The Role of Accounting in Organizational Control

THE ROLE OF ACCOUNTING IN ORGANIZATIONAL CONTROL By: Abdur Rehman (IIUI) Table of Contents Introduction 3 Accounting4 Key Theories in Organizational Control6 * Key Theories8 Role of Accounting in Organization’s Control10 * Working of Control Theory at Workplace12 Conclusion13 References 14 online sources15 Introduction Accounting is the gathering and compilation of information for decision makers – including investors, managers, lenders, public and the regulators. Accounting schemes influence performance and management and have affects on organizations, departments, and even countries.Information controlled within an accounting scheme has the influence to manipulate actions. Accounting information systems are largely strong behavioral drivers in the perspective of a corporation – where bottom line and profits are every day concerns. In this paper we will discuss the role of accounting plays in organizations control. Accounting has impact on each and every aspect of the organization from record keeping to manufacturing and from sales to further investment.It is accounting concepts and details in shape of Financial Accounting, Cost Accounting, Management Accounting or whatever you may call which gives insight of an organization for the stake holders to think and plan for future. Accounting prompt owners to expand, managers to plan, equity investors to invest, bankers to lend, tax officials to calculate and buyers to spend etc. we will briefly discuss different calculation concepts which are based on the accounting data which effect investment and leads to ultimate organizations control. Organizations are sets of agreements among ersons or groups who are aggravated by self-interest for the collection of resources and contribute to the information to achieve control in the organization. Accounting and control provides a similar utility in organizations. We shall try to highlight some control theories and role of accounting in up-coming pages. A ccounting Accounting is the main means of serving managers of an organization, potential equity investors, creditors and bond holders, equity investors, creditors and bond holders of an organization, customers and suppliers of an organization and other stake holders to take decisions.Accounting is somewhat to facilitate people in their individual lives a lot as it affects organizations. We all use bookkeeping thoughts when we map what we are going to accomplish with our wealth. We have to chart out how much of it we will use and how much we will keep. We may engrave a chart, recognized as a financial plan, or we may merely maintain it in our minds. Though, when people usually talk about accounting it means the kind used by trade and other business organizations who also keep a record of it which is then used by different stakeholders including owners, buyers, investors, tax officials, banks etc. or a variety of rationales. (Frank Wood, 1997, p 4) Generally speaking Accounting can be defined as â€Å"distinguishing, manipulative and matching economic information to let well-informed selection and decision by users of the information. (Frank Wood, 1997, p 253) Accounting can also be defined as â€Å"the language of business† because it is the essential device for recording, coverage, and evaluating economic proceedings and dealings that influence business ventures.Accounting procedures gather all the aspects of a business's financial presentation including from capital expenses, payroll costs, and other responsibility to sales proceeds and owners' equity. Financial data enclosed in accounting documents is then interpreted and used as essential in reaching to the actual picture of a business's true financial comfort. Equipped with said meaningful knowledge, businesses can make suitable financial, managerial and strategic decisions about their future opportunities; on the other hand, unfinished or incorrect accounting data can destroy a company, no matter its volume or orientation.Accounting's significance as an indicator of business health in provisions of precedent, current, and upcoming and tool of business steering is replicated in the language of the American Institute of Certified Public Accountants (AICPA), which distinct accounting as a â€Å"service activity. † Accounting, said the AICPA, is projected â€Å"to provide quantitative information, largely financial in character, about economic actions that are intentional to be helpful in reaching at economic decisions, making logical choices among alternative courses of action. † Key Theories in Organizational ControlOrganizations are sets of agreements among persons or groups who are aggravated by self-interest for the collection of resources and contribute to the information to achieve control in the organization. All wonders of the world are physical substance which we could see or touch. Civilization has always calculated the deeds of its evolutions by â€Å" hardware,† from the Pyramids to the Great Wall of China. But, the truth is that world is never ruled by the brilliant engineers rather these greatest achievements of human kind are due to the organization who made these Pyramids or the Great Wall possible.Similarly bravery of Alexander the Great is legend, but it was organizational skill that made him so great his skills with the sword or horse. Organization plan and process are the unacknowledged â€Å"software† of human evolution. Combining a lot of individuals for a common purpose is virtually impossible in-spite of their engineering skills without such software when those individuals are free to select and proceed for their own goals. Accounting and control are the key factors in organization in terms of how they are put together, function, sustain, transformed, and suspend.Recurring to the computer symbol, if humans and capital are the hardware of organizations, accounting and control are their operating software. Software unites different parts of the computer in their proper association, and makes it possible to function. Accounting and control provides a similar utility in organizations. Considering organizations as a group of agreements among people crafts it easier to see the enabling utility of accounting and control. As organizations are in hundreds of sizes and forms, so the same exists for accounting and control systems.A broad survey of their mixture and degree is a fine starting point to construct a theory of accounting and control in organizations. A distinct feature of control theory as compare to other personal theories is that it is being derived from the working of mechanical processes. Current modern control theory was originated by Norbet Wiener’s 1948 Cybernetics. Initially it was applied to physical systems however same was applicable to human behavior. It connects multiple disciplines like mathematics, neurophysiology, anthropology, psychology, biology, electrical engineering, etc. t has the capacity to link all such field combine to describe how a system work as a whole. Same is done by dividing the system into specific pieces and studying the relations and connections among them. For the control theory to be applicable on humans, it is important to show that same kind of direct similarities exists among the concepts linked with machines that can also be associated with humans. Certain assumptions are required for control theory to work including; I. Human beings have a system in themselvesII. Society has also a system III. These systems are self amendable This overview shows a simple and easy look of control theory, however the functioning of the theory is quite complex in reality. Therefore, it is necessary to outline and review the components for a greater considerate as well as to be able to relate the theory in helpful, practical, and appropriate ways. Researchers have recognized that control processes are the main and essential ways in which organizations work.As control theory exploration ranges on many years and based on a number of rich customs, hypothetical limits have reserved it from producing reliable and interpretable practical conclusions and from reaching agreement relating to the character of key associations. However it reveals that we can overcome such problems by synthesizing varied, yet matching, flow of control researches into an academic framework and experiential tests that can more explain the variety of control methods (e. g. the use of norms, rules, monitoring or direct supervision) meant for specific control targets (e. . behavior, input, output etc. ) are applied within specific kinds of control systems (i. e. integrative, bureaucracy, clan, market). Before going into further details, firstly we shall look upon the characteristics of an effective control system. Any organization must develop control systems which contribute to the organizational goals and resources. There are several charac teristics, some of which are described as below; I. Focus on vital points: Vital points include all the fields of an organization’s operations that directly contribute to the success of its operations.For example, controls must be applied where costs can’t be exceeded from a certain amount. II. Incorporation into recognized procedures: For effective controls, procedures must be in line and not disturb operations. III. Acceptance by internal customers: Involvement of employee during making of control design increase its acceptance among internal customers. IV. Timely availability of information: Project costs, deadlines, time period to complete the project, associated costs, priority, etc. must be available in devising any control mechanism.Any deviation or uninformed workers mostly contribute for failures. V. Economic feasibility: Effective organization control systems gave answers to questions such as, â€Å"What it costs? † â€Å"How much it saves? † or â€Å"What is the return on the investment? † Shortly, comparison of costs with benefits guarantees that the payback of controls prevail over the costs. VI. Correctness: Efficient control systems give truthful information which is valid, reliable, consistent and useful. VII. Unambiguousness: Controls must be easy and simple to recognize.Control Theory is somewhat difficult to recognize in a hypothetical wisdom and trying to use it for the examination or relevance for problem solving in actual world conditions is likewise complex. Control theory has some similarity with goal setting theory, but this similarity is a more complex as it can help us to link and understood concepts easily with the more complicated aspects of Control Theory, but at the same time it causes confusion on the requirement or utilization of Control Theory. Similarly, jargons used in Control Theory is absent in other similar theories which may become a source of confusion.Furthermore, Control Theory may l ooks to be like goal setting theory, however many people have expressed their concern that control theory is more mechanistic to be applied to humans. When examining Control Theory, it must be helpful to kept in mind the similarities, but at the same time linking of concepts must also be kept examined along-with conditions or theory apparatus with care as they may become the basis for more perplexity than the advantage such comparisons produce. Simply Control Theory is a theory in itself, not just Goal Setting Theory. Below are the Key Theories of Organizational Control.Key Theories Baligaand Jaeger (1984) relates control to procedures â€Å"where a person / group or organization decide or deliberately influence what the other person, group or organization will do†. In their words, the monitoring process is at the center of control because in such situation targets are set and performance is evaluated. Control is required in order to make people act according to the best inte rest of the organization. Ouchi (1979) and his colleagues developed the dominant key theory of organization control which forecasts the option among control methods as a utility of the task perspective.Ouchi divides control mechanisms into three categories: markets, bureaucracies and clans. However it has two limitations; I. It predicts the options of control alternates but not the effect of those alternates on performance. II. It restricts those choices to a single control alternate in a specific context and do not allow the choice of a combination or portfolio of control alternates. Merchant’s (1985) define three parts classification where results and action controls looks like to some degree of bureaucratic controls and clan controls overlap with social controls.Results control is normally discussed in financial results whereas action controls include physical and administrative behavioral restraint. Social group influence, preparation, individual self-control, rewards as well as assortment and appointment procedures comprise of personnel controls. Simons (1995) has divided in four categories for control systems including use, Beliefs and boundary systems and diagnostic and interactive control. From such division, diagnostic controls looks like the traditional results controls or accounting, where measures may be non-financial.However, accounting information can also be used by the managers interactively through highlighting the most important events in usual interaction with subordinates. Boundary systems can be taken somewhat similar to action controls, and Merchant’s (1985) social controls can comprise of belief systems, which are a form of positive control posing no limitations but presenting possibility for prospect development. Klein explains the integrated control theory which defines that adding cognitive communication between the comparator and the effecter.Moreover it is the communication among the two which depicts that in humanâ₠¬â„¢s the sensor, principles, or behavior and not a set piece. Given below is the list of the main components of the integrated control theory model. I. Goal II. Behavior III. Performance IV. Feedback (Sensor) V. Comparator VI. Error VII. Attribution Search VIII. Subjective Expected Utility of Goal Attainment IX. Individual and Situational Characteristics, Goal Choice and Cognitive Change X. Behavior Change Role of Accounting in Organization’s ControlAlthough management control encompasses a broad range of mechanisms and practices besides accounting, however accounting form an essential element of the control systems. Accounting can waive off the gap between a headquarters and its subsidiaries, as it is a communication which can make local problems a global substance Therefore, accounting explains operations making performance evident and consequently handy. However, the accountability system is habitually quite a means for headquarters to keep an eye on and get involved in s ubsidiaries’ dealings than a means to chart the subsidiaries according to their individual judgment.This means that accounting also produces and constructs distance in stipulations of gap and time by signifying the accountability structures. In addition, accounting can defend a subsidiary in opposition to the headquarters as long as the exposure necessities are correctly met. Accounting and control systems plays an important role in acquisitions and mergers which is mainly used to expand business to foreign countries. In the words of Jones (1985) accounting controls may be of significant importance during the initial phase of the business after an acquisition.According to him, a lot of changes in accounting systems and control may cause a negative impact and results in undesirable outcomes. Granlund (2003) suggests that accounting system acts as important factor in a merger, as an integrator of the organizations. However Roberts (1990), advises that accounting controls may im pede the development of a profitable longer term strategy. In a business's accounting system when the reports are prepared in different formats and essential data is maneuver or recapitulate in different customs to make possible decision making.Accountants unite these data objects in a variety of ways to give information to internal as well as external users. Said data holds information potentially pertinent to a wide range of groups. Along-with business owners, who depend on accounting data to measure their enterprise's financial development, accounting data can convey relevant and important information to creditors, investors, managers, and others who coordinate with the business. Accounting is mainly used for three major purposes; I. External reporting: these reports are used by the creditors, investors, outside parties and government authorities.II. Routine internal reporting: Periodically generated reports which are used by the managers to take internal decisions. III. Non-rout ine internal reporting: Such kind of reports is generated to support decisions and different projects which come-up due to need. It is pertinent to mention that while origination of different kind of reports by using different formats, basic is summarized and manipulated to facilitate for decision making. However accounting can be broadly divided into following distinct divisions: * Financial Accounting * Management AccountingFinancial accounting is a branch of accounting that provides people outside the business—such as investors or loan officers—with qualitative information regarding an enterprise's economic resources, obligations, financial performance, and cash flow. Financial accounting measures and records business transactions and provides financial statements that are based on generally accepted accounting principles (GAAP). Executive compensation is tied to profit figures reported in the financial statements and equity share valuation is also based to a large extent on these financial statements.Management accounting, on the other hand, refers to accounting data used by business owners, supervisors, and other employees of a business to measure their enterprises’ strength and operating styles. Management accounting as a discipline focuses on accounting information that facilitates decision making by managers of the organization. Accounting data is also used in Cost Accounting which is defined as a type of accounting procedure that  aspire to incarcerate a company's expenses of manufacture by appraising the contribution costs of every step of manufacture as well as predetermined or fixed costs such as depreciation in capital equipment.Cost accounting firstly gauge and witness these costs independently, then make a contrast of input results to output or actual results to aid company management in measuring financial performance. Accounting has many roles in today’s modern business. It helps business to determine the profit o r loss for doing business in a specific time period. It helps businesses to determine the taxes which become due. Accounting is information which is used by the investors to decide whether to invest in a specific firm or not.In current world, standard accounting practices commonly known as Generally Accepted Accounting Principles (GAAP) are used to ensure compliance to all laws and to avoid fraud. From the basic accounting principles, it includes balance sheet equation of assets =  liabilities + stockholder's equity where source of information for the balance sheet is the income statement which is derives by revenues – expenses = net income (loss). Such data is used to prepare financial statements including statement of retained earnings.These financial accounting reports work as barometer to identify the taxes along with other ratios which are used by the investors including current ratio, debt to total assets ratio. All of these reports contribute an important role in tod ay's businesses irrespective of the fact that whether it may be a multi-billion dollar organization or a small store. These financial reports are not only helpful in providing valuable information to the investors but also gave a snap shot of the business to the owners at any point in time. Working of Control Theory at WorkplaceControl theory has multiple applications at workplace. For example, to improve employee working, managers must ensure to have specific and challenging goals which result in improved performance than ambiguous goals. Vague goals like â€Å"try harder† or â€Å"do your best† give no fine relative standard and through feedback. With no specific standard and apparent opinion, an employee will not be able to identify errors and then will not employ in behavior alteration that improves performance. Conclusion Accounting is related with gathering, examining and corresponding economic information.Accounting information supports in many important decisio ns, like assisting users in making well managed decisions, in relation to the effective allocation of scarce resources. Accounting has a long history and it is being seen as socially constructed i. e. it is practiced by people for people and therefore it is more of an art rather than a science. Unlike other professions, which have a body of hypothetical knowledge to depend on to make choices, accounting has evolved as a craft with few rules and little to no theoretical knowledge underpinning its practice and function.Due to alteration in social and economic movement, accounting has been bared to criticism for failing to be more responsive and adaptable. As a result the profession has moved forward to restore accountings position in society by taking a number of initiatives to put into practice theory.

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

Family Culture And Traditions Essay

In our family, there are several cultures and traditions that play a very important role in defining our family values and cultures. These traditions and cultures have been passed to our parents from our grandparents. The two most vivid cultural patterns that are present in our family have been inherited by our parents from the blending of the culture from my maternal grandparents and my fraternal grandparents. The cultural patterns Every Sunday, our father always prepares supper for the family and he makes sure that we eat together as one family. This tradition has been there in our family for a long period and helps us to bond together as a family unit. Every night on the eve of my birthday or that of my siblings, our parents usually sneaks into our bed room and fills it with balloons which are stuffed with money and toys. My father usually writes several bunches of poems and leaves the written poems on our table. When we wake up in the morning, our mother usually prepares for us a birthday morning cake which we enjoy together as a family as we read the bunch of poems. Different roles in upholding the traditions During our Sunday’s family dinner/supper, my parents dwells on teaching us on the way forward in regards to behaviour and especially the use of courteous words such as ‘please’ and ‘thank you’. They discourage talking when the mouth is full as well as placing of the elbows on the table. Also during this time, my parents taught us the developmental skills like literacy skills when story telling during family conversations. During these conversations, my parents would learn more on our attitudes and interests. From these meals, my parents gauge our moods and needs thus helping us solve our problems in the end. My parents oversee the family to ensure that everybody attends to maintain the unity and stability in the family. The dinner table or birthday party is a significant place for socialization especially for the children. These act as prime setting for socialization regarding the norms and rules on values of the family and the acceptable behaviour. From the nutritional perspective, the children master what is considered acceptable; basically the foods and non-food materials. From the family dinners and birthdays, my siblings and I have discovered manners and behaviour restraints that the wider world requires. Through conversations of the family during birthday parties and family dinner, we learned of our parents’ interests and attitudes in relations to the world. We always help our father to prepare for the family dinner on Sundays. As the eldest, I helped my father prepare the foods and especially the desert and vegetables while my other younger siblings have inevitably prepared the table. The Purpose As part of our tradition, the family meal is a symbol of a shared family life. On Sunday’s, family supper/dinner acts to bringing us together in the family. This greatly leads to our social well-being at the same time providing predictable structure to our Sundays which is often reassuring especially psychologically. In our family, everybody is involved in this activity and this applies to the buying of food, preparations of food done by my father, making and laying the table, and finally serving of food. With all this participation, it is not a surprise that the provision of this  family meal is a classic demonstration that we love and care for our family stability and unity. From the initial stages of shopping to clearing the table, each member of the family participates in this exercise responsibly and this promotes family solidarity. The Reflection on the Tradition Though we have a happy family, we also experience our setbacks. At the end of the day, the members of the family who are already exhausted after a busy day at school or work and probably maybe irascible meet for a family meal or party. Hostility may arise perceived at the dinner/party table maybe because of the injustices and behaviour which is unacceptable. Refusal to eat, complaining about bad cooking or lack of appreciation on whatever served on the table are some of the things that can lead to these conflict at the dinning/party table. Therefore, family dinners and birthday parties have many positive virtues that are sometimes fought with strain and negative consequences which greatly depend on the styles of parenting. As a socially combining role, when we share a meal during birthdays or Sunday dinner, it brings people together in a network of reciprocal commitments and shared social relationships. As a tradition routine, Sunday family dinner prepared by my father has been most frequent planned ritual activity in our family which usually takes place in our family house. The family meal and birthday parties in specific have come to represent the dynamics of the family and overtime generations are complaining on its downfall. In the times of change, family meals and parties represented solidity and perhaps the complaint of the lost family may in fact be the response to feared exchange in the arrangements and frameworks of families. These family traditions still influence me to date. This is due to the fact that they provide a source of identity on top of strengthening the family bond. I believe that the families that engage in frequent traditional  practices report stronger relationship and unity than families that haven’t accepted rituals together. I will carry my family traditions in future because I view them as a way of offering comfort and security. This is because our family beliefs and rituals are the cure to the feeling that comes from our world which is fast-paced and ever-changing. It’s relieving to have a few constants in one’s life. Am also for the idea that these family traditions teach values and this is achieved by for instance through family stories where the value of education, life-long learning and reading is instilled; and through regular family dinners or parties, the centrality of familial togetherness is instilled. With all this in mind, I will definitely carry these traditions in the future.

Trending and predicting movements of economic indicators Dissertation

Trending and predicting movements of economic indicators - Dissertation Example One financial instrument that is normally used by governments is the issuance of treasury bills or government bonds wherein the earning interest rates will generally be followed by the banks of that country. By using the interest rates that will define the treasury-bill holder’s earnings will slowly influence the financial market to adjust its interest rates. In the absence of other economic indicators, the treasury-bill interest rates will not only be adopted by the banks in their own financial transactions but it will also be used as the bench mark for the amount of money that will be available to borrowers. In theory, if the interest rates are low more people will borrow money from the banks. If the interest rates are high, the theory sustains that little to no borrower will loan money from the banks and most economic activity will be financed from in-house sources. Other instruments or means of conducting monetary policy includes making the government as the lender of last resort wherein the government will be the source of funds that will be available to borrowers normally a function provided by banks and other financial institutions. Another means of conducting monetary policy includes changing the reserve requirements in banks in order for them to operate. Another is where the government announces its intent to reduce or control inflation or by simply indicating the interest rates it wants for the money it intends to loan out. And last but not the least is moral suasions.... One financial instrument that is normally used by governments is the issuance of treasury bills or government bonds wherein the earning interest rates will generally be followed by the banks of that country. By using the interest rates that will define the treasury-bill holder’s earnings will slowly influence the financial market to adjust its interest rates. In the absence of other economic indicators the treasury-bill interest rates will not only be adopted by the banks in their own financial transactions but it will also be used as the bench mark for the amount of money that will be available to borrowers. In theory, if the interest rates are low more people will borrow money from the banks. If the interest rates are high the theory sustains that little no borrower will loan money from the banks and most economic activity will be financed from in-house sources. Other instruments or means of conducting monetary policy includes making the government as the lender of last reso rt wherein the government will be the source of funds that will be available to borrowers normally banks and other financial institutions. Another means of conducting monetary policy includes changing the reserve requirements in banks in order for them to operate. Another is where the government announces its intent to reduce or control inflation or by simply indicating the interest rates it wants for the money it intends to loan out. And last but not the least is moral suasions or influencing financial institutions about their operating onuses. This paper was primarily completed using secondary sources and some data made available by the professor. The support and anchor of this paper rests on the theories propounded in the literature review. Literature Review This research is

Privacy and Internet Policy Essay Example | Topics and Well Written Essays - 1250 words - 1

Privacy and Internet Policy - Essay Example This paper addresses how identity theft can be dealt with both on the internet and prohibiting of unauthorized access from outsiders. Protection from identity theft (especially from Internet activities and unauthorized access) Identity deft on the internet occurs in situations where individuals who are not authorized get access to a person’s online account, gets access and carry out activities that are lawful or unlawful. Majority of people who join the internet faces the challenge of identity theft so one has to take all the necessary measure of ensuring that their information is protected (Krishnamurthy and Wills 2009). However, a lot of people do not know how protect their personal information and they end suffering the consequences of not protecting their personal information. Protection of identity theft over the internet In instances where the composer of the email is not known to you, do not bother to have a look at such an email that seem doubtful to you. It is advisab le to only check authentic emails which come from banks, financial firms or your creditors. You should never base on fake emails to update your security information. Doing so, would compromise your security. Currently, there are so many people who are out to con other or sent virus and therefore the best way of avoiding them is to delete the messages they send. Effective firewall is required in order to protect your personal computer from attackers. This ensures that the threats on your personal computers from spywares, hackers and Trojans are minimized. It is further stated by Krishnamurthy and Wills (2009) that the use a Verification Engine assists you in identifying safe websites that you browse at any given time. The information that concern your personal recognition and account numbers should never be stored on a hard disk that can be accessed on the internet. This is because a hard disk that is connected to the internet has a great risk of being reached by interested thieves. It is strongly recommended not to save private information on hard disk even in circumstances of having very strong firewall software being installed. Also there should be credit check run on yourself or your family for at least once in every year to insure that the all records are always in their right format (Moore, 2005). Personal information or PIN numbers should never be given out when an email is received from internet service Providers (ISP) or credit companies requesting for the release of such confidential information. In many instances, identity thieves who pretend to be credit card companies or ISPs normally send emails requesting for personal information that has been updated. As Gina (2011) points out, void posting of information that is private about yourself or that which relates to your family members on the personal web site that can easily be used by a thief. They use this information to come up with fraudulent identity. This implies that one has to be conscious wh en posting personal information. Security Socket Layer (SSL) it is a technology that enables users to set up sessions with internet sites that are secure. This implies that they are exposed to minimal violation of external attackers.

Political Science Syria Government Essay Example | Topics and Well Written Essays - 750 words

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Concept tools Assignment Example | Topics and Well Written Essays - 250 words

Concept tools - Assignment Example In the present study, clinical judgment will be measured by measuring how nurses notice certain data on the patient, how they make sense of the data, and how they intervene in the patient’s issue. An effective measuring tool for clinical judgment that will be employed in the current study is Tanner’s Clinical Judgment tool. In this study, the tool will be used in a pre and posttest design for using clinical judgment in employing JNC guidelines in treating hypertension. Its validity, which is the other thing that makes it suitable for the current study, is underpinned by its wide use in different studies. The results of the tool are credible and this is seen in the reliability of the results of the studies that employed it previously. A limitation with Tanner’s Clinical Judgment model is that it might not capture some aspects of a patient’s condition (Modic, 2013). In summary, Tanner’s Clinical Judgment Model is the appropriate tool to use in the current study because it conceptualizes clinical judgment holistically and does not have many limitations that would hinder its accuracy. The tool has been validated by multiple studies. Ashcraft, A., Opton, L., Bridges, R., Caballero, S., Veesart, A. & Weaver, C. (2013). Simulation Evaluation Using a Modified Lasater Clinical Judgment Rubric. Nursing Education Perspectives, 34(2):

Tylenol Exigencies Business Research Paper Example | Topics and Well Written Essays - 2000 words

Tylenol Exigencies Business - Research Paper Example This in turn, marked the beginning of the company’s integrity in baby business (â€Å"Our Timeline†, n.d., para.3). The company’s efforts to help through provision of relief goods in the aftermath of disasters in the 1900 and 1906 also did a lot to carve its name in the industry. At present times, Johnson & Johnson is in partnership with 250 companies and holds offices in 57 countries distributing their goods in over 175 nations (Johnson & Johnson Corporate, n.d., para.3). In the year 2011, the company’s pharmaceutical sales approximated to $65 billion. Tylenol, McNeil Laboratories & Johnson & Johnson Acetaminophen is the active ingredient in Tylenol. It is widely used for relieving pain and reduction of body temperature for fever. The sales team of McNeil Laboratories came with the term Tylenol by choosing letters from its chemical form N-aceTYL-p-aminophENOL (West, 2009). McNeil Laboratories founded by Robert McNeil was the first to manufacture Tylenol i n 1955 as a prescription drug elixir for children. There was an ongoing issue during that time about the harmful effects of then widely used drug, aspirin. McNeil Laboratories marketed Tylenol as the benign alternative to aspirin (West, 2009). It was in 1959 when Johnson & Johnson acquired McNeil Laboratories and the same year marked Tylenol as a prescription-free drug. From then on, Tylenol has been one of Johnson & Johnson’s most widely sold and most profitable product, bringing an average year to year profit share of 33% (Griese, 2001). The 1982 Tylenol Tampering Widespread scare enveloped the nation when seven deaths have been attributed to Tylenol ingestion in September of 1982. Two off-duty firemen unintentionally hypothesized the connection of Tylenol while listening to police radios and talking about the supposedly mysterious deaths (Kaplan, 2005). Among the victims were a 12 year old student Mary Kellerman; three family members Adam Janus, his brother Stanley and Sta nley’s wife, Theresa, aged 27, 25 and 19 respectively; 27 years old mother of four, Mary Reiner; stewardess Paula Prince, 35 years of age; and 31 years old Mary McFarland. These victims have died within hours after taking Tylenol due to the cyanide content of the capsules, which was stronger by 10,000 times to the cyanide dose the human body can take (Kaplan, 2005). Johnson & Johnson acted promptly. At that time, nobody thought of such tragedy. These murders became the first of its kind. It was revealed in the investigations, that the capsules were tainted by cyanide through the act of an intent person because no specific evidence can link that the Tylenol capsules were laced by cyanide during manufacture (Kaplan, 2005). At the turn of events, Johnson & Johnson decided to put out a major recall for all the Tylenol capsules in the country. The manner in which Johnson & Johnson handled the controversy is regarded by professionals â€Å"to be one of the best in the history of p ublic relations† (Kaplan, 2005). In Griese’s book (2001), she cited the Tylenol tragedy as a befitting example on making the right decisions during time of crisis. Griese enumerated eight steps as guidelines in decision-making. The first step after the identification of the situation is to conduct research. The company McNeil Consumer Healthcare under Johnson & Johnson, which is

Failure Of Materials Essay Example for Free

Failure Of Materials Essay Failure of materials is an analysis in engineering world, to approach and determining about how and why a materials has failed, like iron bar, why it can crocked or porous. Some general causes of failure are structural leading, wear corrosion and latent defects. Failure of materials must be known by an engineer, safety is the first word when engineer working building structure. They can not take random about materials that is used. They must know strength, flexibility, and endure of materials. For adding, with developing and updating software or research that had be done by company, we’ll more easy to know the characteristic of materials that we use. TYPICAL OF FAILURE OF MATERIALS Fatigue Failures Metal fatigue is caused by repeated cycling of the load below its static yield strength. It is a progressive localized damage due to fluctuating stresses and strains on the material. Metal fatigue cracks initiate and propagate in regions where the strain is most severe. The process of fatigue consists of three stages Initial crack initiation, Progressive crack growth across the part, and Final sudden fracture of the remaining cross section. Corrosion Failures Corrosion is chemically induced damage to a material that results in deterioration of the material and its properties. It is most coming from environment. Corrosion is normal, it is can not be removed, but can be minimized with several strategy like proper choice of material, design, coatings, and occasionally by changing the environment. Various types of metallic and nonmetallic coatings are regularly used to protect metal parts from corrosion. If corrosion can be minimized, materials be able to use and more advantage. Ductile and Brittle Metal Failures Ductile metals experience observable plastic deformation prior to fracture. Brittle metals experience little or no plastic deformation prior to fracture. At times metals behave in a transitional manner partially ductile/brittle. Ductile fracture is characterized by tearing of metal and significant plastic deformation. The ductile fracture may have a gray, fibrous appearance. Ductile fractures are associated with overload of the structure or large discontinuities. High Temperature Failures In physics theory, when a materials is being warm, they will expand than original size. We know about boilers, gas turbine engines, and ovens are some of the systems that have components that experience creep.   An understanding of high temperature materials behavior is beneficial Failures involving creep are usually easy to identify due to the deformation that occurs. Failures may appear ductile or brittle. While creep testing is done at constant temperature and constant load actual components may experience damage at various temperatures and loading conditions. in evaluating failures in these types of systems. High temperature progressive deformation of a material at constant stress is called creep. High temperature is a relative term that is dependent on the materials being evaluated. Liquid Metal   and Hydrogen embitterment Failures   Liquid metal embitterment is the decrease in ductility of a metal caused by contact with liquid metal. The decrease in ductility can result in catastrophic brittle failure of a normally ductile material. Very small amounts of liquid metal are sufficient to result in embitterment. The liquid metal can not only reduce the ductility but significantly reduce tensile strength. Liquid metal embitterment is an insidious type of failure as it can occur at loads below yield stress. Thus, catastrophic failure can occur without significant deformation or obvious deterioration of the component. Hydrogen embitterment failures are frequently unexpected and sometimes catastrophic. An externally applied load is not required as the tensile stresses may be due to residual stresses in the material. The threshold stresses to cause cracking are commonly below the yield stress of the material. Very small amounts of hydrogen can cause hydrogen embitterment in high strength steels. Common causes of hydrogen embitterment are pickling, electroplating and welding, however hydrogen embitterment is not limited to these processes. There are causes of materials failure, but not disease that not have medicine. Every materials that we use have a protection, protection come from our knowledge about characteristic of materials as chemical composition, using guide or calculation and appointment the materials for our structure. Addition, many factory,   metal – producing majority, also publish list of their product that have content about characteristic ( strength, elasticity, endure and treatment). Bibliography Omens, J. H., MacKenna, D. A., and McCulloch, A. D. Measurement of Strain and Analysis of Stress in Resting Rat Left Ventricular Myocardium.   J. Biomech Press,1993. TCR Engineering Services Technical Team. White Paper: Investigating Material and Component Failure.   TCR Engineering Services Pvt. Ltd. India, 2004. Haut Donahue, T. L., Gregersen, C., Hull, M. L., and Howell, S. M. Comparison of Viscoelastic, Structural, and Material Properties. ASME, 1994.

Corrosion Properties of Al-B4C Composites

Corrosion Properties of Al-B4C Composites Abstract[SM1] The influences of adding B4C particles on corrosion behavior of Al-2wt.% Cu alloy was studied in 3.5 wt.% NaCl solution at room temperature using linear and cyclic polarization, immersion test and Electrochemical Impedance Spectroscopy (EIS).[SM2] Nano-composites reinforced with 2, 4 and 6 wt. % B4C were produced through mechanical milling and tested to explore the B4C contents effects on the corrosion properties. [SM3]Influences of the grain size were also studied comparing the coarse-grained and milled Al matrix. Results revealed that the corrosion resistance of Al matrix decreases by reducing the particle size. Sample with 2wt.% B4C showed best corrosion resistance amongst all. Key words: Mechanical milling, Nano-composite, Al, B4C, Corrosion Introduction Metal Matrix Composites were remained the focus of attentions in aerospace, automotive and military industries in recent years. These materials offer several advantages including the high strength to weight ratio, excellent wear resistance and high stiffness compared to the original alloys. The commonly used reinforcing materials are; silicon carbide, aluminum oxide and boron carbide. Due to density differences between the reinforcements and the matrix materials, segregation has been found to be a major problem in producing metal matrix composites. Ball milling is considered to be an important technique for producing nano-crystalline composites. Growing interest for this technique is due to preparing materials with unique chemical, physical and mechanical properties. Ball milling process makes uniform distribution of reinforcement particles in the matrix, preventing the segregation which is commonly found in composites fabricated through other methods [1-4]. Effects of B4C particles as reinforcement materials on mechanical properties of aluminum base alloys are existed in the literatures, but studies on corrosion behavior for these composites are rarely reported. Corrosion behavior is a key parameter for assessing the applications of composites in marine environments. All in all, incorporation of the reinforcements into Al alloys increases the corrosion rate of composites in comparison with matrix. Primary corrosion initiation sites in MMCs[SM4] are dependent on electrical conductivity of reinforcement material, reinforcement volume fraction, intermetallic phases and corrosive environment. Grain size has also a major effect on corrosion behavior of the composites [5-9]. Present research aims at studying the corrosion properties of Al-B4C composites. The influence of different B4C contents on corrosion behavior of Al matrix composites was investigated. Coarse-grained Al matrix was also used to explore the effect of grain size on corrosion resistance. Experimental Al–2wt.% Cu and the nano-sized B4C particles were respectively used as matrix and reinforcements in fabricating the specimens. Besides a plain matrix sample, others were synthesized through mechanical alloying of the powder mixtures with 2, 4, and 6 wt.% of B4C. Ball milling was done by a planetary mill, equipped with two tempered steel vials containing Chrome steel balls (φ=20mm). The rotational speed and the ball to powder weight ratio were set at 300 rpm and 10:1, respectively. Milling process was performed at room temperature under argon gas (99.999%) atmosphere protection for 20h[SM5] to achieve steady state condition. Mechanically milled powders were then cold pressed and hot extruded with an extrusion ratio of 10:1 at 550à ¢- ¦C. Reference Al alloy sample was prepared from unmilled aluminum powder using similar pressing and extrusion processes. Electrochemical measurements including linear polarization, cyclic polarization, weight loss and electrochemical impedance spectroscopy tests were applied to 3.5wt.% NaCl solution at room temperature. Three electrodes system, including a working electrode, a platinum counter electrode and a silver-silver chloride electrode (Ag/AgCl) as [SM6]reference electrode were used. The exposed area of samples was polished to 1200 emery paper. Tafel tests were performed at a scan rate of 1 mV/s, from -2000mV to 500mV using a 273A Princeton Applied Research EGG model potentiostat/Galvanostat. Cyclic polarization measurements were carried out under conditions similar to Tafel test. After reaching to the 500mV point, scan direction was reversed. In order to find out the exact protection potential, scan rate of 0.5 mV/s was applied in reverse direction. Disc shape specimens (10mm in diameter and 3mm thick) were immersed in 3.5wt.% NaCl solution in atmosphere for 1, 3, 7, 14 and 28 days. Corroded samples were cleaned according to G1 standard, [SM7]dried and weighed before and after the experiments using a balance (H- Z- K 210 model) with an accuracy of 0.00001 g. The mass losses for samples were finally measured by considering their total surface area. Phase characterization of specimens before and after the immersion were carried out through [SM8]X-ray diffraction (XRD) [SM9]technique on a Phillips X‘Pert Pro diffractometer using monochromatic Cu-KÃŽ ± radiation. Morphology and chemical analysis of samples were also characterized using scanning electron microscopy[SM10] (SEM), SU8040model, equipped with an energy dispersive spectrometer [SM11](EDS). Electrochemical impedance spectroscopy[SM12] (EIS) measurements were done with a frequency ranging from 10 mHz[SM13] to 100 MHz. The results were analyzed by means of Zview2 software. Result and Discussion- The milling part Fig 1 shows the morphology of Al/Cu alloy and the Al/Cu-4wt.% B4C composite after 20 h[SM14] of mechanical milling. By increasing [SM15]milling time, the particle size decreases besides narrower size distributions.[SM16] The nearly equiaxed crystal morphology of particles suggests that, the 20 h[SM17] of milling time was sufficient to reach desired steady-state condition. Results in table 1 demonstrate[SM18] that by increasing the B4C contents, the average particle size decreases. Table 1 shows the influence of B4C content on the crystallite size and the lattice strain of aluminum matrix, according to Williamson–Hall method. As expected, the mechanical milling induced severe plastic deformation, leading to the formation of nano-crystalline metal matrix. The crystallite sizes varied with B4[SM19]C contents, showing the effect of hard particles on grain refining performance of metal matrix[SM20]. It is known that the milling stages introduce plastic deformation of ductile matrix, micro-welding, and the fracture of deformed particles in metal matrix [10, 11]. As compared to mechanical milling of soft powders, the presence of hard particles causes an increase in local deformation of matrix around reinforcement particles, which indeed would enhance the work-hardening rate of metal matrix. Additionally, fracture toughness of composite powders is lower than that of the matrix material [11]. On the other hand, an increase in the content of particles results in more frequent interactions between the dislocations and the hard particles [12], which accelerating the onset of mechanical-milling stage, and contributing to grain-refinement process [11].[SM21] Microstructural examination of as-cast composites revealed that the B4C particles were not distributed uniformly in the matrix and the regional clusters of particles existed. Since the wetting by molten matrix was poor, a uniform distribution of particles could not be observed in composites fabricated by stir casting. In addition, other factors like stirring speed, pouring condition, solidification rate, etc. [SM22]have also had a noticeable influence on particles distribution. In extruded samples, a more even distribution of B4[SM23]C can be observed. Fig. 2 shows the back-scattered electron SEM micrographs of extruded composites used in this study. A uniform distribution of ceramic reinforcements is evident in both composites. In addition, there are no traces of voids in the microstructure which in turn suggests that there was full-densification of composite upon extrusion.[SM24] Result and Discussion- The corrosion part Potentiodynamic Polarization Tests The Potentiodynamic Polarization behaviors of different samples in 3.5 wt% NaCl solutions after 1 hour of testing are given in Fig. 3. Their Ecorr, icorr and ipassive values (obtained from Tafel-type fit) are summarized in Table2[SM25]. Data shows that Al-cast has a lower corrosion rate than Al-milled. As the milled alloy has finer grains, it was expected to be less corrosion resistant because of having more grain boundaries, means higher susceptibility to electrochemical reactions and hence to corrosion. [SM26]It can also be seen that the characteristics of polarization curves for B4C composite samples are quite similar to base the alloy[SM27], indicating that the reactions are similar for both. According to table 2, adding 2wt.% B4C to the base material lowered the corrosion rate slightly, because the ceramic particles may to some extent hindered electrochemical dissolution physically. On the other hand, adding more B4C particle to the composite increases the corrosion rate. In any Al alloy-B4C composites, forming intermetallic compounds plays an important role in any chemical and electrochemical reactions that take place on composite surface in a corrosive environment. Fig.[SM28] 4 shows the X-ray diffraction pattern for Al 6wt.% B4C composite. It can be seen that other than Al matrix, there would be considerable amounts of Al3BC species which were produced when the Al reacted with B4C particles. As Al is more anodic with respect to intermetallic, having more of B4C in matrix dominates the effect of physical blocking of electrochemical reactions for ceramic particles in the solution and corrosion rate increases. Therefore [SM29]other than general corrosion of the matrix, there will be galvanic corrosion between the matrix and intermetallic resulting localized corrosion (pitting) on composite surface. In Al 6wt.% B4C sample, the corrosion rate decreased. This can be explained through passivation point of view as shown in polarization curves in which, the passive current density increases by increasing the B4C content. This may be caused by the formation of more porous and unstable passive layers produced by higher intermetallic particles and also leading to more susceptibility to localized corrosion.[SM30] Weight Loss Measurements Figure 5 represents the weight losses for different samples at different immersion times. Diagram demonstrates that the Al cast has the lowest weight loss, therefore [SM31]the lowest corrosion rate of all samples. B4C composites show higher corrosion rates than Al-milled suggesting that adding B4C to samples increases the corrosion rate.[SM32] As mentioned above, adding B4C to the alloy produces Al3BC intermetallic during corrosion.SEM micrographs of the Al 6wt.% B4C before and after the immersion for 28 days [SM33]are shown in Fig[SM34] 6. Al matrix and Al3BC intermetallic are pointed out in Fig. [SM35]6. EDX analysis results of the intermetallic phase from Figure 4-b is also demonstrated in Fig.[SM36] 7. It reveals that, considerable amounts of the compound exist in the matrix [SM37]which agrees with the XRD results discussed before. Finally, it is observed that the results from immersion and polarization tests are in agreement with each other. It is indicating that besides a general corrosion, there is a galvanic corrosion between the matrix and the particles leading to localized corrosion. Cyclic Polarization Studies: Characteristic potential values such as:[SM38] pitting potential (Epit), corrosion potential (Ecor), and re-passivation potential (Erp) were determined through cyclic polarization studies. As it is observed in Fig[SM39] 8, the nature of potentiodynamic polarization curves in the 3.5% NaCl solution reveals typical characteristics of the material undergoing spontaneous passivation. Reverse scan shows a hysteresis cycle, showing the characteristics of pitting. After reaching to a maximum level, the current begins to decay without any oscillation. Following a linear current-potential relationship is suggesting that an ohmic controlled process was taking place [12-15]. Additional electrochemical parameters given in the table [SM40]are: à °Ã‚ Ã¢â‚¬ ºÃ‚ ¥Epit=Epit-Ecorr, à °Ã‚ Ã¢â‚¬ ºÃ‚ ¥Erp=Epit-Erp. à °Ã‚ Ã¢â‚¬ ºÃ‚ ¥Epit is a measure of the width of passive region on polarization curve, indicating the susceptibility to pitting. à °Ã‚ Ã¢â‚¬ ºÃ‚ ¥Erp is used to assess the repassivation behaviour of propagating pits and hence, the ease with which locally active sites can be eliminated.[SM41] à °Ã‚ Ã¢â‚¬ ºÃ‚ ¥Eprotection represents Erp-Ecorr and indicates the protected region. Pits are formed in this region, repassivation and larger region means more resistance to pitting for composite. [SM42]According to table[SM43] 3, the largest protection region was belonged to Al cast. Milled sample showed a smaller region and because of having more grain boundaries on the surface, by creating smaller nucleation sites for pits, made the sample more prone to pitting. [SM44]Adding B4C to samples confines the region and lowered[SM45] the resistance to pitting corrosion. EIS Studies In order to study the corrosion behavior of B4C composites and the base alloy, EIS measurements were carried out for all specimens at their Ecorr in 3.5% NaCl solution. Figure 9 shows results in the form of nyquist[SM46] diagrams. There is a common characteristic for all curves, i.e. a capacitive semicircle in the high frequency ranges. High frequency capacity loop was mainly related to the characteristics of electrical double layer formed at the interface between the adsorption layer on [SM47]metal surface and the electrolyte [12]. The biggest semicircle was noticed for the Al cast sample, indicating that the alloy has the highest resistance to corrosion. Al-BM sample has an additional semicircle in low frequency range, which may be related to the charge transfer across the metal-electrolyte interface. Another noticeable point is that, inductive loop [SM48]is related to the salt layer formation on the surface. It may also demonstrate that, [SM49]adsorption of an anion like chloride which is presented in electrolyte,[SM50] caused the pitting corrosion. Al-BM also has[SM51] lower resistance to corrosion than Al-Cast. Corrosion resistance for Al 4%B4C sample was higher than the Al-BM. But for the 2% and 6% composites, there were less improvement observed [SM52][16, 17]. Conclusion Results from electrochemical measurements which were carried out on Cast and Milled alloys and the B4C composites[SM53] showed that,[SM54] adding B4C particles to milled alloys will not [SM55]change the corrosion resistance considerably. From corrosion resistance point of view, it would be fair to say that the best sample was[SM56] the Al 2%B4C. [SM1]say kon as phrasal verb kamtar estefadeh koni [SM2]inja be nazaram was studied ro bezar akhare jomle [SM3]The [SM4]inja be nzaram bebenevis MMC mokhafafe chiye,magar inke khayli to mozoe shoma shenakhteh shodeh bashe. [SM5]20 hours [SM6]a [SM7]G1 standard. They were dried and weighed [SM8]inja benazare manbefore and after immersion ro ya toye comma bezar ya biyaresh avale jomle,chon yeho jomlato enghar ghat kardeh. [SM9]inja diffraction bayad capital bashe, magar inke aslan to hozeyeh shoma injori neveshteh mishe. manzuram mesle bala ke toye abstract EIS ro neveshti. [SM10]horofe avale ina bayad capital bashe [SM11]the same as 10SM [SM12]inro hazf kon, chon bala toye abstract neveshti ke mokhafafe chi hastesh. [SM13]inja manzoret mili hertz hastesh? [SM14]20 hours [SM15]the [SM16]besides narrower size distribution, the particle size decreases when the milling time increases. [SM17]20 hours without the [SM18]demonstrates [SM19]subscript [SM20]I think it needs rewriting! [SM21]in jomlehe khayli bolande, hamintor por az information hastesh, behtare beshkanitesh be 2 ta jomle age mishe. [SM22]inja ye comma mikhad [SM23]subscript [SM24]in jomlat nesfesh dar zamane gozashtash nesfesh dar zamane hale! [SM25]fasele beyne table va 2 [SM26] too many information in a sentence, needs rewriting. [SM27]? [SM28]Figure [SM29]a comma here [SM30]too many information in one sentence, needs rewriting. [SM31]comma [SM32].This suggests that adding [SM33]yeja in vasat masata comma mikhad.chon nemidunam chi neveshti nemidunam kojash bezaram [SM34]Figure [SM35]Figure [SM36]Figure [SM37]It reveals that there is considerable amount of the compound in the matrix. [SM38]ino hazf kon [SM39]Figure [SM40]which table? [SM41]needs rewriting [SM42]needs rewriting [SM43]Capital [SM44]too long! [SM45]past or present? [SM46]N [SM47]the [SM48]point is the inductive loop which is [SM49]the [SM50]behtare kole in beyne comma bashe. [SM51]present or past? [SM52]less improvement was observed. [SM53]in behtare beyne 2 ta coma bashe. [SM54]ino delet kon [SM55]does not [SM56]is