{"id":203,"date":"2018-12-06T17:03:34","date_gmt":"2018-12-06T11:33:34","guid":{"rendered":"http:\/\/www.physicsacademyonline.com\/blog\/?p=203"},"modified":"2021-02-25T14:03:45","modified_gmt":"2021-02-25T08:33:45","slug":"tips-to-solve-physics-jee-advanced-2018-questions-ii","status":"publish","type":"post","link":"https:\/\/www.physicsacademyonline.com\/blog\/jee-advanced-preparation\/tips-to-solve-physics-jee-advanced-2018-questions-ii.html","title":{"rendered":"Tips to Solve Physics JEE (Advanced) 2018 Questions &#8211; II"},"content":{"rendered":"<h3 style=\"font-size: 1.20em; font-weight: normal !important;\"><i>Let <em><del>Us<\/del> <\/em><\/i><span style=\"font-family: impact; letter-spacing: 1.5px; font-size: 24px; color: #000;\">YOU<\/span><i> Solve Physics JEE (Advanced 2018) Questions:<\/i><\/h3>\n<p>As you will recall, we plan to demonstrate here in a series of blogs, how a combination of sound theoretical knowledge and sharp problem-solving skill can ensure your success in any Board or Competitive Examination. We continue with JEE (Advanced) 2018, Paper-1, earlier known as IIT-JEE, held on 20 May 2018. There must be a large number of websites and printed books which dish out solutions to such question papers every year with unfailing regularity. But how much does a student benefit by passively memorising the steps of a solution done by somebody else? Not much, really. A student, with serious aspiration for success in an exam, must put themselves in the position of an examinee. They must test their own skill in quickly connecting a question to the relevant theories and formulas, and in applying them in proper order without errors. So we are here to let YOU solve the questions and problems, with minimal guidance from our end. That will make the desired difference in your preparation. Having said that, you must check your answers from our video solutions given along with.<\/p>\n<p>&nbsp;<\/p>\n<p>Returning to Section 2 of Paper-1, there are eight problems in it. Each problem&#8217;s answer is a numerical value without unit (unit, if any, is already mentioned in the question). The correct numerical value has to be reported up to the second decimal place, after rounding off if necessary. There is no partial marking: full three marks for correct answer, zero marks otherwise.<\/p>\n<p>&nbsp;<\/p>\n<p><u>Question 7<\/u>:\u00a0\u00a0 Two vectors <strong>A<\/strong> and <strong>B<\/strong> are defined as: <strong>A<\/strong> = a<strong>i\u00a0 <\/strong>and <strong>B<\/strong> = a (cos<i style=\"font-family: serif;\">\u03c9<\/i>t <strong>i<\/strong> + sin<i style=\"font-family: serif;\">\u03c9<\/i>t <strong>j<\/strong>), where a is a constant and <i style=\"font-family: serif;\">\u03c9<\/i> = <i style=\"font-family: serif;\">\u03c0<\/i>\/6 rad\/s. If \u2502<strong>A<\/strong> + <strong>B<\/strong>\u2502=\u00a0 \u00a0<span style=\"font-family: serif;\">\u221a<\/span>3 .\u2502<strong>A<\/strong> &#8211; <strong>B<\/strong>\u2502 at time t = <i style=\"font-family: serif;\">\u03c4<\/i> \u00a0for the first time, the value of <i style=\"font-family: serif;\">\u03c4<\/i> \u00a0in second is\u00a0\u00a0 ______________ .<\/p>\n<p>&nbsp;<\/p>\n<p>This problem involves Vectors, an introductory chapter in Physics, and a bit of Trigonometry. A vector can be expressed either as magnitude-and-direction or in unit-vector notation, which is the case here. How to add up or subtract two vectors expressed in terms of <strong>i<\/strong>, <strong>j<\/strong> (and <strong>k <\/strong>)? Subscribe to\u00a0 <a href=\"http:\/\/www.physicsacademyonline.com\">www.physicsacademyonline.com<\/a>, and watch this lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Vectors; Video Name \u2013 Resolution of a Position Vector in Space.<\/span><\/i><\/p>\n<p>Hence find <strong>A<\/strong> + <strong>B<\/strong> and\u00a0 <strong>A<\/strong> &#8211; <strong>B<\/strong>. How to find the magnitude (denoted by modulus sign \u2502\u2502) of a vector expressed in terms of unit vectors? The same lecture mentioned above will be useful.<\/p>\n<p>Hence find \u2502<strong>A<\/strong> + <strong>B<\/strong>\u2502 and \u2502<strong>A<\/strong> &#8211; <strong>B<\/strong>\u2502 . Use the condition stated in the problem to get a trigonometric equation (involving sub- multiple angles). Solve for t (general solution of trigonometric equations). From a series of possible values of t, pick the smallest value <em>t <\/em>.<\/p>\n<p><em>Only when you have completed<\/em>, confirm your answer from our MP4 video solution below.<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/TpfnZeEcyp0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><u>Question 9<\/u>:\u00a0 A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60\u00b0 with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2 &#8211; <i style=\"font-family: serif;\">\u221a<\/i>3) \/ <i style=\"font-family: serif;\">\u221a<\/i>10 s, then the height of the top of the inclined plane in metres is ______________ . Take g = 10 m\/s\u00b2.<\/p>\n<p>&nbsp;<\/p>\n<p>What is the general formula for the centre-of-mass acceleration, a<sub>cm\u00a0 <\/sub>, of <em>any<\/em> round body rolling without slipping down an incline? Subscribe to <a href=\"http:\/\/www.physicsacademyonline.com\">www.physicsacademyonline.com<\/a>, and avail of this lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Rotational Mechanics; Video Name \u2013 Rolling Motion of a Rigid Body down an Incline.<\/span><\/i><\/p>\n<p>Notice that a<sub>cm <\/sub>depends on the moment of inertia, I<sub>cm\u00a0 <\/sub>, of the body which varies from one shape to another. Now, what are the values of I<sub>cm\u00a0 <\/sub>\u00a0for a ring and a disc? Visit the lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Rotational Mechanics; Video Name \u2013 Moment of Inertia, and Its Calculation for Various Bodies.<\/span><\/i><\/p>\n<p>Back with this information, find a<sub>cm<\/sub> for a ring and a disc. Alternatively, if you advance a little more in the first lecture, you will find a Table already sitting pretty with your last results!<\/p>\n<p>Express the length of the incline in terms of the assumed height, h, of its top. So you have a ring and a disc, both starting from rest,\u00a0 their centres of mass traversing the same length with different accelerations. What formula do you use to find their times of descent? Go through this lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Motion in One and Two Dimensions; Video Name \u2013 Rectilinear Motion with Constant Acceleration.<\/span><\/i><\/p>\n<p>Once you get there, use the condition given in the problem to solve for h. A little play with surds will be necessary, and the rounded value of g will help.<\/p>\n<p>A word of advice for sincere students. I&#8217;m sure you can already feel the power of the theories. Don&#8217;t borrow a formula blindly from a lecture; see its derivation. Trust me, it enhances problem-solving skill. There may be other interesting results in the vicinity of the formula you are looking for; take mental note of them. You never know what the next year&#8217;s Paper will ask, right?<\/p>\n<p>Below you will find the video solution of the problem.<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/tJ2H4JsTK1s?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><u>Question 10<\/u>:\u00a0 A spring-block system is resting on a frictionless floor, as shown in figure. The spring constant is 2.0 N\/m, and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass 1.0 kg, moving with a speed of 2.0 m\/s, collides elastically with the first block. The collision is such that the 2.0-kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ______________ .<\/p>\n<p>&nbsp;<\/p>\n<p><a href=\"http:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9.bmp\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone wp-image-205\" src=\"http:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9-300x169.jpg\" alt=\"\" width=\"499\" height=\"281\" srcset=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9-300x169.jpg 300w, https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9-768x432.jpg 768w, https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9-1024x576.jpg 1024w, https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2018\/12\/PIC-Question-9.bmp 1152w\" sizes=\"(max-width: 499px) 100vw, 499px\" \/><\/a><\/p>\n<p>In an elastic head-on collision between two blocks of known masses (remember the spring is massless) and initial velocities, what are the final velocities of the blocks immediately after the collision?\u00a0 Subscribe to <a href=\"http:\/\/www.physicsacademyonline.com\">www.physicsacademyonline.com<\/a>, to watch the lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Impulse, Collision, and Centre of Mass; Video Name \u2013 Mathematical Study of One-Dimensional Collision.<\/span><\/i><\/p>\n<p>How do you interpret a &#8220;frictionless floor&#8221;?<\/p>\n<p>While the unattached block moves in straight line after the collision, the block attached to the spring performs simple harmonic motion about its initial position. What is the time period of this SHM? The following lectures are suggested: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Work and Energy; Video Name \u2013 Work Done by Spring Force\u00a0 &amp;\u00a0\u00a0 Chapter Name \u2013 Mechanics; Category \u2013 Advanced; Topic Name \u2013 Gravitation; Video Name \u2013 Advanced-Level Problems on Gravitation II (problem 2).<\/span><\/i><\/p>\n<p>After how long does the 2.0-kg block return to its initial position for the first time? How far has the 1.0-kg block travelled by then?<\/p>\n<p><em>Only after you have completed<\/em>, check your answer from our MP4 video solution below.<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/m0F3FT_WYpE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Let Us YOU Solve Physics JEE (Advanced 2018) Questions: As you will recall, we plan to demonstrate here in a series of blogs, how a combination of sound theoretical knowledge&#8230;<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[25],"tags":[],"_links":{"self":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/203"}],"collection":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/comments?post=203"}],"version-history":[{"count":14,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/203\/revisions"}],"predecessor-version":[{"id":406,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/203\/revisions\/406"}],"wp:attachment":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/media?parent=203"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/categories?post=203"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/tags?post=203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}