The Man I Killed Essay Example for Free

The Man I Killed Essay â€Å"The Man I Killed† the author Tim O Brien is the character in the story but the story does not use first person. This is because the story is not revolved around him but revolved on the man he has just killed in the Vietnam war. The character in this story focuses on the dead mans physicality and the story he has fabricated for him. The character in this story seems to be in shock because he does not speak or stop looking at the dead soldier, â€Å"Kiowa shook his head. There was some silence before he said, â€Å"Stop staring Talk to me† (p. 797-798) In the story â€Å"The Lives of the Dead†, the narrator is Tim O Brien who tells the story of his first love who happens to be his first death. This story is in first person and he describes how he keeps those he has known alive by telling stories about them, â€Å"When I write about her now, three decades later, its tempting to dismiss it as a crush, an infatuation of childhood, but I know for a fact that what we felt for each other was as deep and rich as love can ever get. † Tim also describes his experiences in the war, â€Å"I remember the smell of burnt straw; I remember broken fences and torn-up trees and heaps of stone and brick and pottery. † (p. 799) The repetition of dialogue in â€Å"The Man I Killed† gives the readers a sense of truth in the story and imagery of what the dead man looked like for the character. As the character repeats over and over again what the man looked like and what his life was like before the war, it gives the readers a sense of the psychological affects the war had on men. Using these techniques of imager and dialogue repetition in this story allowed me to get a deeper feeling of what shock and guilt felt like for these men. Although the reader may not have intended for me to feel these emotions this is what I felt during the reading. It allowed me to understand how the character consoled himself and also punished himself. For me if I imagined the youngs mans life before the war and I took that away, I would consider this as a form of punishment. Imagery is very detailed in both stories because one of them describes the man he has just killed, â€Å"His jaw was in his throat, his upper lip and teeth were gone, his one eye was shut, his other eye was a star-shaped hole † (p. 795) and the soldiers fabricated life story, â€Å"He had been born, maybe, in 1946 in the village of My Khe near the central coastline of Quan Ngai Province, where his parents farmed † (p. 795) In the other story he describes his first live, her death  at age 9, and his experiences in the war especially with death, â€Å"The place was deserted-no people, no animals- and the only confirmed kill was an old man who lay face up-near a pigpen at the center of the village. † (p. 799) The imagery of the people he has lost and his experiences with them allow the readers to get a feeling of how many people the narrator has lost or has seen died. We are allowed to get a feeling about death and the certain ways those around him coped with it and how he did too. In the story â€Å"The Man I Killed† the character copes with his first kill in war by providing him with a story. He gives the young man a story which in turn gives him symbolically life. His life becomes to have meaning and he is no longer thought as a dead man but a person who had dreams, a wife, and longing to be far away from war, â€Å"He had no stomach for violence. He loved mathematics And as he waited, in his final year at the university, he fell in love with a classmate, a girl of seventeen The use of Linda in the war story, â€Å"The Lives of the Dead† is to show his readers how he coped with his fisrt death who happened to be his first love. Linda, a girl at the age of 9 died from a brain tumor which devastates Tim O Brien who than begins to dream about her at night which provides him comfort knowing that in his dreams she still lived. This is how Brien coped with the deaths he experienced in war, â€Å"But ths too is true: stories can save us. Theyre all dead. But in a story, which is kind of dreaming, the dead sometimes smile and sit uo and return to the world. † (p. 799) The author story tells, metafiction, to keep those that have lost their lives from truly dying. Through the form of story he keeps them living and although these are stories of fiction, to the author it is preserving their lives. In the story â€Å"The Man I killed† he preserves the fallen soldier by inventing a fictitious life for him. For Linda and his fellow comrades he gives them stories too. â€Å"But in a story I can steal her soul. I can revive, at least briefly, that which is absolute and unchanging. Its not the surface that matters, its the identity that lives inside. † Just as he imagines them and dreams of them, his stories become their new lives where they continue to live on. He knows if he continues to create these stories, they will never die and through this we can see the power of storytelling. â€Å"and sometimes I can even see Timmy skating with Linda under the yellow floodlights. Im young and happy. Ill never die. †

West Indies :: essays research papers

  Ã‚  Ã‚  Ã‚  Ã‚  Countless of years ago a great mountain range stretched north from what is now the topmost coast of South America, the range was in a constant state of upheaval, lashed by continuous rains, swept by storms, with fire spouting from every peak finally the mountains dropped beneath the sea, quieted most of the volcanoes. The exposed peaks were covered with verdure of fantastic beauty, and left these peaks above the sea to form the chain of West Indian islands as we know today. Although they were islands surrounded by the Caribbean Sea, and are nowhere near Asian India, they were still considered the West Indian islands. Then why the concept of West Indian, we ask? Christopher Columbus, who discovered these islands, can surely explain why he gave such a name to islands that were never Indian descent.   Ã‚  Ã‚  Ã‚  Ã‚  Discovered in 1492 by Christopher Columbus the West Indies were given this name through his mistaken belief that he had reached the Indies, and he himself wrote of them as Las Yndias Ocidentales, referred to as the accidental Indies. After the mistake was realized they were later called them West Indies to distinguish them from the East Indies and at the time in the sixteenth century they were known as the Little Indies, while the East Indies were called the Great Indies. The native inhabitants of the West Indies and America were called Indians as a result of the same error. To distinguish them from the inhabitants of India they were to be called Amerindians or Red Indians. The islands are divided into three major groups: the Bahamas, the Greater Antilles, and the Lesser Antilles. The Greater Antilles consist of Cuba, Hispaniola, Puerto Rico, and Jamaica, and all the rest, except the Bahamas, are included in the group of Lesser Antilles, and were also called the C aribee Islands.   Ã‚  Ã‚  Ã‚  Ã‚  The name West Indies is often loosely applied to the mainland territories of South and Central America (the Spanish Main) and in the past was even applied to those in North America. The name America has been used as including the West Indies. The British use of Windward Islands and Leeward Islands has brought up confusion. The Spaniards correctly called all the eastern islands of the West Indian chain the Windward Islands, Islas de Balovento, and the small islands close to the northern shores of South America the Leeward Islands, Islas de Sotavento.

Surface Areas and Volumes

Question Bank In Mathematics Class X (Term–II) 13 SURFACE AREAS AND VOLUMES A. SUMMATIVE ASSESSMENT (c) Length of diagonal = TH G (a) Lateral surface area = 4l2 (b) Total surface area = 6l2 (c) Length of diagonal = 3 l 3. Cylinder : For a cylinder of radius r and height h, we have : (a) Area of curved surface = 2? rh BR O ER 2. Cube : For a cube of edge l, we have : O YA L TEXTBOOK’S EXERCISE 13. 1 Unless stated otherwise, take ? = 22 . 7 Q. 1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. [2011 (T-II)] 1 S l 2 ? b2 ? h2 5. Sphere : For a sphere of radius r, we have : Surface area = 4? 2 6. Hemisphere (solid) : For a hemisphere of radius r we have : (a) Curved surface area = 2? r2 (b) Total surface area = 3? r2 PR (a) Lateral surface area = 2h(l + b) (b) Total surface area = 2(lb + bh + lh) (d) Total surface area of hollow cylinder = 2? h(R + r) + 2? (R2 – r2) 4. Cone : For a cone of height h, radius r and sla nt height l, we have : (a) Curved surface area = ? rl = ? r h 2 ? r 2 (b) Total surface area = ? r2 + ? rl = ? r (r + l) Sol. Let the side of cube = y cm Volume of cube = 64 cm3 Then, volume of cube = side3 = y3 As per condition ? y3 = 64 ? y3 = 4 3 AK AS HA 13. SURFACE AREA OF A COMBINATION OF SOLIDS 1. Cuboid : For a cuboid of dimensions l, b and h, we have : (b) Total surface area = 2? r2 + 2? rh = 2? r(r + h) (c) Curved surface area of hollow cylinder = 2? h(R – r), where R and r are outer and inner radii N Q. 3. A toy is in the form of a cone of radius 3. 5 cm mounted on a hemisphere of same radius. The total height of the toy is 15. 5 cm. Find the total surface area of the toy. [2011 (T-II)] Sol. Radius of the cone = Radius of hemisphere = 3. 5 cm Total height of the toy = 15. 5 cm ? Height of the cone = (15. 5 – 3. 5) cm = 12 cm Slant height of the cone (l ) = G O Diameter of the hollow cylinder = 14 cm 14 Radius of the hollow hemisphere = cm = 7 cm 2 ? Radius o f the base of the hollow cylinder = 7 cm Total height of the vessel = 13 cm ? Height of the hollow cylinder = (13 – 7) cm = 6 cm Inner surface area of the vessel = Inner surface area of the hemisphere + Inner surface area of the hollow cylinder = 2? (7)2 cm2 + 2? (7) (6) cm2 = 98 cm2 + 84 cm2 = (98 + 84) cm2 22 = 182? cm2 = 182 ? cm2 = 26 ? 22 cm2 7 = 572 cm2. PR AK = Q. 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm.Find the inner surface area of the vessel. [2011 (T-II)] Sol. ? Diameter of the hollow hemisphere = (3. 5)2 ? (12) 2 cm = 156. 25 cm = 12. 5 cm Total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone = 2? (3. 5)2 cm2 + (3. 5) (12. 5) cm2 = 24. 5? cm2 + 43. 65 cm2 = 68. 25? cm2 = O TH ER YA L BR S Q. 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemi sphere can have? Find the surface area of the solid. [2011 (T-II)] Sol. Side of cubical block = 7 cm Side of cubical block = Diameter of hemisphere = 7 cm ? R = 7 7 ? R = cm 2 Surface area of solid = Surface area of the cube – Area of base of hemisphere + C. S. A. of hemisphere 2 – ? R2 + 2? R2 = 6 ? side = 6 (7)2 cm2 + ? R2 22 7 7 2 = 6 ? 7 ? 7 cm2 + ? ? cm 7 2 2 7? ? = ? 6 ? 49 ? 11? ? cm2 2? ? 77 ? ? ? 588 ? 77 ? 2 = ? 294 ? ? cm2 = ? ? cm . 2? 2 ? ? ? r 2 ? h2 2 AS 665 cm2 = 332. 50 cm2 2 HA 68. 25 ? 22 cm2 = 214. 5 cm2. 7 ? y = 4 cm Hence, side of cube is 4 cm. For the resulting cuboid length (l ) = 4 + 4 = 8 cm breadth (b) = 4 cm height (h) = 4 cm ? Surface area of the resulting cuboid = 2(lb + bh + hl ) = 2(8 ? 4 + 4 ? 4 + 4 ? 8) cm2 = 2(32 + 6 + 32) cm2 = 2(80) cm2 = 160 cm2. N Q. 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface are a of the remaining solid. Sol. Diameter of the hemisphere = l = Side of the cube = 45? mm2 + 25? mm2 = (45 + 25) mm2 = 70? mm2 22 = 70 ? mm2 = 220 mm2. 7 Hence, surface area of capsule = 220 mm2 O Q. 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure below). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm.Find its surface area. TH ER YA L BR Sol. Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm 5 Radius of the hemisphere = r = mm 2 Height of the cylinder = [14 – (2. 5 + 2. 5)] mm = 9 mm Surface area of the capsule = Surface area of cylinder + 2 Surface area of hemisphere G O S = l2 ? ? ? 24 ? . 4 = 2? (2) (2. 1) m2 + (2) (2. 8) m2 = (8. 4? + 5. 6) m2 22 2 = 14? m2 = 14 ? m = 44 m2 7 ? Cost the canvas of the tent at the rate of Rs 500 per m2 = Rs 44 ? 500 = Rs 22000 Hence, cost of the canvas is Rs 22000. Q. 8. From a solid cylinder whose height is 2. 4 cm a nd diameter 1. cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. Sol. Height of cylinder = 2. 4 cm Height of cone = 2. 4 cm Radius of cylinder = r = Radius of cone = 0. 7 cm Slant height, of the cone l= 3 PR ?l? ?l 2 2 ? 6l 2 = ? ? ? ? 6l = 4 ? 2? 2 Radius of the cylinder = 2 m Total surface area of the tent = Curved surface area of the cylinder + Curved surface area of the cone AK ?l? ?l? 2 = 2? ? ? ? 6l ? ? ? ? ? 2? ?2? 2 2 AS l 2 Surface area of the remaining solid = Surface area of hemisphere + Surface area of cube – Area of base of hemisphere ?Radius of the hemisphere = Q. 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2. 1 m and 4 m respectively and the slant height of the top is 2. 8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2 (note that the base of the tent will not be covered with canvas. ) Sol. Radius of the cone = 2 m ? ? 5? 2 ? ? 5? 2 2 = 2? ? ? (9) mm + 2 ? 2? ? ? ? mm ? 2? ? 2? ? ? ? ? (0. 7)2 ? (2. 4) 2 cm = 2. 5 cm HA 1. 4 cm = 0. 7 cm 2 N Q. 9.A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3. 5 cm, find the total surface area of the article. Sol. Height of cylinder = 10 cm Total surface area of the remaining solid = C. S. A. of cylinder + C. S. A. of cone + Area of base = 2? rh + ? rl + ? r2 = ? r (2 h + l + r) 22 = ? 0. 7 ? (2 ? 2. 4 + 2. 5 + 0. 7) cm2 7 22 7 = ? (4. 8 + 3. 2) cm2 7 10 22 7 = ? ? 8. 0 cm2 7 10 176 = cm2 = 17. 6 cm2 10 Hence, total remaining surface area = 17. 6 cm2 = 18 cm2. Radius of cylinder = 3. cm Total surface area of the article = C. S. A of cylinder + 2 C. S. A. of hemisphere = 2? (3. 5 (10) cm2 + 2 [2? (3. 5)2] cm2 = 70 cm2 + 49 ? cm2 = (70 + 49) cm2 22 2 = 119? cm2 = 119 ? cm 7 = 17 ? 22 cm2 = 374 cm2. OTHER IMPORTANT QUESTIONS Q. 1. A cylindrical pencil sharpened at one edge is the combination of : (a) a cone and a cylinder (b) frustum of a cone and a cylinder (c) a hemisphere and a cylinder (d) two cylinders Sol. (a) The given shape is a combination of a BR O TH ER S PR AK Its surface area = 6 ? YA L AS Increase in surface area = ? Per cent increase = cone and a cylinder. G O Q. . If each edge of a cube is increased by 50%, the percentage increase in the surface area is : (a) 25% (b) 50% (c) 75% (d) 125% Sol. (d) Let the edge of the cube be a. Then, its surface area = 6a2 150a 3a New edge = = . 100 2 4 Q. 3. The total surface area of a hemisphere of radius 7 cm is : [2011(T-II)] (a) 447 ? cm2 (b) 239 ? cm2 (c) 147 ? cm2 (c) 174 ? cm2 Sol. (c) Total surface area of the hemisphere = 3? r2 = 3 ? ? 49 cm2 = 147? cm2 Q. 4. If two solid hemispheres of same base radius r are joined together along their bases, HA 9a 2 27a 2 = 4 2 27a 2 15a 2 – 6a2 = 2 2 15a 2 100 ? 2 = 125% 6a 2 N hen curved surface area of this new solid is : (a) 4? r2 (b) 6? r2 2 (c) 3? r (d) 8? r2 Sol. (a) The resulting solid will be a sphere of radius r. ? Its curved surface area = 4? r2. Q. 9. The total surface area of a top (lattu) as shown in the figure is the sum of total surface area of hemisphere and the total surface area of cone. Is it true? Sol. No, the statement is false. Total surface area of the top (lattu) is the sum of the curved surface area of the hemisphere and the curved surface area of the cone. Sol. (d) We have ? 2 6a1 2 6a2 a13 a2 3 = AS 4 64 a1 ? = 3 a2 27 HA Q. 5. Volumes of two cubes are in the ratio 64 : 27.The ratio of their surface areas is : (a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9 Q. 10. Two cones with the same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed. N 32 Sol. True. Since the curved surface area taken t ogether is same as the sum of curved surface areas measured separately. G O ?r r 2 ? h2 ? 2? rh . Is it true? YA L Q. 7. If a solid cone of base radius r and height h is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the shape is BR . . . Radius of the hemispherical toy, r = 3. cm Curved surface area of the toy = 2? r2 22 =2? ? (3. 5)2 cm2 = 77 cm2 7 Total surface area of the toy = 3? r2 22 =3? ? (3. 5)2 cm2 = 115. 50 cm2. 7 O TH ER Q. 8. Two identical solid cubes of side a are joined end to end. Then find the total surface area of the resulting cuboid. Sol. The resulting solid is a cuboid of dimensions 2a ? a ? a. ? Total surface area of the cuboid = 2 (lb + bh + hl) = 2 (2a ? a + a ? a + a ? 2a) = 10a2. 5 S Q. 6. The diameter of a solid hemispherical toy is 7 cm. Find its curved surface area and total surface area. Sol. Diameter of the hemispherical toy = 7 cm. Q. 11. A tent of height 8. 5 m is in the form of a right circular cylinder with diameter of base 30 m and height 5. 5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs 45 per m2. Sol. PR 22 ? 8 ? 17 cm2 7 = 854. 85 cm2 = 855 cm2 (approx. ) = 2 (? rl) = 2 ? Height of the tent = 8. 25 m. Height of the cylindrical part = 5. 5 m . . . Height of the conical part = (8. 25 – 5. 5) m = 2. 75 m. 30 Base radius of the tent = m = 15 m. 2 . . . Slant height of the conical part (15)2 + (2. 75)2 m = = 15. 25 m. = AK = 42 16 = = 16 : 9 9 Sol. Slant height of each cone = 82 ? 152 cm 64 ? 225 cm = 17 cm. ? Surface area of the resulting shape 225 + 7. 5625 m Curved surface area of the tent = curved surface area of the cylindrical part + curved surface area of the conical part = 2? rh + ? rl = ? r (2h + l) 22 = ? 15 (2 ? 5. 5 + 15. 25) m2 7 ? 22 ? = ? ? 15 ? 26. 25? m 2 ? 7 ? = 1237. 50 m2. Rate of the canvas = Rs 45 per m2 . . . Cost of the canvas = Rs (1237. 50 ? 45) = Rs 55687. 50. Sol. Slant height of the cone = = = AS and height of the cone = 14 cm BR = 22 ? 7 ( 7 5 + 7) cm2 7 O TH = 7 ? 14 cm = 245 cm = 7 5 cm. Total surface area of the cone = ? rl + ? r2 = ? r (l + r) 2 2 ER Slant height of the cone = r 2 ? h2 = 154 ( 5 + 1) cm2 Surface area of the cube = 6 ? 142 cm2 = 1176 cm2 ? Surface area of the remaining solid left out after the cone is carved out = surface area of the cube – area of base of the cone + curved surface area of the cone 22 2 ? ? = ? 1176 ? ? 7 ? 154 5 ? cm2 7 ? ? YA L = ? 1022 ? 154 5 ? cm2. ? ? Q. 13. A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. [2007, 2011 (T-II)] 6 G O S Q. 14. A solid is in the form of a right circular cylinder with hemispherical ends.The total height of the solid is 58 cm and the diameter of the cylinder is 28 cm. Find the total surface area of the solid. [2006] Sol. Q. 15. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the PR Q. 12. A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out. Sol. Diameter of the cone = 14 cm = 625 cm = 25 cm ? Total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone = 2? r2 + ? rl = ? r (2r + l) =Radius of the each hemisphere = base radius of the cylinder = 14 cm Total height of the toy = 58 cm ? Height of the cylinder = [58 – (14 + 14)] cm = 30 cm ? Total surface area of the solid = 2? r2 + 2? rh + 2? r2 = 2? r (2r + h) 22 =2? ? 14 (2 ? 14 + 30) cm2 7 = 88 ? 58 cm2 = 5104 cm2. AK 22 ? 7 (14 + 25) cm = 858 cm2. 7 HA N Height of the toy = 31 cm Base radius of the cone = radius of the hemisphere = 7 cm ? Height of the cone = (31 – 7) cm = 24 cm r 2 ? h2 72 ? 242 cm 49 ? 576 cm cylindrical part a re 5 cm and 13 cm respectively. The radii of the hemisphercial and conical parts are the same as that of the cylindrical part.Find the surface area of the toy if the total height of the toy is 30 cm. [2002] Sol. = 2? r2 + 2? rh + ? rl = ? r (2r + 2h + l ) = = 22 ? 5 (2 ? 5 + 2 ? 13 + 13) cm2 7 22 ? 5 ? 49 cm2 = 770 cm2. 7 TH PRACTICE EXERCISE 13. 1 A Choose the correct option (Q 1 – 7) : 1. A funnel is the combination of : (a) a cone and a cylinder (b) frustrum of a cone and a cylinder (c) a hemisphere and a cylinder (d) a hemisphere and a cone. 2. A plumbline (shahul) is the combination of : (a) a cone and a cylinder (b) a hemisphere and a cone (c) frustrum of a cone and a cylinder (d) a sphere and a cylinderO ER = 144 ? 25 cm = 13 cm. Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cylinder + curved surface area of the cone BR 3. A shuttle cock used for playing badminton has the shape of the combination of : [2011 (T-II)] ( a) a cylinder and a cone (b) a cylinder and a hemisphere (c) a sphere and a cone (d) frustrum of a cone and a hemisphere 4. The height of a conical tent is 14 m and its floor area is 346. 5 m2. The length of 1. 1 m wide 7 G O YA L S canvas required to built the tent is : (a) 490 m (b) 525 m (c) 665 m (d) 860 m 5.The ratio of the total surface area to the lateral surface area of a cylinder with base diameter 160 cm and height 20 cm is : (a) 1 : 2 (b) 2 : 1 (c) 3 : 1 (d) 5 : 1 6. The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is : (a) 30? cm2 (b) 65? cm2 2 (c) 80? cm (d) none of these 7. A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter : (a) r cm (b) 2r cm (c) h cm (d) 2h cm 8. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6? r2. Is it true? PR Slant height of the cone = 122 ? 52 cm = 22 ? 612. 75 cm2 = 1925. 78 cm2. 7 ? Required cost of painting = Rs 5. 25 ? 1925. 78 = Rs 1010. 38. AK Radius of the cone = Radius of the cylinder = radius of the hemisphere = 5 cm. Total height of the toy = 30 cm Height of the cylinder h = 13 cm ? Height of the cone = [30 – (13 + 5)] cm = 12 cm. Internal radius (r) of the vessel = 12 cm Total surface area of the vessel = 2? R2 + 2? r2 + ? (R2 – r2) = [2 ? (12. 5)2 + 2 ? 122 + (12. 52 – 122)] cm 2 = [312. 5 + 288 + 12. 25] cm 2 AS HA Q. 16. The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively.If the cost of painting 1 cm2 of the surface area is Rs 5. 25, find the total cost of painting the vessel all over. [2001] Sol. External radius (R) of the vessel = 12. 5 cm N ER 16. A rocket is in the form of a cone of height 28 cm, surmounted over a right circular cylinder of height 112 cm. The radius of the bases of cone and cylinder are equal, each being 21 cm. Find the total surface area of the rocket. ? ? = ? ? ? ? 7? 22 G 13. 2 VOLUME OF A COMBINATION OF SOLIDS 1. Volume of a cuboid of dimensions l, b and h = l ? b ? h. 2. Volume of a cube of edge l = l3. 3. Volume of a cylinder of base radius r and height h = ? 2h. O YA L BR 4. Volume of a cone of base radius r and height 1 h = ? r2h. 3 4 3 5. Volume of a sphere of radius r = ? r . 3 2 6. Volume of a hemisphere of radius r = ? r3. 3 TEXTBOOK’S EXERCISE 13. 2 22 . 7 O TH Unless stated otherwise, take ? = Q. 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of . 8 S PR Sol. AK 9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape formed is 4? h + 4? r2. Is it true? 10. A solid ball is exactly fitted inside the cubical box of side a. Surface area of the ball is 4? a2. Is it true? 11. From a solid cylinder whose height is 2. 4 cm and diameter 1. 4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. 12. A decorative block shown below, is made of two solids – a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter 4. 2 cm. Find the total surface area of the block. 22 ? ? = ? . ? 7? [2011 (T-II)] 3. A tent of height 3. 3 m is in the form of a right circular cylinder of diameter 12 m and height 2. 2 m, surmounted by a right circular cone of the same diameter. Find the cost of canvas of the tent at the rate of Rs 500 per m2. 15. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 108 cm and the diameter of hemispherical ends is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. AS HA 14. Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. N O YA L BR O 1 ? 3 ? 2 cm = ? cm3. 3 3 ? Q. 2.Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same. ) Sol. = ? ?+ ? TH For conical portion : Radius of the base (r) = G Height of cone (h1) = 2 cm 3 cm = 1. 5 cm 2 1 2 ? r h 3 9 We know that, volume of cone = ER 22 3 cm = 66 cm3 7 Hence, the volume of the air contained in the model that Rachel made is 66 cm3. 21 ? S Q. 3. A gulab jamun, contained sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2. 8 cm (see figure). [2011 (T-II)] Sol. Gulab jamun is in the shape of cylinder with two hemispherical ends. Diameter of cylinder = 2. 8 cm ? Radius of cylinder = 1. 4 cm Height of cylindrical part = (5 – 1. 4 – 1. 4) cm = (5 – 2. 8) cm = 2. 2 cm PR AK AS Radius of the hemisphere = Radius of cone = 1 cm Height of cone = h = 1 cm 2 2 ? Volume of hemisphere = ? r3 = ? (1)3 cm3 3 3 2 = ? m3 .. (i) 3 1 1 ? Volume of cone = ? r2h = ? (1)2 (1) cm3 3 3 1 = ? cm3 .. (ii) 3 Volume of the solid = Volume of the hemisphere + Volume of cone Volume of cone OAB = = 1 2 ? r h1 3 1 (1. 5)2 (2) ? cm3 = 1. 5? cm3 †¦ (i) 3 1 Volume of cone A? B? O? = ? r2h1 3 1 = (1. 5)2 ? (2) ? cm3 = 1. 5? cm3 †¦ (ii) 3 For cylindrical portion : Radius of the base (r) = 1. 5 cm Height of cylinder h2= 12 cm – (2 + 2) cm = 8 cm ? Volume of cylinder = ? r2h2 = ? (1. 5)2 (8) cm3 = 18? cm3 .. (iii) Adding equations (i), (ii) and (iii), we have Total volume of the model = volume of the two co nes + volume of the cylinder. = 1. 5? cm3 + 1. ? cm3 + 18? cm3 = 21? cm3 HA N Volume of a gulab jamun 2 2 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2) cm3 + ? (1. 4)3cm3 3 3 = = 1 22 14 3 ? ? 0. 25 ? cm 3 7 10 ER 4 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2)cm3 3 ? 4 ? 1. 4 ? ? 2. 2 ? cm3 = ? (1. 4)2 ? 3 ? ? ? 5. 6 ? 6. 6 ? = ? (1. 96) ? ? cm3 3 ? ? ? (1. 96) (12. 2) = cm3 3 ? Volume of 45 gulab jamuns ? (1. 96) (12. 2) = 45 ? cm3 3 = 15? (1. 96) (12. 2) cm3 22 ? 1. 96 ? 12. 2 cm3 = 15 ? 7 = 15 ? 22 ? 0. 28 ? 12. 2 = 1127. 28 cm3 30 ? Volume of syrup = 1127. 28 ? cm3 100 = 338. 184 = 338 cm3 (approximately) 11 cm3 30 ? Volume of four conical depressions 11 3 22 3 cm = cm = 1. 7 cm3 30 15 ? Volume of the wood in the pen stand = (525 – 1. 47) cm3 = 523. 53 cm3. =4? S PR Q. 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0. 5 cm are drop ped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. Sol. Radius of cone = 5 cm Height of cone = 8 cm Volume of cone = = AK = = O YA L BR Q. 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.The dimensions of the cuboid are 15 cm by 10 cm by 3. 5 cm. The radius of each of the depressions is 0. 5 cm and the depth is 1. 4 cm. Find the volume in the entire stands. (See figure). TH O Radius of spherical lead shot, r1 = 0. 5 cm ? Volume of a spherical lead shot G Sol. Length of cuboid, l = 15 cm Width of cuboid, b = 10 cm Height of cuboid, h = 3. 5 cm Volume of the cuboid = 15 ? 10 ? 3. 5 cm3 = 525 cm3 Volume of a conical depression = 4 3 4 ? 3 ? r = ? (0. 5)3 cm3 = cm 3 1 3 6 ? Volume of water that flows out = 1 ? (0. 5)2 (1. 4) cm3 3 10 AS 1 ? volume of the cone 4 1 ? 200? ? 50? cm3 ? ? = 4? 3 ? 3 HA 2 1 ? r h = ? (5)2 8 cm3 3 3 200 ? cm3 3 N Let the number of lead shots dropped in the vessel be n. Volume of n lead shots = As per condition, ? n? cm3 6 n? 50? = 6 3 = 31680? cm3 + 3840 cm3 = 35520 cm3 = 35520 ? 3. 14 cm3 = 111532. 8 cm3 ? Mass of the pole = 111532. 8 ? 8 g = 892262. 4 g = 892. 26 kg Hence, the mass of the pole is 892. 26 kg (approximately). BR O TH ER S Sol. Diameter of cylinder ABCD = 24 cm 24 cm3 2 = 12 cm Height of cylinder ABCD (h) = 220 cm ? Volume of cylinderABCD = ? r2h = (12)2 (220)cm3 = 31680? cm3 Base radius of cylinder A? B? C? D? , R = 8 cm Height of cylinder A? B? C?D? (H) = 60 cm ? Volume of cylinder A? B? C? D? = ? R2h = (8)2 (60) cm3 = 3840? cm3 ? Volume of solid iron pole = Volume of the cylinder ABCD + Volume of the cylinder A? B? C? D? Base radius of cylinder ABCD, r = YA L PR Q. 6. A solid iron pole consist of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use ? = 3. 14) Radius of the cone OAB (r) = 60 cm Height of cone OAB (h1) = 120 cm ? Volume of cone OAB 1 2 1 ? r h1 = ? (60)2 (120) cm3 3 3 = 144000? m3 Radius of the hemisphere (r) = 60 cm = ? Volume of hemisphere = = = Radius of the cylinder (r) = Height of cylinder (h2) = ? Volume of cylinder = 11 G O AK AS 50? 6 ? ? n = 3 ? ? n = 100 Hence, the number of lead shots dropped in the vessel is 100. Q. 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. Sol. HA N 2 3 ? r 3 2 ? (60)3 cm3 3 144000? m3 60 cm 180 cm ? r2h2 So, r = OTHER IMPORTANT QUESTIONS Q. 1. Volume of the largest right circular cone that can be cut out from a cube of edge 4. 2 cm is : (a) 9. 7 cm3 (b) 77. 6 cm3 3 (c) 58. 2 cm (d) 19. 4 cm3 O TH YA L BR O Sol. (d) Radius of the cone = 4. 2 cm = 2. 1 cm. 2 ER 8. 5 cm 2 S Sol. Diameter of sphere = 8. 5 cm 4 ? 3. 14 ? 4. 25 ? 4. 25 ? 4. 25 cm3 + 8 ? 3. 14 cm3 3 = 321. 39 cm3 + 25. 12 cm3 = 346. 51 cm3 = Hence, she is correct. The correct volume is 346. 51 cm3. remains unfilled. Then the number of marbles that the cube can accommodate is : (a) 142296 (b) 142396 (c) 142496 (d) 142596 Sol. a) Volume of the cube = 223 cm3 = 10648 cm3 Space which remains unfilled G Height of the cone = 4. 2 cm. 1 ? Volume of the cone= ? r2h 3 = PR Q. 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8. 5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and ? = 3. 14. Amount of water it holds = 4 ? 8. 5 ? ? ? ? cm3 + ? 12 (8) cm3 3 ? 2 ? 10648 cm3 = 1331 cm3 8 Remaining space = (10648 – 1331) cm3 = = 9317 cm3 1 22 ? ? 2. 1 ? 2. 1 ? . 2 cm3 = 19. 404 cm3. 3 7 Q. 2. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0. 5 cm and it is assumed that 1 space of the cube 8 12 4 ? (0. 25)3 cm3 3 Let n marbles can be accommodated. Volume of 1 marble = Then, n ? AK 3 4 22 ? ? (0. 25)3 = 9317 3 7 AS HA = ? (60)2 (180) cm3 = 648000? cm3 ? Volume of water left in the cylinder = Volume of the cylinder – [Volume of the cone + Volume of the hemisphere] = 648000? cm3 – [144000? + 144000? ] cm3 = 648000 cm3 – 288000? cm3 = 360000 cm3 360000? = m3 = 0. 36? m3 100 ? 100 ? 100 22 3 = 0. 36 ? m = 1. 131 m3 (approx. 7 Radius of cylindrical neck = 1 cm Height of cylindrical neck = 8 cm N ?n= 9317 ? 3 ? 7 4 ? 22 ? (0. 25) 3 = 142296. Q. 3. A medicine capsule is in the shape of a cylinder of diameter 0. 5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is : (a) 0. 36 cm3 (c) 0. 34 cm3 Sol. (a) (b) 0. 35 cm3 ( d) 0. 33 cm3 Q. 5. The volume of a sphere (in cu. cm) is equal to its surface area (in sq. cm). The diameter of the sphere (in cm) is : [2011 (T-II)] (a) 3 (b) 6 (c) 2 (d) 4 4 3 ? r = 4? r 2 3 ? r = 3 ? d = 2r = 2 ? 3 = 6 cm Sol. (b) BR = 22 ? ? ? (0. 25)2 ? ? 0. 25 ? 1. 5? cm3 3 7 ? ? O TH Height of the cylindrical part = (2 – 0. 5) cm = 1. 5 cm Radius of each hemispherical part = Radius of the cylindrical part = 0. 25 cm. ? Capacity of the capsule 4 ? 4 ? = ? r3 + ? r2h = ? r2 ? r ? h ? 3 ? 3 ? Q. 7. The ratio between the radius of the base and the height of the cylinder is 2 : 3. If its volume is 1617 cm3, the total surface area of the cylinder is : [2011 (T-II)] (a) 208 cm2 (b) 77 cm2 (c) 707 cm2 (d) 770 cm2 Sol. (d) Let the radius and height of the cylinder be 2x and 3x respectively. Then, volume of the cylinder = ? r2h 22 ? 1617 = ? 2x)2 ? 3x 7 YA L = 22 ? 5. 5 ? ? (0. 25)2 ? ? cm3 = 0. 36 cm3 7 ? 3 ? ER Q. 4. A solid piece of iron in the form of a cuboid of dimensions 49 cm ? 33 cm ? 24 cm is moulded to form a solid sphere. The radius of the sphere is : [2011 (T-II)] (a) 25 cm (b) 21 cm (c) 19 cm (d) 23 cm Sol. (b) Volume of sphere = Volume of cuboid S PR 4 3 ? r1 r 8 2 3 = ? 1 = 4 3 27 r2 3 ? r 3 2 ? Ratio between surface areas = 4 : 9 1617 ? 7 343 = 22 ? 4 ? 3 8 ? x = 3. 5 cm. ? Total surface area of the cylinder = 2? r (h + r) ? x3 = G O AK ? 4 3 ? r = (49 ? 33 ? 24) cm3 = 38808 cm3 3 38808 ? 3 ? 7 cm 3 = 9261 cm 3 4 ? 22 ? r3 = r = 21 cm Q. 8. On increasing each of the radius of the base and the height of a cone by 20%, its volume will be increased by : (a) 25% (b) 40% (c) 50% (d) 72. 8% 13 AS Q. 6. The ratio of the volumes of two spheres is 8 : 27. The ratio between their surface areas is : [2011 (T-II)] (a) 2 : 3 (b) 4 : 27 (c) 8 : 9 (d) 4 : 9 Sol. (d ) 22 ? 7 (10. 5 + 7) cm2 7 = 44 ? 17. 5 cm2 = 770 cm2. =2? HA N Sol. (d) Volume of the original cone = New radius New height 1 2 ? r h 3 = 6r 120r = = 5 100 6h 120h = = 5 100 2 4 3 3 2 3 ? = = 3 2? 2? 2? 6: ? 2? 3 ? = 6 ? Hence, ratio of the volume of sphere to that of cube = cm. Then, volume of the metallic solid cylinder of 91 2 ? r h. 375 ? Per cent increase in volume = AK ? 216 ? 125 ? 2 = ? ? ? r h ? 375 ? height 10 = BR Q. 9. A sphere and a cube have the same surface. Show that the ratio of the volume of sphere to that of the cube is 6: ? O 91? 100 ? 3 = 72. 8%. 375 TH ER = 91 2 100 ? r h ? 1 2 375 ? r h 3 2 cm. 3 = Volume of the metal in the spherical shell 32 4 2 = ? 53 ? 33 ? r ? 3 3 32 2 4 r = (125 ? 27) ? 3 3 3 4 ? ? 98 ? r2 = 32 3 49 7 ? r = cm ? r2 = 4 2 Hence, the diameter of the base of the cylinder AS ( Increase in volume = 72 2 1 ? r h – ? r2h 3 125 2011 (T-II)] Sol. Let the radius of the sphere be r and the edge YA L O of the cube be x. Whole surface area of sphere = 4? r2 and whole surface area of cube = 6Ãâ€"2. According to question, ? S Q. 11. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4 3 ? a . Is it true? 3 PR = 7 cm. Sol. Diameter of the ball = side of the cube ? Radius of the ball = ? Volume of the ball = G 4? r2 = 6Ãâ€"2. r2 x 2 = 6 3 r = ? = 4? 2? x 3 2? 4 3 ? r Volume of sphere 3 Now, = Volume of cube x3 = Hence, the statement is false. 4 ? r? 4 ? r? r ? = ? ? 3 ? x? 3 ? x? x 3 2 Q. 12.From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. 14 HA ) a 2 1 ? 6r ? 6h New volume = ? ? ? ? 3 ? 5 ? 5 72 2 = ? r h. 125 Q. 10. The internal and external radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted to form a solid 2 cylinder of height 10 cm, find the diameter of 3 the cylinder. [2011 (T-II)] Sol. Let the radius of the base of the cylinder be 4 a3 ? a3 = 3 8 6 N Sol. Volume of the cube = 73 cm3 = 343 cm3 Sol. 1 ? ? 32 ? 7 cm3 3 = 66 cm3 ? Volume of the remaining solid = (343 – 66) cm3 Volume of the cone = = 277 cm3.AK = = Q. 13. The difference betw een the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in making cylinder is 176 cm3, find the outer and inner diameters of the cylinder. [2010] Sol. Let the inner and outer radii of the cylinder be r cm and R cm respectively. Then, the height of the cylinder = 14 cm. Inner surface of the cylinder = 2? r ? 14 cm2 = 28? r cm2 Outer surface of the cylinder = 2? R ? 14 cm2 = 28? R cm2 Difference of the two surfaces = (28? R – 28? r) ? 88 = 28? (R – r) ? AS Radius of the hemispherical portion = 5 cm = radius of the cone. Height of the conical portion = (10 – 5) cm = 5 cm. Capacity of the shape = PR TH (R – r) = 88 ? 7 =1 28 ? 22 ER 1 2 ? r (2r + h) 3 1 22 = ? ? 5 ? 5 (2 ? 5 + 5) cm3 3 7 2750 22 ? 25 = ? 15 cm3 = cm3. 7 21 ? R–r= 1 †¦ (i) Volume of the metal used in making the cylinder = ? (R2 – r2) ? 14 cm3 . .. 176 = ? (R + r) (R – r) ? 14 BR O S 1 2750 ? cm 3. 6 7 ? Required volume of the ice cream Space which remains unfilled = ? 2750 2750 ? ? = ? ? cm3 6? 7 ? ? 7 2750 5 ? cm3 = 327. 4 cm3. 7 6 ? ? (R + r) = YA L 176 ? 7 =4 22 ? 1 ? 14 †¦ (ii) R = 2. 5 cm and G Solving (i) and (ii), we have r = 1. cm Hence, inner and outer diameters of the cylinder are 3 cm and 5 cm respectively. Q. 14. An ice cream cone, full of ice cream is having radius 5 cm and height 10 cm as shown. Calculate the volume of ice cream provided that its 1 part is left unfilled with ice cream. 6 O R+r= 4 Q. 15. A solid toy is in the form of a hemisphere surmounted by a right-circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. Sol.Volume of the toy = Volume of the cone + Volume of the hemisphere = 1 2 2 1 ? r h + ? r3 = ? r2 (h + 2r) 3 3 3 15 HA 2 3 1 2 ? r + ? r h 3 3 N = 1 22 1408 ? ? 4 ? 4 (4 + 8) cm3 = cm3. 3 7 7 Sol. Capacity of the box = 16 ? 8 ? 8 cm3 = 1024 cm3 Volume of the 16 glass spheres 4 = 16 ? ?r3 3 4 22 = 16 ? ? ? 2 ? 2 ? 2 cm3 3 7 11264 = cm3 21 Volume of water filled in the box 11264 ? ? 10240 = ? 1024 ? cm3 ? cm3 = 21 ? ? 21 A cube circumscribes this toy, hence edge of the cube = 8 cm. Volume of the cube = 83 cm3 = 512 cm3 ? Required difference in the volumes of the toy and the cube = 487. 61 cm3. 1408 ? ? = ? 512 ? ? cm3 7 ? ? 2176 cm3 = 310. 6 cm3. 7 Total surface area of the toy = curved curface area of the cone + curved surface area of the hemisphere = 2 2 2 = ? r h ? r ? 2? r 2 ? 2 ? = ? r ? h + r + 2 r ? ? ? = YA L 22 ? 4 ? 16 ? 16 ? 2 ? 4 ? cm2 ? ? 7 BR O TH ER diameter of the dome is equal to its total height above the floor, find the height of the building. [2001] Sol. Let the internal height of the cylindrical part be h and the internal radius be r. Then, total height of the building =h+r Also, 2r = h + r ? h = r. Now, volume of the building = Volume of the cylindrical part + Volume of the hemispherical part ? ? ? ? S PR and contains 41 O 22 ? 4 ? ? 4 2 ? 8 ? cm2 = ? 7 88 ? 4 = 7 ? 2 ? 2 cm2 ? G 88 ? 4 = ? 3. 41 cm2 = 171. 47 cm2. 7 Q. 16. 16 glass spheres each of radius 2 cm are packed into a cubical box of internal dimensions 16 cm ? 8 cm ? 8 cm and then the box is filled with water. Find the volume of water filled in the box. 16 880 ? 3 ? 7 =8 21? 5 ? 22 ? r =2 Hence, height of the building = h + r r3 = = (2 + 2) m = 4 m. AK 41 Q. 17. A building is in the form of a cylinder surmounted by a hemispherical valuted dome 19 m3 of air. If the internal 21 2 880 = ? r3 + ? r3 [? r = h] 3 21 5? r 3 880 = 21 3 AS 2 19 = ? r2h + ? r3 3 21 HA N Q. 18. A godown building is in the form as shown in the figure.The vertical cross section parallel to the width side of the building is a rectangle 7 m ? 3 m, mounted by a semicircle of radius 3. 5 m. The inner measurements of the cuboidal portion of the bu ilding are 10 m ? 7 m ? 3 m. Find the volume of the godown and the total interior surface area excluding the floor 22 ? ? (base). ? ? = ? . ? 7 ? ? 1 2? = 2 ? ?r ? = ? r2 ? 2 ? 22 ? (3. 5) 2 m2 = 38. 5 m2 7 Total interior surface area excluding the base floor = area of the four walls = = 250. 5 m2. Sol. The godown building consists of cuboid at the bottom and the top of the building is in the form of half of the cylinder.Length of the cuboid = 10 m, Breadth of the cuboid = 7 m Height of the cuboid = 3 m Volume of the cuboid = lbh = 10 ? 7 ? 3 m3 = 210 m3. Radius of the cylinder = 3. 5 m Length of the cylinder = 10 m 1 2 Volume of the half of the cylinder = ? r h 2 1 22 = ? ? (3. 5)2 ? 10 m3 2 7 = 192. 5 m3 Volume of the godown = volume of the cuboid + volume of the half cylinder = (210 + 192. 5) m3 = 402. 5 m3 Interior surface area of the cuboid = Area of four walls = 2 (l + b) h = 2(10 + 7) 3 m2 = 102 m2 Interior curved surface area of half of the cylinder 22 = ? rh = ? 3. 5 ? 10 m 2 = 110 m2 7 YA L BR O TH ER Q. 19.A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2. 1 m and 4 m respectively and the slant height of the top is 2. 8 m, find the area of canvas used for making the tent. Find the cost of the canvas of the tent at the rate of Rs 550 per m2. Also, find the volume of air enclosed in the tent. [2008C] Sol. O S G PR Height of the cone, H = AK ? 2. 8 ? 2 ? 22 = 7. 84 ? 4 m = 1. 95 m Area of canvas required for making the tent = Curved surface area of the tent = Curved surface area of the cylindrical part + curved surface area of the conical part = 2? rh + ? l = ? r (2h + l ) = Interior area of two semicircles 17 22 ? 2 (2 ? 2. 1 + 2. 8) m2 7 AS m HA 1 (curved surface area of the cylinder) 2 + 2 (area of the semicircle) = (102 + 110 + 38. 5) m2 + N 44 ? 7 m2 = 44 m2. 7 Cost of canvas = Rs 500 ? 44 = Rs 22000. Volume of the air enclosed in the tent = Volume of the cylindrical part + Vo lume of the conical part = = ? r2h + = = 88 8. 25 3 ? m = 34. 57 m3. 7 3 ER Q. 20. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimal.Also find the total surface area of the remaining solid. (Take ? = 3. 14) [2008, 2011 (T-II)] Q. 21. A juice seller serves his customers using a glass as shown in the figure. The inner diamater of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use ? = 3. 14) [2009] Sol. Radius of the cylindrical glass r = 2. 5 cm Radius of the cylinder = radius of the cone = 6 cm. Height of the cylinder = height of the cone = 8 cm. Volume of the remaining solid 1 2 = ? 2h – ? r2h = ? r2h 3 3 2 = ? 3. 1416 ? 36 ? 8 cm3 3 = 603. 19 cm3 Slant height of the cone, l O YA L BR O TH Sol. G S Q. 22. A cylindrical vessel with internal diamater 10 cm and height 10. 5 cm is full of water. A solid cone of the diameter 7 cm and height of 6 cm is completely immersed in water. Find the volume of (i) water displaced out of the cylindrical vessel. (ii) water left in the cylindrical vessel. [Take ? = 18 PR Height of the glass = 10 cm Apparent capacity of the glass = ? r2h = 3. 14 ? 2. 5 ? 2. 5 ? 10 cm3 = 196. 25 cm3 Volume of the hemispherical portion 2 2 = ? r3 = ? 3. 14 ? 2. 5 ? 2. 5 ? 2. 5 cm3 3 3 = 32. 71 cm3 ?Actual capacity of the glass = (196. 25 – 32. 71) cm3 = 163. 54 cm3. AK AS 22 ] 7 HA 1. 95 ? 22 ? ? 22 ? 2. 1 ? m3 3 ? 7 ? ? N H? 1 2 ? ?r H = ? r2 ? ? h ? ? 3? 3 ? = 36 ? 64 cm = 10 cm Total surface area of the remaining solid = curved surface area of the cylinder + area of top + curved surface area of the cone = 2? rh + ? r2 + ? rl = ? r (2h + r + l) = 3. 14 ? 6 (16 + 6 + 10) cm2 = 18. 84 ? 32 cm2 = 602. 88 cm2. = r 2 ? h2 [2009] Sol. Radius of the cylinder, r = 5 cm Height of the cylinder, h = 10. 5 cm Capacity of the vessel = ? r2h 22 = ? 5 ? 5 ? 10. 5 cm3 = 825 cm3 7 1 Volume of the cone = ? r2h 3 1 22 = ? ? 3. 5 ? 3. 5 ? 6 cm3 = 77 cm3. 7 (i) Water displaced out of the cylinder = Volume of the cone = 77 cm3 (ii) Water left in the cylindrical vessel = Capacity of the vessel – Volume of the cone = (825 – 77) cm3 = 748 cm3. 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0. 5 cm and depth is 2. 1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. Sol. Volume of a cuboid = 10 ? 5 ? 4 cm3 = 200 cm3. Volume of the conical depression Choose the correct option (Q 1 – 5) : 1. The surface area of a sphere is 154 cm2. The volume of the sphere is : 2 1 (a) 179 cm3 (b) 359 cm3 3 2 2 3 1 (c) 1215 cm (d) 1374 cm3 3 3 2.The ratio of the volumes of two spheres is 8 : 27. The ratio between their surfa ce areas is : (a) 2 : 3 (b) 4 : 27 (c) 8 : 9 (d) 4 : 9 3. The curved surface area of a cylinder is 264 m2 and its volume is 924 m3. The height of the cylinder is : (a) 3 m (b) 4 m (c) 6 m (d) 8 m 4. The radii of the base of a cylinder and a cone of same height are in the ratio 3 : 4. The ratio between their volumes is : (a) 9 : 8 (b) 9 : 4 (c) 3 : 1 (d) 27 : 16 TH ER PRACTICE EXERCISE 13. 2A 5. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the figure is : (a) ? 2h (b) ? r 2 ? 3h ? 2r ? 3 ? r 2 ? 3h ? 2r ? (c) 3 YA L BR O S 6. Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. G O 7. Marbles of diameter 1. 4 cm are dropped into a cylindrical beaker of diameter 7 cm containing 19 PR Q. 23. A pen stand made of wood is in the shape of a cuboid with fo ur conical depressions and a cubical depression to hold pens and pins respectively. The dimensions of the cuboid are 4 22 ? ? (0. 5)2 ? 2. cm3 3 7 = 2. 2 cm3 Volume of cubical depression = 33 cm3 = 27 cm3. ? Volume of wood in the entire stand = [200 – (2. 2 + 27)] cm3 = 170. 8 cm3. = (d) ?r 3 (3h + 4r ) 3 AK AS HA 1 2 1 22 ? r h = ? ? (0. 5)2 ? 2. 1 cm3 3 3 7 Volume of 4 conical depressions = N 11. An ice cream cone consists of a right circular cone of height 14 cm and the diameter of the circular top is 5 cm. It has a hemispherical scoop of ice cream on the top with the same diameter as of the circular top of the cone. Find the volume of ice cream in the cone. 12. A solid toy is in the form of a hemisphere surmounted by a right circular cone.Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover. [2011 (T-II)] 13. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6. 75 cm. What is the radius of the ball? 13. 3 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER TEXTBOOK’S EXERCISE 13. 3 22 , unless stated otherwise. 7 Q. 1. A metallic sphere of radius 4. 2 cm is melted and recast into the shape of a cylinder of Take ? = 20 G O YA L BRO TH ER S 16. A heap of rice is in the form of a cone of diameter 9 m and height 3. 5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap? 17. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0. 04 m3. 18. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm respectively.If the slant height of th e conical portion is 5 cm, find the total surface area and volume of the rocket. (Take ? = 3. 14) radius 6 cm. Find the height of the cylinder. Sol. Radius of sphere = 4. 2 cm ? Volume of sphere = PR some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5. 6 cm. 8. A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3. 5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water.If the radius of the cylinder is 5 cm and height 10. 5 cm, find the volume of water left in the cylindrical tub. 9. The largest possible sphere is carved out from a solid cube of side 7 cm. Find the volume of the sphere. 10. A cylindrical boiler, 2 m high, is 3. 5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered 22 ? ? by the lid. ? ? = ? ? 7 ? 14. From a solid cylinder of height 12 cm and base diameter 10 cm, a conical cavity with the same height and diameter is carved out. Find the volume of the remaining solid. 15.A building is in the form of a cylinder surmounted by a hemispherical dome as shown in the figure. The base diameter of the dome is equal 2 of the total height of the building. Find the 3 height of the building, if it contains 67 1 m3 of 27 to AK AS air. HA N [2011 (T-II)] 4 3 4 ? r = ? (4. 2)3 cm3 3 3 Volume of cylinder = ? R2H = ? (6)2H cm3 As per condition, Volume of the sphere = Volume of the cylinder 4 ? ? (4. 2)3 = ? (6)2H 3 ? ? Radius (r) = 7 m 2 2 Depth (h) = 20 m Volume of sphere of radius 6 cm 4 = ? (6)3 cm3 3 Volume of sphere of radius 8 cm ? †¦ (i) Hence, the height of the platform is 2. m. = As per condition, G ? ? 4 3 4 4 4 ? R = ? (6)3 + ? (8)3 + ? (10)3 3 3 3 3 3 = (6)3 + (8)3 + (10)3 R R3 = 1728 O YA L 4 3 3 ? R cm 3 BR 4 ? (10)3 cm3 †¦ (iii) 3 Let the radius of the resulting sphere be R cm. T hen volume of the resulting sphere = TH ER 4 ? (8)3 cm3 3 Volume of sphere of radius 10 cm = †¦ (ii) Q. 4. A well of diameter 3 m dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [2011 (T-II)] Sol. For well : S PR O †¦ (iv) 3 m 2 Depth of well (h) = 14 m ?Volume of earth taken out = ? r2h Radius of well (r) = AK H = Sol. We know that, volume of the sphere = 4 3 ? r 3 AS Q. 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. 245? 245 ? 22 ? H= ? H= 2. 5 308 308 ? 7 Diameter = 3 m 63 ? 3? = ? ? ? (14) m3 = ? m3 2 ? 2? Width of the embankment = 4 m Let the height of the embankment be H m. ? Radius of the well with embankment, R ? R = 3 1728 ? R = 12 Hence, the radius of the resulting sphere is 12 cm. Q. 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is

World Economy Changed - Free Essay Example

Sample details Pages: 5 Words: 1418 Downloads: 9 Date added: 2017/06/26 Category Economics Essay Type Argumentative essay Did you like this example? Scenario of the world economy changed very rapidly after Second World War. In the post world war scene, United States of America emerged as a dominant economic power where as the United Kingdom and France lost their leading roles in economic activities. The pattern of economic growth also changed a lot in post war era. It was large corporations in USA, engaged in manufacturing activities, which contributed in generation of millions of stable jobs, especially in manufacturing space. These big corporations formed the center of fast and stable economic growth over the years. Economy of scale, division of labor, specialized jobs, and economic might of these corporations made these large establishments unbeatable for years. Rest of the world was just follower of this model of industrial growth of United States of America. In later half of the twentieth century Whole world seen the emergence and growth of these large industrial organizations. Few of these industrial est ablishments grew in size enormously and established their bases in multiple countries. (U.S. Department of State. 2009). This was a result of search for new markets and cheaper resources, so as to maximize the profits. These big corporations drove the American economy as well as world economy for whole later half of the twentieth century. But with the beginning of the twenty first century, scenario started changing and symptom also began to be visible for trouble for these large organizations. American corporations, which were mainly in the manufacturing domain, began to feel the heat of changing economic scenario in the wake of cheap labor availability in East Asian countries. (Ted Fishman. 2005). It also came to fore that owing to smaller size, newly emerging corporations of smaller size, worked with more dynamism and efficiency. Overheads of these smaller corporations were also less as compared to large organizations. Emergence of service sector as a full fleshed industry also co ntributed in changing service conditions, compensation methods and pension benefits. This led to changes in financial markets as well. All these factors led to a situation where established financial market, job market and investment and monetary scenario came under transition towards a new set up and became somewhat unstable in that process. Tremors of this instability are felt far and wide in every part of the world. Stock markets fell in almost every country amidst fear of mounting losses. Banks failed in many parts of the world resulting failure of insurance companies along with these. A scenario emerged in United States where industry shown a final decisive shift from manufacturing to services. It also established that era of stable long lasting jobs is over now. People have to cope up with fast changing service and retail industry, where average tenure for a job is comparatively less as compared to a manufacturing industry job. Size and dimensions of Global Financial Crisis were unprecedented and caused failure of many large establishments like banks, insurance companies, manufacturing establishments, severe crisis in stock markets and escalation in unemployment threatening instability to whole financial mechanism in few countries. Governments of few countries had to come out with huge financial assistance packages to finance markets so as to maintain stability and seize any industrial unrest as a result. Though it was a complex web of many factors which led to global financial crisis, but few important ones can be listed as following. Economy of United States of America shifted from being a manufacturing as base to service at focus. This resulted in restructuring or reshaping of established large corporations in few cases whereas failure of few in others. (Gerald F. Davis. 2009). This caused a panic in stock markets, where people invested heavily in these large corporations seeing their profits and growth in previous years. Shift from manufactu ring based economy to services based economy resulted in reducing availability of long term jobs as in service industry jobs average tenure for job remains substantially low as compared to manufacturing industry jobs. This resulted in newer set of service conditions emerging for labor markets. One of the primary results was portable pension. It was to facilitate the worker, who had to change job frequently on account of fast changing service industry where average tenure for a job was somewhere near 3 years in general. (Department of Agriculture. 2009). But portable pension, required professional pension fund managing companies. These companies operated in professionally and to get higher returns invested these funds in stock markets with higher returns. Formulation of multiple and sensitive nature products out of these pension funds, created chaos, once stock markets started u underperforming as these funds were actually not meant for short term investments. When these pension f unds came in stock markets through institutional investors, these created in new perspective for stock prices. Focus shifted from performance of the corporations in deciding the price of share. But in the race of profit maximizing, rates were decided not only company profitability but tradability of company shares. (Gerald F. Davis 2009). This was not a sound practice and it complicated the whole stock market scenario, once market started falling. Financial market underwent deregulation, which let to development of few tradable instruments. These financial instruments came in to existence by securitization of mortgages and credit card debt etc. Common man having property which can be mortgaged became issuer of financial instruments. (Gerald F. Davis. 2009). At the other hand same person became investor in other type of instruments. This led to emergence of a very complex system of financial market, which was not only unclear in totality but unstable also. Inflated cost of houses was taken as basis for securitization, leading to wrong valuation of resulting financial instruments. Businesses have become very dynamic in every sphere now days. Internet has changed the way business was done traditionally. Now companies based in California are outsourcing their jobs from a location thousands miles away in some other part of the world. Similarly a company generating services at one end of the world is selling it at other end of the world. So world economy has become more integrated in current times than ever before. It has created a scene, where economy of one country gets affected by happening in some other country. But even then some steps can be taken so as to minimize the resulting impact of such type of financial crisis in future. Financial markets must have strong regulation, where valuation of every financial instruments is done with due deliberations. There should not be scope of inflated valuation of securities by few stake holders for their benefit s. There should also be categorization of assets which can be used for securitization. For example credit card debt is not a credible asset which should be allowed to be converted in security. Pension funds management by private professional fund managers are reality in current times. Scope and dimensions of their operations are bound to increase in future. But there must be brought a very strong regulatory mechanism, which regulate operations of funds management with prudent norms in place. These fund managers should not be allowed to invest with just an eye on maximizing the wealth of investors, but should also look for stability of the operations. Lending operations of the banks and other financial institutions should also be kept under check. Subprime lending not only reduces the stability of the operations, but also put the financial system at risk of losses. So only credit worthy individuals or institutions should be allowed to raise debt and that even within prudent lendi ng norms. There must be check on the exposure limit of individual banks or other non banking financial institutions to any single business house or industrial segments. Today business world is fact changing. Any innovation in particular industrial segment may alter whole established scenario, causing losses to big corporations. In such cases, if financial institutions have large exposure to such industrial house or industry, it causes problem of stability. Some time losses may be in the tune of even causing failure of bank. To avoid such scenario, it is all the time better to take preventive step of putting ceiling caps on exposure of banking and non banking financial institutions to any single industry of industrial house. There must strict regulations so as to govern stock markets, so as institutional investors are not in a position to manipulate prices of the stocks to their benefit. It has been observed that in few countries, institutional investors have manipulated the st ock prices resulting in losses to general investor community, erasing confidence of investors in market. Don’t waste time! Our writers will create an original "World Economy Changed" essay for you Create order

Being a Successful Employee - 978 Words

Intrapersonal and interpersonal perceptions are vital to succeed at university and at work. They both work together in order to improve productivity in the work force. However this can only be achieved when the individual has a positive emotion and is ‘happy’ towards their negotiator (Govan, Neale Overback, 2010). Similarly, the intrapersonal perception that an individual has, impacts the job satisfaction of employees. Likewise, the intrapersonal and interpersonal perceptions influence the success of academic performance in university that contributes to later at work. An individuals interpersonal and intrapersonal perceptions are influenced by the emotions of the negotiators in the university or at work (Govan, Neale Overback, 2010). Intrapersonal is where an individual uses their own perspectives and experiences to succeed in their life. People with intrapersonal perceptions usually have a high internal locus of control. Interpersonal is others perception of oneself which is influenced by one’s intrapersonal perception. If a manager has a negative view on a situation this will lead the employers to also have a negative view and reaction (Govan, Neale Overback, 2010). As a result of this there is less productivity and success and it can even impact the way customers view the company. Due to the negative environment at work it can lead employees to higher absenteeism and turnover, which would be a waste of resources for the company. An angry manager will change theShow MoreRelatedBeing A Successful Employee : An Integrative Framework For T he Development Of Social Skills1394 Words   |  6 PagesThere are numerous skills that are required to be in one’s personality to be a successful employee. Social skill is one of the dominant enterprises that enhance career contentment. Primarily, social skill is interpersonal, which is mainly described as the mechanism of properly interacting with other individuals. 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Providing such an atmosphere and a place of retreat for their employees shows that Delta cares about the health and well being of their staff. According to Delta’s S.V.P Allison Ausband stated the following: making sure our employees are well taken care of and have the tools to look and perform their best is something we’re proud to do for our people. We know working at theRead MoreIntegrative Paper1526 Words   |  7 Pagesindividuals. Organizational Behavior and Management concept and The Heart of Change gave indicators that where successful interpretation of one another perceptions. The two books where a complement to one another while establishing each other theories. Increase Urgency Increase urgency action is the first step presented in The Heart of Change book which presents eight steps for a successful large-scale transformation to create a sense of urgency that the change is necessary. Urgency helps motivate

New Belgium Brewing Marketing Strategy to Expand free essay sample

New entries to brewing have a relative ease in creating home micro-breweries, which is aided by the simplicity of the brewing process. Micro-breweries are measured as producing less than 15,000 barrels of beer per year with 75% or more of the beer sold off-site. However, in order to mass produce beers to competitive market levels, more expensive technologies are needed. This provides some protection to established firms in the industry. In addition, the expert knowledge of the brew-masters in the established firms takes years of hands-on experience to acquire, knowledge which is not readily available to most new entrants.The bargaining power in the industry belongs to the customers due to the variety of craft beers throughout the country and specialty, regional beers. Substitutes for beer include liquors, wines, sodas, teas, juices, and sports drinks. The wide variety of alternative options for consumers is a general threat to the brewing industry. Supplies are generally bought by brewers, with New Belgium providing an example by getting their raw malt materials from the United States and Canada, hops from the Pacific Northwest, and packaging material from Colorado. We will write a custom essay sample on New Belgium Brewing Marketing Strategy to Expand or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Then the rivalries within the industry are pretty intense, with the three major players in the domestic market (Budweiser, Miller, Coors), mid-major players (Sam Adams New Belgium) and the craft/homemade breweries. At any level, a brewing company will have heated competition for consumers. The brewing industry remains to be an attractive industry for the owners to invest in because of the worldwide appeal of the product, the longevity of beer’s position as a beverage for humanity, and the United States’ cultural inclusion of beers.New Belgium has found a niche for higher-quality, craft brewing. In addition, the demand for these specially made craft brews have increased despite an overall decline in beer sales in the United States. Growth of the craft brewing industry in 2010 was 11% by volume and 12% by dollars compared to growth in 2009 of 7. 2% by volume and 10. 3% by dollars. Overall, U. S. beer sales were down an estimated 1. 0% by volume in 2010. New Belgium states that they opened distribution in six new states in 2009, and increased production by 15%, despite the declining economic climate.Thus craft brewers, like New Belgium, are a threat to the established powers of the beer industry, the Big Three (Bud, Coors, Miller). Craft Beer Sales Growth BEFORE 2010 New Belgium ought to be more aggressive in their desire for growth. The intense rivalry of specifically craft-breweries, not to mention the beer industry as a whole, demands aggressive growth policies. With the continued growth of craft breweries in the domestic market, New Belgium has the potential to gain market share with their mass-production capabilities already in place, and their niche as a specialty, craft beer.Particularly, they must be more aggressive in their marketing campaign. Facebook and Twitter are essential to any Internet advertising campaign for a company nowadays. New Belgium has been proficient in gaining attention through Facebook, as the numbers of â€Å"likes† for New Belgium Brewery’s Facebook has grown from 63,000 in 2010 to its current number of 179,714 (as of June 27, 2011). The brewery’s Twitter account currently has 27,415 followers, and then the company is following well over 15,000 of their customers and other businesses in the area.These connections give them a huge opportunity to communicate directly with consumers. In order to maximize this opportunity, they should promote their existing events like the Tour de Fat bike race and beer festivals across the country. New Belgium should also open a second brewery in the eastern half of the United States to accommodate more growth into uncharted territories. Putting the new brewery in an already established eastern market like Indiana would allow New Belgium to solidif y their hold on current markets while giving them the potential to grow into neighboring territories. Introduction to the Northeast is critically important because as of now they are ceding the whole region to rival competitors like Samuel Adams. The pluses to this second brewery include additional capacity for production of beers, wider reach of distribution, and ultimately higher sales due the territorial growth. The minuses include a possible loss of company culture at the new brewery due to its distance from Fort Collins. Another minus would be the loyalty current consumers have for regional craft beers already in the market. However these minuses can be addressed by implementing a couple different measures.First, in order to acquaint New Belgium products into new markets, the company should employ a specialized sales team that will go into local bars and pubs and buy New Belgium products for locals. This will begin the grassroots campaign which New Belgium has thrived upon throughout their history. In order to keep the company culture in the new eastern brewery, the co-founders Jeff and Kim Lebesch should be sent to head the new branch. Because the culture and values are already entrenched in the Colorado brewery, the New Belgium culture would be more easily maintained there without the founders being present.Then the founders could personally oversee the implementation of the culture in the new branch. The market positioning of a product is characterized by its pricing, quality, service, distribution, and packaging. For the Fat Tire Ale is the flagship beer of the New Belgium Brewery. It is priced competitively with other specialty beers in the industry, with a bottled 12-pack sold for $13. 99. This makes it less expensive than imported beers, but more expensive than mass-produced products of the Big Three. New Belgium considers it â€Å"one f the most affordable luxuries around†. This also implies that the quality of Fat Tire is superior to its competition. New Belgium assures the quality of its beers by using suppliers who meet their expectation according to the criteria of quality, sustainable practices of the company, supplier support and price. Having these expectations ensure that the highest quality materials are used to go into their brands, Fat Tire included. Service-wise, they do not ship beers to individual vendors, but only to distributors and retailers of beers.The sustainability practice of New Belgium illustrates how they serve their community and the environment, which then appeals to community-conscious consumers. In distribution, New Belgium currently supplies Fat Tire in 26 states as of 2010 and plan on expanding it into additional states in 2011. Consumers can access it from alcohol retailers, in local pubs and bars, or straight from the brewery in Fort Collins, Colorado. Fat Tire’s packaging is central to New Belgium’s marketing efforts, as the bike featured on Fat Tire bottles is now part of the official company logo.The bicycle image is meant to convey New Belgium’s â€Å"whimsical approach to business, the history of [the company’s] beginnings, and their emphasis on sustainability. † The design of the label on the beer gives a vintage and rustic look to Fat Tire, which appeals to its inspiration from the bike trails of Belgium. In addition, the bottles are all made in Colorado for New Belgium, which is in line with the company’s pledge to support the local community. One recommendation for the Fat Tire brand is to elevate its marketing to a mor e mainstream level. This would push Fat Tire from simply being a craft beer with a strong cult following, to a top-tier, household-name brand. This would entail more traditional marketing endeavors such television and magazine ads, as well as a greater Internet presence through advertisements. The advertisements would focus on the story behind the Fat Tire and its superior quality compared the mass-produced beers of the Big Three. The pluses to this sort of re-positioning would include initiating the marketing push into the mind of the average consumer, thus creating opportunity for similar growth with other brands.The minus to this would be if they miss the mark on their marketing campaign, then their loyal followers could be turned away from the Fat Tire brand, thinking that the company has lost touch with its roots. An interesting note would be the question of how competitors would react to a mainstream Fat Tire. Competitors could either anticipate a successful New Belgium launch and react accordingl y, or try to stomp out the mainstreaming efforts of New Belgium to keep them in their current place in the market. New Belgium appeals more to the consumer customer in their marketing focus.They list their core audience and target market as â€Å"craft beer drinkers†¦college educated, more affluent than the average beer drinker, and in their career years (30-50 years old). † Additional markets that New Belgium appeals to include â€Å"fans of Belgian style brews, folks who admire the Colorado lifestyle, and people with more socially aware buying habits. † In this, they can try to focus their marketing efforts towards these particular people then allow them to spread the word about New Belgium to other segments of the market.New Belgium’s transactional customers include smaller bars and pubs, and alcohol retailers within their distribution areas. Typically, New Belgium targets retailers who support the â€Å"slow food movement† and other sustainable practices in line with New Belgium ideals. The marketing program at New Belgium is currently based on mostly grassro ots efforts, with some additional online support through Facebook and Twitter. They generally do not advertise strongly in television or print media such as magazines as of now.They heavily promote company events, beer festivals and their annual Tour de Fat bicycle race. In order to become more competitive and stand out from the competition, they must more aggressively market their products to the consumer. We recommend that New Belgium host a second Tour de Fat bike race in the Eastern United States, where New Belgium has yet to expand to. The Appalachian Mountains provide a similar environmental backdrop to the race, which would be ideal in expanding the company’s sustainability principles to the new markets.The pluses to this include expansion into new markets and the continuation to spreading their core values and culture to communities in the area. The minus is the relative lack of brand recognition in the area, thus increasing the risk of an unsuccessful turn-out for the event. It would be interesting to see if the people of Appalachia are as environmentally conscious as those from the Rocky Mountain area where New Belgium comes from. In order to broaden the base and accessibility of the New Belgium brands, in is essential for marketing to play a greater role in the company than it currently does.This expansion would require television exposure, print advertisements in environmentally-conscious magazines, and expanded usage of the Facebook and Twitter platforms. In television, they reach many more target consumers. Their target consumers (ages 30-50) watch TV on average 4 hours in a 24 hour span, and having these advertisements would put their brands on the consumers’ minds. The magazine ads would be effective because they could pick and choose which magazines to place ads, which focuses on readers who are active and environmentally minded.Then the internet resources of Facebook and Twitter ought to include contests and giveaways for followers of Ne w Belgium. For instance, New Belgium could host a â€Å"Green Project Contest† in which Twitter followers can tweet green project ideas to support and maintain ecosystems, and then New Belgium would award the winner $10,000 to implement their project idea. This would both advertise New Belgium commitment to sustainability and boost the traffic to their Twitter page.The pluses of the television campaign include more exposure to consumers and additional legitimacy to their brand name. The repetition of the ads would also drive the brand name into the consciousness of the consumer mind. The minus of the television ads would the relative high costs of putting it on air as well as creating the commercial, which needs to coincide with the core values and ideals of the company. The pluses of the magazine are the choice of magazine in which to place ads as well as spreading the detailed imagery of the New Belgium brand.The minuses include a possible contradiction to the environmental ideals by using additional paper to promote their ‘sustainable’ measures. The pluses of the internet campaign include encouraging interaction with consumers to post or ‘tweet’ their experiences with New Belgium and their products. The Green Project Contest would add to their sustainability efforts while at the same time promoting the brand. The minus of these internet efforts would possible over-exposure to loyal followers and the uncontrollable variables of the internet.

Amara Holdings Limited

Question: Discuss about the Amara Holdings Limited. Answer: Introduction: The company owns hotels and the different others residential as well as the commercial properties throughout different parts of the Southeast Asia. One of its hotels in Saigon was sold in the year 2008 (Amaraholdings.com, 2016). The current trends throughout the market segments revealed that the demands of the properties are increasing in large numbers and the firm has the interest in two of the office buildings, five specialty restaurants, a shopping complex, three apartments and different residential properties.Moreover, the firm is looking forward to its expansion throughout different geographical areas with the help of developing and capturing new properties. The firm was incorporated in the year 1970 and traced its roots in the 1930s. Some of the top competitors of the firm are Keppel Corporation limited, Hotel properties limited and Wheelock properties limited. Company and industry analysis Sales have been considered to be one of the most crucial factors which help in the establishment of any of the firm concerned. There need to be effective strategies from the firm in order to entry into the new market segments for establishing its business processes (Ryz?ko, 2011). The pricing from the competitors and the selection of the geographical location for the development of the properties plays a major role in increasing the profitability in this particular industry. Therefore, Amara Holdings Limitedis planning to capture different locations in order to enhance the growth and the expansion of the business processes of the firm. References Amaraholdings.com. (2016).Amara Holdings. [online] Available at: https://www.amaraholdings.com/ [Accessed 13 Nov. 2016]. Ryzko, D. (2011).Emerging intelligent technologies in the industry. 1st ed. Berlin: Springer.