{"id":86996,"date":"2025-06-01T04:25:07","date_gmt":"2025-06-01T04:25:07","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=86996"},"modified":"2025-06-01T04:25:07","modified_gmt":"2025-06-01T04:25:07","slug":"a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/","title":{"rendered":"A steel rod having radius r and length L gets stretched along its leng"},"content":{"rendered":"

A steel rod having radius r and length L gets stretched along its length by \u0394L, when a force F is applied to it. If another rod made of the same material having radius 2r and length L is subjected to the same force F, then the elongation observed for the second rod is<\/p>\n

[amp_mcq option1=”4\u0394L” option2=”2\u0394L” option3=”\u0394L \/4″ option4=”\u0394L \/2″ correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nAccording to Hooke’s Law, for a material under tensile stress, Young’s Modulus (Y) is given by Y = Stress \/ Strain.
\nStress = F\/A, where F is the force applied and A is the cross-sectional area.
\nStrain = \u0394L\/L, where \u0394L is the elongation and L is the original length.
\nSo, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L).
\nRearranging for elongation, \u0394L = (F * L) \/ (A * Y).
\nThe cross-sectional area of a rod with radius r is A = \u03c0r\u00b2.
\nFor the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y).
\nFor the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F.
\nThe cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2.
\nThe elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y).
\nSubstituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082:
\n\u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081.
\nThus, the elongation observed for the second rod is \u0394L\/4.
\n<\/section>\n
\nElongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.
\n<\/section>\n
\nThis problem assumes the material behaves elastically and obeys Hooke’s Law. Young’s modulus is a property of the material. The force applied must be within the elastic limit of the material.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A steel rod having radius r and length L gets stretched along its length by \u0394L, when a force F is applied to it. If another rod made of the same material having radius 2r and length L is subjected to the same force F, then the elongation observed for the second rod is [amp_mcq … <\/p>\n

Detailed SolutionA steel rod having radius r and length L gets stretched along its leng<\/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":[1091],"tags":[1105,1160,1128],"class_list":["post-86996","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1105","tag-physical-properties-of-materials","tag-physics","no-featured-image-padding"],"yoast_head":"\nA steel rod having radius r and length L gets stretched along its leng<\/title>\n<meta name=\"description\" content=\"According to Hooke's Law, for a material under tensile stress, Young's Modulus (Y) is given by Y = Stress \/ Strain. Stress = F\/A, where F is the force applied and A is the cross-sectional area. Strain = \u0394L\/L, where \u0394L is the elongation and L is the original length. So, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L). Rearranging for elongation, \u0394L = (F * L) \/ (A * Y). The cross-sectional area of a rod with radius r is A = \u03c0r\u00b2. For the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y). For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F. The cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2. The elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y). Substituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082: \u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081. Thus, the elongation observed for the second rod is \u0394L\/4. Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.\" \/>\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-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A steel rod having radius r and length L gets stretched along its leng\" \/>\n<meta property=\"og:description\" content=\"According to Hooke's Law, for a material under tensile stress, Young's Modulus (Y) is given by Y = Stress \/ Strain. Stress = F\/A, where F is the force applied and A is the cross-sectional area. Strain = \u0394L\/L, where \u0394L is the elongation and L is the original length. So, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L). Rearranging for elongation, \u0394L = (F * L) \/ (A * Y). The cross-sectional area of a rod with radius r is A = \u03c0r\u00b2. For the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y). For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F. The cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2. The elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y). Substituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082: \u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081. Thus, the elongation observed for the second rod is \u0394L\/4. Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:25:07+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A steel rod having radius r and length L gets stretched along its leng","description":"According to Hooke's Law, for a material under tensile stress, Young's Modulus (Y) is given by Y = Stress \/ Strain. Stress = F\/A, where F is the force applied and A is the cross-sectional area. Strain = \u0394L\/L, where \u0394L is the elongation and L is the original length. So, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L). Rearranging for elongation, \u0394L = (F * L) \/ (A * Y). The cross-sectional area of a rod with radius r is A = \u03c0r\u00b2. For the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y). For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F. The cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2. The elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y). Substituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082: \u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081. Thus, the elongation observed for the second rod is \u0394L\/4. Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.","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-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/","og_locale":"en_US","og_type":"article","og_title":"A steel rod having radius r and length L gets stretched along its leng","og_description":"According to Hooke's Law, for a material under tensile stress, Young's Modulus (Y) is given by Y = Stress \/ Strain. Stress = F\/A, where F is the force applied and A is the cross-sectional area. Strain = \u0394L\/L, where \u0394L is the elongation and L is the original length. So, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L). Rearranging for elongation, \u0394L = (F * L) \/ (A * Y). The cross-sectional area of a rod with radius r is A = \u03c0r\u00b2. For the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y). For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F. The cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2. The elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y). Substituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082: \u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081. Thus, the elongation observed for the second rod is \u0394L\/4. Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:25:07+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/","name":"A steel rod having radius r and length L gets stretched along its leng","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:25:07+00:00","dateModified":"2025-06-01T04:25:07+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"According to Hooke's Law, for a material under tensile stress, Young's Modulus (Y) is given by Y = Stress \/ Strain. Stress = F\/A, where F is the force applied and A is the cross-sectional area. Strain = \u0394L\/L, where \u0394L is the elongation and L is the original length. So, Y = (F\/A) \/ (\u0394L\/L) = (F * L) \/ (A * \u0394L). Rearranging for elongation, \u0394L = (F * L) \/ (A * Y). The cross-sectional area of a rod with radius r is A = \u03c0r\u00b2. For the first rod: \u0394L\u2081 = (F * L) \/ (\u03c0r\u00b2 * Y). For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F. The cross-sectional area of the second rod is A\u2082 = \u03c0(2r)\u00b2 = 4\u03c0r\u00b2. The elongation for the second rod is \u0394L\u2082 = (F * L) \/ (A\u2082 * Y) = (F * L) \/ (4\u03c0r\u00b2 * Y). Substituting the expression for \u0394L\u2081 into the equation for \u0394L\u2082: \u0394L\u2082 = (1\/4) * [(F * L) \/ (\u03c0r\u00b2 * Y)] = (1\/4) * \u0394L\u2081. Thus, the elongation observed for the second rod is \u0394L\/4. Elongation under a given force is inversely proportional to the cross-sectional area of the rod, assuming the material, length, and force are constant. Area is proportional to the square of the radius.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-steel-rod-having-radius-r-and-length-l-gets-stretched-along-its-leng\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"A steel rod having radius r and length L gets stretched along its leng"}]},{"@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\/86996","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=86996"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86996\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=86996"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=86996"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=86996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}