{"id":93371,"date":"2025-06-01T11:49:13","date_gmt":"2025-06-01T11:49:13","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=93371"},"modified":"2025-06-01T11:49:13","modified_gmt":"2025-06-01T11:49:13","slug":"what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/","title":{"rendered":"What is the length of a simple pendulum, which has a frequency of 0.5"},"content":{"rendered":"

What is the length of a simple pendulum, which has a frequency of 0.5 Hz (take g = 10 m\/s2<\/sup>)?<\/p>\n

[amp_mcq option1=”1 m” option2=”2 m” option3=”1.5 m” option4=”3 m” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is A) 1 m.
\n<\/section>\n
\nThe frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.
\n<\/section>\n
\nThe period (T) is the reciprocal of the frequency (T = 1\/f = 1\/0.5 = 2 seconds). The period formula is T = 2\u03c0\u221a(L\/g). Squaring both sides gives T\u00b2 = 4\u03c0\u00b2(L\/g), so L = gT\u00b2 \/ (4\u03c0\u00b2). Using T=2s, g=10m\/s\u00b2, and \u03c0\u00b2\u224810: L = 10 * (2)\u00b2 \/ (4 * 10) = 10 * 4 \/ 40 = 40 \/ 40 = 1 meter. This confirms the result.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

What is the length of a simple pendulum, which has a frequency of 0.5 Hz (take g = 10 m\/s2)? [amp_mcq option1=”1 m” option2=”2 m” option3=”1.5 m” option4=”3 m” correct=”option1″] This question was previously asked in UPSC CISF-AC-EXE – 2024 Download PDFAttempt Online The correct answer is A) 1 m. The frequency (f) of a … <\/p>\n

Detailed SolutionWhat is the length of a simple pendulum, which has a frequency of 0.5<\/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":[1089],"tags":[1103,1129,1128],"class_list":["post-93371","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nWhat is the length of a simple pendulum, which has a frequency of 0.5<\/title>\n<meta name=\"description\" content=\"The correct answer is A) 1 m. The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.\" \/>\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\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is the length of a simple pendulum, which has a frequency of 0.5\" \/>\n<meta property=\"og:description\" content=\"The correct answer is A) 1 m. The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:49:13+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":"What is the length of a simple pendulum, which has a frequency of 0.5","description":"The correct answer is A) 1 m. The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.","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\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/","og_locale":"en_US","og_type":"article","og_title":"What is the length of a simple pendulum, which has a frequency of 0.5","og_description":"The correct answer is A) 1 m. The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:49:13+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\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/","url":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/","name":"What is the length of a simple pendulum, which has a frequency of 0.5","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:49:13+00:00","dateModified":"2025-06-01T11:49:13+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is A) 1 m. The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 \/ (2\u03c0\u221a(L\/g)). Rearranging this formula to solve for L gives L = g \/ (4\u03c0\u00b2f\u00b2). Given f = 0.5 Hz and g = 10 m\/s\u00b2, and using the common approximation \u03c0\u00b2 \u2248 10 when g \u2248 9.8 or 10 m\/s\u00b2, we get L = 10 \/ (4 * 10 * (0.5)\u00b2) = 10 \/ (40 * 0.25) = 10 \/ 10 = 1 meter.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/what-is-the-length-of-a-simple-pendulum-which-has-a-frequency-of-0-5\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"What is the length of a simple pendulum, which has a frequency of 0.5"}]},{"@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\/93371","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=93371"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/93371\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=93371"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=93371"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=93371"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}