{"id":6095,"date":"2023-09-27T12:41:58","date_gmt":"2023-09-27T12:41:58","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6095"},"modified":"2023-10-03T04:43:55","modified_gmt":"2023-10-03T04:43:55","slug":"ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 13 Surface Areas and Volumes <\/strong>Exercise 13.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 13 Surface Areas and Volumes<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.4<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-5\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.5<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong><span class=\"fontstyle0\">Unless stated otherwise, take <\/span><span class=\"fontstyle2\">\u03c0 <\/span><span class=\"fontstyle0\">= <\/span><\/strong><strong><span class=\"fontstyle0\">22\/7<\/span><\/strong><strong><span class=\"fontstyle0\">.<\/span> <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. A solid is in the shape of a cone standing on a hemisphere, with both their radii being equal to 1 cm and the height of the cone being equal to its radius. Find the volume of the solid in terms of \u03c0.<br \/>\n<\/strong><strong>Solution &#8211;<\/strong>\u00a0The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6680\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"187\" height=\"159\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-1.png 304w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-1-300x255.png 300w\" sizes=\"auto, (max-width: 187px) 100vw, 187px\" \/><br \/>\nHere r = 1 cm<br \/>\nh = 1 cm.<br \/>\nNow, Volume of solid = Volume of conical part + Volume of hemispherical part<br \/>\nWe know the volume of cone = \u2153 \u03c0r<sup>2<\/sup>h<br \/>\nAnd,<br \/>\nThe volume of the hemisphere = \u2154\u03c0r<sup>3<br \/>\n<\/sup>Volume of the solid = volume of the conical part + volume of the hemispherical part <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/3 \u03c0r<sup>2<\/sup>h + 2\/3 \u03c0r<sup>3<br \/>\n<\/sup>= 1\/3 \u03c0r<sup>3<\/sup>\u00a0+ 2\/3 \u03c0r<sup>3<\/sup> [Since, h = r] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>3<br \/>\n<\/sup>= \u03c0 (1cm)<sup>3<br \/>\n<\/sup>= \u03c0 cm<sup>3<br \/>\n<\/sup>Thus,\u00a0the volume of the solid is \u03c0 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm, and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model are nearly the same.)<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6683 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3-300x116.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"116\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3-300x116.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3.png 621w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nHeight of cylinder = 12 \u2013 2 &#8211; 2 = 8 cm<br \/>\nRadius = 3\/2 = 1.5 cm<br \/>\nHeight of cone = 2 cm<br \/>\nNow, the total volume of the air contained will be = Volume of cylinder + 2 \u00d7 (Volume of the cone)<br \/>\n\u2234 Total volume = \u03c0r<sup>2<\/sup>h + [2 \u00d7 (\u2153 \u03c0r<sup>2<\/sup>h )]<br \/>\n= 18 \u03c0 + 2(1.5 \u03c0)<br \/>\n= 66 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm (see figure).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6682\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Que-3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"178\" height=\"258\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6683\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"286\" height=\"111\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3.png 621w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-3-300x116.png 300w\" sizes=\"auto, (max-width: 286px) 100vw, 286px\" \/><br \/>\nIt is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.<br \/>\nSo, the total height of a gulab jamun = 5 cm.<br \/>\nDiameter = 2.8 cm<br \/>\nSo, radius = 1.4 cm<br \/>\n\u2234 The height of the cylindrical part = 5 cm \u2013 (1.4 + 1.4) cm<br \/>\n= 2.2 cm<br \/>\nNow, the total volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres<br \/>\n= \u03c0r<sup>2<\/sup>h + (4\/3)\u03c0r<sup>3<br \/>\n<\/sup>= 4.312\u03c0 + (10.976\/3) \u03c0<br \/>\n= 25.05 cm<sup>3<br \/>\n<\/sup>We know that the volume of sugar syrup = 30% of the total volume<br \/>\nSo, the volume of sugar syrup in 45 gulab jamuns = 45 \u00d7 30%(25.05 cm<sup>3<\/sup>)<br \/>\n= 45 \u00d7 7.515<br \/>\n= 338.184 cm<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm, and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6684\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Que-4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"272\" height=\"240\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Que-4.png 354w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Que-4-300x264.png 300w\" sizes=\"auto, (max-width: 272px) 100vw, 272px\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The volume of the cuboid = length \u00d7\u00a0 width \u00d7 height<br \/>\nWe know the cuboid\u2019s dimensions as 15 cm \u00d7 10 cm \u00d7\u00a03.5 cm<br \/>\nSo, the volume of the cuboid = 15 \u00d7 10 \u00d7\u00a03.5<br \/>\n= 525 cm<sup>3<br \/>\n<\/sup>Here, depressions are like cones, and we know,<br \/>\nVolume of cone = (\u2153)\u03c0r<sup>2<\/sup>h<br \/>\nGiven, radius (r) = 0.5 cm<br \/>\ndepth (h) = 1.4 cm<br \/>\n\u2234 Volume of 4 cones = 4 \u00d7 (\u2153) \u03c0r<sup>2<\/sup>h<br \/>\n= 4 \u00d7 (\u2153) \u00d7 22\/7 \u00d7 0.5 \u00d7 0.5 \u00d7 1.4<br \/>\n= 1.46 cm<sup>2<br \/>\n<\/sup>Now, the volume of wood = Volume of the cuboid \u2013 4 \u00d7volume of the cone<br \/>\n= 525 &#8211; 1.46<br \/>\n= 523.54 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6685 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-5-300x174.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"174\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-5-300x174.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-5-768x446.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-5-850x494.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-5.png 977w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nFor the cone,<br \/>\nRadius = 5 cm,<br \/>\nHeight = 8 cm<br \/>\nAlso,<br \/>\nRadius of sphere = 0.5 cm<br \/>\nVolume of water in the vessel = Volume of the conical vessel<br \/>\nVolume of all lead shots dropped into the vessel = 1\/4 \u00d7 Volume of the water in the vessel<br \/>\nNumber of lead shots = 1\/4 \u00d7 volume of the water in the vessel \u00f7 volume of each lead shot<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">We will find the volume of the water in the vessel and lead shot by using formulae;<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of the sphere\u00a0= 4\/3 \u03c0r<sup>3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>(where r is the radius of the sphere)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of the cone\u00a0= 1\/3 \u03c0R<sup>2<\/sup>h\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(where R and h are the radius and height of the cone respectively)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Height of the conical vessel, h = 8 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of the conical vessel, R = 5 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of the spherical lead shot, r = 0.5 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Number of lead shots = 1\/4 \u00d7 volume of the water in the vessel \u00f7 volume of each lead shot <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1 \/4\u00a0\u00d7 (1\/3) \u03c0R<sup>2<\/sup>h \u00f7 4 \u03c0r<sup>3<\/sup>\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0R<sup>2<\/sup>h\/12 \u00d7 3\/4\u03c0r<sup>3<br \/>\n<\/sup>= R<sup>2<\/sup>h \/ 16r<sup>3<br \/>\n<\/sup>= (5cm \u00d7 5 cm \u00d7 8 cm) \/ (16 \u00d7 0.5 cm \u00d7 0.5 cm \u00d7 0.5 cm) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 100 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus,\u00a0the number of lead shots dropped in the vessel is 100.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm<sup>3<\/sup> of iron has approximately 8 g mass.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6686 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-6-166x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"166\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-6-166x300.png 166w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-6.png 294w\" sizes=\"auto, (max-width: 166px) 100vw, 166px\" \/><br \/>\nGiven the height of the big cylinder (H) = 220 cm<br \/>\nThe radius of the base (R) = 24\/2 = 12 cm<br \/>\nSo, the volume of the big cylinder = \u03c0R<sup>2<\/sup>H<br \/>\n= \u03c0(12)<sup>2\u00a0<\/sup>\u00d7 220 cm<sup>3<br \/>\n<\/sup>= 99565.8 cm<sup>3<br \/>\n<\/sup>Now, the height of the smaller cylinder (h) = 60 cm<br \/>\nThe radius of the base (r) = 8 cm<br \/>\nSo, the volume of the smaller cylinder = \u03c0r<sup>2<\/sup>h<br \/>\n= \u03c0(8)<sup>2 <\/sup>\u00d7 60 cm<sup>3<br \/>\n<\/sup>= 12068.5 cm<sup>3<br \/>\n<\/sup>\u2234 The volume of iron = Volume of the big cylinder+ Volume of the small cylinder<br \/>\n= 99565.8 + 12068.5<br \/>\n=111634.5 cm<sup>3<br \/>\n<\/sup>We know,<br \/>\nMass = Density \u00d7 volume<br \/>\nSo, the mass of the pole = 8 \u00d7 111634.5<br \/>\n= 893 Kg (approx.)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder if the radius of the cylinder is 60 cm and its height is 180 cm.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6687 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-7-278x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"278\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-7-278x300.png 278w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-7-300x324.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-7.png 336w\" sizes=\"auto, (max-width: 278px) 100vw, 278px\" \/><br \/>\nHere, the volume of water left will be = Volume of the cylinder \u2013 Volume of solid<br \/>\nGiven,<br \/>\nRadius of cone = 60 cm,<br \/>\nHeight of cone = 120 cm<br \/>\nRadius of cylinder = 60 cm<br \/>\nHeight of cylinder = 180 cm<br \/>\nRadius of hemisphere = 60 cm<br \/>\nNow,<br \/>\nThe total volume of solid = Volume of Cone + Volume of the hemisphere<br \/>\nVolume of cone = 1\/3\u03c0r<sup>2<\/sup>h<br \/>\n= 1\/3 \u00d7 \u03c0 \u00d7 60<sup>2 <\/sup>\u00d7 120cm<sup>3<\/sup><br \/>\n= 144\u00d710<sup>3<\/sup>\u03c0 cm<sup>3<br \/>\n<\/sup>Volume of hemisphere = (\u2154) \u00d7 \u03c0 \u00d7 60<sup>3\u00a0<\/sup>cm<sup>3<\/sup><br \/>\n= 144 \u00d7 10<sup>3<\/sup>\u03c0 cm<sup>3<br \/>\n<\/sup>So, total volume of solid = 144 \u00d7 10<sup>3<\/sup>\u03c0 cm<sup>3<\/sup> + 144 \u00d7 10<sup>3<\/sup>\u03c0 cm<sup>3<\/sup><br \/>\n= 288 \u00d7 10<sup>3<\/sup>\u03c0 cm<sup>3<br \/>\n<\/sup>Volume of cylinder = \u03c0 \u00d7 60<sup>2 <\/sup>\u00d7 180<br \/>\n= 648000<br \/>\n= 648 \u00d7 10<sup>3<\/sup>\u00a0\u03c0 cm<sup>3<br \/>\n<\/sup>Now, the volume of water left will be = Volume of the cylinder \u2013 Volume of solid<br \/>\n= (648 &#8211; 288) \u00d7 10<sup>3 <\/sup>\u00d7 \u03c0<br \/>\n= 1.131 m<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm<sup>3<\/sup>. Check whether she is correct, taking the above as the inside measurements and \u03c0 = 3.14.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6688 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-8-180x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"180\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-8-180x300.png 180w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.2-Ans-8.png 291w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><br \/>\nFor the cylinder part, Height (h) = 8 cm and Radius (R) = (2\/2) cm = 1 cm<br \/>\nFor the spherical part, Radius (r) = (8.5\/2) = 4.25 cm<br \/>\nNow, volume of this vessel = Volume of cylinder + Volume of sphere<br \/>\n= \u03c0 \u00d7 (1)<sup>2 <\/sup>\u00d7 8 + (4\/3)\u03c0(4.25)<sup>3<br \/>\n<\/sup>= 346.51 cm<sup>3<br \/>\n<\/sup>Hence, the child\u2019s calculation is not correct.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\" target=\"_blank\" rel=\"noopener\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 13 Surface Areas and Volumes Exercise 13.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 13 Surface Areas and Volumes NCERT Class [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1109,1111,1044,1049,1048],"class_list":["post-6095","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-13-surface-areas-and-volumes-solutions","tag-ncert-class-10-mathematics-exercise-13-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.6) - 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