{"id":88706,"date":"2025-06-01T07:18:45","date_gmt":"2025-06-01T07:18:45","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88706"},"modified":"2025-06-01T07:18:45","modified_gmt":"2025-06-01T07:18:45","slug":"a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/","title":{"rendered":"A metallic wire having resistance of 20 \u03a9 is cut into two equal parts"},"content":{"rendered":"

A metallic wire having resistance of 20 \u03a9 is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to<\/p>\n

[amp_mcq option1=”20 \u03a9” option2=”10 \u03a9” option3=”5 \u03a9” option4=”15 \u03a9” correct=”option3″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-2 – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9.
\n<\/section>\n
\nThe resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
\n<\/section>\n
\nFor two resistors R\u2081 and R\u2082 in parallel, the equivalent resistance can also be calculated as R\u209a = (R\u2081 * R\u2082) \/ (R\u2081 + R\u2082). In this case, R\u209a = (10 \u03a9 * 10 \u03a9) \/ (10 \u03a9 + 10 \u03a9) = 100 \/ 20 = 5 \u03a9.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A metallic wire having resistance of 20 \u03a9 is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to [amp_mcq option1=”20 \u03a9” option2=”10 \u03a9” option3=”5 \u03a9” option4=”15 \u03a9” correct=”option3″] This question was previously asked in UPSC NDA-2 – 2020 Download PDFAttempt Online … <\/p>\n

Detailed SolutionA metallic wire having resistance of 20 \u03a9 is cut into two equal parts<\/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":[1094],"tags":[1288,1201,1128],"class_list":["post-88706","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1288","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nA metallic wire having resistance of 20 \u03a9 is cut into two equal parts<\/title>\n<meta name=\"description\" content=\"The original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9. The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.\" \/>\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-metallic-wire-having-resistance-of-20-\u03c9-is-cut-into-two-equal-parts\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A metallic wire having resistance of 20 \u03a9 is cut into two equal parts\" \/>\n<meta property=\"og:description\" content=\"The original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9. The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-\u03c9-is-cut-into-two-equal-parts\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:18:45+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 metallic wire having resistance of 20 \u03a9 is cut into two equal parts","description":"The original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9. The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.","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-metallic-wire-having-resistance-of-20-\u03c9-is-cut-into-two-equal-parts\/","og_locale":"en_US","og_type":"article","og_title":"A metallic wire having resistance of 20 \u03a9 is cut into two equal parts","og_description":"The original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9. The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-\u03c9-is-cut-into-two-equal-parts\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:18:45+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\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/","name":"A metallic wire having resistance of 20 \u03a9 is cut into two equal parts","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:18:45+00:00","dateModified":"2025-06-01T07:18:45+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The original metallic wire has a resistance R = 20 \u03a9. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R\u2081 = R\u2082 = R\/2 = 20 \u03a9 \/ 2 = 10 \u03a9. When these two parts are connected in parallel, the equivalent resistance (R\u209a) is given by the formula 1\/R\u209a = 1\/R\u2081 + 1\/R\u2082. Substituting the values, 1\/R\u209a = 1\/10 \u03a9 + 1\/10 \u03a9 = 2\/10 \u03a9 = 1\/5 \u03a9. Therefore, the equivalent resistance R\u209a = 5 \u03a9. The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-metallic-wire-having-resistance-of-20-%cf%89-is-cut-into-two-equal-parts\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"A metallic wire having resistance of 20 \u03a9 is cut into two equal parts"}]},{"@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\/88706","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=88706"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88706\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}