Super 20 Sample Papers for Tamil Nadu Board for 2022 Combined Book Class 10

ANSWERS TO UNSOLVED SAMPLE PAPERS Sample Paper-4 Part - I 1. (d) trainee 2. (a) separated 3. (d) development 4. (b) certain 8. (a) -ness 5. (b) agreement 6. (b) durable 7. (c) loaves 11. (b) ware 9. (c) Business Process Outsourcing 10. (d) stands for 12. (c) off 13. (b) would have 14. (b) in spite of Part – II Section – I 15. The order had come from Berlin to teach only German in the schools of Alsace and Lorraine. The new master was to come the next day. Hence Mr. Hamel said it was the last French lesson. 16. To complete the journey with ultimate honesty was the quality of skipper which helped to bring out a successful expedition. Rather than the destination, it was the journey which was the most important. So her contention was to make sure that they go by the rules of circumnavigation which say that one need not use any auxiliary means of repulsion and take anybody else’s assistance. 17. Dragon Dictate helps Alisha overcome the difficulty in typing. 18. Sanyal recited the poem in the tea shop earlier since he recognised Aditya Narayan Chowdhury, from the mole on his right cheek, who cheated him of his prize for poetry recitation. Section – II 19. (a) The poet means that the past was very dull and glum. (b) One can travel cheerfully with a happy heart. 20. (a) The molten metal is hammered. (b) The meaning of the word, ‘gauged’ is measured. 21. (a) Summer here means development. (b) By the phrase ‘spring will come again’ , the poet means that the future will be better. 22. (a) A machine cannot love, show pity or forgive. (b) If a mistake is done by man, they will be destroyed and put to death. Section – III 23. The company sells cars. 24. He said that he knew a better restaurant. 25. ‘What was the idee of all them cops tarryhootin’ round the house last night?’ 26. I ate some rice along with a cup of soup. 27. (a) Science has armed man with inventions which are not less than miracles. (b) But science has also given man deadly weapons of warfare. Section – IV 28. • Go straight and take the first right. • Walk past the bank on your left and take the left and you will enter the main road. • Go straight. You will find the medical store on your left. B-85

MATHEMATICS QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to Total be Answered Marks Part-I 1 14 14 Multiple Choice Questions Part-II (Totally 14 questions will be given. Answer any Ten. 2 10 20 Any one question should be answered compulsorily) Part-III (Totally 14 questions will be given. Answer any Ten. 5 10 50 Any one question should be answered compulsorily) Part-IV 8 2 16 Total Marks 100 Weightage of Marks S.No. Purpose Weightage 1. 2. Knowledge 30% 3. 4. Understanding 40% Application 20% Skill/Creativity 10% C-1

1SAMPLE PAPER – (SOLVED) Time: 3 Hours Maximum Marks: 100 PART - I I. Choose the correct answer. Answer all the questions. [Answers are in bold] [14 × 1 = 14] 1. A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is ......................... . (1) 8 (2) 20 (3) 12 (4) 16 2. The sum of the exponents of the prime factors in the prime factorization of 1729 is ............... . (1) 1 (2) 2 (3) 3 (4) 4 3. If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is ................ . (1) 0 (2) 6 (3) 7 (4) 13 4. Which of the following should be added to make x4 + 64 a perfect square .............. . (1) 4x2 (2) 16x2 (3) 8x2 (4) –8x2 5. The number of points of intersection of the quadratic polynomial x2 + 4x + 4 with the X axis is ..................... . (1) 0 (2) 1 (3) 0 or 1 (4) 2 A 6. In the adjacent figure ∠BAC = 90° and AD ^ BC then ......... . (1) BD·CD = BC2 (2) AB·AC = BC2 (3) BD·CD = AD2 BD C (4) AB·AC = AD2 7. If (5, 7), (3, p) and (6, 6) are collinear, then the value of p is............. . (1) 3 (2) 6 (3) 9 (4) 12 8. If the ratio of the height of a tower and the length of its shadow is 3 : 1 , then the angle of elevation of the sun has measure ................... . (1) 45° (2) 30° (3) 90° (4) 60° 9. The total surface area of a cylinder whose radius is 1 of its height is ............ . 3 (1) 9 π h2 sq. units 8 (2) 24ph2 sq.units (3) 8 π h2 sq.units (4) 56 π h2 sq. units 9 9 C-3

10. The probability of getting a job for a person is x 3 . If the probability of not getting the job is 2 3 then the value of x is ................. . (1) 2 (2) 1 (3) 3 (4) 1.5 11. The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is ................. . (1) q + r (2) p + r (3) p + q r p+r r p+q p+ q p+q+ (4) p + q + 12. If there are 28 relation from a set A = {2, 4, 6, 8} to a set B, then the number of elements in B is. ................. . (1) 7 (2) 14 (3) 5 (4) 4 13. If a1, a2, a3 . . . . are in A.P. such that a4 = 3 , then the 13th term of the AP is ................. . (1) 23 (2) 0 a7 2 (3) 12a1 (4) 14a1 14. The X-intercept of the line 2x – 3y + 5 = 0 is .................. . (1) 5 (2) −5 (3) 2 (4) −2 2 25 5 PART - II [10 × 2 = 20] II. Answer any ten questions. Question No. 28 is compulsory. 15. If B × A = {(−2, 3),(−2, 4),(0, 3),(0, 4), (3, 3),(3, 4)} find A and B. Ans. B × A = {(–2, 3)(–2, 4) (0, 3) (0, 4) (3, 3) (3, 4)} A = {3, 4} B = {–2, 0, 3} 16. Let A = {1, 2, 3, 4, ..., 45} and R be the relation defined as “is square of a number” on A. Write R as a subset of A × A. Also, find the domain and range of R. Ans. A = {1, 2, 3, 4 . . . .45} T he relation is defined as \"is square of a number\" R = {(1, 1) (2, 4) (3, 9) (4, 16) (5, 25) (6, 36)} Domain of R = {1, 2, 3, 4, 5, 6} Range of R = {1, 4, 9, 16, 25, 36} 17. What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case? Ans. Find the L.C.M of 35, 56, and 91 35 = 5 × 7 56 = 2 × 2 × 2 × 7 91 = 7 × 13 C-4 Mathematics–X

L.C.M = 23 × 5 × 7 × 13 = 3640 Since it leaves remainder 7 The required number = 3640 + 7 = 3647 The smallest number is = 3647 18. Find the 19th term of an A.P. –11, –15, –19, ... Ans. First term (a) = – 11 C ommon difference (d) = –15 – (–11) = –15 + 11 = –4 n = 19 tn = a + (n – 1) d tn = –11 + 18 (–4) = –11 – 72 t19 = –83 19th term of an A.P. is – 83 19. Find the GCD of 12(x4 – x3), 8(x4 − 3x3 + 2x2) whose LCM is 24x3 (x –1)(x – 2) Ans. p(x) = 12(x4 – x3) 2 = 12x3(x – 1) –2 –1 g(x) = 8(x4 – 3x3 + 2x2) = 8x2(x2 – 3x + 2) = 8x2(x – 2) (x – 1) L.C.M. = 24x3 (x – 1) (x – 2) G.C.D. = p ( x) × g ( x) L.C.M. = 12x3 ( x −1) × 8x2 ( x − 2)( x −1) 24x3 ( x −1)( x − 2) G.C.D. = 4x2(x – 1) Sample Paper-1 C-5

20. Find the square root of the polynomial x4 – 12x3 + 42x2 – 36x + 9 by division method x2 – 6x + 3 Ans. x2 x4 – 12x3 + 42x2 – 36x + 9 x4 (–) 2x2 – 6x – 12x3 + 42x2 –12x3 –12x3 + 36x2 = – 6x 2x2 (+) (–) 2x2 – 12x + 3 6x2 – 36x + 9 6x2 6x2 – 36x + 9 =3 2x2 (–) (+) (0) 0 x4 −12x3 + 42x2 − 36x + 9 = |x2 – 6x + 3| 21. Write expression α + 3 + β+3 in terms of α + β and αβ. β α Ans. α + 3 + β+ 3 = α (α + 3) + β (β + 3) β α αβ α 2 + 3α + β2 + 3β = αβ = α 2 + β2 + 3α + 3β [a2 + b2 = (a + b)2 – 2ab ] αβ = (α + β )2 − 2αβ + 3(α + β) αβ 22. In the given diagram show that DPST ~ DPQR Ans. In ∆PST and ∆PQR , P 24 PS = 2= 2 , PT = 4 =2 ST PQ 2+1 3 PR 4+2 3 2 1R Thus, PS = PT and ∠P is common PQ PR Q Therefore, by SAS similarity, ∆PST ~ ∆PQR C-6 Mathematics–X

23. Find the value of \"a\" for which the given points (2, 3), (4, a) and (6, –3) are collinear. Ans. Let the points be A (2, 3), B(4, a) and C(6, –3). Since the given points are collinear. 1 Area of a triangle = 0 12 4 6 2 2 2 3 a −3 3 ( x1 y2 + x2 y3 + x3 y1 ) − ( x2 y1 + x3 y2 + x1y3 ) = 0 1 ( 2a − 12 + 18) − (12 + 6a − 6) = 0 2 2a + 6 – (6 + 6a) = 0 2a + 6 – 6 – 6a = 0 0 =0 –4a = 0 ⇒ a = 4 The value of a = 0 24. The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building. Ans. Let the height of the first building AD be \"x\" m E ∴ EC = 120 − x 45° In the right ∆ CDE, CE (120 - x) tan 45° = CD D 45° 1 = 120 − x ⇒ 70 = 120 − x C 70 70 m x x 120 m x = 50 m ∴ The height of the first building is 50 m A 70 m B 25. Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice. Ans. Sample space (S) = {(1, 1, 1) (1, 1, 2) (1, 1, 3) . . . . (6, 6, 6) n(S) = 63 = 216 Let A be the event of getting the same number on all the three dice A = {(1, 1, 1) (2, 2, 2) (3, 3, 3) (4, 4, 4) (5, 5, 5) (6, 6, 6) n(A) = 6 P(A) = n(A) = 6 = 1 n (S) 216 36 26. The radius and height of a cylinder are in the ratio 2:7. If the curved surface area of the cylinder is 352 sq.cm. Find its radius. Ans. Let the radius be \"2x\" and the height be \"7x\" Curved surface area = 352cm2 2πrh = 352 2 × 22 × 2x × 7x = 352 7 Sample Paper-1 C-7

44 × 2x2 = 352 x2 = 4345×22 = 176 =4 44 x = 4 = 2 Radius of the cylinder (2 × 2) = 4 cm 27. A number is selected at random from integers 1 to 100. Find the probability that it is not a perfect cube. Ans. Sample space = {1, 2, 3, . . . . 100} n(S) = 100 Let A be the event of getting a perfect cube. A = (1, 8, 27, 64} n(A) = 4 n(A) P(A) = n(S) P(A) = 4 = 1 100 25 The probability that the selected number is not a perfect cube is ( )P A = 1 – P(A) = 1 – 1 25 = 24 25 28. Which term of the arithmetic sequence 24, 23 1 , 22 1 , 21 3 . . . . . . is 3? 424 Ans. The given AP is 24, 23 1 , 22 1 , 21 3 . . . . 3 424 Here a = 24, d = 23 1 – 24 = – 3 , l = 3 44 n = l − a + 1 d 3 − 24 = − 3 + 1 4 = –21 × −4 +1 3 = 28 + 1 = 29 29th term of the arithmetic sequence is 29 C-8 Mathematics–X

PART - III III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50] 29. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that A × (B − C) = (A × B) − (A × C) Ans. A = {1, 2, 3, 4, 5, 6, 7}; B = {2, 3, 5,7} and C = {2} B – C = {2, 3, 5, 7} – {2} = {3, 5, 7} A × (B – C) = {1, 2, 3, 4, 5, 6, 7} × {3, 5, 7} = {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5) (2, 7) (3, 3) (3, 5) (3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5) (5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} .... (1) A × B = {1, 2, 3, 4, 5, 6, 7} × {2, 3, 5,7} = {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3) (2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7) (4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5) (5, 7) (6, 2) (6, 3) (6, 5) (6, 7) (7, 2) (7, 3) (7, 5) (7, 7)} A × C = {1, 2, 3, 4, 5, 6, 7} × {2} = {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (7, 2)} (A × B) – (A × C) = {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5) (2, 7) (3, 3) (3, 5) (3, 7) (4, 3) (4, 5) (4, 7) (5, 3) 5, 5) (5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} .... (2) From (1) and (2) we get A × (B – C) = (A × B) – (A × C) 30. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that (A ∩ B) × C = (A × C) ∩ (B × C) Ans. A = {1, 2, 3, 4, 5, 6, 7} B = {2, 3, 5,7} C = {2} (A ∩ B) × C = (A × C) ∩ (B × C) A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7} = {2, 3, 5, 7} ( A ∩ B) × C = {2, 3, 5, 7} × {2} = {(2, 2) (3, 2) (5, 2) (7, 2)} ....(1) A × C = {1, 2, 3, 4, 5, 6, 7} × {2} = {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (7, 2)} B × C = {2, 3, 5, 7} × {2} = {(2, 2) (3, 2) (5, 2) (7, 2)} (A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)}. . . . (2) From (1) and (2) we get (A ∩ B) × C = (A × C) ∩ (B × C) Sample Paper-1 C-9

31. In an A.P., sum of four consecutive terms is 28 and their sum of their squares is 276. Find the four numbers. Ans. Let us take the four terms in the form (a – 3d), (a – d), (a + d) and (a + 3d). Since sum of the four terms is 28, a − 3d + a − d + a + d + a + 3d = 28 4a = 28 gives a = 7 Similarly, since sum of their squares is 276, (a − 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 276. a2 − 6ad + 9d 2 + a2 − 2ad + d 2 + a2 + 2ad + d 2 + a2 + 6ad + 9d 2 = 276 4a2 + 20d 2 = 276 ⇒ 4(7)2 + 20d 2 = 276. d 2 = 4 gives d = ± 2 If a = 7, d = 2 then the four numbers are 7 – 3(2), 7 – 2, 7 + 2, 7 + 3(2) That is the four numbers are 1, 5, 9 and 13. If a = 7, d = −2 then the four numbers are 13, 9, 5 and 1 Therefore, the four consecutive terms of the A.P. are 1, 5, 9 and 13. 32. Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36? Ans. The greatest number of 6 digits is 999999. The greatest number must be divisible by L.C.M of 24, 15 and 36 24 15 36 2 12 3 5 2 18 26 29 23 33 24 = 23 × 3 15 = 3 × 5 360) 999999 (2777 36 = 22 × 32 720 2799 L.C.M = 23 × 32 × 5 = 360 2520 2799 To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360 2520 The greatest number in 6 digits = 999999 – 279 2799 = 999720 2520 279 C-10 Mathematics–X

33. There are 12 pieces of five, ten and twenty rupee currencies whose total value is `105. But when first 2 sorts are interchanged in their numbers its value will be increased by `20. Find the number of currencies in each sort. Ans. Let the number of `5 currencies be \"x\" Let the number of `10 currencies be \"y\" and the number of `20 currencies be \"z\" By the given first condition x + y + z = 12 ..... (1) By the given second condition 5x + 10y + 20z = 105 x + 2y + 4z = 21 (÷5) ..... (2) By the given third condition 10x + 5y + 20z = 105 + 20 10x + 5y + 20z = 125 2x + y + 4z = 25 ..... (3) (1) × 4 ⇒ 4x + 4y + 4z = 48 .... (1) (2) × 1 ⇒ x + 2y + 4z = 21 .... (2) _ (_–_)__(–_)____(–_)____(_–_)__ (1) – 2 ⇒ 3x + 2y + 0 = 27 3x + 2y = 2 7 .... (4) Subtracting (2) and (3) (2) ⇒ x + 2y + 4z = 21 (3) ⇒ 2x + y + 4z = 25 _ __(_–_)_(_–_)__(–_)_____(–_)___ – x + y + 0 = – 4 x – y = 4 .... (5) (4) × 1 ⇒ 3x + 2y = 27 .... (4) (5) × 2 ⇒ 2x – 2y = 8 .... (5) ___________________ (4) + (5) ⇒ 5x + 0 = 35 x = 35 =7 5 Substituting the value of x = 7 in (5) 7 – y = 4 ⇒ – y = 4 – 7 – y = – 3 ⇒ y = 3 Substituting the value of x = 7, y = 3 in .... (1) Sample Paper-1 C-11 7 + 3 + z = 12 z = 12 – 10 = 2 x = 7, y = 3, z = 2

Number of currencies in ` 5 = 7 Number of currencies in ` 10 = 3 Number of currencies in ` 20 = 2 34. Find the square root of the expression 4x2 + 20 x + 13 − 30 y + 9 y2 y2 y x x2 Ans. 2x + 5− 3y yx 2x 4x2 + 20x +13 − 30 y + 9y2 y y2 y x x2 4x2 (–) y2 4x + 5 20x +13 (–) y y 20x + 25 y 4x +10 − 3y −12 − 30 y + 9y2 yx x x2 (–) Hence, 4x2 + 20x +13 − 30 y + 9y2 = 2x +5− 3y −12 − 30 y + 9y2 y2 y x x2 y x x x2 0 35. If α, β are the roots of 7x2 + ax + 2 = 0 and if β − α = −13 . Find the values of a. 7 Ans. α and β are the roots of 7x2 + ax + 2 = 0 α+β= −a ; αβ = G72 i ven β – α = − 13 7 13 ⇒ α–β = 77 Squaring on both sides (α – β)2 =  173 2 α2 + β2 – 2αβ = 169 49 (α + β)2 – 2αβ – 2αβ = 169 49 (α + β)2 – 4αβ = 169 49 C-12 Mathematics–X

 a  2 4  72 169 a2 − 8 169 7 49 49 7 49 − − = ⇒ = a2 = 169 + 8 = 169 + 56 49 49 7 49 a2 = 225 ⇒ a2 = 225 × 49 49 49 49 a2 = 225 ⇒ a = ± 225 = ± 15 The value of a = 15 or –15 36. State and prove Thales theorem Statement A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio. A Proof : In DABC, D is a point on AB and E is a point on AC. 3 Given D1 2E To prove : AD = AE 1 2 DB EC B C Construction : Draw a line DE || BC No. Statement Reason 1. ∠ABC = ∠ADE = ∠1 Corresponding angles are equal because DE || BC 2. ∠ACB = ∠AED = ∠2 Corresponding angles are equal because DE || BC 3. ∠DAE = ∠BAC = ∠3 Both triangles have a common angle. By AAA similarity 4. DABC ∼ DADE Corresponding sides are proportional AB = AC AD AE Split AB and AC using the points D and E. AD + DB = AE + EC AD AE On simplification 1+ DB = 1+ EC AD AE Cancelling 1 on both sides DB = EC AD AE Taking reciprocals AD = AE Hence proved DB EC Sample Paper-1 C-13

37. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (–4, –2), (–3, k), (3, –2) and (2, 3) Ans. Let the vertices A (–4, –2), B (–3, k), C (3, –2) and D (2, 3) 1 Area of the Quadrilateral = 28 sq. units 2 = 28 ( x1 y2 + x2 y3 + x3 y4 + x4 y1 ) − ( x2 y1 + x3 y2 + x4 y3 + x1y4 ) 1 −4 −3 3 2 −4 k −2 3 −2 = 28 2 −2 1 ( −4k + 6 + 9 − 4) − (6 + 3k − 4 −12) = 28 2 = 28 1 ( 11) (3k 10) 2 −4k + − − –4k + 11 – 3k + 10 = 56 –7k + 21 = 56 –7k = 56 – 21 –7k = 35 ⇒ 7k = – 35 k = – 35 = –5 ∴ The value of k = –5 7 38. A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45° . The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30° . Determine the speed at which the bird flies. ( 3 = 1. 732) Ans. A is the initial position of the bird B is the final position of the bird Let the speed of the bird be \"s\" A 2s B E Distance = speed × time 80 m 45° ∴ AB = 2 s 80 m 30° Let CD be x C x D 2s ∴ CE = x + 2s AD CD In the ∆ CDA, tan 45° = 1 = 8x0 x = 80 ...(1) In the ∆ BCE = BE tan 30° CE 1 = x 80 3 + 2s x + 2 s = 80 3 x = 80 3 − 2 s ...(2) C-14 Mathematics–X

From (1) and (2) we get 80 3 − 2 s = 80 80 3 − 80 = 2 s ⇒ 80 ( 3 − 1) = 2 s s = 80( 3 −1) = 40 (1.732 − 1) = 40 × 0. 732 = 29. 28 2 Speed of the flying bird = 29. 28 m / sec 39. Some boys are playing a game, in which the stone thrown by them landing in a circular region (given in the figure) is considered as win and landing other than the circular region is considered as loss. What is the probability to win the game? (π = 3.14) Area of a rectangle = l × b sq. units 4 feet = 3 × 4 sq. feet = 12 sq. feet 3 feet sample space (S) = 12 1 feet n (S) = 12 Let A be the event of getting the stone landing in a circular region n (A) = Area of a circle = πr2 = π × 1 × 1 (radius of a circle = 1 feet) = π n(A) P (A) = n(S) = π = 22 12 7×12 11 11 = 7×6 = 42 11 157 Probability to win the game = 42 (or) 600 40. The perimeters of the ends of frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum be 8 cm, find the whole surface area of the frustum. [Use π = 3.14] Ans. Let the radii of circular ends be R and r [R > r] R Perimeter of circular ends are 207.24 cm and 169.56 cm ∴2 π R = 207.24 cm ⇒ R = 207.24 = 207.24 = 33 l 2π 2 × 3.14 r ⇒ R = 33 m and 2 π r = 169.56 Sample Paper-1 C-15

⇒ r = 169.56 = 169.56 = 27 2π 2 × 3.14 ∴ r = 27 m Slant-height of the frustum l = h2 + (R − r)2 = 82 + (33 − 27)2 = 64 + 36 = 10 cm The whole surface area of the frustum = π [(R2 + r2) + (R + r) l ] ∴ Required whole surface area of the frustum = 3.14 [332 + 272 + (33 + 27) × 10] cm2 = 3.14 [1089 + 729 +600] cm2 = 3.14 [2418] cm2 = 7592.52 cm2 41. A jar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marbles at random from the jar is 1 and the probability of selecting 3 a green marble at random is 4 . How many white marbles does the jar contain? 9 Ans. n(S) = 54 Let the number of blue marble, be 'x' Let A be the event of getting blue marbles n(A) = x n(A) P(A) = n(S) 1 = x 3 54 3x = 54 x = 54 = 18 3 N umber of blue marble is 18. Let the number of green marble be \"y\". Let B be the event of getting green marbles. n(B) P(B) = n(S) 4 = 5y4 9 9y = 54 × 4 y = 54 × 4 = 24 9 C-16 Mathematics–X

Blue marbles + green marbles + white marbles = 54 18 + 24 + white marbles = 54 ∴ Number of white marbles = 54 – 42 = 12 The jar contain 12 white marbles. 42. One - fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountain and the remaining 15 camels were seen on the bank of a river. Find the total number of camels. Ans. Le the total number of camels be 'x' x Number of camels seen in the forest = 4 Number of camels gone to the mountain = 2 x Number of camels on the bank of river = 15 Total number of camels = x + 2 x + 15 4 x x = 4 + 2 x + 15 4x = x + 8 x + 60 (multiply by 4) 3x – 8 x – 60 = 0 Let x = a2 3a2 – 8 a2 – 60 = 0 3a2 – 8a – 60 = 0 3a (a – 6) + 10(a – 6) = 0 (3a + 10) (a – 6) = 0 3a + 10 = 0 or a – 6 = 0 a = −10 or a=6 3 2 x =  −10  or 62 (x = a2)  3 Number of camels can-not be a fractionx = 1090 (or) 36 ∴ Number of camels = 36 Sample Paper-1 C-17

PART - IV IV. Answer all the questions. P' [2 × 8 = 16] P 43. (a) Construct a triangle similar to a given triangle PQR with its sides equal to 7 of the corresponding sides of the triangle PQR 4 Ans. Given a triangle PQR, we are required to QR R' construct another triangle whose sides are 7 Q1 x 4 Q2 Q3 Q4 Q5 Q6 Q7 of the corresponding sides of the triangle PQR. Steps of construction 1. Construct a ∆PQR with any measurement. 2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P. 3. Locate 7 points (the greater of 7 and 4 in 7 ) 4 Q1, Q2, Q3, Q4, Q5, Q6 and Q7 on QX so that QQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q4Q5 = Q5Q6 = Q6Q7 7 4 4. Join Q4 (the 4th point, 4 being smaller of 4 and 7 in ) to R and draw a line through Q7 parallel to Q4 R, intersecting the extended line segment QR at R'. 5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P' Then ∆P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of ∆PQR. [OR] C-18 Mathematics–X

(b) Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point. Rough Diagram Ans. Radius = 4 cm; Distance = 11 cm A 4 cm O 11 cm P 4 cm 4 cmB A 10.2 cm P O M 11 cm 4 cm 10.2 cm B Steps of construction: (i) With O as centre, draw a circle of radius 4 cm. (ii) Draw a line OP = 11 cm. (iii) Draw a perpendicular bisector of OP, which cuts OP at M. (iv) With M as centre and MO as radius, draw a circle which cuts previous circle A and B. (v) Join AP and BP. AP and BP are the required tangents. This the length of the tangents PA = PB = 10.2 cm V erification: In the right angle triangle OAP PA2 = OP2 – OA2 = 112 – 42 = 121 – 16 = 105 PA = 105 = 10.2 cm Length of the tangents = 10.2 cm Sample Paper-1 C-19

44. (a) Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0. Ans. (i) Draw the graph of y = x2 + 3x + 2 preparing the table of values as below. x –4 –3 –2 –1 0 1 2 3 4 16 x2 16 9 4 1 0 1 4 9 12 2 3x –12 –9 –6 –3 0 3 6 9 30 2 22222222 y 6 2 0 0 2 6 12 20 (ii) Plot the points (–4, 6), (–3, 2), (–2, 0), (–1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30). (iii) To solve x2 + 2x + 1 = 0 subtract x2 + 2x + 1 = 0 from y = x2 + 3x + 2 y = x2 + 3x + 2 0 = x2 + 2x + 1 (–) (–) (–) y = x + 1 (iv) Draw the graph of y = x +1 from the table x –4 –2 –1 02 4 y –3 –1 0 13 5 The equation y = x + 1 represent a straight line. This line intersect the curve at only one point (–1, 0). The solution set is (–1). C-20 Mathematics–X

Scale y (4, 30) x - axis: 1cm = 1 unit y - axis: 1cm = 2 units 30 – 28 – 26 – 24 – y = x2 + 3x + 2 22 – 20 – (3, 20) 18 – 16 – 14 – 12 – (2, 12) 10 – (–4, 6) 8– y=x+1 6 – (1, 6) (4, 5) (–3, 2) 4– 4 5 6 x' –6 –5 –4 –3 2 – (0, 2) (2, 3) (–4, –3) (–2, 0) (–1, 0) (0, 1) (2, 1) x 2 3 [OR] –2 –1 O 1 (–2, –1) – 2– – 4– – 6– y' Sample Paper-1 C-21

(b) Graph the following quadratic equation and state their nature of solutions: (2x − 3) (x + 2) = 0 Ans. y = (2x − 3) (x + 2) = 2x2 + 4x – 3x – 6 = 2x2 + x – 6 (i) Prepare a table of values for y from x – 4 to 4 x –4 –3 –2 –1 0 1 2 3 4 x2 16 9 4 1 0 1 4 9 16 2x2 32 18 8 2 0 2 8 18 32 x –4 –3 –2 –1 0 1 2 3 4 –6 –6 –6 –6 –6 –6 –6 –6 –6 –6 y 22 9 0 –5 –6 –3 4 15 30 (ii) Plot the points (–4, 22) (–3, 9) (–2, 0) (–1, –5) (0, –6) (1, –3), (2, 4), (3, 15) and (4, 30). (iii) Join the points by a free hand smooth curve. (iv) The curve intersect the X - axis at (–2, 0) and 1 1 , 0 . ∴ The solution set is  −2, 112 2 (v) Since there are two points of intersection with X - axis, the quadratic equation has real and un - equal roots. C-22 Mathematics–X

Scale y (4, 30) x - axis: 1cm = 1 unit y - axis: 1cm = 2 units 30 – 28 – 26 – (–4, 22) 24 – y = 2x2 + x – 6 22– 20– 18– 16– (3, 15) 14– 12– (–3, 9) 10– 8– 6– 4– (2, 4) 2– (–2, 0) 5 6 x x' –6 –5 –4 –3 –2 –1 O 1 2 3 4 (–1, –5) – 2– (1, –3) – 4– – 6– (0, –6) y' Sample Paper-1 C-23

4SAMPLE PAPER – (UNSOLVED) Time: 3 Hours Maximum Marks: 100 PART - I I. Choose the correct answer. Answer all the questions. [14 × 1 = 14] 1. If A = {1,2}, B = {1, 2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ....................... . (1) (A × C) ⊂ (B × D) (2) (B × D) ⊂ (A × C) (3) (A × B) ⊂ (A × D) (4) (D × A) ⊂ (B × A) 2. The range of the relation R = {(x, x2) | x is a prime number less than 13} is ............... . (1) {2, 3, 5, 7} (2) {2, 3, 5, 7, 11} (3) {4, 9, 25, 49, 121} (4) {1, 4, 9, 25, 49, 121} 3. Given F1 = 1 , F2 = 3 and Fn = F n – 1 + F n – 2 then F5 is .................. . (1) 3 (2) 5 (3) 8 (4) 11 4. The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P .............. . (1) 4551 (2) 10091 (3) 7881 (4) 13531 5. The solution of the system x + y − 3z = −6 , − 7y + 7z = 7, 3z = 9 is ............... . (1) x = 1, y = 2, z = 3 (2) x = −1, y = 2, z = 3 (3) x = −1, y = –2, z = 3 (4) x = 1, y = 2, z = 3 6. If (x – 6) is the HCF of x2 – 2x – 24 and x2 – kx – 6 then the value of k is .............. . (1) 3 (2) 5 (3) 6 (4) 8 7. The slope of the line which is perpendicular to a line joining the points (0, 0) and (– 8, 8) is ............. . 1 (3) 3 (1) –1 (2) 1 (4) – 8 8. The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the pole is 60° . The height of the pole (in metres) is equal to ......................... . b bb (1) 3 b (2) 3 (3) 2 (4) 3 9. If two solid hemispheres of same base radius r units are joined together along their bases, then curved surface area of this new solid is ............ . (1) 4p r2 sq. units (2) 6p r2 sq. units (3) 3p r2 sq. units (4) 8p r2 sq. units C-66

10. The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is ................. . q (2) p r p+q (4) p+r (1) p + q + r p+q+ (3) p + q + r p+q+r 11. Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is 1 ................. . 9 , then the number of tickets bought by Kamalam is (1) 5 (2) 10 (3) 15 (4) 20 12. If a and b are the zeros of the polynomial p(x) = 4x2 + 3x + 7 then 1 + 1 is equal to . . . . . αβ (1) 73 (2) − 73 (3) 73 (4) −3 7 13. If a, b, c are in A.P then a − b is equal to ................. . b−c (1) ba (2) bc (3) a (4) 1 c 14. If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to .................. . (1) 1200 π cm2 (2) 600 πcm2 (3) 300 πcm2 (4) 600 cm2 PART - II [10 × 2 = 20] II. Answer any ten questions. Question No. 28 is compulsory. 15. Let A = {x ∈ N | 1 < x < 4}, B = {x ∈ W | 0 ≤ x < 2} and C = {x ∈ N | x < 3}. Then verify that A × (B ∩ C) = (A × B) ∩ (A × C) 16. If the Highest Common Factor of 210 and 55 is expressible in the form 55x –325, find x. 17. Determine the general term of an A.P. whose 7th term is –1 and 16th term is 17. 18. If nine times ninth term is equal to the fifteen times fifteenth term, Show that six times twenty fourth term is zero. 19. Find the excluded values of the expression x2 + 6x + 8 x2 + x − 2 20. Solve 3p2 + 2 5 p – 5 = 0 by formula method. 21. Write the expression α + 3 + β+3 in terms of α + β and αβ. β α 22. Rhombus PQRB is inscribed in DABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus. 23. Find the equation of a straight line which is parallel to the line 3x – 7y = 12 and passing through the point (6 , 4) Sample Paper-4 C-67

24. Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 3 m. 25. Two coins are tossed together. What is the probability of getting different faces on the coins? 26. Three dice are thrown simultaneously. Find the probability of getting the same number. 27. If the curved surface area of a solid hemisphere is 2772 sq. cm, then find its total surface area. 28. The line 4x + 3y – 12 = 0 intersect the X, Y - axis at A and B respectively. Find the area of ∆AOB PART - III [10 × 5 = 50] III. Answer any ten questions. Question No. 42 is compulsory. 29. Represent the given relations {(x, y) | x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4} by (a) an arrow diagram, (b) a graph and (c) a set in roster form. 30. Find the H.C.F. of 396, 504 and 636 using Euclid's division algorithm. 31. The Sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms. 32. There are 12 pieces of five, ten and twenty rupee currencies whose total value is `105. But when first 2 sorts are interchanged in their numbers its value will be increased by `20. Find the number of currencies in each sort. 33. Simplify 2a2 + 5a + 3 ÷ a2 + 6a + 5 2a2 + 7a + 6 −5a2 − 35a − 50 34. A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains. 35. ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD. 36. Find the equation of the perpendicular bisector of the line joining the points A(–4,2) and B(6, –4). 37. From the top of a tower 50 m high, the angles of depression of the top and bottom of a tree and observed to be 30° and 45° respectively. Find the height of the tree. ( 3= 1.732) 38. Arul has to make arrangements for the accommodation of 150 persons for his family function. For this purpose, he plans to build a tent which is in the shape of cylinder surmounted by a cone. Each person occupies 4 sq. m of the space on ground and 40 cu. meter of air to breathe. What should be the height of the conical part of the tent if the height of cylindrical part is 8 m? 39. A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls. C-68 Mathematics–X

40. The internal bisector of A of ∆ABC meets BC at D and the external bisector of A meets BC produced at E. Prove that BD = CD . BE CE 41. The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator determine the fraction. 42. If α and β are the roots of the equation 3x2 – 5x + 2 = 0 form the equation whose roots are α 2 and β2 βα PART - IV IV. Answer all the questions. [2 × 8 = 16] 43. (a) Draw a circle of diameter 6 cm from a point P, which is 8 cm away from its centre. Draw the two tangents PA and PB to the circle and measure their lengths. [OR] (b) Construct a triangle ∆PQR such that QR = 5 cm, ∠P = 30° and the altitude from P to QR is of length 4.2 cm. 44. (a) Draw the graph of y = x2 and hence solve x2 – 4x – 5 = 0. [OR] (b) Draw the graph of y = x2 + 3x + 2 and use it to solve the equation x2 + 2x + 4 = 0. Sample Paper-4 C-69

ONE MARK & TWO MARK ANSWERS FOR UNSOLVED SAMPLE PAPERS SAMPLE PAPER – 4 1. (1) (A × C) ⊂ (B × D) 2. (3) {4, 9, 25, 49, 121} 3. (4) 11 4. (3) 7881 5. (1) x = 1, y = 2, z = 3 6. (2) 5 7. (2) 1 8. (3) b 9. (1) 4p r2 sq. units 2 10. (2) p 11. (3) 15 −3 p+q+r 12. (4) 7 13. (4) 1 14. (2) 600 πcm2 15. A × (B ∩ C) = (A × B) ∩ (A × C) B ∩ C = {0,1} ∩ {1,2} = {1} A × (B ∩ C) = {2, 3} × {1} = {(2,1), (3,1)} .... (1) A × B = {2, 3} × {0,1} = {(2, 0), (2,1), (3, 0), (3,1)} A × C = {2, 3} × {1, 2} = {(2, 1),(2, 2),(3, 1),(3, 2)} (A × B) ∩ (A × C) = {(2, 0), (2,1), (3, 0), (3,1)} ∩ {(2, 1), (2, 2), (3, 1), (3, 2)} = {(2, 1),(3, 1)} .... (2) From (1) and (2), A × (B ∩ C) = (A × B) ∩ (A × C) is verified. 16. Using Euclid’s Division Algorithm, let us find the HCF of given numbers 210 = 55 × 3 + 45 55 = 45 × 1 + 10 45 = 10 × 4 + 5 10 = 5 × 2 + 0 The remainder is zero. So, the last divisor 5 is the Highest Common Factor (HCF) of 210 and 55. Since, HCF is expressible in the form 55x −325 = 5 55x = 330 x = 6 17. Let the A.P. be t1, t2, t3, t4, . . . . It is given that t7 = –1 and t16 = 17 a + (7 – 1) d = –1 and a + (16 – 1) d = 17 a + 6d = –1 .... (1) a + 15d = 17 .... (2) Subtracting equation (1) from equation (2), we get 9d = 18 ⇒ d = 2 Putting d = 2 in equation (1), we get a + 12 = –1 So, a = –13 Hence, general term tn = a + (n – 1) d = –13 + (n – 1) × 2 = 2n – 15 C-94

18. tn = a + (n – 1)d 9 times 9th term = 15 times 15th term 9t9 = 15 t15 9 [a + 8d] = 15[a + 14d] 9a + 72d = 15a + 210d 9a – 15a + 72 d – 210 d = 0 – 6a – 138 d = 0 6a + 138 d = 0 6 [a + 23 d] = 0 6 [a + (24 – 1)d] = 0 6 t24 = 0 ∴ Six times 24th terms is 0. 19. x + 6x + 8 = (x + 4) (x + 2) x2 + x –2 = (x + 2) (x – 1) x2 + 6x +8 = (x + 4)( x + 2) = x +4 8 –2 x2 +x−2 (x + 2)( x − 1) x −1 4 2 x+4 2 –1 x −1 The expression is undefined when x – 1 = 0 ⇒ x = 1 The excluded value is 1 20. Compute 3p2 + 2 5 p – 5 = 0 with the standard form ax2 + bx + c = 0 a = 3, b = 2 5 , c = –5 p = −b ± b2 − 4ac 2a Substituting the values of a, b and c in the formula we get, ( )−2 5 ± 2 5 2 − 4(3)(−5) = −2 5± 80 = − 5±2 5 2(3) 6 3 21. p= Therefore, x = 5 ,− 5 3 α + 3 + β+ 3 = α (α + 3) + β (β + 3) β α αβ α 2 + 3α + β2 + 3β = αβ Answers C-95

= α 2 + β2 + 3α + 3β [a2 + b2 = (a + b)2 – 2ab ] αβ = (α + β )2 − 2αβ + 3(α + β) αβ 22. Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC By basic proportionality theorem AP = PQ ⇒ 12 − x = x AB BC 12 6 12x = 6 (12 – x) A 12x = 72 – 6x x| Q x 12x + 6x = 72 12 cm P 18x = 72 ⇒ x = 72 = 4 x x| R 18 10 3 m | 6 cm | Side of a rhombus = 4 cm PQ = RB = 4 cm 23. Equation of the straight line, parallel to B C 3x – 7y – 12 = 0 is 3x – 7y + k = 0 Since it passes through the point (6,4) 3(6) – 7(4) + k = 0 k = 28 −18 = 10 Therefore, equation of the required straight line is 3x – 7y +10 = 0. 24. Height of the tower (AC) = 10 3 m Distance between the base of the tower and point of observation (AB) = 30 m Let the angle of elevation ∠ ABC be θ In the right ∆ ABC, tan θ = AC C AB 10 3 3 = 30 = tan θ = 1 = tan 30° 3 3 θ A 30 m ∴ Angle of inclination is 30° B 25. When two coins are tossed together, the sample space is S = {HH,HT,TH,TT}; n(S) = 4 Let A be the event of getting different faces on the coins. A = {HT,TH}; n(A) = 2 Probability of getting different faces on the coins is P(A) = n(A) = 2 = 1 n (S) 4 2 C-96 Mathematics–X

26. Sample space = {(1, 1, 1) (1, 1, 2) (1, 1, 3) . . . . (6, 6, 6)} n(S) = 216 Let A be the event of getting the same number on all the three dice A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4) (5, 5, 5) (6, 6, 6)} n(A) = 6 P(A) = n(A) = 6 = 1 n(S) 216 36 27. Curved surface area of a hemisphere = 2772 sq.cm 2πr2 = 2772 πr2 = 2772 = 1386 2 T.S.A of the hemisphere = 3πr2 sq. units = 3 × 1386 cm2 = 4158 cm2 28. The equation of the line AB is 4x + 3y – 12 = 0 B 4x + 3y = 12 ( ÷ by 12) 4x + 3y = 12 O A 12 12 12 (0, 0) x + y = 1 3 4 ∴ The point A is (3, 0) and B is (0, 4) Area of ∆AOB = 1 [x1y2 + x2y3 + x3y1 – (x2y1 + x3y2 + x1y3)] 2 = 1 3 0 0 3 2 0 4 0 0 = 1 [12 + 0 + 0 – (0 + 0 + 0)] 2 = 1 × 12 2 = 6 sq.unit Answers C-97

SCIENCE QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to be Total Marks Answered Part-I 1 12 12 Multiple Choice Questions Part-II 2 7 14 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-III 4 7 28 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-IV 73 21 Total Marks 75 S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity D-1

Time: 3 Hours 1SAMPLE PAPER – Maximum Marks: 75 (SOLVED) PART - I (i) Answer all the questions. [Answers are in Bold ] [12 × 1 = 12] (ii) Choose the most suitable answer and write the code with the corresponding answer. 1. Inertia of a body depends on .................. . (a) weight of the object (b) acceleration due to gravity of the planet (c) mass of the object (d) Both a & b 2. The speed of light in vacuum or air is ................... . (a) 3 × 109 m/s (b) 3 × 108 m/s (c) 3 × 10–9 m/s (d) 3 × 10–8 m/s  3. When a sound wave travels through air, the air particles ..................... . (a) vibrate along the direction of the wave motion (b) vibrate but not in any fixed direction (c) vibrate perpendicular to the direction of the wave motion (d) do not vibrate 4. The number of neutrons is 8O16 is............. . (a) 8 (b) 16 (c) 32 (d) 24 5. Which of the following is a triatomic molecule? (a) Glucose (b) Helium (c) Carbon-di-oxide (d) Hydrogen 6. Photolysis is a decomposition reaction caused by............... . (a) heat (b) electricity (c) light (d) mechanical energy 7. Water which is absorbed by roots is transported to aerial parts of the plant through .................. (a) phloem (b) epidermis (c) cortex (d) xylem 8. Metastasis is associated with ................. . (a) Malignant tumour (b) Benign tumour (c) Both (a) and (b) ( d) Crown gall tumour 9. Oxygen is produced at what point during photosynthesis? (a) when ATP is converted to ADP (b) when CO2 is fixed (c) when H2O is splitted (d) All of these 10. The phenomenon by which carbohydrates are oxidised to release ethyl alcohol is ............... . (a) Glycolysis (b) Kreb’s cycle (c) Photosynthesis (d) Fermentation 11. Which one is referred as “Master Gland”? (a) Pituitary gland (b) Thyroid gland (c) Adrenal gland (d) Pineal gland 12. 9 : 3 : 3 : 1 ratio is due to ................. . (a) Segregation (b) Crossing over (c) Independent assortment (d) Recessiveness D-3

PART - II Answer any seven questions. (Q. No: 22 is compulsory) [7 × 2 = 14] 13. State Boyle’s law. When the temperature of a gas is kept constant, the volume of a fixed mass of gas is inversely proportional to its pressure. P ∝ 1 V 14. Differentiate mass and weight. S. No. Mass Weight (i) The quantity of matter contained in the The gravitation force exerted on it body due to the Earth's gravity alone. (ii) Scalar quantity Vector quantity (iii) Unit: kg Unit: N (iv) Constant at all the places Variable with respect to gravity. 15. What is refraction of light? When a ray of light travels from one transparent medium into another obliquely, the path of the light undergoes deviation. This deviation of ray of light is called refraction. 16. What is Concentrated Solution? Two solutions having same solute and solvent, the one which contains higher amount of solute per the given amount of solvent is said to be ‘Concentrated Solution’. 17. Give any two examples for heterodiatomic molecules. Heterodiatomic molecules – E.g: HCl, NaCl. 18. Match the following: (i) Pleura (a) Brain (ii) Meninges (b) Kidney (iii) Pericardium (c) Heart (iv) Capsule (d) Lungs (a) (ii) (b) (iv) (c) (iii) (d) (i) 19. Which hormone requires iodine for its formation? What will happen if intake of iodine in our diet is low? The hormones secreted by the thyroid gland are: (a) Triiodothyronine (T3) (b) Tetraiodothyronine or Thyroxin (T4), which need an amino acid tyrosine and Iodine for its formation. If the intake of Iodine in our diet is low or due to the inadequate supply of iodine in our diet leads to the enlargement of thyroid gland, which protrudes, as swelling in the neck and is called as goitre. 20. Define Triple fusion. During double fertilization, one sperm fuses with the egg and forms a diploid zygote. The other sperm fuses with the secondary nucleus to form the primary endosperm nucleus which is triploid in nature. This is called Triple fusion. D-4    Science – X

21. Define Cancer. Cancer is an abnormal and uncontrolled division of cells that invade and destroy surrounding tissue forming a tumor or neoplasm. It is a heterogenous group of cells that do not respond to the normal cell division. 22. The potential difference between two conductor is 110 V. How much work moving 5C charge from one conductor to the other? Given, V = 110 V , Charge q = 5C V= W q W=V×q W = 110 × 5 = 550 J ∴ Work done W = 550 J [7 × 4 = 28] PART - III Answer any seven questions (Q.No: 32 is compulsory) 23. Describe the rocket propulsion. (i) Propulsion of rockets is based on the law of conservation of linear momentum as well as Newton’s III law of motion. (ii) Rockets are filled with a fuel (either liquid or solid) in the propellant tank. When the rocket is fired, this fuel is burnt and a hot gas is ejected with a high speed from the nozzle of the rocket, producing a huge momentum. (iii) To balance this momentum, an equal and opposite reaction force is produced in the combustion chamber, which makes the rocket project forward. (iv) While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out. (v) Since, there is no net external force acting on it, the linear momentum of the system is conserved. (vi) The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket. ( vii) At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity. 24. Write any five points of properties of light. 1. Light is a form of energy. 2. Light always travels along a straight line. 3. Light does not need any medium for its propagation. It can even travel through vacuum. 4. Different coloured lights have different wavelength and frequency. 5. The speed of light in vacuum or air is, c = 3 × 108 m/s. 25. (a) Calculate the current and the resistance of a 100 W, 200 V electric bulb in an electric circuit. Given, P = 100 W, V = 200 V Power P = V I So, Current, I = P = 100 = 0.5A V 20 Sample Paper - 1    D-5

Resistance, R = V = 200 = 400 Ω I 0.5 (b) What is heating effect of current? A part of the energy from the source can be converted into useful work and the rest will be converted into heat energy. Thus, the passage of electric current through a wire, results in the production of heat. This phenomenon is called heating effect of current. 26. (a) Write the physical properties of Ethanol. 1. Ethanol is a colourless liquid, having a pleasant smell and a burning taste. 2. It is a volatile liquid. Its boiling point is 78°C (351K), which is much higher than that of its corresponding alkane, i.e. ethane (Boiling Point = 184 K). 3. It is completely miscible with water in all proportions. (b) Write the uses of Ethanol. 1. Ethanol is used in medical wipes, as an antiseptic. 2. It is used as as anti-freeze in automobile radiators. 3. It is used for effectively killing micro organisms like bacteria, fungi, etc., by including it in many hand sanitizers. 27. (a) Explain the various types of binary solutions based on the physical states of solute and solvent table. Solute Solvent Example Solid solution Solid Solid Copper dissolved in gold (Alloys) Liquid Solid Mercury with sodium (Amalgam) Liquid solution Solid Liquid Sodium chloride dissolved in water Liquid Liquid Ethyl chloride dissolved in water Gas Liquid Carobn-di-oxide dissolved in water (soda water) Gaseous solution Liquid Gas Water vapour in air (cloud) Gas Gas Mixture of Helium - Oxygen gases. (b) State Henry’s law? Henry’s law states that, the solubility of a gas in a liquid, is directly proportional to the pressure of the gas over the solution at a definite temperature. 28. Write the differences between endocrine and exocrine gland. S.No. Endocrine Exocrine 1. Secretion of endocrines are hormones. Secretion of exocrines are enzymes. 2. Endocrine glands do not have specific Exocrine glands have specific duct to duct hence ductless gland. carry their secretions. 3. Example: Pituitary gland Example: Salivary gland D-6    Science – X

29. (a) What are Allosomes? Allosomes are chromosomes which are responsible for determining the sex of an individual. There are two types of sex chromosomes. X and Y chromosomes. Human male have one X chromosome and one Y chromosomes and the human female have two X chromosomes. (b) What do you understand by the term phenotype and genotype? External appearance of a particular trait is known as phenotype. Example: Tall, dwarf Genotype is the genetic expression of an organism. Example: TT, tt 30. (a) What is Aerobic respiration? Aerobic respiration is a type of cellular respiration in which organic food is completely oxidized with the help of oxygen into carbon dioxide, water and energy. It occurs in most plants and animals. (b) What is respiratory quotient? The ratio of volume of carbon dioxide liberated and the volume of oxygen consumed, during respiration is called Respiratory Quotient (R.Q) R.Q. = Volume of CO2 liberated Volume of O2 consumed 31. Trace the pathway followed by water molecules from the time it enters a plant root to the time it escapes into the atmosphere from a leaf. Root hair cell → Root cortex cell → Xylem vessels Once the water enters the root hairs, it moves to the cortical cells by osmosis and then reaches the xylem. From there the water is transported to the stem and leaves. If excess, the water escapes from the leaf by transpiration. Root hair cell Root cortex cell Xylem vessels T. S. of the root showing movement of water from soil to xylem 32. (a) State Newton’s third law of motion and explain with examples: Law: ‘For every action, there is an equal and opposite reaction. They always act on two different bodies’. FB = – FA Examples: (i) When birds fly: Push the air downwards with their wings - Action Air push the bird upwards - Reaction (ii) When firing bullet: Sample Paper - 1    D-7

Gun recoils backward and the bullet is moving forward - Action Gun equals this forward action by moving backward - Reaction. (b) How will you identify saturated and unsaturated compounds? • Take the given sample solution in a test tube. • Add a few drops of bromine water • If the given compound is unsaturated, it will decolourise bromine water. • If the given compound is saturated it will not decolourise bromine water. PART - IV [3 × 7 = 21] (1) Answer all the questions. (2) Each question carries seven marks. (3) Draw diagram wherever necessary. 33. (a) (i) What is the Power of Accommodation? The ability of the eye lens to focus nearby as well as the distant objects is called power of accommodation of the eye. (ii) Explain about the types of defects in human eye. There are four types: 1. Myopia 2. Hypermeteropia 3. Presbyopia 4. Astigmatism 1. Myopia: Myopia, also known as short sightedness, occurs due to the lengthening of eye ball. Myopia affected people can see the near objects clearly but distant objects cannot be seen clearly. The focal length of eye lens is reduced or the distance between eye lens and retina increases. Due to this, the image of distant objects are formed before the retina. This defect can be corrected using a concave lens. xy x–y The focal length of the required concave lens is, f = 2. Hypermeteropia: Hypermeteropia, also known as long sightedness, occurs due to the shortening of eye ball. With this defect, distant objects can be seen clearly but nearby objects cannot be seen clearly. The focal length of eye lens is increased or the distance between eye lens and retina decreases. Due to this, the image of nearby object are formed behind the retina. This defect can be corrected by using a convex lens. The focal length of the required convex lens is, f = dD d–D 3. Presbyopia: Due to ageing, ciliary muscles become weak and the eye-lens become rigid and so the eye loses its power of accommodation. Some persons may have both the defects of vision - myopia as well as hypermetropia. This can be corrected by ‘bifocal lenses’. In which, upper part consists of concave lens (to correct myopia) used for distant vision and the lower part consists of convex lens (to correct hypermetropia) used for reading purposes. 4. Astigmatism: In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses. [OR] D-8    Science – X

(b) (i) A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta. (Ratio of momenta = 1:2) Given: Let mb , mt are the masses of truck and bike. mt = 4mb ..........(1) Here kinetic energies of both truck and bike are same 1 mtvt2 = 1 mb vb2 ..........(2) 2 2 1 × 4mbvt 2 = 1 mbvb2 2 2 vt2 = vb2 vt2 = 1 vb2 4 vt = 12 vb ............(3) To find ratio of the momenta, (i.e) mt vt : mb vb Ptruck = mt vt Pbike mbvb = 4× 1 = 2 = 2:1 2 1 Pbike = 1 = 1 : 2 (or) Ptruck 2 (ii) State the law of reflection of sound. Laws of reflection: • The incident wave, the normal to the reflecting surface and the reflected wave at the point of incidence lie in the same plane. • The angle of incidence ∠i is equal to the angle of reflection ∠r. 34. (a) (i) Calculate the number of Molecules in 54 g of H2O. Number of Molecules = Avogadro’s number × Gram mass Gram molecular mass = 6.023 × 1023 × 54 18 = 18.069 × 1023 Molecules (ii) Calculate the pH of 0.001 M NaOH. [OH –] = Normality = Molarity × acidity = 0.001 × 1 = 1 × 10 –3 Sample Paper - 1    D-9

pOH = –log10[OH–] = –log10 1× [10 –3] = – log101 – log1010 –3 = 0 – (– 3 log1010) pOH = 3 pH + pOH = 14 pH = 14 – pOH = 14 – 3 pH = 11 (iii) Calculate the mass of 0.01 mole of H2O. n = 0.01 mole; µ = 18 g mol–1 ; W = ? W = n × M = 0.01 × 18 = 0.18 g [OR] (b) (i) In what way hygroscopic substances differ from deliquescent substances. Difference between hygroscopic and deliquescent substances is in the extent to which each material can absorb moisture. This is because both of these terms are very much related to each other and they refer to the property of observing and the retention of moisture from the air. However, they differ in the extent of absorption of moisture where hygroscopic materials absorb moisture but not the extent the original substance dissolves in it, which is the casewith deliquescence. Therefore deliquescence can be regarded as an extreme condition of hygroscopic activity. Hygroscopic substances Deliquescence substance When exposed to the atmosphere at When exposed to the atmospheric air at ordinary temperature, they absorb ordinary temperature, they absorb moisture and do not dissolve. moisture and dissolve. Hygroscopic substances do not change Deliquescent substances change its its physical state on exposure to air. physical state on exposure to air. Hygroscopic substances may be Deliquescent substances are crystalline amorphous solids or liquids. solids. (ii) Define hydration. When ionic substances are dissolved in water to make their saturated aqueous solution, their ions attract water molecules which then attached chemically in certain ratio. This process is called hydration. 35. (a) (i) Enumerate the functions of blood. Functions of blood: 1. Transport of respiratory gases (O2 and CO2). D-10    Science – X

2. Transport of digested food materials to the different body cells. 3. Transport of hormones. 4. Transport of nitrogenous excretory products like ammonia, urea and uric acid. 5. It is involved in protection of the body and defense against diseases. 6. It acts as buffer and also helps in regulation of pH and body temperature. 7. It maintains proper water balance in the body. (ii) Differentiate Aerobic from Anaerobic respiration. Aerobic Anaerobic 1. Aerobic respiration takes place with the Anaerobic respiration takes place without help of oxygen. oxygen. 2. Glucose is converted into water, CO2 and Glucose is converted into ethanol or lactic energy acid. [OR] (b) Differentiate light dependent reaction and light independent reaction? Light dependent reaction Light independent reaction 1. It is called Hill reaction or Light reaction. It is called Dark reaction or Bio synthetic pathway or Calvin cycle. 2. The reaction is carried out in Thylakoid This reaction is carried out in the stroma membranes (Grana) of the chloroplast. of the chloroplast. 3. Photosynthetic pigments absorb the CO2 is reduced into carbohydrates with light energy and convert it into chemical the help of light generated ATP and energy ATP and NADPH2. NADPH2. 4. It is carried out in the presence of light. It is carried out in the absence of light. Sample Paper - 1    D-11

4SAMPLE PAPER – Maximum Marks: 75 Time: 3 Hours (UNSOLVED) PART - I Choose the most suitable answer. Answer all the questions. [12 × 1 = 12] 1. Newton's III law is applicable .............................. . (a) for a body is at rest (b) for a body in motion (c) both (a) & (b) (d) only for bodies with equal masses 2. Amount of light entering into the eye is controlled by ........................... . (a) retina (b) iris (c) pupil (d) medulla oblongata 3. When sound travels from air to water, which parameter does not change? (a) Wavelength (b) Frequency (c) Velocity (d) Temperature 4. One percent mole of any substance contains ...................... molecules. (a) 6.023 × 1023 (b) 6.023 × 10–23 (c) 3.0115 × 1023 (d) 12.046 × 1023 5. Photolysis is a decomposition reaction caused by .......................... . (a) heat (b) electricity (c) light (d) mechanical energy 6. The pH of a solution is 3. Its [OH–] concentration is .......................... . (a) 1 × 10–3 M (b) 3M (c) 1 × 10–11M (d) 11M 7. Casparian strips are present in the .......................... of the root. (a) cortex (b) pith (c) pericycle (d) endodermis 8. ....................... are the largest of the leucocytes. (a) Monocytes (b) Lymphocytes (c) Basophils (d) Neutrophils 9. Root hairs are .......................... . (a) cortical cell (b) p rojection of epidermal cell (c) u nicellular (d) both b and c 10. Pusa Komal is a disease resistant variety of ............................... . (a) sugarcane (b) rice (c) cow pea (d) maize 11. The term Ethnobotany was coined by................ . (a) Khorana (b) J.W. Harshberger (c) Ronald Ross (d) Hugo de Vries 12. The number of chromosomes found in human beings are ................... . (a) 22 pairs of autosomes and 1 pair of allosomes. (b) 2 2 autosomes and 1 allosome (c) 46 autosomes (d) 46 pairs of autosomes and 1 pair of allosomes PART - II [7 × 2 = 14] Answer any seven questions. (Q. No: 22 is compulsory) 13. While catching a cricket ball the fielder lowers his hands backwards. Why? 14. Differentiate concave lens and convex lens. 15. State the law of volume. 16. Correct the mistake: (True or False) ( a) Two elements sometimes can form more than one compound. (b) The gram atomic mass of an element has no unit. D-30

17. Differentiate between combination and decomposition reaction. 18. Write the reaction for photosynthesis. 19. Which hormone requires iodine for its formation? What will happen if intake of iodine in our diet is low? 20. Mention the function of endosperm. 21. Distinguish between Somatic gene therapy and germ line gene therapy. 22. Calculate the number of moles in 2 g of NaOH. PART - III Answer any seven questions (Q. No: 32 is compulsory) [7 × 4 = 28] 23. Deduce the equation of a force using Newton's second law of motion. 24. Explain the structure of the eye and power of accommodation. 25. (i) Define one Calorie. (ii) Write the characteristics features of heat energy transfer. 26. (i) State Avogadro's Hypothesis. (ii) In what way hygroscopic substances differ from deliquescent substances. 27. Explain the classification based on the direction of the reactions. 28. (i) What is respiratory quotient? (ii) Differentiate Aerobic and Anaerobic respiration. 29. (i) What are Okazaki fragments? (ii) What is the important of valves in the heart? 30. What are synthetic auxins? Give examples. 31. What is the importance of rainwater harvesting? 32. (i) For a person with hypermeteropia, the near point has moved to 1.5m. Calculate the focal length of the correction lens in order to make his eyes normal. (ii) Identify the types of chemical reactions? (a) AB → A + B (b) AB + CD → AD + CB PART - IV Answer all the questions. [3 × 7 = 21] 33. (a) (i) Compare the properties of alpha, beta and gamma radiation. (ii) State Soddy and Fajan's displacement law. (OR) (b) (i) 92U235 experiences one α - decay and one β - decay. Find number of neutrons in the final daughter nucleus that is formed. (ii) Draw a ray diagram of formation of images by a convex lens. 34. (a) (i) Calculate the % of each element in Calcium Carbonate. (ii) Calculate the number of moles in 1.51 × 1023 molecules of NH4Cl. (iii) Find the percentage of Nitrogen in ammonia. (OR) (b) (i) Explain solid solution, liquid solution and gaseous solution. (ii) What are the methods of preventing corrosion. 35. (a) Discuss the importance of biotechnology in the field of medicine. (OR) (b) What are the phases of menstrual cycle? Indicate the changes in the ovary and uterus. Sample Paper - 4    D-31

One-mark & Two-mark Solutions for Unsolved Sample Papers Sample Paper-4 1. (c) both (a) & (b) 5. (c) light PART - I 2. (b) iris 9. (d) both b and c 3. (b) Frequency 4. (a) 6.023 × 1023 6. (c) 1 × 10–11M 10. (c) cow pea 7. (d) endodermis 11. (b) J.W. Harshberger 8. (a) Monocytes 12. (a) 22 pairs of autosomes and 1 pair of allosomes PART - II 13. While catching a cricket ball the fielder lowers his hands backwards. So increase the time during which the velocity of the cricket ball decreases to zero. Therefore the impact of force on the palm of the fielders will be reduced. 14. Convex Lens Concave Lens The lens which is thicker at the centre than at The lens which is thinner at the centre than at the edges. the edges. It is a converging lens. It is a diverging lens. It produces mostly real images. It produces virtual images. it is used to treat Hypermetropia. It is used to treat Myopia 15. When the pressure of gas is kept constant, the volume of a gas is directly proportional to the temperature of the gas. (i.e) V α T, V/T = constant. 16. (a) True (b) False. Correct statement: The gram atomic mass of an element is expressed in the unit ‘grams’. 17. Combination reactions Decomposition reactions A combination reaction is a reaction in which In a decomposition reaction, a single compound two or more reactants combine to form a splits into two or more simpler substances under compound. suitable conditions. Light C6H12O6 + 6H2O + 6O2↑ Glucose + Water + Oxygen 18. 6CO2 + 12H2O Chlorophyll Carbon dioxide + Water 19. Thyroxine hormone requires iodine for its formation. Goitre will occur due to the inadequate supply of iodine in our diet. Goitre is the enlargement of the thyroid gland. 20. Endosperm provides nourishment to the developing embryo. D-48

21. Somatic gene therapy Germ line gene therapy Somatic gene therapy is the replacement Germ line gene therapy is the replacement of of defective gene in somatic cells. defective gene in germ cell (egg and sperm). 22. Number of moles = mass 2 = 0.05 Molecular mass = 40 Answers  D-49

SOCIAL SCIENCE QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to be Total Marks Answered Part-I Multiple Choice Questions 1 14 14 Part-II 2 10 20 (Totally 14 questions will be given. Answer any Ten. 5 10 50 Any one question should be answered compulsorily) Part-III (Totally 13 questions will be given. Answer any Ten. Any one question should be answered compulsorily) Part-IV 82 16 Total Marks 100 S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity E-1

1Sample Paper- Time: 3 Hours (SOLVED) Maximum Marks: 100 PART- I [14 × 1 = 14] Answer all the questions. Choose the correct answer. [Answers are in Bold] 1. ...................... was in competition with Germany and United States. (a) Japan (b) England (c) Africa (d) India 2. Who was the Commander-in-Chief responsible for the new military regulations in Vellore Fort? (a) Col. Fancourt (b) Major Armstrong (c) Sir John Cradock (d) Colonel Agnew 3. The NAM held its first conference at .................................. in 1961. (a) Egypt (b) Belgrade (c) Ghana (d) Indonesia 4. Who declared that “Land belongs to God” and collecting rent or tax on it was against divine law? (a) Titu Mir (b) Sidhu (c) Dudu Mian (d) Shariatullah 5. In which session of the Indian National Congress was Non-Cooperation approved? (a) Bombay (b) Madras (c) Lucknow (d) Nagpur 6. The Southern most point of India is .................... . (a) Andaman (b) Kanyakumari (c) Indira point (d) Kavaratti 7. We wear cotton during ..................... . (a) Summer (b) Winter (c) Rainy (d) Northeast monsoon 8. Which of the following organization has divided the Indian soils into 8 major groups? (a) Indian Council of Agricultural Research (b) Indian Meteorological Department (c) Soil Survey of India (d) Indian Institute of Soil Science 9. The longitudinal extent of Tamil Nadu is ................. . (a) 76°18'E to 80°20'E (b) 76°18'W to 80°20'W (c) 86°18'E to 10°20'E (d) 86°18'W to 10°20'W 10. ...................... is the most common trigger of a landslide. (a) Air (b) Fire (c) Water (d) Land 11. If the fundamental rights of Indian citizen are violated, they possess the right to have an access to ........................ . (a) The Parliament (b) The Attorney General (c) T he President of India (d) The Supreme Court of India 12. Non–military issues are ................. . (a) Energy security (b) Water security (c) Pandemics (d) All the above 13. Tamil Nadu Integrated Nutrition Programme was started in ..................... . (a) 1980 (b) 1975 (c) 1955 (d) 1985 14. India is .................. large producer in agricultural product. (a) 1st (b) 3rd (c) 4th (d) 2nd E-3

PART - II [10 x 2 = 20] Answer any 10 questions. Question No. 28 is compulsory. 15. Discuss Mahadev Govind Ranade’s contribution to social reforms. (i) Mahadev Govind Ranade was a great social reformer. He advocated for inter-caste dining, inter-caste marriage, widow remarriage and improvement of women and depressed classes. (ii) He founded the Widow Marriage Association, the Poona Sarvajanik Sabha and Decean Education society. 16. Who were the three prominent dictators of the post World War I? The three prominent dictators of the post-World War I were Mussolini (Italy), Hitler (Germany) and Franco (Spain). 17. Identify the Palayams based on the division of east and west. Among the 72 Palayakkarars, there were two blocks namely the eastern and the western Palayams. (i) The eastern Palayams were – Sattur, Nagalapuram, Ettayapuram and Panchalam Kurichi. (ii) The western Palayams were – Uttrumalai, Thalavankottai, Naduvakurichi, Singampatti and Seithur. 18. What was the conflict between the Swarajists and no-changers? • In due course of time the Congress got divided into two groups—pro-changers (swarajists) and no-changers. Some of the Congressmen led by Motilal Nehru and C. R. Das wanted to contest the elections and enter the legislature. • They argued that the national interest could be promoted by working in the Legislative Councils under Dyarchy and weakening the colonial government from within. These Congressmen were called the pro-changers. • On the other hand, no-changers like Vallabhbhai Patel, C Rajagopalachari and other followers of Gandhi wanted to continue non-cooperation with the government. 19. Write a short note on ‘Monsoon Wind’. • The word ‘monsoon’has been derived from the Arabic word ‘Mausim’which means ‘season’. • These winds appear to blow from southwest for six months and from northeast for another six months. • In India it is used to refer to the winds which reverse their directions in summer and winter. 20. What is Multipurpose project? A comprehensive river valley project which serves a number of purposes simultaneously is called a “Multipurpose project”. 21. Define: Disaster Risk Reduction. The concept and practice of disaster risks through systematic efforts to analyse and manage the causal factors of disasters including thorough reduced exposure to hazards, lessened vulnerability of people and property, wise management of land and environment and improved preparedness for adverse events. 22. What is the importance of the Governor of a state? (i) The Governor is the Constitutional head of the State Executive. The administration of a State is carried on in the name of the Governor. (ii) He directly rules a State when there is the imposition of the President’s rule in the State. He is an integral part of the State legislative. E-4    Social Science – X

23. How do you assess the importance of Chabahar agreement? i. A trilateral agreement called the Chabahar Agreement was signed between India, Afghanistan and Iran, which has led to the establishment of transit and transport corridor among three countries using Chabahar port. ii. This port is seen as golden gateway for India to access landlocked markets of Afghanistan and Central Asia by passing Pakistan. 24. What is globalization? Globalization is the process of integrating various economies of the world without creating any barriers in the free flow of goods and services, technology, capital and even labour or human capital. 25. Why should a developing economy diversify out of agriculture? i. As an economy grows and incomes increase, consumers tend to spend a lesser share of their income on products from the agricultural sector. ii. There are limits to the ability of agriculture to absorb labour due to the declining marginal productivity of land. iii. Due to this, there is a need for an economy’s production and employment base to diversify away from agriculture. 26. During cyclone, how does the Meterological department warn the fishermen? (i) During disturbed weather over the seas, the ports likely to be affected are warned by concerned ACWCs / CWCs by advising the port authorities through port warnings to hoist appropriate Storm Warning Signals. The Department also issue “Fleet forecast” for Indian Navy. (ii) Coastal Bulletins for Indian coastal areas covering up to 75 km from the coast line and sea area bulletins for the sea areas beyond 75 km. The special warnings are issued for fishermen four times a day in normal weather and every three hourly in accordance with the four stage warning in case of disturbed weather. (iii) The general public, the coastal residents and fishermen are warned through state government officials and broadcast of warnings through All India Radio and National Television telecast programmes in national and regional hook up. (iv) A sytem of warning dissemination for fishermen through World Space Digital Based radio receiver is being planned. 27. Mention any three industrial development agencies in Tamil Nadu and their role. SIPCOT (State Industries Promotion Corporation of Tamil Nadu), 1971 : It was formed in the year 1971 to promote industrial growth in the state by setting up industrial estates. TANSIDCO (Tamil Nadu Small Industries Development corporation), 1970 : TANSIDCO is a state-agency of the state of Tamil Nadu established in the year 1970 to promote small-scale industries in the state. It gives subsidies and provide technical assistance for new firms in the small scale sector. TIDCO: (TamilNadu Industrial Development Corporation), 1965 : TIDCO is another government agency to promote industries in the state and to establish industrial estates. 28. Write some name of the nutrition programmes in Tamil Nadu. • Purachi Thalaivar M.G.R. Nutrition Meal Programme • National Programme of Nutritional Support to Primary Education • General ICDS Projects and World Bank Assisted Integrated Child Development Services Sample Paper - 1    E-5

• Pradhan Manthri Gramodaya Yojana Scheme (PMGYS) [10 x 5 = 50] • Tamil Nadu Integrated Nutrition Programme • Mid-Day Meal Programme PART - III Answer any 10 questions. Question No. 41 is compulsory. 29. Fill in the blanks: (i) With the collapse of ............................ , the idea of non-alignment lost relevance. (ii) ................ soil is suitable for the cultivation of tea and coffee plants. (iii) Our tradition and national ethos is to practice ................ . (iv) .............. play an important role in the supply of quality goods at responsible rates to common people. (v) .............. is the third largest airport in India after Mumbai and Delhi. Answers (i) Soviet Union (ii) Laterite (iii) Disarmament (iv) Consumer co-operatives (v) Chennai International Airport 30. Match the following: (i) Thalamuthu - Widows Remarriage Reform Act (ii) Subedar Shik Adam - Bihar (iii) Kunwar Singh - M. N. Roy (iv) Communist Party of India - Vellore Revolt (v) Iswarchandra Vidyasagar - Anti-Hindi agitation Answer (i) Thalamuthu - Anti-Hindi agitation (ii) Subedar Shik Adam - Vellore Revolt (iii) Kunwar Singh - Bihar (iv) Communist Party of India - M. N. Roy (v) Iswarchandra Vidyasagar - Widows Remarriage Reform Act 31. Match the following: (i) Tidel energy - Impact of Urbanization (ii) Urban sprawl - India with Lahore (iii) Major arterial roads - October to December (iv) North east monsoon - 234 (v) Assembly constituencies - Thiruvananthapuram Answer (i) Tidel energy - Thiruvananthapuram (ii) Urban sprawl - Impact of Urbanization (iii) Major asterial roads - India with Lahore (iv) North east monsoon - October to December (v) Assembly constituencies - 234 E-6    Social Science – X