{"id":88489,"date":"2025-06-01T07:12:12","date_gmt":"2025-06-01T07:12:12","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88489"},"modified":"2025-06-01T07:12:12","modified_gmt":"2025-06-01T07:12:12","slug":"a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/","title":{"rendered":"A ball is released from rest and rolls down an inclined plane, as show"},"content":{"rendered":"

A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane :
\nWhich one of the following is the correct value of angle \u03b8 that the plane makes with the horizontal? (g = 1000 cm\/s\u00b2)<\/p>\n

[amp_mcq option1=”\u03b8 = sin\u207b\u00b9 (1\/9\u20228)” option2=”\u03b8 = sin\u207b\u00b9 (1\/20)” option3=”\u03b8 = sin\u207b\u00b9 (1\/80)” option4=”\u03b8 = sin\u207b\u00b9 (1\/100)” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80).
\n<\/section>\n
\n– The ball is released from rest, so initial velocity u = 0.
\n– The distance covered along the plane is s = 100 cm in time t = 4 s.
\n– The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2.
\n– Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2.
\n– 100 = (1\/2)a(16) => 100 = 8a.
\n– The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2.
\n– The acceleration of an object rolling down an inclined plane (without slipping, but here it’s just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination.
\n– Given g = 1000 cm\/s\u00b2.
\n– So, 12.5 = 1000 sin \u03b8.
\n– sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80.
\n– \u03b8 = sin\u207b\u00b9 (1\/80).
\n<\/section>\n
\n– The acceleration due to gravity along an inclined plane is g sin \u03b8, and perpendicular to the plane is g cos \u03b8. The net force along the plane determines the acceleration.
\n– The value of g is given as 1000 cm\/s\u00b2, which is equivalent to 10 m\/s\u00b2.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane : Which one of the following is the correct value of angle \u03b8 that the plane makes with the horizontal? (g = 1000 cm\/s\u00b2) … <\/p>\n

Detailed SolutionA ball is released from rest and rolls down an inclined plane, as show<\/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":[1094],"tags":[1114,1129,1128],"class_list":["post-88489","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1114","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA ball is released from rest and rolls down an inclined plane, as show<\/title>\n<meta name=\"description\" content=\"The correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80). - The ball is released from rest, so initial velocity u = 0. - The distance covered along the plane is s = 100 cm in time t = 4 s. - The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2. - Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2. - 100 = (1\/2)a(16) => 100 = 8a. - The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2. - The acceleration of an object rolling down an inclined plane (without slipping, but here it's just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination. - Given g = 1000 cm\/s\u00b2. - So, 12.5 = 1000 sin \u03b8. - sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80. - \u03b8 = sin\u207b\u00b9 (1\/80).\" \/>\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-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A ball is released from rest and rolls down an inclined plane, as show\" \/>\n<meta property=\"og:description\" content=\"The correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80). - The ball is released from rest, so initial velocity u = 0. - The distance covered along the plane is s = 100 cm in time t = 4 s. - The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2. - Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2. - 100 = (1\/2)a(16) => 100 = 8a. - The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2. - The acceleration of an object rolling down an inclined plane (without slipping, but here it's just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination. - Given g = 1000 cm\/s\u00b2. - So, 12.5 = 1000 sin \u03b8. - sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80. - \u03b8 = sin\u207b\u00b9 (1\/80).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:12:12+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 ball is released from rest and rolls down an inclined plane, as show","description":"The correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80). - The ball is released from rest, so initial velocity u = 0. - The distance covered along the plane is s = 100 cm in time t = 4 s. - The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2. - Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2. - 100 = (1\/2)a(16) => 100 = 8a. - The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2. - The acceleration of an object rolling down an inclined plane (without slipping, but here it's just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination. - Given g = 1000 cm\/s\u00b2. - So, 12.5 = 1000 sin \u03b8. - sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80. - \u03b8 = sin\u207b\u00b9 (1\/80).","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-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/","og_locale":"en_US","og_type":"article","og_title":"A ball is released from rest and rolls down an inclined plane, as show","og_description":"The correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80). - The ball is released from rest, so initial velocity u = 0. - The distance covered along the plane is s = 100 cm in time t = 4 s. - The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2. - Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2. - 100 = (1\/2)a(16) => 100 = 8a. - The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2. - The acceleration of an object rolling down an inclined plane (without slipping, but here it's just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination. - Given g = 1000 cm\/s\u00b2. - So, 12.5 = 1000 sin \u03b8. - sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80. - \u03b8 = sin\u207b\u00b9 (1\/80).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:12:12+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-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/","name":"A ball is released from rest and rolls down an inclined plane, as show","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:12:12+00:00","dateModified":"2025-06-01T07:12:12+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct value of angle \u03b8 that the plane makes with the horizontal is sin\u207b\u00b9 (1\/80). - The ball is released from rest, so initial velocity u = 0. - The distance covered along the plane is s = 100 cm in time t = 4 s. - The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1\/2)at\u00b2. - Substitute the values: 100 cm = (0)(4 s) + (1\/2)a(4 s)\u00b2. - 100 = (1\/2)a(16) => 100 = 8a. - The acceleration down the inclined plane is a = 100 \/ 8 = 12.5 cm\/s\u00b2. - The acceleration of an object rolling down an inclined plane (without slipping, but here it's just about acceleration along the plane component of gravity) is a = g sin \u03b8, where g is the acceleration due to gravity and \u03b8 is the angle of inclination. - Given g = 1000 cm\/s\u00b2. - So, 12.5 = 1000 sin \u03b8. - sin \u03b8 = 12.5 \/ 1000 = 125 \/ 10000 = 1 \/ 80. - \u03b8 = sin\u207b\u00b9 (1\/80).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-is-released-from-rest-and-rolls-down-an-inclined-plane-as-show\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"A ball is released from rest and rolls down an inclined plane, as show"}]},{"@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\/88489","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=88489"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88489\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88489"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88489"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88489"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}