{"id":87849,"date":"2025-06-01T06:55:24","date_gmt":"2025-06-01T06:55:24","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87849"},"modified":"2025-06-01T06:55:24","modified_gmt":"2025-06-01T06:55:24","slug":"the-time-period-of-a-1-m-long-pendulum-approximates-to","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-time-period-of-a-1-m-long-pendulum-approximates-to\/","title":{"rendered":"The time period of a 1 m long pendulum approximates to"},"content":{"rendered":"

The time period of a 1 m long pendulum approximates to<\/p>\n

[amp_mcq option1=”6 s” option2=”4 s” option3=”2 s” option4=”1 s” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe time period of a 1 m long pendulum approximates to 2 s.
\n<\/section>\n
\nThe time period ($T$) of a simple pendulum for small oscillations 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.
\n<\/section>\n
\nGiven length $L = 1$ m. Taking the standard value of $g \\approx 9.8$ m\/s\u00b2 and $\\pi \\approx 3.14$: $T = 2 \\times 3.14 \\times \\sqrt{\\frac{1}{9.8}} \\approx 6.28 \\times \\sqrt{0.102} \\approx 6.28 \\times 0.319 \\approx 2.005$ s. A common approximation for calculation purposes is sometimes $\\pi^2 \\approx g$, which gives $T = 2\\pi \\sqrt{\\frac{1}{\\pi^2}} = 2\\pi \\times \\frac{1}{\\pi} = 2$ s. Both calculations yield a value very close to 2 seconds.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The time period of a 1 m long pendulum approximates to [amp_mcq option1=”6 s” option2=”4 s” option3=”2 s” option4=”1 s” correct=”option3″] This question was previously asked in UPSC NDA-1 – 2022 Download PDFAttempt Online The time period of a 1 m long pendulum approximates to 2 s. The time period ($T$) of a simple pendulum … <\/p>\n

Detailed SolutionThe time period of a 1 m long pendulum approximates to<\/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":[1093],"tags":[1108,1129,1128],"class_list":["post-87849","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1108","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe time period of a 1 m long pendulum approximates to<\/title>\n<meta name=\"description\" content=\"The time period of a 1 m long pendulum approximates to 2 s. 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