{"id":119311,"date":"2022-05-04T14:41:35","date_gmt":"2022-05-04T09:11:35","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119311"},"modified":"2022-05-04T14:41:35","modified_gmt":"2022-05-04T09:11:35","slug":"chapter-13-surface-areas-and-volumes-questions-and-answers-ncert-solutions-for-class-10-maths","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-13-surface-areas-and-volumes-questions-and-answers-ncert-solutions-for-class-10-maths","title":{"rendered":"Chapter 13 – Surface Areas and Volumes Questions and Answers: NCERT Solutions for Class 10 Maths"},"content":{"rendered":"

Exercise: 13.1<\/h2>\n

1. 2 cubes each of volume 64 cm3are joined end to end. Find the surface area of the resulting cuboid.<\/h2>\n

Answer:
\nThe diagram is given as:<\/h3>\n

\"\"<\/p>\n

Given,
\nThe Volume (V) of each cube is = 64 cm3
\nThis implies that a3= 64 cm3
\n\u2234 a = 4 cm
\nNow, the side of the cube = a = 4 cm
\nAlso, the length and breadth of the resulting cuboid will be 4 cm each. While its height will be 8 cm.
\nSo, the surface area of the cuboid = 2(lb+bh+lh)
\n= 2(8\u00d74+4\u00d74+4\u00d78) cm2
\n= 2(32+16+32) cm2
\n= (2\u00d780) cm2= 160 cm2<\/h3>\n

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.<\/h2>\n

Answer:
\nThe diagram is as follows:<\/h3>\n

\"\"<\/p>\n

Now, the given parameters are:
\nThe diameter of the hemisphere = D = 14 cm
\nThe radius of the hemisphere = r = 7 cm
\nAlso, the height of the cylinder = h = (13-7) = 6 cm
\nAnd, the radius of the hollow hemisphere = 7 cm
\nNow, the inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part
\n(2\u03c0rh+2\u03c0r2) cm2= 2\u03c0r(h+r) cm2
\n2\u00d7(22\/7)\u00d77(6+7) cm2= 572 cm2<\/h3>\n

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.<\/h2>\n

Answer:
\nThe diagram is as follows:<\/h3>\n

\"\"<\/p>\n

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7\/2 cm
\nThe total height of the toy is given as 15.5 cm.
\nSo, the height of the cone (h) = 15.5-3.5 = 12 cm<\/h2>\n

\"\"<\/p>\n

\u2234 The curved surface area of cone = \u03c0rl
\n(22\/7)\u00d7(7\/2)\u00d7(25\/2) = 275\/2 cm2
\nAlso, the curved surface area of the hemisphere = 2\u03c0r2
\n2\u00d7(22\/7)\u00d7(7\/2)2
\n= 77 cm2
\nNow, the Total surface area of the toy = CSA of cone + CSA of hemisphere
\n= (275\/2)+77 cm2
\n= (275+154)\/2 cm2
\n= 429\/2 cm2= 214.5cm2
\nSo, the total surface area (TSA) of the toy is 214.5cm2<\/h3>\n

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.<\/h2>\n

Answer:
\nIt is given that each side of cube is 7 cm. So, the radius will be 7\/2 cm.<\/h3>\n

\"\"<\/p>\n

We know,
\nThe total surface area of solid (TSA) = surface area of cubical block + CSA of hemisphere \u2013 Area of base of hemisphere
\n\u2234 TSA of solid = 6\u00d7(side)2+2\u03c0r2-\u03c0r2
\n= 6\u00d7(side)2+\u03c0r2
\n= 6\u00d7(7)2+(22\/7)\u00d7(7\/2)\u00d7(7\/2)
\n= (6\u00d749)+(77\/2)
\n= 294+38.5 = 332.5 cm2
\nSo, the surface area of the solid is 332.5 cm2<\/h3>\n

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.<\/h2>\n

Answer:
\nThe diagram is as follows:<\/h3>\n

\"\"<\/p>\n

Now, the diameter of hemisphere = Edge of the cube = l
\nSo, the radius of hemisphere = l\/2
\n\u2234 The total surface area of solid = surface area of cube + CSA of hemisphere \u2013 Area of base of hemisphere
\nTSA of remaining solid = 6 (edge)2+2\u03c0r2-\u03c0r2
\n= 6l2+ \u03c0r2
\n= 6l2+\u03c0(l\/2)2
\n= 6l2+\u03c0l2\/4
\n= l2\/4(24+\u03c0) sq. units<\/h3>\n

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.<\/h2>\n

\"\"<\/p>\n

Answer:
\nTwo hemisphere and one cylinder are shown in the figure given below.<\/h3>\n

\"\"<\/p>\n

Here, the diameter of the capsule = 5 mm
\n\u2234 Radius = 5\/2 = 2.5 mm
\nNow, the length of the capsule = 14 mm
\nSo, the length of the cylinder = 14-(2.5+2.5) = 9 mm
\n\u2234 The surface area of a hemisphere = 2\u03c0r2= 2\u00d7(22\/7)\u00d72.5\u00d72.5
\n= 275\/7 mm2
\nNow, the surface area of the cylinder = 2\u03c0rh
\n= 2\u00d7(22\/7)\u00d72.5\u00d79
\n(22\/7)\u00d745 = 990\/7 mm2
\nThus, the required surface area of medicine capsule will be
\n= 2\u00d7surface area of hemisphere + surface area of the cylinder
\n= (2\u00d7275\/7) \u00d7 990\/7
\n= (550\/7) + (990\/7) = 1540\/7 = 220 mm2<\/h3>\n

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)<\/h2>\n

Answer:
\nIt is known that a tent is a combination of cylinder and a cone.<\/h3>\n

\"\"<\/p>\n

From the question we know that
\nDiameter = 4 m
\nSlant height of the cone (l) = 2.8 m
\nRadius of the cone (r) = Radius of cylinder = 4\/2 = 2 m
\nHeight of the cylinder (h) = 2.1 m
\nSo, the required surface area of tent = surface area of cone + surface area of cylinder
\n= \u03c0rl+2\u03c0rh
\n= \u03c0r(l+2h)
\n= (22\/7)\u00d72(2.8+2\u00d72.1)
\n= (44\/7)(2.8+4.2)
\n= (44\/7)\u00d77 = 44 m2
\n\u2234 The cost of the canvas of the tent at the rate of \u20b9500 per m2will be
\n= Surface area \u00d7 cost per m2
\n44\u00d7500 = \u20b922000
\nSo, Rs. 22000 will be the total cost of the canvas.<\/h3>\n

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
\nsame height and same diameter is hollowed out. Find the total surface area of the
\nremaining solid to the nearest cm2.<\/h2>\n

Answer:
\nThe diagram for the question is as follows:<\/h3>\n

\"\"<\/p>\n

From the question we know the following:
\nThe diameter of the cylinder = diameter of conical cavity = 1.4 cm
\nSo, the radius of the cylinder = radius of the conical cavity = 1.4\/2 = 0.7
\nAlso, the height of the cylinder = height of the conical cavity = 2.4 cm<\/h3>\n

\"\"<\/p>\n

Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder
\n= \u03c0rl+(2\u03c0rh+\u03c0r2)
\n= \u03c0r(l+2h+r)
\n= (22\/7)\u00d7 0.7(2.5+4.8+0.7)
\n= 2.2\u00d78 = 17.6 cm2
\nSo, the total surface area of the remaining solid is 17.6 cm2<\/h3>\n

Exercise: 13.2<\/h2>\n

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 is equal to its radius. Find the volume of the solid in terms of \u03c0.<\/h2>\n

Solution:
\nHere r = 1 cm and h = 1 cm.
\nThe diagram is as follows.<\/h3>\n

\"\"<\/p>\n

Now, Volume of solid = Volume of conical part + Volume of hemispherical part
\nWe know the volume of cone = \u2153 \u03c0r2h
\nAnd,
\nThe volume of hemisphere = \u2154\u03c0r3
\nSo, volume of solid will be<\/h3>\n

\"\"<\/p>\n

= \u03c0 cm3<\/p>\n

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 aluminum 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.)<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nHeight of cylinder = 12\u20134 = 8 cm
\nRadius = 1.5 cm
\nHeight of cone = 2 cm
\nNow, the total volume of the air contained will be = Volume of cylinder+2\u00d7(Volume of cone)
\n\u2234 Total volume = \u03c0r2h+[2\u00d7(\u2153 \u03c0r2h )]
\n= 18 \u03c0+2(1.5 \u03c0)
\n= 66 cm3.<\/h3>\n

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 length 5 cm and diameter 2.8 cm (see figure).<\/h2>\n

\"\"<\/p>\n

Solution:<\/h3>\n

\"\"<\/p>\n

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.
\nSo, the total height of a gulab jamun = 5 cm.
\nDiameter = 2.8 cm
\nSo, radius = 1.4 cm
\n\u2234 The height of the cylindrical part = 5 cm\u2013(1.4+1.4) cm
\n=2.2 cm
\nNow, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres
\n= \u03c0r2h+(4\/3)\u03c0r3
\n= 4.312\u03c0+(10.976\/3) \u03c0
\n= 25.05 cm3
\nWe know that the volume of sugar syrup = 30% of total volume
\nSo, volume of sugar syrup in 45 gulab jamuns = 45\u00d730%(25.05 cm3)
\n= 45\u00d77.515 = 338.184 cm3<\/h3>\n

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.).<\/h2>\n

\"\"<\/p>\n

Solution:
\nVolume of cuboid = length x width x height
\nWe know the cuboid\u2019s dimensions as 15 cmx10 cmx3.5 cm
\nSo, the volume of the cuboid = 15x10x3.5 = 525 cm3
\nHere, depressions are like cones and we know,
\nVolume of cone = (\u2153)\u03c0r2h
\nGiven, radius (r) = 0.5 cm and depth (h) = 1.4 cm
\n\u2234 Volume of 4 cones = 4x(\u2153)\u03c0r2h
\n= 1.46 cm2
\nNow, volume of wood = Volume of cuboid \u2013 4 x volume of cone
\n= 525-1.46 = 523.54 cm2<\/h3>\n

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.<\/h2>\n

Solution:
\nFor the cone,
\nRadius = 5 cm,
\nHeight = 8 cm
\nAlso,
\nRadius of sphere = 0.5 cm
\nThe diagram will be like<\/h3>\n

\"\"<\/p>\n

It is known that,
\nVolume of cone = volume of water in the cone
\n= \u2153\u03c0r2h = (200\/3)\u03c0 cm3
\nNow,
\nTotal volume of water overflown= (\u00bc)\u00d7(200\/3) \u03c0 =(50\/3)\u03c0
\nVolume of lead shot
\n= (4\/3)\u03c0r3
\n= (1\/6) \u03c0
\nNow,
\nThe number of lead shots = Total Volume of Water over flown\/ Volume of Lead shot
\n= (50\/3)\u03c0\/(\u2159)\u03c0
\n= (50\/3)\u00d76 = 100<\/h3>\n

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 cm3of iron has approximately 8 g mass.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given, the height of the big cylinder (H) = 220 cm
\nRadius of the base (R) = 24\/2 = 12 cm
\nSo, the volume of the big cylinder = \u03c0R2H
\n= \u03c0(12)2\u00d7 220 cm3
\n= 99565.8 cm3
\nNow, the height of smaller cylinder (h) = 60 cm
\nRadius of the base (r) = 8 cm
\nSo, the volume of the smaller cylinder = \u03c0r2h
\n= \u03c0(8)2\u00d760 cm3
\n= 12068.5 cm3
\n\u2234 Volume of iron = Volume of the big cylinder+ Volume of the small cylinder
\n= 99565.8 + 12068.5
\n=111634.5 cm3
\nWe know,
\nMass = Density x volume
\nSo, mass of the pole = 8\u00d7111634.5
\n= 893 Kg (approx.)<\/h3>\n

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.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Here, the volume of water left will be = Volume of cylinder \u2013 Volume of solid
\nGiven,
\nRadius of cone = 60 cm,
\nHeight of cone = 120 cm
\nRadius of cylinder = 60 cm
\nHeight of cylinder = 180 cm
\nRadius of hemisphere = 60 cm
\nNow,
\nTotal volume of solid = Volume of Cone + Volume of hemisphere
\nVolume of cone = 1\/3\u03c0r2h = 1\/3 \u00d7 \u03c0\u00d7602\u00d7120cm3= 144\u00d7103\u03c0 cm3
\nVolume of hemisphere = (\u2154)\u00d7\u03c0\u00d7603cm3=144\u00d7103\u03c0 cm3
\nSo, total volume of solid =144\u00d7103\u03c0 cm3+144\u00d7103\u03c0 cm3= 288\u00d7103\u03c0 cm3
\nVolume of cylinder = \u03c0\u00d7602\u00d7180 = 648000 = 648\u00d7103\u03c0 cm3
\nNow, volume of water left will be = Volume of cylinder \u2013 Volume of solid
\n= (648-288) \u00d7 103\u00d7\u03c0 = 1.131 m3<\/h3>\n

8. A 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 cm3. Check whether she is correct, taking the above as the inside measurements, and \u03c0 = 3.14.<\/h2>\n

Solution:
\nGiven,
\nFor the cylinder part, Height (h) = 8 cm and Radius (R) = (2\/2) cm = 1 cm
\nFor the spherical part, Radius (r) = (8.5\/2) = 4.25 cm<\/h3>\n

\"\"<\/p>\n

Now, volume of this vessel = Volume of cylinder + Volume of sphere
\n= \u03c0\u00d7(1)2\u00d78+(4\/3)\u03c0(4.25)3
\n= 346.51 cm3<\/h3>\n

Exercise: 13.3<\/h2>\n

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.<\/h2>\n

Solution:
\nIt is given that radius of the sphere (R) = 4.2 cm
\nAlso, Radius of cylinder (r) = 6 cm
\nNow, let height of cylinder = h
\nIt is given that the sphere is melted into a cylinder.
\nSo, Volume of Sphere = Volume of Cylinder
\n\u2234 (4\/3)\u00d7\u03c0\u00d7R3= \u03c0\u00d7r2\u00d7h.
\nh = 2.74 cm<\/h3>\n

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.<\/h2>\n

Solution:
\nFor Sphere 1:
\nRadius (r1) = 6 cm
\n\u2234 Volume (V1) = (4\/3)\u00d7\u03c0\u00d7r13
\nFor Sphere 2:
\nRadius (r2) = 8 cm
\n\u2234 Volume (V2) = (4\/3)\u00d7\u03c0\u00d7r23
\nFor Sphere 3:
\nRadius (r3) = 10 cm
\n\u2234 Volume (V3) = (4\/3)\u00d7 \u03c0\u00d7 r33
\nAlso, let the radius of the resulting sphere be \u201cr\u201d
\nNow,
\nVolume of resulting sphere = V1+V2+V3
\n(4\/3)\u00d7\u03c0\u00d7r3= (4\/3)\u00d7\u03c0\u00d7r13+(4\/3)\u00d7\u03c0\u00d7r23+(4\/3)\u00d7\u03c0\u00d7r33
\nr3= 63+83+103
\nr3= 1728
\nr = 12 cm<\/h3>\n

3. A 20 m deep well with diameter 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.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m
\nSo, radius = 7\/2 m
\nAlso, Depth (h) = 20 m
\nVolume of the earth dug out will be equal to the volume of the cylinder
\nLet the height of the platform = H
\nVolume of soil from well (cylinder) = Volume of soil used to make such platform
\n\u03c0\u00d7r2\u00d7h = Area of platform \u00d7 Height of the platform
\nWe know that the dimension of the platform is = 22\u00d714
\nSo, Area of platform = 22\u00d714 m2
\n\u2234 \u03c0\u00d7r2\u00d7h = 22\u00d714\u00d7H
\n\u21d2 H = 2.5 m<\/h3>\n

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.<\/h2>\n

Solution:
\nThe shape of the well will be cylindrical as given below.<\/h3>\n

\"\"<\/p>\n

Given, Depth (h1) of well = 14 m
\nDiameter of the circular end of the well =3 m
\nSo, Radius (r1) = 3\/2 m
\nWidth of the embankment = 4 m
\nFrom the figure, it can be said that the embankment will be a cylinder having an outer radius (r2) as 4+(3\/2) = 11\/2 m and inner radius (r1) as 3\/2m
\nNow, let the height of embankment be h2
\n\u2234 Volume of soil dug from well = Volume of earth used to form embankment
\n\u03c0\u00d7r12\u00d7h1= \u03c0\u00d7(r22-r12) \u00d7 h2
\nSolving this, we get,
\nThe height of the embankment (h2) as 1.125 m.<\/h3>\n

5. A container shaped like a right circular cylinder having diameter 12 cm and height 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.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Number of cones will be = Volume of cylinder \/ Volume of ice cream cone
\nFor the cylinder part,
\nRadius = 12\/2 = 6 cm
\nHeight = 15 cm
\n\u2234 Volume of cylinder = \u03c0\u00d7r2\u00d7h = 540\u03c0
\nFor the ice cone part,
\nRadius of conical part = 6\/2 = 3 cm
\nHeight = 12 cm
\nRadius of hemispherical part = 6\/2 = 3 cm
\nNow,
\nVolume of ice cream cone = Volume of conical part + Volume of hemispherical part
\n= (\u2153)\u00d7\u03c0\u00d7r2\u00d7h+(\u2154)\u00d7\u03c0\u00d7r3
\n= 36\u03c0 +18\u03c0
\n= 54\u03c0
\n\u2234 Number of cones = (540\u03c0\/54\u03c0)
\n= 10<\/h3>\n

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?<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

It is known that the coins are cylindrical in shape.
\nSo, height (h1) of the cylinder = 2 mm = 0.2 cm
\nRadius (r) of circular end of coins = 1.75\/2 = 0.875 cm
\nNow, the number of coins to be melted to form the required cuboids be \u201cn\u201d
\nSo, Volume of n coins = Volume of cuboids
\nn \u00d7 \u03c0 \u00d7 r2\u00d7 h1= l \u00d7 b \u00d7 h
\nn\u00d7\u03c0\u00d7(0.875)2\u00d70.2 = 5.5\u00d710\u00d73.5
\nOr, n = 400<\/h3>\n

7. A cylindrical bucket, 32 cm high and with radius of base 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.<\/h2>\n

Solution:
\nThe diagram will be as-<\/h3>\n

\"\"<\/p>\n

Given,
\nHeight (h1) of cylindrical part of the bucket = 32 cm
\nRadius (r1) of circular end of the bucket = 18 cm
\nHeight of the conical heap ((h2) = 24 cm
\nNow, let \u201cr2\u201d be the radius of the circular end of the conical heap.
\nWe know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
\n\u2234 Volume of sand in the cylindrical bucket = Volume of sand in conical heap
\n\u03c0\u00d7r12\u00d7h1= (\u2153)\u00d7\u03c0\u00d7r22\u00d7h2
\n\u03c0\u00d7182\u00d732 = (\u2153)\u00d7\u03c0 \u00d7r22\u00d724
\nOr, r2= 36 cm
\nAnd,
\nSlant height (l) = \u221a(362+242) = 12\u221a13 cm.<\/h3>\n

8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km\/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?<\/h2>\n

Solution:
\nIt is given that the canal is the shape of a cuboid with dimensions as:
\nBreadth (b) = 6 m and Height (h) = 1.5 m
\nIt is also given that
\nThe speed of canal = 10 km\/hr
\nLength of canal covered in 1 hour = 10 km
\nLength of canal covered in 60 minutes = 10 km
\nLength of canal covered in 1 min = (1\/60)x10 km
\nLength of canal covered in 30 min (l) = (30\/60)x10 = 5km = 5000 m
\nWe know that the canal is cuboidal in shape. So,
\nVolume of canal = lxbxh
\n= 5000x6x1.5 m3
\n= 45000 m3
\nNow,
\nVolume of water in canal = Volume of area irrigated
\n= Area irrigated x Height
\nSo, Area irrigated = 56.25 hectares
\n\u2234 Volume of canal = lxbxh
\n45000 = Area irrigatedx8 cm
\n45000 = Area irrigated x (8\/100)m
\nOr, Area irrigated = 562500 m2= 56.25 hectares.<\/h3>\n

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?<\/h2>\n

Solution:
\nConsider the following diagram-<\/h3>\n

\"\"<\/p>\n

\"\"<\/p>\n

Volume of water that flows in t minutes from pipe = t\u00d70.5\u03c0 m3
\nRadius (r2) of circular end of cylindrical tank =10\/2 = 5 m
\nDepth (h2) of cylindrical tank = 2 m
\nLet the tank be filled completely in t minutes.
\nVolume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
\nVolume of water that flows in t minutes from pipe = Volume of water in tank
\nt\u00d70.5\u03c0 = \u03c0\u00d7r22\u00d7h2
\nOr, t = 100 minutes<\/h3>\n

Exercise: 13.4<\/h2>\n

1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

 <\/p>\n

Radius (r1) of the upper base = 4\/2 = 2 cm
\nRadius (r2) of lower the base = 2\/2 = 1 cm
\nHeight = 14 cm
\nNow, Capacity of glass = Volume of frustum of cone
\nSo, Capacity of glass = (\u2153)\u00d7\u03c0\u00d7h(r12+r22+r1r2)
\n= (\u2153)\u00d7\u03c0\u00d7(14)(22+12+ (2)(1))
\n\u2234 The capacity of the glass = 102\u00d7(\u2154) cm3<\/h3>\n

2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nSlant height (l) = 4 cm
\nCircumference of upper circular end of the frustum = 18 cm
\n\u2234 2\u03c0r1= 18
\nOr, r1= 9\/\u03c0
\nSimilarly, circumference of lower end of the frustum = 6 cm
\n\u2234 2\u03c0r2= 6
\nOr, r2= 3\/\u03c0
\nNow, CSA of frustum = \u03c0(r1+r2) \u00d7 l
\n= \u03c0(9\/\u03c0+3\/\u03c0) \u00d7 4
\n= 12\u00d74 = 48 cm2<\/h3>\n

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.<\/h2>\n

\"\"<\/p>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nFor the lower circular end, radius (r1) = 10 cm
\nFor the upper circular end, radius (r2) = 4 cm
\nSlant height (l) of frustum = 15 cm
\nNow,
\nThe area of material to be used for making the fez = CSA of frustum + Area of upper circular end
\nCSA of frustum = \u03c0(r1+r2)\u00d7l
\n= 210\u03c0
\nAnd, Area of upper circular end = \u03c0r22
\n= 16\u03c0
\nThe area of material to be used for making the fez = 210\u03c0 + 16\u03c0 = (226 x 22)\/7 = 710 2\/7
\n\u2234 The area of material used = 710 2\/7 cm2<\/h3>\n

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nr1= 20 cm,
\nr2= 8 cm and
\nh = 16 cm
\n\u2234 Volume of the frustum = (\u2153)\u00d7\u03c0\u00d7h(r12+r22+r1r2)
\n= 1\/3\u00d73.14 \u00d716((20)2+(8)2+(20)(8))
\n=1\/3\u00d73.14 \u00d716(400 + 64 + 160) = 10449.92 cm3= 10.45 lit
\nIt is given that the rate of milk = Rs. 20\/litre
\nSo, Cost of milk = 20\u00d7volume of the frustum
\n= 20\u00d7 10.45
\n= Rs. 209
\nNow, slant height will be<\/h3>\n

\"\"<\/p>\n

l = 20 cm
\nSo, CSA of the container = \u03c0(r1+r2)\u00d7l<\/h3>\n

\"\"<\/p>\n

= 1758.4 cm2
\nHence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)
\n= 1758.4+\u03c0r2= 1758.4+\u03c0(8)2
\n= 1758.4+201 = 1959.4 cm2
\n\u2234 Total cost of metal = Rs. (8\/100) \u00d7 1959.4 = Rs. 157<\/h3>\n

5. A metallic right circular cone 20 cm high and whose vertical angle is 60\u00b0 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1\/16 cm, find the length of the wire.<\/h2>\n

Solution:
\nThe diagram will be as follows<\/h3>\n

\"\"<\/p>\n

Consider AEG<\/h3>\n

\"\"<\/p>\n

Radius (r1) of upper end of frustum = (10\u221a3)\/3 cm
\nRadius (r2) of lower end of container = (20\u221a3)\/3 cm
\nHeight (r3) of container = 10 cm
\nNow,
\nVolume of the frustum = (\u2153)\u00d7\u03c0\u00d7h(r12+r22+r1r2)<\/h3>\n

\"\"<\/p>\n

Solving this we get,
\nVolume of the frustum = 22000\/9 cm3
\nThe radius (r) of wire = (1\/16)\u00d7(\u00bd) = 1\/32 cm
\nNow,
\nLet the length of wire be \u201cl\u201d.
\nVolume of wire = Area of cross-section x Length
\n= (\u03c0r2)xl
\n= \u03c0(1\/32)2x l
\nNow, Volume of frustum = Volume of wire
\n22000\/9 = (22\/7)x(1\/32)2x l
\nSolving this we get,
\nl = 7964.44 m<\/h3>\n

Exercise: 13.5<\/h2>\n

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given that,
\nDiameter of cylinder = 10 cm
\nSo, radius of the cylinder (r) = 10\/2 cm = 5 cm
\n\u2234 Length of wire in completely one round = 2\u03c0r = 3.14\u00d75 cm = 31.4 cm
\nIt is given that diameter of wire = 3 mm = 3\/10 cm
\n\u2234 The thickness of cylinder covered in one round = 3\/10 m
\nHence, the number of turns (rounds) of the wire to cover 12 cm will be-<\/h3>\n

\"\"<\/p>\n

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds
\n40 x 31.4 cm = 1256 cm
\nRadius of the wire = 0.3\/2 = 0.15 cm
\nVolume of wire = Area of cross-section of wire \u00d7 Length of wire
\n= \u03c0(0.15)2\u00d71257.14
\n= 88.898 cm3
\nWe know,
\nMass = Volume \u00d7 Density
\n= 88.898\u00d78.88
\n= 789.41 gm<\/h3>\n

2. A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \u03c0 as found appropriate)<\/h2>\n

Solution:
\nDraw the diagram as follows:<\/h3>\n

\"\"<\/p>\n

Let us consider the ABA
\nHere,
\nAS = 3 cm, AC = 4 cm
\nSo, Hypotenuse BC = 5 cm
\nWe have got 2 cones on the same base AA\u2019 where the radius = DA or DA\u2019
\nNow, AD\/CA = AB\/CB
\nBy putting the value of CA, AB and CB we get,
\nAD = 2\/5 cm
\nWe also know,
\nDB\/AB = AB\/CB
\nSo, DB = 9\/5 cm
\nAs, CD = BC-DB,
\nCD = 16\/5 cm
\nNow, volume of double cone will be<\/h3>\n

\"\"<\/p>\n

Solving this we get,
\nV = 30.14 cm3
\nThe surface area of the double cone will be<\/h3>\n

\"\"<\/p>\n

= 52.75 cm2<\/h3>\n

3. A cistern, internally measuring 150 cm \u00d7 120 cm \u00d7 100 cm, has 129600 cm3of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each being 22.5 cm \u00d7 7.5 cm \u00d7 6.5 cm?<\/h2>\n

Solution:
\nGiven that the dimension of the cistern = 150 \u00d7 120 \u00d7 110
\nSo, volume = 1980000 cm3
\nVolume to be filled in cistern = 1980000 \u2013 129600
\n= 1850400 cm3
\nNow, let the number of bricks placed be \u201cn\u201d
\nSo, volume of n bricks will be = n\u00d722.5\u00d77.5\u00d76.5
\nNow as each brick absorbs one-seventeenth of its volume, the volume will be
\n= n\/(17)\u00d7(22.5\u00d77.5\u00d76.5)
\nFor the condition given in the question,
\nThe volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern
\nOr, n\u00d722.5\u00d77.5\u00d76.5 = 1850400+n\/(17)\u00d7(22.5\u00d77.5\u00d76.5)
\nSolving this we get,
\nn = 1792.41<\/h3>\n

4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.<\/h2>\n

Solution:
\nFrom the question, it is clear that
\nTotal volume of 3 rivers = 3\u00d7[(Surface area of a river)\u00d7Depth]
\nGiven,
\nSurface area of a river = [1072\u00d7(75\/1000)] km
\nAnd,
\nDepth = (3\/1000) km
\nNow, volume of 3 rivers = 3\u00d7[1072\u00d7(75\/1000)]\u00d7(3\/1000)
\n= 0.7236 km3
\nNow, volume of rainfall = total surface area \u00d7 total height of rain<\/h3>\n

\"\"<\/p>\n

= 0.7280 km3
\nFor the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.
\nBut, 0.7280 km3= 0.7236 km3
\nSo, the question statement is true.<\/h3>\n

5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).<\/h2>\n

\"\"<\/p>\n

Solution:
\nGiven,
\nDiameter of upper circular end of frustum part = 18 cm
\nSo, radius (r1) = 9 cm
\nNow, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder
\nSo, r2= 8\/2 = 4 cm
\nNow, height (h1) of the frustum section = 22 \u2013 10 = 12 cm
\nAnd,
\nHeight (h2) of cylindrical section = 10 cm (given)
\nNow, the slant height will be-<\/h3>\n

\"\"<\/p>\n

Or, l = 13 cm
\nArea of tin sheet required = CSA of frustum part + CSA of cylindrical part
\n= \u03c0(r1+r2)l+2\u03c0r2h2
\nSolving this we get,
\nArea of tin sheet required = 782 4\/7 cm2<\/p>\n

6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.<\/h2>\n

Solution:
\nConsider the diagram<\/h3>\n

\"\"<\/p>\n

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1and r2are the radii of the frustum ends of the cone and h be the frustum height.
\nNow, consider the \u0394ABG and \u0394ADF,
\nHere, DF||BG
\nSo, \u0394ABG ~ \u0394ADF<\/h3>\n

\"\"<\/p>\n

Now, by rearranging we get,<\/h3>\n

\"\"<\/p>\n

\"\"<\/p>\n

The total surface area of frustum will be equal to the total CSA of frustum + the area of upper circular end + area of the lower circular end
\n= \u03c0(r1+r2)l+\u03c0r22+\u03c0r12
\n\u2234 Surface area of frustum = \u03c0[(r1+r2)l+r12+r22]<\/h3>\n

7. Derive the formula for the volume of the frustum of a cone.<\/h2>\n

Solution:
\nConsider the same diagram as the previous question.<\/h3>\n

\"\"<\/p>\n

Now, approach the question in the same way as the previous one and prove that
\n\u0394ABG ~ \u0394ADF
\nAgain,<\/h3>\n

\"\"<\/p>\n

Now, rearrange them in terms of h and h1<\/h3>\n

\"\"<\/p>\n

The total volume of frustum of the cone will be = Volume of cone ABC \u2013 Volume of cone ADE
\n= (\u2153)\u03c0r12h1-(\u2153)\u03c0r22(h1\u2013 h)
\n= (\u03c0\/3)[r12h1-r22(h1\u2013 h)]<\/h3>\n

\"\"<\/p>\n

Now, solving this we get,
\n\u2234 Volume of frustum of the cone = (\u2153)\u03c0h(r12+r22+r1r2)<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Maths NCERT book solutions for Chapter 13 – Surface Areas and Volumes Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119311","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119311","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119311"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119311\/revisions"}],"predecessor-version":[{"id":119399,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119311\/revisions\/119399"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119311"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119311"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}