Organizational Behavior Essay Example | Topics and Well Written Essays - 1750 words - 5

Organizational Behavior - Essay Example f organizational behavior leads to good relationships among the employees, poor organizational behavior management can lead to poor performance among the individual employees and groups (McGinnis, 2006). Group dynamics affect the operations of organizations. In most cases, group dynamics negatively affect the employee performance. Various theories explain factors responsible for the various behavioral types employees exhibit towards each other. The variance theory explains the various causes of behavior in an organizational context. The organizational structure, people, technology and the working environment are the most important factors influencing organizational behavior (McGinnis, 2006). If there is the absence of joy and happiness in the workplace, diagnosing the causes of these incidences is the only way of solving these problems. Meetings are an important component for effective management. Despite the escalated issues affecting the organization bovver time, and issues in the meetings, the organization has not done away with the meetings. The role played by meetings in the management process and policy formulation strategies has considerable effects in any organization. They help in the policy implementation process by designing the way and direction the organizations should take. In an effort to achieve normal operations in the organization and meetings serve their intended purpose, discipline is essential. Calling of names, flaming emails between co-workers, official charges brought to HR, uncivil behavior in meetings leading, sometimes, to screaming, name-calling, and physical threats is an indication of unmotivated, stressed and an organizational work that lacks morals. Changing the organizational structure is one of the ways of achieving the change (Hofstede, et al., 1997). According to systems theory, organizations are organs comprised of different parts performing a similar task. However, the criteria of organizing these parts determine the

Biography of Fernando Botero Essay Example for Free

Biography of Fernando Botero Essay Fernando Botero, also named the most Colombian of Colombian artists, has developed a style the world notices as his own. Fernando was born in 1932 in Medellin, Colombia. Fernando came across heavy schooling as a child, which isolated him from traditional art in museums and such cultural institutions. It was his strict school however that brought art to be an interest for Fernando, the school Botero attended was run by Jesuits who were strict and brought little enjoyment into Botero’s life. To find enjoyment Botero began to draw at a young age. At that young age his inspiration was anything that interest him such as bullfighting. Fernando was a great fan of bullfights so he would paint scenes of this then sold them in front of the arena for 5 pesos. He spent nearly two years painting this subject. He had a growing interest in art his entire life; he shared his thoughts, and studied. When Botero was seventeen he worked for the Medellin newspaper, El Colombiano, titled Picasso and the Nonconformity of Art, which showed Botero’s mind and how it is linked with art. Botero is so well known because of his signature style, robust and round objects and characters. Botero tells critics that he is simply attracted to his form without knowing why. He claims that artists never know why artists use a form, he claims the style is intuitive and that the explanation for their style can be rationalized after it’s adopted. Botero is difficult to understand, as he doesn’t share his opinion with his art or even explaining his art. â€Å"He shares his vision with us but not telling us how to feel about it. Navas- Nieves says. We know his works are personal as some of his famous works depict his youth, â€Å"The Bishop†, â€Å"The Nun†, â€Å"The Bullfighter†, and â€Å"The Widow†. It is unknown to all except Botero however if these drawings reflect the beauty with these no proportional, bright colored, exaggerated sized or the turmoil. Botero eliminated brushwork and texture in his paintings as he favored a smoother look. Botero’s works are abstract and are educated by a Columbian upbringing and social commentary. When Colombian children go to church they see all these Madonnas, so clean and perfect. In South America china-like perfection is very much a part of the ideal toilet of beauty. More so even than the polychrome wood sculptures in Spain, Latin American sculptures look like porcelain. So, in contrast to Europe or North America, you connect the notions of art and beauty at a very early age. I grew up with the idea that art is beauty. All my life Ive been trying to produce art that is beautiful to discover all the elements that go to make up visual perfection. When you come from my background you can’t be spoilt by beauty, because youve never really seen it. If youre born in Paris, say, you can see art everywhere, so by the time you come to create art yourself you’re spoilt – youre tired of beauty as such and want to do something else. With me it was quite different. I wasnt tired of beauty; I was hungering for it. † Botero said this knowing he was not raised with art, which is his beauty. The lack of art in his life made him develop such an original style. The church influences Botero, his baroque style was adopted from the church. Botero is the most Colombian of Colombian artists because of his insulation from international trends. Botero once remarked â€Å"A painter can do things a photographer can’t do, because a painter can make the invisible visible. † This thinking lead Botero to create a series of serious paintings about prisoner abuse in Abu Ghraib. Botero, through his paintings in this series allowed us to feel the prisoners suffering. The paintings brought us to the agony of the victims, the humiliation they felt and all the pain was transferred from the painting to us, making us the victim. This series proved moreover Botero’s skill of art. Many critics thought the series was out of Botero’s league claiming that Botero’s style cannot do justice to the seriousness of the subject. Botero proved the critics wrong when he captures the emotion of the victims. The difference between photographs and paintings are that photographs come with a story. Just something that happened. They can be beautiful but paintings show affects and emotion and thoughts and different depictions. Much more thought goes into a painting, which sets the two apart.

Microbes in the Antarctic Continent Essay -- Biology, Microorganism

Microbes in the Antarctic continent play an important role in the ecosystem function and sustainability. The variation in snow and ice covers in Antarctic continent markedly affects all ecological variables including the composition of microbial assemblages and their function (Quayle et al., 2002). Minor changes in the global climate could significantly affect the ice and snow melt regimen in this continent. Therefore the Antarctic ecosystem is considered to be a good ecological indicator for the global climate change (Walther et al., 2002). Since 1961, an increase in 1Â ºC temperature from -11Â ºC to approximately -10 Â ºC) in Schirmacher Oasis has been recorded (http://south.aari.nw.ru/data/data.asp?lang=0&station=1). This may have resulted in an increased melting of snow, glacial and continental ice thereby affecting the lacustrine systems in Schirmacher Oasis. It is possible that Lake Tawani(P), which initially existed as a low-catchment depression, progressively filled with w ater from glacial ice and snow melts through visible surface channels and eventually become a permanent landlocked freshwater lake with a thriving microbial ecosystem (Figure 1). Although over 100 freshwater lakes in Schirmacher Oasis harbor a rich microbial consortium, only a few lakes have been subjected to the study of the microbial diversity. Our results revealed that analysis of the 16S rRNA gene sequence of the culture-independent community DNA had a better coverage of the diversity of the bacteria in the samples. Overall, the 16S rRNA gene analysis resulted in the identification of 8 different phyla, 20 different genera including two clones from the Candidate OP 10 groups. Interestingly two genera, Sphingomonas and Janthinobacterium were found to be comm... ...intermixing of some of the lake waters with the snow and ice melt through channels that connect them. Therefore the dynamic microbial ecosystem at the Schirmacher Oasis lakes is driven by the interaction between the annual weather events and microorganisms inhabiting in these lakes. In this study, we describe the bacterial diversity in a previously unexplored freshwater Lake Tawani (projected) [described here in Lake Tawani(P)] using culture-based and culture-independent methods. We have targeted the bacterial conserved segments of the 16S Small Subunit ribosomal RNA gene (described here in 16S rRNA) and the rpoB gene that code for the ÃŽ ²-subunit bacterial RNA polymerase. Investigating the microbial diversity of the lakes especially those connected through channels will help understand the dynamic nature of the freshwater lake ecosystems on the Antarctic continent.

What is a Class Diagram?

A class diagram models the static structure of a system. It shows relationships between classes, objects, attributes, and operations.Basic Class Diagram Symbols and NotationsClassesClasses represent an abstraction of entities with common characteristics. Associations represent the relationships between classes.Illustrate classes with rectangles divided into compartments. Place the name of the class in the first partition (centered, bolded, and capitalized), list the attributes in the second partition (left-aligned, not bolded, and lowercase), and write operations into the third. Active ClassesActive classes initiate and control the flow of activity, while passive classes store data and serve other classes. Illustrate active classes with a thicker border. VisibilityUse visibility markers to signify who can access the information contained within a class. Private visibility, denoted with a – sign, hides information from anything outside the class partition. Public visibility, denoted with a + sign, allows all other classes to view the marked information. Protected visibility, denoted with a # sign, allows child classes to access information they inherited from a parent class. AssociationsAssociations represent static relationships between classes. Place association names above, on, or below the association line. Use a filled arrow to indicate the direction of the relationship. Place roles near the end of an association. Roles represent the way the two classes see each other.Multiplicity (Cardinality)Place multiplicity notations near the ends of an association. These symbols indicate the number of instances of one class linked to one instance of the other class. For example, one company will have one or more employees, but each employee works for just one company.Composition and AggregationComposition is a special type of aggregation that denotes a strong ownership between Class A, the whole, and Class B, its part. Illustrate composition with a filled diamond. Use a hollow diamond to represent a simple aggregation relationship, in which the â€Å"whole† class plays a more important role than the â€Å"part† class, but the two classes are not dependent on each other. The diamond ends in both composition and aggregation relationships point toward the â€Å"whole† class (i.e., the aggregation).GeneralizationGeneralization is another name for inheritance or an â€Å"is a† relationship. It refers to a relationship between two classes where one class is a specialized version of another. For example, Honda is a type of car. So the class Honda would have a generalization relationship with the class car. In real life coding examples, the difference between inheritance and aggregation can be confusing. If you have an aggregation relationship, the aggregate (the whole) can access only the PUBLIC functions of the part class. On the other hand, inheritance allows the inheriting class to access both the PUBLIC and PROTECTED functions of the superclass. https://www.smartdraw.com/uml-diagram/5. Describe the six (6) different relationship notation that exists in UML Class Diagram? (6 Marks) Answer: Relationships in Class DiagramsClasses are interrelated to each other in specific ways. In particular, relationships in class diagrams include different types of logical connections. The following are such types of logical connections that are possible in UML: †¢ Association †¢ Directed Association †¢ Reflexive Association †¢ Multiplicity †¢ Aggregation †¢ Composition †¢ Inheritance/Generalization †¢ Realization Associationis a broad term that encompasses just about any logical connection or relationship between classes. For example, passenger and airline may be linked as above: Directed Associationrefers to a directional relationship represented by a line with an arrowhead. The arrowhead depicts a container-contained directional flow. Reflexive AssociationThis occurs when a class may have multiple functions or responsibilities. For example, a staff member working in an airport may be a pilot, aviation engineer, a ticket dispatcher, a guard, or a maintenance crew member. If the maintenance crew member is managed by the aviation engineer there could be a managed by relationship in two instances of the same class. Multiplicityis the active logical association when the cardinality of a class in relation to another is being depicted. For example, one fleet may include multiple airplanes, while one commercial airplane may contain zero to many passengers. The notation 0..* in the diagram means â€Å"zero to many† Aggregationrefers to the formation of a particular class as a result of one class being aggregated or built as a collection. For example, the class â€Å"library† is made up of one or more books, among other materials. In aggregation, the contained classes are not strongly dependent on the lifecycle of the container. In the same example, books will remain so even when the library is dissolved. To show aggregation in a diagram, draw a line from the parent class to the child class with a diamond shape near the parent class. CompositionThe composition relationship is very similar to the aggregation relationship. with the only difference being its key purpose of emphasizing the dependence of the contained class to the life cycle of the container class. That is, the contained class will be obliterated when the container class is destroyed. For example, a shoulder bag's side pocket will also cease to exist once the shoulder bag is destroyed. Inheritance / Generalizationrefers to a type of relationship wherein one associated class is a child of another by virtue of assuming the same functionalities of the parent class. In other words, the child class is a specific type of the parent class. To show inheritance in a UML diagram, a solid line from the child class to the parent class is drawn using an unfilled arrowhead. Realizationdenotes the implementation of the functionality defined in one class by another class. To show the relationship in UML, a broken line with an unfilled solid arrowhead is drawn from the class that defines the functionality to the class that implements the function. In the example, the printing preferences that are set using the printer setup interface are being implemented by the printer. https://creately.com/blog/diagrams/class-diagram-relationships/ 6. Provide the list of six (6) â€Å"Multiplicity† constraint? ANSWER:MultiplicityMultiplicity is a definition of cardinality – i.e. number of elements – of some collection of elements by providing an inclusive interval of non-negative integers to specify the allowable number of instances of described element. Multiplicity interval has some lower bound and (possibly infinite) upper bound:multiplicity-range ::= [ lower-bound ‘..' ] upper-bound lower-bound ::= natural-value-specification upper-bound ::= natural-value-specification | ‘*'Lower and upper bounds could be natural constants or constant expressions evaluated to natural (non negative) number. Upper bound could be also specified as asterisk ‘*' which denotes unlimited number of elements. Upper bound should be greater than or equal to the lower bound

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

Free Essays on Prepare For Takeoff

Prepare For Take-Off Since the tragedy of September 11th, our nation’s airports have undergone several security changes to prevent such a tragedy from happening again. Some of the changes are more obvious than others, like the presence of the National Guard in our airports. However, there are some changes that are not seen nor felt by the public until they are presented with them, like not being allowed to bring drinking containers down into the terminal. Because most people are not aware of the changes, things tend to run a little slower now in our airports than normal. Louis Armstrong International Airport, formerly New Orleans International Airport, has been no exception. For 2 and a half years I was employed at Armstrong Airport by one of the venders and witnessed first hand the affects of September 11th. Because of the new security regulations, passengers had to wait longer to board their planes, which made my job more of a hassle. The best way to make a flight experience more enjoyable starts before one boards the plane. It is now suggested that passengers arrive two hours before their flight’s departure time. Do arrive early as suggested. When arriving at the airport, there will always be men and women with uniforms on asking â€Å"Do you need a skycap?† These ladies and gentlemen are there to be at a passengers service to help them check in their luggage, receive their boarding confirmation without the hassle of having to go to the ticket counter. This service is called â€Å"Curb-Side Check In†. It is quick and allows the passenger to head straight down to the terminal. However, this service is for domestic flights only. International flights have to go to the ticket counter but can use the service to help with the luggage. Don’t forget to tip the skycaps because they can be out there up to twelve hours a day and they are only paid minimum wage. If one has to go to their airline’s ticket counter, th... Free Essays on Prepare For Takeoff Free Essays on Prepare For Takeoff Prepare For Take-Off Since the tragedy of September 11th, our nation’s airports have undergone several security changes to prevent such a tragedy from happening again. Some of the changes are more obvious than others, like the presence of the National Guard in our airports. However, there are some changes that are not seen nor felt by the public until they are presented with them, like not being allowed to bring drinking containers down into the terminal. Because most people are not aware of the changes, things tend to run a little slower now in our airports than normal. Louis Armstrong International Airport, formerly New Orleans International Airport, has been no exception. For 2 and a half years I was employed at Armstrong Airport by one of the venders and witnessed first hand the affects of September 11th. Because of the new security regulations, passengers had to wait longer to board their planes, which made my job more of a hassle. The best way to make a flight experience more enjoyable starts before one boards the plane. It is now suggested that passengers arrive two hours before their flight’s departure time. Do arrive early as suggested. When arriving at the airport, there will always be men and women with uniforms on asking â€Å"Do you need a skycap?† These ladies and gentlemen are there to be at a passengers service to help them check in their luggage, receive their boarding confirmation without the hassle of having to go to the ticket counter. This service is called â€Å"Curb-Side Check In†. It is quick and allows the passenger to head straight down to the terminal. However, this service is for domestic flights only. International flights have to go to the ticket counter but can use the service to help with the luggage. Don’t forget to tip the skycaps because they can be out there up to twelve hours a day and they are only paid minimum wage. If one has to go to their airline’s ticket counter, th...

Use of Trademark Names in Fiction

Use of Trademark Names in Fiction Use of Trademark Names in Fiction Use of Trademark Names in Fiction By Mark Nichol A couple of years ago, a site visitor asked about the necessity of obtaining permission to refer to trademarked products by name in fiction. Here’s the specific query: How is copyright dealt with in fiction writing? For example, if I sell a story where I wrote that a character jogged to Burger King in his new Reeboks, would there be copyright infringement? Do I need to get approval from the holders of the copyright to use their names in my stories? And, how would I go about doing that? How do I find out what is copyright protected and what isn’t? And here, better late than never, is the response: Fortunately for the multitudes of authors who write fiction (and the innumerable publishing companies that print their books), writers are free, for the most part, to include trademarked names in their stories. The passage in question is especially innocuous, because the references to Burger King and Reeboks are benign: Nobody in the novel dies from eating a Whopper, and no character is fatally run over in traffic because his running shoes are defective. But even if the author had implicated one of these brands in someone’s death, legal retribution would be unlikely; the sheer volume of media overwhelms any one corporation’s efforts to monitor for and suppress defamatory references to their products. But risk is relative: If a writer with the stature of, say, J.K. Rowling had resorted to the plot device of a deadly hamburger or a dangerous pair of running shoes, her publisher would likely be sent a cease-and-desist letter. This terse request from the trademark owner would call on the publishing company to refrain from associating the company’s delicious and nutritious WhopperR brand beef-patty sandwiches or light but sturdy and comfortable ReeboksR brand athletic shoes with anyone’s death. (Side note: The registered trademark symbol is never required; in commercial publications, it is often inserted to imbue one’s products with a protective aura or to refer to those of others, as a courtesy, to encourage reciprocity.) To avoid such a consequence, an astute editor would likely request that Ms. Rowling excise such libelous references before submitting the final manuscript, thus avoiding the arrival of a letter referring to â€Å"possible recourse to further legal options to protect our valuable intellectual property rights.† But your editor would likely do the same, perhaps suggesting that instead, you call the fast food franchise Hamburger Prince or the shoes Teezoks. Interestingly, assigning closely similar names, or describing companies or products that resemble real ones but are not named in their honor (or, often, dishonor), is fair play. Want to improve your English in five minutes a day? Get a subscription and start receiving our writing tips and exercises daily! Keep learning! Browse the Fiction Writing category, check our popular posts, or choose a related post below:7 Examples of Passive Voice (And How To Fix Them)Latin Words and Expressions: All You Need to Know35 Synonyms for Rain and Snow

A Brief History of Kosovo Independence

A Brief History of Kosovo Independence Following the demise of the Soviet Union and its domination over Eastern Europe in 1991, the constituent components of Yugoslavia began to dissolve. For some time, Serbia, retaining the name of the Federal Republic of Yugoslavia and under control of the genocidal Slobodan Milosevic, forcefully retained possession of nearby provinces. History of Kosovo Independence Over time, places such as Bosnia and Herzegovina and Montenegro gained independence. The southern Serbian region of Kosovo, however, remained part of Serbia. The Kosovo Liberation Army fought Milosevic’s Serbian forces and a war of independence took place from about 1998 through 1999. On June 10, 1999, the United Nations Security Council passed a resolution which ended the war, established a NATO peacekeeping force in Kosovo, and provided for some autonomy which included a 120-member assembly. Over time, Kosovo’s desire for full independence grew. The United Nations, the European Union, and the United States worked with Kosovo to develop an independence plan. Russia was a major challenge for Kosovo independence because Russia, as a U.N. Security Council member with veto power, promised they would veto and plan for Kosovo independence that did not address Serbia’s concerns. On February 17, 2008, ​the Kosovo Assembly unanimously (109 members present) voted to declare independence from Serbia. Serbia declared that the independence of Kosovo was illegal and Russia supported Serbia in that decision. However, within four days of Kosovo’s declaration of independence, fifteen countries (including the United States, United Kingdom, France, Germany, Italy, and Australia) recognized the independence of Kosovo. By mid-2009, 63 countries around the world, including 22 of the 27 members of the European Union had recognized Kosovo  as independent. Several dozen countries have established embassies or ambassadors in Kosovo. Challenges remain for Kosovo to obtain full international recognition and over time, the de facto status of Kosovo as independent will likely spread so that almost all of the world’s countries will recognize Kosovo as independent. However, United Nations membership will likely be held up for Kosovo until Russia and China agree to the legality of Kosovo’s existence. Kosovo is home to approximately 1.8 million people, 95% of whom are ethnic Albanians. The largest city and capital are Pristina (about half a million people). Kosovo borders Serbia, Montenegro, Albania, and the Republic of Macedonia.

Recommendation of Ambercrombie & Fitch Essay Example | Topics and Well Written Essays - 500 words - 1

Recommendation of Ambercrombie & Fitch - Essay Example Due to the experience, the company will have a benchmark of success for what is achievable. CEO from outside will bring a new perspective and cutting-edge approach. It has been proven through research that outside CEOs with new ideas spend more resources on research and development, and their decisions shows commitment to innovation. Outside talent is necessary so as to breathe fresh life into the firm (Blumberg, et al. 2013). By going outside for a new CEO, the company will send a strong message to its entire workforce and the board that the company recognizes the importance of change, and is ready to invest in its future. A new CEO with new ideas isn’t burdened by the past and is able quickly make a big impact to the company. Hiring a new CEO from outside is what the company requires so as to reach the desired growth. Companies’ changes are ever constant in recent times, as accelerating technology, economic volatility, and globalization are indicators that times for business as usual are long gone. As a result, a lot of companies have been compelled to innovate, pivot, or rethink their strategies entirely. Ambercrombie & Fitch can also go the same route by changing its cultures. Leaders have discovered that in some situations, these steps are challenging and some are even impossible to perform without altering the culture that is associated with the company. A different strategy may call upon the change of the entire culture mind-set of the organization; ‘the way we do things around here’. It is necessary for the leadership of the company to own and be engaged in the process. Only through ownership and engagement can the leadership ensure that Ambercrombie & Fitch is fully aligned and there is the right organizational structure in place, a system that is relevant, the correct management practices, and the required talent in place so as to grow in the desired direction. The

History of Botswana Essay Example | Topics and Well Written Essays - 1250 words

History of Botswana - Essay Example There was a high expectation that the northern region of what was to become the Republic of Botswana was under the British colonies that was protecting it. They were located in the north of the Molopo River and their intention was to merge the protectorate with the south of the region. The southern part of that region was under the colony of Cape. The British controlled the area called Bechuanaland after a request to put it under its protection was influenced by the then tribal leader, Khama III. Khama traveled to the United Kingdom to request the British to put their land under the protection of the crown so that it was not disturbed by their hostile neighbors. To but their land under their protectorate, the colonial secretary requested that it allows the British to construct railway line across their land that later came to be an economic opportunity to their land as it had opened them up to trade. In the early twentieth century, particularly 1910, the Cape politicians wanted to jo in the Bechuanaland and this was influenced by the Union of South Africa, but it would later become a challenge to join the state. Before the Cape politicians could join the Bechuanaland, there was to be a signed agreement between the rulers of Bechuanaland and the British. The grandson of Khama III by the name of Seretse Khama went to the United Kingdom to study at the Oxford University. At that time, their neighbors, South Africa, had introduced apartheid laws that restricted the intermarriage and other fundamental rights between different races.

Biomolecular techniques (bitter taste perception of Research Paper - 1

Biomolecular techniques (bitter taste perception of phenylthiocarbamide (PTC) - Research Paper Example The project was done to determine the phenotypic and genotypic characteristics of students, and the results that were obtained were matched to those of European and Sub-Saharan cohorts. It was found allele combination of homozygous tasters, heterozygous tasters, and homozygous non-tasters were similar to those of the European cohort. This implied that the experiment was largely successful and accurate for the determination of phenotypes and genotypes of the PTC gene. The results can be used in making informed decisions with regard to dietary intake of foods rich in anti-oxidants, in planning of alternative nutrient rich meals for children that are sensitive tasters and finally it can be used by clinicians in the treatment plan of cancer or cardiovascular complications patients. Every individual is different from another, and this is attributed to their genetic make-up. The sense of taste also varies between different categories of people where, for example, some people can sense the taste of some chemicals while others are not able to. One such chemical is phenylthiocarbamide (PTC) or phenylthiourea an organic compound that tastes very bitter to some people while others are unable to taste it at all. Studies done in the past have indicated that polymorphisms in sensory receptor genes in humans can alter the perception in individuals through the coding for receptor types that are functionally distinct (Bufe et al., 2005). The ability of an individual to taste PTC depends primarily on their genetic makeup and is controlled by the PTC gene known as the TASR238 taste receptor gene, located on the chromosome 7 (7q34) and is about 1003 bp long. There are three coding SNPs that are non-synonymous within the taster TAS2R38 gene which are: rs713598–G145C, Ala49Pro; rs1726866–T785C, Val262Ala; rs10246939 – A886G, Ile296Val, and are responsible several haplotypes (Kim, et al 2005; Bufe , 2005). PTC sensitivity is a Mendelian

Land study Essay Example | Topics and Well Written Essays - 1500 words

Land study - Essay Example According to Covanta Energy (2011), the Rookery South EfW Generating Station will be able to â€Å"convert approximately 585,000 tonnes of residual waste per year into 65MWe of electricity, of which 55MWe would be exported to the national grid. That's enough electricity equivalent to meet the needs of Bedford and the Marston Vale;† meaning, approximately 82,500 houses (About the Project, para. 2). As far as the size and construction, Covanta Energy did not want to hinder or takeaway from any long-distance views. Therefore, the designers decided on a building made up of boxes that interlock (See Figure 1), to reduce the height of the structure as much as possible; the other option was a 3-stack building. Still, with 9 times the total capacity per annum of the Isle of Man EfW facility, the current design is approximately 2m lower than â€Å"the visible height above ground level of the Isle of Man EfW Facility† (Covanta Energy, 2011). ... Assets including designated and undesignated archaeological sites and historic buildings such as parks and gardens, monuments, areas of conservation, and registered battlefields are the focus of the assessment. Since the Rookery South EfW Generating Station will be built in the location of a former clay pit, there is very little, if any potential for negative impact on historical assets. The only potential for disturbance Covanta Energy (2011) has acknowledged is to â€Å"currently unrecorded sub-surface archaeological features during those construction works that take place outside of the pit of minor significance† (Cultural Heritage, para. 2). The residents and local decision-makers feel different from Covanta about the impact on the historical area of the proposed development. Residents of the Marston Vale and surrounding areas near the proposed site of the project want to hold on to one of twelve community forests in England (Ampthill, 2011). The resounding message to Cova nta is the pollution generated from the facility combined with the eyesore of a massive processing plant viewable from long distances will ruin the historical beauty of the land (Ampthill, 2011). Although Covanta recognizes the fact that the facility will be visible from long distances, potentially causing negative reactions like those conveyed by the residents in the surrounding communities, they feel they have taken adequate steps to minimize any adverse impact. The steps they refer to include the low-profile design, as opposed to the 3-stack alternative, and the selection of material finishes for the buildings, which they feel are as aesthetically pleasing as a waste processing plant can get; all aspects of the design were developed after consulting with Central Bedfordshire Council,

Design a Marketing Metrics system for a commercial organisation of the Essay

Design a Marketing Metrics system for a commercial organisation of the group's choice - Essay Example The authors have aptly summarized the difference between yesteryears and the current day marketing where it has become a science rather than an art. Art and skills are definitely required but the scientific factor is a much needed factor. Here comes the metrics portion of the marketing activity. â€Å"A metric is a measuring system that quantifies a trend, dynamic, or characteristic. In virtually all disciplines, practitioners use metrics to explain phenomena, diagnose causes, share findings, and project the results of future events† (Farris et al., 2006). The executive summary gives a comprehensive and compact picture of the importance of Marketing Metrics in today’s business world. If the marketers are unable to quantify their projections they would not even know what they are looking forward and then all planning and marketing activity becomes vague. By quantifying it is meant that marketing requires numbers like percentage of customers requiring their products, judg e the change in market conditions, explain customer habits and purchase trends. All the quantitative measures can be put in the form of metrics which will define the future course of marketing activity in an organization. This marketing metrics is being done for Coca Cola who needs no introduction in the beverage market where they have a huge presence for decades. Even though Coca Cola has established its brand over the years but still it requires a dynamic marketing team to work round the clock to ensure that it reaches out to its customers in a market which is volatile and can change in a matter of no time. Therefore marketing metrics is required to constantly analyze the changes in trend, the purchasing habitat of people, customer tastes, and inclination to reward systems or accumulation of points on coca cola consumption. The report also emphasizes the background of the company, its non financial resources and its financial resources. The non financial resources include market s hare, customer relationship management, awareness, competitor analysis, product lifecycle and customer satisfaction. The financial resources include Sales and Revenue as well as the Return on Investment (ROI). A lot of planned marketing metrics is done by Coca Cola Company so that it can stay competitive in the market. Marketing metrics involves quantitative analysis based on which important business judgments are made and the marketing managers needs to be spot on and accurate in this analysis to use it to the good effect of the organization as a whole. Introduction Marketing metrics is a complex phenomenon and corporate like Coca Cola needs a range of metric to come to a definite and logical analysis. To achieve the marketing metrics a range of parameters and phenomenon needs to be derived and appropriate data input and analysis needs to be done. Customer satisfaction survey is one of the key input areas where customers quantify their satisfaction level. Similarly trend analysis i s done with customer survey which gives an insight to the company as to the market trends and habits of the consumers. This section gives a comprehensive insight into the method and manner of producing a marketing metric which would serve the purpose of getting quality and quantitative information of the market situation which will enable the marketing department to come to marketing decisions based on analysis. â€Å"The challenge, of course is knowing what to measure and exactly how to measure it. That is where Marketing

Land study Essay Example | Topics and Well Written Essays - 1500 words

Land study - Essay Example According to Covanta Energy (2011), the Rookery South EfW Generating Station will be able to â€Å"convert approximately 585,000 tonnes of residual waste per year into 65MWe of electricity, of which 55MWe would be exported to the national grid. That's enough electricity equivalent to meet the needs of Bedford and the Marston Vale;† meaning, approximately 82,500 houses (About the Project, para. 2). As far as the size and construction, Covanta Energy did not want to hinder or takeaway from any long-distance views. Therefore, the designers decided on a building made up of boxes that interlock (See Figure 1), to reduce the height of the structure as much as possible; the other option was a 3-stack building. Still, with 9 times the total capacity per annum of the Isle of Man EfW facility, the current design is approximately 2m lower than â€Å"the visible height above ground level of the Isle of Man EfW Facility† (Covanta Energy, 2011). ... Assets including designated and undesignated archaeological sites and historic buildings such as parks and gardens, monuments, areas of conservation, and registered battlefields are the focus of the assessment. Since the Rookery South EfW Generating Station will be built in the location of a former clay pit, there is very little, if any potential for negative impact on historical assets. The only potential for disturbance Covanta Energy (2011) has acknowledged is to â€Å"currently unrecorded sub-surface archaeological features during those construction works that take place outside of the pit of minor significance† (Cultural Heritage, para. 2). The residents and local decision-makers feel different from Covanta about the impact on the historical area of the proposed development. Residents of the Marston Vale and surrounding areas near the proposed site of the project want to hold on to one of twelve community forests in England (Ampthill, 2011). The resounding message to Cova nta is the pollution generated from the facility combined with the eyesore of a massive processing plant viewable from long distances will ruin the historical beauty of the land (Ampthill, 2011). Although Covanta recognizes the fact that the facility will be visible from long distances, potentially causing negative reactions like those conveyed by the residents in the surrounding communities, they feel they have taken adequate steps to minimize any adverse impact. The steps they refer to include the low-profile design, as opposed to the 3-stack alternative, and the selection of material finishes for the buildings, which they feel are as aesthetically pleasing as a waste processing plant can get; all aspects of the design were developed after consulting with Central Bedfordshire Council,

Year-Round Education Research Paper Example | Topics and Well Written Essays - 1000 words

Year-Round Education - Research Paper Example It is important to know the difference between this modified calendar and traditional system to compare the benefits and drawbacks of this development. Unlike the traditional system where a multiple month vacation (mostly in summer) is given to students, year round schools offer a cumulative holidays of about 2 months distributed over the year. These year round schools can further offer single track, multi track or extended year format. When this new system was introduced in the mid 80s, a lot of controversies erupted that highlighted concerns over certain issues. In order to analyze and compare these two systems it is mandatory to dissect those issues. Traditional system had served its purpose well for a very long time. Now if it was to change it could have some positive or negative effect on the individuals associated with it, most importantly, the students and the teachers (Haser & Ilham 2005). So while comparing these two systems focus should be on observing any change in their a ttitude. For example, a student from a year round school system can be disturbed looking at the kids of his age enjoying their long summer vacations. Other factors that are important to consider are effect on academic performance, cost and changing a norm of a society. Year round Education and Academic Achievement: Various researches have been conducted around the world to compare the two education calendars. Lindsay-Brown, 2010, investigated and compared the impact of year round school calendar and traditional school calendar on the academic achievements in North Central South Carolina. In this study, 256 elementary students from four different schools, each pair following different calendars, were tested using the Palmetto Achievement Challenge Test (PACT). English language arts and mathematics were the subjects tested in the study. After all the fine adjustments were done to the obtained data, the result showed that there was no significant difference in the academic achievements when students from year round schools versus traditional schools were compared. Another review paper of 39 studies (Cooper et al 2003) also indicated a weak and insignificant effect of modified school calendar on academic achievement. Most of the think tanks that are in the favor of this modified calendar place their arguments on the basis of superior academic achievements. These results, however, is in direct contradiction to their theory. Some groups such as â€Å"Summer Matters† believe that traditional school system is best for the students and society. The argument is based on philosophy of family ties and relationships. They believe that long summer breaks allow families to come together and cherish the moments with their children. This social support catalyzes children efforts and stimulates them to perform better in the field of education. Moreover, they insist that summer vacations provide an excellent opportunity for kids to learn and discover new things. Why Year round Education Calendar? So if year round education does not help students to excel in their academics, then why this system has gained popularity over the years? Parents, faculty and administrator weigh certain other factors as well such as growing school enrollments, working parents and shrinking budget (Sheilds & Stevens, 2000). These problems are partially solved by introducing this education calendar. For example, multi-track year round schooling divides students into groups and rotates them in an organized manner. This multi tracking can increase the capacity of school by 30%. Therefore, the total number of students accommodated in 4 traditional schoo

Elections and Media Essay Example for Free

Elections and Media Essay The most influential part of American society is the media. Because of it, people have been witness to numerous historical events such as inaugurations, assassinations, and acts of terrorism. We would not have been informed if it had not been for the excessive coverage from the television; however, at the same time, the television has been a continuous barrier within the political world because it emphasizes materialistic items and meaningless ideas rather than the important tasks at hand that can affect an entire nation. The media has an arguable hold on the politics of this nation. As told in Source C, by Menand, many attribute Kennedys victory in the close election of 1960 to the presentation he made in two televised debates in the final months of his campaign; however, the people who listened to the debates on the radio, and did not see televised images, believed it was a draw between Nixon and Kennedy. The majority who watched the debates on the television thought that Kennedy had a crisper image than the badly postured Nixon. This evidence supports the fact that the television makes the voters focus on the image of the candidate rather than ideals and their intellectual responses to the questions at hand. In the 1960 election, the television had won the nation away from sound to images. It is not fair to other candidates because the other candidates could be better off running the country than a person who was won over by materialistic views. Television may make a candidate appear one way, but when they are in office they become two-faced, thus promoting the fact that more and more candidates are pursuing images more than issues. An example of this can bee seen in Source B, by Hart and Triece, where it states that Bill Clinton showed his boxers on MTV because he believed that was what the audience was looking for in order to support him as president. The media, mainly television, has the ability to control the way a nation can think as well as the way a nation can vote. Television is like any other form of entertainment. It has the job of telling the citizens what they want to hear, rather than being honest and telling the truth. That is why there is not one single news broadcasting station that will voice both liberal and conservative views. As told by Ted Koppel in Source F, networks influence the nation by cropping and pulling out the best parts of events in the presidential election that make their political party stand out from the rest. This is a reason why so many elections are full of slandering because each broadcast network tries to dig up information from the past or present to use against a candidate. For instance, during George Bushs election in 2000, the press dug up information about his daughters at college and their partying ways. This was a way to try and sabotage his chances of presidency, and it almost worked. The nation started to view him as an unworthy candidate, but that information had no relevance on his ability to be president. Fortunately it did not affect Bush winning the election, although it is still being seen today. The media continuously shoots down the major issues at hand and puts the focus on non-important ideas. Television has a way of being intimate with people and making them feel like they are getting valuable information, but what they do not know is that they are not getting the facts and information from direct sources, furthering the question of who can the American people trust to get accurate information. The inaccurate information can be the deciding factor between who a new voter will vote for, which in return can affect the outcome of an election or presidential term. A prime example of this can be seen in Source E, by Ranney. In this article, it states that President Lyndon Johnson was supporting the fact that the war in Vietnam could and would be won. At that time the nation believed him; however, when the news network took it upon them to see if Johnson was true to his word, they came back with the reciprocal of what Johnson stated. CBS news network reported that Vietnam was a bloody scene and there was no military victory in the future. Can you guess who the public believed? Of course the media! Basically in a single second, all of the citizens contradicted their previous beliefs, based by Johnson, and supported the facts brought in by CBS. This further upholds the idea that television and other forms of entertainment have stolen the idea of self thought and independent opinions. It somewhat contradicts freedom of speech and belief because what we are hearing is what the media itself picked out, rather than our own intelligence. Unfortunately due to this hold the television has over us, Johnson was overwhelmingly fought against, and he decided to end the army and navy bombardment in Vietnam, as well as not run for another term. The media had beaten him, and there was no way to stop it. In conclusion, all sources of entertainment, such as television, have altered the ways Americans think and view politics. The television has a power over the people by showing the bad sides of each candidate in an election rather than what good they have done for our society. By focusing on the negativity and worthless aspects of politics, television can be considered a prominent problem against politics in the United States. Once the nation wants to decide to support self-opinions, there will never be a presidential election or event that will not be decided by television.

Natural vs Synthetic Fiber Reinforced Polymer

Natural vs Synthetic Fiber Reinforced Polymer Concrete technology as a branch discipline of technology requires increase in the degree of specialization and consolidation of the fiber material in the cement matrix form composite materials. It requires knowledge of the concepts related to the interaction between the fiber and adhesive cement, mortar or matrix concrete that influence the production and nature of the product. The scientists and engineers have been actively exploring to find the materials that will be used as replacement of conventional materials that can provide a feature best new design and innovation to enhance the material. The development of fiber technology is in the line with the development of knowledge of the material. Following the high demand and innovation in applying fiber, the fiber technology has produced various kinds of fibers potential for commercialization. Participation fiber reinforcement in concrete, mortar and cement adhesive work to improve the engineering properties of many based materials such as fracture resistance, bending strength and resistance to fatigue, impact, thermal shock or chipping. Consolidation of materials in the form of cement mortar or concrete has become an attraction as a building material because it is inexpensive, has the resilience and has a compressive strength and stiffness sufficient for restructuring. However, the disadvantages are located on fragile nature, tensile strength and impact of the weak as well as receptive to moisture movement. Hence, reinforced by fibers that have enhanced capabilities offer a suitable alternative, practical and economical to overcome the lack of features of conventional concrete or mortar. Elements of a fiber is a continuous filament in the form of an express term sheet or spreadsheet form. Fibers can generally be categorized into three types : synthetic fibers, natural fibers and mineral fibers. Synthetic fibers are man-made fibers. It is based chemicals such as petrochemicals and synthetic fibers derived mostly from nylon, polyster, aerylic polymer and polyacrylonitrile fibers used to make fiberglass. There is also a bundle of fibers that make the polymer chain is as strong as aramid and chain bond length as dyneema. While natural fiber derived from natural sources, from plants and animals. Plant fibers are cellulose and lignin-based stacks such as cotton, jute, coir, oil palm bunches, flax and so on. It can be obtained from seeds (cotton, kapok), leaf (pineapple, banana), leather plant (jute, kenaf, rattan, hemp), fruit (coconut, palm) and straw (rice, wheat, barley, grass). Next, animal fibers derived from protein particles like silk and wool. For mineral fibers, it derived from the earths crust and it happens naturally. It is based on asbestos fibers (chrysotile, amosite, crocidolite, tremolite, actinolite, anthophyllite), ceramic fibers (glass wool, quartz, aluminum oxide, silicon carbide) and fiber-metal (steel, aluminum). However, both of natural and synthetic fiber reinforced concrete have their own challenges and weakness. Nothing is being done without deficiencies. Synthetic fiber however has more challenges than natural fiber because of its production. Future development of natural and synthetic fiber reinforced polymer concrete will make us want to investigate more about them. OBJECTIVES To know about polymer concrete and why fiber being reinforced in it. To describe the characteristics of natural and synthetic fiber reinforced polymer concrete. To describe the challenges in environment while using both composite materials in construction. To describe the future development in both composite materials. LITERATURE REVIEW POLYMER CONCRETE AND ITS CHARACTERISTICS Polymer concrete is a composite material in which the binder consists entirely of a synthetic organic polymer. It is variously known as synthetic resin concrete, simply resin concrete or plastic resin concrete. Because the use of a polymer instead of Portland cement represents a substantial increase in cost, polymers should be used only in applications in which the higher cost can be justified by superior properties, low labor cost or low energy requirements during processing and handling. It is therefore important that architects and engineers have some knowledge of the capabilities and limitations of polymer concrete materials in order to select the most appropriate and economic product for a specific application. Polymer concrete consists of a mineral filler such as an aggregate and a polymer binder which may be a thermoplastic, but more frequently, it is a thermosetting polymer. When sand is used as a filler, the composite is referred to as a polymer mortar. Other fillers include chalk, gravel, limestone, crushed stone, condensed silica fume (silica flour, silica dust), quartz, clay, granite, expanded glass, and metallic fillers. Generally, any dry, non-absorbent, solid material can be used as a filler. To produce polymer concrete, a monomer or a pre-polymer which mean a product resulting from the partial polymerization of a monomer, a hardener (cross-linking agent) and a catalyst are mixed with the filler. Other ingredients added to the mix include plasticizers and fire retardants. Sometimes, silane coupling agents are used to increase the bond strength between the polymer matrix and the filler. To achieve the full potential of polymer concrete products for certain applications, various fiber reinforcements are used. These include glass fiber, glass fiber-based mats, fabrics and metal fiber. Setting times and times for development of maximum strength can be readily varied from a few minutes to several hours by adjusting the temperature and the catalyst system. The amount of polymer binder used is generally small and is usually determined by the size of the filler. Normally the polymer content will range from 5 to 15 percent of the total weight, but if the filler is fine, up to 30 p ercent may be required. Polymer concrete composites have generally good resistance to attack by chemicals and other corrosive agents, good resistance to abrasion, have very low water sorption properties and good marked freeze-thaw stability. Also, the greater strength of polymer concrete in comparison to that of Portland cement concrete permits the use of up to 50 percent less material. This puts polymer concrete on a competitive basis with cement concrete in certain special applications. The chemical resistance and physical properties are generally determined by the nature of the polymer binder to a greater extent than by the type and the amount of filler. In turn, the properties of the matrix polymer are highly dependent on time and the temperature to which it is exposed. The viscoelastic properties of the polymer binder give rise to high creep values. This is a factor in the restricted use of polymer concrete in structural applications. Its deformation response is highly variable depending on formulation ; the elastic moduli may range from 20 to about 50 GPa, the tensile failure strain being usually 1 percent. Shrinkage strains vary with the polymer used. For example, high for polyester and low for epoxy-based binder. It must be taken into account in an application. A wide variety of monomers and pre-polymers are used to produce polymer concrete. The polymers most frequently used are based on four types of monomers or pre-polymer systems : methyl methacrylate (MMA), polyester pre-polymer-styrene, epoxide pre-polymer hardener (cross-linking monomer) and furfuryl alcohol. Table 1 : General Characteristics And Applications of Polymer Concrete Products Poly (methylmethacrylate) General Characteristics Low tendency to absorb water. As a result, high freeze-thaw resistance ; low rate of shrinkage during and after setting. Outdoor durability and good chemical resistance. Typical Applications Used in the manufacture of faà §ade plates, stair units and sanitary products for curbstones. Polyester General Characteristics Good adhesion to other materials, relatively strong, good chemical and freeze-thaw resistance but have high-setting and post-setting. Typical Applications Because of lower cost, widely used in panels for public and commercial pipes, buildings, floor tiles, stairs, various precast and cast-in applications in construction works. Epoxy General Characteristics Strong adhesion to most building materials, low shrinkage, good creep and fatigue resistance, superior chemical resistance and low water sorption. Typical Applications Epoxy polymer products are relatively costly. They are mainly used in special applications including use in mortar for industrial flooring, skid-resistant overlays in highways, epoxy plaster for exterior walls and resurfacing of deteriorated structures. Furan-based polymer General Characteristics Composite materials with high resistance to chemicals which most acidic or basic aqueous media, strong resistance to polar organic liquids such as ketones, aromatic hydrocarbons and chlorinated compounds. Typical Applications Furan polymer mortars and grouts are used for brick such as carbon brick and red shale brick, floors and linings that are resistant to chemicals, elevated temperatures and thermal shocks. Source : Blaga, A. and Beaudoin, J.J., (1985). Polymer Concrete. Canadian Building Digest published November 1985. THE CHARACTERISTICS OF FIBER AND TOUGHNESS CONCEPT Characteristics of fiber in use to hardened concrete : Fibers should be significantly stiffer than the matrix which has a higher modulus of elasticity than the matrix. Fiber content by volume must be adequate. There must be a good fiber-matrix bond. Fiber length must be sufficient. Fibers must have a high aspect ratio. Means that they must be long relative to their diameter. Toughness is defined as the area under a load-deflection (or stress-strain) curve. Adding fibers to concrete greatly increases the toughness of the material. That is, fiber-reinforced concrete is able to sustain load at deflections or strains much greater than those at which cracking first appears in the matrix. NATURAL FIBER REINFORCED POLYMER CONCRETE AND ITS CHARACTERISTICS Potential use of natural fiber reinforced concrete in the application of natural fibers has long attracted the attention of researchers. Various researches has been conducted in many countries for a variety of mechanical properties, physical performance and durability of materials reinforced by natural fibers. Natural fibers are categorized as organic waste from plants such as fiber coconut, sisal, bagasse, jute, wood dust and so on. Natural fiber reinforced concrete is essentially a special concrete where it contains fibers with a small diameter, independently and randomly distributed in the cement matrix. Uniform distribution in the cement matrix, contributing to an increase in the tensile and resistance to cracking, impact and improved the ductility values à ¢Ã¢â€š ¬Ã¢â‚¬ ¹Ãƒ ¢Ã¢â€š ¬Ã¢â‚¬ ¹for the good aspects of energy absorption. Although many types of fibers were used as reinforce material in concrete, the use of natural fibers had long been in existence and there is a lot of evidence of the usage of these fibers in the history of civilization. Nature has given human the fiber reinforced material in the form of wood, bamboo and other plants. The use of straw in mud bricks and horse hair in the mortar has the potential of natural fibers. Only in the late 1960s and early 1970s, research began to study the potential use of various types of natural fibers as reinforcement material in the slab concrete and cement-based composite materials. Natural fiber reinforced cement or concrete products that use fibers such as coir, sisal, sugar bagasse, bamboo and so on have been produced and tested in more than 40 countries. For economic reasons in developing countries where natural fibers is so much available, it is demanding for construction industry players to enhance the usefulness of these resources in an effective and economical as to introduce composite materials for residential use and others. Basic needs use of natural fibers as reinforcement material in concrete matrix is à ¢Ã¢â€š ¬Ã¢â‚¬ ¹Ãƒ ¢Ã¢â€š ¬Ã¢â‚¬ ¹tensile strength and high elastic modulus, the bond between the matrix and fiber, good chemical composition, stable geometry and good durability. SYNTHETIC FIBER REINFORCED POLYMER CONCRETE AND ITS CHARACTERISTICS Synthetic fibers are man-made fibers resulting from research and development in the petrochemical and textile industries. There are two different physical fiber forms: monofilament fibers, and fibers produced from fibrillated tape. Currently there are two different synthetic fiber volumes used in application, namely low-volume percentage (0.1 to 0.3% by volume) and high-volume percentage (0.4 to 0.8% by volume). Most synthetic fiber applications are at the 0.1% by volume level. At this level, the strength of the concrete is considered unaffected and crack control characteristics are sought. Fiber types that have been tried in concrete matrices include : acrylic, aramid, carbon, nylon, polyester, polyethylene and polypropylene. The characteristics is depend on the types of synthetics used to reinforced with polymer concrete. Different fiber has different properties. Adding carbon fiber decreased the unit weight of polymer concrete. Carbon fiber provides much higher compressive strength, flexure strength and ductility of polymer concrete. PVC and polypropylene fibers did not significantly influence the compressive strength and gave the lowest pulse velocities and modulus. NATURAL VS SYNTHETIC FIBER REINFORCED POLYMER CONCRETE : CHALLENGES The challenges of polymer concrete are the monomers of polymer can be volatile, combustible and toxic. Initiators, which are used as catalysts, are combustible and harmful to human skin. The promoters and accelerators are also dangerous. Natural fibers are emerging as lightweight, low cost, and more environmentally rather than synthetic fibers in composites. This is because : natural fiber production has lower environmental impacts compared to synthetic fiber production. natural fiber composites have higher fiber content for equivalent performance, reducing more polluting base polymer content. the light-weight natural fiber composites improve fuel efficiency and reduce emissions in the use phase of the component, especially in auto applications. end of life incineration of natural fibers results in recovered energy and carbon credits. A compound reinforced with natural fibers is not only low density, low-cost, and abrasion resistant, it also offers an absence of toxicity and better dimensional stability. Polyester raw material releases high amounts of carbon dioxide. This rapidly increases global warming, which is why polyester and other synthetic fabrics are widely discouraged. The other reason is that some synthetic fabrics come from non-renewable resources such as oil. Eventually rise of these synthetic fibers usage have been causing environmental problems such as dumping and recycling. In addition, glass fiber can cause acute irritation of the skin, eyes, and respiratory tract. Mainly concerns have been raised for long term disease such as cancer and lung scarring. Moreover, when released, glass fiber does not decompose and hence again it results in environmental pollutions, as well as, threaten animal life and nature along. Therefore, one of the solutions is using natural fibers instead of synthetic fibers in developing composites materials as they are renewable. Also the consumption of renewable resources would provide positive image for sustainability of green environment. Natural fibers are less harmful to the environment and the society because they are derived from plants and animals which are more eco-friendly. Products which manufactured from natural fabric eventually dissolves into the earth. Plant and animal based fabrics are a part of the evolutionary process of life. They return to the earth to return once more to life. Synthetic fibers are more harmful to the environment because they are enhanced with chemicals. Polyester and nylon fabrics are made from a substance which creates nitrous oxide. Materials that are labeled petrochemical, flame retardants, nylon, acetate and non-wrinkle are all chemically treated. Chemicals which used for the manufacture of synthetic fabrics is harmful and can enter into the water supply and cause health problems. Also workers who are continuously exposed to dangerous chemicals are at risk for developing auto-immune diseases and disease of the lung. Products made from petrochemicals take years to break down, creating a constant need for landfills. Synthetic products that are disposed into the ocean are a threat to marine life. The threat to aquatic animals will eventually precipitate a food shortage. Although, synthetic fibers may offer softer fabrics and more durable materials, the long term effect on the environment far outweigh any advantages. The high cost of petrol along with global awareness of how natural fibers improve overall quality of life will help motivate manufacturers to find more innovative ways to utilize natural fibers. NATURAL VS SYNTHETIC FIBER REINFORCED POLYMER CONCRETE : FUTURE DEVELOPMENT Synthetic : Although not investigated extensively, the use of two or more fiber types in the same concrete mix is considered promising. The decision to mix two fibers may be based on the properties that they may individually provide or simply based on economics. Considerable improvement in the load deflection response was observed mixing steel with polypropylene fibers. In a more recent study, steel micro-fibers (25 microns in diameter and 3 mm long) and carbon micro-fibers (18 microns in diameter and 6 mm long) both in mono- and hybrid- forms were investigated. In the mono-form, steel fiber provided better strengthening than the carbon fiber and carbon fiber provided better toughening than the steel fiber. Interestingly, in the hybrid form (in combination), they both retained their individual capacities to strengthen and toughen. It appears possible, therefore, that by properly controlling fiber properties and combining them in appropriate proportions, one can actually tailor-make h ybrid fiber composites for specifically designed applications. Natural : Environmental awareness and depletion of the petroleum resources are among vital factors that motivate a number of researchers to explore the potential of reusing natural fiber as an alternative composite material in industries such as packaging, automotive and building constructions. However, their applications are still limited due to several factors like moisture absorption, poor wettability and large scattering in mechanical properties. Among the main challenges on natural fibers reinforced matrices composite is their inclination to entangle and form fibers agglomerates during processing due to fiber-fiber interaction. So, the research on natural fiber is being done by mercerization treatment on mechanical properties enhancement of natural fiber reinforced composite or so-called bio composite. It specifically discussed on mercerization parameters, and natural fiber reinforced composite mechanical properties enhancement. It was found that the most parameters used in merc erization treatment were alkali concentration, fiber soaking temperature and fiber soaking duration. Although similar types of reinforced fiber are used, it could give different values in its final composite mechanical properties due to different parameter setting during a mercerization treatment process. Therefore, there is a significant need to conduct further work focusing on main effect and interaction effect of mercerization parameters setting toward enhancement of natural fiber reinforce composite mechanical properties. CONCLUSION In conclusion, natural fiber reinforced polymer concrete has more environmentally characteristics than the synthetic one. But, in the context of advantages, synthetic fiber reinforced polymer concrete has more than natural. Both of them have their own advantages and disadvantages. Because of several characteristics of natural fibers such as moisture absorption, poor wettability and large scattering in mechanical properties, thus it makes reinforcement with polymer concrete a bit less advantageous. Future works will be needed to improve the properties of both natural and synthetic fiber reinforced polymer concrete included with environment impacts.