{"id":90872,"date":"2025-06-01T10:39:20","date_gmt":"2025-06-01T10:39:20","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90872"},"modified":"2025-06-01T10:39:20","modified_gmt":"2025-06-01T10:39:20","slug":"a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/","title":{"rendered":"A wire of resistance R is cut into four equal parts. These parts are t"},"content":{"rendered":"

A wire of resistance R is cut into four equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio $\\frac{\\text{R’}}{\\text{R}}$ is :<\/p>\n

[amp_mcq option1=”$\\frac{1}{16}$” option2=”$\\frac{1}{4}$” option3=”4″ option4=”16″ correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe ratio $\\frac{\\text{R’}}{\\text{R}}$ is $\\frac{1}{16}$.
\n<\/section>\n
\nLet the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $\\frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = \\frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R’ of resistances connected in parallel is given by the formula: $\\frac{1}{R’} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\frac{1}{R_3} + \\frac{1}{R_4}$. Substituting the values, we get: $\\frac{1}{R’} = \\frac{1}{R\/4} + \\frac{1}{R\/4} + \\frac{1}{R\/4} + \\frac{1}{R\/4} = \\frac{4}{R} + \\frac{4}{R} + \\frac{4}{R} + \\frac{4}{R} = \\frac{4+4+4+4}{R} = \\frac{16}{R}$. Therefore, $R’ = \\frac{R}{16}$. The required ratio $\\frac{\\text{R’}}{\\text{R}}$ is $\\frac{R\/16}{R} = \\frac{1}{16}$.
\n<\/section>\n
\nThe resistance of a wire is directly proportional to its length. Cutting a wire into four equal parts reduces the length of each part to one-fourth of the original length, thus reducing the resistance of each part to R\/4. Connecting resistors in parallel decreases the total equivalent resistance compared to the individual resistances. This principle is used in electrical circuits to control current flow.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A wire of resistance R is cut into four equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio $\\frac{\\text{R’}}{\\text{R}}$ is : [amp_mcq option1=”$\\frac{1}{16}$” option2=”$\\frac{1}{4}$” option3=”4″ option4=”16″ correct=”option1″] This question was previously asked in UPSC CAPF – 2023 Download PDFAttempt Online The ratio $\\frac{\\text{R’}}{\\text{R}}$ … <\/p>\n

Detailed SolutionA wire of resistance R is cut into four equal parts. These parts are t<\/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":[1085],"tags":[1105,1201,1128],"class_list":["post-90872","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1105","tag-electric-current","tag-physics","no-featured-image-padding"],"yoast_head":"\nA wire of resistance R is cut into four equal parts. These parts are t<\/title>\n<meta name=\"description\" content=\"The ratio $frac{text{R'}}{text{R}}$ is $frac{1}{16}$. Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R' of resistances connected in parallel is given by the formula: $frac{1}{R'} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} + frac{1}{R_4}$. Substituting the values, we get: $frac{1}{R'} = frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} = frac{4}{R} + frac{4}{R} + frac{4}{R} + frac{4}{R} = frac{4+4+4+4}{R} = frac{16}{R}$. Therefore, $R' = frac{R}{16}$. The required ratio $frac{text{R'}}{text{R}}$ is $frac{R\/16}{R} = frac{1}{16}$.\" \/>\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-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A wire of resistance R is cut into four equal parts. These parts are t\" \/>\n<meta property=\"og:description\" content=\"The ratio $frac{text{R'}}{text{R}}$ is $frac{1}{16}$. Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R' of resistances connected in parallel is given by the formula: $frac{1}{R'} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} + frac{1}{R_4}$. Substituting the values, we get: $frac{1}{R'} = frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} = frac{4}{R} + frac{4}{R} + frac{4}{R} + frac{4}{R} = frac{4+4+4+4}{R} = frac{16}{R}$. Therefore, $R' = frac{R}{16}$. The required ratio $frac{text{R'}}{text{R}}$ is $frac{R\/16}{R} = frac{1}{16}$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:39:20+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 wire of resistance R is cut into four equal parts. These parts are t","description":"The ratio $frac{text{R'}}{text{R}}$ is $frac{1}{16}$. Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R' of resistances connected in parallel is given by the formula: $frac{1}{R'} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} + frac{1}{R_4}$. Substituting the values, we get: $frac{1}{R'} = frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} = frac{4}{R} + frac{4}{R} + frac{4}{R} + frac{4}{R} = frac{4+4+4+4}{R} = frac{16}{R}$. Therefore, $R' = frac{R}{16}$. The required ratio $frac{text{R'}}{text{R}}$ is $frac{R\/16}{R} = frac{1}{16}$.","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-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/","og_locale":"en_US","og_type":"article","og_title":"A wire of resistance R is cut into four equal parts. These parts are t","og_description":"The ratio $frac{text{R'}}{text{R}}$ is $frac{1}{16}$. Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R' of resistances connected in parallel is given by the formula: $frac{1}{R'} = frac{1}{R_1} + frac{1}{R_2} + frac{1}{R_3} + frac{1}{R_4}$. Substituting the values, we get: $frac{1}{R'} = frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} + frac{1}{R\/4} = frac{4}{R} + frac{4}{R} + frac{4}{R} + frac{4}{R} = frac{4+4+4+4}{R} = frac{16}{R}$. Therefore, $R' = frac{R}{16}$. The required ratio $frac{text{R'}}{text{R}}$ is $frac{R\/16}{R} = frac{1}{16}$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:39:20+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-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/","name":"A wire of resistance R is cut into four equal parts. These parts are t","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:39:20+00:00","dateModified":"2025-06-01T10:39:20+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The ratio $\\frac{\\text{R'}}{\\text{R}}$ is $\\frac{1}{16}$. Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $\\frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = \\frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R' of resistances connected in parallel is given by the formula: $\\frac{1}{R'} = \\frac{1}{R_1} + \\frac{1}{R_2} + \\frac{1}{R_3} + \\frac{1}{R_4}$. Substituting the values, we get: $\\frac{1}{R'} = \\frac{1}{R\/4} + \\frac{1}{R\/4} + \\frac{1}{R\/4} + \\frac{1}{R\/4} = \\frac{4}{R} + \\frac{4}{R} + \\frac{4}{R} + \\frac{4}{R} = \\frac{4+4+4+4}{R} = \\frac{16}{R}$. Therefore, $R' = \\frac{R}{16}$. The required ratio $\\frac{\\text{R'}}{\\text{R}}$ is $\\frac{R\/16}{R} = \\frac{1}{16}$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-resistance-r-is-cut-into-four-equal-parts-these-parts-are-t\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"A wire of resistance R is cut into four equal parts. These parts are t"}]},{"@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\/90872","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=90872"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90872\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90872"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90872"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90872"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}