{"id":424,"date":"2019-05-08T22:31:54","date_gmt":"2019-05-08T17:01:54","guid":{"rendered":"https:\/\/www.physicsacademyonline.com\/blog\/?p=424"},"modified":"2021-02-25T14:02:41","modified_gmt":"2021-02-25T08:32:41","slug":"solve-jee-main-2018-with-a-smile-iii","status":"publish","type":"post","link":"https:\/\/www.physicsacademyonline.com\/blog\/jee-main\/solve-jee-main-2018-with-a-smile-iii.html","title":{"rendered":"Solve JEE (MAIN) 2018 With A Smile – III"},"content":{"rendered":"

Let Us<\/del> <\/em><\/i>YOU<\/span> Solve Physics JEE (MAIN) 2018 Physics Paper<\/i><\/h3>\n

In a series of blogs, we @ www.PhysicsAcademyOnline.com<\/a> encourage our readers (students preparing for engineering and medical entrance exams) to solve on their own<\/em> Physics questions from recent examination papers. We believe, and we want our readers to believe, a solid grip on the theories and quick problem-solving skill are the success keys to a high rank in a keenly contested exam. Presently, we are discussing JEE (Main) 2018, Paper-1, held on 08 April 2018.<\/p>\n

Question 8<\/u>:<\/strong> A particle is moving with a constant speed in a circular orbit of radius R under a central force inversely proportional to the nth<\/sup> power of R. If the period of revolution of the particle is T, then:<\/p>\n

(1) T\u221d <\/i>R 3\/2<\/sup> for any n
\n(2) T\u221d <\/i>Rn\/2 + 1<\/sup>
\n(3) T\u221d <\/i>R(n + 1)\/2<\/sup>
\n(4) T\u221d <\/i>Rn\/2<\/sup><\/p>\n

Using the condition given in the question, express the central force, Fc<\/sub> , in terms of orbit radius, R. What is the acceleration of the particle? Discussed in detail here: Chapter Name \u2013 Mechanics; Category \u2013 Basic; Topic Name \u2013 Circular Motion; Video Name \u2013 Derivation of Centripetal Acceleration.<\/span><\/i><\/p>\n

For an alternative expression, you may see: Chapter Name \u2013 Mechanics; Category \u2013 Basic;\u00a0 <\/span><\/i>Topic Name \u2013 Circular Motion; Video Name \u2013 Relations between Linear and Angular Quantities.<\/span><\/i><\/p>\n

Use Newton’s second law to express Fc<\/sub> . Compare with its earlier expression. How do you introduce time period, T, in your last result? The relevant formula is here: Chapter Name \u2013 Mechanics; Category \u2013 Basic; <\/span><\/i>Topic Name \u2013 Circular Motion; Video Name \u2013 Angular Quantities in Circular Motion.<\/span><\/i><\/p>\n

Little rearrangement leads you to to the correct option.<\/p>\n

Question 9:<\/u><\/strong> A solid sphere of radius r, made of a soft material of bulk modulus K, is surrounded by a liquid in a cylindrical container. A massless piston of area “a” floats on the surface of the liquid, covering the entire cross section of the container. When a mass m is placed on top of the piston to compress the liquid, the fractional decrement in the radius of the sphere, \u2502dr\/r\u2502, is:<\/p>\n

(1) Ka\/mg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2) Ka\/3mg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (3) mg\/3Ka\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4) mg\/Ka<\/p>\n

What is the formula for bulk modulus, K, of a material? The one using calculus notation is preferred. Watch the lecture: Chapter Name \u2013 Elasticity and Fluid Mechanics; Category \u2013Basic;\u00a0 <\/span><\/i>Topic Name \u2013 Elasticity; Video Name \u2013 Bulk Modulus & Relations among Elastic Constants.<\/span><\/i>
\nAs the mass m is placed on the piston, what is the differential change in pressure? Therefore, what is the fractional change in volume of the sphere?<\/p>\n

Now for a sphere, volume, V = 4 \u03c0<\/i>r3<\/sup>\/3. Taking logarithm on both sides, ln V = ln (4\u03c0<\/i>\/3) + 3 ln r. Differentiating both sides, dV\/V = 3 dr\/r. The fractional change in radius follows. By the way, how is “fractional decrement” different from “fractional change”?<\/p>\n","protected":false},"excerpt":{"rendered":"

Let Us YOU Solve Physics JEE (MAIN) 2018 Physics Paper In a series of blogs, we @ www.PhysicsAcademyOnline.com encourage our readers (students preparing for engineering and medical entrance exams) to…<\/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\/424"}],"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=424"}],"version-history":[{"count":25,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/424\/revisions"}],"predecessor-version":[{"id":478,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/posts\/424\/revisions\/478"}],"wp:attachment":[{"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/media?parent=424"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/categories?post=424"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.physicsacademyonline.com\/blog\/wp-json\/wp\/v2\/tags?post=424"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}