitors are charged by a battery of emf V by connecting between terminal A and B. After long time battery is disconnected and inductor of inductance L is connected between A and B at time t = 0 so that an oscillatory circuit is formed. Now at an instant t0 switch S is closed, then find the amplitude of charge oscillations of the remaining capacitors.

C

C

L

C

S

A B

(A)

20

3cos tLC1CV 1

6 3

(B)

20

3cos tLCCV 1 1

3 5 3

(C)

20

3cos tLC1CV 1

4 3

(D) CV3

Sol. A



17

3. A gas has been subjected to a cycle of isochoric, isothermal

processes 1 – 2 – 3 – 4 – 1 as shown in the figure. Plot the graph for this cycle on PT diagram.

1

2 3

4

T

V

(A)

2

3

4

1

T

p

(B)

2

3

4

1

T

p

(C)

2

3

4

1

T

p

(D) 2 3

4 1

T

p

Sol. B 4. A disk is rotated in the horizontal plane with a constant angular speed equal

to /2 rad/sec. A body A with a mass m kg is moved in a frictionless slot at a uniform speed of 1 m/sec relative to the platform radially inwards. What is the net impulse developed on the body as it goes from r = 2m to r = 1 m. Assume mass of the disk is negligible.

r

(A) 2

21 m 1 ( 1)2 2

N-sec (B) 2

2m 1 ( 1)2

N-sec

(C) 2

23 m 1 ( 1)2 2

N-sec (D) 2

22m 1 ( 1)2

N-sec

Sol. B Impulse = change in momentum

p

= m ˆ ˆ ˆ ˆi j m i j2

|p

| = m2

21 ( 1)2

N- sec.

Y Y Y

X

X

/2

X

r = 1m

r = 2 m



18

5. In the given figure graph is drawn between

PVnT

and P at different temperature. Where n

is the number of moles and R is universal constant. Then which of the following is correct option.

(A) T1 < T2 <T3

(B) T1 < T2 >T3 (C) T1 > T2 >T3

(D) T1 > T2 <T3

Ideal gas

T1

T2

T3

200 400 600 800

8.314

(pV/

nT)[

J m

ol1

K

1 ]

0

P(atm) Sol. C 6. Even a photon experiences appreciable gravitational pull due to a massive star, due to this the

wavelength of the photon changes. The change in wavelength of light of wavelength emitted from a massive star of mass M and radius R is (Consider a photon of frequency f has energy hf (h

is plank constant) and mass 2

hfc

. (All dimensions are in SI unit) Assume GM << Rc2.

(A) 2

GM3Rc (B) 2

GMRc

(C) 2

2 GMRc (D) 2

GM2Rc

Sol. B

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. 7 to 9 A circuit containing capacitors C1 and C2, shown in the figure is in the stead state with key K1 closed. At the instant t = 0, K1 is opened and k2 is closed. 7. Find the angular frequency of oscillations of the L.C.

circuit. (A) 3 104 rad/sec (B) 4 104 rad/sec (C) 5 104 rad/sec (D) 6 104 rad/sec

k1

20 V

L = 0.2 mH

C1 = 2F

C2 = 2F

R

k2

Sol. C

= 4

1

1 5 10LC



19

8. Determine the first instant t(in sec), when energy in the inductor becomes one third of that in the capacitor C1.

(A) 10 sec3 (B) 20 sec

3

(C) 40 sec3 (D) 10 sec

Sol. A 9. Calculate the charge on the plates of the capacitor at that instant. (A) 10 3 C (B) 20 3 C (C) 30 3 C (D) 40 3 C Sol. A

22 20

1 1 1

q1 1 q 1 q 13 2 C 2 C 2 C Where q0 = C1V1

q = 10 3 C and using q = q0 cos t

t = 103 sec.

Paragraph for Question Nos. 10 to 12

A U tube containing two different liquid of density and 2 is fixed vertically on a rotating table as shown in the figure. Initially table is at rest and table can rotate about a vertical axis passing through the centre of the table. The interface of two liquids of densities and 2 respectively lies at the point A in a U tube at rest. The height of liquid

column above A is 8 a3

, where AB = 2a. The cross-sectional area of

the tube is S. Now the table is whirled with angular velocity about a vertical axis as shown in figure such that the interface of the liquids

shifts towards B by 2 a3

, then at this instant find

10. Find PB PA.

2

a 2a

A B

(A) ga (B) 2ga (C) 3ga (D) 4ga Sol. B PB – PA = (2a) 2g – (2ag) = 2ag 11. Find the force exerted by liquid on liquid 2 at the interface.

(A) 9Sag4

(B) 29Sag3

(C) 9Sag10

(D) 29Sag10

Sol. A



20

12. Find the value of .

(A) 9g4a

(B) 9g32a

(C) 9g8a

(D) 9g16a

Sol. B

S(Px – PA) =5a/3 2 2 2

a

8Sxdx S a9

. . . (1)

similarly S(PB – Px) = 2 256 S a9 . . . (2)

solving (1) and (2)

Fx = 9 Sag4

And = 9g32a

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:


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. A mechanism with two sliders is shown in

the figure. Slider A at the instant shown has a speed of 3 m/sec and is accelerating at the rate of 15m/s2 towards right at the given instant. If rod AB has a length of 3m and is 60 at the instant shown. Match the following.

R = 3 m

B

A

v Column I Column II

(A) Angular velocity of rod ( in rad/sec)

(p) 3

(B) Angular acceleration of the rod (in rad/sec2)

(q) 2( 3 1)

(C) Velocity of particle B (in m/sec) (r) 5 (D) Radial acceleration of particle B

(in m/sec2) (s) 3 (2 3 1)

2

(t) 3 Sol. (A) (p) (B) (q) (C) (t) (D) (t)



21

2. Match the following Column – I Column – II

(A) Particle A has non zero acceleration (p)

A

dB/dt = K

Variable magnetic field is present in a cylindrical region. A charged particle A is placed at rest in the magnetic field and then released.

(B) Particle A is moving with constant speed

(q)

q, m B

A

E

x

y i

j

V0

A constant electric field E and constant magnetic field B is present in j and k direction respectively. A particle of charge q and mass m is placed in x-y plane and released with velocity v0 in i direction. It is given that E = v0B.

(C) Locus of particle A is straight line. (r) A m

Beaker is filled with non viscous liquid. A particle of mass m is released from a point above the liquid. Density of particle is more than that of liquid. Consider the elastic collision of the particle with the base of the vessel.

(D) Locus of particle A is circle. (s) D

B

O

0

C

Particle A

The figure shows two rods BC and OD hinged from B and O respectively. Rod BC has a groove in which particle A can slide smoothly. Rod BC is rotating with constant angular speed 0.

(t)

A

Sun

Particle A is revolving around sun in elliptical orbit

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



22

SECTION –C

Integer Answer Type

This section contains 5 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. Due to a vertical temperature gradient in the atmosphere the index of refraction varies. Suppose

index of refraction varies as n = n0 1 ay where n0 is the index of refraction at the surface and a = 2.0 106 per metre. A person of height h = 2.0 m stands on a level surface. Actual distance from the person of the farthest point on the road beyond which nothing is visible to him is equal to 1000 meter. Then find the value of .

Sol. 2 Using snell’s law

0 0n (1 ay) cos n sin2

2

1(1 ay) 1dy1dx

, where tan = dy/dx

h 1000

0 0

dy dxay

= 2m.

(0, 0)

1000

h

x

y

(x, y)

2. A jet of water from a nozzle which if fixed

on the cart is deflected by 90 as shown in figure. Water jet is entering with velocity 1 cm/s with respect to cart and cross section area of jet is A = 2 cm2. Find the horizontal resisting force F (in dyne), acting on the cart to keep it moving on the horizontal road with constant horizontal speed of 1 cm/s.

A

Sol. 6 2 2

1 2 1F A V V AV where, V1 is the speed of cart and V2 is speed of water jet with respect to cart.



23

3. Four identical coherent sound sources S1, S2, S3 & S4 are placed

collinearly at a separation d from each other as shown in the figure. Each one can produce sound of intensity I and wavelength . When the separation d is equal to minimum possible value d1, no sound was detected at a point P on the line joining the sources. But when the separation d is equal to minimum possible value d2, the intensity of sound at P is 3I. Find the value of (4d1 – 6d2)

S1 S2 S3 S4 P

d d d

Sol. 0 For d1 phase difference between two sound sources should be 90o and for d2 phase difference

should be 60o

12 d

2

and 22 d

3

(4d1 – 6d2) = 0 4. A, B and C are three identical parallel conducting plates. Plates A and

C are rigidly attached to insulating fixed support and plate B is attached to insulating spring as shown in figure. Spring and support is very small in size and not affecting the induced charges between the plates. Plates are separated by equal distance d. If plate B is doing S.H.M. for small displacement x as compared to distance d. Time period of oscillation of plate B is T = (2 a)sec. Then find the value of a. Given that spring constant K = 2 N/m, mass of plate B is 8 kg, emf of battery is . Assume the area of plates is much larger then the separation between the plates. (Assume whole system is placed in gravity free space)

A B C

d d

Sol. 2 5. Electromagnetic radiation falls on a metallic body whose work function is 2eV. For a particular

radiation of frequency , the maximum kinetic energy of the photoelectron is found to be 4 eV.

Find the maximum kinetic energy (in eV) of photoelectron for the radiation of frequency 53

.

Sol. 8 hv = + Kmax hv = 2 + 4 = 6 eV

Now, 'max

5h 2 K3

'max

5K 6 23

'maxK 8eV

PAPER-1 (09.05.2017)€¦ · Integer Answer Type This section contains 8 questions. The answer to...

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