{"id":88063,"date":"2025-06-01T07:00:44","date_gmt":"2025-06-01T07:00:44","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88063"},"modified":"2025-06-01T07:00:44","modified_gmt":"2025-06-01T07:00:44","slug":"a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/","title":{"rendered":"A uniform meter scale of mass 0.24 kg is made of steel. It is kept on"},"content":{"rendered":"

A uniform meter scale of mass 0.24 kg is made of steel. It is kept on two wedges, W\u2081 and W\u2082, in a horizontal position. W\u2081 is at a distance of 0.2 m from one of its ends, while W\u2082 is at distance of 0.4 m from the other end. If the force on the scale is N\u2081 due to W\u2081 and N\u2082 due to W\u2082, then : (take g = 10.0 m s\u207b\u00b2)<\/p>\n

[amp_mcq option1=”N\u2081 = 1.6 N and N\u2082 = 0.8 N” option2=”N\u2081 = 0.8 N and N\u2082 = 1.6 N” option3=”N\u2081 = 0.6 N and N\u2082 = 1.8 N” option4=”N\u2081 = 1.8 N and N\u2082 = 0.6 N” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N.
\n<\/section>\n
\nFor a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 – 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 – 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 – 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 – 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 – 1.8 = 0.6 N.
\n<\/section>\n
\nThis problem is a classic example of applying the conditions for static equilibrium: zero net force and zero net torque. Choosing the pivot point for calculating torques at one of the unknown force locations simplifies the calculation by eliminating that force from the torque equation.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A uniform meter scale of mass 0.24 kg is made of steel. It is kept on two wedges, W\u2081 and W\u2082, in a horizontal position. W\u2081 is at a distance of 0.2 m from one of its ends, while W\u2082 is at distance of 0.4 m from the other end. If the force on the … <\/p>\n

Detailed SolutionA uniform meter scale of mass 0.24 kg is made of steel. It is kept on<\/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":[1103,1129,1128],"class_list":["post-88063","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA uniform meter scale of mass 0.24 kg is made of steel. It is kept on<\/title>\n<meta name=\"description\" content=\"The correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N. For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 - 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 - 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 - 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 - 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 - 1.8 = 0.6 N.\" \/>\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-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A uniform meter scale of mass 0.24 kg is made of steel. It is kept on\" \/>\n<meta property=\"og:description\" content=\"The correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N. For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 - 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 - 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 - 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 - 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 - 1.8 = 0.6 N.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:00:44+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 uniform meter scale of mass 0.24 kg is made of steel. It is kept on","description":"The correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N. For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 - 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 - 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 - 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 - 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 - 1.8 = 0.6 N.","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-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/","og_locale":"en_US","og_type":"article","og_title":"A uniform meter scale of mass 0.24 kg is made of steel. It is kept on","og_description":"The correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N. For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 - 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 - 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 - 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 - 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 - 1.8 = 0.6 N.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:00:44+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-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/","name":"A uniform meter scale of mass 0.24 kg is made of steel. It is kept on","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:00:44+00:00","dateModified":"2025-06-01T07:00:44+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct option is C. Calculating the forces using equilibrium conditions yields N\u2081 = 0.6 N and N\u2082 = 1.8 N. For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W\u2081 is at 0.2 m from one end (say, the 0m mark), and W\u2082 is at 0.4 m from the other end (the 1m mark), placing it at 1.0 - 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N\u2081 + N\u2082 = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W\u2081 (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 - 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N\u2082 (at 0.6m) creates an anticlockwise torque N\u2082 * (0.6 - 0.2) = N\u2082 * 0.4 Nm. Setting the sum of torques to zero: 0.72 - 0.4 N\u2082 = 0, which gives N\u2082 = 0.72 \/ 0.4 = 1.8 N. Substituting N\u2082 into the force equation: N\u2081 + 1.8 = 2.4, which gives N\u2081 = 2.4 - 1.8 = 0.6 N.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-uniform-meter-scale-of-mass-0-24-kg-is-made-of-steel-it-is-kept-on\/#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":"A uniform meter scale of mass 0.24 kg is made of steel. It is kept on"}]},{"@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\/88063","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=88063"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88063\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88063"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88063"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}