pcm test-3 advanced paper-2-key & sol

Batch: TARGET-14REPEATERS PCM TEST [3]

[ Advanced Paper - II ]Ex.Date: 31-10-13

ANSWERKEY

PART-I: PHYSICS1. (A) 2. (C) 3. (B) 4. (C)

5. (C) 6. (C) 7. (B) 8. (B)

9. (A,B,D) 10. (B,C,D) 11. (B,C) 12. (A,C)

13. (5) 14. (7) 15. (0) 16. (7)

17. (6) 18. (2) 19. A P, B S, C R, D Q

20. A s; B p, r; C s; D p, r

CHEMISTRY21. (D) 22. (B) 23. (A) 24. (A)

25. (C) 26. (C) 27. (D) 28. (C)

29. (AC) 30. (C,D) 31. (A,B,C) 32. (B,C,D)

33. (5) 34. (4) 35. (1) 36. (3)

37. (0) 38. (6) 39. A p, B t, C s, D r

40. Apq; B r; Cps; Dr

MATHEMATICS41. (C) 42. (B) 43. (B) 44. (B)

45. (B) 46. (C) 47. (C) 48. (B)

49. (B, C) 50. (A,D) 51. (A,B) 52. (A, C, D)

53. (2) 54. (2) 55. (2) 56. (8)

57. (1) 58. (5) 59. Aq; B r; Cr; Dr

60. As; Bp; C r; Dq

Rao IIT Academy/ Target Batch -2014 / Repeaters PCM Test [1] Advanced Level Paper [2] / SOLUTION

website: www.raoiit.comwebsite: www.raoiit.com 2

RIIT- Repeaters PCM Test[3] - Advanced Paper [2]_SOLUTION

PART-I: PHYSICS [Solution]= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =1. (A)

2 2ˆ ˆ ˆ ˆw xy i yx j dx i dy j

2 2xy dx yx dy

2 21 d x y2

2 2x y2

(conservative force)

Topic :- WPE/ Sub Topic :- / Level :- 2/ Sr. Rep. Batch-2014 - DMp Sir

2. (C)

dp ˆ ˆF & p 2cos t i 2sin t jdt

ˆ ˆ ˆ ˆF P 2sin t i 2cos t j 2cos t i 2sin t j

= 0

F P

Topic :- NLM/ Sub Topic :- / Level :- 2/ Sr. Rep. Batch-2014 - DMp Sir

3. (B)As tension in right side string in more, it will move downwards.

Topic :- NLM/ Sub Topic :- / Level :- 2/ Sr. Rep. Batch-2014 - DMp Sir

4. (C)

As minv 5gR at the bottom-must point, the complete vertical circular motion, for h <2·5R, circular

motion will cease.Topic :- W.E. / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

5. (C)

ˆ ˆ2 2drv a bt i jdt

1t

ˆ ˆ2 2v a b i j

ˆ ˆ cos30v j v j

2 2 32 2 22

a b




223

a b

Topic :- Kinematics / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

6. (C)

Apply a = vdvdx

Topic :- Kinematics / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

7. (B)Topic :- Kinematics / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

8. (B)For the sphere,

0

dQ d msdt dt

30

43

dQ dR sdt dt

As 00 constant 0

ddt

34

3

dQ dR sdt dt

From Stefan’s law, 4 2 44

dQ AT R Tdt

Thus,

2 4 3443

dR T R sdt

1

ddt R s

Topic :- Heat / Sub Topic :- .................. / Level :- 2 / Sr. Rep. Batch-2014 - GSp Sir

9. (A,B,D)

Work done CA is cA A cw nR T T Specific heat capacity

321 13

vR RC C R

n R




23

2

2, ,3

P const pv const nT

92

C R

Topic :- Heat / Sub Topic :- .................. / Level :- 2 / Sr. Rep. Batch-2014 - DMp Sir

10. (B,C,D)In equilibrium average kinetic energy.

21 32 2rmsmu KT

Topic :- Heat / Sub Topic :- .................. / Level :- 2 / Sr. Rep. Batch-2014 - DMp Sir

11. (B,C)15 , 10k

/ 1.5 /v k m s

2 2 0.210

mk

Topic :- Wave / Sub Topic :- .................. / Level :- 1 / Sr. Rep. Batch-2014 - DMp Sir

12. (A,C)At any point of the string, maximum speed = A

102 110 10Va a f

3

3

1 1 102 2 10 2

f Hza

V f

22 10V mf


13. (5)Spring is compressed by x. Block stops and will not return back if kx mg

0.4 1 100.4

10mgx mk

Work done against friction =

2 20

1 12.5252 2

mg x mv kx

0.4 1 10 0.4 2.525

20

1 11 10 0.162 2

v




2023.40 1.6v

0 5 / secv m

Topic :- W.E. / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

14. (7)

22 2 2 sin 30sin3 32 2

BB g h huhg g

2 24 3 sin 30 sin 30Bgiven h h

33

4 4h

74 4

h 7h m


15. (0)

2 23 6 9x t t t

2 6dxv tdt

0, 6At v

6, 6At t v

Initial 21 6 182

KE m m

Final 21 6 182

KE m m

Work done 18 18 0KE m m


16. (7)Loss in KE = Work done against retarding force F offered by target.

221 1

2 2 2vmv m Fx

...(1)

and 21 02

mv Fx ....(2)

Find eqs. (1) and (2), ' 4

3xx

4 4' . 21 253 3

x x cm




' 7x x cm


17. (6)

1 3 1 8 3 4 20v

5 / secv m

For block 2 21 39, 1 5 82 2fA W J

For block 2 21 27, 3 5 42 2fB W J

Net work done by friction = – 6JTopic :- W.E. / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir

18. (2)For A,T, = kx ...(1)

For , 22kxB T mg ...(2)

From eqs., (1) and (2)

2 3 1025 5 10mgxk

= 1.2 m

If v is the speed of block A2 2

2 21 1 1 12 2 2 2 2 2

v xmv m kx k

2xmg

On solving 1/228 3

3 2 8gx Kxv

m

1/210 1.2 3 10 1.448

3 2 8 3

2 2 24.2 2.1 /v I in m s where I = 2




19. A P, B S, C R, D Q

/PV V dy dx

PV will be negative when slope is positive and slope is positive at E.

/dP B dy dx

dP will be zero at points where slope is zero.B and D are thus the points.If dP is positive, nearby points are in compression.dP is positive where slope is negative and slope is negative at C and so particle near C will be undercompression.For similar arguments, particle near E will be under tension.


20. A S; B P,R; C S; D P,R

If 1 2m m

1 1m g T m a

or1

1

m g Tam

...(1)

and 2 2T m g m a

or2

2

T m gam

...(2)

From eqs. (1) and (2), 1 2

1 2

m g T T m gm m

or 1 2

1 2

2m m gTm m

But 1 1T K X

So, 1 2

11 1 2

2m m gXK m m

and 1 2

22 1 2

2m m gXK m m

If 2 1m m then 2 1

2 1

m g T T m gm m

It will give same value of

1 2

1 2

2m m gTm m

and same values of 1X and 2XTopic :- W.E. / Sub Topic :- .................. / Level :-2 / Sr. Rep. Batch-2014 - DMp Sir




PART-II: CHEMISTRY [Solution]= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =21. (D)

Sol: From real gas behaviour idealP V b RT [1 mole of real gas at low pressure of 0.01 atm.]

0.0821 300 246.3 .0.01ideal

RTV b ltP

0.95 0.05ideal

ideal ideal

V b bZV V

0.05 246.3b lt b is excluded volume and is 4 times actual volume of gaseous molecules. Consider

344 12.315 .3

b r lt

33 3

23

16 12.315 103 6.023 10

r m

3 27 312.315 30 106.023 16

r m

1/3

912.31 30 10 1.076.023 16

r m nm

2 2.14r nmTopic: ( Gaseous state ) Subtopic: ( Real gases )_L-II_ Repeater Batch_Advaced_P2_AUch sir.

22. (B)is correct because

O(g) + e (g)– O (g)– egH = –140.9 kJ/mol–

exothermic

O (g) + e (g)– – O (g)–2 egH = 744 kJ/mol–

endothermicTopic: (Periodic Properties) ; Sub Topic: ( ) : Level : 1_Repeater Batch_Advaced_P2_KRch Sir

23. (A)Sol: For isoentropic process 0systemS

2 1.

1 2

ln ln 0P mT PnC nRT P

25 600ln ln2 300

P

1.75atmTopic: ( Thermo dynamics ) Subtopic: ( Entropy )_L-IRepeater Batch_Advaced_P2_AUch sir.




24. (A)Conceptual

Topic: (Periodic Properties) ; Sub Topic: ( ) : Level : 2_Repeater Batch_Advaced_P2_KRch Sir

25. (C)

Sol: 181.63 10E hcv J 1st line of Lyman series

Translational K.E. of H-atom = 32 A

R TN

1823

3 8.314 1.63 102 6 10

T

47.84 10T K Topic: Atomic st. + Gaseous state_Subtopic:Bohr’s model + KTG)_L-I_Repeater Batch_Advaced_P2_AUch sir.

26. (C)Sodium is geting oxidized and water is reduced to hydrogen gas.

Topic: Chemical equilibrium_Subtopic: Introduction_L-I_ Repeater Batch_Advaced_P2_KRch sir.

27. (D)Conceptual

Topic: Alkane_Subtopic: Preparation of Alkane_L-I_ Repeater Batch_Advaced_P2_KRch sir.

28. (C)

Sol: At constant volume 1 2

1 2

P PT T

2300 31200 2

P

and 1 24.63V Lfor single phase

dG Vdp SdT

300

2

200

. 2 10 .G V P T dT 210 500001231.5 200

2

781.5JTopic: Themodynamics_Subtopic: Free energy_L-I_Repeater Batch_Advaced_P2_AUch sir.29. (AC)

Anion of (ii) become aromatic. Amide (iv) is very less basic so it does not exist as zwitter ion. Only (iii) exitsas zwitter ion.When P site accept H+ it gives resonance stabilised species.

H N2 COOH

+NH2

Topic:GOC _Subtopic: Acid strength_L-I_ _Repeater Batch_Advaced_P2_AUch sir.




30. (CD)Sol: Le chatelier’s principle is not quantitative. If both stress would cause the same direction of shift, the shift is

determinable. If the two stress would cause shifts in opposite direction, no qualitative deduction is possible.Topic: (Chemical equillibrium ) Subtopic: (Le chatelier’s)_L-I__Repeater Batch_Advaced_P2_AUch sir.

31. (ABC)Conceptual

Topic: GOC_Subtopic: Mixed_L-I_ _Repeater Batch_Advaced_P2_AUch sir.

32. (BCD)Sol: ConceptualTopic: ( Atomic structure ) Subtopic: (Wave function)_L-I__Repeater Batch_Advaced_P2_AUch sir.

33. (5)

12

H X Ndou C

88 12

= 5Topic: GOC_Subtopic: Degree of unsaturation (dou)_L-I_ _Repeater Batch_Advaced_P2_KRch sir.

34. (4)900 32 4100 72

x

Topic: ( Mole concept ) Subtopic: (Avagadro hypothesis)_L-I__Repeater Batch_Advaced_P2_AUch sir.

35. (1)

Sol: we have 2

2 21 2

1 1 1 1 344 16 4H H HE z R hc R hc R hc

n n

This is possible for H atom for n1 & n2 as 1 & 2 respectively.

2 1 1n n n Topic: ( Atomic structure ) Subtopic: ( Bohr’s model )_L-I_ _Repeater Batch_Advaced_P2_AUch sir.

36. (3)

Br Br Br

Topic: Alkanes_Subtopic: Bromonation_L-I_ _Repeater Batch_Advaced_P2_KRch sir.

37. (0)Sol: 40.63 1000 108.8 373.4 0G H T S Topic: ( Thermo dynamics ) Subtopic: ( Gibs free energy )_L-I_ _Repeater Batch_Advaced_P2_AUch sir.




38. (6)

OO /Zn3+

Cl /hv2CH -CH=O3

Cl

2

*

1

Cl

*

Cl2

Cl1

Topic: Alkanes_Subtopic: Chloronation_L-II_ _Repeater Batch_Advaced_P2_KRch sir.

39. , , ,A p B t C s D r Topic: Chemical bonding_ Subtopic: Stucture_L-I__Repeater Batch_Advaced_P2_AUch sir.

40. (A-pq; B-r; C-ps; D-r)Topic: GOC_ Subtopic: Regents_L-I_ _Repeater Batch_Advaced_P2_AUch sir.




PART-III: MATHEMATICS [Solution]= = = = = = = = = = = = = = = = = = = = == = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =41. (C)

Let General term = tn

2

22

1 111

nt n n

2 22 2

22

1 11

n n n nn n

24 3 2 2

222

2 3 2 1 111

nn n n n n nt

n nn n

2 1 11

1 1nn ntn n n n

11

1nS tn n

1 1 1

1 1n nn nn n n n

1 1 9999991 999 1

1 1000 1000n

n

Topic: Sequence & Series_Sub Topic: Special Series_L-Tricky_SR-14(Repeaters Batch) Advance - II_GPm

Sir_ 31 Oct. 2013

42. (B)

1

sin 22 sin 2sin tan 2 tancos cos 3 2cos 2 cos cos 2 cos

t

Similarly, 2 tan 3 tan 2t

3 tan 4 tan 3t ..........................................................................................

t tan 1 tann n n

tan 1 tannS n

Topic: Trigonometry_Sub Topic: Special Series_L-Tricky_SR-14(Repeaters Batch) Advance - II_GPm Sir_ 31Oct. 2013

43. (B)

2 2

31 1

7 6 1 3 2x x

x x x x x

– + – +

–3 –1 2




1 0 13 2x x

x x

3, 1 2,x

The smallest integral value of x is –2Topic: Quadratic Equation_Sub Topic:Wavy Curve Method_L-Easy_SR-14(Repeaters Batch) Advance - II_GPm

Sir_ 31 Oct. 2013

44. (B)

The normal at P(t1) on y2 = 4x is y – 2t1 = –t1 21x t ....(1)

The normal at Q(t2) is y – 2t2 = 22 2t x t ....(2)

Suppose (1) and (2) cut at (t2, 2t) on the parabola, then 1 21 2

2 2t t tt t

1 21 2 1 2

1 2

22

t tt t t t

t t

The equation of the chord PQ is y (t1+t2) = 2x + 2t1t2

1 22 4 0x y t t

which obviously passes through the intersection of 2x + 4 = 0 and y = 0i.e. (–2, 0) for all values of t1 and t2.

Topic: Parabola_Sub Topic: Normals_L-Tricky_SR-14(Repeaters Batch) Advance - II_GPm Sir_ 31 Oct. 2013

45. (B)

3 36 6 2 2sin cos sin cosx x x x

2 2 4 2 2 4sin cos sin sin cos cosx x x x x x

22 2 2 21 sin cos 3sin cosx x x x

= 1 – 3sin2 x cos2x

231 sin 24

x

1sin2 2

3x

Now solution exists if 1 0

& 11 2 1

3

4 11 & 1 13 4

1 , 14

Topic: Trigonometry_L - Easy_SR-14(Repeaters Batch) Advance - II_GPm Sir_ 31 Oct. 2013




46. (C)

1 1a b b cb b

ab bc

2 2

1 1b ab a b b c b b c

ab bc

2 21 11 1

a b c bab bc

abc a c abc

2abc a c

1 12ba c

1 1, &ba c

are in AP

1, &a cb

are in HP..

Topic: Sequence & Series; Sub Topic: AP & HP_L - Easy_SR-14(Repeaters Batch) Advance - II_GPm Sir_ 31Oct. 2013

47. (C)

1 12 n n nr r rC C C

1 121 1

n nr r

n nr r

C C r n rC C n r r

22 22 1nr r r n r r n r n

22 22 4 4 2n n nr r n r

Topic: Binomial Theorem; Sub-Topic: Sequences & series in BT_L - Tricky_SR-14(Repeaters Batch) Advance -II_GPm Sir_ 31 Oct. 2013

48. (B)If we number the rows, the last integer in the nth row is :

11 2 3 4 ......

2n n

n

If nth row contains : 2014

12014 4028 1

2n n

n n

Now, 62 63 3906 & 63 64 4032

63 Bn

Topic Sequence & Series / Sub Topic :- Special series/ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced - I/GPm Sir/ 31 Oct. 2013




49. (B, C)

P, q, r are in H.P. 2 222 or 2prq q p r prp r

also, q2 = –2pr

q2(p+r)2 = q4

q2[(p+r)2 – q2] = 0

(p + r + q) (p + r – q) = 0

p + r = q

p + q + r 0, ,

2qp r are in A.P. or 2p, q, 2r are in A.P..

q2[p2+r2+2pr] = (–2 pr)2 = q4

q2 [p2 + r2 – q2] = q4 q2 [p2 + r2 – 2q2] = 0 or p2 + r2 = 2q2

p2, q2, r2 are in A.P..Topic Sequence & Series / Sub Topic :- A.P./ Level :- Tricky / SR.-14 / (Repeaters Batch) Advanced - I /GPm

Sir/ 31 Oct. 2013

50. (AD)

1 1 2 2andm n m and n are positive

Also 2 4m n and 2 4n m

4 2 416 16 4m n m m

4 and 4m n

If 8 4m n m n acceptable

If 9 , 4, 5 or 5, 4m n m n not acceptable

If 10 , 5, 5 or 6, 4 or 4, 6m n m n not acceptable

If 11 , 7, 4 or 4, 7 or 6, 5 or 5, 6 also 6, 5 and 5, 6m n m n sat isfies thecondition of integral roots m + n can be equal to 8 or 11.

Topic Quadratic Equations / Sub Topic :- Integral roots/ Level :- Tricky / SR.-14 / (Repeaters Batch) Advanced- I /GPm Sir/ 31 Oct. 2013

51. (AB)Third side makes equal angles with the given lines.

If slope is m, then 7 1

1 7 1m m

m m

(m – 7) (1– m) + (1 + m) (1 + 7m) = 06m2 + 16m – 6 = 0 m = –3 or 1/ 3

Topic Straight lines / Sub Topic :-Slopes, Isosceles triangle/ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced- I /GPm Sir/ 31 Oct. 2013




52. (A, C, D)Equation of the circle be x2 + y2 + 2gx + 2fy + c = 0, it passes through (2a, 0) 4a2 + 4ag + c = 0Radical axis is 2gx + 2fy + c + a2 = 0 which is 2x – a = 0

220,

1 0g f c a f ag c a

a

2 0,a ag c c g a and the equation of the circle is x2 + y2 – 2ax = 0Centre is (a, 0), passes through (0, 0) and (a, a)

Topic Circles / Sub Topic :-Radical Axis/ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31Oct. 2013

53. [2]2 3y x .........(1)

22 3dyy xdx

slope of tangent at 2 3

2

4 ,8

3 32p m m

dy xP mdx y

equation of tangent at P : y – 8m3 = 3m (x – 4m2) y = 3mx - 4m3 ................(2)it cuts curve again at point Q, solving (1) and (2), we get x = 4m2, m2

put x = m2 in equation (2)

2 3 3 2 33 4 ,y m m m m Qis m m

slope of tangent

2 3

4

3,

3 322m m

mdyQ mdx m

slope of normal at

1 23/ 2 3

Qm m

since tangent at P is normal at Q

22 3 9 23

m mm

Topic Normal to a Curve_ Level :- Tricky / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31 Oct. 2013

54. [2]Let BC be the side parallel to XY, h is the length of the altitude from A and r be the inradius, then

2XY h rBC h

Also by area of triangles, 12

r s BC h

A C

B

x

y

12 2pr BC h




ph rBC

2p BC BCXY XY

p

is maximum for 4pBC

and has value 28p

Topic: Straight Lines_ Level :- Tricky / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31 Oct. 2013

55. [2]Let 180 and 180A C B D

A

B

CD

B and D lies inside the circle through other three pointsand A and C lie outside the circle through other three points.

Topic: Circles_ Level :- Tricky / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31 Oct. 2013

56. [8]599 = 5(26 – 1)49 = 5 (26m –1) = 130m – 5 = 130m – 13 + 8 = 13 (10m – 1) +8 where m is any integer.Therefore remainder is 8.

Topic: Binomial Theorem_ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31 Oct. 2013

57. [1]

23 9 3 3 33 1d d d d

Topic: Progressions_ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir/ 31 Oct. 2013

58. [5]Given relation implies

3 3 4 41 2 1 2 1 2 1 24 2 4 2 0x x x x x x x x

But 1 2 1 21,x x x x c

gives 24 5 0 As 0, 4 5c c c c Topic: Quadratic Equation_ Level :- Easy / SR.-14 / (Repeaters Batch) Advanced - I /GPm Sir.

59. (A-q; B-r; C-r; D-r)

(A) 1 1 2sin cos2

x x

1 2 1cos cosx x 2 0,1.x x x

(B)1 1sin sin 2x y

x y

1sin xx

is increasing 0x and decreasing for 0x




1 1sin sin1 and 1x yx y

1 1sin sin 2x yx y

has no solution.

(C) cos cos sin sin cos sin2

x x x

cos 2 sin2

x n x

cos sin 22

x x n

no solution.

(D) tan 2 tan6

x x

let tan x y

22 3 1 0y y

no solution.Topic :- Trigonometry / Sub Topic :- _____________ / Level :- Tricky / SR.-14 / (Repeaters Batch)

Advanced - I /GPm Sir/ 31 Oct. 2013

60. As; Bp; C r; Dq

(A) Let the desired line by 2 1 2 0,x y x y which is perpendicular to 3 5 0,x y

so, 5 .2

Thus the line is 3 4 0x y

(B) any point on the line 2 1x y is 1 2 , .t t The image of the point in 0x y is given by, 1 2x t y t eliminating t we get 1 2 2 1y x x y

(C) in right angled triangles ACB and OCN, CBA CON

yB C(x,y)

A NxO the trianlges are similar

34

CN AC yON BC x

(D) the third vertex C is (h, k) then centroid 1 1 2 53 and 3 7.

3 3h kx h x y k y

now 2 2 6 3 7 2 2 3h k x y x y

pcm test-3 advanced paper-2-key & sol

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