{"id":87168,"date":"2025-06-01T04:30:31","date_gmt":"2025-06-01T04:30:31","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87168"},"modified":"2025-06-01T04:30:31","modified_gmt":"2025-06-01T04:30:31","slug":"a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/","title":{"rendered":"A particle moves in a circle of radius 4 m. Its linear speed is given"},"content":{"rendered":"<p>A particle moves in a circle of radius 4 m. Its linear speed is given by 4\u221a3 t, where t is the time measured in seconds. At t = 1 s, the angle made by the resultant acceleration &#8211; r\u0302 with direction (+ r\u0302 is the radial direction) is given by \u03c6. Which one among the following is the correct value of \u03c6 ?<\/p>\n<p>[amp_mcq option1=&#8221;tan\u207b\u00b9 (1\/\u221a3)&#8221; option2=&#8221;tan\u207b\u00b9 (1\/\u221a2)&#8221; option3=&#8221;tan\u207b\u00b9 (2\/\u221a3)&#8221; option4=&#8221;tan\u207b\u00b9 (\u221a3)&#8221; correct=&#8221;option1&#8243;]<\/p>\n<div class=\"psc-box-pyq-exam-year-detail\">\n<div class=\"pyq-exam\">\n<div class=\"psc-heading\">This question was previously asked in<\/div>\n<div class=\"psc-title line-ellipsis\">UPSC Geoscientist &#8211; 2024<\/div>\n<\/div>\n<div class=\"pyq-exam-psc-buttons\"><a href=\"\/pyq\/pyq-upsc-geoscientist-2024.pdf\" target=\"_blank\" class=\"psc-pdf-button\" rel=\"noopener\">Download PDF<\/a><a href=\"\/pyq-upsc-geoscientist-2024\" target=\"_blank\" class=\"psc-attempt-button\" rel=\"noopener\">Attempt Online<\/a><\/div>\n<\/div>\n<section id=\"pyq-correct-answer\">\nIn circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry.<br \/>\n<\/section>\n<section id=\"pyq-key-points\">\nThe linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.<br \/>\n<\/section>\n<section id=\"pyq-additional-information\">\ntan(\u03c6) = (4\u221a3) \/ 12 = \u221a3 \/ 3 = 1\/\u221a3. Therefore, \u03c6 = tan\u207b\u00b9(1\/\u221a3).<br \/>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>A particle moves in a circle of radius 4 m. Its linear speed is given by 4\u221a3 t, where t is the time measured in seconds. At t = 1 s, the angle made by the resultant acceleration &#8211; r\u0302 with direction (+ r\u0302 is the radial direction) is given by \u03c6. Which one among &#8230; <\/p>\n<p class=\"read-more-container\"><a title=\"A particle moves in a circle of radius 4 m. Its linear speed is given\" class=\"read-more button\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/#more-87168\">Detailed Solution<span class=\"screen-reader-text\">A particle moves in a circle of radius 4 m. Its linear speed is given<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1103,1129,1128],"class_list":["post-87168","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>A particle moves in a circle of radius 4 m. Its linear speed is given<\/title>\n<meta name=\"description\" content=\"In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry. The linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A particle moves in a circle of radius 4 m. Its linear speed is given\" \/>\n<meta property=\"og:description\" content=\"In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry. The linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:30:31+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A particle moves in a circle of radius 4 m. Its linear speed is given","description":"In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry. The linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/","og_locale":"en_US","og_type":"article","og_title":"A particle moves in a circle of radius 4 m. Its linear speed is given","og_description":"In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry. The linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:30:31+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/","name":"A particle moves in a circle of radius 4 m. Its linear speed is given","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:30:31+00:00","dateModified":"2025-06-01T04:30:31+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry. The linear speed is v = 4\u221a3 t. The tangential acceleration is a_t = dv\/dt = d(4\u221a3 t)\/dt = 4\u221a3 m\/s\u00b2. At t = 1 s, v = 4\u221a3 m\/s. The radial acceleration is a_r = v\u00b2\/r = (4\u221a3)\u00b2 \/ 4 = (16*3) \/ 4 = 48\/4 = 12 m\/s\u00b2. The radial direction +r\u0302 points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle \u03c6 between the resultant acceleration and the radial direction (+r\u0302, pointing inward) satisfies tan(\u03c6) = a_t \/ a_r.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-in-a-circle-of-radius-4-m-its-linear-speed-is-given\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"A particle moves in a circle of radius 4 m. Its linear speed is given"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87168","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87168"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87168\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87168"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87168"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87168"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}