This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
मध्यकालीन साहित्य की चर्चा कीजिए । |
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Answer» उत्तर भारत में मध्ययुग की शुरुआत संस्कृत साहित्य से हुई थी ।
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| 2. |
मध्यकाल की प्रादेशिक भाषाओं की संक्षिप्त में जानकारी दीजिए । |
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Answer» प्रादेशिक भाषाओं को प्रादेशिक राजाओं ने वेग प्रदान किया था ।
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| 3. |
अथर्ववेद में कौन-सी जानकारी दी जाती है ? |
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Answer» अथर्ववेद में अनेक प्रकार के कर्मकांडों और संस्कारों का वर्णन किया जाता है । |
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| 4. |
तक्षशिला विद्यापीठ की जानकारी दीजिए । |
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Answer» वर्तमान समय में पाकिस्तान के रावलपिंडी के पश्चिम में प्राचीन तक्षशिला विद्यापीठ था ।
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| 5. |
यजुर्वेद की जानकारी दीजिए । |
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Answer» यजुर्वेद को यज्ञ का वेद कहते हैं । यह वेद गद्य और पद्य स्वरूप में लिखा गया है । यज्ञ के समय बोले जानेवाले मंत्रों, क्रियाओं और विधि विधानों का इसमें वर्णन किया गया है । |
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| 6. |
नालंदा विद्यापीठ की जानकारी दीजिए । |
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Answer» बिहार के पटना जिले में बडगाँव में प्राचीन नालंदा विद्यापीठ आया हुआ था ।
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| 7. |
How is inbreeding depression got rid off? |
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Answer» By mating the selected animals from one population with unrelated superior animals of the same breed. |
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| 8. |
राजतरंगिणी ग्रन्थ की रचना किसने की थी ?(A) सोमदेव(B) कल्हण(C) कालिदास(D) अबुलफजल |
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Answer» सही विकल्प है (B) कल्हण |
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| 9. |
यु. एन. श्वांग ने नालंदा विद्यापीठ का वर्णन किस प्रकार किया ? |
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Answer» महाविद्यालय में सात बड़े खंड थे, जिनमें व्याख्यान के लिए तीन खंड़ थे । |
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| 10. |
कृष्णदेव राय ………. और ……… भाषा का विद्वान था । |
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Answer» सही उत्तर है संस्कृत, तेलुगु |
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| 11. |
आदिपुराण किसने और किस पर लिखा है ? |
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Answer» आदिपुराण कवि पंपा ने जैन धर्म पर लिखी थी । |
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| 12. |
राजतरंगिणी का फारसी में ……….. ने अनुवाद किया । |
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Answer» सही उत्तर है जैनुअबिदिन |
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| 13. |
वलभी को ….…. से …….. के दरम्यान वहाँ के शासकों ने आश्रय दिया । |
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Answer» ई.स. 480, ई.स. 775 |
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| 14. |
बंगाल के सुलतानों ने ………. को आश्रय देकर बंगाली में साहित्य सर्जन करवाया । |
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Answer» सही उत्तर है कृतिदास |
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| 15. |
किसके आश्रय से वाराणसी का सारनाथमठ प्रसिद्ध विद्याकेन्द्र बना ?(A) अज्ञातशत्रु(B) अशोक(C) गुणसेन(D) भीमसेन |
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Answer» सही विकल्प है (B) अशोक |
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| 16. |
वाराणसी (काशी) के किस राजा ने तत्त्वज्ञान और विद्या को आश्रय दिया ?(A) अज्ञातशत्रु(B) गुणसेन(C) वेदव्यास(D) अशोक |
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Answer» (A) अज्ञातशत्रु |
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| 17. |
What is the perpendicular distance of the point P (6, 7, 8) from xy-plane?(A) 8 (B) 7 (C) 6 (D) None of these |
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Answer» Let L be the foot of perpendicular drawn from the point P (6, 7, 8) to the xy-plane and the distance of this foot L from P is z-coordinate of P, i.e., 8 units. |
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| 18. |
L is the foot of the perpendicular drawn from a point P (6, 7, 8) on the xy-plane. What are the coordinates of point L?(A) (6, 0, 0) (B) (6, 7, 0) (C) (6, 0, 8)(D) none of these |
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Answer» Since L is the foot of perpendicular from P on the xy-plane, z-coordinate is zero in the xy-plane. Hence, coordinates of L are (6, 7, 0). |
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| 19. |
L is the foot of the perpendicular drawn from a point (6, 7, 8) on x-axis. The coordinates of L are(A) (6, 0, 0) (B) (0, 7, 0)(C) (0, 0, 8) (D) none of these |
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Answer» Since L is the foot of perpendicular from P on the x- axis, y and z-coordinates are zero. Hence, the coordinates of L are (6, 0, 0). |
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| 20. |
If the point P lies on z-axis, then coordinates of P are of the form ________. |
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Answer» On the z-axis, x = 0 and y = 0. So, the required coordinates are of the form (0, 0, z). |
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| 21. |
The equation of z-axis, are ________. |
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Answer» Any point on the z-axis is taken as (0, 0, z). So, for any point on z-axis, we have x = 0 and y = 0, which together represents its equation |
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| 22. |
A line is parallel to x-axis if all the points on the line have equal ________. |
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Answer» A line is parallel to x-axis if each point on it maintains constant distance from y-axis and z-axis. So, each point has equal y and z-coordinates. . |
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| 23. |
What is the locus of a point for which y = 0, z = 0?(A) equation of x-axis (B) equation of y-axis(C) equation of z-axis (D) none of these |
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Answer» Locus of the point y = 0, z = 0 is x-axis, since on x-axis both y = 0 and z = 0. |
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| 24. |
A line is parallel to xy-plane if all the points on the line have equal ________. |
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Answer» A line is parallel to xy-plane if each point P(x, y, z) on it is at same distance from xy-plane. Distance of point P from xy plane is ‘z’ So, line is parallel to xy-plane if all the points on the line have equal z-coordinate |
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| 25. |
L, is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xz-plane. What are the coordinates of point L ?(A) (3, 0, 0) (B) (0, 4, 5) (C) (3, 0, 5)(D) (3, 4, 0) |
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Answer» Since L is the foot of perpendicular segment drawn from the point P (3, 4, 5) on the xz-plane. Since the y-coordinates of all points in the xz-plane are zero, coordinate of the foot of perpendicular are (3, 0, 5). |
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| 26. |
A line is parallel to xy-plane if all the points on the line have equal _____. |
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Answer» A line parallel to xy-plane if all the points on the line have equal z-coordinates |
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| 27. |
The equation x = b represents a plane parallel to _____ plane. |
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Answer» Since x = 0 represent yz-plane, therefore x = b represent a plane parallel to yz -plane at a unit distance b from the origin. |
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| 28. |
Perpendicular distance of the point P (3, 5, 6) from y-axis is ________ |
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Answer» Since M is the foot of perpendicular from P on the y-axis, therefore, its x and z-coordinates are zero. The coordinates of M is (0, 5, 0). Therefore, the perpendicular distance of the point P from y-axis √(32 + 62) = 45 . |
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| 29. |
L is the foot of perpendicular drawn from the point P (3, 4, 5) on zxplanes. The coordinates of L are ________. |
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Answer» Since L is the foot of perpendicular from P on the zx-plane, y-coordinate of every point is zero in the zx-plane. Hence, coordinate of L are (3, 0, 5). |
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| 30. |
The length of the foot of perpendicular drawn from the point P (a, b, c) on z-axis is _____. |
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Answer» The coordinates of the foot of perpendicular from the point P (a, b, c) on z-axis is (0, 0,c). The distance between the point P (a, b, c) and (0, 0, c) is √(a2 +b2) . |
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| 31. |
In a three dimensional space the equation x2 – 5x + 6 = 0 represents A. points B. planes C. curves D. pair of straight lines |
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Answer» Given: x2 – 5x + 6 = 0 x2 – 5x + 6 = 0 ⇒ x2 – 3x – 2x + 6 = 0 ⇒ x(x – 3) – 2(x – 3) = 0 ⇒ (x – 3)(x – 2) = 0 ⇒ (x – 3) = 0 or (x – 2) = 0 ⇒ x = 3 or x = 2 Both the results represents planes which are parallel to yz-plane Hence, x2 – 5x + 6 = 0 represents planes |
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| 32. |
Find the value of p for which the line through the points A (4, 1, 2) and B (5, p, 0) is perpendicular to the line through the points C (2, 1, 1) and D (3, 3, -1). |
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Answer» It is given that Line joining A (4, 1, 2) and B (5, p, 0) is written as AB = i + (p – 1)j – 2k Line joining C (2, 1, 1) and D (3, 3, -1) is written as CD = i + 2j – 2k In order prove that these two lines are perpendicular we must show that angle between these two lines is π/2 Dot product AB . CD = 0 Substituting the values (i + (p – 1)j – 2k). (i + 2j – 2k) = 0 So we get 1 + 2 (p – 1) + 4 = 0 p = -3/2 Hence, value of p is -3/2. |
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| 33. |
Show that the points A (2, 3, 4), B (-1, -2, 1) and C (5, 8, 7) are collinear. |
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Answer» It is given that A (2, 3, 4), B (-1, -2, 1) and C (5, 8, 7) AB = – 3i – 5j – 3k So the point on the line AB with A on the line R = (2, 3, 4) + a (- 3, -5, -3) Consider C = (2 – 3a, 3 – 5a, 4 – 3a) So we get (5, 8, 7) = (2 – 3a, 3 – 5a, 4 – 3a) Here if a = -1 we get LHS = RHS So the point C lies on the line joining AB Therefore, the three points are collinear. |
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| 34. |
Check whether the statements is True or False. The y-axis and z-axis, together determine a plane known as yz-plane. |
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Answer» Answer is True |
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| 35. |
Check whether the statements is True or False. The point (4, 5, – 6) lies in the VIth octant |
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Answer» Answer is False the point (4, 5, – 6) lies in the Vth octant, |
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| 36. |
Prove that a straight line and parabola cannot intersect at more than two points. |
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Answer» Let the standard equation of parabola y2 = 4ax …..(1) Equation of line be y = mx + c …(2) Solving (1) & (2) (mx + c)2 = 4ax ⇒ mx2 + 2mcx + c2 – 4ax = 0 ⇒ mx2 + 2x(mc – 2a) + c2 = 0 This equation can not have more than two solutions and hence a line and parabola cannot intersect at more than two points. |
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| 37. |
Find a polynomial equation of minimum degree with rational coefficients, having 2 + √3i as a root. |
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Answer» Given roots is (2 + √3i) The other root is (2 – √3i), since the imaginary roots with real co-efficient occur as conjugate pairs. x2 – x(S.O.R) + P.O.R = 0 ⇒ x2 – x(4) + (4 + 3) = 0 ⇒ x2 – 4x + 7 = 0 |
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| 38. |
Find a polynomial equation of minimum degree with rational coefficients, having √5 – √3 as a root. |
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Answer» The given one roots of the polynomial equation are (√5 – √3) The other roots are (√5 + √3), (-√5 + √3) and (- √5 – √3). The quadratic factor with roots (√5 – √3) and (√5 + √3) is = x2 – x(S.O.R) + P.O.R = x2 – x(2√5) + (√5 – √3) (√5 + √3) = x2 – 2√5 x + 2 The other quadratic factors with roots (-√5 + √3) (-√5 – √3) is = x2 – x (S.O.R) + P.O.R = x2 – x (-2√5 ) + (5 – 3) = x2 + 2√5x + 2 To rationalize the co-efficients with minimum degree (x2 – 2√5 x + 2) (x2 + 2√5 x + 2) = 0 ⇒ (x2 + 2)2 – (2√5 x)2 = 0 ⇒ x4 + 4 + 4x2 – 20x2 = 0 ⇒ x4 – 16x2 + 4 = 0 |
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| 39. |
Find a polynomial equation of minimum degree with rational co-efficients having 1 – i as a root. |
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Answer» Given root is 1 – i The other root is 1 + i Sum of the roots: 1 – i + 1 + i = 2 product of the roots: (1 – i) (1 + i) = (1)2 + (1)2 ⇒ 1 + 1 = 2 ∴ The required polynomial equation of minimum degree with rational coefficients is x2 – x (S.R.) + (P.R.) = 0 x2 – 2x + 2 = 0 |
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| 40. |
A well of inner diameter 14 m is dug to a depth of 12 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of the embankment so formed. |
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Answer» Given that, Inner diameter = 14 cm Therefore, Radius = 7 cm Also, Depth = 12 m Therefore, Volume of earth dug out = πr2h = \(\frac{22}{7}\times7\times7\times12\) = 1848 m3 It is also given that, Width of embankment = 7 m Therefore, Total radius = 7 + 7 = 14 m Volume of embankment = Total volume – Inner volume = πro2h – πr12h = πh (ro2 – r12) = \(\frac{22}{7}\)h (142 - 72) = \(\frac{22}{7}\)h x 147 = 21 × 22h = 462 × h m3 Since, Volume of embankment = Volume of earth dug out Therefore, 1848 = 462 h h = \(\frac{1848}{462}\) h = 4 m Therefore, Height of the embankment = 4 m |
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| 41. |
A cube is made by arranging 64 cubes having side of 1 cm, find total surface area of cube so formed. |
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Answer» Side of one cube = 1 cm. ∴Volume of one cube = a3 = (1)3 = 1 cm3 ∴ Volume of 64 cubes = 64 x 1 = 64 cm3 Let the side of new cube = x cm. ∴ Volume of new cube = Volume of 64 cubes ∴ x3 = 64 => x = (64)1/3 = 4 cm. On comparing x = 4 ∴ Total surface area of new cube = 6a2 = 6 x (4)2 = 6 x 16 = 96 cm.2 |
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| 42. |
Fill in the blanks:1. Space occupied by any solid is called its___2. The quantity of liquid in a three dimensional pot is called___3. Area of four walls of a room = 2 x (l + b) x ___4. Surface area of a solid is equal to___of all areas of its faces.5. 1 m3 =___litre. |
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Answer» 1. Volume 2. Capacity 3. h 4. Sum 5. 1000 |
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| 43. |
The area of three adjacent faces of a cuboid are p, q and r. The volume of cuboid is V. Prove that V2 = pqr. |
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Answer» Let l, b and h be the dimensions of cuboid respectively. Then, p = lb q = bh r = hl ∴ pqr = (lb) (bh) (hl) = l2b2h2 = (lbh)2 = V2 ⇒V2 = pqr |
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| 44. |
Determine side of a cube whose total surface area is 1014 cm2. Find its volume also. |
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Answer» Let the side of cube be a cm. Then the total surface area of cube = 6a2 .sq. m According to question 6a2 = 1014 a2 = 1014/6 = 169 a = √169 = 13 cm Hence the side of cube = 13 cm Volume of cube = (side)3 Hence 13 x 13 x 13 = 2197 cm3 |
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| 45. |
Determine side of a cube whose total surface area is 600 square cm. |
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Answer» Let a be the length of side of cube. Then, total surface area of cube = 6a2 cm2 According to question, 6 a2 = 600 a2 = 600/6 = 100 a = √100 = 10 cm Hence, required length of side of cube is 10 cm. |
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| 46. |
State whether True or False:1. 1 cm3 = 1 mL2. 1 L = 1000 cm33. The 2/3 part of volume of a tank of dimension 6m x 5m x 4m is 80 m3.4. The volume of a cylindrical poll, whose height is 7 m and diameter is 12 cm, is π (6)2 x 7 cm3. |
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Answer» 1. True 2. True 3. True 4. False. |
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| 47. |
Dimensions of a cuboidal ice is 50 cm x 30 cm x 20 cm. Find its weight in kilogram. If weight of 1000 cm3 ice is 900 gram. |
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Answer» Volume of ice plate = 50 x 30 x 20 = 30000 cm3 ∴Weight of ice plate = 30000 x 600/1000 = 27000 gram = 27000/1000 = 27 kg. |
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| 48. |
To make a road plain a roller has to complete 750 rounds. If the diameter of roller is 84 cm and length 1 meter then find the area of road. |
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Answer» For Rollar Diameter = 84 cm. Radius (r) = 84/2 cm. = 42 cm. Length (h) = 1 meter = 100 cm. ∴ Curved surface area = 2πrh = 2 . 22/7 . 42 . 100 = 26400 cm2 ∴ Area of road made plane in one round = 26400 cm2 ∴ Area of road = 26400 x 750 = 19800000 cm2 = 19800000/10000 m2 = 1980 m2 |
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| 49. |
From a cylindrical milk tanker of radius 1.5 m and length 7 m how many polythene of one liter can be packed? (1 m3 = 1000 liter). |
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Answer» For tanker Radius (r) = 1.5 meter Height (h) = 7 meter ∴ Volume = πr2h = 22/7 . (1.5) (1.5) x 7 = 49.5 m3 = 49.5 x 1000 litre = 49500 litre Hence, the required number of polythene packs are 49500. |
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| 50. |
A cylindrical pipe, which is open from both sides, is made of iron sheet of thickness 2 cm. If external diameter is 16 cm and height is 100 cm, then find the total iron used in making the pipe? |
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Answer» External diameter = 16 cm. External radius (R) = 16/2 cm. = 8 cm. Thickness of sheet = 2 cm. ∴ Inner radius (r) = 8 – 2 = 6 cm. Height (h) = 100 cm. ∴ Used iron = External Volume – Internal Volume = πR2h – πr2h = π (R2 – r2) h = 22/7 {(8)2 – (6)2} 100 = 22/7 x 28 x 100 = 8800 cm3 |
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