{"id":88560,"date":"2025-06-01T07:14:50","date_gmt":"2025-06-01T07:14:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88560"},"modified":"2025-06-01T07:14:50","modified_gmt":"2025-06-01T07:14:50","slug":"a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/","title":{"rendered":"A pendulum clock is lifted to a height where the gravitational acceler"},"content":{"rendered":"

A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g\/2. The time period of the second pendulum would be :
\n(in terms of period T of the first pendulum)<\/p>\n

[amp_mcq option1=”$\\sqrt{2}$ T” option2=”$\\frac{1}{\\sqrt{2}}$ T” option3=”$2\\sqrt{2}$ T” option4=”T” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is A) $\\sqrt{2}$ T.
\n<\/section>\n
\nThe time period (T) of a simple pendulum is given by the formula $T = 2\\pi \\sqrt{\\frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum.
\nFor the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2\\pi \\sqrt{\\frac{L}{g}} = T$.
\nFor the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2\\pi \\sqrt{\\frac{L_2}{g_2}} = 2\\pi \\sqrt{\\frac{L}{g\/2}} = 2\\pi \\sqrt{\\frac{2L}{g}}$.
\nWe can rewrite $T_2$ in terms of $T_1$: $T_2 = \\sqrt{2} \\times (2\\pi \\sqrt{\\frac{L}{g}}) = \\sqrt{2} T_1 = \\sqrt{2} T$.
\n<\/section>\n
\nThe time period of a simple pendulum is independent of the mass and amplitude (for small oscillations) of the bob. It depends only on the length of the string and the local acceleration due to gravity. The question tests the understanding that mass does not influence the time period and how the time period scales with changes in gravitational acceleration.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g\/2. The time period of the second pendulum would be : (in terms of … <\/p>\n

Detailed SolutionA pendulum clock is lifted to a height where the gravitational acceler<\/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":[1119,1129,1128],"class_list":["post-88560","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1119","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA pendulum clock is lifted to a height where the gravitational acceler<\/title>\n<meta name=\"description\" content=\"The correct answer is A) $sqrt{2}$ T. The time period (T) of a simple pendulum is given by the formula $T = 2pi sqrt{frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum. For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2pi sqrt{frac{L}{g}} = T$. For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2pi sqrt{frac{L_2}{g_2}} = 2pi sqrt{frac{L}{g\/2}} = 2pi sqrt{frac{2L}{g}}$. We can rewrite $T_2$ in terms of $T_1$: $T_2 = sqrt{2} times (2pi sqrt{frac{L}{g}}) = sqrt{2} T_1 = sqrt{2} T$.\" \/>\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-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A pendulum clock is lifted to a height where the gravitational acceler\" \/>\n<meta property=\"og:description\" content=\"The correct answer is A) $sqrt{2}$ T. The time period (T) of a simple pendulum is given by the formula $T = 2pi sqrt{frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum. For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2pi sqrt{frac{L}{g}} = T$. For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2pi sqrt{frac{L_2}{g_2}} = 2pi sqrt{frac{L}{g\/2}} = 2pi sqrt{frac{2L}{g}}$. We can rewrite $T_2$ in terms of $T_1$: $T_2 = sqrt{2} times (2pi sqrt{frac{L}{g}}) = sqrt{2} T_1 = sqrt{2} T$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:14:50+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 pendulum clock is lifted to a height where the gravitational acceler","description":"The correct answer is A) $sqrt{2}$ T. The time period (T) of a simple pendulum is given by the formula $T = 2pi sqrt{frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum. For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2pi sqrt{frac{L}{g}} = T$. For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2pi sqrt{frac{L_2}{g_2}} = 2pi sqrt{frac{L}{g\/2}} = 2pi sqrt{frac{2L}{g}}$. We can rewrite $T_2$ in terms of $T_1$: $T_2 = sqrt{2} times (2pi sqrt{frac{L}{g}}) = sqrt{2} T_1 = sqrt{2} T$.","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-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/","og_locale":"en_US","og_type":"article","og_title":"A pendulum clock is lifted to a height where the gravitational acceler","og_description":"The correct answer is A) $sqrt{2}$ T. The time period (T) of a simple pendulum is given by the formula $T = 2pi sqrt{frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum. For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2pi sqrt{frac{L}{g}} = T$. For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2pi sqrt{frac{L_2}{g_2}} = 2pi sqrt{frac{L}{g\/2}} = 2pi sqrt{frac{2L}{g}}$. We can rewrite $T_2$ in terms of $T_1$: $T_2 = sqrt{2} times (2pi sqrt{frac{L}{g}}) = sqrt{2} T_1 = sqrt{2} T$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:14:50+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-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/","name":"A pendulum clock is lifted to a height where the gravitational acceler","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:14:50+00:00","dateModified":"2025-06-01T07:14:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is A) $\\sqrt{2}$ T. The time period (T) of a simple pendulum is given by the formula $T = 2\\pi \\sqrt{\\frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum. For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2\\pi \\sqrt{\\frac{L}{g}} = T$. For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g\/2$. Time period $T_2 = 2\\pi \\sqrt{\\frac{L_2}{g_2}} = 2\\pi \\sqrt{\\frac{L}{g\/2}} = 2\\pi \\sqrt{\\frac{2L}{g}}$. We can rewrite $T_2$ in terms of $T_1$: $T_2 = \\sqrt{2} \\times (2\\pi \\sqrt{\\frac{L}{g}}) = \\sqrt{2} T_1 = \\sqrt{2} T$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-pendulum-clock-is-lifted-to-a-height-where-the-gravitational-acceler\/#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 pendulum clock is lifted to a height where the gravitational acceler"}]},{"@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\/88560","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=88560"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88560\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88560"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88560"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88560"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}