CBSE Class 10 Maths Chapter-13 Surface Areas and Volumes – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 13 – Surface Areas and Volumes prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-13 Surface Areas and Volumes Important Questions
CBSE Class 10 Maths Important Questions Chapter 13 – Surface Areas and Volumes
2 Mark Questions
Unless stated otherwise, take 
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Ans. Abbreviation: CSA = Curved Surface Area TSA = Total Surface Area
V = Volume
Volume of cube = (Side)3
According to question, (Side)3 = 64
(Side)3 = 43
Side = 4 cm
For the resulting cuboid, length 
= 4 + 4 = 8 cm, breadth 
= 4 cm and height 
= 4 cm
Surface area of resulting cuboid = 
= 2 (8 x 4 + 4 x 4 + 4 x 8)
= 2 (32 + 16 + 32)
= 2 x 80 = 160 cm2
Unless stated otherwise, take
2. 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 
Ans. Abbreviation: CSA = Curved Surface Area TSA = Total Surface Area
V = Volume
For hemisphere, Radius 
= 1 cm
Volume = 
= 
= 
cm3
For cone, Radius of the base = 1 cm
Height = 1 cm
Volume = 
= 
= 
cm3
Volume of the solid = V of hemisphere + V of cone
= + = 
cm3
3. 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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).
Ans. Volume of the cuboid = 
= 15 x 10 x 3.5 = 525 cm3
Volume of conical depression =
= 
= 
cm3
Volume of four conical depressions = 
= 1.47 cm3
Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm3
Unless stated otherwise, take
4. 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.
Ans. Abbreviation: CSA = Curved Surface Area TSA = Total Surface Area
V = Volume
For sphere, Radius = 4.2 cm
Volume = 
= 
cm3
For cylinder, Radius (R) = 6 cm
Let the height of the cylinder be H cm.
Then, Volume = 
= 
cm3
According to question, Volume of sphere = Volume of cylinder
=
H = 
H = 2.74 cm
5. 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.
Ans. Let the volume of resulting sphere be 
cm.
According to question,
= 216 + 512 + 1000
= 1728
= 12 cm
6. 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.
Ans. Diameter of well = 7 m
Radius of well = 
m
And Depth of earth dug = 20 m
Length of platform = 22 m, Breadth of platform = 14 m
Let height of the platform be 
m
According to question,
Volume of earth dug = Volume of platform
= 2.5 m
7. 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.
Ans. Diameter of well = 3 m
Radius of well = 
m
and Depth of earth dug = 14 m
Width of the embankment = 4 m
Radius of the well with embankment 
= 
m
Let the height of the embankment be m
According to the question,
Volume of embankment = Volume of the earth dug
= 
= 
= 1.125 m
8. 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.
Ans. For right circular cylinder, Diameter = 12 cm
Radius = 
= 6 cm and height = 15 cm
For cone, Diameter = 6 cm
Radius 
= 
= 3 cm and height 
= 12 cm
Let 
cones be filled with ice cream.
Then, According to question,
Volume of cones = Volume of right circular cylinder
=
= 15
9. 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 x 10 cm x 3.5 cm?
Ans. For silver coin, Diameter = 1.75 cm
Radius 
= 
cm and Thickness = 2 mm = 
cm
For cuboid, Length = 5.5 cm, Breadth = 10 cm and Height 
= 3.5 cm
Let coins be melted.
Then, According to question,
Volume of coins = Volume of cuboid
= 
= 400
Unless stated otherwise, take
10. 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.
Ans. Abbreviation: CSA = Curved Surface Area
TSA = Total Surface Area
V = Volume
Here, 
= 2 m, 
= 1 m and 
= 14 m
Capacity of the glass = 
= 
= 
= 
cm2
11. 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.
Ans. Let 
cm and 
cm be the radii of the ends 
of the frustum of the cone.
Then, 
= 4 cm
= 18 cm
= 9 cm
= 6 cm
= 3 cm
Now, CSA of the frustum = 
= 
= (9 + 3) x 4 = 48 cm2
12. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). 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.
Ans. Here, = 10 cm, = 4 cm and = 15 cm
Surface area = 
= 
= 
= 
cm2
13. 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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. 
Ans. Here, = 20 cm, = 8 cm and = 16 cm
Volume of container =
= 
= 
= 
= 10449.92 cm3 = 10.44992 liters
Cost of the milk = 10.44992 x 20 = Rs. 208.8894 = Rs. 209
Now, surface area = + 
= 
+
= 
+ 
= 
= 1158.4 + 200.96
= 1959.36 cm2
Area of the metal sheet used = 1959.36 cm2
Cost of metal sheet = 
= 156.7488 = Rs. 156.75
14. 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.
Ans. Abbreviation: CSA = Curved Surface Area
TSA = Total Surface Area
V = Volume
Number of rounds to cover 12 cm, i.e. 120 mm = 
= 40
Here, Diameter = 10 cm, Radius 
cm
Length of the wire in completing one round = 
cm
Length of the wire in completing 40 rounds = 
cm
Radius of the copper wire = mm = 
cm
Volume of wire = 
= 
cm3
Mass of the wire = 
= 787.98 g
15. 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 as found appropriate)
Ans. Hypotenuse = 
= 5 cm
In figure, 
ADB
CAB [AA similarity]
AD = 
cm
Also, 
DB = 
cm
CD = BC – DB = 
cm
Volume of the double cone = 
= 
= 30.14 cm3
Surface area of the double cone = 
= 
= 
= 52.75 cm2
16. A cistern, internally measuring 150 cm x 120 cm x 110 cm has 129600 cm3 of 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 x 7.5 cm x 6.5 cm?
Ans. Volume of cistern = 150 x 120 x 110 = 1980000 cm3
Volume of water = 129600 cm3
Volume of cistern to be filled = 1980000 – 129600 = 1850400 cm3
Volume of a brick = 22.5 x 7.5 x 6.5 = 1096.875 cm3
Let bricks be needed.
Then, water absorbed by bricks = 
cm3
= 1792 (approx.)
17. 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.
Ans. Volume of rainfall = 
= 9.728 km3
Volume of three rivers = 
= 0.7236 km3
Hence, the two are not approximately equivalent.
18. 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 figure).
Ans. Slant height of the frustum of the cone 
= 
= 13 cm
Area of the tin sheet required = CSA of cylinder + CSA of the frustum
= 
= 
= 
= 
= 
cm2
19. Determine the ratio of the volume of a cube to that of a sphere which with exactly fit inside the cube.
Ans. Let the radius of the sphere which fits exactly into a cube be r units. Then length of each edge of cube = 2r units
Let V1 and V2 be the volumes of the cube and sphere
Then 
20. Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r.
Ans. Radius of cone = radius of hemisphere = r
Height of cone = radius of hemisphere
Volume of cone = 
21. The height of a right circular cone is 12 cm and the radius of its base is 4.5 cm. Find its slant height.
Ans. 
Slant height 
22. How many balls, each of radius 1cm, can be made from a solid sphere of lead of radius 8 cm.
Ans. Number of balls = 
23. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameter of its two circular ends are4cm and 2cm. Find the capacity of the glass.
Ans. 
Capacity of glass 
24. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 cm. What is the length of wire?
Ans. Radius of sphere 
Volume of sphere 
Let length of the wire = 
cm
R= Radius of the wire = 1 cm
Volume of wire 
25. An iron pipe 20 cm long has exterior diameter equal to 25cm. If the thickness of the pipe is 1cm. Find the whole surface area of the pipe.
Ans. 
Total surface area of pipe 
26. Find the ratio of the volumes of two circular cones. If 
and 
Ans. Ratio of volumes of two cones
27. A solid iron pole consists of a cylinder of height 110 cm and of base diameter 24 cm which is surmounted by a cone 9 cm high, find the mass of the pole. Given that 1 cm3 of iron has 8g mass approx. 
Ans. For cylinder 
For cone 
Volume of pipe = volume of cylindrical portion + volume of cone
28. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Ans. Two cubes joined end to end, we get cuboid
Surface area of cuboid 
29. Kuldeepmade a bird bath for his garden in the shape of a cylinder with a hemi spherical depression at one end. The height of the cylinder is 1.45 cm and its radius is 30 cm. Find the total surface area of the bird-bath.
Ans. Let h be the height of cylinder and r be the common radius of the cylinder and hemisphere.
Total surface area of bird bath= C.S.A. of cylinder + C.S.A. of hemisphere
30. A vessel is in the form of an inverted cone. Its height is 8 cm and radius of its top, which is open, is 5 cm it is filled with water up to brim. When lead shots each of which is a sphere of radius 0.5 cm, are dropped into the vessel. One-forth of the water flows out. Find the number of lead shots dropped.
Ans. Radius of lead shot = 0.5 cm
Radius of cone = 5 cm
Let 
be numbers of lead shots are dropped
x × volume of one lead shot = 
volume of the cone
100 lead shots are dropped
31. A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. Find the volume of the cone.
Ans. 
Volume of cone 
32. 2 cubes each of volume 216cm3are joined end to end. Find the surface area of the resulting cuboid.
Ans. Two cubes joined end to end, we get cuboid
Surface area of cuboid
33. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14cm and the total height of the Vessel is 13cm. Find inner surface area.
Ans. Inner surface area
[
Radius of base of the cylinder = radius of hemisphere]
34. A spherical ball of diameter 21 cm is melted and recast into cubes each of side 1cm. Find the number of cubes thus formed.
Ans. Volume of spherical ball
Volume of each cube 
Required number of cubes
= 4851