{"id":87675,"date":"2025-06-01T06:51:03","date_gmt":"2025-06-01T06:51:03","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87675"},"modified":"2025-06-01T06:51:03","modified_gmt":"2025-06-01T06:51:03","slug":"two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/","title":{"rendered":"Two metallic wires A and B are made using copper. The radius of wire A"},"content":{"rendered":"

Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across the wire A, causing power dissipation, P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is applied across it causing power dissipation, P\u2081. Which one of the following is the correct relationship between P and P\u2081?<\/p>\n

[amp_mcq option1=”P = 2P\u2081” option2=”P = P\u2081\/2″ option3=”P = 4P\u2081” option4=”P = P\u2081” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct relationship is P = P\u2081\/2.
\n<\/section>\n
\nThe resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R.
\nFor wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l).
\nFor wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)).
\nPower P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l).
\nComparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)].
\nSince P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.
\n<\/section>\n
\nThe resistance of wire B is half the resistance of wire A (R_B = (\u03c1l \/ (2\u03c0r\u00b2)) compared to R_A = (\u03c1l \/ (\u03c0r\u00b2))). Since power is inversely proportional to resistance for a constant voltage (P = V\u00b2\/R), lower resistance leads to higher power dissipation. Therefore, wire B dissipates more power than wire A when the same voltage is applied.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across the wire A, causing power dissipation, P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is … <\/p>\n

Detailed SolutionTwo metallic wires A and B are made using copper. The radius of wire A<\/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":[1119,1201,1128],"class_list":["post-87675","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1119","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo metallic wires A and B are made using copper. The radius of wire A<\/title>\n<meta name=\"description\" content=\"The correct relationship is P = P\u2081\/2. The resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R. For wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l). For wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)). Power P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l). Comparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)]. Since P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.\" \/>\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-a-and-b-are-made-using-copper-the-radius-of-wire-a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two metallic wires A and B are made using copper. The radius of wire A\" \/>\n<meta property=\"og:description\" content=\"The correct relationship is P = P\u2081\/2. The resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R. For wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l). For wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)). Power P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l). Comparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)]. Since P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:51:03+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 A and B are made using copper. The radius of wire A","description":"The correct relationship is P = P\u2081\/2. The resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R. For wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l). For wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)). Power P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l). Comparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)]. Since P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.","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-a-and-b-are-made-using-copper-the-radius-of-wire-a\/","og_locale":"en_US","og_type":"article","og_title":"Two metallic wires A and B are made using copper. The radius of wire A","og_description":"The correct relationship is P = P\u2081\/2. The resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R. For wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l). For wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)). Power P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l). Comparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)]. Since P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:51:03+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-a-and-b-are-made-using-copper-the-radius-of-wire-a\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/","name":"Two metallic wires A and B are made using copper. The radius of wire A","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:51:03+00:00","dateModified":"2025-06-01T06:51:03+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct relationship is P = P\u2081\/2. The resistance of a wire is given by R = \u03c1 * (l \/ A), where \u03c1 is resistivity, l is length, and A is the cross-sectional area (A = \u03c0r\u00b2). Power dissipated is P = V\u00b2\/R. For wire A: R_A = \u03c1 * (l \/ (\u03c0r\u00b2)). Power P = V\u00b2 \/ R_A = V\u00b2 \/ [\u03c1 * (l \/ (\u03c0r\u00b2))] = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l). For wire B: radius is 2r, length is 2l. R_B = \u03c1 * (2l \/ (\u03c0(2r)\u00b2)) = \u03c1 * (2l \/ (4\u03c0r\u00b2)) = \u03c1 * (l \/ (2\u03c0r\u00b2)). Power P\u2081 = V\u00b2 \/ R_B = V\u00b2 \/ [\u03c1 * (l \/ (2\u03c0r\u00b2))] = (2V\u00b2\u03c0r\u00b2) \/ (\u03c1l). Comparing P and P\u2081: P\u2081 = 2 * [(V\u00b2\u03c0r\u00b2) \/ (\u03c1l)]. Since P = (V\u00b2\u03c0r\u00b2) \/ (\u03c1l), we have P\u2081 = 2P. This relationship can be rewritten as P = P\u2081\/2.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-metallic-wires-a-and-b-are-made-using-copper-the-radius-of-wire-a\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"Two metallic wires A and B are made using copper. The radius of wire A"}]},{"@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\/87675","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=87675"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87675\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87675"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87675"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87675"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}