{"id":128521,"date":"2023-09-12T17:30:00","date_gmt":"2023-09-12T12:00:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128521"},"modified":"2023-11-29T10:29:29","modified_gmt":"2023-11-29T04:59:29","slug":"rd-sharma-class-10-solutions-chapter-16-exercise-16-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-1\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128529\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.1-scaled.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1\" width=\"2048\" height=\"1152\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.1-scaled.jpg 2048w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.1-768x432.jpg 768w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.1-1536x864.jpg 1536w\" sizes=\"(max-width: 2048px) 100vw, 2048px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1:\u00a0<\/strong>This exercise comprises questions involving solid conversions from one type to another. Students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> to clear up any questions they have about this chapter. Also available for download is the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-surface-areas-and-volumes\/\"><strong>RD Sharma Class 10 Solutions Chapter 16<\/strong><\/a> Exercise 16.1 PDF.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d059e84bc12\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d059e84bc12\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-1\/#where-can-i-get-the-correct-rd-sharma-class-10-solutions-chapter-16-exercise-161\" title=\"Where can I get the correct RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1?\">Where can I get the correct RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-1\/#is-the-rd-sharma-class-10-solutions-chapter-16-exercise-161-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 available on the Kopykitab website?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-16-exercise-161-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-16-Exercise-16.1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-16-Exercise-16.1.pdf\">RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-16-exercise-161-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q1. <\/strong><\/p>\n<p><strong>How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>A solid sphere of radius, R = 8 cm<\/p>\n<p>With this sphere, we have to make spherical balls of radius r = 1 cm<\/p>\n<p>Let\u2019s assume that the number of balls made is n<\/p>\n<p>Then, we know that<\/p>\n<p>Volume of the sphere = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>The volume of the solid sphere = the sum of the volumes of n spherical balls.<\/p>\n<p>n x 4\/3 \u03c0r<sup>3<\/sup>\u00a0= 4\/3 \u03c0R<sup>3<\/sup><\/p>\n<p>n x 4\/3 \u03c0(1)<sup>3<\/sup>\u00a0= 4\/3 \u03c0(8)<sup>3<\/sup><\/p>\n<p>n = 8<sup>3<\/sup>\u00a0= 512<\/p>\n<p>Therefore, 512 balls can be made of radius 1 cm each with a solid sphere of radius 8 cm.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q2. <\/strong><\/p>\n<p><strong>How many spherical bullets each 5 cm in diameter can be cast from a rectangular block of metal 11dm x 1 m x 5 dm?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>A metallic block of dimension 11dm x 1m x 5dm<\/p>\n<p>The diameter of each bullet = 5 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the sphere = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>Since, 1 dm = 10<sup>-1<\/sup>m = 0.1 m<\/p>\n<p>The volume of the rectangular block = 1.1 x 1 x 0.5\u00a0= 0.55 m<sup>3<\/sup><\/p>\n<p>Radius of the bullet = 5\/2 = 2.5 cm<\/p>\n<p>Let the number of bullets made from the rectangular block be n.<\/p>\n<p>Then from the question,<\/p>\n<p>The volume of the rectangular block = sum of the volumes of the n spherical bullets<\/p>\n<p>0.55 = n x 4\/3 \u03c0(2.5)<sup>3<\/sup><\/p>\n<p>Solving for n, we have<\/p>\n<p>n = 8400<\/p>\n<p>Therefore, 8400 can be cast from the rectangular block of metal.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q3. <\/strong><\/p>\n<p><strong>A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two balls are 2 cm and 1.5 cm respectively. Determine the diameter of the third ball.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of the spherical ball = 3 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of the sphere = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>So, its the volume (V) = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>That the ball is melted and recast into 3 spherical balls.<\/p>\n<p>Volume (V<sub>1<\/sub>) of first ball = 4\/3 \u03c0 1.5<sup>3<\/sup><\/p>\n<p>Volume (V<sub>2<\/sub>) of second ball = 4\/3 \u03c02<sup>3<\/sup><\/p>\n<p>Let the radius of the third ball = r cm<\/p>\n<p>The volume of the third ball (V<sub>3<\/sub>) = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>The volume of the spherical ball is equal to the volume of the 3 small spherical balls.<\/p>\n<p><img class=\"alignnone size-full wp-image-128559\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a-4.png\" alt=\"a\" width=\"265\" height=\"98\" \/><\/p>\n<p>Now,<\/p>\n<p>Canceling out the common part from both sides of the equation we get,<\/p>\n<p>(3)<sup>3<\/sup>\u00a0= (2)<sup>3<\/sup>\u00a0+ (1.5)<sup>3\u00a0<\/sup>+ r<sup>3<\/sup><\/p>\n<p>r<sup>3\u00a0<\/sup>= 3<sup>3<\/sup>\u2013 2<sup>3<\/sup>\u2013 1.5<sup>3<\/sup>\u00a0cm<sup>3<\/sup><\/p>\n<p>r<sup>3\u00a0<\/sup>= 15.6 cm<sup>3<\/sup><\/p>\n<p>r = (15.6)<sup>1\/3<\/sup>\u00a0cm<\/p>\n<p>r = 2.5 cm<\/p>\n<p>As diameter = 2 x radius = 2 x 2.5 cm<\/p>\n<p>= 5.0 cm.<\/p>\n<p>Thus, the diameter of the third ball is 5 cm<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q4. <\/strong><\/p>\n<p><strong>2.2 cubic dm of brass is to be drawn into a cylindrical wire of 0.25 cm in diameter. Find the length of the wire.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>2.2 dm<sup>3<\/sup>\u00a0of brass is to be drawn into a cylindrical wire of Diameter = 0.25 cm<\/p>\n<p>So, the radius of the wire (r) = d\/2<\/p>\n<p>= 0.25\/2 = 0.125*10<sup>-2\u00a0<\/sup>cm<\/p>\n<p>Now, 1 cm = 0.01 m<\/p>\n<p>So, 0.1cm = 0.001 m<\/p>\n<p>Let the length of the wire be (h)<\/p>\n<p>We know that,<\/p>\n<p>The volume of the cylinder = \u03c0r<sup>2<\/sup>h<\/p>\n<p>It\u2019s understood that,<\/p>\n<p>The volume of cylindrical wire = Volume of brass of 2.2 dm<sup>3<\/sup><\/p>\n<p><img class=\"alignnone size-full wp-image-128560\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/b-3.png\" alt=\"b\" width=\"298\" height=\"46\" \/><\/p>\n<p>h = 448 m<\/p>\n<p>Therefore, the length of the cylindrical wire drawn is 448 m<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q5. <\/strong><\/p>\n<p><strong>What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm, and thickness 2.5 mm?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The diameter of the solid cylinder = 2 cm<\/p>\n<p>Length of hollow cylinder = 16 cm<\/p>\n<p>The solid cylinder is recast into a hollow cylinder of length 16 cm, external diameter of 20 cm, and thickness of 2.5 mm = 0.25 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of a cylinder = \u03c0r<sup>2<\/sup>h<\/p>\n<p>The radius of the solid cylinder = 1 cm<\/p>\n<p>So,<\/p>\n<p>Volume of the solid cylinder = \u03c01<sup>2<\/sup>h = \u03c0h cm<sup>3<\/sup><\/p>\n<p>Let\u2019s assume the length of the solid cylinder as h<\/p>\n<p>And,<\/p>\n<p>Volume of the hollow cylinder = \u03c0h(R<sup>2<\/sup>\u2013 r<sup>2<\/sup>)<\/p>\n<p>Thickness of the cylinder = (R \u2013 r)<\/p>\n<p>0.25 = 10 \u2013 r<\/p>\n<p>So, the internal radius of the cylinder is 9.75 cm<\/p>\n<p>Volume of the hollow cylinder = \u03c0 \u00d7 16 (100 \u2013 95.0625)<\/p>\n<p>Hence, it\u2019s understood that<\/p>\n<p>The volume of the solid cylinder = volume of the hollow cylinder<\/p>\n<p>\u03c0h = \u03c0 \u00d7 16(100 \u2013 95.06)<\/p>\n<p>h = 79.04 cm<\/p>\n<p>Therefore, the length of the solid cylinder is 79.04 cm.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q6. <\/strong><\/p>\n<p><strong>A cylindrical vessel having a diameter equal to its height is full of water which is poured into two identical cylindrical vessels with a diameter of 42 cm and a height of 21 cm which are filled completely. Find the diameter of the cylindrical vessel.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The diameter of the cylinder = the height of the cylinder<\/p>\n<p>\u21d2 h = 2r, where h \u2013 the height of the cylinder and r \u2013 radius of the cylinder<\/p>\n<p>We know that,<\/p>\n<p>The volume of a cylinder = \u03c0r<sup>2<\/sup>h<\/p>\n<p>So, volume of the cylindrical vessel = \u03c0r<sup>2<\/sup>2r\u00a0= 2\u03c0r<sup>3<\/sup> (as h = 2r)\u2026.. (i)<\/p>\n<p>Now,<\/p>\n<p>The volume of each identical vessel = \u03c0r<sup>2<\/sup>h<\/p>\n<p>Diameter = 42 cm, so the radius = 21 cm<\/p>\n<p>Height = 21 cm<\/p>\n<p>So, the volume of two identical vessels = 2 x \u03c0 21<sup>2<\/sup>\u00a0\u00d7 21 \u2026.. (ii)<\/p>\n<p>Since the volumes on equations (i) and (ii) are equal<\/p>\n<p>On equating both equations, we have<\/p>\n<p>2\u03c0r<sup>3<\/sup>= 2 x \u03c0 21<sup>2<\/sup>\u00a0\u00d7 21<\/p>\n<p>r<sup>3<\/sup>\u00a0= (21)<sup>3<\/sup><\/p>\n<p>r = 21 cm<\/p>\n<p>So, d = 42 cm<\/p>\n<p>Therefore, the diameter of the cylindrical vessel is 42 cm.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q7. <\/strong><\/p>\n<p><strong>50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>50 circular plates each with a diameter of 14 cm<\/p>\n<p>The radius of circular plates = 7cm<\/p>\n<p>Thickness of plates = 0.5 cm<\/p>\n<p>As these plates are one above the other, the total thickness of all the plates = 0.5 x 50 = 25 cm<\/p>\n<p>So, the total surface area of the right circular cylinder formed = 2\u03c0r \u00d7 h + 2\u03c0r<sup>2<\/sup><\/p>\n<p>= 2\u03c0r (h + r)<\/p>\n<p>= 2(22\/7) x 7 x (25 + 7)<\/p>\n<p>= 2 x 22 x 32 = 1408 cm<sup>2<\/sup><\/p>\n<p>Therefore, the total surface area of the cylinder is 1408 cm<sup>2<\/sup><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q8. <\/strong><\/p>\n<p><strong>25 circular plates, each of radius 10.5 cm and thickness 1.6 cm are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>250 circular plates each with a radius of 10.5 cm and a thickness of 1.6 cm.<\/p>\n<p>As the plates are placed one above the other, the total height becomes = 1.6 x 25 = 40 cm<\/p>\n<p>We know that,<\/p>\n<p>The curved surface area of a cylinder = 2\u03c0rh<\/p>\n<p>= 2\u03c0 \u00d7 10.5 \u00d7 40 = 2640 cm<sup>2<\/sup><\/p>\n<p>And, the volume of the cylinder = \u03c0r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 \u00d7 10.5<sup>2\u00a0<\/sup>\u00d7 40 = 13860 cm<sup>3<\/sup><\/p>\n<p>Therefore,<\/p>\n<p>The curved surface area of the cylinder is 2640 cm<sup>2<\/sup>\u00a0and the volume of the cylinder is 13860 cm<sup>3<\/sup><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q9. <\/strong><\/p>\n<p><strong>Find the number of metallic circular discs with a 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter of 4.5 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Radius of each circular disc = r = 1.5\/2 = 0.75 cm<\/p>\n<p>Height of each circular disc = h = 0.2 cm<\/p>\n<p>Radius of cylinder = R = 4.5\/ 2 = 2.25 cm<\/p>\n<p>Height of cylinder = H = 10 cm<\/p>\n<p>So, the number of metallic discs required is given by n<\/p>\n<p>n = Volume of cylinder\/volume of each circular disc<\/p>\n<p>n = \u03c0R<sup>2<\/sup>H\/ \u03c0r<sup>2<\/sup>h<\/p>\n<p>n = (2.25)<sup>2<\/sup>(10)\/ (0.75)<sup>2<\/sup>(0.2)<\/p>\n<p>n = 3 x 3 x 50 = 450<\/p>\n<p>Therefore, 450 metallic discs are required.<\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Q10. <\/strong><\/p>\n<p><strong>How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm \u00d7 42 cm \u00d7 21 cm?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Radius of each spherical lead shot = r = 4.2\/ 2 = 2.1 cm<\/p>\n<p>The dimensions of the rectangular lead piece = 66 cm x 42 cm x 21 cm<\/p>\n<p>So, the volume of a spherical lead shot = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>= 4\/3 x 22\/7 x 2.1<sup>3<\/sup><\/p>\n<p>And, the volume of the rectangular lead piece = 66 x 42 x 21<\/p>\n<p>Thus,<\/p>\n<p>The number of spherical lead shots = Volume of rectangular lead piece\/ Volume of a spherical lead shot<\/p>\n<p>= 66 x 42 x 21\/ (4\/3 x 22\/7 x 2.1<sup>3<\/sup>)<\/p>\n<p><strong>=\u00a0<\/strong>1500<\/p>\n<p><strong>11. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of each spherical lead shot = r = 4\/2 = 2 cm<\/p>\n<p>Volume of each spherical lead shot = 4\/3 \u03c0r<sup>3<\/sup>\u00a0= 4\/3 \u03c0 2<sup>3<\/sup>\u00a0cm<sup>3<\/sup><\/p>\n<p>Edge of the cube = 44 cm<\/p>\n<p>Volume of the cube = 44<sup>3<\/sup>\u00a0cm<sup>3<\/sup><\/p>\n<p>Thus,<\/p>\n<p>Number of spherical lead shots = Volume of cube\/ Volume of each spherical lead shot<\/p>\n<p>= 44 x 44 x 44\/ (4\/3 \u03c0 2<sup>3<\/sup>)<\/p>\n<p>= 2541<\/p>\n<p><strong>12. Three cubes of a metal whose edges are in the ratio\u00a03:\u00a04: 5 are melted and converted into a single cube whose diagonal is\u00a012\u221a3\u00a0cm. Find the edges of the three cubes.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the edges of three cubes (in cm) be 3x, 4x, and 5x respectively.<\/p>\n<p>So, the volume of the cube after melting will be = (3x)<sup>3<\/sup>\u00a0+ (4x)<sup>3<\/sup>\u00a0+ (5x)<sup>3<\/sup><\/p>\n<p>= 9x<sup>3<\/sup>\u00a0+ 64x<sup>3<\/sup>\u00a0+ 125x<sup>3<\/sup>\u00a0= 216x<sup>3<\/sup><\/p>\n<p>Now, let a be the edge of the new cube so formed after melting<\/p>\n<p>Then we have,<\/p>\n<p>a<sup>3<\/sup>\u00a0= 216x<sup>3<\/sup><\/p>\n<p>a = 6x<\/p>\n<p>We know that,<\/p>\n<p>Diagonal of the cube =\u00a0<strong>\u221a<\/strong>(a<sup>2<\/sup>\u00a0+ a<sup>2<\/sup>\u00a0+ a<sup>2<\/sup>) = a<strong>\u221a<\/strong>3<\/p>\n<p>So, 12<strong>\u221a<\/strong>3 = a<strong>\u221a<\/strong>3<\/p>\n<p>a = 12 cm<\/p>\n<p>x = 12\/6 = 2<\/p>\n<p>Thus, the edges of the three cubes are 6 cm, 8 cm, and 10 cm respectively.<\/p>\n<p><strong>13. A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Radius of metallic sphere = R = 10.5 cm<\/p>\n<p>So, its volume = 4\/3 \u03c0R<sup>3<\/sup>\u00a0= 4\/3 \u03c0(10.5)<sup>3<\/sup><\/p>\n<p>We also have,<\/p>\n<p>Radius of each cone = r = 3.5 cm<\/p>\n<p>Height of each cone = h = 3 cm<\/p>\n<p>And, its volume = 1\/3 \u03c0r<sup>2<\/sup>h = 1\/3 \u03c0(3.5)<sup>2<\/sup>(3)<\/p>\n<p>Thus,<\/p>\n<p>The number of cones = Volume of metallic sphere\/ Volume of each cone<\/p>\n<p>= 4\/3 \u03c0(10.5)<sup>3<\/sup>\u00a0\/ 1\/3 \u03c0(3.5)<sup>2<\/sup>(3)<\/p>\n<p>= 126<\/p>\n<p><strong>14. The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having a uniform cross-section. Find the length of the wire.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of the sphere = 9\/2 cm<\/p>\n<p>So, its volume = 4\/3 \u03c0r<sup>3<\/sup>\u00a0= 4\/3 \u03c0(9\/2)<sup>3<\/sup><\/p>\n<p>And, the radius of the wire = 2 mm = 0.2 cm<\/p>\n<p>Let the length of the wire = h cm<\/p>\n<p>So, the volume of wire = \u03c0r<sup>2<\/sup>h = \u03c0(0.2)<sup>2<\/sup>h<\/p>\n<p>Then, according to the question we have<\/p>\n<p>The volume of wire = Volume of a sphere<\/p>\n<p>\u03c0(0.2)<sup>2<\/sup>h = 4\/3 \u03c0(9\/2)<sup>3<\/sup><\/p>\n<p>h = 4 x 729\/ (3 x 8 x 0.01) = 12150 cm<\/p>\n<p>Therefore, the length of the wire = 12150 cm<\/p>\n<p><strong>15. An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1\/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the radius of the big ball be x cm<\/p>\n<p>Then, the radius of the small ball = x\/4 cm<\/p>\n<p>And, let the number of balls = n<\/p>\n<p>Then according to the question, we have<\/p>\n<p>The volume of n small balls = Volume of the big ball<\/p>\n<p>n x 4\/3 \u03c0(x\/4)<sup>3<\/sup>\u00a0= 4\/3 \u03c0x<sup>3<\/sup><\/p>\n<p>n x (x<sup>3<\/sup>\/ 64) = x<sup>3<\/sup><\/p>\n<p>n = 64<\/p>\n<p>Therefore, the number of small balls = 64<\/p>\n<p>Next,<\/p>\n<p>The surface area of all small balls\/ surface area of big ball = 64 x 4\u03c0(x\/4)<sup>2<\/sup>\/ 4\u03c0(x)<sup>2<\/sup><\/p>\n<p>= 64\/16 = 4\/1<\/p>\n<p>Thus, the ratio of the surface area of the small balls to that of the original ball is 4:1<\/p>\n<p><strong>16. A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of the copper sphere = 3 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the sphere = 4\/3 \u03c0 r<sup>3<\/sup><\/p>\n<p>= 4\/3 \u03c0 \u00d7 3<sup>3<\/sup>\u00a0\u2026.. (i)<\/p>\n<p>Also, given that the copper sphere is melted and recast into a right circular cone<\/p>\n<p>Height of the cone = 3 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of the right circular cone = 1\/3 \u03c0 r<sup>2<\/sup>h<\/p>\n<p>= 1\/3 \u03c0 \u00d7 r<sup>2\u00a0<\/sup>\u00d7 3 \u2026.. (ii)<\/p>\n<p>On comparing equations (i) and (ii) we have,<\/p>\n<p>4\/3 \u03c0 \u00d7 3<sup>3<\/sup>\u00a0= 1\/3 \u03c0 \u00d7 r<sup>2\u00a0<\/sup>\u00d7 3<\/p>\n<p>r<sup>2<\/sup>\u00a0= 36<\/p>\n<p>r = 6 cm<\/p>\n<p>Therefore, the radius of the base of the cone is 6 cm.<\/p>\n<p><strong>17. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. <\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Diameter of the copper wire = 1 cm<\/p>\n<p>So, radius of the copper wire = 1\/2 cm = 0.5 cm<\/p>\n<p>Length of the copper rod = 8 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of the cylinder = \u03c0 r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 \u00d7 0.5<sup>2\u00a0<\/sup>\u00d7 8 \u2026\u2026. (i)<\/p>\n<p>Length of the wire = 18 m = 1800 cm<\/p>\n<p>The volume of the wire = \u03c0 r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 r<sup>2<\/sup>\u00a0\u00d7 1800 \u2026.. (ii)<\/p>\n<p>On equating both equations, we have<\/p>\n<p>\u03c0 \u00d7 0.5<sup>2\u00a0<\/sup>\u00d7 8 = \u03c0 r<sup>2<\/sup>\u00a0\u00d7 1800<\/p>\n<p>r<sup>2<\/sup>\u00a0= 2 \/1800 = 1\/900<\/p>\n<p>r = 1\/30 cm<\/p>\n<p>Therefore, the diameter of the wire is 1\/15 cm i.e. 0.67 mm which is the thickness of the wire.<\/p>\n<p><strong>18. The diameters of the internal and external surfaces of a hollow spherical shell are 10cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of 8\/3, find the diameter of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The internal diameter of the hollow sphere = 6 cm<\/p>\n<p>So, the internal radius of the hollow sphere = 6\/2 cm = 3 cm = r<\/p>\n<p>External diameter of the hollow sphere = 10 cm<\/p>\n<p>So, the external radius of the hollow sphere = 10\/2 cm = 5 cm = R<\/p>\n<p>We know that,<\/p>\n<p>Volume of the hollow spherical shell = 4\/3 \u03c0 \u00d7 (R<sup>3<\/sup>\u00a0\u2013 r<sup>3<\/sup>)<\/p>\n<p>= 4\/3 \u03c0 \u00d7 (5<sup>3<\/sup>\u00a0\u2013 3<sup>3<\/sup>) \u2026.. (i)<\/p>\n<p>And given, the length of the solid cylinder = 8\/3 cm<\/p>\n<p>Let the radius of the solid cylinder be r cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the cylinder = \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>= \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 8\/3 \u2026.. (ii)<\/p>\n<p>Now equating both (i) and (ii), we have<\/p>\n<p>4\/3 \u03c0 \u00d7 5<sup>3<\/sup>\u00a0\u2013 3<sup>3<\/sup>\u00a0= \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 8\/3<\/p>\n<p>4\/3 x (125 \u2013 27) = r<sup>2<\/sup>\u00a0\u00d7 8\/3<\/p>\n<p>98\/2 = r<sup>2<\/sup><\/p>\n<p>r<sup>2<\/sup>\u00a0= 49<\/p>\n<p>r = 7<\/p>\n<p>So, d = 7 x 2 = 14 cm<\/p>\n<p>Therefore, the diameter of the cylinder is 14 cm<\/p>\n<p><strong>19. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm x 10 cm x 7 cm?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Diameter of the coin = 1.75 cm<\/p>\n<p>So, its radius = 1.74\/2 = 0.875 cm<\/p>\n<p>Thickness or the height = 2 mm = 0.2 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of the cylinder (V<sub>1<\/sub>) = \u03c0r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 0.875<sup>2<\/sup>\u00a0\u00d7 0.2<\/p>\n<p>And, the volume of the cuboid (V<sub>2<\/sub>) = 11 \u00d7 10 \u00d7 7 cm<sup>3<\/sup><\/p>\n<p>Let the number of coins needed to be melted be n.<\/p>\n<p>So, we have<\/p>\n<p>V<sub>2<\/sub>\u00a0= V<sub>1<\/sub>\u00a0\u00d7 n<\/p>\n<p>11 \u00d7 10 \u00d7 7 = \u03c0 0.875<sup>2<\/sup>\u00a0\u00d7 0.2 x n<\/p>\n<p>11 \u00d7 10 \u00d7 7 = 22\/7 x 0.875<sup>2<\/sup>\u00a0\u00d7 0.2 x n<\/p>\n<p>On solving we get, n = 1600<\/p>\n<p>Therefore, the number of coins required is 1600<\/p>\n<p><strong>20. The surface area of a solid metallic sphere is 616 cm<sup>2<\/sup>. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed. <\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The height of the cone = 28 cm<\/p>\n<p>The surface area of the solid metallic sphere = 616 cm<sup>3<\/sup><\/p>\n<p>We know that,<\/p>\n<p>The surface area of the sphere = 4\u03c0r<sup>2<\/sup><\/p>\n<p>So, 4\u03c0r<sup>2<\/sup>\u00a0= 616<\/p>\n<p>r<sup>2<\/sup>\u00a0= 49<\/p>\n<p>r = 7<\/p>\n<p>The radius of the solid metallic sphere = 7 cm<\/p>\n<p>Let\u2019s assume r to be the radius of the cone<\/p>\n<p>We know that,<\/p>\n<p>Volume of the cone = 1\/3 \u03c0r<sup>2<\/sup>h<\/p>\n<p>= 1\/3 \u03c0r<sup>2\u00a0<\/sup>(28) \u2026.. (i)<\/p>\n<p>Volume of the sphere = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>= 4\/3 \u03c07<sup>3<\/sup> \u2026\u2026\u2026. (ii)<\/p>\n<p>On equating equations (i) and (ii), we have<\/p>\n<p>1\/3 \u03c0r<sup>2\u00a0<\/sup>(28) = 4\/3 \u03c07<sup>3<\/sup><\/p>\n<p>Eliminating the common terms, we get<\/p>\n<p>r<sup>2\u00a0<\/sup>(28) = 4 x 7<sup>3<\/sup><\/p>\n<p>r<sup>2\u00a0<\/sup>= 49<\/p>\n<p>r =7<\/p>\n<p>So, diameter of the cone = 7 x 2 = 14 cm<\/p>\n<p>Therefore, the diameter of the base of the cone is 14 cm<\/p>\n<p><strong>21.<\/strong>\u00a0<strong>A cylindrical bucket, 32 cm high and 18 cm in radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Height of the cylindrical bucket = 32 cm<\/p>\n<p>The radius of the cylindrical bucket = 18 cm<\/p>\n<p>Height of conical heap = 24 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of cylinder = \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>And, volume of cone = 1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>Then, from the question<\/p>\n<p>The volume of the conical heap = Volume of the cylindrical bucket<\/p>\n<p>1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 24 = \u03c0 \u00d7 18<sup>2<\/sup>\u00a0\u00d7 32<\/p>\n<p>r<sup>2<\/sup>\u00a0= 18<sup>2<\/sup>\u00a0x 4<\/p>\n<p>r = 18 x 2 = 36 cm<\/p>\n<p>Now,<\/p>\n<p>The slant height of the conical heap (l) is given by<\/p>\n<p>l = \u221a(h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>)<\/p>\n<p>l = \u221a(24<sup>2<\/sup>\u00a0+ 36<sup>2<\/sup>) = \u221a1872<\/p>\n<p>l = 43.26 cm<\/p>\n<p>Therefore, the radius and slant height of the conical heap is 36 cm and 43.26 cm respectively.<\/p>\n<p><strong>22. A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number of cones made be n<\/p>\n<p>Given,<\/p>\n<p>The radius of the metallic sphere = 5.6 cm<\/p>\n<p>The radius of the cone = 2.8 cm<\/p>\n<p>Height of the cone = 3.2 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of a sphere = 4\/3 \u03c0 \u00d7 r<sup>3<\/sup><\/p>\n<p>So, V<sub>1<\/sub>\u00a0= 4\/3 \u03c0 \u00d7 5.6<sup>3<\/sup><\/p>\n<p>And,<\/p>\n<p>Volume of cone = 1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>V<sub>2<\/sub>\u00a0= 1\/3 \u03c0 \u00d7 2.8<sup>2<\/sup>\u00a0\u00d7 3.2<\/p>\n<p>Thus, the number of cones (n) = Volume of the sphere\/ Volume of the cone<\/p>\n<p>n = 4\/3 \u03c0 \u00d7 5.6<sup>3<\/sup>\u00a0\/ (1\/3 \u03c0 \u00d7 2.8<sup>2<\/sup>\u00a0\u00d7 3.2)<\/p>\n<p>n = (4 x 5.6<sup>3<\/sup>)\/ (2.8<sup>2<\/sup>\u00a0\u00d7 3.2)<\/p>\n<p>n = 28<\/p>\n<p>Therefore, 28 such cones can be formed.<\/p>\n<p><strong>23. A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of the pipe are 8 cm and 7 cm respectively. Find the length of the pipe.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the length of the pipe be h cm.<\/p>\n<p>Then, Volume of cuboid = (53 x 40 x 15) cm<sup>3<\/sup><\/p>\n<p>Internal radius of the pipe = 7\/2 cm = r<\/p>\n<p>External radius of the pipe = 8\/2 = 4 cm = R<\/p>\n<p>So, the volume of iron in the pipe = (External Volume) \u2013 (Internal Volume)<\/p>\n<p>= \u03c0R<sup>2<\/sup>h \u2013 \u03c0r<sup>2<\/sup>h<\/p>\n<p>= \u03c0h(R<sup>2<\/sup>\u2013 r<sup>2<\/sup>)<\/p>\n<p>= \u03c0h(R \u2013 r) (R + r)<\/p>\n<p>= \u03c0(4 \u2013 7\/2) (4 + 7\/2) x h<\/p>\n<p>= \u03c0(1\/2) (15\/2) x h<\/p>\n<p>Then from the question it\u2019s understood that,<\/p>\n<p>The volume of iron in the pipe = volume of iron in a cuboid<\/p>\n<p>\u03c0(1\/2) (15\/2) x h = 53 x 40 x 15<\/p>\n<p>h = (53 x 40 x 15 x 7\/22 x 2\/15 x 2) cm<\/p>\n<p>h = 2698 cm<\/p>\n<p>Therefore, the length of the pipe is 2698 cm.<\/p>\n<p><strong>24. The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The internal diameter of the hollow spherical shell = 6 cm<\/p>\n<p>So, the internal radius of a hollow spherical shell = 6\/2 = 3 cm = r<\/p>\n<p>External diameter of hollow spherical shell = 10 cm<\/p>\n<p>So, the external diameter of the hollow spherical shell = 10\/2 = 5 cm = R<\/p>\n<p>Diameter of the cylinder = 14 cm<\/p>\n<p>So, the radius of cylinder = 14\/2 = 7 cm<\/p>\n<p>Let the height of the cylinder be taken as h cm<\/p>\n<p>Then, according to the question we have<\/p>\n<p>The volume of the cylinder = Volume of a spherical shell<\/p>\n<p>\u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h = 4\/3 \u03c0 \u00d7 (R<sup>3\u00a0<\/sup>\u2013 r<sup>3<\/sup>)<\/p>\n<p>\u03c0 \u00d7 7<sup>2<\/sup>\u00a0\u00d7 h = 4\/3 \u03c0 \u00d7 (5<sup>3\u00a0<\/sup>\u2013 3<sup>3<\/sup>)<\/p>\n<p>h = 4\/3 x 2<\/p>\n<p>h = 8\/3 cm<\/p>\n<p>Therefore, the height of the cylinder = 8\/3 cm<\/p>\n<p><strong>25. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The internal diameter of a hollow sphere = 4 cm<\/p>\n<p>So, the internal radius of a hollow sphere = 2 cm<\/p>\n<p>External diameter of hollow sphere = 8 cm<\/p>\n<p>So, the external radius of a hollow sphere = 4 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the hollow sphere 4\/3 \u03c0 \u00d7 (4<sup>3<\/sup>\u00a0\u2013 2<sup>3<\/sup>) \u2026 (i)<\/p>\n<p>Also given,<\/p>\n<p>Diameter of the cone = 8 cm<\/p>\n<p>So, the radius of the cone = 4 cm<\/p>\n<p>Let the height of the cone be x cm<\/p>\n<p>Volume of the cone 1\/3 \u03c0 \u00d7 4<sup>2<\/sup>\u00a0\u00d7 h \u2026.. (ii)<\/p>\n<p>As the volume of the hollow sphere and cone are equal. We can equate equations (i) and (ii)<\/p>\n<p>So, we get<\/p>\n<p>4\/3 \u03c0 \u00d7 (4<sup>3<\/sup>\u00a0\u2013 2<sup>3<\/sup>) = 1\/3 \u03c0 \u00d7 4<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>4 x (64 \u2013 8) = 16 x h<\/p>\n<p>h = 14<\/p>\n<p>Therefore, the height of the cone so obtained will have a height of 14 cm<\/p>\n<p><strong>26<\/strong>.<strong>\u00a0A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The internal radius of a hollow sphere = 2 cm<\/p>\n<p>The external radius of the hollow sphere = 4 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the hollow sphere 4\/3 \u03c0 \u00d7 (4<sup>3<\/sup>\u00a0\u2013 2<sup>3<\/sup>) \u2026 (i)<\/p>\n<p>Also given,<\/p>\n<p>The base radius of the cone = 4 cm<\/p>\n<p>Let the height of the cone be x cm<\/p>\n<p>Volume of the cone 1\/3 \u03c0 \u00d7 4<sup>2<\/sup>\u00a0\u00d7 h \u2026.. (ii)<\/p>\n<p>As the volume of the hollow sphere and cone are equal. We can equate equations (i) and (ii)<\/p>\n<p>So, we get<\/p>\n<p>4\/3 \u03c0 \u00d7 (4<sup>3<\/sup>\u00a0\u2013 2<sup>3<\/sup>) = 1\/3 \u03c0 \u00d7 4<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>4 x (64 \u2013 8) = 16 x h<\/p>\n<p>h = 14<\/p>\n<p>Now,<\/p>\n<p>The slant height of the cone (l) is given by<\/p>\n<p>l = \u221a(h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>)<\/p>\n<p>l = \u221a(14<sup>2<\/sup>\u00a0+ 4<sup>2<\/sup>) = \u221a212<\/p>\n<p>l = 14.56 cm<\/p>\n<p>Therefore, the height and slant height of the conical heap is 14 cm and 14.56 cm respectively.<\/p>\n<p><strong>27. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of the spherical ball = 3 cm<\/p>\n<p>We know that,<\/p>\n<p>The volume of the sphere = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>So, its the volume (V) = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>That the ball is melted and recast into 3 spherical balls.<\/p>\n<p>Volume (V<sub>1<\/sub>) of first ball = 4\/3 \u03c0 1.5<sup>3<\/sup><\/p>\n<p>Volume (V<sub>2<\/sub>) of second ball = 4\/3 \u03c02<sup>3<\/sup><\/p>\n<p>Let the radius of the third ball = r cm<\/p>\n<p>The volume of the third ball (V<sub>3<\/sub>) = 4\/3 \u03c0r<sup>3<\/sup><\/p>\n<p>The volume of the spherical ball is equal to the volume of the 3 small spherical balls.<\/p>\n<p><img class=\"alignnone size-full wp-image-128561\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/c-1.png\" alt=\"c\" width=\"265\" height=\"98\" \/><\/p>\n<p>Now,<\/p>\n<p>Canceling out the common part from both sides of the equation we get,<\/p>\n<p>(3)<sup>3<\/sup>\u00a0= (2)<sup>3<\/sup>\u00a0+ (1.5)<sup>3\u00a0<\/sup>+ r<sup>3<\/sup><\/p>\n<p>r<sup>3\u00a0<\/sup>= 3<sup>3<\/sup>\u2013 2<sup>3<\/sup>\u2013 1.5<sup>3<\/sup>\u00a0cm<sup>3<\/sup><\/p>\n<p>r<sup>3\u00a0<\/sup>= 15.6 cm<sup>3<\/sup><\/p>\n<p>r = (15.6)<sup>1\/3<\/sup>\u00a0cm<\/p>\n<p>r = 2.5 cm<\/p>\n<p>As diameter = 2 x radius = 2 x 2.5 cm<\/p>\n<p>= 5.0 cm.<\/p>\n<p>Thus, the diameter of the third ball is 5 cm<\/p>\n<p><strong>28. A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel are required to graze the path to a depth of 20 cm?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The diameter of the circular pond = is 40 m<\/p>\n<p>So, the radius of the pond = 40\/2 = 20 m = r<\/p>\n<p>Thickness (width of the path) = 2 m<\/p>\n<p>The whole view of the pond looks like a hollow cylinder.<\/p>\n<p>And the height will be 20 cm = 0.2 m<\/p>\n<p>So,<\/p>\n<p>Thickness (t) = R \u2013 r<\/p>\n<p>2 = R \u2013 20<\/p>\n<p>R = 22 m<\/p>\n<p>Volume of the hollow cylinder = \u03c0 (R<sup>2<\/sup>\u2013 r<sup>2<\/sup>) \u00d7 h<\/p>\n<p>= \u03c0 (22<sup>2<\/sup>\u2013 20<sup>2<\/sup>) \u00d7 0.2<\/p>\n<p>= 52.8 m<sup>3<\/sup><\/p>\n<p>Therefore, the volume of the hollow cylinder is the required amount of sand needed to spread across to a depth of 20 m.<\/p>\n<p><strong>29. A 16 m deep well with a diameter of 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7m. Find the height of the platform.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us assume the well to be a solid right-circular cylinder<\/p>\n<p>Radius(r) of the cylinder = 3.5\/2 m = 1.75 m<\/p>\n<p>Depth of the well or height of the cylinder (h) = 16 m<\/p>\n<p>We know that,<\/p>\n<p>The volume of the cylinder (V<sub>1<\/sub>) = \u03c0r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 \u00d7 1.75<sup>2<\/sup>\u00a0\u00d7 16<\/p>\n<p>Given,<\/p>\n<p>The length of the platform (l) = 27.5 m<\/p>\n<p>Breadth of the platform (b) =7 m<\/p>\n<p>Now, let the height of the platform be x m<\/p>\n<p>We know that,<\/p>\n<p>The volume of the rectangle = l*b*h<\/p>\n<p>V<sub>2<\/sub>\u00a0= 27.5*7*x<\/p>\n<p>As the earth dug up is spread evenly to form the platform<\/p>\n<p>The volumes of both, the well and the platform should be the same.<\/p>\n<p>V<sub>1<\/sub>\u00a0= V<sub>2<\/sub><\/p>\n<p>\u03c0 \u00d7 1.75 \u00d7 1.75 \u00d7 16 = 27.5 \u00d7 7 \u00d7 x<\/p>\n<p>x = 0.8 m = 80 cm<\/p>\n<p>Therefore, the height of the platform is 80 cm.<\/p>\n<p><strong>30. A well of diameter 2 m is dug 14 m deep. The earth taken out of it is evenly spread all around it to form an embankment of height 40 cm. Find the width of the embankment.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The radius of the circular cylinder (r) = 2\/2 m = 1 m<\/p>\n<p>Height of the well (h) = 14 m<\/p>\n<p>We know that,<\/p>\n<p>The volume of the solid circular cylinder = \u03c0 r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 \u00d7 1<sup>2<\/sup>\u00d7 14 \u2026. (i)<\/p>\n<p>And,<\/p>\n<p>The height of the embankment (h) = 40 cm = 0.4 m<\/p>\n<p>Let the width of the embankment be (x) m.<\/p>\n<p>The embankment is a hollow cylinder with an external radius = 1 + x and an internal radius = 1<\/p>\n<p>Volume of the embankment = \u03c0 \u00d7 r<sup>2\u00a0<\/sup>\u00d7 h<\/p>\n<p>= \u03c0 \u00d7 [(1 + x)<sup>2<\/sup>\u00a0\u2013 (1)<sup>2<\/sup>]\u00d7 0.4 \u2026.. (ii)<\/p>\n<p>As the well is spread evenly to form an embankment then the volumes will be the same.<\/p>\n<p>So, on equating equations (i) and (ii), we get<\/p>\n<p>\u03c0 \u00d7 1<sup>2<\/sup>\u00a0\u00d7 14 = \u03c0 \u00d7 [(1 + x)<sup>2<\/sup>\u00a0\u2013 (1)<sup>2<\/sup>]\u00a0x 0.4<\/p>\n<p>14\/0.4 = 1 + x<sup>2\u00a0<\/sup>+ 2x \u2013 1<\/p>\n<p>35 = x<sup>2<\/sup>\u00a0+ 2x<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 2x \u2013 35 = 0<\/p>\n<p>Solving by factorization method, we have<\/p>\n<p>(x + 7) (x \u2013 5) = 0<\/p>\n<p>So, x = 5 m can only be the solution as it\u2019s a positive value.<\/p>\n<p>Therefore, the width of the embankment is 5 m.<\/p>\n<p><strong>31. A well with an inner radius of 4 m is dug up and 14 m deep. Earth taken out of it has spread evenly all around a width of 3 m to form an embankment. Find the height of the embankment.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The inner radius of the well = 4 m<\/p>\n<p>Depth of the well = 14 m<\/p>\n<p>We know that,<\/p>\n<p>The volume of the cylinder = \u03c0 r<sup>2<\/sup>h<\/p>\n<p>= \u03c0 \u00d7 4<sup>2\u00a0<\/sup>\u00d7 14 \u2026. (i)<\/p>\n<p>From the question, it\u2019s told that<\/p>\n<p>The earth taken out from the well is evenly spread all around it to form an embankment<\/p>\n<p>And, the width of the embankment = 3 m<\/p>\n<p>So, the outer radius of the well = 3 + 4 m = 7 m<\/p>\n<p>We know that,<\/p>\n<p>Volume of the hollow embankment = \u03c0 (R<sup>2<\/sup>\u00a0\u2013 r<sup>2<\/sup>) \u00d7 h<\/p>\n<p>= \u03c0 \u00d7 (7<sup>2<\/sup>\u00a0\u2013 4<sup>2<\/sup>) \u00d7 h \u2026\u2026 (ii)<\/p>\n<p>On equating both the equations (i) and (ii), we get<\/p>\n<p>\u03c0 \u00d7 4<sup>2\u00a0<\/sup>\u00d7 14 = \u03c0 \u00d7 (7<sup>2<\/sup>\u00a0\u2013 4<sup>2<\/sup>) \u00d7 h<\/p>\n<p>h = 4<sup>2\u00a0<\/sup>\u00d7 14 \/ (33)<\/p>\n<p>h = 6.78 m<\/p>\n<p>Therefore, the height of the embankment so formed is 6.78 m.<\/p>\n<p><strong>32. A well of diameter 3 m is dug up to 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Diameter of the well = 3 m<\/p>\n<p>So, the radius of the well = 3\/2 m = 1.5 m<\/p>\n<p>Depth of the well (h) = 14 m<\/p>\n<p>Width of the embankment (thickness) = 4 m<\/p>\n<p>So, the radius of the outer surface of the embankment = (4 + 1.5) m = 5.5 m<\/p>\n<p>Let the height of the embankment be taken as h m<\/p>\n<p>We know that the embankment is a hollow cylinder<\/p>\n<p>Volume of the embankment = \u03c0 (R<sup>2<\/sup>\u00a0\u2013 r<sup>2<\/sup>) \u00d7 h<\/p>\n<p>= \u03c0 (5.5<sup>2<\/sup>\u00a0\u2013 1.5<sup>2<\/sup>) \u00d7 h \u2026.. (i)<\/p>\n<p>Volume of earth dug out = \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>= \u03c0 \u00d7 2<sup>2<\/sup>\u00a0\u00d7 14 \u2026.. (ii)<\/p>\n<p>On equating both (i) and (ii) we get,<\/p>\n<p>\u03c0 (5.5<sup>2<\/sup>\u00a0\u2013 1.5<sup>2<\/sup>) \u00d7 h\u00a0= \u03c0 \u00d7 (3\/2)<sup>2<\/sup>\u00a0\u00d7 14<\/p>\n<p>(30.25 \u2013 2.25) x h = 9 x 14\/ 4<\/p>\n<p>h = 9 x 14\/ (4 x 28)<\/p>\n<p>h = 9\/8 m<\/p>\n<p>Therefore, the height of the embankment is 9\/8 m<\/p>\n<p><strong>33. Find the volume largest right circular cone that can be cut out of a cube whose edge is 9 cm.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The side of the cube = 9 cm<\/p>\n<p>The largest cone that can be cut from the cube will have the base diameter = side of the cube<\/p>\n<p>2r = 9<\/p>\n<p>r = 9\/2 cm = 4.5 cm<\/p>\n<p>And,<\/p>\n<p>Height of cone = side of the cube<\/p>\n<p>So, the height of cone (h) = 9 cm<\/p>\n<p>Thus,<\/p>\n<p>Volume of the largest cone to fit in = 1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>= 1\/3 \u03c0 \u00d7 4.5<sup>2<\/sup>\u00a0\u00d7 9<\/p>\n<p>= 190.93 cm<sup>3<\/sup><\/p>\n<p>Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm<sup>3<\/sup><\/p>\n<p><strong>34. A cylindrical bucket, 32 cm high and 18 cm in radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Height of the cylindrical bucket = 32 cm<\/p>\n<p>The radius of the cylindrical bucket = 18 cm<\/p>\n<p>Height of conical heap = 24 cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of cylinder = \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>And, volume of cone = 1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>Then, from the question<\/p>\n<p>The volume of the conical heap = Volume of the cylindrical bucket<\/p>\n<p>1\/3 \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 24 = \u03c0 \u00d7 18<sup>2<\/sup>\u00a0\u00d7 32<\/p>\n<p>r<sup>2<\/sup>\u00a0= 18<sup>2<\/sup>\u00a0x 4<\/p>\n<p>r = 18 x 2 = 36 cm<\/p>\n<p>Now,<\/p>\n<p>The slant height of the conical heap (l) is given by<\/p>\n<p>l = \u221a(h<sup>2<\/sup>\u00a0+ r<sup>2<\/sup>)<\/p>\n<p>l = \u221a(24<sup>2<\/sup>\u00a0+ 36<sup>2<\/sup>) = \u221a1872<\/p>\n<p>l = 43.26 cm<\/p>\n<p>Therefore, the radius and slant height of the conical heap is 36 cm and 43.26 cm respectively.<\/p>\n<p><strong>35. Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Length of the rectangular surface = 6 m = 600 cm<\/p>\n<p>Breadth of the rectangular surface = 4 m = 400 cm<\/p>\n<p>Height of the perceived rain = 1 cm<\/p>\n<p>So,<\/p>\n<p>The volume of the rectangular surface = length * breadth * height<\/p>\n<p>= 600*400*1 cm<sup>3<\/sup><\/p>\n<p>= 240000 cm<sup>3<\/sup> \u2026\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>Also given,<\/p>\n<p>The radius of the cylindrical vessel = 20 cm<\/p>\n<p>Let the height of the cylindrical vessel be taken as h cm<\/p>\n<p>We know that,<\/p>\n<p>Volume of the cylindrical vessel = \u03c0 \u00d7 r<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>= \u03c0 \u00d7 20<sup>2<\/sup>\u00a0\u00d7 h \u2026\u2026\u2026.. (ii)<\/p>\n<p>As all the rainwater is transferred to the cylindrical vessel<\/p>\n<p>We can equate both (i) and (ii) for equal volumes,<\/p>\n<p>240000 = \u03c0 \u00d7 20<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>h = 190.9 cm<\/p>\n<p>Therefore, the height of the cylindrical vessel is nearly 191 cm.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-16-exercise-161\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631792774783\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-16-exercise-161-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631792911511\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-the-correct-rd-sharma-class-10-solutions-chapter-16-exercise-161\"><\/span>Where can I get the correct RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>On the Kopykitab website, students can find appropriate and accurate RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1. The best reference materials are RD Sharma textbooks, which give students and teachers a wide choice of sample questions to answer. The solutions are carefully crafted to ensure that students grasp the concepts and perform well on their board exams.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631792991626\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-16-exercise-161-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1:\u00a0This exercise comprises questions involving solid conversions from one type to another. Students can use the RD Sharma Class 10 Solutions to clear up any questions they have about this chapter. Also available for download is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.1 PDF. &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-1\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128529,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128521"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=128521"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128521\/revisions"}],"predecessor-version":[{"id":513787,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128521\/revisions\/513787"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128529"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128521"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128521"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}