{"id":173099,"date":"2021-10-27T17:03:46","date_gmt":"2021-10-27T11:33:46","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=173099"},"modified":"2021-10-27T17:03:46","modified_gmt":"2021-10-27T11:33:46","slug":"sphere-questions-for-nmat-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/sphere-questions-for-nmat-pdf\/","title":{"rendered":"Sphere Questions for NMAT &#8211; Download [PDF]"},"content":{"rendered":"<h1>Sphere Questions for NMAT &#8211; Download [PDF]<\/h1>\n<p>Download Sphere Questions for NMAT PDF. Top 6 very important Sphere Questions for NMAT based on asked questions in previous exam papers.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/downloads\/13582\" target=\"_blank\" class=\"btn btn-danger  download\">Download Sphere Questions for NMAT<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/pay\/b5hNV\" target=\"_blank\" class=\"btn btn-info \">Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n<p>Take <a href=\"https:\/\/cracku.in\/nmat-mocks\" target=\"_blank\" rel=\"noopener noreferrer\">NMAT mock test<\/a><\/p>\n<p><b>Question 1:\u00a0<\/b>A cricket ball of radius 20cm is covered in ice of uniform thickness. The ice is melting at a consistent rate of 50cm<sup>3<\/sup> everyday. When the thickness of the ice is 10cm, what is the rate of decrease of the thickness of the ice sphere per day?<\/p>\n<p>a)\u00a011%<\/p>\n<p>b)\u00a01%<\/p>\n<p>c)\u00a0.01%<\/p>\n<p>d)\u00a00.001%<\/p>\n<p><b>Question 2:\u00a0<\/b>A solid right circular cone of height 12cm and base radius of 9cm is cut at the top to form a frustum. The cut part is melted and used completely to form a sphere of radius 3cm. What is the height of the frustum?<\/p>\n<p>a)\u00a0$4\\sqrt[3]{3}$<\/p>\n<p>b)\u00a0$3\\sqrt[3]{3}$<\/p>\n<p>c)\u00a0None of the above<\/p>\n<p>d)\u00a0Can\u2019t be determined<\/p>\n<p><b>Question 3:\u00a0<\/b>A sphere of radius 7 cm is melted to form regular tetrahedrons of side $2\\sqrt2$ cm each. What volume of the material is wasted? ($\\pi = 22\/7$)<\/p>\n<p>a)\u00a02 $cm^3$<\/p>\n<p>b)\u00a01.27 $cm^3$<\/p>\n<p>c)\u00a03.77 $cm^3$<\/p>\n<p>d)\u00a00 $cm^3$<\/p>\n<div class=\"a-single a-22\"><a href=\"https:\/\/cracku.in\/cat-8-months\/e?utm_source=blog&utm_medium=banner&utm_campaign=catbanners\"><img decoding=\"async\" src=\"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2026\/03\/CAT-2026-Tejas-Batch-Starts-on-9th-Feb-Mon-1.png\" \/><\/a><\/div>\n<p><b>Question 4:\u00a0<\/b>Assuming the earth to be a perfect sphere and suppose the equator (0 degree latitude) is 25,000 miles in length. What is the approximate length of the 60 degree latitude?<\/p>\n<p>a)\u00a015,000 miles<\/p>\n<p>b)\u00a016,667 miles<\/p>\n<p>c)\u00a012,500 miles<\/p>\n<p>d)\u00a0None of these<\/p>\n<p><b>Question 5:\u00a0<\/b>A cylinder of radius 12 cm was filled to the brim. A sphere of radius 3 cm was completely immersed into the cylinder and removed out. By how much has the height of water gone down in the cylinder?<\/p>\n<p>a)\u00a01 cm<\/p>\n<p>b)\u00a00.5 cm<\/p>\n<p>c)\u00a00.25 cm<\/p>\n<p>d)\u00a0None of these<\/p>\n<p><b>Question 6:\u00a0<\/b>A bronze sphere of radius 4 cm was melted and the liquid was used to make spheres of radius 1 cm each. How many smaller spheres are made in total if 50% of the bronze is wasted in the process?<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/t.me\/MBAWithCracku\" target=\"_blank\" class=\"btn btn-info \">Join 7K MBA Aspirants Telegram Group<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.cracku.app&amp;hl=en\" target=\"_blank\" class=\"btn btn-alone \">Download Highly Rated CAT preparation App<\/a><\/p>\n<p><span style=\"text-decoration: underline;\"><strong>Answers &amp; Solutions:<\/strong><\/span><\/p>\n<p><strong>1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>$\\frac{4}{3}\\pi\\ \\left(R_{ball}+R_{ice_n}\\right)^3$We are given with a cricket ball of 20 cm covered in ice of unknown cm in a shape of sphere.<br \/>\nwhen we have 10 of ice;<\/p>\n<p>volume of total sphere would be = volume of (ball+ice)<\/p>\n<p>= $\\frac{4}{3}\\pi\\ \\left(R_{ball}+R_{ice}\\right)^3$<\/p>\n<p>=$\\frac{4}{3}\\pi\\ \\left(20+10\\right)^3$<\/p>\n<p>=$\\frac{4}{3}\\pi\\ 2700$<\/p>\n<p>=$3600\\pi\\ $<\/p>\n<p>=113040 $cm^3$<\/p>\n<p>Melting volume rate =\u00a0$\\ 50\\ cm^3$<\/p>\n<p>New volume after melting one day = 113040-50 = 112990<\/p>\n<p>$\\frac{4}{3}\\pi\\ \\left(R_{ball}+R_{ice_n}\\right)^3$ = 112990<\/p>\n<p>$\\ \\left(R_{ball}+R_{ice_n}\\right)^3=\\frac{112990}{4\\pi\\ }3\\cdot$<\/p>\n<p>$\\ R_{ice_n}=\\left(\\frac{112990}{4\\pi\\ }3\\right)^{\\frac{1}{3}}-R_{ball}\\cdot$<\/p>\n<p>$\\ R_{ice_n}=29.99 &#8211; R_{ball}\\cdot$<\/p>\n<p>New thickness of ice = 9.99<\/p>\n<p>Rate of change = .01\/10 =.001%<\/p>\n<p><strong>2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>The volume of the cut part=volume of sphere = $(4\/3)\\pi (3^3)$ = $36 \\pi$.<br \/>\nHence, the volume of the cut-out cone is $36 \\pi$.<br \/>\nLet h be the height of the cut-out cone and r be the radius of the base.<br \/>\nHence, $h\/r = 12\/9 =&gt; h=4\/3 r$.<br \/>\nThus, the volume of cut-out cone $= (\\pi)\/3 * r^2 * (4\/3) r = 36 \\pi$<br \/>\nSo, $r^3 = 81$<br \/>\nSo, $r = 3\\sqrt [3]{3}$. Hence $h = 4\\sqrt[3]{3}$.<br \/>\nHence, length of frustum $h_f= 12 &#8211; 4\\sqrt[3]{3}$.<br \/>\nHence, correct option is none of the above.<\/p>\n<p><strong>3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>The volume of the sphere = (4\/3)*(22\/7)*(343)<br \/>\nThe volume of the tetrahedron = $\\sqrt2*2 \\sqrt2*2 \\sqrt2*2\\sqrt2 \/12$<\/p>\n<p>The volume of the sphere= n x the volume of tetrahedron.<\/p>\n<p>Number of tetrahedrons that can be formed = [(4\/3)*(22\/7)*(343)]\/[$\\sqrt2*2 \\sqrt2*2 \\sqrt2*2\\sqrt2 \/12$] = 539.<\/p>\n<p>Here n is an integer that imply whole volume is used up to make 539 tetrahedron.<\/p>\n<p>Therefore, volume wasted = 0.<\/p>\n<p><strong>4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>The radius of the circular plane containing the 60-degree latitude is (cos 60) * radius of the earth.<br \/>\nSo, length of the 60 degree latitude is 1\/2 the length of equator = 12,500 miles<\/p>\n<p><strong>5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>3.14 * 12 * 12 * h = 4\/3 * 3.14 * 3 * 3 * 3<br \/>\nSo, h = 0.25 cm<\/p>\n<p><b>6)\u00a0Answer:\u00a032<\/b><\/p>\n<p>Without wastage, the total number of spheres possible is $4^{3} = 64$.<br \/>\nAs, 50% of the bronze is wasted, the number of smaller sphere possible is 32.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/pay\/b5hNV\" target=\"_blank\" class=\"btn btn-danger \">Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n<div class=\"a-single a-22\"><a href=\"https:\/\/cracku.in\/cat-8-months\/e?utm_source=blog&utm_medium=banner&utm_campaign=catbanners\"><img decoding=\"async\" src=\"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2026\/03\/CAT-2026-Tejas-Batch-Starts-on-9th-Feb-Mon-1.png\" \/><\/a><\/div>\n<p>We hope this Direction Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Sphere Questions for NMAT &#8211; Download [PDF] Download Sphere Questions for NMAT PDF. Top 6 very important Sphere Questions for NMAT based on asked questions in previous exam papers. Take NMAT mock test Question 1:\u00a0A cricket ball of radius 20cm is covered in ice of uniform thickness. The ice is melting at a consistent rate [&hellip;]<\/p>\n","protected":false},"author":32,"featured_media":173873,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,169,125,4977],"tags":[4979,5117,5121],"class_list":{"0":"post-173099","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-downloads","9":"category-featured","10":"category-nmat","11":"tag-nmat-2021","12":"tag-sphere","13":"tag-sphere-questions"},"better_featured_image":{"id":173873,"alt_text":"NMAT Sphere Questions","caption":"NMAT Sphere Questions","description":"NMAT Sphere 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