Othello Game Strategy Essays - Abstract Strategy Games, Go, Reversi

Othello: Game Strategy OTHELLO PROJECT Othello is a game very similar to tic tac toe. In both games you must flank your opponent diagonally, vertically, horizontally, or orthogonally (any row or line.) A strategy without voluntary pass is described here. (voluntary pass is where you give up one move, or lose a turn.) THE COMPUTER AND ITS MOVES The computer will at the beginning of the game make short moves, which are moves which include less than 5 stones changing hands. These are setup moves. You, on the other hand, make long moves for points, while the computer is sitting, waiting patiently for you to finish. So, do not make very long moves at the beginning of the game. Keep yourself to making short moves, and NEVER put a stone where the computer has another one on the end. Example:You see a row of 5 black stones, with a white stone on the end, and beside that white stone a black. You of course, would put a white stone at the end with the 5 black stones, but that is a bad mistake. Go to the side with the one black (a setup move so the computer cannot capitalize when you turn 5 black stones white, and the computer your seven whites black. Always play within your set "square". If you had on the bottom row 3 white, and on the second row 2 blacks, you would only play within that designated square, which is 3 x 3 stones. (see diagram.) This ensures that the computer can only play one stone past the square at any time, which will help you in predicting where the computer will move and a basic strategy. Always try to keep yourself to one move to win (if you could have two moves in a row, you would win something to zero.) Don't have long rows of white connecting to one black. If you do, the computer can get up to 6 in one move. GAME OVERVIEW STRATEGY FIRST MOVE:Always move north or south. North and south means top or bottom, and east and west means horizontally. Then calculate where your square will be and wait for the computer to make its move. SETUP MOVES:For the next 10 or so moves, or until the board is about half full, restrict yourself to only short moves. These will setup yourself for long ones. Also, keep yourself to one move to win. The one move to win system is effective for winning close to moderately close games. Eventually the computer will make one long move that will mess up your short setup moves, but after that go for blocking the computer instead of points. If you can successfully block the computer, going for points will be unnecessary. The points will make themselves if you block properly. Also, once in a while fall behind to setup moves if you are stuck. Be careful that you do not fall behind too much though! MOVES FOR POINTS:After the board is about half full, or you've made about 10 or so moves, it is time for getting some points. If you have setup yourself properly, you should see several long moves which will effectively obliterate the computer. After this stage, you should be leading by at least fifteen points or more. When you have reached that goal (I would go for more, at least twenty, if possible) there should be less than 7 spaces on the board. If there isn't, keep going until there is. FINAL BLOW:Now you should just watch yourself. Don't make foolish moves where the computer can connect for more than 5 points. If it was inevitable, just leave it. If you then keep blocking properly, you should win easily. NOTES You are using the same strategy the computer would use on you. (basically the same.) The computer has the advantage of passing moves whenever it wants, but it will not likely unless the game is close or hopeless, or there are no legal moves for it to make. You can make yourself lose something to zip. Completed short moves by the computer can easily be changed into long ones, by the same method described above. (page one, paragraph two.)

Brazillian Pepper essays

Brazillian Pepper essays The word exotic has always called to mind visions of the fascinating, unusual, curious, sophisticated, and even the bizarre (Kramer, 1998:5). This term seems to allure people to investigate, comprehend, and even collect exotic paraphernalia-including plants. The phrase exotic plant does sound glamorous, but not to the native plants in Florida. Native plants live well with each other, sharing food, water, shelter, and space. When brought to a different environment these now non-native plants have trouble adapting with native plants. They take control of the space provided and begin to overcrowd the area, taking out many natives. Non-native species also lack natural controls such as disease and predation, which helps to keep a balance between species. Exotic plants were first introduced to south Florida in the late 1800s and the population of them has been increasing ever since (Public Affairs Office, 1997a, 1997b). Most nonindigenous plants feel welcomed in the Sunshine States partly tropical climate. They are able to stand the heavy rainfalls and Floridas partial droughts (Kramer, 1998). One non-native species that has grown throughout south Florida, and has even been called the Florida Holly, is quickly spreading (Public Affairs Office, 1997b). The south Florida Water Management District surveyed an area of 284,708 Hectares taken over by Brazilian pepper in 1993. This evergreen shrub had the largest range out of six other species surveyed and has almost completely displaced native understory plants (Kramer, 1997). The Brazilian pepper (schinus terebinthiflius) was introduced in the nineteenth century as an ornamental plant, and did not begin to be pervasive until the 1900s-50 years later (Kramer 1997; Public Affairs Office, 1997b). Local distribution of its seeds is primarily by racoons and opossums, while birds spread the seeds long-distances. The Brazilian pepper does not require bar...

Brazillian Pepper essays

Brazillian Pepper essays The word exotic has always called to mind visions of the fascinating, unusual, curious, sophisticated, and even the bizarre (Kramer, 1998:5). This term seems to allure people to investigate, comprehend, and even collect exotic paraphernalia-including plants. The phrase exotic plant does sound glamorous, but not to the native plants in Florida. Native plants live well with each other, sharing food, water, shelter, and space. When brought to a different environment these now non-native plants have trouble adapting with native plants. They take control of the space provided and begin to overcrowd the area, taking out many natives. Non-native species also lack natural controls such as disease and predation, which helps to keep a balance between species. Exotic plants were first introduced to south Florida in the late 1800s and the population of them has been increasing ever since (Public Affairs Office, 1997a, 1997b). Most nonindigenous plants feel welcomed in the Sunshine States partly tropical climate. They are able to stand the heavy rainfalls and Floridas partial droughts (Kramer, 1998). One non-native species that has grown throughout south Florida, and has even been called the Florida Holly, is quickly spreading (Public Affairs Office, 1997b). The south Florida Water Management District surveyed an area of 284,708 Hectares taken over by Brazilian pepper in 1993. This evergreen shrub had the largest range out of six other species surveyed and has almost completely displaced native understory plants (Kramer, 1997). The Brazilian pepper (schinus terebinthiflius) was introduced in the nineteenth century as an ornamental plant, and did not begin to be pervasive until the 1900s-50 years later (Kramer 1997; Public Affairs Office, 1997b). Local distribution of its seeds is primarily by racoons and opossums, while birds spread the seeds long-distances. The Brazilian pepper does not require bar...

Family Business Essay Example | Topics and Well Written Essays - 2000 words

Family Business - Essay Example The paper tells that the family business is an enterprise governed and managed for the purpose of pursuing and developing the vision of the business â€Å"held by a dominant coalition controlled by members of the same family or a small number of families† for its potential sustainability across generations of the family, sometimes in business association with some of its extended families. Most definitions of family business centre around the kinship of family members owning and managing a venture, state Rogoff and Heck. According to Habbershon, Williams and MacMillan, it is the intersection between family members, the family, and the business that epitomises the distinctive set of characteristics that explain performance differences between family and nonfamily businesses. The intersection may also be a source of conflict within the family and the business. In the domain of family business leadership transition, Hartel, Bozer and Levin consider an adaptation of executive coac hing to be helpful in the process. Family systems are important in family enterprises. Further, family businesses form the foundation stones of evolving economies state Gomez-Mejia, Takacs, Nunez-Nickel et al as well as Hunter and Wilson. Two-thirds of organisations are family-owned and managed. The fact that this segment of the economy is facing an impending crisis is disquieting, particularly in view of the fact that there is little governmental assistance for family business owners to resolve the emerging threat successfully. Thesis Statement: The purpose of this paper is to investigate the theory of family business, examine the crucial role of the family system in family business, and determine whether the adaptation of executive coaching can successfully support leadership transition in family enterprises. The Theory of Family Business The lenses through which academic research and literature on family businesses are viewed and interpreted is the theory explaining different asp ects of family business management. Research helps to reinforce theory. For example, Sharma, Chrisman and Chua (2003) use the theory of planned behaviour to help explain the extent to which family businesses engage in succession planning. At the same time, a broad theory of family business is more important because it will help in setting the parameters for research in the domain of family businesses. A broad theory will also function as a tool for retaining, expanding, and propagating knowledge on the field. The theory of the family firm explains the concept of the family business, the reason for the existence of the same, and the determinants of their scale and scope. It is a comparatively new area of study in relation to rigorous theoretical investigations. However, there are identifiable trends in defining family business, and in differentiating between family firms and non-family firms, thereby addressing the existence of family firms, the factors that support their survival, g rowth, and the creation of long-lasting economic and non-economic value. Habbershon (2006) as well as Chrisman, Chua and Litz (2003) reiterate that the family business exists because of the reciprocal economic and non-economic value created through the integration of family and business systems. The joining of the two systems leads to capabilities of â€Å"familiness† that cannot be duplicated, and which contribute to the survival and growth of family businesses. A resource-based view of the family business explains how it identifies and develops capabilities of familiness, how they transfer those capabilities to new leaders and new family business structures, and their methods of renewing their capabilities during the transformation in circumstances and conditions. On the other hand,

The Visit to Kenya Essay Example | Topics and Well Written Essays - 1000 words

The Visit to Kenya - Essay Example The country had a remarkable and beautiful climate. The wide distribution of green plants all over the place shocked me. The trees and birds were beautiful against clear bright blue skies, this contrasted sharply with Saudi Arabia, and I realized that Africa is so much blessed with a remarkable natural environment. They do not require artificially controlled environments in their homes. The temperatures are exceptionally cool at all times of the day, and I appreciate my cousin for insisting that I visit him in this wonderful part of the globe. I had always been fascinated by the majestic lions since I learned about them in Pre-School and I now had an opportunity to spot one in the wild. After traveling for several hours in the Tsavo plains, our tour guide suddenly told our driver to slow down, and he pointed towards a large tree. The sight was beautiful; I saw the largest lion I have ever seen resting under the tree. We saw very many other beautiful animals in the wild. The clouds opened on our way from the recreational area. It was such a superb thing with a mixture of so many pleasant smells around me. I was taken back to reality when our vehicle skidded and came to a sudden stop, and our driver declared that we were mud stuck. We alighted from the vehicle and Mabrouk solemnly declared that only a tow truck could remove the tour van from the muddy pothole. After several minutes, a number of Africans came clad in bright red clothing and after talking with the driver, for a minute, they succeeded in removing the van from the pothole.

U.S History Review Essay Example for Free

U.S History Review Essay It is important to study the history of the U. S because it helps us to learn about the people of America and even about American societies. It also helps us understand how these people or the societies usually behave. This is because history tends to base evidence on analyzing and contemplating about how societies function not only nowadays, but from the past. Examining the history of the U. S can help us understand the changes of the society from the very past up to the state they are in today (Crabtree, 1993). Further, studying U. S history can make us understand Americas’ political progress from the past to date. We thus can be informed on the shape of the politics in the past and the major developments. Moreover we can even look into the factors that caused changes in the various American political parties. Conversely, historical analyses of voter turnout in key American elections and the associated evolution can make us understand some of the problems American people face today. This can also enable us to understand the present political health of the U. S (Steele, 2009). Moreover, we can understand economic changes in America and factors or policies that the government used to see their economy rising. This information can also help any other nation to improve their economy (Crabtree, 1993). Conversely, the cultural values of the American people can be understood by studying the history of U. S. We can understand the issues of the past by having a look at how people used to live in the past ages which serve as a sense of excitement and beauty (Steele, 2009). Studying the history of America can make one grow psychologically. This can broaden one’s mind besides making somebody to understand how they can solve problems as they come using facts that were used by philosophers in the past (The Manhattan Institute). References Crabtree, D. (1993). The Importance of History. Retrieved 8th June 2010, from http://www. mckenziestudycenter. org/society/articles/history. html Steele, D. D. (2009). The Importance of Local History. Charleston, SC: BiblioBazaar, LLC. The Manhattan Institute. Why Study War? Victor Davis Hanson, City Journal Summer 2007. Retrieved 8th June 2010, http://www. city-journal. org/html/17_3_military_history. html

French Essays Egalitarian Political Regimes

French Essays Egalitarian Political Regimes Explain and Discuss the Fragility of Egalitarian Political Regimes, as Represented in BOTH the Lettres Persanes AND the Contrat Social. Though The Spirit of Laws is probably the best-known work of Charles de Secondat, Baron de Montesquieu, his Lettres Persanes (Persian Letters) is another famous work in which he explores, with perhaps more depth, the notion of equality and egalitarian political rule. A generation later, John Jacques Rousseau would appear on the political landscape and present his own ideas on the same topic. Chief to be explored among his writings will be the Contrat Social (Social Contract) in which Rousseau lays out with some detail a discussion of the nature of egalitarian political regimes and explores various strengths and weaknesses of them. Montesquieu and the Fragility of Egalitarianism In the beginning of the 89th letter, Montesquieu claims that â€Å"A Paris rà ¨gnent la libertà © et l’à ©galità ©.† Birthrights, social ranks, and even military victories did not set men apart (in terms of class distinctions) in Paris during his writing. This was a thing to be praised by Montesquieu. He saw too much in the world that lent itself away from egalitarianism, at least insofar as the right of persons to be equal is concerned. It will be beneficial here to take a moment to set up Montesquieu’s views on the republic to better lay a foundation for his comments on equality. In Book 11 of the Spirit of the Laws, Montesquieu explores the (then) unique situation in England of a monarchy controlled, to an extent, by a constitution, and it that portion of the Spirit of the Laws Montesquieu is chiefly impressed by and concerned with the Englishman’s â€Å"liberty.† As regards the very nature of a republic, Montesquieu argues in the Spirit of t he Laws that there are three basic types of governmental systems. The despot rules by inculcating fear in the people. The monarch does better and rules by a sense of honor and by â€Å"fixed established laws.† Both of these types of governing are fairly stable. One does not need to necessarily think of them as intrinsically fragile in the sense in which, say, the last political option (i.e., the republic) may be thought to be fragile. The despot, so long as he maintains fear amongst the peoples, has nothing to fear himself. Apparently for Montesquieu, it is the monarchy which is the first and primary type of government. He writes in Letter 131 of the Lettres Persanes, â€Å"Les premiers gouvernements du monde furent monarchiques.† Coming on the heels of this original type of government would be both the despotic rule and the republic, the latter of which comes by â€Å"chance,† he indicates. Apparently, despotism amounts to little more than a degeneration of an original monarchy. But, the republic is a genuine advancement of the Greeks. However, this advancement brings with it an intrinsic tendency toward reversion to that which preceded it, either monarchy or despotism, and this fact may be due to the complexity of the republic in both its nature and principles. For Montesquieu, one of the things that may typify the fragile nature of the republic is that it â€Å"cannot survive without what Montesquieu calls political virtue.† It is this requirement that the citizens must embody this political virtue (without which the republic could not endure) that lends to the fragile nature of republics. If the people cease persisting in this virtue, the republic could not endure, for the republic exists and continues only so long as the habits and eventual character of political virtue are exemplified in the people. In the republic, there is no one-to-one correspondence with what exists in despotism or a monarchy: a strong central authority. Therefore, the people must, by loving egalitarianism and the laws, arrange a situation for themselves wherein the needs of the good are served, even if at the expense of the needs of the many. This is exactly what Greece did, he argues, and it is incumbent upon any subsequent attempts at a republic to do the same. â€Å"L’amour de la libertà ©, la haine des rois, conserva longtemps la Grà ¨ce dans l’indà ©pendance, et à ©tendit au loin le gouvernement rà ©publicain.† Rousseau and the Fragility of Egalitarianism One could hardly resist beginning the discussion on Rousseau with his famous opening to chapter one of the Contrat Social. â€Å"Lhomme est nà © libre, et partout il est dans les fers.† How this particular situation came to be, Rousseau does not attempt to answer. Rather, he focuses his attention on how it is that man can get back to his original (or perhaps â€Å"primal†) state of freedom. If man in a state of servitude obeys his masters, he does well. However, if he can break free from that state, he does better still because to be free is man’s natural and original state, seen most evidently within the rites of passage intrinsic to family life. Although it could not be rightly said that Rousseau takes no points of departure from the thought of Montesquieu, there are nevertheless significant points of agreement between them on the idea of the republic. Rousseah offers as his main contribution to the discussion over the republic that a return to the ancient (i.e., Greek) polis is the most advisable course of action. Yet, an intrinsic tension to this suggestion is that Rousseau simultaneously advocates the idea of the â€Å"natural law† quite strongly. According to Helena Rosenblatt, for Rousseau the natural law is a very self-interested concept, which is at least prima facie at odds with the republican ideal of each person being grounded in virtue and community as that which adheres the republic together and maintains it. The more refined concept of the â€Å"general will† complicates the matter further and makes egalitarianism a la republicanism an even more fragile thing. Rousseau’s â€Å"General Will† In his writings prior to the Social Contract, Rousseau had explicitly indicated that he denied that man was naturally and easily a sociable creature. No, man’s first inclinations are not toward the public good, but in the direction of particular self-interests and this is evident by the historical facts that â€Å"les longs dà ©bats, les dissensions, le tumulte, annoncent lascendant des intà ©rà ªts particuliers et le dà ©clin de lEtat.† So, what takes place amidst the social contract is the necessity of all citizens when laying down public policy to not act in merely self-interested ways. The good of the many, the common good, was to be the overriding concern of all citizens in this regard, and this is the â€Å"general will† of Roussea, which he explores and elaborates in great throughout the Social Contract. But, what makes this concept of the â€Å"general will† even more tense and lending to the creation of a fragile situation for egalitarianism i s the paradoxical idea related to literally enforcing that citizens act in accord with the general will. The general will is not merely reducible to the â€Å"will of all people combined.† No, it is the â€Å"right† will which ever seeks the good of the whole State and never acts in a merely self-interested way. It is basically the will of God then, which must ever be right and, since God is omnibenevolent and always has the interests of everyone in mind, this is in line with the general will as Rousseau explicates it here. He writes, â€Å"Afin donc que le pacte social ne soit pas un vain formulaire, il renferme tacitement cet engagement qui seul peut donner de la force aux autres, que quiconque refusera dobà ©ir à   la volontà © gà ©nà ©rale y sera contraint par tout le corps: ce qui ne signifie autre chose sinon quon le forcera dà ªtre libre.† This is the key to the whole enterprise. It prevents the social contract from becoming, as he says, â€Å"un vain formulaire† (an empty formula). But, of course, although such an aspect of the overall contract is certainly sensible, how it is appropriated lends itself to fragility. The line is not always so clear when one is acting merely in his own self-interest and when he is acting in respect to the common good (or both simultaneously, which would apparently not violate the general will). It is not necessarily contradictory in its premise, but it is certainly paradoxical, as Rousseau surely felt. Conclusion Both Montesquieu and Rousseau in their respective days were vastly aware with the attending problems associated with the reintroduction of the ancient ideas of the republic and egalitarianism. However, they each firmly believed that whatever problems may accompany the advent of such in Modernity, it would certainly be worth it. For both of them, as most Westerners today would greatly sympathize, any form of egalitarianism via a republic, whatever fragility may accompany it, would be greatly preferable to either a monarchy or (especially) a despotic State. Works Consulted Krause, Sharon. The Politics of Distinction and Disobedience: Honor and the Defense of Liberty in Montesquieu, Polity 31, 3 (1999): 469-99. Grant, Ruth Weissbourd. Hypocrisy and Integrity : Machiavelli, Rousseau, and the Ethics of Politics. Chicago: University of Chicago Press, 1999. Morris, Christopher W. The Social Contract Theorists : Critical Essays On Hobbes, Locke, and Rousseau. Lanham, Md.: Rowman Littlefield, 1999. Riesenberg, Peter N. Citizenship in the Western Tradition : Plato to Rousseau. Chapel Hill, NC: University of North Carolina Press, 1992. Rosenblatt, Helena. Rousseau and Geneva : From the First Discourse to the Social Contract, 1749-1762. Cambridge ; New York: Cambridge University Press, 1997. Shklar, Judith. The Spirit of the Laws: necessity and freedom. In Montesquieu, pp. 93-110. Oxford: Oxford University Press, 1987.

Sales Tools

This is a report to understand the role of personal selling within the overall marketing strategy such as Promotion mix: personal and impersonal communication; objectives of promotional activity, push-pull strategies; integrating sales with other promotional activities; evaluating promotion; allocation of promotion budget and Understanding buyer behaviour: consumer and organisational purchase decision-making. And also this report involved the role of the sales team: definition and role of personal selling; types of selling; characteristics and have link with LEVI’S case study.After reading this report reader could understand to how to apply the principles of the selling process to a product or service such as customer-oriented approach; objective setting; preparation and rehearsal; opening remarks; techniques and personal presentation; need for identification and stimulation; presentation; product demonstration and use of visual aids; handling and pre-empting objections; techn iques and proposals for negotiation; buying signals; closing techniques; post sale follow-up; record keeping; customer relationship marketing (CRM)And also reader could be able to understand the role and objectives of sales management such as, Controlling sales output: purpose and role of the sales budget; performance standards: performance against targets (financial, volume, call-rate, conversion, pioneering); appraisals; self-development plans; customer care.Reader also will have Be able to understand how to plan sales activity for a product or service in ways such sales settings: sales channels (retailers, wholesalers, distributors, agents multi-channel and online retailers); importance of market segmentation: business-to-business (BTB) selling; industrial selling; selling to public authorities; selling for resale; telesales; selling services; pioneering; systems selling; selling to project teams or groups international selling: role of agents and distributors; sources, selection and appointment of agents/distributors; agency contracts; training and motivating agents/distributors; use of expatriate versus local sales personnel; role, duties and characteristics of the export sales team; coping in different cultural environments;the role of ICT in communicating with an international sales team Exhibitions and trade fairs: role, types and locations of trade fairs and exhibitions; how trade fairs and exhibitions fit in with corporate strategy and objectives; setting objectives for participation in an exhibition; audience profile and measurement; qualification and follow-up of exhibition leads; evaluation of exhibition attendance; setting budgets; financial assistance or exhibition attendance etc.. to a certain extend.Table of Content Personal selling, promotion mix, buyer behaviour and the decision making process in different situations ,role of sales teams within marketing strategy Role of sales teams within marketing strategy ow sales strategies are developed in line with corporate objectives Importance of recruitment and selection procedure Role of motivation, remuneration and training in sales management Sales plan for a levi’s male cap Conclusion Bibliography Personal selling, promotion mix, buyer behaviour and the decision making process in different situations ,role of sales teams within marketing strategy Personal selling is a promotional method in which one party (e. g. , salesperson) uses skills and techniques for building personal relationships with another party (e. g. , those involved in a purchase decision) that results in both parties obtaining value.In most cases the â€Å"value† for the salesperson is realized through the financial rewards of the sale while the customer’s â€Å"value† is realized from the benefits obtained by consuming the product. However, getting a customer to purchase a product is not always the objective of personal selling. For instance, selling may be used for the purpose of simply delivering information. Marketers have at their disposal four major methods of promotion. Taken together these comprise the promotion mix. Those four promotion methods are 1. Advertising 2. Sales Promotion 3. Public Relations 4. Personal Selling So, Personal selling comes under promotional mix. That means personal selling is a part of promotional mix.In LEVI’S case also LEVI’S company might definitely use promotional mix and they can improve the promotional mix by improving personal selling that both aspects have a positive relationship. Buyer behaviour consists of activities/process followed in making any buying decision of goods as well as a service. In LEVI’S case buyer behaviour would be 1. Type One: Traditionalist – probably over 45; buys from department stores; buys polyester suits and trousers; shops with his wife. 2. Type Two: Classic Independent – a real ‘clothes horse’; 21% of the market but buys 46% of wool blend suits; shops at independent stores; has expensive tastes. 3. Type Three: Utilitarian – wears jeans for work and leisure; 26% of the market; a Levis loyalist. 4.Type Four: Trendy Casual – buys ‘designer’ high fashion labels; might buy 501’s but usually considers Levi’s too mass market; 19% of the market. 5. Type Five: Price Shopper – buys the lowest price product wherever they may be; no potential for Levi; 14% of the market. There are five types of buyer behaviours and they have different types of decision making in different situation. One thing needs to be highlighted here is that consumer behaviour does not end with purchase of goods or service, but also post purchase activities are included in consumer decision making. The below image shows the different type of buyer behaviour. Role of sales teams within marketing strategy Marketing Strategy is something that helps companies achieves Marketing objectives.Marketing objectives help achieve corporate objectives and corporate objectives aim to achieve a competitive advantage over rival organizations. Firstly, a Managing Director or senior management team, or executive board of directors (who ever is in charge) decides on overall corporate objectives. One corporate objective might be to increase sales by some percentage. In order to achieve this percentage sales team always work towards that. So the role of sales team would be work to corporate objective goals and marketing strategies. LEVI’S case study team have adopted to the marketing strategies and LEVI’S ‘ marketing team responded to this information by deciding to focus on jackets and trousers only for the launch and let suits ‘slipstream’.The Director of Consumer Marketing stated: ‘The thing that is going to overcome Levi’s image for casual only apparel is a suit made by Levi that doesn’t look like anything else we have ever made. Once that gets on the r acks people will put an asterisk on the image that says Oh, they can also make a good suit when they put their mind to it’. So sales responded according to the marketing strategies ‘Soon after this decision, salesmen started contacting retail buyers. After 4 months of selling to the trade it was clear sales targets for the range would not be met.A subsequent price reduction failed to redeem the situation and Tailored Classics achieved only 65% of its sales targets. so role of sales team would be always adopted to mass marketing strategy. ’ ow sales strategies are developed in line with corporate objectives Whenever a person wanted to develop sales strategies he/she should always should develop in line with corporate objectives, so, he/she should keep these below mentioned tips before they make sales strategies. Do you have the time and resources necessary to interact? Can you competitively position yourself to overpower the competitor? Do you have sufficient uniq ue selling points to change the rules and do you have time to accomplish this? Do you have access to the key decision makers to do this? Can you sufficiently quantify your unique selling points to prevent price pressure?Can you neutralize the competitor's strengths? Do you have the capacity to deliver? What would it take to make this a viable opportunity? Importance of recruitment and selection procedure The recruitment is a process of finding and hiring the best-qualified candidate (from within or outside of an organization) for a job opening, in a timely and cost effective manner. The recruitment process includes analysing the requirements of a job, attracting employees to that job, screening and selecting applicants, hiring, and integrating the new employee to the organization. Importance of the recruitment are listed below * Recruitment is the process which links the employers with the employees.Increase the pool of job candidates at minimum cost. * Help increase the success rat e of selection process by decreasing number of visibly under qualified or overqualified job applicants. * Help reduce the probability that job applicants once recruited and selected will leave the organization only after a short period of time. * Meet the organizations legal and social obligations regarding the composition of its workforce. Begin identifying and preparing potential job applicants who will be appropriate candidates. * Increase organization and individual effectiveness of various recruiting techniques and sources for all types of job applicantsRole of motivation, remuneration and training in sales management Motivation means Internal and external factors that stimulate desire and energy in people to be continually interested and committed to a job, role or subject, or to make an effort to attain a goal. Motivation results from the interaction of both conscious and unconscious factors such as the * Intensity of desire or need, * Incentive or reward value of the goal, a nd * Expectations of the individual and of his or her peers. These factors are the reasons one has for behaving a certain way. Role of motivation in sales management is motivate sales people towards bringing business to the organization. The most widely recognised method of attracting and retaining marketing employees is through remuneration packages.This is because a good emuneration package can have a positive effect on a person's standard of living. Remuneration refers to money or substitutes for money. This may include wages, salaries, commissions and bonuses, incentive plans and allowances. The total value of all these items is called a remuneration package. Sales plan for a levi’s male cap A sales plan LEVI’S cap is basically should be a strategic and tactical plan for achieving market because of the higher price objectives. It should be a step-by-step and detailed process that will show how you will acquire new business for LEVI’S caps; and how you will g ain more business from the existing customer base.It involves making and/or exceeding our sales quota within our sales territory how to penetrate to international business. Here are the steps to help you develop a sales plan for LEVI’S male caps: 1. Segment the target audience – The first step is to clearly identify the target markets. Who are more likely to buy LEVI’S caps. The more defined the target market, the better. Target market can be defined as high-income men ages 30-60 who loves to buy the latest electronic gadgets. If the target market is done, the next step is to prioritize the target market to ensure that resources are directed towards your key target market. Prospects are more likely to purchase if LEVI’S can talk features of male cap. 2.Cap industry – Current trends in industry, and how the business fares with these trends. LEVI’S must look at the range of similar products now available, and compare how the product stacks up to competitor products. Take a bigger picture of the industry and find out prospects. 3. Develop sales strategies – LEVI’S sales strategies must include determining how LEVI’S can reach the sales quota, how can get more sales from existing customers, and how can raise awareness in the marketplace and community about your business. Your sales strategies also involve making a decision on who is actually going to do the selling. Will you do it skills to do it? how to improve confidence and presentation skill etc†¦ 4. Think through sales plan.This is the meat of the sales plan. LEVI’S need to write down the sales strategies based on the analysis you’ve done of your business and what you can do. Will you attend trade shows or do cold calling? Will they use search engine advertising or other forms of online advertising? How they going to reach exactly to target audience? It is important that LEVI’S break down the sales strategies into quantif iable activities . 5. Measure and improve. A sales plan is not something that LEVI’S create and then hide inside the drawer. Should follow the activities and tasks outlined in sales plan. All the planning in the world is worthless if they cannot or do not implement any of it.Keep track of how they are performing vis-à  -vis sales target.. Conclusion This report have explained the role of personal selling within the overall marketing strategy such as Promotion mix: personal and impersonal communication; objectives of promotional activity, push-pull strategies; integrating sales with other promotional activities; evaluating promotion; allocation of promotion budget and Understanding buyer behaviour: consumer and organisational purchase decision-making. And also this report involved the role of the sales team: definition and role of personal selling; types of selling; characteristics and have link with LEVI’S case study.Report also explained how to plan sales activity fo r LEVI’S male caps in ways such sales settings: sales channels (retailers, wholesalers, distributors, agents multi-channel and online retailers); importance of market segmentation: business-to-business (BTB) selling; industrial selling; selling to public authorities; selling for resale; telesales; selling services; pioneering; systems selling; selling to project teams or groups international selling: role of agents and distributors; sources, selection and appointment of agents/distributors; agency contracts; training and motivating agents/distributors; use of expatriate versus local sales personnel; role, duties and characteristics of the export sales team; coping in different cultural environments; the role of ICT in communicating with an international sales team Exhibitions and trade fairs: role, types and locations of trade fairs and exhibitions; how trade fairs and exhibitions fit in with corporate strategy and objectives; setting objectives for participation in an exhibi tion; audience profile and measurement; qualification and follow-up of exhibition leads; evaluation of exhibition attendance; setting budgets; financial assistance or exhibition attendance etc..

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

Romanesque Art essays

Romanesque Art essays During the Romanesque period, art most often portrayed biblical events or depicted saints and other religious imagery. There were also several different styles of Romanesque art. The development of these styles can be credited to the geographic area in which the art was made, the audience that the art was made for, and the message that the art was trying to portray. There are three images from the Romanesque time period that illustrate these styles very well. The first image is Virgin and Child from the Auvergne region of France (c.1150-1200), a sculptural piece whose style was developed because of audience. The next piece of art is the South Portal of the Priory Church of Saint-Pierre in France (c.1115-30), which was commissioned by Abbot Roger and portrays a message. The final piece of art which displays a style based on where it was made, is Creation and Fall (c.1106-20) in the Modena Cathedral in Emilia, Italy. It was crafted by the sculptor Wiligelmus. Virgin and Child was made from oak and originally painted, but now there are only faint traces of color on the surface. It was made to attract audiences who were on pilgrimages or crusades. During this time period, churches received their funding from visitors. They had to find a way to attract these people to the church, and one way to draw visitors was to utilize the relics in the churchs possession. Virgin and Child shows the Virgin Mary as Christs throne, which was known as the Throne of Wisdom (Stockstad). Jesus sits on her lap with a book entitled The Word of God and he is raising his left hand as if he is giving praise to those who view the statue. The sculpture also has two cavities; one is located in Marys chest and the other is behind her shoulder, both of which might have contained holy relics at one time. Besides being a container for church relics, this sculpture was probably used in church processions as well. A...

Clark And Menefee, Architects Essays - Menefee, Formwork, Plywood

Clark And Menefee, Architects Essays - Menefee, Formwork, Plywood Clark And Menefee, Architects Maggie Cookman September 27, 2000 The Reid House was designed by W.G. Clark and Charles Menefee and built in Johns Island, SC in 1986. Menefee and Clark designed primarily in the American South. Clark and Menefee are known for their tripartite vertical organization. The base level normally consists of secondary bedroom(s)/studio spaces and services. The First floor is a piano nobile of principal rooms with a double-height living space. The attic level usually consists of the master bedroom and bath. The Reid House is set up in this fashion. The house is located in a modest setting, surrounded by house trailers and cheaply built houses. The image of the house was derived from vernacular farm buildings as well as from more formal Palladian structures. One author described the setting as Johns Island, a peaceful landscape where truck farmers tend tomato fields carved out of scrub-pine and dwarf-cedar forests, and where the front yards of shacks are littered with junked cars, rusting agricultural machinery, and other dec aying impedimenta of the Industrial Revolution. The house is a three-story tower with two components. The first is a 20 ft. sq. section made of concrete block, housing the living and bedrooms, referred to as the served space(s). The second part, referred to as the serving space(s), is a wood-frame shed that holds the kitchen and the bathrooms. These two components are joined at the fireplace and chimney, around which the stair winds. The materials used for the house are inexpensive, in keeping with the surrounding structures. One section is made of concrete blocks, exposed on the inside and covered with waterproofing paint on the outside. The other part of the house is sheathed in plywood and battens and its roof is covered in asphalt shingle. The floors are painted pine, the interior partitions, painted plywood. The total cost of the house was $102,000, only $2,000 over the budget that the Reids had set. They wanted the house built because they wanted to move their two small children out of a trailer home, and they wanted to have a larger space in which they could manage their 120-acre horse farm. The total area of the house is only 1600 sq. ft. One author noted that the house [reconciles] lofty aspirations and modest means. W.G. Clark is not a native to Charleston. He worked for six years for Robert Venturi before going to work with Charles Menefee on the Middleton Inn for Charles Duell. This project was Clarks first major work, and was more in tune with the work of Peter Eisenman. Charles Duell, a Middleton descendent, dreamed up the idea of the Middleton Inn, 15 miles outside of Charleston. He envisioned a guesthouse and conference center, and planned on seasonal guests who came for flower festivals and other annual events. The Inn was remote from city tourist attractions, and Clark capitalized on this and made it a rural retreat in the woods. The Inn was filled with Charleston details, which helped to bridge the gap between the city and the rural hideaway in the woods. These details included terra-cotta chimney pots, wooden shutters, stick-style furniture, special stucco called slave coat, and Charleston Green paint, which accentuated the building in the midst of the trees and growth in the surroundi ng woods. Clark and Menefee exemplified an uncommon American virtue, restraint. Their structures had a simple and clear formal order, and were compact in plan. Their belief was that generosity was achieved in section. In describing their architecture, one critic notes that Clark and Menefees buildings distil a didactic language through which both formal meaning and construction can be revealed and understood. It was also said that their houses were idealized pavilions sitting solidly on the site in the classical manner. Their designs were small and succinct, and interior finishes were sometimes rough, but their craft was excellent. Clark and Menefee succeeded in practical designs, while economizing on budgets and space.

Operations Management Essay Example | Topics and Well Written Essays - 1250 words - 1

Operations Management - Essay Example The Concept of Operational Management The operations management as a business function has many issue related to it that needs to be satisfied to increase the efficiency and productivity of a business concern. These are capacity requirements, technology, facilities, workforce, supply and distribution, quality, production planning and organizational structure. According to Shim (1999, Pg.4)â€Å"Operation policy is concerned with setting broad policies and plans for using the production resources of the firm to best support the firm’s long term competitive strategy†. However, the four basic issues of operation management strategy are as follows. a) Cost : The cost of a product or service decides the profit or loss of a company. In every market segment, there is a class of people who opt for low -cost product or service. So to compete on this basis, a company should produce goods and services which are of low- cost. b) Quality : Needless to mention, quality is a prioritiz ed element of any industry. Quality is classified into product quality and process quality. Product quality ensures better customer satisfaction and process quality make sure that products are defect free and is produced with total quality management. c) Speed of delivery : The speed with which a company delivers products and services to its customers decides the purchasing decision of a customer. The capability of a business firm to deliver products and services on a fast basis decides the price of their product and the company reputation. d) Flexibility : By the term flexibility, we mean that a company should offer various types of products and services to its customers. It also means how quickly a company can convert old products to new one to meet customer demands. These issues play an important role in understanding the challenges an operational manager can face in this competitive world. Challenges faced by Operational Managers a) Absence of Capital – Capital is the mai n resource of any business and this strengthen the foundation of a business. Capital is required from the launch of a business firm and al the strategic plans and operation of a business largely depend on capital availability. An operations manager needs enough capital to work in a full fledge way. An absence or lack of capital can restrict an operational manager in many ways. An operational manager with less available capital will contribute inferiorly to a business. Absence of capital can restrict an organization from many perspectives. A company with inefficient capital can offer fewer choices and solution to an operations manager. If a company is financially weak, the managers can dwindle on decision- making and planning proposals. b) Lack of efficient planning - Planning is crucial to business as it starts with proper intelligent planning. If planning is not proper, managers cannot align their functions with the goal of the firm. Planning needs to be extensive and if organizati onal leaders do not plan ahead of execution, then the business can doom to failure. If the basic decision - making is defective then an operations manager can do less about organizing and operating a company. An operations manager has the full responsibility of business functioning, if the planning is not well versed then the flow of business is interrupted

Marine Dolphins Research Paper Example | Topics and Well Written Essays - 1250 words

Marine Dolphins - Research Paper Example Other dolphins have a varied diet that may include lobsters, crabs, squid, shrimps and fish. The dolphins bear live young ones and feed the young ones with milk. Many marine dolphins are facing extinction due to human activities, but some laws have been formulated to protect their existence (Nakamura 64). Types of Marine Dolphins The major types of marine dolphins are pacific bottlenose, rough-toothed, spotted dolphin and spinner dolphin. The bottlenose dolphins are the most famous of all cetacean species. The scientific name for bottlenose dolphin is Tursiops truncatus. Rough-toothed dolphin has vertical grooves that run from the gum line to the tip. The scientific name for rough-toothed dolphin is Steno bredanensis. Spotted dolphins are of smaller size than rough-toothed or bottlenose dolphins. They have a dark cape that extends from their forehead to the dorsal fin. The scientific name for spotted dolphin is Stenella attenuata. Spinner dolphins are smaller than the spotted dolphin s. The spinner dolphins got their name from their spinning behavior. Their scientific name is Stenella longirostris (Shirihai 77). Adaptations of Marine Dolphins Like other marine animals, dolphins have become adapted to the sea life. Anatomically, the bodies of the dolphins are streamlined to enable them to move more efficiently in the aquatic environment. The streamlined body ensures fast movement of the dolphins as it reduces the resistance of water when the dolphin is moving. The hind limbs of the dolphins have disappeared, and front limbs are developed into flippers (Gordon 26). This helps the dolphin in steering balance and changing direction. The dolphins have a powerful tail, which helps in propulsion. The speed under which the dolphins move at is closely related to the feeding habit of the dolphin. Research has shown that those dolphins that feed on slow moving prey do not exceed a speed of 10 mph. Other dolphins that feed on fast moving fish species usually attain a speed of up to 15 mph. In addition, dolphins have a smooth skin. The dolphin’s skin is thick, hairless and lacks glands. It is kept smooth by the constant slough and replacements. For example, the bottlenose dolphin replaces its outermost skin layer after every two hours. This smooth skin of the dolphin increases the speed at which the dolphin moves. Like other marine mammals, the dolphins have a thick layer of blubber below their skin. This layer of blubber is important since it insulates the dolphins against heat loss. The dolphins rely on their sense of hearing to detect any danger around them. The sense of hearing in dolphin is a very complicated behavior referred to as echolocation. Echolocation is where, a dolphin emits a series of split clicks by its blowhole focusing the sound to the melon and the melon directs the sound pulses in a directional beam. When the directed sound waves hit an object on its path, it is reflected back to the dolphin, and the dolphin detects an obst ruction ahead on its way. Through echolocation, a dolphin is capable of detecting the distance, size, shape and direction of movement of objects in water (Smolker 54). Factors Affecting Marine Dolphin Population Tuna and Dolphin Issue Dolphins and Tuna often swim together. The tuna tries to take advantage of food finding tactics of