{"id":28921,"date":"2021-07-14T18:22:07","date_gmt":"2021-07-14T12:52:07","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28921"},"modified":"2021-07-14T18:35:28","modified_gmt":"2021-07-14T13:05:28","slug":"ncert-solutions-for-class-10-maths-chapter-13-ex-13-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2<\/h2>\n

\"NCERT<\/p>\n

Unless stated otherwise, take \u03c0 = \\(\\frac { 22 }{ 7 }\\)<\/p>\n

Question 1.
\nA 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 is equal to its radius. Find the volume of the solid in terms of n.
\nSolution:
\nRadius of cone = 1 cm. and radius of hemisphere is also = 1 cm.
\nVolume of solid = Volume of cone + Volume of hemisphere
\n\"NCERT<\/p>\n

Question 2.
\nRachel, 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 to be nearly the same.)
\nSolution:
\nVolume of air contained in the model = Total volume of the solid
\nDiameter of base of each cone = 3 cm
\n\u2234 Radius of base of each cone = \\(\\frac { 3 }{ 2 }\\)
\nHeight of each cone = 2 cm
\n\"NCERT
\nVolume of the air inside the model = Volume of air inside = Volume of cone + volume of cylinder + volume of other cone.
\n\"NCERT
\nVolume of the model = 66 cm\u00b3<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA 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 length 5 cm and diameter 2.8 cm (see figure).
\n\"NCERT
\nSolution:
\nVolume of one piece of gulab jamun = Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8
\nRadius of each hemispherical portion = \\(\\frac { 2.8 }{ 2 }\\) = 1.4 cm
\n\"NCERT
\nRadius of gulab jamun = r = \\(\\frac { 2.8 }{ 2 }\\) = 1.4 cm and
\nheight = 5 cm so, height of cylinder (h) = 5 – (2.8) = 2.2.
\nVolume 45 gulab jamuns = 25.05 x 45 = 1127.279
\n30% of its volume = \\(\\frac{1127.279 \\times 30}{100}\\)
\n= 338.18 = 338 cm\u00b3.<\/p>\n

Question 4.
\nA 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.
\nFind the volume of wood in the entire stand (see figure).
\n\"NCERT
\nSolution:
\nPen stand is made in the shape of a cuboid whose length, breadth and height are respectively = 15 cm, 10 cm, 3.5 cm.
\nSo, the volume of pen stand = length x breadth x height
\n= 15 cm x 10 cm x 3.5 cm = 525 cm\u00b2
\nEach hole, is in the shape of a cone, so the volume of cone
\n= \\(\\frac { 1 }{ 3 }\\)r\u00b2h
\nRadius of hole = 0 0.5 cm, and height = 1.4 cm.
\n\u2234 Volume of 4 (holes) cone = 4 x \\(\\frac { 1 }{ 3 }\\)r\u00b2h
\n= \\(\\frac { 4 }{ 3 }\\) x \\(\\frac { 22 }{ 7 }\\) 0.5 x 0.5 x 1.4 cm\u00b3
\n= \\(\\frac{4 \\times 4.4 \\times 0.25}{3}\\) = \\(\\frac { 4 }{ 3 }\\) = 1.466 cm\u00b3
\nVolume of wood in the entire stand = Volume of cuboid – Volume 4 holes.
\n= (525 – 1.466) cm\u00b3.
\n= 523.533 cm\u00b3.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA 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.
\nSolution:
\n\"NCERT
\nWhen lead shots are dropped into the vessel, then
\nVolume of water flows out = Volume of leads shots
\n\"NCERT
\nNumber of lead shots dropped in the vessel = 100.<\/p>\n

Question 6.
\nA 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 cm3 of iron has approximately 8 g mass. (Use \u03c0 = 3.14)
\n\"NCERT
\nSolution:
\nGiven: radius of 1st<\/sup> cylinder = 12 cm
\nand height of 1st<\/sup> cylinder = 220 cm
\n\u2234 Volume of 1st<\/sup> cylinder = \u03c0r\u00b2h
\n= \u03c0(12)\u00b2 (220) cm\u00b3
\n= 144 x 220\u03c0 cm\u00b3
\n= 144 x 220 x 3.14 cm\u00b3
\n= 99475.2 cm\u00b3 … (i)
\nGiven: radius of 2nd<\/sup> cylinder = 8 cm
\nand height of 2nd<\/sup> cylinder = 60 cm
\n\u2234 Volume of 2nd<\/sup> cylinder = \u03c0r\u00b2h
\n= \u03c0(8)\u00b2 (60) cm\u00b3 = 64 x 60\u03c0 cm\u00b3
\n= 64 x 60 x 3.14 cm\u00b3
\n= 12057.6 cm\u00b3 … (ii)
\nTotal volume of solid = Volume of 1st<\/sup> cylinder + Volume of 2nd<\/sup> cylinder
\n= 99475.2 cm\u00b3 + 12057.6 cm\u00b3 = 111532.8 cm\u00b3
\nGiven: mass of 1 cm\u00b3 of iron = 8 g
\n\u2234 Mass of 111532.8 cm\u00b3 of iron = 111532.8 x 8 g
\n= 892262.4 g = 892.262 kg<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nA 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.
\nSolution:
\nHeight of right circular cone = 120 cm.
\nand radius of hemisphere and cone = 60 cm.
\nRadius of cylinder = 60 cm, and height = 180 cm.
\n\"NCERT
\nVolume of water, left in the cylinder = Volume of cylinder – (Volume of cone + volume hemisphere)
\n\"NCERT
\nSo the left in the cylinder = 1,31 m\u00b3. (approx).<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nA spherical glass vessel has a cylindrical neck 8 cm long, 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\u00b3. Check whether she is correct, taking the above as the inside measurements, and \u03c0 = 3.14.
\nSolution:
\nVolume of water the glass vessel can hold = 345 cm\u00b3 (Measured by the child)
\nRadius of the cylindrical part = \\(\\frac { 2 }{ 2 }\\) = 1 cm
\n\"NCERT
\nHeight of the cylindrical part = 8 cm
\n\u2234 Volume of the cylindrical part = \u03c0r\u00b2h
\n= 3.14 x (1)\u00b2 x 8 cm\u00b3
\nDiameter of the spherical part Radius = 8.5 cm
\n\u2234 Radius = \\(\\frac { 8.5 }{ 2 }\\) cm
\n= \\(\\frac{7713.41}{24}\\)
\n= 321.392 cm\u00b3.
\nTotal volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
\n= 25.12 cm\u00b3 + 321.39 cm\u00b3 = 346.51 cm\u00b3
\nVolume measured by child is 345 cm\u00b3, which is not correct. Correct volume is 346.51 cm\u00b3.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 Unless stated otherwise, take \u03c0 = Question 1. A solid is in the shape of …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. 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