NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

 Exercise 13.1

Q.1) 2 cubes each of volume 64 𝑐𝑚3 are joined end to end. Find the surface area of the resulting cuboid.
Sol.1) Side of cube = 3√𝑣𝑜𝑙𝑢𝑚𝑒
= 3√64 = 4𝑐𝑚
Length of new cuboid = 8 𝑐𝑚, height = 4 𝑐𝑚, width = 4 𝑐𝑚
Surface Area can be calculated as follows:
= 2(𝑙𝑏 + 𝑙ℎ + 𝑏ℎ)
= 2(8 × 4 + 8 × 4 + 4 × 4)
= 2 × 80 = 160𝑐𝑚3
Alternate Method:
Surface area of cube = 6 × 𝑠𝑖𝑑𝑒2
When two cubes are joined end to end, then out of 12 surfaces; two surfaces are lost due to joint. Thus, we need to take surface area of 10 surfaces and hence surface area can be given as follows:
= 10 × 𝑠𝑖𝑑𝑒2
= 10 × 42 = 160 𝑐𝑚2

Q.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.
Sol.2) Radius = 7 𝑐𝑚
Height of cylindrical portion = 13 – 7 = 6 𝑐𝑚
Curved surface are of cylindrical portion can be calculated as follows:
= 2𝜋𝑟ℎ
= 2 × 22/7 × 7 × 6
= 264𝑐𝑚2
Curved surface area of hemispherical portion can be calculated as follows:
= 2𝜋𝑟2
= 2 × 22/7 × 7 × 7
= 308𝑐𝑚2
Total surface are = 308 + 264 = 572 𝑠𝑞. 𝑐𝑚

Q.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.
Sol.3) Radius of cone = 3.5 𝑐𝑚, height of cone = 15.5 – 3.5 = 12 𝑐𝑚
Slant height of cone can be calculated as follows:

= 22/7 × 3.5 × 12.5
= 137.5𝑐𝑚2
Curved surface area of hemispherical portion can be calculated as follows:
= 2𝜋𝑟2
= 2 × 22/7 × 3.5 × 3.5
= 77𝑐𝑚2
Hence, total surface area = 137.5 + 77 = 214.5 𝑠𝑞. 𝑐𝑚

Q.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.
Sol.4) The greatest diameter = side of the cube = 7 𝑐𝑚
Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere +
Curved Surface Area of hemisphere
Surface Area of Cube = 6 × 𝑆𝑖𝑑𝑒2
= 6 × 7 × 7 = 294 𝑠𝑞. 𝑐𝑚
Surface Area of Base of Hemisphere
= 𝜋𝑟2
= 22/7 × 3.52 = 38.5𝑐𝑚2
Curved Surface Area of Hemisphere = 2 × 38.5 = 77 𝑠𝑞 𝑐𝑚
Total Surface Area = 294 – 38.5 + 77 = 332.5 𝑠𝑞. 𝑐𝑚

Q.5) A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Sol.5) This question can be solved like previous question. Here the curved surface of the hemisphere is a depression, unlike a projection in the previous question
Total Surface Area = 6𝑑2 − 𝜋 (𝑑/2)2 + 2𝜋 (𝑑/2)2 
= 6𝑑2 − 𝜋 (𝑑/2)2 
= (1/4) 𝑑2(𝜋 + 24)

Q.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.
Sol.6) Height of Cylinder = 14 – 5 = 9 𝑐𝑚, radius = 2.5 𝑐𝑚
Curved Surface Area of Cylinder= 2𝜋𝑟ℎ
= 2𝜋 × 2.5 × 9
= 45𝜋𝑐𝑚2
Curved Surface Area of two Hemispheres = 4𝜋𝑟2
= 4𝜋 × 2.52
= 2𝜋 𝑐𝑚2
Total Surface Area= 45𝜋 + 25𝜋
= 70𝜋 = 220 𝑐𝑚2

Q.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.)
Sol.7) Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 𝑚
Curved Surface Area of cylindrical portion = 2𝜋𝑟ℎ

= 2𝜋 × 2 × 2.1
= 8.4𝜋𝑚2
Curved surface area of conical portion = 𝜋𝑟𝑙
= 8.4𝜋 × 2 × 2.8
= 14 × 22/7 = 44𝑚2
Cost of canvas = 𝑅𝑎𝑡𝑒 × 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎
= 500 × 44 = 𝑅𝑠. 22000

Q.8) From a solid cylinder whose height is 2.4 𝑐𝑚 and diameter 1.4 𝑐𝑚, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest 𝑐𝑚2.
Sol.8) Radius = 0.7 𝑐𝑚 and height = 2.4 𝑐𝑚
Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder
+ Curved Surface Area of Cone
Curved Surface Area of Cylinder = 2𝜋𝑟ℎ
= 2𝜋 × 0.7 × 2.4
= 3.36 𝜋 𝑐𝑚2
Area of top = 𝜋𝑟2
= 𝜋 × 0.72
= 0.49𝜋 𝑐𝑚2
Slant height of cone can be calculated as follows:

Curved surface area of cone = 𝜋𝑟𝑙
= 𝜋 × 0.7 × 2.5
= 1.75𝜋 𝑐𝑚2
Hence, remaining surface area of structure = 3.36𝜋 + 0.49𝜋 + 1.75𝜋
= 5.6𝜋 = 17.6 𝑐𝑚2
= 18 𝑐𝑚2 (𝑎𝑝𝑝𝑟𝑜𝑥. )

Q.9) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 𝑐𝑚, and its base is of radius 3.5 𝑐𝑚, find the total surface area of the article.
Sol.9) Radius = 3.5 𝑐𝑚, height = 10 𝑐𝑚
Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres
Curved Surface Area of Cylinder = 2𝜋𝑟ℎ
= 2𝜋 × 3.5 × 10
= 70𝜋 𝑐𝑚2
Surface area of sphere = 4𝜋𝑟2
= 4𝜋 × 3.52
= 49𝜋
Total surface area = 70𝜋 + 49𝜋
= 119𝜋
= 119 × 22/7 = 374𝑐𝑚2

Exercise 13.2

Q.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 𝜋.
Sol.1) radius = 1 𝑐𝑚, height = 1 𝑐𝑚

Q.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 to be nearly the same.)
Sol.2) Height of cylinder = 12 – 4 = 8 𝑐𝑚, radius = 1.5 𝑐𝑚, height of cone = 2 𝑐𝑚
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 1.52 × 8
= 18𝜋 𝑐𝑚3

Q.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.
Sol.3) Length of cylinder = 5 – 2.8 = 2.2 𝑐𝑚,
radius = 1.4 𝑐𝑚
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 1.42 × 2.2
= 4.312 𝜋 𝑐𝑚3

= 7.515 𝑐𝑚3
Volume of syrup in 45 gulab jamuns = 45 × 7.515 = 338.184 𝑐𝑚3

Q.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.
Sol.4) Dimensions of cuboid = 15 𝑐𝑚 × 10 𝑐𝑚 × 3.5 𝑐𝑚,

Q.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.
Sol.5) radius of cone = 5 𝑐𝑚, height of cone = 8 𝑐𝑚, radius of sphere = 0.5 𝑐𝑚

Q.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 1𝑐𝑚3 of iron has approximately 8g mass
Sol.6) radius of bigger cylinder = 12 𝑐𝑚, height of bigger cylinder = 220 𝑐𝑚
Radius of smaller cylinder = 8 𝑐𝑚, height of smaller cylinder = 60 𝑐𝑚
Volume of bigger cylinder = 𝜋𝑟2ℎ
= 𝜋 × 122 × 220
= 31680𝜋 𝑐𝑚3
Volume of smaller cylinder = 31680𝜋 + 3840𝜋
= 35520 𝜋 𝑐𝑚3
Mass = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 × 𝑉𝑜𝑙𝑢𝑚𝑒
= 8 × 35520𝜋
= 892262.4 𝑔𝑚
= 892.3 𝑘𝑔

Q.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.
Sol.7) Radius of cone = 60 𝑐𝑚, height of cone = 120 𝑐𝑚
Radius of hemisphere = 60 𝑐𝑚
Radius of cylinder = 60 𝑐𝑚, height of cylinder = 180 𝑐𝑚

Volume of cone = (1/3)𝜋𝑟2ℎ
= (1/3)𝜋 × 602 × 120
= 144000 𝜋 𝑐𝑚3
Volume of hemisphere = (144000 + 144000)𝜋
= 288000𝜋 𝑐𝑚3
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 602 × 180
= 648000𝜋 𝑐𝑚3
Volume of water left in the cylinder = (648000 − 288000)𝜋
= 360000𝜋 = 1130400 𝑐𝑚3

Q.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 𝜋 = 3.14.
Sol.8) Radius of cylinder = 1 𝑐𝑚, height of cylinder = 8 𝑐𝑚, radius of sphere = 8.5 𝑐𝑚
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 12 × 8
= 8𝜋 𝑐𝑚

Exercise 13.3

Q.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.
Sol.1) Radius of sphere = 4.2 𝑐𝑚, radius of cylinder = 6 𝑐𝑚

Q.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.
Sol.2) Radii of spheres = 6 𝑐𝑚, 8 𝑐𝑚, 10 𝑐𝑚

Q.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.
Sol.3) Radius of well = 3.5 𝑚, depth of well = 20 𝑚
Dimensions of rectangular platform = 22 𝑚 × 14 𝑚
Volume of earth dug out = 𝜋𝑟2ℎ
= 𝜋 × 3.52 × 20
= 770 𝑚3
Area of top of platform = Area of Rectangle – Area of Circle
(because circular portion of mouth of well is open)
= 22 × 14 − 𝜋 × 3.52
= 308 − 38.5 = 269.5 𝑚2
Height = 𝑉𝑜𝑙𝑢𝑚𝑒/𝐴𝑟𝑒𝑎
= 770/269.5 = 2.85 𝑚

Q.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.
Sol.4) Radius of well = 1.5 𝑚, depth of well = 14 𝑚,
width of embankment = 4 𝑚
Radius of circular embankment = 4 + 1.5 = 5.5 𝑚
Volume of earth dug out = 𝜋𝑟2ℎ
= 𝜋 × 1.52 × 14
= 31.5 𝜋 𝑚3
Area of top of platform = (Area of bigger circle – Area of smaller circle)
= 𝜋(𝑅2 − 𝑟2)
= 𝜋(5.52 − 1.52) 
= 28𝜋
Height = 𝑉𝑜𝑙𝑢𝑚𝑒/𝐴𝑟𝑒𝑎
= 31.5𝜋/28𝜋
= 1.125 𝑚

Q.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.
Sol.5) Radius of cylinder = 6 cm, height of cylinder = 15 cm
Radius of cone = 3 cm, height of cone = 12 cm
Radius of hemispherical top on ice cream = 3 𝑐𝑚
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 62 × 15
= 540 𝜋 𝑐𝑚3
Volume of cone = 1/3 × 𝜋 × 32 × 12
= 36𝜋 𝑐𝑚3
Volume of hemisphere = 2/3 𝜋𝑟3
= 2/3 × 𝜋 × 33
= 18𝜋 𝑐𝑚3
Volume of ice cream = (36 + 18)𝜋
= 54𝜋 𝑐𝑚3
Hence, number of ice creams = Volume of cylinder/Volume of ice cream
= 540𝜋/54𝜋 = 10

Q.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 × 10 𝑐𝑚 × 3.5 𝑐𝑚?
Sol.6) Radius of coin = 0.875 cm, height = 0.2 cm
Dimensions of cuboid = 5.5 𝑐𝑚 × 10 𝑐𝑚 × 3.5 𝑐𝑚
Volume of coin = 𝜋𝑟2ℎ
= 𝜋 × 0.8752 × 0.2
= 0.48125 𝑐𝑚3
Volume of cuboid = 5.5 × 10 × 3.5 = 192.5 𝑐𝑚3
Number of coins = 192.5/0.48125 = 400

Q.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.
Sol.7) Radius of cylinder = 18 𝑐𝑚, height = 32 𝑐𝑚
Height of cone = 24 𝑐𝑚
Volume of cylinder = 𝜋𝑟2ℎ
= 𝜋 × 182 × 32
Volume of cone = Volume of cylinder

Q.8) Water in a canal, 6 m wide an area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Sol.8) Depth = 1.5 m, width = 6 𝑚, height of standing water = 0.08 𝑚
In 30 minutes, length of water column = 5 𝑘𝑚 = 5000 𝑚
Volume of water in 30 minutes = 1.5 × 6 × 5000 = 45000 𝑐𝑢𝑏𝑖𝑐 𝑚
Area = 𝑉𝑜𝑙𝑢𝑚𝑒/𝐻𝑒𝑖𝑔ℎ𝑡
= 45000/0.08 = 562500 𝑚2

Q.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?
Sol.9) Radius of pipe = 10 𝑐𝑚 = 0.1 𝑚, length = 3000 𝑚/ℎ
Radius of tank = 5 𝑚, depth = 2 𝑚
Volume of water in 1 hr through pipe = 𝜋𝑟2ℎ
= 𝜋 × 0.12 × 3000
= 30𝜋 𝑚3
Volume of tank = 𝜋𝑟2ℎ
= 𝜋 × 52 × 2
= 50𝜋 𝑚3

Exercise 13.4

Q.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.
Sol.1) We have 𝑅 = 2, 𝑟 = 1 𝑐𝑚 and ℎ = 14 𝑐𝑚
Volume of frustum

Q.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 curved surface area of the frustum.
Sol.2) Slant height 𝑙 = 4 𝑐𝑚, perimeters = 18 𝑐𝑚 and 6 𝑐𝑚
Radii can be calculated as follows:

Q.3) A fez, the cap used by the Turks, is shaped like the frustum of a cone. 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.

Sol.3) 𝑅 = 10 𝑐𝑚, 𝑟 = 4 𝑐𝑚, slant height = 15 𝑐𝑚
Curved surface area of frustum = 𝜋(𝑅 + 𝑟)𝑙
= 𝜋(𝑅 + 𝑟)𝑙
= 𝜋(10 + 4)15
= 22/7 × 14 + 15 = 660
Area of upper base = 𝜋𝑟2 = 𝜋 × 42

= 16𝜋 = 50(2/7)
Hence, total surface area = 660 + 50(2/7)
= 710(2/7)𝑐𝑚2

Q.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.
Sol.4) Height of frustum = 16 𝑐𝑚, 𝑅 = 20 𝑐𝑚, 𝑟 = 8 𝑐𝑚
Volume of frustum

Surface area of frustum = 𝜋(𝑅 + 𝑟)𝑙 + 𝜋𝑟2
= 𝜋[(𝑅 + 𝑟)𝑙 + 𝑟2]
= 𝜋[(20 + 8)20 + 82]
= 𝜋(560 + 64)
= 22/7 × 624 = 1959.36
Cost of metal sheet at Rs. 8 per 100 sq. cm = 19.5936 × 8 = 𝑅𝑠. 156.75

Q.5) A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 𝑐𝑚, find the length of the wire.
Sol.5) Volume of frustum will be equal to the volume of wire and by using this relation we can
calculate the length of the wire.
In the given figure; AO = 20 cm and
hence height of frustum 𝐿𝑂 = 10 𝑐𝑚
In triangle AOC we have angle CAO = 30° (half of vertical angle of cone BAC)
Therefore;
tan 30° = 𝑂𝐶/𝐴𝑂
1/√3 = 𝑂𝐶/20
𝑂𝐶 = 20/√3

Using similarity criteria in triangles AOC and ALM it can be shown that 𝐿𝑀 = 10/√3 (because LM bisects the cone through into height)
Similarly, LO = 10 cm
Volume of frustum can be calculated as follows:

Exercise 13.5

Q.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.
Sol.1) For copper wire: Diameter = 3 𝑚𝑚 = 0.3 𝑐𝑚
For cylinder; length ℎ = 12 𝑐𝑚, 𝑑 = 10 𝑐𝑚
Density of copper = 8.88 𝑔𝑚 𝑐𝑚3
Curved surface area of cylinder can be calculated as follows:
= 2𝜋𝑟ℎ
= 𝜋 × 10 × 12
= 120 𝜋 𝑐𝑚2
Length of wire can be calculated as follows:
𝐿𝑒𝑛𝑔𝑡ℎ = 𝐴𝑟𝑒𝑎/𝑊𝑖𝑑𝑡ℎ
= 120𝜋/0.3 = 400𝜋
= 1256 𝑐𝑚
Now, volume of wire can be calculated as follows:
𝑉 = 𝜋𝑟2ℎ
= 3.14 × 0.152 × 1256
= 88.7364 𝑐𝑚3
Mass can be calculated as follows:
𝑀𝑎𝑠𝑠 = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 × 𝑉𝑜𝑙𝑢𝑚𝑒
= 8.88 × 88.7364
= 788 𝑔 (𝑎𝑝𝑝𝑟𝑜𝑥. )

Q.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.

Sol.2) In triangle ABC;
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = 32 + 42
𝐴𝐶2 = 9 + 16 = 25
𝐴𝐶 = 5 𝑐𝑚
In triangle ABC and triangle BDC;

𝐴𝐵𝐶 = 𝐵𝐷𝐶 (right angle)
𝐵𝐴𝐶 = 𝐷𝐵𝐶
Hence, Δ𝐴𝐵𝐶~Δ𝐵𝐷𝐶
So, we get following equations:
𝐴𝐵/𝐵𝐶 = 𝐵𝐷/𝐵𝐶 
3/5 = 𝐵𝐷/4
𝐵𝐷 = 3×4/5 = 2.4
In triangle BDC;
𝐷𝐶2 = 𝐵𝐶2– 𝐵𝐷2
𝐷𝐶2 = 42 – 2.42
𝐷𝐶2 = 16 – 5.76 = 10.24
𝐷𝐶 = 3.2
From above calculations, we get following measurements for the double cone formed:
Upper Cone: 𝑟 = 2.4 𝑐𝑚, 𝑙 = 3 𝑐𝑚, ℎ = 1.8 𝑐𝑚
Volume of cone = 1/3 𝜋𝑟2ℎ
= 1/3 × 3.14 × 2.42 × 1.8
= 10.85184 𝑐𝑚3
Curved surface area of cone = 𝜋𝑟𝑙
= 3.14 × 2.4 × 3
= 22.608 𝑐𝑚2
Lower Cone: 𝑟 = 2.4 𝑐𝑚, 𝑙 = 4 𝑐𝑚, ℎ = 3.2 𝑐𝑚
Volume of cone = 1/3
𝜋𝑟2ℎ = 1/3 × 3.14 × 2.42 × 3.2
= 19.2916 𝑐𝑚3
Curved surface area of cone = 𝜋𝑟𝑙
= 3.14 × 2.4 × 4 = 30.144 𝑐𝑚2
Total volume = 19.29216 + 10.85184 = 30.144 𝑐𝑚3
Total surface area = 30.144 + 22.608 = 52.752 𝑐𝑚2

Q.3) A cistern measuring 150 𝑐𝑚 × 120 𝑐𝑚 × 110 𝑐𝑚 has 129600 𝑐𝑚3 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 brick being 22.5 𝑐𝑚 × 7.5 𝑐𝑚 × 6.5 𝑐𝑚?
Sol.3) Volume of cistern = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × 𝑑𝑒𝑝𝑡ℎ
= 150 × 120 × 110
= 1980000 𝑐𝑚3
Vacant space = Volume of cistern – Volume of water
= 1980000 – 129600 = 1850400 𝑐𝑚3
Volume of brick = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡
= 22.5 × 7.5 × 6.5
Since the brick absorbs one seventeenth its volume hence remaining volume will be equal to 16/17
the volume of brick
Remaining volume
= 22.5 × 7.5 × 6.5 × 16/17
Number of bricks = Remaining volume of cistern/remaining volume of brick
= 1850400×7/22.5×7.5×6.5×16
= 31456800/17550 = 1792
Therefore, 1792 bricks were placed in the cistern.

Q.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 97280 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.
Sol.4) Area of the valley= 𝐴 = 97280 𝑘𝑚2
Level in the rise of water in the valley= ℎ = 10 𝑐𝑚 𝑜𝑟 (1/10000) 𝑘𝑚 amount of rain fall in 14 day = 𝐴ℎ = 97280 × (1/10000) = 9.828 𝑘𝑚3
Amount of rain fall in 1 day = 9.828/14 = 0.702𝑘𝑚3
Volume of water in 3 𝑟𝑖𝑣𝑒𝑟𝑠 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡
= 1072 × 75 × 3
= 1072 × (75/1000) × (3/1000)
= 0.2412 × 3
= 0.7236 𝑘𝑚3

Q.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 diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
Sol.5) Curved surface area of cylinder = 2𝜋𝑟ℎ
= 𝜋 × 8 × 10
= 80𝜋
Slant height of frustum can be calculated as follows:

Curved surface area of frustum = 𝜋(𝑟1 + 𝑟2)𝑙
= 𝜋(9 + 4) × 13 = 169𝜋
Total curved surface area = 169𝜋 + 80𝜋
= 249 + 3.14 = 781.86 𝑐𝑚2

Q.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.
Sol.6) Let h be the height, l the slant height and r and r the radii of the circular bases of the frustum ABB’ A’ shown in Fig. such that 𝑟1 > 𝑟2 .
Let the height of the cone 𝑉𝐴𝐵 be ℎ1 and its slant height be i.e., 𝑉𝑂 = ℎ1 and 𝑉𝐴 =
𝑉𝐵 = 𝑙1
∴ 𝑉𝐴’ = 𝑉𝐴 – 𝐴𝐴’ = 𝑙 – 𝑙 and 𝑉𝑂’ = 𝑉𝑂 – 𝑂𝑂’ = ℎ – ℎ
Here, 𝛥𝑉𝑂𝐴 ~ 𝛥𝑉𝑂‘𝐴’

Curved surface area of the frustum = 𝜋(𝑟1 + 𝑟2 )𝑙
Total surface area of the frustum = Lateral (curved) surface area + Surface area of circular bases = 𝜋 (𝑟1 + 𝑟2 )𝑙 + 𝜋𝑟12 + 𝜋𝑟22
= 𝜋 {(𝑟1 + 𝑟2 ) 𝑙 + 𝑟12 + 𝑟22 }.

Q.7) Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Sol.7) If 𝐴1 𝑎𝑛𝑑 𝐴2 are the surface areas of two circular bases, then

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