{"id":6096,"date":"2023-10-03T04:44:59","date_gmt":"2023-10-03T04:44:59","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6096"},"modified":"2023-10-03T04:44:59","modified_gmt":"2023-10-03T04:44:59","slug":"ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3"},"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 13.3 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-13-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.2<\/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; 13.3<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Given, the radius of the sphere (R) = 4.2 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cylinder (r) = 6 cm<\/span><br \/>\n<span style=\"color: #000000;\">let the height of the cylinder = h<br \/>\nthe volume of the sphere = Volume of the cylinder<br \/>\n\u21d2 (4\/3)\u00d7\u03c0\u00d7R<sup>3\u00a0 <\/sup>= \u03c0\u00d7r<sup>2<\/sup>\u00d7h<br \/>\n\u21d2 (4\/3) \u00d7 4.2 \u00d7 4.2 \u00d7 4.2 = 6 \u00d7 6 \u00d7 h<br \/>\n\u21d2 4 \u00d7 1.4 \u00d7 4.2 \u00d7 4.2 = 6 \u00d7 6 \u00d7 h<br \/>\n\u21d2 h = (4 \u00d7 1.4 \u00d7 4.2 \u00d7 4.2) \/ (6 \u00d7 6)<br \/>\n\u21d2 h = 2.74 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>For Sphere 1:<br \/>\n<\/strong>Radius (r<sub>1<\/sub>) = 6 cm<br \/>\n\u2234 Volume (V<sub>1<\/sub>) = (4\/3) \u00d7 \u03c0 \u00d7 r<sub>1<\/sub><sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>For Sphere 2:<br \/>\n<\/strong>Radius (r<sub>2<\/sub>) = 8 cm<br \/>\n\u2234 Volume (V<sub>2<\/sub>) = (4\/3) \u00d7 \u03c0 \u00d7 r<sub>2<\/sub><sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>For Sphere 3:<br \/>\n<\/strong>Radius (r<sub>3<\/sub>) = 10 cm<br \/>\n\u2234 Volume (V<sub>3<\/sub>) = (4\/3) \u00d7 \u03c0 \u00d7 r<sub>3<\/sub><sup>3 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Also, let the radius of the resulting sphere be \u201cr\u201d<br \/>\nNow,<br \/>\nThe volume of the resulting sphere = V<sub>1 <\/sub>+ V<sub>2 <\/sub>+ V<sub>3<br \/>\n<\/sub>\u21d2 (4\/3) \u00d7 \u03c0 \u00d7 r<sup>3<\/sup> = (4\/3)\u00d7\u03c0\u00d7r<sub>1<\/sub><sup>3 <\/sup>+ (4\/3)\u00d7\u03c0\u00d7r<sub>2<\/sub><sup>3<\/sup> + (4\/3)\u00d7\u03c0\u00d7r<sub>3<\/sub><sup>3<br \/>\n<\/sup>\u21d2 (4\/3) \u00d7 \u03c0 \u00d7 r<sup>3<\/sup> = (4\/3)\u00d7\u03c0 [r<sub>1<\/sub><sup>3<\/sup> + r<sub>2<\/sub><sup>3\u00a0 <\/sup>+ r<sub>3<\/sub><sup>3 <\/sup>]<br \/>\n\u21d2 r<sup>3\u00a0<\/sup>= 6<sup>3 <\/sup>+ 8<sup>3 <\/sup>+ 10<sup>3<br \/>\n<\/sup>\u21d2 r<sup>3\u00a0<\/sup>= 1728<br \/>\n\u21d2 r = 12 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. A 20 m deep well with a diameter of 7 m is dug, and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given, the height of deep well which form a cylinder (h) = 20 m<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6700 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-3-300x149.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"149\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-3-300x149.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-3-768x382.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-3-850x423.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-3.png 869w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nDepth of the cylindrical well, <em>h<\/em> = 20 m<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cylindrical well, <em>r<\/em> = 7\/2 m<\/span><br \/>\n<span style=\"color: #000000;\">Length of the cuboidal platform, <em>l<\/em> = 22 m<\/span><br \/>\n<span style=\"color: #000000;\">Breadth of the cuboidal platform, <em>b<\/em> = 14 m<\/span><br \/>\n<span style=\"color: #000000;\">Let the height of the cuboidal platform = <em>H<br \/>\n<\/em>Volume of the cylindrical well = Volume of the cuboidal platform<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 <em>\u03c0r<sup>2<\/sup>h = lbH<br \/>\n<\/em>\u21d2 H = \u03c0r<sup>2<\/sup>h \/ lb<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 H = (22\/7 \u00d7 7\/2 \u00d7 7\/2 \u00d7 20) \/ (22 \u00d7 14)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 H = 5\/2 m<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 H = 2.5 m<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the height of the platform will be 2.5 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The shape of the well will be cylindrical, as given below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6701 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4-1024x488.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"403\" height=\"192\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4-1024x488.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4-300x143.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4-768x366.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4-850x405.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-4.png 1146w\" sizes=\"auto, (max-width: 403px) 100vw, 403px\" \/><br \/>\nGiven, depth (h<sub>1<\/sub>) of well = 14 m<br \/>\nDiameter of the circular end of the well = 3 m<br \/>\nSo, Radius (r<sub>1<\/sub>) = 3\/2 m<br \/>\nWidth of the embankment = 4 m<br \/>\nOuter radius of the embankment, R = Inner radius + Width<\/span><br \/>\n<span style=\"color: #000000;\">R = 1.5 m + 4 m <\/span><br \/>\n<span style=\"color: #000000;\">= 5.5 m<\/span><br \/>\n<span style=\"color: #000000;\">Let the height of embankment be <em>h<\/em> <\/span><br \/>\n<span style=\"color: #000000;\">Volume of the cylindrical well = Volume of the hollow cylindrical embankment\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u03c0r\u00b2h<sub>\u2081<\/sub> = \u03c0R\u00b2h &#8211;\u00a0\u03c0r\u00b2h <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u03c0r\u00b2h<sub>\u2081<\/sub>\u00a0= \u03c0h (R\u00b2 &#8211;\u00a0r\u00b2)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r\u00b2h<sub>\u2081<\/sub>\u00a0= h (R &#8211; r )(R + r)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 h\u00a0= [(r\u00b2h<sub>\u2081<\/sub>)\/(R &#8211; r)(R + r)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 h = [((1.5)\u00b2 \u00d7 14)\/(5.5 &#8211; 1.5)(5.5 + 1.5)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 h = (2.25 \u00d7 14)\/(4 \u00d7 7)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 h = 1.125 m<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the height of the embankment will be 1.125 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let the height and radius of ice cream container (cylinder) be h and r.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6702\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5-300x176.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"176\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5-300x176.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5-1024x602.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5-768x452.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5-850x500.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-5.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nThe number of cones will be = Volume of cylinder\/Volume of ice cream cone<br \/>\n<strong>For the cylinder part<\/strong>,<br \/>\nRadius = 12\/2 = 6 cm<br \/>\nHeight = 15 cm<br \/>\n\u2234 Volume of cylinder = \u03c0\u00d7r<sup>2<\/sup>\u00d7h<br \/>\n= \u03c0\u00d76\u00d76\u00d715<br \/>\n= 540\u03c0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>For the ice cone part,<br \/>\n<\/strong>Radius of conical part = 6\/2 = 3 cm<br \/>\nHeight = 12 cm<br \/>\nRadius of hemispherical part = 6\/2 = 3 cm<br \/>\nThe volume of the ice cream cone = Volume of the conical part + Volume of the hemispherical part<br \/>\n= (\u2153)\u00d7\u03c0\u00d7r<sup>2<\/sup>\u00d7h + (\u2154)\u00d7\u03c0\u00d7r<sup>3<br \/>\n<\/sup>= (\u2153)\u00d7\u03c0\u00d73\u00d73\u00d712 + (\u2154)\u00d7\u03c0\u00d73\u00d73\u00d73<br \/>\n= 36\u03c0 + 18\u03c0<br \/>\n= 54\u03c0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u2234 Number of cones = (540\u03c0\/54\u03c0)<br \/>\n= 10<br \/>\nTherefore, 10 ice cream cones can be filled with the ice cream in the container.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm \u00d7 10 cm \u00d7 3.5 cm?<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>It is known that the coins are cylindrical in shape.<br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6703\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-6-300x73.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"73\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-6-300x73.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-6.png 540w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/strong>height (h<sub>1<\/sub>) of the cylinder = 2 mm = 0.2 cm<br \/>\nRadius (r) of circular end of coins = 1.75\/2 = 0.875 cm<br \/>\nthe number of coins to be melted to form the required cuboids be \u201cn\u201d<br \/>\nVolume of n coins = Volume of cuboids<br \/>\nn \u00d7 \u03c0 \u00d7 r<sup>2\u00a0<\/sup>\u00d7 h<sub>1<\/sub> = l \u00d7 b \u00d7 h<br \/>\nn \u00d7 \u03c0 \u00d7 (0.875)<sup>2 <\/sup>\u00d7 0.2 = 5.5 \u00d7 10 \u00d7 3.5<br \/>\nn \u00d7 22\/7 \u00d7 (0.875)<sup>2 <\/sup>\u00d7 0.2 = 5.5 \u00d7 10 \u00d7 3.5<br \/>\nn = (5.5 \u00d7 10 \u00d7 3.5)\/(22\/7 \u00d7 (0.875)<sup>2 <\/sup>\u00d7 0.2)<br \/>\nn = (5.5 \u00d7 10 \u00d7 3.5 \u00d7 7)\/(22 \u00d7 0.875 \u00d7 0.875<sup>\u00a0<\/sup>\u00d7 0.2)<br \/>\nn = 400<br \/>\nTherefore, the number of coins melted to form such a cuboid is 400.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A cylindrical bucket, 32 cm high and with a radius of a base of 18 cm, 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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram will be as-<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6704\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7-300x139.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"139\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7-300x139.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7-1024x474.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7-768x356.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7-850x394.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-7.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nGiven,<br \/>\nHeight (h<sub>1<\/sub>) of cylindrical part of the bucket = 32 cm<br \/>\nRadius (r<sub>1<\/sub>) of circular end of the bucket = 18 cm<br \/>\nHeight of the conical heap (h<sub>2<\/sub>) = 24 cm<br \/>\nNow, let r<sub>2 <\/sub>be the radius of the circular end of the conical heap.<br \/>\nWe know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.<br \/>\n\u2234 The volume of sand in the cylindrical bucket = Volume of sand in the conical heap<br \/>\n\u21d2 \u03c0\u00d7r<sub>1<\/sub><sup>2<\/sup>\u00d7h<sub>1<\/sub>\u00a0= (\u2153)\u00d7\u03c0\u00d7r<sub>2<\/sub><sup>2<\/sup>\u00d7h<sub>2<br \/>\n<\/sub>\u21d2 \u03c0\u00d718<sup>2<\/sup>\u00d732 = (\u2153)\u00d7\u03c0 \u00d7r<sub>2<\/sub><sup>2<\/sup>\u00d724<br \/>\n\u21d2 r<sub>2<\/sub><sup>2<\/sup> = (18 \u00d7 18 \u00d7 32 \u00d7 3) \/\u00a0 24<br \/>\n\u21d2 r<sub>2<\/sub><sup>2<\/sup> = 1296<br \/>\n\u21d2 r<sub>2<\/sub>= 36 cm<br \/>\nSlant height (l) = \u221a(36<sup>2 <\/sup>+ 24<sup>2<\/sup>)<br \/>\n= 12\u221a13 cm.<br \/>\nThus, the radius and slant height of the conical heap is 36 cm and 12\u221a13\u00a0cm respectively.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Water in a canal, 6 m wide and 1.5 m deep, flows at a speed of 10 km\/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>It is given that the canal is the shape of a cuboid with dimensions as:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6705\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-8-300x132.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"132\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-8-300x132.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-8-768x337.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-8.png 773w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nBreadth (b) = 6 m<br \/>\nHeight (h) = 1.5 m<br \/>\nThe speed of canal = 10 km\/hr<br \/>\nLength of canal covered in 1 hour = 10 km<br \/>\nLength of canal covered in 60 minutes = 10 km<br \/>\nLength of canal covered in 1 min = (1\/60) \u00d7 10 km<br \/>\nLength of canal covered in 30 min (l) = (30\/60) \u00d7 10 = 5km = 5000 m<br \/>\nWe know that the canal is cuboidal in shape.<br \/>\nThe volume of the canal = l\u00d7b\u00d7h<br \/>\n= 5000 \u00d7 6 \u00d7 1.5 m<sup>3<br \/>\n<\/sup>= 45000 m<sup>3<br \/>\n<\/sup>Now, The volume of water in the canal = Volume of area irrigated<br \/>\n= Area irrigated \u00d7 Height<br \/>\nArea irrigated = 56.25 hectares<br \/>\n\u2234 The volume of the canal = l\u00d7b\u00d7h<br \/>\n45000 = Area irrigated \u00d7 8 cm<br \/>\n45000 = Area irrigated \u00d7 (8\/100)m<br \/>\nArea irrigated = 562500 m<sup>2\u00a0<\/sup>= 56.25 hectares.<br \/>\nTherefore, 562500 m<sup>2<\/sup> or 56.25 hectares area will be irrigated in 30 minutes.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km\/h, in how much time will the tank be filled?<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Consider the following diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6706 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9-1024x426.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"431\" height=\"179\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9-1024x426.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9-300x125.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9-768x320.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9-850x354.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.3-Ans-9.png 1146w\" sizes=\"auto, (max-width: 431px) 100vw, 431px\" \/><br \/>\nRadius of the cylindrical tank, R = 10 \/ 2 m = 5 m<\/span><br \/>\n<span style=\"color: #000000;\">Depth of the cylindrical tank, H = 2 m <\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cylindrical pipe, r = 20\/2 cm = 10\/100 m = 0.1 m<\/span><br \/>\n<span style=\"color: #000000;\">Length of the water flowing through the pipe in 1 hour (60 minutes) = 3 km<\/span><br \/>\n<span style=\"color: #000000;\">Length of the water flowing through the pipe in 1 minute, h = 3 km\/60 = (3 \u00d7 1000 m) \/60 = 50 m<\/span><br \/>\n<span style=\"color: #000000;\">Volume of water flowing through pipe in &#8216;t&#8217; minutes = volume of water in cylindrical tank <\/span><br \/>\n<span style=\"color: #000000;\">t \u00d7 \u03c0r<sup>2<\/sup>h = \u03c0R<sup>2<\/sup>H<\/span><br \/>\n<span style=\"color: #000000;\">t = R<sup>2<\/sup>H\u00a0\/ r<sup>2<\/sup>h <\/span><br \/>\n<span style=\"color: #000000;\">t = (5 \u00d7 5 \u00d7 2 ) \/ (0.1 \u00d7 0.1 \u00d7 50 )<\/span><br \/>\n<span style=\"color: #000000;\">t = 100<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the cylindrical tank will be filled in 100 minutes.<\/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 13.3 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":[1045,1109,1112,1044,1049,1048],"class_list":["post-6096","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-13-3-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 v27.4) - 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