Ftre 2013 Class 8 Paper 1

FIITJEE

Talent Reward Exam

for student presently in

Class 8

Time: 3 Hours

CODE 8A

Instructions:

Note:

Check all the sheets of this question paper. Please ensure the same SET is marked on header of all the sheets inside as indicated above ‘Maximum Marks’ of this page. In case SET marked is not the same on all pages, immediately inform the invigilator and CHANGE the Questions paper.

Registration Number

Name of the Candidate : _____________________________________________________________ Test Centre

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.

1. This Question Paper Consists of 7 Comprehension Passages based on Physics, Chemistry and Mathematics which has total 29 objective type questions.

2. All the Questions are Multiple Choice Questions having only one correct answer. Each question from Q. 1 to 9 carries +6 marks for correct answer and –2 marks for wrong answer. Each question from Q. 10 to 29 carries +8 marks for correct answer and –3 marks for wrong answer.

3. Answers have to be marked on the OMR sheet.

4. The Question Paper contains blank spaces for your rough work. No additional sheets will be provided for rough work.

5. Blank papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed.

6. Before attempting paper write your Name, Registration number and Test Centre in the space

provided at the bottom of this sheet.

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Straight Objective Type

COMPREHENSION – 1 (Question No. 1 - 3)

If a body is permanently at rest either no force is acting on it or resultant of all forces acting on it is equal to zero. Newton’s third law of motion says that forces always, exist in pairs, known as action and reaction. Action and reaction do not act on the same body, i.e. they act on two different bodies due to their mutual interaction. Action and reaction forces are equal in magnitude and opposite in direction, this does not mean that any two forces of same magnitude and having opposite direction are action reaction pair. If a body is accelerating resultant of all forces acting on it is equal to the product of its

mass and acceleration.

In the adjacent figure a ball A is suspended from a fixed support by a thin inextensible mass less string. As long as the ball is connected to the string it remains at rest.

A

1. If tension in the string is equal to 30 N, then the weight of the ball is equal to

(A) 30 N upward (B) 30 N downward

(C) 60 N upward (D) 60 N downward

Ans. B

Sol. Since bob is at rest, so tension must be balance by the weight. 2. The value of the resultant force on the ball is equal to

(A) 30 N upward (B) 60 N downward

(C) 30 N downward (D) zero

Ans. D

Sol. Since bob is at rest, so the value of the resultant force on the ball is equal to zero.

3. The string is cut. Assuming interaction between the ball A and the earth only, and the take the mass of the earth equal to 6  1024 kg. The acceleration of the earth just after cutting the string is equal to

(A) 5  1024 m/s2 upward (B) 5  1024 m/s2 downward

(C) 10 m/s2 (D) 10 m/s2 downward

Ans. A

Sol. a weight of ball mass of earth  = 24 24 30 5 10 6 10     m/s 2 upward

COMPREHENSION – 2 (Question No. 4 - 6)

Since long time, man has been able to fulfil his basic needs from substances derived from natural sources. The rapid development of science and technology as well as the population explosion has led to the synthesis of many man-made materials. Fabrics are made using fibres obtained from natural or artificial sources. Fibres are also used for making a large variety of household items like face wash, cold

drink bottles, chocolate wrappers, handkerchief, cellophane tape, fishing net, broken toys, sweater, socks, shirts etc.

The resources behind all these household items include both natural as well as man-made. Now based on the data given above, answer the following questions.

4. Which of the following cannot be a synthetic fibre?

(A) Cold drink bottles (B) Chocolate wrappers

(C) Handkerchief (D) Cellophane tape

Ans. C

Sol. Handkerchief is generally made up of cotton, while others are made up of synthetic fibres. 5. Which of the following will come under the category of plastics?

(A) Fishing net (B) Sweater

(C) Broken toys (D) Socks

Ans. C

Sol. Toys are broken that means they are made up of thermosetting plastics. 6. Which of the following will contain ester linkage?

(A) Shirt (B) Fishing net

(C) Sweater (D) Socks

Ans. A

Sol. Shirts are made up of polyesters which contain ester linkage.

COMPREHENSION – 3 (Question No. 7 - 9)

The absolute value of an integer is its numerical value irrespective of its sign (or nature). The absolute value of an integer ‘x’ is written as |x| and is defined as

x, if x 0 x x if x 0        .

7. Solution set of the equation |x  2| = 5 is

(A) {3,  7} (B) { 3, 7} (C) {3, 6} (D) none of these Ans. B Sol. |x  2| = 5 x  2 =  5 x = 7 x  2 =  5 x =  3 x  { 3, 7}.

8. The maximum value of the expression 27  |9x  8| is

(A) 27 (B) 17

Ans. A

Sol. 27  |9x  8| is maximum when |9x  8| = 0. 9. If x < 0, then | x| is equal to

(A) x (B)  x

(C) both (D) none of these

Ans. B

Sol. If x < 0, then  x > 0 and hence | x| =  x.

COMPREHENSION – 4 (Question No. 10 - 14)

You see your face daily in a plane mirror. You might have noticed that image formed by a plane mirror is identical to the object in all respects except for lateral inversion i.e., left hand side of the object becomes the right hand side of image and vice versa. The object distance and image distance from the plane mirror are always equal

10. You are standing at a distance of 3 m from a large vertical plane mirror. The distance between you and your image is equal to

(A) 3 m (B) 6 m

(C) 9 m (D) 12 m

Ans. B

Sol. Image distance and object distance are same in magnitude for plane mirror. 11. Now you move 2 m away from the mirror. The height of your image

(A) will inverse (B) will decrease

(C) will remain same (D) depends on the size of the mirror.

Ans. C

Sol. Height of image will not change

12. If you move towards the mirror with the speed of 2 m/s, the distance between you and your image (A) will increase at the rate of 2 m/s (B) will decrease at the rate of 2 m/s

(C) will increase at the rate of 4 m/s (D) will decrease at the rate of 4 m/s

Ans. D

Sol. If you move towards the mirror with the speed of 2 m/s, the image also move towards you with speed 2 m/s, so the distance between you and your image will decrease at the rate of 4 m/s. 13. In the adjacent figure ‘M’ is a plane mirror and doted line is

perpendicular to it. AB represents an object making an angle of 60 with the dotted line. The angle made by the image of AB, (AB) with the same dotted line is equal to

(A) 30 (B) 90 (C) 120 (D) 150 M A B 60

Ans. C Sol. M A B 60 A B 120

14. If the end B is at a distance of 10 cm from the mirror and length of AB is equal to 20 cm, the distance of A from its own image A is equal to

(A) 20 cm (B) 30 cm (C) 40 cm (D) 60 cm Ans. C Sol. BC20 cos 60 10 cm AA '40 cm M A B 60 A B 120 C

COMPREHENSION – 5 (Question No. 15 - 19)

Though most metals undergo similar kind of reactions, the “vigour” with which they react is not the same. Some are more reactive than the others. Metals along with hydrogen (a non-metal) are arranged in order of their reactivity in a series called activity series

Decreasing order of activity

The chemical activity of metals decreases from top to bottom in series. The metal higher in line series is more reactive than the metal lower in the series. More reactive metal displaces less reactive metal from its salt solution. Cation of less reactive metal is discharged first at the cathode.

15. Which of the following metals has no reaction even with steam

(A) Sodium (B) Calcium

(C) Iron (D) Silver

Ans. D

Sol. Silver is least reactive among these four. K Na Ca Mg Al Zn Fe H Cu Hg Ag Au

16. The gas which is produced when sodium reacts with water

(A) Hydrogen (B) Oxygen

(C) Water gas (D) Ozone

Ans. A

Sol. Metal when reacts with water, hydrogen gas is produced.

17. Beakers A, B and C contain zinc sulphate, silver nitrate and iron (II) sulphate solutions respectively. Copper pieces are added to each beaker. Blue colour will appear in case of

(A) Beaker A (B) Beaker B

(C) Beaker C (D) All the beakers

Ans. B

Sol. Blue colour will appear in beaker B as copper is more reactive than silver.

18. Four hypothetical metallic elements A, B, C and D form soluble nitrates having formulae ANO3, B(NO3)2, CNO3 and D(NO3)3.

Strips of each of these four metals were immersed in solutions (same concentrations) of the remaining three metal nitrates separately, and the following observations were recorded.

1. Metal B underwent reaction in all solutions. 2. Metal A only reacted with CNO3.

The order of increasing reactivity of the metals is

(A) C, A, D and B (B) B, C, D and A

(C) A, D, C and B (D) B, D, A and C

Ans. A

Sol. As metal B underwent reaction in all solutions so it is most reactive metal while metal A only reacted with CNO3 that means A is more reactive with C only.

19. A manufacturer uses electroplating of an iron nail with zinc to enhance its look and market value by putting it in an electrolytic cell as per the following diagram. Observe the diagram and identify the electrode X

(A) Iron (Anode) (B) Iron (Cathode)

(C) Zinc (Anode) (D) Zinc (Cathode)

Iron Nail

Electrolyte

Ans. C

Sol. Zn++ will be released from anode and will get deposited on iron nail which acts as a cathode.

COMPREHENSION – 6 (Question No. 20 - 24)

An identity is an equation consisting of variables which is true for all values of the variables. (i) (a + b) (a  b) = a2  b2

(ii) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (iii) (a + b)3 = a3 + b3 + 3ab (a + b)

(iv) (a  b)3 = a3  b3  3ab (a  b)

(v) (x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc + ca) x + abc (vi) a3 + b3 = (a + b) (a2  ab + b2)

(vii) a3  b3 = (a  b) (a2 + ab + b2)

(viii) (a + b + c) (a2 + b2 + c2  ab  bc  ca) = a3 + b3 + c3  3abc 20. If x  1 3 x  2, then 3 3 1 x x  is equal to (A)  8 (B) 63 8  (C) 8 63  (D) none of these Ans. B Sol. Since, 3 3 3 1 1 1 1 x x 3x x x x x x                    3 3 3 3 1 3 x 3 2 x 2                   3 3 27 1 9 x 8 x 2     3 3 1 27 9 x 8 2 x     = 27 36 63 8 8     .

21. If p  2q = 6 and pq = 8, then p3  8q3 is equal to

(A) 405 (B) 508 (C) 504 (D) 210 Ans. C Sol. (p  2q)3 = p3  (2q)3  3 p 2q p 

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Ans. A Sol. (a + b)3 = a3 + b3 + 3ab (a + b)  (5)3 = 35 + 3ab (5)  125  35 = 15ab ab = 90 6 15  . 23. (a  3b)3 + (3b  5c)3 + (5c  a)3 is equal to (A) (a  3b) (3b  5c) (5c  a) (B) 2 (a  3b) (3b  5c) (5c  a) (C) 3(3a  b) (3b  5c) (5c  2a) (D) 3(a  3b) (3b  5c) (5c  a)

Ans. D Sol. Since, a  3b + 3b  5c + 5c  a = 0  (a  3b)3 + (3b  5c)3 + (5c  a)3 = 3 (a  3b) (3b  5c) (5c  a). 24. If x 1 5 x   and 2 3 1 x 8 x

  , then the value of 3

2 1 x x  is (A) 215 (B) 256 (C) 125 (D) 525 Ans. C Sol. Suppose, 3 2 1 x k x    2 3 2 3 1 1 x x 8 k x x       2 3 1 1 1 x 2 x 3 x 8 k x x x                            (5)2  2 + (5)3  3 (5) = 8 + k  23 + 125  15 = 8 + k  23  8 + 110 = k  15 + 110 = k k = 125.

COMPREHENSION – 7 (Question No. 25 - 29)

A triangle ABC has been inscribed in a circle. The bisectors of A, B and C meet the circle at P, Q and R respectively and if BAC = 50.

25. BQP is equal to

(A) 50 (B) 25

(C) 130 (D) 75

Sol. BAP =  BQP = 25. R A Q B P C 25 25 26. 1 ABC 2 is equal to (A) 2QRC (B) 2QBC (C) QRC (D) none of these Ans. C Sol. 1 2ABC = QBC = QRC 27. RQP is equal to

(A) ACB + 20 (B) ACB + 50

(C) ACB + 25 (D) 1 2ACB + 25 Ans. D Sol. RQP = RQB + BQP = 1 2ACB + 25. 28. 2QRP is equal to (A) 1 2ABC + 25 (B) ABC + 50 (C) 1 2ABC + 50 (D) ABC + 25 Ans. B Sol. 2QRP = 2 (QPC + CRP) = 2 1 ABC 25 2          = ABC + 50.

29. QPR is equal to

(A) 130 (B) 115

(C) 75 (D) 65

Ans. D

Sol. QPR = QPA + APR = 1

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2   

= 1

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