{"id":342,"date":"2019-02-16T08:43:40","date_gmt":"2019-02-16T03:13:40","guid":{"rendered":"https:\/\/www.physicsacademyonline.com\/blog\/?p=342"},"modified":"2021-02-25T14:02:54","modified_gmt":"2021-02-25T08:32:54","slug":"solve-jee-main-2018-with-a-smile-ii","status":"publish","type":"post","link":"https:\/\/www.physicsacademyonline.com\/blog\/jee-main\/solve-jee-main-2018-with-a-smile-ii.html","title":{"rendered":"Solve JEE (MAIN) 2018 With A Smile &#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 (MAIN) 2018 Physics Paper<\/i><\/h3>\n<p>We are into a new blog, and as a continuation from the previous blog, we guide you to solve more Physics questions from JEE (Main) 2018, Paper-1. These are all MCQs with a single correct option, and you crack them based on the theories learnt fresh from the oven \u2013 oops! From the video lectures @ <a href=\"https:\/\/www.physicsacademyonline.com\" target=\"_blank\" rel=\"noopener\">www.PhysicsAcademyOnline.com<\/a>.<\/p>\n<p><a href=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/02\/JEE-MAIN-2018-question-3.png\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-416\" src=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/02\/JEE-MAIN-2018-question-3.png\" alt=\"\" width=\"214\" height=\"197\" \/><\/a><\/p>\n<p><strong><u>Question 3<\/u>:<\/strong>\u00a0 Two masses m<sub>1<\/sub> = 5 kg and m<sub>2<\/sub> = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum mass\u00a0 m that should be put on top of m<sub>2<\/sub> to stop the motion is:<\/p>\n<p>(1) 18.3 kg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) 27.3 kg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 (3) 43.3 kg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(4) 10.3 kg<\/p>\n<p>Initially, when (m<sub>1<\/sub> + m<sub>2<\/sub>) system moves, what are the forces causing and resisting its motion? Subscribe to <a href=\"https:\/\/www.physicsacademyonline.com\" target=\"_blank\" rel=\"noopener\">www.PhysicsAcademyOnline.com<\/a>, and watch the lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Newton&#8217;s Laws of Motion; Video Name \u2013 Free Body Diagram<\/span><\/i>.<\/p>\n<p>Draw the free body diagrams of m<sub>1<\/sub> and m<sub>2<\/sub> . Write their respective equations of motion, and hence the equation of motion for the system. When mass m is placed on m<sub>2<\/sub> , which force changes and to what value? Open the lecture: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Friction; Video Name \u2013 Static and Kinetic Friction<\/span><\/i>.<\/p>\n<p>What condition must be fulfilled if the motion has to stop eventually?<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/h5lJJ_XzLXo?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><strong><u>Question 5<\/u>:\u00a0<\/strong>\u00a0 In a collinear collision, a particle with an initial speed v<sub>0<\/sub> strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is:<\/p>\n<p>(1) v<sub>0<\/sub>\/4 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) <i style=\"font-family: serif; font-size: 15px;\">\u221a<\/i>2. v<sub>0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sub>(3) v<sub>0<\/sub>\/2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4) v<sub>0<\/sub>\/<i style=\"font-family: serif; font-size: 15px;\">\u221a<\/i>2<\/p>\n<p>This is an uncommon &#8220;superelastic&#8221; collision, in which the final kinetic energy of the particles exceeds the initial kinetic energy. But energy remains conserved or not, which quantity will remain conserved during the collision? Study 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 Principle of Conservation of Linear Momentum, and Some Applications<\/span><\/i>.<\/p>\n<p>Apply the conservation principle to get an equation involving initial speed v<sub>0<\/sub> and final speeds v<sub>1<\/sub> and v<sub>2<\/sub> . What is the formula for kinetic energy of a particle? It&#8217;s here: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Work and Energy; Video Name \u2013 Concept of Energy, and Derivation of Work-Energy Theorem<\/span><\/i>.<\/p>\n<p>Use the condition given in the question to obtain one more equation involving v<sub>0<\/sub> , v<sub>1<\/sub> and v<sub>2 <\/sub>. Manipulate the two equations to find relative velocity, (v<sub>2<\/sub> &#8211; v<sub>1<\/sub>). The video solution below shows a little algebraic jugglery!<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/cDo7TBPuWZM?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><a href=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/02\/Jee-main-2018with-smile-2.png\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-391\" src=\"https:\/\/www.physicsacademyonline.com\/blog\/wp-content\/uploads\/2019\/02\/Jee-main-2018with-smile-2.png\" alt=\"\" width=\"237\" height=\"187\" \/><\/a><\/p>\n<p><strong><u>Question 6<\/u>:\u00a0<\/strong>\u00a0 Seven identical discs, each of mass M and radius R, are welded symmetrically, as shown in figure. The moment of inertia of the arrangement about the axis normal to the plane of the discs and passing through the point P is:<\/p>\n<p>(1) 19 MR<sup>2<\/sup>\/2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) 55 MR<sup>2<\/sup>\/2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (3) 73 MR<sup>2<\/sup>\/2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4) 181 MR<sup>2<\/sup>\/2<\/p>\n<p>What is the formula for moment of inertia of a uniform disc about a normal through its centre? Find from the lecture:\u00a0 <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>Now, what is the moment of inertia of the central disc about O? What is the moment of inertia of each of the six surrounding discs about O? You will require to watch: <i><span style=\"color: #8270a4;\">Chapter Name \u2013 Mechanics; Category \u2013\u00a0 Basic; Topic Name \u2013 Rotational Mechanics; Video Name \u2013 Two Important Theorems on Moment of Inertia<\/span><\/i>.<\/p>\n<p>What is the total moment of inertia of the seven-disc system about O? Hence, what is the total M.I. about P? Confirm your answer from the video solution.<\/p>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" width=\"740\" height=\"416\" src=\"https:\/\/www.youtube.com\/embed\/GB7-zCYySDg?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","protected":false},"excerpt":{"rendered":"<p>Let Us YOU Solve Physics JEE (MAIN) 2018 Physics Paper We are into a new blog, and as a continuation from the previous blog, we guide you to solve more&#8230;<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[26],"tags":[],"_links":{"self":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/342"}],"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=342"}],"version-history":[{"count":21,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/342\/revisions"}],"predecessor-version":[{"id":647,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/342\/revisions\/647"}],"wp:attachment":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/media?parent=342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/categories?post=342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/tags?post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}