{"id":90684,"date":"2025-06-01T10:33:38","date_gmt":"2025-06-01T10:33:38","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90684"},"modified":"2025-06-01T10:33:38","modified_gmt":"2025-06-01T10:33:38","slug":"a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/","title":{"rendered":"A solid spherical ball made of iron is melted and two new balls are ma"},"content":{"rendered":"

A solid spherical ball made of iron is melted and two new balls are made whose diameters are in the ratio of 1:2. The ratio of the volume of the smaller new ball to the original ball is<\/p>\n

[amp_mcq option1=”1:3″ option2=”1:5″ option3=”2:9″ option4=”1:9″ correct=”option4″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC CAPF – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
Let the radius of the original ball be $R$. Its volume is $V_{orig} = \\frac{4}{3}\\pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = \\frac{4}{3}\\pi r_1^3$ and $V_2 = \\frac{4}{3}\\pi r_2^3 = \\frac{4}{3}\\pi (2r_1)^3 = \\frac{4}{3}\\pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = \\frac{4}{3}\\pi r_1^3 + \\frac{4}{3}\\pi (8r_1^3) = \\frac{4}{3}\\pi (r_1^3 + 8r_1^3) = \\frac{4}{3}\\pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $\\frac{V_1}{V_{orig}} = \\frac{\\frac{4}{3}\\pi r_1^3}{\\frac{4}{3}\\pi (9r_1^3)} = \\frac{r_1^3}{9r_1^3} = \\frac{1}{9}$. The ratio is 1:9.<\/section>\n
When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V \\propto r^3$).<\/section>\n
If two spheres have radii in the ratio $r_a:r_b = k:l$, their volumes are in the ratio $V_a:V_b = r_a^3:r_b^3 = k^3:l^3$. In this problem, the ratio of radii of the two new spheres is 1:2, so their volumes are in the ratio $1^3:2^3 = 1:8$. The total volume of the two new spheres is proportional to $1+8=9$ units. The smaller new sphere has a volume proportional to 1 unit. The original sphere’s volume is equal to the sum of the volumes of the new spheres, which is proportional to 9 units. Thus, the ratio of the smaller new sphere’s volume to the original sphere’s volume is 1:9.<\/section>\n","protected":false},"excerpt":{"rendered":"

A solid spherical ball made of iron is melted and two new balls are made whose diameters are in the ratio of 1:2. The ratio of the volume of the smaller new ball to the original ball is [amp_mcq option1=”1:3″ option2=”1:5″ option3=”2:9″ option4=”1:9″ correct=”option4″] This question was previously asked in UPSC CAPF – 2022 Download … <\/p>\n

Detailed SolutionA solid spherical ball made of iron is melted and two new balls are ma<\/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":[1108,1102],"class_list":["post-90684","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1108","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nA solid spherical ball made of iron is melted and two new balls are ma<\/title>\n<meta name=\"description\" content=\"Let the radius of the original ball be $R$. Its volume is $V_{orig} = frac{4}{3}pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = frac{4}{3}pi r_1^3$ and $V_2 = frac{4}{3}pi r_2^3 = frac{4}{3}pi (2r_1)^3 = frac{4}{3}pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = frac{4}{3}pi r_1^3 + frac{4}{3}pi (8r_1^3) = frac{4}{3}pi (r_1^3 + 8r_1^3) = frac{4}{3}pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $frac{V_1}{V_{orig}} = frac{frac{4}{3}pi r_1^3}{frac{4}{3}pi (9r_1^3)} = frac{r_1^3}{9r_1^3} = frac{1}{9}$. The ratio is 1:9. When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V propto r^3$).\" \/>\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-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A solid spherical ball made of iron is melted and two new balls are ma\" \/>\n<meta property=\"og:description\" content=\"Let the radius of the original ball be $R$. Its volume is $V_{orig} = frac{4}{3}pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = frac{4}{3}pi r_1^3$ and $V_2 = frac{4}{3}pi r_2^3 = frac{4}{3}pi (2r_1)^3 = frac{4}{3}pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = frac{4}{3}pi r_1^3 + frac{4}{3}pi (8r_1^3) = frac{4}{3}pi (r_1^3 + 8r_1^3) = frac{4}{3}pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $frac{V_1}{V_{orig}} = frac{frac{4}{3}pi r_1^3}{frac{4}{3}pi (9r_1^3)} = frac{r_1^3}{9r_1^3} = frac{1}{9}$. The ratio is 1:9. When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V propto r^3$).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:33:38+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 solid spherical ball made of iron is melted and two new balls are ma","description":"Let the radius of the original ball be $R$. Its volume is $V_{orig} = frac{4}{3}pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = frac{4}{3}pi r_1^3$ and $V_2 = frac{4}{3}pi r_2^3 = frac{4}{3}pi (2r_1)^3 = frac{4}{3}pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = frac{4}{3}pi r_1^3 + frac{4}{3}pi (8r_1^3) = frac{4}{3}pi (r_1^3 + 8r_1^3) = frac{4}{3}pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $frac{V_1}{V_{orig}} = frac{frac{4}{3}pi r_1^3}{frac{4}{3}pi (9r_1^3)} = frac{r_1^3}{9r_1^3} = frac{1}{9}$. The ratio is 1:9. When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V propto r^3$).","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-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/","og_locale":"en_US","og_type":"article","og_title":"A solid spherical ball made of iron is melted and two new balls are ma","og_description":"Let the radius of the original ball be $R$. Its volume is $V_{orig} = frac{4}{3}pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = frac{4}{3}pi r_1^3$ and $V_2 = frac{4}{3}pi r_2^3 = frac{4}{3}pi (2r_1)^3 = frac{4}{3}pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = frac{4}{3}pi r_1^3 + frac{4}{3}pi (8r_1^3) = frac{4}{3}pi (r_1^3 + 8r_1^3) = frac{4}{3}pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $frac{V_1}{V_{orig}} = frac{frac{4}{3}pi r_1^3}{frac{4}{3}pi (9r_1^3)} = frac{r_1^3}{9r_1^3} = frac{1}{9}$. The ratio is 1:9. When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V propto r^3$).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:33:38+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-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/","name":"A solid spherical ball made of iron is melted and two new balls are ma","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:33:38+00:00","dateModified":"2025-06-01T10:33:38+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Let the radius of the original ball be $R$. Its volume is $V_{orig} = \\frac{4}{3}\\pi R^3$. This ball is melted to make two new balls whose diameters are in the ratio 1:2. Let their radii be $r_1$ and $r_2$. The ratio of radii is the same as the ratio of diameters, so $r_1 : r_2 = 1:2$, or $r_2 = 2r_1$. The volumes of the new balls are $V_1 = \\frac{4}{3}\\pi r_1^3$ and $V_2 = \\frac{4}{3}\\pi r_2^3 = \\frac{4}{3}\\pi (2r_1)^3 = \\frac{4}{3}\\pi (8r_1^3)$. Since the volume is conserved, $V_{orig} = V_1 + V_2 = \\frac{4}{3}\\pi r_1^3 + \\frac{4}{3}\\pi (8r_1^3) = \\frac{4}{3}\\pi (r_1^3 + 8r_1^3) = \\frac{4}{3}\\pi (9r_1^3)$. The question asks for the ratio of the volume of the smaller new ball ($V_1$) to the original ball ($V_{orig}$). This ratio is $\\frac{V_1}{V_{orig}} = \\frac{\\frac{4}{3}\\pi r_1^3}{\\frac{4}{3}\\pi (9r_1^3)} = \\frac{r_1^3}{9r_1^3} = \\frac{1}{9}$. The ratio is 1:9. When a solid is melted and recast into new shapes, the total volume remains constant. The volume of a sphere is proportional to the cube of its radius ($V \\propto r^3$).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-solid-spherical-ball-made-of-iron-is-melted-and-two-new-balls-are-ma\/#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 solid spherical ball made of iron is melted and two new balls are ma"}]},{"@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\/90684","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=90684"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90684\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90684"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90684"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90684"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}