{"id":88010,"date":"2025-06-01T06:59:35","date_gmt":"2025-06-01T06:59:35","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88010"},"modified":"2025-06-01T06:59:35","modified_gmt":"2025-06-01T06:59:35","slug":"a-mass-is-attached-to-a-spring-that-hangs-vertically-the-extension-pr","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-is-attached-to-a-spring-that-hangs-vertically-the-extension-pr\/","title":{"rendered":"A mass is attached to a spring that hangs vertically. The extension pr"},"content":{"rendered":"

A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be :<\/p>\n

[amp_mcq option1=”6 cm” option2=”1 cm” option3=”0 cm” option4=”36 cm” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nAccording to Hooke’s Law, the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. The force stretching the vertical spring is the weight of the attached mass. The weight of the mass is proportional to the acceleration due to gravity.
\n<\/section>\n
\n– Hooke’s Law: Force (F) = spring constant (k) \u00d7 extension (x), or F = kx.
\n– Weight (W) = mass (m) \u00d7 acceleration due to gravity (g).
\n– On Earth: $W_{Earth} = mg_{Earth} = kx_{Earth}$.
\n– On the Moon: $W_{Moon} = mg_{Moon} = kx_{Moon}$.
\n<\/section>\n
\nGiven: Extension on Earth $x_{Earth} = 6$ cm.
\nAcceleration due to gravity on the Moon $g_{Moon} = \\frac{1}{6} g_{Earth}$.
\nFrom Hooke’s Law on Earth: $mg_{Earth} = k \\times 6$ cm. So, $\\frac{mg_{Earth}}{k} = 6$ cm.
\nOn the Moon: $mg_{Moon} = kx_{Moon}$.
\nSubstitute $g_{Moon}$: $m \\left(\\frac{g_{Earth}}{6}\\right) = kx_{Moon}$.
\n$\\frac{1}{6} (mg_{Earth}) = kx_{Moon}$.
\nSubstitute $mg_{Earth} = k \\times 6$ cm: $\\frac{1}{6} (k \\times 6 \\text{ cm}) = kx_{Moon}$.
\n$k \\times 1 \\text{ cm} = kx_{Moon}$.
\n$x_{Moon} = 1$ cm.
\nThe extension is directly proportional to the weight, and thus directly proportional to gravity. Since gravity on the Moon is one-sixth of Earth’s, the extension will also be one-sixth.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be : [amp_mcq … <\/p>\n

Detailed SolutionA mass is attached to a spring that hangs vertically. The extension pr<\/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":[1105,1129,1128],"class_list":["post-88010","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA mass is attached to a spring that hangs vertically. The extension pr<\/title>\n<meta name=\"description\" content=\"According to Hooke's Law, the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. The force stretching the vertical spring is the weight of the attached mass. The weight of the mass is proportional to the acceleration due to gravity. - Hooke's Law: Force (F) = spring constant (k) \u00d7 extension (x), or F = kx. - Weight (W) = mass (m) \u00d7 acceleration due to gravity (g). - On Earth: $W_{Earth} = mg_{Earth} = kx_{Earth}$. - On the Moon: $W_{Moon} = mg_{Moon} = kx_{Moon}$.\" \/>\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-mass-is-attached-to-a-spring-that-hangs-vertically-the-extension-pr\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A mass is attached to a spring that hangs vertically. 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The weight of the mass is proportional to the acceleration due to gravity. - Hooke's Law: Force (F) = spring constant (k) \u00d7 extension (x), or F = kx. - Weight (W) = mass (m) \u00d7 acceleration due to gravity (g). - On Earth: $W_{Earth} = mg_{Earth} = kx_{Earth}$. - On the Moon: $W_{Moon} = mg_{Moon} = kx_{Moon}$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-mass-is-attached-to-a-spring-that-hangs-vertically-the-extension-pr\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:59:35+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 mass is attached to a spring that hangs vertically. 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The extension pr","og_description":"According to Hooke's Law, the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. The force stretching the vertical spring is the weight of the attached mass. 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