tag:blogger.com,1999:blog-50416226269707203922024-03-06T04:18:59.167+05:30Compilation of my math/physics answers from YA!Yahoo Answers is my favorite site giving me an opportunity to help students and letting me keep in touch with math and physics which are the subjects of my interest. Reading the answers of challenging problems given by subject experts adds to my knowledge. This blog is a compilation of my selected answers given in YA!Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.comBlogger469125tag:blogger.com,1999:blog-5041622626970720392.post-58933547463349316032013-02-04T22:18:00.002+05:302013-02-04T22:20:12.097+05:30Q.469. Indefinite integration.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 469.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Integrate cosec^5 (5x) dx.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 469.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let 5x = u</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 5 dx = du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Integrand</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1/5) ∫ cosec^5 u du ... ( 1 )</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>∫ cosec^3 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= ∫ cosecu * cosec^2 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Using integration by parts</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= cosecu ∫ cosec^2 u du - ∫ [d/du(cosecu) ∫ cosec^2u du] du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosecu cotu - ∫ cosecu cot^2 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosecu cotu - ∫ cosecu (cosec^2 u - 1) du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosecu cotu - ∫ cosec^3 u du + ∫ cosecu du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>2 ∫ cosec^3 u du </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosecu cotu + ln ltan(u/2)l + 2c</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> ∫ cosec^3 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l + c ... ( 2 )</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Now, ∫ cosec^5 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= ∫ cosec^3 u * cosec^2 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Integrating by parts,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= cosec^3 u ∫ cosec^2 u du - ∫ [d/du(cosec^3 u) ∫ cosecu du] du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosec^3 u * cotu - ∫ 3 cosec^3 u * cot^2u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosec^3 u * cotu - 3 ∫ cosec^3u (cosec^2u - 1) du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosec^3 u * cotu - 3 ∫ cosec^5 u du + 3 ∫ cosec^3 u du</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 4 ∫ cosec^5 u du </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosec^3 u * cotu + 3 ∫ cosec^3 u du</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>[Plugging the value of ∫ cosec^3 u du from ( 2 ) above]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - cosec^3 u * cotu + 3 [ - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l ] + 4c</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> ∫ cosec^5 u du </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - (1/4) cosec^3 u * cotu - (3/8) cosecu cotu + (3/8) ln ltan(u/2)l ] + c</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>[Plugging in ( 1 ) above]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Integrand</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - (1/20) cosec^3 (5x) * cot(5x) - (3/40) cosec(5x) cot(5x) </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong> + (3/40) ln ltan(5x/2)l + c'. [c' = c/5]</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Confirmation that the above answer is correct as verified by Wolfram Alpha:</strong></span><br />
<a href="http://www.wolframalpha.com/input/?i=Differentiate%3A+-+%281%2F20%29+cosec%5E3+%285x%29+*+cot%285x%29+-+%283%2F40%29+cosec%285x%29+cot%285x%29+%2B+%283%2F40%29+ln+tan%285x%2F2%29"><span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Wolfram Alpha Link</strong></span></a><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-88887133011391274852013-02-04T13:47:00.001+05:302013-02-04T13:53:33.803+05:30Q.468. Static Equilibrium, Application of Lami's Theorem<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 468.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>An object of mass 3kg is suspended by two light, inextensible strings. The strings make angles of 30degrees and 40degrees to the horizontal.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Find the magnitude of the tension in each spring.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 468.</strong></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmrq9HZE1uqljzW0wH_RDvKlg5f83QwZSC3Et6wAoxaEPHdc8W_3F65KJ2-bn7JyePjtH7t3O-kGqexzDel3ZzSPK4Fj6-gdwEIFY5FRwPHIqv0EeocmmgcMJvGG94LcxbJRqZ-3er3KE/s1600/Lami's+theorem.JPG" imageanchor="1" style="clear: right; cssfloat: right; float: right; margin-bottom: 1em; margin-left: 1em;"><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"></span></a><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Refer to the figure as shown.</strong></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMuNyvMCWXEXpFu0cMPVCZDzRoVDYEVe9XOexZzLFVMIfNTOZRHCu93Meawwv5H1VUWVpocb9yEcdgktkIsppZwZxwmtEIjKpCxDu0hETgHJEt2fGYErOm6yA1hACeIB79tEgHZ_p_VuI/s1600/Lami's+theorem.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" ea="true" height="222" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMuNyvMCWXEXpFu0cMPVCZDzRoVDYEVe9XOexZzLFVMIfNTOZRHCu93Meawwv5H1VUWVpocb9yEcdgktkIsppZwZxwmtEIjKpCxDu0hETgHJEt2fGYErOm6yA1hACeIB79tEgHZ_p_VuI/s400/Lami's+theorem.JPG" width="400" /></a></div>
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>T = tension in the string making an angle of 30° to the horizontal</strong></span></div>
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>T ' = tension in the string making an angle of 40° to the horizontal</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>By Lami's theorem,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>T/sin(90° + 40°) = T '/sin(90° + 30°) = 3 * 9.81/sin(180° - 40° - 30°)</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> T = (3 * 9.81) * cos40°/sin70° N = 24 N</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and T ' = (3 * 9.81) * cos30°/sin70° N = 27.1 N.</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For Lami's theorem, please refer to the following link:</strong></span><br />
<a href="http://en.wikipedia.org/wiki/Lami's_theorem"><span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Lami's Theorem</strong></span></a><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-28745744147671716902013-02-01T17:47:00.001+05:302013-02-01T17:47:32.170+05:30Q.467. Friction - Minimum force to pull a sled against friction.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 467.</strong></span><br />
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<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Imagine you are dragging your sibling on a sled across a flat, snowy surface at a constant speed by hauling on a rope attached to the front of the sled. The rope makes an angle θ with respect to the ground. If the coefficient of kinetic friction μk between the sled and the snow is 0.1, then what should θ be in radians if you want to exert the least amount of force necessary to keep the sled going?</strong></span></div>
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<div _yuid="yui_3_1_1_6_135972068706261" class="content">
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 467.</strong></span></div>
<div _yuid="yui_3_1_1_6_135972068706261" class="content">
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let F = force applied<br />and R = normal reaction from the ground<br />=> Fsinθ + R = mg<br />=> R = mg - Fsinθ<br /><br />Horizontal component of the force balances the kinetic frictional force for the sled to move with constant velocity<br />=> Fcosθ = (mg - Fsinθ) * 0.1<br />=> F (cosθ + 0.1 sinθ) = 0.1 mg<br /><br />F will be minimum when cosθ + 0.1 sinθ is maximum<br />cosθ + 0.1 sinθ <br />= r [(1/r) cosθ + (0.1/r) sinθ], where r = √[1 + (0.1)^2]<br />= r [cos(α - θ)], where cosα = 1/r<br />Maximum value of cosθ + 0.1 sinθ will be for <br />θ = α<br />= arccos(1/r) <br />= arccos 1/√[1 + (0.1)^2] <br />= 5.71°.</strong></span></div>
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-22134867682509612902013-01-31T10:26:00.001+05:302013-01-31T22:48:12.955+05:30Q.466. Electrostatic equilibrium<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 466.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Three 3.37-g Styrofoam balls of radius 2 cm are coated with carbon black to make them conducting and then are tied to 1.27-m-long threads and suspended freely from a common point. Each ball is given the same charge, q. At equilibrium, the balls form an equilateral triangle with sides of length 24.38 cm in the horizontal plane. Determine the absolute value of q.</strong></span> <br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 466.</strong></span> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Consider any one of the three balls.</strong></span> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Three forces act on it.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>i ) its weight, mg, downwards</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>ii ) tension in the thread, T, in the direction from the ball </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong> to the point of suspension and</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>iii ) the resultant of the two forces of repulsion from the remaining two balls.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let O be the point of suspension, A the position of the ball </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and C the center of the equilateral triangle formed by the three balls.</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>OA = 1.27 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>AC cos30° = 24.38/2 cm = 0.1219 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> AC = (0.1219)/cos30° = 0.141 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> OC = √[(1.27)^2 - (0.141)^2] m = 1.262 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> m∠AOC = arctan[(0.141)/(1.262)] = 6.38°</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Vertical component of T balances weight mg</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Tcos6.38° = 3.37 x 10^-3 x 9.81</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> T = 0.03327 N</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Horizontal component of T balances the resultant of the two forces of repulsion </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>from the remaining two balls</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 0.03327sin6.38° = 2 * kq^2/(0.2438)^2 * cos30°</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> q^2 = (0.2438)^2 * 0.03327 sin6.38° / (2 * 9 x 10^9 x cos30°)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> q^2 = 1.409669 x 10^(-14) C</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> q = 0.1187 μC.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-85762007922565498422013-01-26T06:16:00.003+05:302013-01-26T06:17:17.215+05:30Q.465. Remainder in a large division.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 465.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>What is the remainder if 5^2009 + 13^2009 is divided by 18 ?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 465.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Using,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For an odd positive integer n,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>x^n + y^n</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (x + y) [x^(n-1) - x^(n-2)y + x^(n-2)y^2 - ... + y^(n-1)]</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>5^2009 + 13^2009</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (5 + 13) [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 18 * [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 5^2009 + 13^2009 is divisible by 18 with zero remainder.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-49550743864711238062013-01-20T15:56:00.000+05:302013-01-20T15:57:36.704+05:30Q.464. To find the equation of the in-circle given vertices of a triangle<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 464.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Find the equation of circle inscribed in the triangle with vertices </strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>(-7,-10),(-7,15) and (5,-1).</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 464.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let A (-7, -10), B (-7, 15) and C (5, -1) be the given vertices of ΔABC</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> a = BC = √[(5+7)^2 + (-1-15)^2] = 20, </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>b = CA = √[(5+7)^2 + (-1+10)^2] = 15</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and c = AB = √(-7+7)^2 + (15+10)^2] = 25</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> coordinates of the in-center are</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>x = 1/(a+b+c) * (-7a -7b+5c) </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1/(20+15+25) * (-7*20 - 7*15 + 5*25) = - 2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and y = 1/(a+b+c) * (-10a +15b-c) </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1/(20+15+25) * (-10*20 +15*15 - 25) = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> incenter = (-2, 0)</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Radius of the in-circle</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= perpendicular distance from (-2, 0) to the line through AB </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= -2 + 7 = 5</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> eqn. of the inscribed circle is</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(x + 2)^2 + y^2 = 5^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^2 + y^2 + 4x - 21 = 0.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-68361069467216119902013-01-15T23:04:00.001+05:302013-01-15T23:04:45.351+05:30Q.463. de Broglie wavelength of electron in hydrogen atom<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 463.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>What is the wavelength of electron in n'th Bohr orbit ?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 463.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For the electron of hydrogen atom,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>mv^2/r = ke^2/r^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> mv^2 * r = ke^2</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>ByBohr's first hypothesis,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>mvr = nh/2π</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Taking ratio,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>v = 2πke^2/nh</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> mv = 2mπ * ke^2/nh</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>de Broglie wavelength of the elctron</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= h/mv</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= nh^2 / 2 π mke^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 2 π n * [(1/mk) * (h/2π*e)^2] </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 2 π n * [1/(9.1 x 10^-31 x 9 x 10^9) * (6.62 x 10^-34/2π x 1.6 x 10^-19)^2 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 2 π n * 0.00529 x 10^8 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 2 π n * 0.529 angstrom units ... [1 angstrom = 10^-10 m].</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-8184666331838433362012-11-28T11:05:00.004+05:302012-11-28T11:05:30.953+05:30Q.462. Integration.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 462.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Find (x^3) / [(x^2) + 4x + 8] dx.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 462.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Note that the numerator is a polynomial of degree 3 and the denominator is of degree 2.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Hence, the first step is to perform a division and express the function in the form of a quotient + remainder/denominator so that the remainder is a polynomial of degree less than the denominator. Integration follows thereafter. Thus,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>x^3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= x (x^2 + 4x + 8) - 4x^2 - 8x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= x (x^2 + 4x + 8) - 4(x^2 + 4x + 8) + 8x + 32</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^3/(x^2 + 4x + 8)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (x - 4) + 8 (x + 4) / (x^2 + 4x + 8)</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Integral</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= ∫ (x - 4) dx + 4 ∫ (2x + 4 + 4) / (x^2 + 8x + 8) dx</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= x^2/2 - 4x + 4 ∫ (d(x^2 + 8x + 8) / (x^2 + 8x + 8) + 16 ∫ dx / [(x + 2)^2 + 2^2]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= x^2/2 - 4x + 4 log(x^2 + 8x + 8) + 8 tan^-1 [(x + 2)/2] + c.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-36191225307665636952012-11-23T16:18:00.001+05:302012-11-23T16:21:21.669+05:30Q.461. Fluid dynamics - Bernoulli's principle<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 461.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>A liquid of density 1190 kg/m3 flows with speed 1.44m/s into a pipe of diameter 0.23m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 8.04 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.3 atm. The acceleration of gravity is 9.8 m/s2 and </strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Patm = 1.013 × 105 Pa.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 461.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Vb </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1.44 * (0.23/0.05)^2 </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 30.4704 m/s </strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>According to Bernoulli's principle,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Pa + (1/2) ρVa^2 + ρgHa = Pb + (1/2) ρVb^2 + ρgHb</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Pa - Pb</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= ρ * [(1/2) (Vb^2 - Va^2) + g(Hb - Ha)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1190) * [(1/2) ((30.4704)^2 - (1.44)^2) + 9.81 * 8.04]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1190) * (463.1858 + 78.8724) N/m^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 645049 N/m^2 </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 645049 * 9.86923267 × 10-6 atm</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 6.366 atm</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> Pa = 6.366 + 1.3 = 7.666 atm.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For conversion of N/m^2 to atm, refer to the link as under.</strong></span><br />
<a href="https://www.google.co.in/#hl=en&sclient=psy-ab&q=N%2Fm%5E2+to+atm&oq=N%2Fm%5E2+to+atm&gs_l=hp.3..0l4.36719.40453.1.40844.12.10.0.0.0.0.2219.6562.3-2j1j7-1j0j2.6.0.les%3Bepmergend..0.0...1.1.c9SMUV6A6Bo&pbx=1&fp=1&bpcl=38625945&biw=1144&bih=664&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&cad=b"><span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Conversion Link:</strong></span></a><br />
<br />
<a href="http://in.answers.yahoo.com/question/index;_ylt=AkXekvY.JVreyVVmDqOxumORHQx.;_ylv=3?qid=20121114211202AAbpuJl"><span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Link to YA!</strong></span></a></div>
Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com1tag:blogger.com,1999:blog-5041622626970720392.post-53660412841561381502012-11-23T11:40:00.000+05:302012-11-23T11:52:45.651+05:30Q.460. Wave on a stretched string.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 460.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>The mass of a string is 6.8 × 10-3 kg, and it is stretched so that the tension in it is 220 N. A transverse wave traveling on this string has a frequency of 190 Hz and a wavelength of 0.61 m. What is the length of the string?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 460.</strong></span> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Speed of the transverse wave on the string,</strong></span> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>v = λ f</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v = 0.61 * 190 m/s</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>If l = length of the string,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>v = √[T/(m/l)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 0.61 * 190 = √[220/(6.8 x 10^-3/l)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 220 l / (6.8 x 10^-3) = (0.61 * 190)^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> length of the string,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>l = 6.8 x 10^-3 * (0.61 * 190)^2 / 220 m</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> l = 0.4152 m = 41.52 cm.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-75201608253718712222012-11-23T11:32:00.003+05:302012-11-23T11:52:16.088+05:30Q.459. Concave Mirror Problem.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 459.</strong></span><br />
<span style="font-family: inherit;"><span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Suppose your height is h, and you are standing in front of a concave mirror of focal length f where your real image is height h/8.73. What is your distance from the mirror in terms of f.</strong></span> </span><br />
<span style="font-family: inherit;"><br /></span>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 459. </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(h/8.73)/h = v/u </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v/u = 1/(8.73) ... ( 1 )</strong></span><br />
<br /><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>1/v + 1/u = 1/f</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v = fu/(u - f)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v/u = f/(u - f) ... ( 2 )</strong></span><br />
<br /><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>From ( 1 ) and ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>f/(u - f) = 1/(8.73)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u - f = 8.73f</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u = 9.73f.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-6394884670975414672012-11-16T13:04:00.000+05:302012-11-16T13:04:51.024+05:30Q.458. Circular Motion under gravity<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 458.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>A particle moves from rest at point A on the surface of a smooth circular cylinder of radius R. At B the particle leaves the cylinder. Find the equation relating θ1 and θ2.</strong></span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixTVO223NMMxclS1RPZOTTNQ0j02MXLaEGNS9U_ddbAFOfFe0GXlt9v7zuxQ9CS-Sd41gg5PbQQlODXQSnH1X_BiZSIgkBXySnkiig9aGN8KWBUIYKZdBQGcTx2vJyDRlZvU_mvWEQHf4/s1600/untitled.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="381" rea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixTVO223NMMxclS1RPZOTTNQ0j02MXLaEGNS9U_ddbAFOfFe0GXlt9v7zuxQ9CS-Sd41gg5PbQQlODXQSnH1X_BiZSIgkBXySnkiig9aGN8KWBUIYKZdBQGcTx2vJyDRlZvU_mvWEQHf4/s400/untitled.JPG" width="400" /></a></div>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 458.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Potential energy at A = mgRcosθ1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Potential energy at B = mgRsinθ2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Gain of kinetic energy = loss of potential energy</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (1/2) mv^2 = mgR (cosθ1 - sinθ2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> mv^2/R = 2mg (cosθ1 - sinθ2) ... ( 1 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>At point B, the particle will lose contact if the centripetal force = radial component of weight</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> mv^2/R = mgsinθ2 ... ( 2 )</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>From ( 1 ) and ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>mgsinθ2 = 2mg (cosθ1 - sinθ2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2cosθ1 - 3sinθ2 = 0.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-90798483208405485552012-11-15T11:26:00.003+05:302012-11-15T11:27:29.159+05:30Q.457. Trigonometric inequality.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 457.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Prove that (sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B ) for all 0 < a,b < π/2.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 457.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B )</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> (sin^3 A cosB + cos^A sinB) / sinB cosB ≥ 1/cos(A - B)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> (sin^3 A * cosB + cos^3 A * sinB) * cos(A - B) ≥ sinB cosB</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> (sin^2 A * 2sinA cosB + cos^2 A * 2cosA sinB) * cos(A - B) ≥ 2sinB cosB</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> [sin^2 A {sin(A+B) + sin(A-B)} + cos^2 A {sin(A+B) - sin(A-B)}] * cos(A - B) ≥ sin2B</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> [(sin^2 A + cos^2 A) sin(A + B) - (cos^2 A - sin^2 A) sin(A - B)] * cos(A - B) ≥ sin2B</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin(A + B) cos(A - B) - cos2A sin(A - B) cos(A - B) ≥ sin2B</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> 2sin(A + B) cos(A - B) - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin2A + sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin2A - sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> 2cos(A + B) sin(A - B) - cos2A * 2sin(A - B) cos(A - B) ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin(A - B) [cos(A + B) - cos2A cos(A - B)] ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin(A - B) [2cos(A + B) - 2cos2A cos(A - B)] ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin(A - B) [2cos(A + B) - cos(3A - B) - cos(A + B)] ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> sin(A - B) [cos(A + B) - cos(3A - B)] ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><=> 2sin2A sin^2 (A - B) ≥ 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>which is true</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> the given inequality is true.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-4444129715084596692012-11-14T10:26:00.000+05:302012-11-14T10:28:02.755+05:30Q.456. Elastic head-on collision.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 456.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>A .45kg ice puck, moving east with a speed of 3m/s has a head on elastic collision with a .9 kg puck initially at rest. What will be the speed and direction of each object after the collision?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 456.</strong></span> <br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For elastic collision, both the linear momentum as well as </strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>kinetic energies before and after collision are conserved.</strong></span> <br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let u = velocity of the 0.45 kg ice puck </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong> after the collision towards the east</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and v = velocity of the 0.9 kg ice puck </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong> after the collision towards the east</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>By the law of conservation of linear momentum, </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>initial linear momenta of the pucks before the collision </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= momenta after the collision</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (0.45) * 3 + 0.9 * 0 = 0.45 u + 0.9 v</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u + 2v = 3 ... ( 1 )</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>By the law of conservation of kinetic energy,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(1/2) * (0.45) * 3^2 + 0 = (1/2) * (0.45) u^2 + (1/2) * (0.9) v^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u^2 + 2v^2 = 9 ... ( 2 )</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging u = 3 - 2v from ( 1 ) in ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (3 - 2v)^2 + 2v^2 = 9</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 6v^2 - 12v = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v (v - 2) = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v = 0 or 2 m/s</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>v cannot be zero => v = 2 m/s</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and u = 3 - 2v = 3 - 2v = - 1 m/s</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer:</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Velocity of 0.45 kg puck is 1 m/s towards the west and</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>velocity of the 0.9 kg puck is 2 m/s towards the east.</strong></span><br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-44111973651652261172012-11-09T05:48:00.002+05:302012-11-09T06:10:15.492+05:30Q.455. Geometry challenge<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 455.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>A circle (radius = r), and an equilateral triangle (side = 2r), fit perfectly in a square, as shown in the diagram.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>What is (length CD) divided by (height of triangle)?</strong></span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi81oS3nrRQdeTSaO49Mr2iSdICVLDYdiBnOy31a42W3Ty-SQYYeRTBpXAxL3wbinjcjFexPvWX7JBYh2VBv5qG-RLFACREsCNlEs9QFlz00vOM76OZBU5KZGu54LAX2LIwFCYlP5O1_Is/s1600/CTinSquare.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" rea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi81oS3nrRQdeTSaO49Mr2iSdICVLDYdiBnOy31a42W3Ty-SQYYeRTBpXAxL3wbinjcjFexPvWX7JBYh2VBv5qG-RLFACREsCNlEs9QFlz00vOM76OZBU5KZGu54LAX2LIwFCYlP5O1_Is/s400/CTinSquare.jpg" width="400" /></a></div>
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<div style="border-bottom: medium none; border-left: medium none; border-right: medium none; border-top: medium none;">
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 455.</strong></span></div>
<div style="border-bottom: medium none; border-left: medium none; border-right: medium none; border-top: medium none;">
<u><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Trigonometric Proof:</strong></span></u></div>
<div style="border-bottom: medium none; border-left: medium none; border-right: medium none; border-top: medium none;">
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Refer to the figure:</strong></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRO5RA8SmF3uSCOvYvHnnRbahohUqpgZCSVBRLxFGyoCV3PrGzLbzt1vLyA2piL4bAkw5Js3e8EellrFdBz_uXfN5gHhjhQ0APq0BnVclG3x9LBHnE39953Wtt3hUNLQRcv_6JnfrpMOA/s1600/Copy+of+CTinSquare.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" rea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiRO5RA8SmF3uSCOvYvHnnRbahohUqpgZCSVBRLxFGyoCV3PrGzLbzt1vLyA2piL4bAkw5Js3e8EellrFdBz_uXfN5gHhjhQ0APq0BnVclG3x9LBHnE39953Wtt3hUNLQRcv_6JnfrpMOA/s400/Copy+of+CTinSquare.jpg" width="400" /></a></div>
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let O be the center of the circle.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>With r = 1, (The required ratio is independent of the value of r.)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>vertical side of the square</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1 + ODcos60° + CDcos30°</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1 + 1/2 + (√3/2) CD</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 3 + (√3/2) CD ... ( 1 )</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Horizontal side of the square</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1 + ODcos30° + (BC - CDcos60°)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1 + √3/2 + 2 - CD/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 3 + √3/2 - (1/2) CD ... ( 2 )</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Equatting ( 1 ) and ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>3/2 + (√3/2) CD = 3 + √3/2 - (1/2) CD</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (√3 - 1)/2 CD = (√3 - 3)/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> CD = √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and Height of the triangle = 2 cos30° = √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> CD/Height of the triangle </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= √3 / √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=======================================…</strong></span><br />
<u><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Proof using Co-ordinate Geometry:</strong></span></u><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Refer to the figure:</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwSBbX-yAbynayd6vKdwyu-Au4bR1Tyoc3k2z-goCvJIJJ6SHqayv0mqmSD2r4VIBlP1E_AH-h4CwvdyweFD_y1lHk_pr38H6XxICa-qnRxo9QbH5yo_dqpaRP3IgMu8hKanZ-QYZPFWo/s1600/8166370669_fb8ebeb77d_z.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong><img border="0" height="392" rea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwSBbX-yAbynayd6vKdwyu-Au4bR1Tyoc3k2z-goCvJIJJ6SHqayv0mqmSD2r4VIBlP1E_AH-h4CwvdyweFD_y1lHk_pr38H6XxICa-qnRxo9QbH5yo_dqpaRP3IgMu8hKanZ-QYZPFWo/s400/8166370669_fb8ebeb77d_z.jpg" width="400" /></strong></span></a></div>
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<div style="border-bottom: medium none; border-left: medium none; border-right: medium none; border-top: medium none;">
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Consider the given drawn inverted as above.</strong></span></div>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>The required ratio, being independent of the radius, let r = 1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> The eqn. of the circle with its center as origin is</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>x^2 + y^2 = 1 ... ( 1 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let the length of side of the square = a</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> B = (a-1, a-1), C = (a-3, a-1) and A = (a-2, a-1-√3)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Slope of AC = - √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let the eqn. of the tangent AC be y = - √3x + c</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> c = r √(1 + m^2) = 2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> eqn. of AC is y = - √3x + 2 ... ( 2 )</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Solving eqn. ( 1 ) and ( 2 ) gives </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>x^2 + (-√3x + 2)^2 = 1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 4x^2 - 4√3x + 3 = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (2x - √3) = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x-coordinate of D is √3/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging in eqn. ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>y-coordinate is 1/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> D = (- √3/2, 1/2)</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging coordinates of C in eqn. ( 2 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>a -1 = - √3 (a - 3) + 2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> a = (3√3 + 3) / (√3 + 1) = 3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> C = (0, 2)</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>CD^2 = (0 + √3/2)^2 + (3/2)^2 = 3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> CD = √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and Height of the triangle = 2 cos30° = √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> CD/Height of the triangle </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= √3 / √3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 1.</strong></span> <br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-75022931275371164462012-11-06T12:56:00.002+05:302012-11-06T12:58:32.202+05:30Q.454. Application of differentiation.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 454.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4.00 p.m.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 454.</strong></span> <br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let the position of ship B at noon be at the origin</strong></span> <br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> ship A has x-coordinate = - 100 km </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and B has x-coordinate = 0 km</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>At 4.00 p.m., ship A is at x = - 100 + 4 * 35 = 40 km</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and B is at x = 0 and y = 4 * 25 = 100 km</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Distance between them at 4.00 p.m. </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= √[(100)^2 + (40)^2] = 20√(29) km</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Distance between the ships,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>s^2 = x^2 + y^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2s ds/dt = 2x dx/dt + 2y dy/dt</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> rate of change of distance between the ships, ds/dt </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (x dx/dt + y dy/dt) / s</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= [40 * 35 + 100 * 25] / [20√(29)] km/hr</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1400 + 2500) / [20√(29)] km/hr</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 195/√(29) km/hr</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>≈ 36.21 km/hr.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-78377715315076790242012-11-04T19:11:00.003+05:302012-11-04T19:12:53.798+05:30Q.453. Rotational Motion.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 453.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>The moment of inertia of a hollow sphere of mass m and radius r about an axis through its centre is (2mr^2)/3. A hollow sphere is bowled across a horizontal surface. The sphere initially slides with a velocity of u. After a short interval, it starts to roll without slipping.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Show that the velocity of the ball when it starts to roll is √(3/5) u.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 453.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Kinetic energy of the ball when it slides without rolling</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1/2) mu^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>When it starts rolling without slipping, it has both linear as well as rotational motion and the above kinetic energy gets converted into kinetic energy of linear and rotational motion</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>If v = velocity when it rolls without slipping, then</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(1/2) mu^2 = (1/2)mv^2 + (1/2) I ω^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> mu^2 = mv^2 + (2/3) mr^2ω^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u^2 = v^2 + (2/3) v^2 ... [because ωr = v]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> u^2 = 5/3v^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v^2 = 3/5u^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> v = √(3/5) u.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-60097972003962240002012-11-04T15:56:00.002+05:302012-11-04T15:57:36.330+05:30Q.452. Trigonometric inequality.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 452.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>In any triangle ABC, prove that : sin A sin B sin C ≤ 3√3/8.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 452.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>f = sinA sinB sinC</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= sinA sinB sin[π - (A+B)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= sinA sinB sin(A+B)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= sinA sinB (sinA cosB + cosA sinB)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= sin^2 A sinB cosB + sin^2 B sinA cosA</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>To find the maximum value,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>δf/δA = 0 </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2sinA cosA sinB cosB + sin^2 B * cos2A = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sin2A cosB + cos2A sinB = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sin(2A + B) = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2A + B = π ... ( 1 ) [cannot be zero or higher multiple of π for traingle ABC]</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Similarly,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>δf/δB = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2B + A = π ... ( 2 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Solving ( 1 ) and ( 2 ), </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>A = B = π/3 => C = π/3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sinA sinB sinC = [sin(π/3)]^3 = (3√3)/8</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>This can be shown to be maximum and not minimum by taking some arbitrary values of A, B and C.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Taking A = π/2, B = C = π/4</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sinA sinB sinC = 1/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>which shows that (3√3)/8 is the maximum value of sinA sinB sinC</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sinA sinB sinC ≤ (3√3)/8.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=======================================…</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>For the benefit of the readers, I reproduce the elegant solution given on page 4-66 of the following link.</strong></span><br />
<br />
<span style="color: #660000;"><span style="font-family: Arial, Helvetica, sans-serif;"><strong>Click to open link</strong></span><span style="font-family: Arial, Helvetica, sans-serif;"><strong>.</strong></span></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Consider three points P(A, sinA), Q(B, sinB) and R(C, sinC) </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>on the curve y = sinx such that A + B + C = π</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Centroid of ΔPQR = [π/3, (1/3)(sinA + sinB + sinC)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Therefore, centroid lies on x = π/3 and is inside the triangle,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (1/3) (sinA + sinB + sinC) ≤ sin(π/3) = √3/2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (sinA sinB sinC)^(1/3) ≤ √3/2 ... [ GM ≤ AM ]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sinA sinB sinC ≤ (3√3)/8.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-5273506642976552822012-11-03T13:39:00.003+05:302012-11-03T13:41:17.579+05:30Q.451. Challenging area problem of geometry.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 451.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Equilateral triangle PQR is on square ABCD, as at the picture:</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>For given areas of right triangles (22 and 23) find unknown trapezoidal area, exact value.</strong></span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgM8E7oYjkdhd4pllZvxPRReEsv7b6iWusdjshmMQXkuQFj4aLe7vWy9Yl1LE6F8WvLw1xdpOCRoAy1Tfi4k3S_sUV_h6trqNgSxhK1ZMTalvkTzZxicpsNGCu1uLDEpVnrRSBrB9iMVqY/s1600/untitled.GIF" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" qea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgM8E7oYjkdhd4pllZvxPRReEsv7b6iWusdjshmMQXkuQFj4aLe7vWy9Yl1LE6F8WvLw1xdpOCRoAy1Tfi4k3S_sUV_h6trqNgSxhK1ZMTalvkTzZxicpsNGCu1uLDEpVnrRSBrB9iMVqY/s400/untitled.GIF" width="400" /></a></div>
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 451.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let ∠ DRP = x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> ∠ CRQ = 120° - x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let a = length of the side of the equilateral triangle</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (1/2) acosx * asinx = 22</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> a^2/4 sin2x = 22 ... ( 1 )</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Similarly,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>a^2/4 sin(240° - 2x) = 23 ... ( 2 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Taking ratio ( 2 ) to ( 1 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>sin(240° - 2x) / sin2x = 23/22</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 22 * [- (√3/2) cos2x + (1/2) sin2x] = 23 * sin2x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> - 11√3 cos2x = 12sin2x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> tan2x = - (11√3)/12</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sin2x = (11√3)/13√3 = 11/13</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and cos2x = sin2x/tan2x = - 12/(13√3)</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging sin2x in ( 1 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>a^2 = 88 / (11/13) = 104</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Area of triangle = (√3/4) a^2 = 26√3</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Length of side of the square</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= acosx + acos(120° - x)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= acosx - (a/2)cosx + (a√3/2) sinx</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (a/2) cosx + (a√3/2) sinx</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Area of the square</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= [(a/2) cosx + (a√3/2) sinx]^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= a^2 [(1/4) cos^2 x + (3/4) sin^2 x + √3 sinx cosx]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (104) * [(1/8)(1 + cos2x) + (3/8)(1 - cos2x) + (√3/2) sin2x]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (104) * [(1/2) - (1/4) cos2x + (√3/2) sin2x]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (104) * [(1/2) + (1/4) (12/13√3) + (√3/2) * (11/13)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 52 + 30√3</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> area of the trapezium</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= area of the square - area of equilateral triangle - areas of the two right triangles</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 52 + 30√3 - 22 - 23 - 26√3</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 7 + 4√3.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-77483969872082662072012-11-02T21:53:00.002+05:302012-11-03T14:08:57.301+05:30Q.450. To find the equation of a non-standard parabola.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 450.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Determine the general equation for the parabola with its focus is located on (6,4) and that its directrix equation is 4x + y -6 = 0.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 450.</strong></span> <br />
<strong><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;">P</span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;">arabola is defined as a set of points equidistant from a given point, called focus and a given line, called directrix.</span></strong> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>If P (x, y) is a point on the parabola,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>its distance from the focus (6, 4) = √[(x - 6)^2 + (y - 4)^2]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>and its distance from the directrix 4x + y - 6 = 0 is </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>l4x + y - 6l / √(4^2 + 1^2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> √[(x - 6)^2 + (y - 4)^2] = l4x + y - 6l / √(4^2 + 1^2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 17 [(x - 6)^2 + (y - 4)^2] = (4x + y - 6)^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 17 (x^2 + y^2 - 12x - 8y + 52) = (16x^2 + y^2 + 36 + 8xy - 12y - 48x)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^2 + 16y^2 - 8xy - 156x - 124y + 848 = 0.</strong></span><br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-8075242720394087352012-10-30T18:06:00.000+05:302012-11-01T09:39:51.502+05:30Q.449. Area of a quadrilateral.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 449.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>In quadrilateral ABCD it's known that:</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>AC * BD = x and AB² - BC² + CD² - DA² = y</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Find area of quadrilateral in terms of x and y.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 449.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Refer to the following figure.</strong></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWDHOmDbeQFvJOeDGkMmlFFneR3Ki2LJbv3s_lET7cDsAFuv8wb_vUJCYpxmj_eE5Pv2UD7Dt2Yu9lcB0jdOS_049JrdOV3n9M6izfs124q5vMvngABAFE9YjD3FkuUPHBwCENIkr1Qaw/s1600/Q.450.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="305" qea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWDHOmDbeQFvJOeDGkMmlFFneR3Ki2LJbv3s_lET7cDsAFuv8wb_vUJCYpxmj_eE5Pv2UD7Dt2Yu9lcB0jdOS_049JrdOV3n9M6izfs124q5vMvngABAFE9YjD3FkuUPHBwCENIkr1Qaw/s320/Q.450.jpg" width="320" /></a></div>
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let A(0, 0) be at the origin.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let X-axis be along AC and C = (m, 0)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Let B = (a, b) and D = (c, d)</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>AC * BD = x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> m * BD = x</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> BD^2 = x^2/m^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (a - c)^2 + (b - d)^2 = x^2/m^2 ... ( 1 )</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>AB^2 - BC^2 + CD^2 - DA^2 = y</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> a^2 + b^2 - (m - a)^2 - b^2 + (m - c)^2 + d^2 - c^2 - d^2 = y</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2m (a - c) = y</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging (a - c) = y/2m in ( 1 ),</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>y^2/4m^2 + (b - d)^2 = x^2/m^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> l b - d l = √[x^2/m^2 - y^2/4m^2]</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Area of the quadrilateral</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1/2) AC * l b - d l</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1/2) m * √[x^2/m^2 - y^2/4m^2]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (1/2) √[x^2 - y^2/4)].</strong></span><br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-13858469919474083162012-10-28T12:33:00.003+05:302012-11-01T09:41:08.691+05:30Q.448. Rate of change (Application of Differentiation)<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 448.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>At what rate is the viewing angle changing when the observer is 50 feet from the tunnel if they are traveling at 6 feet per second? The picture is as under.</strong></span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6iSIygcYopUZN0r0E8EWNnkRBJiDb6iVD5ZXQ27Jvl1WLwhVYMsQ77bBsD0uNVtYrc7XEd6qr768Rv3pIzAJwQn1KrEclXWsVMRzyTv_ta57PhSZDaWJW1uDWAtL3EuxC7NM1db9pazQ/s1600/Rate+of+change.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="202" oea="true" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6iSIygcYopUZN0r0E8EWNnkRBJiDb6iVD5ZXQ27Jvl1WLwhVYMsQ77bBsD0uNVtYrc7XEd6qr768Rv3pIzAJwQn1KrEclXWsVMRzyTv_ta57PhSZDaWJW1uDWAtL3EuxC7NM1db9pazQ/s400/Rate+of+change.JPG" width="400" /></a></div>
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 448.</strong></span> <br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>θ</strong></span> <span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= arctan(25/50) - arctan(15/50)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= arctan(1/2 - 3/10) / (1 + 3/20)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= arctan (4/23)</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>If the observer were x ft. away,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>θ</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= arctan(25/x - 15/x)/(1 + 375/x^2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= arctan [10x/(x^2 + 375)]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> tanθ = 10x/(x^2 + 375)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> sec^2 θ dθ/dt = [(x^2 + 375) * 10 - 10x * 2x]/(x^2 + 375)^2 dx/dt ... ( 1 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>sec^2 θ = 1 + tan^2 θ = 1 + 4/23 = 27/23</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Plugging sec^2 θ = 27/23, dx/dt = 6 and x = 50 in ( 1 )</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> dθ/dt</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= [(2500+375)*10 - 20*2500)] * 6/[(27/23) * (2500+375)^2] rad/sec</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (28750 - 50000) * 6 / [(27/23) * (8265625) rad/sec</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - (21250 * 23) * 6 / (27 * 8265625) rad/s</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= - 0.01314 rad/s.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>[Negative sign indicates that the angle is decreasing.]</strong></span><br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-58267221925187521212012-10-26T23:43:00.003+05:302012-11-01T09:42:05.895+05:30Q.447. Elimination of θ from two equations.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 447.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Eliminate θ from (x/a) cos θ + (y/b) sin θ = 1 and xsin θ - y cos θ = √(a² sin² θ + b² cos²θ).</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 447.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>(x/a) cosθ + (y/b) sinθ = 1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (x/a)^2 cos^2 θ + 2(xy/ab) sinθ cosθ + (y/b)^2 sin^2 θ = 1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = sec^2 θ</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = 1 + tan^2 θ </strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> [(y/b)^2 - 1] tan^2 θ + 2(xy/ab) tanθ + [(x/a)^2 - 1] = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> [(y^2 - b^2)/(x^2 - a^2)] * (a/b)^2 tan^2 θ + [2xy/(x^2 - a^2)] * (a/b) tanθ + 1 = 0 ... ( 1 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>xsinθ - ycosθ =√(a² sin² θ + b² cos² θ)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^2 sin^2 θ - 2xysinθ cosθ + y^2 cos^2θ = a^2 sin^2 θ + b^2 cos^2 θ</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^2 tan^2 θ - 2xy tanθ + y^2 = a^2 tan^2 θ + b^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (x^2 - a^2) tan^2 θ - 2xy tanθ + (y^2 - b^2) = 0</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> [(x^2 - a^2)/(y^2 - b^2)] tan^2 θ - [2xy / (y^2 - b^2)] tanθ + 1 = 0 ... ( 2 )</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Comparing eqns. ( 1 ) and ( 2 ),</strong></span><br />
<span style="font-family: Arial, Helvetica, sans-serif;"><br /><strong><span style="color: #073763;"></span></strong></span><span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>using coefficients of tanθ and constant terms,</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>[2xy/(x^2 - a^2)] * (a/b) / [- 2xy / (y^2 - b^2)] = 1</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> [1/(x^2 - a^2)] * (a/b) = - 1/(y^2 - b^2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> - a (y^2 - b^2) = b(x^2 - a^2)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> bx^2 + ay^2 = ba^2 + ab^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> x^2/a + y^2/b = a + b.</strong></span><br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com0tag:blogger.com,1999:blog-5041622626970720392.post-49474385589751520532012-10-23T14:19:00.003+05:302012-11-01T09:43:32.253+05:30Q.446. Trigonometric Equation.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 446. </strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Solve 2sin(10º)sin(20º + θ) = sin(θ) for 0º < θ < 90º.</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 446.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>2sin(10º)sin(20º + θ) = sin(θ)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> cos(θ + 10º) - cos(θ + 30º) = sinθ</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> cos(θ + 10º) = cos(θ + 30º) + cos(90º - θ)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> cos(θ + 10º) = 2cos60º cos(θ - 30º)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> cos(θ + 10º) = cos(θ - 30º)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> θ + 10º = ± (θ - 30º)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> 2θ = 20º ... [taking the -ve sign on RHS as +ve sign gives no result]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> θ = 10º.</strong></span><br />
<br />
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Madhukar Daftaryhttp://www.blogger.com/profile/05025591560470587737noreply@blogger.com2tag:blogger.com,1999:blog-5041622626970720392.post-47338706285915254982012-10-22T13:12:00.002+05:302012-11-01T09:32:44.903+05:30Q.445. Work/energy.<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>Question 445.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>An 18 kg girl slides down a playground slide that is 3.6 m high.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>When she reaches the bottom of the slide, her speed is 1.3 m/s.</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>a)How much energy was dissipated by friction?</strong></span><br />
<span style="color: #660000; font-family: Arial, Helvetica, sans-serif;"><strong>b)If the slide is inclined at 20°, what is the coefficient of friction between the girl and the slide?</strong></span><br />
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<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Answer 445.</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>a)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong></strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Energy dissipated by friction</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= her initial PE - her final KE</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= mgh - (1/2)mv^2</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 18 * [9.81 * 3.6 - 0.50 * (1.3)^2]</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 18 * (35.316 - 0.845) J</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= 629.48 J</strong></span><br />
<br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>b)</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>Normal force = 18 * 9.81 * cos20°</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> frictional force = μ * 18 * 9.81 * cos20°</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> work done by the frictional force</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>= (3.6 / sin20°) * μ * 18 * 9.81 * cos20°</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> (3.6 / sin20°) * μ * 18 * 9.81 * cos20° = 629.48</strong></span><br />
<span style="color: #073763; font-family: Arial, Helvetica, sans-serif;"><strong>=> μ = 629.48 * tan20° / (3.6 * 18 * 9.81)</strong></span><br />
=> μ = 0.36.<br />
<br />
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