{"id":92285,"date":"2025-06-01T11:19:58","date_gmt":"2025-06-01T11:19:58","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=92285"},"modified":"2025-06-01T11:19:58","modified_gmt":"2025-06-01T11:19:58","slug":"a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/","title":{"rendered":"A wheel of circumference 2 m rolls on a circular path of radius 80 m."},"content":{"rendered":"<p>A wheel of circumference 2 m rolls on a circular path of radius 80 m. What is the angle made by the wheel about the centre of the path, if it rotates 64 times on its own axis?<\/p>\n<p>[amp_mcq option1=&#8221;1.6 rad&#8221; option2=&#8221;1.4 rad&#8221; option3=&#8221;1.2 rad&#8221; option4=&#8221;0.8 rad&#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 CBI DSP LDCE &#8211; 2023<\/div>\n<\/div>\n<div class=\"pyq-exam-psc-buttons\"><a href=\"\/pyq\/pyq-upsc-cbi-dsp-ldce-2023.pdf\" target=\"_blank\" class=\"psc-pdf-button\" rel=\"noopener\">Download PDF<\/a><a href=\"\/pyq-upsc-cbi-dsp-ldce-2023\" target=\"_blank\" class=\"psc-attempt-button\" rel=\"noopener\">Attempt Online<\/a><\/div>\n<\/div>\n<section id=\"pyq-correct-answer\">\nThe wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference.<br \/>\nThe wheel rotates 64 times on its own axis.<br \/>\nTotal distance covered by the wheel rolling on the circular path = Number of rotations $\\times$ Circumference of the wheel<br \/>\nTotal distance = $64 \\times 2$ m = 128 m.<\/p>\n<p>This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($\\theta$) subtended at the center (in radians) is $s = R \\times \\theta$.<br \/>\nHere, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m.<br \/>\nSo, $128 = 80 \\times \\theta$.<br \/>\nSolve for $\\theta$:<br \/>\n$\\theta = \\frac{128}{80}$ radians.<\/p>\n<p>Simplify the fraction:<br \/>\n$\\theta = \\frac{128 \\div 16}{80 \\div 16} = \\frac{8}{5}$ radians.<br \/>\n$\\frac{8}{5} = 1.6$ radians.<br \/>\n<\/section>\n<section id=\"pyq-key-points\">\nWhen a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = R\\theta$, where $R$ is the radius and $\\theta$ is the angle subtended at the center *in radians*.<br \/>\n<\/section>\n<section id=\"pyq-additional-information\">\nThe radius of the wheel itself ($\\frac{1}{\\pi}$ m) is not directly needed to solve this problem, as the distance covered is given by the circumference and the number of rotations. The problem asks for the angle subtended at the *centre of the path*, not the centre of the wheel.<br \/>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>A wheel of circumference 2 m rolls on a circular path of radius 80 m. What is the angle made by the wheel about the centre of the path, if it rotates 64 times on its own axis? [amp_mcq option1=&#8221;1.6 rad&#8221; option2=&#8221;1.4 rad&#8221; option3=&#8221;1.2 rad&#8221; option4=&#8221;0.8 rad&#8221; correct=&#8221;option1&#8243;] This question was previously asked in UPSC &#8230; <\/p>\n<p class=\"read-more-container\"><a title=\"A wheel of circumference 2 m rolls on a circular path of radius 80 m.\" class=\"read-more button\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/#more-92285\">Detailed Solution<span class=\"screen-reader-text\">A wheel of circumference 2 m rolls on a circular path of radius 80 m.<\/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":[1086],"tags":[1105,1102],"class_list":["post-92285","post","type-post","status-publish","format-standard","hentry","category-upsc-cbi-dsp-ldce","tag-1105","tag-quantitative-aptitude-and-reasoning","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 wheel of circumference 2 m rolls on a circular path of radius 80 m.<\/title>\n<meta name=\"description\" content=\"The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference. The wheel rotates 64 times on its own axis. Total distance covered by the wheel rolling on the circular path = Number of rotations $times$ Circumference of the wheel Total distance = $64 times 2$ m = 128 m. This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($theta$) subtended at the center (in radians) is $s = R times theta$. Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m. So, $128 = 80 times theta$. Solve for $theta$: $theta = frac{128}{80}$ radians. Simplify the fraction: $theta = frac{128 div 16}{80 div 16} = frac{8}{5}$ radians. $frac{8}{5} = 1.6$ radians. When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = Rtheta$, where $R$ is the radius and $theta$ is the angle subtended at the center *in radians*.\" \/>\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-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A wheel of circumference 2 m rolls on a circular path of radius 80 m.\" \/>\n<meta property=\"og:description\" content=\"The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference. The wheel rotates 64 times on its own axis. Total distance covered by the wheel rolling on the circular path = Number of rotations $times$ Circumference of the wheel Total distance = $64 times 2$ m = 128 m. This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($theta$) subtended at the center (in radians) is $s = R times theta$. Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m. So, $128 = 80 times theta$. Solve for $theta$: $theta = frac{128}{80}$ radians. Simplify the fraction: $theta = frac{128 div 16}{80 div 16} = frac{8}{5}$ radians. $frac{8}{5} = 1.6$ radians. When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = Rtheta$, where $R$ is the radius and $theta$ is the angle subtended at the center *in radians*.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:19:58+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 wheel of circumference 2 m rolls on a circular path of radius 80 m.","description":"The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference. The wheel rotates 64 times on its own axis. Total distance covered by the wheel rolling on the circular path = Number of rotations $times$ Circumference of the wheel Total distance = $64 times 2$ m = 128 m. This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($theta$) subtended at the center (in radians) is $s = R times theta$. Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m. So, $128 = 80 times theta$. Solve for $theta$: $theta = frac{128}{80}$ radians. Simplify the fraction: $theta = frac{128 div 16}{80 div 16} = frac{8}{5}$ radians. $frac{8}{5} = 1.6$ radians. When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = Rtheta$, where $R$ is the radius and $theta$ is the angle subtended at the center *in radians*.","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-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/","og_locale":"en_US","og_type":"article","og_title":"A wheel of circumference 2 m rolls on a circular path of radius 80 m.","og_description":"The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference. The wheel rotates 64 times on its own axis. Total distance covered by the wheel rolling on the circular path = Number of rotations $times$ Circumference of the wheel Total distance = $64 times 2$ m = 128 m. This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($theta$) subtended at the center (in radians) is $s = R times theta$. Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m. So, $128 = 80 times theta$. Solve for $theta$: $theta = frac{128}{80}$ radians. Simplify the fraction: $theta = frac{128 div 16}{80 div 16} = frac{8}{5}$ radians. $frac{8}{5} = 1.6$ radians. When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = Rtheta$, where $R$ is the radius and $theta$ is the angle subtended at the center *in radians*.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:19:58+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-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/","name":"A wheel of circumference 2 m rolls on a circular path of radius 80 m.","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:19:58+00:00","dateModified":"2025-06-01T11:19:58+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The wheel has a circumference of 2 m. When the wheel rotates once on its own axis, the distance it covers rolling on a surface is equal to its circumference. The wheel rotates 64 times on its own axis. Total distance covered by the wheel rolling on the circular path = Number of rotations $\\times$ Circumference of the wheel Total distance = $64 \\times 2$ m = 128 m. This distance is an arc length on the circular path. The formula relating arc length (s), radius (R), and the angle ($\\theta$) subtended at the center (in radians) is $s = R \\times \\theta$. Here, the arc length $s = 128$ m and the radius of the circular path $R = 80$ m. So, $128 = 80 \\times \\theta$. Solve for $\\theta$: $\\theta = \\frac{128}{80}$ radians. Simplify the fraction: $\\theta = \\frac{128 \\div 16}{80 \\div 16} = \\frac{8}{5}$ radians. $\\frac{8}{5} = 1.6$ radians. When a wheel rolls, the distance covered in one rotation is equal to its circumference. The arc length ($s$) of a sector in a circle is given by the formula $s = R\\theta$, where $R$ is the radius and $\\theta$ is the angle subtended at the center *in radians*.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wheel-of-circumference-2-m-rolls-on-a-circular-path-of-radius-80-m\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CBI DSP LDCE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cbi-dsp-ldce\/"},{"@type":"ListItem","position":3,"name":"A wheel of circumference 2 m rolls on a circular path of radius 80 m."}]},{"@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\/92285","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=92285"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92285\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=92285"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=92285"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=92285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}