FILE NO. IIT- 260-27 CPT_25_(12 STD) NEW_MPC_17-11-08_Q.P

FILE NO. IIT- 260/27(15/11/2008)CODE SHAKUNTAL

Common Practice Test – 25 XII STD NEW [89 Batches]

Test Date: 17 t h Nov. 2008.

Time: 3 hours Maximum marks: 246

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

INSTRUCTIONSA. General:

1. This booklet is your Question Paper containing 69 questions. The booklet has 17 pages.

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

3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers, and electronic gadgets in any form are not allowed to be carried inside the examination hall.

4. The answer sheet, a machine–readable Objective Response Sheet (ORS), is provided separately.

5. DO NOT TAMPER WITH / MUTILATE THE ORS OR THE BOOKLET.

B. Filling the ORS6. On the lower part of the ORS, write in ink, your name in box L1,

your Registration No. in box L2 and Name of the Centre in box L3. Do not write these anywhere else.

7. Write your Registration No. in ink, in the box L4 provided in the lower part of the ORS and darken the appropriate bubble UNDER each digit of your Registration No. with a good quality HB pencil.

8. The ORS has a CODE printed on its lower and upper parts.9. Make sure the CODE on the ORS is the same as that on this booklet

and put your signature in ink in box L5 on the ORS affirming that you have verified this.

10. IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET.

C. Question paper format:11. The question paper consists of 3 parts (Part I: Mathematics, Part

II: Physics, Part III: Chemistry). Each part has 4 sections.12. Section I contains 6 multiple choice questions. Each question has 4

choices (A), (B), (C) and (D), out of which only one is correct.

13. Section II contains 4 multiple correct answer type questions. Each questions has 4 choices (A), (B), (C) and (D), out of which one or more answers are correct.

14. Section III contains 4 Reasoning type questions. Each question contains STATEMENT–1 AND STATEMENT–2.Bubble (A) if both the statements are TRUE and STATEMENT–2

is the correct explanation of STATEMENT–1.Bubble (B) if both the statements are TRUE but STATEMENT–2

is NOT the correct explanation of STATEMENT–1.Bubble (C) if STATEMENT–1 is TRUE and STATEMENT–2 is

FALSE.Bubble (D) if STATEMENT–1 is FALSE and STATEMENT–2 is

TRUE.15. Section IV contains 3 sets of Linked–Comprehension type question.

Each set consists of a paragraph followed by 3 questions. Each question has 4 choices (A), (B), (C) and (D), out which only one is correct.

D. Marking Scheme:16. For each questions in Section I, you will be awarded 3 marks if

you have darkened only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (–1) mark will be awarded.

17. For each question in Section II, you will be awarded 4 marks if you have darkened all the bubble(s) corresponding to the correct answer and zero mark for all other cases. If may be noted that there is no negative marking for wrong answer.

18. For each question in Section III, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (–1) mark will be awarded.

19. For each question in Section IV, you will be awarded 4 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (–1) mark will be awarded

Name of the Student Roll Number

I have read all the instructions and shall abide by them

……..………………………………………………………Signature of the Candidate

I have verified all the information filled in by the Candidate

………………………………………………………..Signature of the Invigilator

South Delhi : 47–B, Kalu Sarai, New Delhi – 110016 Ph.: 011–46080612/13/14North/West Delhi : 15, Central Market, West Punjabi Bagh, New Delhi – 110026 Ph.: 011–45062651/52/53East Delhi : 32-E, Patparganj Village, Mayur Vihar Phase – I, Delhi – 110 091 Ph.: 011–22750043/52Gurgaon : M–21, Old DLF Colony, Sector–14, Gurgaon Ph.: 0124–3217791/92 Dwarka : 317/318, Vikas Surya Galaxy, Main Mkt, Sector–4, Dwarka, New Delhi – 110 075 Ph.: 011–45621724/25Faridabad : SCO-27, IInd Floor, Sector – 16, Faridabad, Haryana – 121002 Ph.: 0122-4074626, 4066938

MATHEMATICSPART – I

SECTION – IStraight Objective Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. In triangle ABC, right-angled at C, tanA + tanB is equal to

(A) a + b (B)

(C) (D)

2. If sin x + sin2x = 1, then the value of cos12x + 3 cos10x + 3 cos8x + cos6x – 1 is equal to (A) 0 (B) 1(C) – 1 (D) 2

3. The value of is

zero if (A) x = 0 (B) y = 0

(C) x = y (D)

4. If sin A = sin B and cos A = cos B, then (A) sin (1/2) (A – B) = 0 (B) sin (1/2) (A + B) = 0(C) cos (1/2) (A – B) = 0 (D) cos (1/2) (A + B) = 0

5. A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is

(A) (B)

(C) (D) none of these

6. There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probabiloity that it is white is (A) 1/4 (B) 1/3

(C) 2/3 (D) 1/5

SECTION – IIMultiple Correct Answer Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

7. If A and B are two independent events such that P(A) = 1/2 and P(B) = 1/5, then (A) P(A B) = 3/5 (B) P(A | B) = 1/2 (C) P(A|A B) = 5/6 (D) P(A B | A B)] = 0

8. For two events A and B, if P(A) = P(A | B) = 1/4 and P(B |A) = 1/2, then (A) A and B are independent (B) A and B are mutually exclusive(C) P(A | B) = 3/4 (D) P(B | A) = 1/2

9. If x = sec tan and y = cosec + cot , then

(A) (B)

(C) (D) x y + x – y + 1 = 0

10. If 0 < , < and cos + cos cos ( + ) = 3/2 then

(A) (B)

(C) = (D)

SECTION – IIIReasoning Type

This section contains 4 reasoning type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

CODE(A) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is a correct explanation for STATEMENT–1(B) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is NOT a correct explanation for STATEMENT–1(C) STATEMENT–1 is True, STATEMENT–2 is False(D) STATEMENT–1 is False, STATEMENT–2 is True

11. Statement – 1: A = {2, 4, 6} , B = {1, 2, 3, } where A & B are the number occuring on a dice, then P(A) + P(B) = 1 Because

Statement – 2: If A1, A2, A3 ... An are all mutually exclusive events, then

P(A1) + P(A2) + ... + P(An) =1.

12. Statement – 1: If P(A/B) P(A) then P(B/A) P(B) Because

Statement – 2: P(A/B) =

13. Statement – 1: Balls are drawn one by one without replacement from a bag containing a white and b black balls, then probability that white balls will be first to exhaust is a/a+b. Because Statement – 2: Balls are drawn one by one without replacement from a bag containing awhite and b black balls then probability that third drawn ball is white is a/a+b.

14. Statement – 1: Out of 5 tickets consecutively numbers, three are drawn at random, thechance that the numbers on them are in A.P. is 2/15.Because Statement – 2: Out of (2n + 1) tickets consectively numbered, three are drawn at random, the

chance that the numbers on them are in A.P. is .

SECTION – IVLinked Comprehension Type

This section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Questions Nos. 15 to 17

, , , are angles in I, II, III and IV quadrant respectively and no one of them is an integral multiple of /2. They form an increasing arithmetic progression.

15. Which one of the following is correct (A) cos ( + ) > 0 (B) cos ( + ) = 0(C) cos ( + ) < 0 (D) cos ( + ) > 0 or cos ( + ) < 0

16. Which one of the following is correct(A) sin ( + ) = sin ( + ) (B) sin ( ) = 3sin ( )(C) tan 2( ) = 4tan ( ) (D) cos ( + ) = 5cos 2

17. If + + + = and = 70º, (A) 400º < < 580º (B) 470º < < 650º

(C) 680º < < 860º (D) 540º < < 900º


A class consists of n students. For 0 k n, let Ek denote the event that exactly k student out of n pass in the examination. Let P(Ek) = pk and let A denote the event that a student X selected at random pass in the examination.

18. If P(Ek) = C for 0 k n, then P(A) equals (A) 1/2 (B) 2/3(C) 1/6 (D) 1/(n + 1)

19. If P(Ek) = C for 0 k n, then the probability that X is the only student to pass the examination is (A) 3/4n (B) 2/(n + 1) (C) 2/n(n + 1) (D) 3n(n + 1)

20. If P(Ek) k for 0 k n, then P(A) equals (A) 3/n(n + 1) (B) (2n + 1)/3n(C) 1/(n + 1) (D) 1/n2


Suppose E1, E2, E3 be three mutually exclusive events such that P(Ei) = pi for i = 1, 2, 3.

21. If p1 = (1 – p), p2 = (1 + 2p) and p3 = (2 + 3p), then p belongs to

(A) (1, 2) (B)

(C) (D)

22. If p1, p2, p3 are the roots of 27x3 – 27x2 + ax – 1 = 0, then value of a is (A) 9 (B) 6

(C) 3 (D)

23. If p1 = (1 + 3p), p2 = (1 – p) and p3 = (1 – 2p), then p belongs to

(A) (B)

(C) (D) (1, 0)

PHYSICSPART – II

SECTION – IStraight Objective Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

24. The displacement of a particle is given at time t by : . Then,

(A) the motion of the particle is SHM with an amplitude of

(B) the motion of the particle is not SHM, but oscillatory with a time period of

(C) the motion of the particle is oscillatory with a time period of

(D) the motion of the particle is a periodic.

25. A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

(A) (B)

(C) (D) k

m

k

26. Two identical rods of length and mass m are welded together at right angles and then suspended from a knife-edge as shown. Angular frequency of small oscillations of the system in its own plane about the point of suspension is

(A) (B)

(C) (D) None of these

27. The horizontal plane is smooth. If the system is slightly displaced and released, it executes SHM. The maximum amplitude, if the upper block does not slip relative to the lower block, is (k is the spring constant and is the coefficient of friction between the two blocks)

(A) (B)

(C) (D)

28. A metal bar of length and area of cross-section is rigidly clamped between two walls. The Young’s modulus of its material is and the coefficient of linear expansion is . The bar is heated so that its temperature increases by . Then the force exerted at the ends of the bar is given by : (A) (B)(C) (D) .

29. Two wires of equal length and crosssection are suspended as shown. Their Young’s Modulii are Y1 and Y2 respectively. The equivalent Young’s modulus will be

(A)Y1 + Y2 (B)

(C) (D)



30. Which of the following functions represents SHM ?(A) (B) (C) (D)

31. The displacement of a particle of mass 100 g from its mean position is given by, y = 0.05 sin(5t + 0.4) [y is in metres]

(A) The period of motion is 0.1 s(B) The maximum acceleration of the particle is m/s2(C) The energy of the particle is joule(D) The force acting on the particle is zero when displacement is 0.05 m from mean position

32. Water is flowing through a long horizontal tube. Let PA and PB be the pressures at two points A and B of the tube(A) PA may be equal to PB

(B) PA may be greater than PB

(C) PA may be smaller than PB

(D) PA = PB only if the cross-sectional area at A and B are equal

33. Two springs of equal lengths and equal cross-sectional areas are made of materials whose Young’s modulii are in the ratio of 2:3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of : (A) (B) 3 : 2(C) (D) 9 : 4.



CODE(A) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is a correct explanation for STATEMENT–1(B) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is NOT a correct explanation for STATEMENT–1(C) STATEMENT–1 is True, STATEMENT–2 is False(D) STATEMENT–1 is False, STATEMENT–2 is True

34. STATEMENT – 1: Frequency of spring mass system in two cases shown in the figure is same. and

STATEMENT – 2: Frequency of a spring mass system , which does not depend

upon acceleration due to gravity.

35. STATEMENT – 1: The graph of total energy of a particle in S.H.M. w.r.t. position is a straight line with zero slope.and

STATEMENT – 2: Total energy of particle in S.H.M. remains constant throughout it’s motion.

36. STATEMENT – 1: When there is a thin layer of water between two glass plates, there is a strong attraction between them.because STATEMENT – 2: The pressure between the plates becomes less than atmospheric pressure because pressure inside a bubble or drop is greater than outside pressure.

37. STATEMENT – 1: Strain causes the stress in an elastic body. because STATEMENT – 2: An elastic rubber is more plastic in nature.




A particle of mass m oscillating under the influence of a Hooke’s law restoring force is given by F = kx exhibits simple harmonic motion

with angular frequency and time period . In the

arrangement shown in the figure, both the springs are relaxed and the blocks are displaced slightly, they perform simple harmonic motion.

38. The frequency of such oscillation is

(A) (B)

(C) (D)

39. The correct condition if the frictional force on block m2 is to act in the direction of its displacement from mean position

(A) (B)

(C) (D)

40. If the spring of spring constant K2 is removed and lower block of mass m2 vibrate with frequency v. Then the maximum amplitude of oscillations that the system can have, if the block is not to slip on the lower block, will be

(A) (B)

(C) (D)


An open rectangular tank 5 m × 4m × 3m high containing water upto a height of 2 m is accelerated horizontally along the longer side as shown in the figure

41. The maximum acceleration that can be given without spilling the water is(A) 2 m/s2 (B) 4 m/s2

(C) 5 m/s2 (D) 3 m/s2

42. If initially, the tank is closed at the top and is accelerated horizontally by 9 m/s2, the gause pressure at the bottom of the front walls of the tank(A) Zero (B) 4.5 × 104 pa(C) 9.0 × 104 pa (D) 2.25 × 104 pa

43. In the above part, the gauge pressure at the bottom the rear walls of the tank(A) Zero (B) 4.5 × 104 pa(C) 9.0 × 104 pa (D) 2.25 × 104 pa


The physical pendulum is just a rigid body, of whatever shape, capable of oscillating about a horizontal axis passing through it. For small oscillations the motion of a physical pendulum is almost as easy as for a simple pendulum. Figure shows a rigid body pivoted about point P and displaced from equilibrium through an angle . The gravitational force provided a restoring torque of magnitude mg l sin about point P and time period is given by

44. The period of oscillation for small angular displacement is

(A) (B)

(C) (D)

45. If the stick is pivoted about a point P, distance x from the center of mass, the period of oscillation is

(A) (B)

(C) (D)

46. For what value of x period of oscillation is minimum

(A) (B)

(C) (D)

CHEMISTRYPART – IIISECTION – I

Straight Objective TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

48. Identify the final product :-

Ph – OH ?

(A) Ph – Me (B) Ph – CH3

(C) Ph – Me (D) none

49. A hydrocarbon (A) of the formula C7H12 on ozonolysis gives a compound (B) which on aldol condensation followed by heating gives 1 – Acetylcyclopentene. A and B are respectively:-

50. The ease of acid-catalysed dehydration of C2H5OH,(CH3)2CHOH and (CH3)3COH follows the order(A)

(B)

(C)

(D)

51. Consider the following transformation

The major product(s) formed is (are)(A) (B)

(C) (D)

52. The major product formed in the reaction

is

(A) (B)

(C) (D)



53. Point out the correct synthesis :

The stable compound [X] may be utilized

56. Which of the following carbonyl compound can form stable hydrate? (A) Cyclopropanone (B) Formaldehyde (C) Hexafluoroacetone (D) Trichloroacetaldehyde (Chloral)



Code(A) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is a correct explanation for STATEMENT–1

(B) STATEMENT–1 is True, STATEMENT–2 is True; STATEMENT–2 is NOT a correct explanation for STATEMENT–1

(C) STATEMENT–1 is True, STATEMENT–2 is False

(D) STATEMENT–1 is False, STATEMENT–2 is True

57. STATEMENT – 1: RCl is hydrolysed to ROH slowly but the reaction proceeds rapidly in the presence of catalytic amount of KI. STATEMENT – 2: I is a strong nucleophile and it reacts with RCl to form RI. I is a better leaving group than Cl and RI is hydrolysed rapidly to ROH.

58. STATEMENT – 1: Iodoform is obtained by the reaction of acetone with hypoiodite and not with iodide. STATEMENT – 2: OI- is an oxidizing agent is well as an iodinating agent.

59. STATEMENT-1: Nucleophilies attack not carbonyl groups at carbon from a direction perpendicular to the plane of carbonyl group but at about 1070 (obtuse angle).

STATEMENT-2: Repulsion from filled or orbital forces nucleophile to attack at obtuse angle.

60. STATEMENT-1: Di-tert. Butyl ether cannot be prepared by Williamson’s ether synthesis.because

STATEMENT-2: Tert. butyl bromide on treatment with sodium tert. butoxide preferentially undergoes substitution by SN2




A young chemist performed three different experiments in the laboratory and got three different results. Experiments – Number – 1.

Experiments – Number – 2.

Experiments – Number – 3.

61. The compound X is

62. The compound Y is

63. The compound Z is


An organic compound ‘X’ having molecular formula C10H16O on reductive ozonolysis yields acetone

and two other compounds Y and Z. Y on treatment with dil. NaOH finally gives , while Z on

treatment with conc. NaOH followed by acidification and heating finally gives .

64. The compound Y is (A) Pentandial (B) 4-oxopentanal (C) Pentan-2, 3-dione (D) Pentan-2, 4-dione

65. The compound X is (A) (H3C)2C = CHCH2CH2C(CH3) = CHCHO

(B) (CH3)2C = CH - - CH2 – CH = CHCHO

(C) (CH3)2C = CH – CH2 – CH2 – CH = - CHO

(D) (H3C)2C = CH – CH2 – CH = - CH2 - -H

66. The compound Z is (A) CHO – CHO (B) CHO – COOH(C) CH3CHO (D) COOH – COOH

Paragraph for Questions Nos. 67 to 69In the following reaction sequence product, I, J and L are formed. K represents a reagent.

Hex – 3 – ynal I J

67. The structure of the product I is :

68. The structures of compound J and K, respectively

69. The structure of product L is

(D) None of these

If you have any genuine question paper problems please contact the following subject concerned persons.Paper Setter : PUNJABI BAGH Mathematics : Mr. Rajiv Ranjan Mobile: 9999069846Physics : Mr. V. B. Singh Mobile: 9990385916Chemistry : Mr. Sanjay Dixit Mobile: 9899699760

Common Practice Test –25XII STD [89 Batches]

ANSWER KEYMATHEMATICS PHYSICS CHEMISTRY

Q. No. Answer Q. No. Answer Q. No. Answers

1 B 24 A 47 B

2 A 25 D 48 B

3 D 26 B 49 A

4 A 27 A 50 A

5 A 28 C 51 C

6 C 29 B 52 A

7 ABCD 30 ABC 53 ACD

8 ACD 31 AC 54 ABD

9 BCD 32 ABD 55 ABC

10 ABC 33 A 56 ABCD

11 C 34 A 57 A

12 A 35 A 58 A

13 D 36 A 59 A

14 D 37 B 60 C

15 A 38 C 61 B

16 A 39 C 62 D

17 C 40 D 63 A

18 A 41 B 64 B

19 C 42 A 65 A

20 B 43 B 66 A

21 D 44 B 67 D

22 A 45 A 68 A

23 B 46 A 69 C

NIIT-DEL-PB-CODE- SHAKUNTAL_PT-25-89 XIISTD - 1

Hints Solution

1. We have

and

2. From sin x + sin2x = 1, we get sin x = cos2x.Now, the given expression is equal to

cos6x (cos6x + 3cos4x + 3 cos2x + 1) – 1 = cos6x (cos2x + 1)3 – 1 = sin3x (sin x + 1)3 – 1= (sin2x + sin x)3 – 1 = 1 – 1 = 0.

3. The given expression is equal to .cos y sin x – sin y cos x + sin y sin x + cos x cos y = sin (x – y) + cos (x – y)

The given expression will have value zero if

4. sin A = sin B sin A – sin B = 0

…..(i)

and cos A = cos B cos B – cos A = 0

…..(ii)

Equations (i) and (ii) are simultaneously true if sin (1/2) (A – B) = 0, while the other factors sin (1/2) (A + B) and cos (1/2) (A + B) cannot both be zero simultaneously.

5. Let E denote the event that a six occurs and A the event that the man reports that it is a six. We have

By Baye’s theorem


6. Let Ei (0 i 2) denote the event that urn contains i white and (2 – i) black balls. Let A denote the event that a white ball is drawn from the urn. We have P(Ei) = 1/3 for i = 0, 1, 2 and

P(A | Ei) = 1/3, P(A | E2) = 2/3 P(A | E3) = 1,By the total probability rule

P(A) = P(E1) P(A | E1) + P(E2) P(A | E2) + P(E3) P(A | E3)

7. We have P(A B) = 1 – P(A B) = 1 – P(A) P(B) [ A and B are

independent]

Next, P(A | B) = P(A) = 1/2, because A and B are independent.

Lastly, since AB = (AB)m P[AB)|(A B)] = 0

8.

A and B are independent. Since P(A B) = P(A) P(B | A) = (1/4) (1/2) = 1/8 0A and B cannot be mutually exclusive. As A and B are independent P(A | B) = P(A) = 1 – P(A) = 1 – 1/4 = 3/4

Since A and B are independent, P(B) = P(B | A) = 1/2

P(B | A) = P(B) = 1 – P(B) = 1/2

9. We have

Multiplying, we get

xy + 1 =

and

Thus x y + x – y + 1 = 0

and


10. The given equation can be written as

…..(i)

Also, since 0 < , < , we have = . Therefore, from (i) we get cos = ½, so that

11. (C)

12. (A) The statement-1 A is true and follows from statement-2

indeed P(A/B) = P(A)

P(B)

P(B/A) P(B)

13. Statement-1 is false. Since if the colour white is first to exhaust then last ball must be black. favourable sample points ((a + b – 1)!)b

req. probability =

(D)

14. (D) 2n+1 = 5, n = 2

P(E) =

For a, b, c are in A.P. a + c = 2b a + c is even a and c are both even or both odd. So, number of ways of choosing a and c is nC2 + n+1C2 = n2 ways.

P(E) =

15,16,170 < < 90º, 90º < < 180º, 180º < < 270º, 270º < < 360º.

270º < + < 450º + lies in the I or IV quadrant and cosine in both is positive.

If d is the common ratio of the A.P.Then = + d, = + 2d, = + 3d. + = + , 2( ) = 2d = . And + = 2. Note = d, = 3d

270º < < 360º 270º < + 3d < 360º


200º < 3d < 290º [ = 70º] 400º < 6d < 580º 680º < 4 + 6d < 860º 680º < + 860º

21. 0 p1, p2, p3 1and 0 p1 + p2 + p3 1

and

22. Use (p1 + p2 + p3) = (p1 p2 p3)1/3 to obtain

p1 = p2 = p3 = 1/3.

24. A

25. (D)

26. (B)

Sol. The distance of centre of mass from point of suspension CS =

M.I. of the system about S = 2/3 m .

27. (A)Sol. Let the amplitude be A

then, the maximum common acceleration when amplitude is A, is given by

kA = 5ma, a =

For the upper block f = mg,

g =

A =

28. (C)

29. (B)


Sol. Equivalent spring constant of a wire is given by

;

or or

30. (A,B,C)

31. (A,C)

32. (A, B, D)

33. (A)

34. (A)Sol. Frequency of a spring mass system does not depend upon acceleration due to gravity.

35. (A)

36. (A)36. The atmospheric pressure from the sides of two plates process them towards each other.

37. (B)

Sol. .

38. (C)

39. (C)

40. (D)

41. (B)

42. (A)

43. (B)

44. (B)

45. (A)

46. (A)



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FILE NO. IIT- 260-27 CPT_25_(12 STD) NEW_MPC_17-11-08_Q.P