1516- Advanced Paper-1 Set-a.pdf

FIITJEE PHYSICS, CHEMISTRY & MATHEMATICS CPT3 - 1

CODE:

Time Allotted: 3 Hours §

BATCHES – 1516

§

PAPER - 1 Maximum Marks: 234

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. You are not allowed to leave the Examination Hall before the end of the test.

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

A. General Instructions 1. 2. 3. 4. 5.

Attempt ALL the questions. Answers have to be marked on the OMR sheets. This question paper contains Three Parts. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics. Each part is further divided into three sections: Section-A, B & C. Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided for rough work. 6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed.

B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR sheet. 2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment No. and write in ink your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Two Parts. (i)

Section-A (01 – 10) contains 10 multiple choice questions which have ONE OR MORE THAN ONE correct answer. Each question carries +3 marks and No negative marking for wrong answer.

(ii) Section-B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2 columns. Statements in the first column have to be matched with statements in the second column. Each question carries +8 marks for all correct answer. For each correct row +2 marks will be awarded. There may be one or more than one correct choice. For each incorrect row –1 mark will be given. (iii) Section-C (01 – 08) contains 8 Numerical based questions with single digit integer as answer, ranging from 0 to 9 (both inclusive) and each question carries +4 marks for correct answer and – 2 mark for wrong answer.

Name of the Candidate :___________________________________________ Batch :____________________ Date of Examination :__________________ Enrolment Number :_______________________________________________

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P PA AR RT T –– II :: P PH HY YS SIIC CS S SECTION – A (Multiple Choice Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE THAN ONE is/are correct. 1.

In a long current carrying cylindrical conductor of radius r, the current is distributed uniformly over its cross section. The magnetic field at a separation x from the axis of the conductor has magnitude B. Then (A) B = 0, at the axis (B) B ∝ x , for 0 ≤ x ≤ r 1 (C) B ∝ , for x > r (D) B is maximum for x = r x

2.

Velocity and acceleration vector of a charged particle moving in a magnetic field at r r some instant are v = 3iˆ + 4ˆj and v = 2iˆ + xjˆ . Select the correct alternative(s). (A) x = –1.5 (B) x = 3 (C) magnetic field is along z-direction (D) kinetic energy of the particle is constant

3.

In figure, R is a fixed conducting ring of negligible resistance and radius a. PQ is a uniform rod of resistance r. It is hinged at the centre of the ring and rotated about this point in clockwise direction with uniform angular velocity ω. There is a uniform magnetic field of strength B pointing inward and r is a stationary resistance. Then (A) current through r is zero

×B P

Q r

R

2Bωa2 5r (C) direction of current in external resistance r is from centre to circumference (D) direction of current in external resistance r is from circumference to centre (B) current through r is

Space for rough work

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4.

E + −

Switch S of the circuit shown in figure is closed at t = 0. If e denotes the induced e.m.f. in L and I is the current flowing through the circuit at time t, which of the following graph is/are correct?

S

R (A)

(C)

6.

i

t (D)

e

O 5.

(B)

e

O

L

O

t

O

t

i

t

During the process AB of an ideal gas (A) work done on the gas is zero (B) density of the gas is constant (C) slope of line A from the T-axis is inversely proportional to the number of moles of the gas (D) slope of line AB from the T-axis is directly proportional to the number of moles of the gas A partition divides a container having insulated walls into two compartments I and II. The same gas fills the two compartments whose initial parameters are given. The partition is a conducting wall which can move freely without friction. Which of the following statements is/are correct with reference to the final equilibrium position? (A) The pressures in the two compartments are equal (B) Volume of compartment I is 3V/5 (C) Volume of compartment II is 12V/5 (D) Final pressure in compartment I is 5P/3

P

B

T

P, V,T

2P,2V,T

I

II

Space for rough work

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7.

Which of the following are correct? (A) An astronaut going from earth to moon along the straight line will experience weightlessness once. (B) When a thin uniform spherical shell gradually shrinks maintaining its shape, the gravitational potential at its centre decreases. (C) In the case of spherical shell, the plot of V versus r is continuous. (D) In the case of spherical shell, the plot of gravitational field intensity I versus r is continuous.

8.

Two satellites S1 and S2 are revolving around the earth in coplanar concentric orbits in the opposite sense. At t = 0, the position of satellites are shown in the diagram. The periods of S1 and S2 are 4 h and 24 h, respectively. The radius of orbit of S1 is 1.28 × 104 km. For this situation mark the correct statement(s).

S1

S2

(A) The angular velocity of S2 as observed by S1 at t = 12 h is 0.468π rads–1. (B) The two satellites are closest to each other for the first time at t = 12 h and then after every 24 h they are closest to each other. (C) The orbital velocity of S1 is 0.64π × 104 km. (D) The velocity of S1 relative to S2 is continuously changing in magnitude and direction both. 9.

A body totally immersed in water is raised in the water by a height h. The density of the body is d and the density of water is d0 while the volume of the body is V and d > d0. Which of the following statements will be true? (A) The net work done by the external agent is Vdgh. (B) The net work done by the external agent is Vgh(d – d0). (C) The potential energy of water is changed by raising the body. (D) The work done in the body by the gravitational force is equal and opposite to the work done by the hydrostatic force so that net work done is zero.

10.

In the figure, an ideal liquid flows through the tube, which is of uniform cross section. The liquid has velocities vA and vB, and pressures PA and PB at the points A and B, respectively. Then ⎯⎯ →

A

⎯⎯ →

B

(A) vA = vB (C) PA = PB

⎯⎯ →

⎯⎯ →

(B) vA > vB (D) PB > PA Space for rough work

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Part 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 labeled A, B, C and D, while the statements in Column II are labeled 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

t

A

p

q

r

s

t

B

p

q

r

s

t

C

p

q

r

s

t

D

p

q

r

s

t

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: 1.

A square loop is placed near a long straight current carrying wire as shown in figure. Match the following column.

i

2.

(A)

Column I If current is increased.

(p)

Column II Induced current in loop is clockwise.

(B)

If current is decreased.

(q)

(C)

If loop is moved away from the wire.

(r)

Induced current in anticlockwise. Wire will attract the loop.

(D)

If loop is moved toward the wire.

(s)

Wire will repel the loop.

loop

is

A piece of metal density ρ1 floats on mercury of density ρ2. The coefficients of expansion of the metal and mercury are γ1 and γ 2 , respectively. The temperatures of both mercury and metal are increased by ΔT. Then match the following column. Column I Column II (A) If γ 2 > γ1 . (p) No effect on submergence. (B)

γ 2 = γ1 .

(q)

(C)

If γ 2 < γ1.

(r)

Fraction of the volume of metal submerged in mercury. The solid sinks.

(D)

( γ 2 − γ1)ΔT .

(s)

The solid lifts up.

Space for rough work

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SECTION−C Integer Answer Type This section contains 8 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 1.

2 kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. Given: specific heat capacities of water and ice are 4.186 kJ K–1kg–1 and 2.092 kJ K–1kg–1, respectively. Latent heat of fusion of ice is 33.47 kJ kg–1 (in kg).

2.

One mole of an ideal mono atomic gas at temperature T0 is expanding slowly while following the law PV–1 = constant. When the final temperature is T0, the heat supplied to the gas is IRT0, where I is an integer. Find the value of I.

3.

Two soap bubbles A and B are kept in a closed chamber where air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. The surface tension of soap water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B, respectively. (Neglect the effect of gravity.)

4.

Two satellites with mass ratio m1 : m2 = 1 : 2 are moving in orbits of radii ratio r1 : r2 = 4 : 1 around a planet. Find the ratio of their time periods. Space for rough work

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5.

A charged particle enters a magnetic field at right angles to the magnetic field. The field exists for a length to 1.5 times the radius of the circular path of the plane. The particle 2π will be deviated from its path by . Find W. W

6.

Two parallel wire carrying equal currents in opposite direction are placed at x = ±a parallel to the y-axis with z = 0. The magnetic field at origin O is B1 and at P(2a, 0, 0) is B2. Then the ratio B1/B2 is

7.

B1 . Find the ratio. B2

In the figure as shown, a conducting wire PQ of length 1 m is moving in a uniform magnetic field B = 3 T with constant velocity v = 2 m.sec towards right R = 2 Ω, C = 1 F and L = 2 H. The currents through the resistance, capacitor and inductor at any time t are I1, I2 and I3, respectively. Find the value of I3 (in A) at time t = 2 sec.

P

R

C

v

L

B Q

8.

In the circuit as shown, charge varies with time t as q = (t2 – 4) where q is in C and time t is in seconds. Find VB A at time t = 3 sec (in V).

q + − 5F

1Ω 0.5H

B

Space for rough work

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P PA AR RT T –– IIII :: C CH HE EM MIIS ST TR RY Y SECTION – A (Multiple Choice Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE THAN ONE is/are correct. 1.

In the reaction OH NaOH aq

( ) ⎯⎯⎯⎯ → Intermediate Br2

the intermediate are (A)

O-

O

(B) Br

Br

(C)

Br O

(D)

Br

O-

Br Br

Space for rough work

2.

In the following reaction

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O

MeNH 2 +

+

excess

H ⎯⎯→ Me N

O

Intermediates can be OH

(A) Me

OH

(B) Me

NH

N OH

(C)

+

OH2 Me

N

O-

(D) Me

N

OH

O-

3.

The state of hybridization in the anionic Part of solid Cl2O6 is (A) sp3d2 (B) sp3d 3 (C) sp (D) sp2

4.

In the reaction: 3Br2 + 6CO32− + 3H2O → 5Br − + Bro3− + 6HCO3− (A) Bromine is oxidised and carbonate is reduced (B) Bromine is oxidised (C) Bromine is reduced (D) It is disproportionation reaction or autoredox change Space for rough work

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5.

CHO

H

OH

HO H

H OH

issubjected toKiliani − Fischer synthesis.

CH2OH The product formed is (A) CHO

(B)

CHO

H

OH

HO

H HO H

OH H OH CH2OH CHO

H

(C)

HO H HO H

H OH

CH2OH (C)

H

CHO HO

OH H OH

H

H H

OH OH

CH2OH

CH2OH 6.

Which of the following statements are correct? (A) Helium diffuses at a rate 8.65 times as much as CO does. (B) Helium escapes at a rate 2.65 times as fast as CO does. (C) Helium escapes at a rate 4 times as fast as CO2 does. (D) Helium escapes at a rate 4 times as fast as SO2 does.

7.

Which of the following is not a polyamide fibre? (A) Nylon -6,6 (B) Nylon-6 (C) Teflon (D) terylene Space for rough work

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8.

O

O

NO 2 NO 2

Br2 ⎯⎯⎯ →A Fe

(Predominant product)

A will be (A)

O

O

(B)

O

O

Br NO 2

(C)

NO 2

Br

O

NO 2 O

(D)

O

NO 2 Br

Br NO 2

NO 2

9.

Pyrophosphorus acid H4P2O5 (A) contains P in +5 oxidation state (B) is dibasic acid (C) is strongly reducing in nature (D) contains one P – O – P bond

10.

CH2Cl2 on heating with NaOH solution gives? (A) CH3OH (B) HCOONa (C) HCHO

NO 2 O

(D) CH 2

Cl

OH

Space for rough work

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Part 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 labeled A, B, C and D, while the statements in Column II are labeled 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

t

A

p

q

r

s

t

B

p

q

r

s

t

C

p

q

r

s

t

D

p

q

r

s

t

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: 1.

2.

Match the Column – I with Column – II Column-I (A) H2(gas) at P = 200 atm,T = 273k (B) H2(gas) at P→O, T = 273k (C) CO2(gas) at P→O, T = 273k (D) Real gas with large molar volume Match the Column – I with Column – II Column-I (A) Aldol condensation (B) Cannizzaro reaction (C) Benzoin condensation (D) Reformatsky reaction

(p) (q) (r) (s)

Column-II Z≠1 Attractive forces Predominate PV = nRT P(V–nb ) = nRT

(p) (q) (r) (s) (t)

Column-II CN– as catalyst Organozinc compound Hydride transfer β – hydroxyl aldehyde Enolate ion

Space for rough work

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SECTION−C Integer Answer Type This section contains 8 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 1.

Find the member of monobromo derivatives in the given reaction. O NBS/hv ⎯⎯⎯⎯ → CCl4

2.

Amongst the following, the total number of compounds which give Hoffmann Bromamide reaction is. CONH 2 CONH 2 , O H3C

CH 2

C

CONHCH 3

,

,

O NH 2 ,

H3C

C

NH 2

3.

A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g/ml. Assuming no change in volume. The molality of a 3.2 molar solution is?

4.

A cylinder containing 5 litre of O2 at 25°C was leaking. When the leakage was detected and checked Pressure inside cylinder was reduced from 8 atm to 2 atm. The ratio of amount of O2 initially Present to that left after leakage is equal to? Space for rough work

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5.

6.

Root mean square speed of a gas is 5 ms-1. If some molecules out of 10 molecules in all are moving with 7 ms-1 and rest all the molecules are moving with 3 ms-1, then number of molecules moving with higher speed is?

N2+ Δ ⎯⎯ → x + co2 + N2

COO-

The degree of unsaturation in X is. 7.

If a mixture containing ethyl acetate and ethyl propanoate is refluxed with C2H5ONa/C2H5OH ester condensation takes place. How many different condensation products would be formed?

8.

How many P – O – P Linkage are Present in uncommon oxide P4O8 Space for rough work

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P PA AR RT T –– IIIIII :: M MA AT TH HE EM MA AT TIIC CS S SECTION – A (Multiple Choice Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE THAN ONE is/are correct. 1.

If a, b, c are rational number (a > b > c > 0) and quadratic equation (a + b – 2c)x2 + (b + c – 2a)x + (c + a – 2b) = 0 has a root in the interval (–1, 0) then which of the following statement (s) is/are correct? (A) a + c < 2b (B) both roots are rational (C) ax2 + 2bx + c = 0 have both roots negative (D) cx2 + 2bx + a = 0 have both roots negative

2.

Let a1, a2, a3, ………. be a sequence of non-zero real number which are in A.P. for k ∈ N . Let fk(x) = akx2 + 2ak+1x + ak+2 (A) fk(x) = 0 has real roots for each k ∈ N . (B) Each of fk(x) = 0 has one root in common. (C) Non-common roots of f1(x) = 0, f3(x) = 0, ………form an A.P. (D) Non-common roots of f1(x) = 0, f3(x) = 0, ………form an H.P.

3.

A bag contains four tickets marked with 112, 121, 211, 222 one ticket is drawn at random from the bag. Let Ei(i = 1,2,3) denote the event that ith digit on the ticket is 2. Then (A) E1 and E2 are independent (B) E2 and E3 are independent (C) E3 and E1 are independent (D) E1, E2, E3 are independent

4.

If (1 +x + x2 + x3)100 = a0 + a1x + a2x2 +……+ a300x300 then which of the following statement(s) is /are correct? (A) a1 = 100 (B) a0 + a1 +a2 +….+300 is divisible by 1024 (C) Coefficients equidistant from beginning and end are equal (D) a0 + a2 +a4 +….+a300 = a1 + a3 +a5 +…..+a299

5.

If a, b, c are in H.P., where a > c > 0, then. 1 1 a+c (A) b > (B) − <0 a−b b−c 2 (C) ac > b2 (D) bc(1 – a), ac(1 – b), ab(1 – c) are in A.P. Space for rough work

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6.

If A(z1), B(z2), C(z3), D(z4) lies on |z| = 4 (taken in order), where z1 + z2 + z3 + z4 = 0 and quadrilateral ABCD have maximum area, then. (A) Maximum area of quadrilateral ABCD = 32 (B) Maximum area of quadrilateral ABCD = 16 (C) The triangle ∆ABC is right angled (D) The quadrilateral ABCD is rectangle

7.

The number of 5 letter words having letters alphabetically increasing from a to z, formed with the letters of the word CALCULUS is divisible by. (A) 2 (B) 3 (C) 5 (D) 11

8.

Let f(x) = ax2 + bx + c, a > 0 and f(2 – x) = f(2 + x) ∀x ∈ R and f(x) = 0 has 2 distinct real roots, then which of the following is true? (A) Atleast one root must be positive (B) f(2) < f(0) > f(1) (C) Minimum value of f(x) is negative (D) 4 a + b = 0

9.

Let z1, and z2 are two non-zero complex number such that |z1 + z2|=|z1| = |z2| then z1 may be z2 (A) 1+ω (B) 1+ω2 (C) ω (D) ω2

10.

The value of a for which inequality |x2 – 1| < a cos x have (a + 1) integral solutions ( a ∈ N) (A) a = 1 (C) a = 3

(B) a = 2 (D) a = 4 Space for rough work

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Part 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 labeled A, B, C and D, while the statements in Column II are labeled 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

t

A

p

q

r

s

t

B

p

q

r

s

t

C

p

q

r

s

t

D

p

q

r

s

t

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: 1.

All letters of the word BREAKAGE are to be jumbled. The number of ways of arranging then so that Column-I Column-II (p) 720 (A) The two A’s are not together (B) The two E’s are together but not two (q) 1800 A’s (C) Neither two A’s nor two E’s are (r) 5760 together (s) 6000 (D) No two vowels are together (t) 7560 Space for rough work

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2.

Match the column. (A) (B) (C) (D)

Column-I The sequence a, b, 10, c, d are in A.P., then a + b + c + d = Six G.M.’s are inserted between 2 and 5, if their product can be expressed as (10)n. Then n = Let a1, a2, a3,…….a10 are in A.P. and h1, h2, h3,……...,h10 are in H.P. such that a1 = h1 = 1 and a10 = h10 = 6, then a4h7 = ∞ 8r The value of ∑ 4 = is equal to r =1 4r + 1

Column-II (p)

6

(q)

2

(r)

3

(s)

20

(t)

40

Space for rough work

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

SECTION−C Integer Answer Type This section contains 8 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 1.

Let P(x) = x6 – x5 – x3 – x2 – x and α, β, γ, δ are the roots of the equation x4 – x3 – x2 – 1 = 0, then P(α) + P(β) + P (γ) + P( δ ) =

2.

In the polynomial function f(x) = (x – 1)(x2 – 2)(x3 – 3).…(x11 – 11) the coefficient of x60 is.

3.

The number of triangle with each side having integral length and the longest side is 11unit is equal to k2 then the value of k is

4.

What is the last digit of 1 + 2 + 3 +….......+ n if the last digit 13 + 23 +..……+ n3 is 1?

5.

Number of ways in which the letters of the word DECISIONS be arranged so that letter N 9 be somewhere to the right of the letter “D” is . Find λ . λ

6.

A real number λ is selected from the interval [0, 10]. The probability that roots of the equation x2 –λx + 4 = 0 are real is p then the value of 5 p is 6

7. 8.

⎛ 5 ⎞ The remainder when ⎜ ∑ 20 C2k −1 ⎟ is divided by 11, is. ⎝ k =1 ⎠ Let z be a complex No satisfying |z – 3| ≤ |z – 1|, |z - 3| ≤ | z – 5|, img(z) ≥ 0,0 ≤ arg z ≤ π / 4 then area of bounded the region is. Space for rough work

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

PHASE TEST-III (PAPER-1) ANSWER KEY SET-A PART – I (PHYSICS) SECTION-A A, B, C, D A, C, D B, D C, D A, B, D A, B, C, D A, B, C A, B, C, D B, C A, D

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

PART – II (CHEMISTRY) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

SECTION-B 1.

A → q, s B → p, r C → p, r D → q, s A→q B→p C→s D→q

2.

6 2 6 8 2 3 6 8

PART – III (MATHS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

SECTION-B 1.

2.

SECTION-C 1. 2. 3. 4. 5. 6. 7. 8.

SECTION-A A,C A,B,C C B,C,D A,C B, D C,D B B, C, D A,B,C

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

3 4 8 4 4 9 6 6

SECTION-A A, B, C, D A, B, D A, B, C A, B, C, D B, C, D A, C, D A, D A, B, C, D C, D A, B SECTION-B

1.

2.

SECTION-C 1. 2. 3. 4. 5. 6. 7. 8.

Paper Code

A→ t B→ q C→ r D→ p A→ s B→ r C→ p D→ q SECTION-C

1. 2. 3. 4. 5. 6. 7. 8.

6 1 6 1 8 3 3 6

HINTS & SOLUTIONS PHYSICS 1.

A, B, C, D

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

For x ≤ r, B =

µ0Ix 2πr 2

Hence, B ∝ x At axis x = 0 ⇒ B = 0 µI For x ≤ r, B = 0 2πx 1 Hence, B ∝ x Also B is maximum at surface (x = r) 2.

A, C, D r r For a charged particle moving in magnetic field a ⊥ v rr Hence, v.a = 0 ⇒ x = −1.5 r The magnetic field should be perpendicular to the a , hence should be perpendicular to x-y plane. The kinetic energy of a moving particle in the magnetic field remains unchanged as magnetic field does so work in the charged particle.

3.

B, D

4.

C, D

5.

A, B, D P-T graph is a straight line passing through origin. Therefore, V = constant. Therefore, Work done on the gas = 0. m 1 Further, ρ = ∝ V V Volume of the gas is constant. Therefore, density of gas is also constant. PV = nRT ⎛ nR ⎞ or P = ⎜ ⎟T ⎝ V ⎠ i.e., slope of P-T line ∝ n.

6.

A, B, C, D

7.

A, B, C For a proper explanation, the earth moon system will be regarded as an isolated system. At a particular point in space, the gravitational force of attraction of the earth on astronaut will be balanced by the gravitational force of attraction of the moon. V = –GM/R. As R decreases, GM/R increases or –GMR decreases. In the case of spherical shell, plot I versus r is discontinuous.

8.

A, B, C, D

9.

B, C

10.

A, D As per continuity equation, Av = constant, vA = vB. As point B is lower level; hence pB > pA.

1.

A → q, s; B → p, r; C → p, r; D → q, s

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2.

A → q; B → p; C → s; D → q

v S ρ1 = v ρ2 As the temperature changes, the fraction of volume submerged ⎞ (1 + γ 2ΔT) ΔfS ⎛ ρ1 ρ2 =⎜ × − 1⎟ = −1 fS ⎝ ρ2 ρ1 ⎠ (1 + γ1ΔT) ( γ − γ1 )ΔT ( γ − γ )ΔT = 2 ;( γ 2 − γ1 )ΔT = 2 1 ;( γ 2 − γ1 )ΔT (1 + γ1ΔT) (1 + γ1ΔT) Fraction of volume submerged fs =

1.

6 2 kg (ice) × mice × 20 + xL = 5 kg (water) × mice × 20 x = 1 kg ∴ final mass of water = 5 + 1 = 6 kg.

2.

2

3.

6

PA = P0 +

4T = 16 N/m2 RA

PB = P0 +

4T = 12 N/m2 RB 3

nB PB ⎛ RB ⎞ = ⎜ ⎟ =6 nA PA ⎝ R A ⎠ 4.

8 For satellite

2πr mv 2 GMm = 2 and v = r T r and T ∝ r 3 . So

T1 ⎛ r1 ⎞ =⎜ ⎟ T2 ⎝ r2 ⎠

3/2

= (4)3/2 = 23 = 8

5.

2

6.

3

7.

6 V = BvL = (3)(2)(1) = 6V VR = VC = VL = 6V (All are in parallel) di VL = L dt

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

6 = (2)

di dt

di = 3 A/sec dt After 2 sec, current will be 6A. 8.

8 q = t2 – 4 At t = 3 sec, q is positive dq di i= = 2t and =2 dt dt At t = 3 sec, q = 5 C and i = 6 A

q di − L − iR = VB C dt q di VA − VB = + L + iR C dt 5 = + (0.5)(0.2) + (6)(1) = 8 V 5 VA −

MATHEMATICS 1.

A, B, C, D (a + b – 2c)x2 + (b + c – 2a)x + (c + a – 2b) = 0 c + a − 2b Have roots 1 and a + b − 2c c + a − 2b as, < 0 ⇒ c + a < 2b a + b − 2c If a, b, c are rational and one root (1) is rational then both roots rational of ax2+bx + c = 0 2b For ax2 + 2bx + c = 0, f(0) = c > 0 and − <0 a So both roots negative. Similarly D is true

2.

A, B, D fk(x) = akx2 + 2ak+1x + ak+2+ fk(–1) =ak – 2ak+1 + ak+2 = 0 − ak + 2 So roots are –1, ak Non common root of f1(x) = 0, f3(x) = 0, ...... −a 3 −a 5 , ,..... a1 a3 a a 1 − 3 ,1 − 5 ,.... a1 a3 −2d −2d , ,... are in H.P. as a1, a3 are in A.P. a1 a3 A, B, C S = {112,121,211,222}

3.

E1 → {211, 222,} P(E1 ) =

1 2

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

1 2 1 E3 → {112, 222,} P(E3 ) = 2 E1 ∩ E2 = {222} 1 As P(E1 ∩ E2 ) = = P(E1 ).P(E2 ) 4 1 P(E1 ∩ E2 ∩ E3 ) = ≠ P(E1).P(E2 ).P(E3 ) 4 A, B, C, D (1 + x + x2 + x3)100 = a0 + a1x + a2x2 + .... +a300x300 a1 = f’(0) = 100 a0 + a1 + a2 +.... + a300 = 4100 = (210)20 is divisible by 210. 1 Replace x by . x (x3 + x2 + x + 1)100 = a0x300 + a1x299 + .... + a300 ∀ i ϵ 1, 2, ...., 300 So ai = a300 –i Put x = –1, a0 – a1 + a2 – a3 +.... + a300 = 0 a0 + a2 + ... + a300 = a1 + a3 + ... + a299 E2 → {121, 222,} P(E2 ) =

4.

5.

B, C, D a, b, c in H.P, a > c > 0 2ac 2ac or a > b= > c , so a > b > c > 0 a+c a+c (i) (a – c)2 > 0 a + c 2ac a+c (a + c)2 > 4ac ⇒ > ⇒ >b 2 a+c 2 (a + c)2 > 4ac ⇒ ac >

1 1 − = a−b b−c

4a2c 2 2

(a + c)

⇒ ac > b2

⎡ 1 1 1 1 ⎤ (a + c) − = (a + c) ⎢ − =− <0 ⎥ 2ac 2ac ac ⎣ a(a − c) c(a − c) ⎦ a− −c a+c a+c

1 1 1 , , → A.P a b c 1 1 1 − 1, − 1, − 1 → A.P a b c 1 ⎛ ⎞ ⎛1 ⎞ ⎛1 ⎞ abc ⎜ − 1⎟ ,abc ⎜ − 1⎟ ,abc ⎜ − 1⎟ → A.P ⎝a ⎠ ⎝b ⎠ ⎝c ⎠

(iv)

6.

7.

A, C, D |z| = 0 and z1 + z2 + z3 + z4 = 0 So points z1, z2, z3, z4 must form rectangle with diagonal 8. 1 A = d1d2 = 32 2

z2

z1

z3

z4

A, D

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

Case 1 (2 identical one kind, 2 identical second kind and one different letter) No of ways = 3c 2 .3c1 =9 Case (2) (2 identical and 3 distinct) No of ways = 3c1 × 4c 3 = 12 Case (3) (all 5 distinct) No of ways = 5c5 = 1 Total No of ways = 22. 8.

A, B, C, D F(2-x) = f(2 + x) F(0) = f(4) c = 16 a + 4b + c 4a + b = 0 so D is true. For x < 2 f(x) is decreasing. So, f(0) > f(1) > f(2).

2 x=2

9.

C, D Use z1 = ω and z2 = ω2 z1 + z2 = -1

10.

A, B

−

1 2

−1

0

1

1 2

−

1 2

−1

0

1

1 2

for a = 1 two integral solution = –1, 1 for a = 2 two integral solution = –1, 0,1 1.

A→ t; B→ q; C→ r; D→ p Word is BREAKAGE. B, R, E, E, A, A, K, G

6! 7c =7560 2! 2 (B) The two E’s are together but not two A’s = 5! 6c2 = 1800 (C) Neither two A’s nor two E’s are together = Total –(Two A’s together but not 2 E’s + two E’s together but not 2 A’s + two A,s as well as two E’s together) 8! = − (5! 6c2 + 5!6c2 + 6!) = 10080 – 2 × 3600 – 720 = 5760 2!2! 4! (D) No two vowels together = 4! 5c 4 = 720 2!2! (A) The two A’s not together =

2.

A→ s; B→ r; C→ p; D→ q (A) a + d = b + c = 10 is a + b + c + d = 20

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

(B) G1 G2 G3 G4 G5 G6 = (ab)3 = 103 ⇒ r = 3 a − a1 5 (C) a1 a2 --- a10 → AP. d = 10 = 9 9 15 24 a4 = a1 + 3d = 1+ = 9 9 h1,h2 – h10 → HP. 1 −1 1 1 1 5 , − → AP. d = 6 =− h1 h2 h10 9 54 1 1 6×5 4 9 = + 6d = 1 − = ⇒ h7 = h7 h1 54 9 4 So, a4h7 = 6. (D)

∞ 8r 8r = ∑ ∑ 4 2 2 r =1 4r + 1 r =1 2r − 2r + 1 2r + 2r + 1 ∞

(

)(

)

∞ ⎛ 1 1 =2 = 2.∑ ⎜ 2 − 2 2r + 2r + 1 R =1 ⎜ 2r − 2r + 1 ⎝

(

1.

) (

)

6 P(x) = x6 – x5 – x3 – x2 – x = x2(x4 - x3 – x2 - 1) + (x4 – x3 – x2 – 1) + (x2 – x + 1) x4 – x3 – x2 – 1 = 0 ∑ α = 1, ∑ αβ = −1 P(α) + P(β) + P(γ)+ P(s) =( ∑ α)2 - 2 ∑ αβ = 1 + 2 -1 + 4 = 6

∑

= ∑ α2 α+

∑

∑

α+

∑

1

1

2.

1 (x – 1) (x2 – 2) (x3 – 3) (x11 – 11) 60 Coefficient of x = (–1)(–2)(–3) + (–4) + (–1)(–5) + (–6) = –6 + 8 + 5 – 6 = 1

3.

6 x y z 11 11 1, 2….11 11 10 2, 3….10 11 9 3, 4….9 11 8 4, 5, 6, 7, 8 11 7 5, 6, 7 11 6 6 Total D = 1 + 3 + 5 +… + 11 = (6)2

4.

No of D 11 9 7 5 3 1

1 2

If last digit of

∑

⎛ n (n + 1) ⎞ n = ⎜ ⎟⎟ is 1 ⎜ 2 ⎝ ⎠ 3

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CPA B-LOT (CPT3-1) PT-3 PAPER-1 (PCM) SET-A

Then last digit of But last digit of

∑

n=

n (n + 1)

n (n + 1) 2

either 1 or 9.

can’t t be 9.

2

5.

8 For word DECISIONS Take two identical demo letters in place of D and N. 9! NO of arrangement of all letters = 2!2!2! Put N right to D by 1 way.

6.

3 x – λx + 4 = 0 D = λ2 − 16 ≥ 0 ⇒ λ ≥ 4 2

Probability of require event = 7.

10 − 4 3 = 10 5

3 6

⎛ 5 ⎞ ⎜ ∑ 202kc−1 ⎟ = 20c1 + 20c3 + 20c5 + 20c7 + 20c9 ⎝ k =1 ⎠ 20c1 + 20c3 + .. + 20c19 = 219

(

+ 20c9 = 218

20c1 + 20c3 + . 6 As ⎛ 5 ⎞ 18 ⎜ ∑ 20c r −1 ⎟ = 2 ⎝ k =1 ⎠

( )

6

= 2108 21

= 23 ( 32 ) = 8 ( 33 − 1) Remainder = 11 – 8 = 3 8.

6

)

21

6 Let z = x + iy |z - 3|≤ |z - 1| ⇒ 2 ≤ x |z 3| ≤ |z - 5| ⇒ x ≤ 4 Img z ≥0 ⇒ y ≥ 0 π 0 ≤ arg z ≤ ⇒ 0 ≤ y ≤ X 4 Area of ABCDEA = 6

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