{"id":84767,"date":"2025-06-01T02:49:33","date_gmt":"2025-06-01T02:49:33","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=84767"},"modified":"2025-06-01T02:49:33","modified_gmt":"2025-06-01T02:49:33","slug":"two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/","title":{"rendered":"Two metallic wires made from copper have same length but the radius of"},"content":{"rendered":"

Two metallic wires made from copper have same length but the radius of wire 1 is half of that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes<\/p>\n

[amp_mcq option1=”2R” option2=”R\/2″ option3=”5R\/4″ option4=”3R\/4″ correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-1 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius.
\nFor wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R.
\nFor wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)].
\nSince R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4.
\nWhen the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2.
\nR_total = R + R\/4 = 5R\/4.
\n<\/section>\n
\n– Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material.
\n– When conductors are in series, their resistances add up (R_total = R1 + R2 + …).
\n<\/section>\n
\nResistivity (\u03c1) is a material property. Since both wires are made of copper, they have the same resistivity. The formula for resistance is derived from Ohm’s law and material properties. In series connection, the current is the same through both components, and the total voltage across the combination is the sum of voltage drops across each component.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two metallic wires made from copper have same length but the radius of wire 1 is half of that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes [amp_mcq option1=”2R” option2=”R\/2″ option3=”5R\/4″ option4=”3R\/4″ correct=”option3″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionTwo metallic wires made from copper have same length but the radius of<\/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":[1087],"tags":[1114,1200,1128],"class_list":["post-84767","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-1","tag-1114","tag-conductivity","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo metallic wires made from copper have same length but the radius of<\/title>\n<meta name=\"description\" content=\"The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius. For wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R. For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)]. Since R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4. When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2. R_total = R + R\/4 = 5R\/4. - Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material. - When conductors are in series, their resistances add up (R_total = R1 + R2 + ...).\" \/>\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\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two metallic wires made from copper have same length but the radius of\" \/>\n<meta property=\"og:description\" content=\"The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius. For wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R. For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)]. Since R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4. When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2. R_total = R + R\/4 = 5R\/4. - Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material. - When conductors are in series, their resistances add up (R_total = R1 + R2 + ...).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T02:49:33+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":"Two metallic wires made from copper have same length but the radius of","description":"The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius. For wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R. For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)]. Since R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4. When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2. R_total = R + R\/4 = 5R\/4. - Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material. - When conductors are in series, their resistances add up (R_total = R1 + R2 + ...).","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\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/","og_locale":"en_US","og_type":"article","og_title":"Two metallic wires made from copper have same length but the radius of","og_description":"The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius. For wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R. For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)]. Since R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4. When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2. R_total = R + R\/4 = 5R\/4. - Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material. - When conductors are in series, their resistances add up (R_total = R1 + R2 + ...).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T02:49:33+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\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/","name":"Two metallic wires made from copper have same length but the radius of","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T02:49:33+00:00","dateModified":"2025-06-01T02:49:33+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The resistance of a wire is given by R = \u03c1 * (l\/A), where \u03c1 is the resistivity, l is the length, and A is the cross-sectional area. The area A = \u03c0r\u00b2, where r is the radius. For wire 1, with radius r1 and length l, the resistance R1 = \u03c1 * l \/ (\u03c0r1\u00b2). We are given R1 = R. For wire 2, with the same length l but radius r2 = 2 * r1 (since r1 is half of r2, r2 is double r1), the resistance R2 = \u03c1 * l \/ (\u03c0r2\u00b2) = \u03c1 * l \/ (\u03c0(2r1)\u00b2) = \u03c1 * l \/ (\u03c0 * 4r1\u00b2) = (1\/4) * [\u03c1 * l \/ (\u03c0r1\u00b2)]. Since R = \u03c1 * l \/ (\u03c0r1\u00b2), we have R2 = R\/4. When the two wires are joined in series, the total resistance is the sum of individual resistances: R_total = R1 + R2. R_total = R + R\/4 = 5R\/4. - Resistance is inversely proportional to the square of the radius (R \u221d 1\/r\u00b2), assuming constant length and material. - When conductors are in series, their resistances add up (R_total = R1 + R2 + ...).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-made-from-copper-have-same-length-but-the-radius-of\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-1\/"},{"@type":"ListItem","position":3,"name":"Two metallic wires made from copper have same length but the radius of"}]},{"@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\/84767","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=84767"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/84767\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=84767"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=84767"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=84767"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}