CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set 01

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. Two identical solid cubes of side \( k \) units are joined end to end. What is the volume, in cubic units, of the resulting cuboid?
(a) \( 2k^3 \)
(b) \( 3k^3 \)
(c) \( 4k^3 \)
(d) \( 6k^3 \)
Answer: (a) \( 2k^3 \)

 

Question. The radius of a sphere (in cm) whose volume is \( 12\pi \text{ cm}^3 \), is 
(a) \( 3 \text{ cm} \)
(b) \( 3\sqrt{3} \text{ cm} \)
(c) \( 3^{2/3} \text{ cm} \)
(d) \( 3^{1/3} \text{ cm} \)
Answer: (c) \( 3^{2/3} \text{ cm} \)

 

Question. A mason constructs a wall of dimensions \( 270 \text{ cm} \times 300 \text{ cm} \times 350 \text{ cm} \) with the bricks each of size \( 22.5 \text{ cm} \times 11.25 \text{ cm} \times 8.75 \text{ cm} \) and it is assumed that \( \frac{1}{8} \) space is covered by the mortar. Then the number of bricks used to construct the wall is 
(a) 11100
(b) 11200
(c) 11000
(d) 11300
Answer: (b) 11200

 

Question. A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm upto its brim. How many children will get the ice cream cones? 
(a) 163
(b) 263
(c) 363
(d) 463
Answer: (c) 363

 

Question. The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is 
(a) \( 9.7 \text{ cm}^3 \)
(b) \( 77.6 \text{ cm}^3 \)
(c) \( 58.2 \text{ cm}^3 \)
(d) \( 19.4 \text{ cm}^3 \)
Answer: (d) \( 19.4 \text{ cm}^3 \)

 

Question. A solid spherical ball fits exactly inside the cubical box of side \( 2a \). The volume of the ball is 
(a) \( \frac{16}{3} \pi a^3 \)
(b) \( \frac{1}{6} \pi a^3 \)
(c) \( \frac{32}{3} \pi a^3 \)
(d) \( \frac{4}{3} \pi a^3 \)
Answer: (d) \( \frac{4}{3} \pi a^3 \)

 

Question. A cube of side length 66 cm is filled with spherical metallic balls of diameter 0.6 cm and it is assumed that \( \frac{5}{8} \) space of the cube remains unfilled. What are number of balls that can be filled in the cube? \( \left( \text{Use } \pi = \frac{22}{7} \right) \) 
(a) 52875
(b) 952875
(c) 1165812
(d) 2264031
Answer: (b) 952875

 

Question. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \( \frac{1}{8} \) space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is 
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Answer: (a) 142296

 

Very Short Answer Questions

 

Question. If the areas of three adjacent faces of a cuboid are \( x, y \) and \( z \) respectively, then find the volume of cuboid.
Answer: Let the length, breadth and height of the cuboid are \( l, b \) and \( h \) respectively.
\( \because x = l \times b; \quad y = b \times h \quad \) and \( \quad z = l \times h \)
\( \implies xyz = (l \times b) \times (b \times h) \times (l \times h) \)
\( = l^2 b^2 h^2 \)
Now, volume of cuboid \( = l \times b \times h \)
\( = \sqrt{xyz} \quad (\because l^2 b^2 h^2 = xyz) \)

 

Question. Two right circular cones have their heights in the ratio \( 1 : 3 \) and radii in the ratio \( 3 : 1 \). What is the ratio of their volumes? 
Answer: Let height of the cones are \( h \) and \( 3h \) respectively and their radii are \( 3r \) and \( r \) respectively.
\( \therefore \) Ratio of volume \( = \frac{\text{Volume of first cone}}{\text{Volume of second cone}} \)
\( = \frac{\frac{1}{3} \pi \times (3r)^2 \times h}{\frac{1}{3} \pi \times r^2 \times 3h} = \frac{9}{3} = \frac{3}{1} \)
\( \therefore \) Required ratio is \( 3 : 1 \).

 

Question. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? 
Answer: As per question
Volume of hemisphere = Surface area of hemisphere
\( \frac{2}{3} \pi r^3 = 3 \pi r^2 \)

\( \implies r = \frac{9}{2} \text{ units} \)
\( \therefore d = 9 \text{ units} \)

 

Question. A solid ball is exactly fitted inside the cubical box of side \( a \). What is the volume of the ball?
Answer: Diameter of the solid ball = edge of the cube \( = a \)
\( \therefore \) Volume of the ball \( = \frac{4}{3} \pi \left( \frac{a}{2} \right)^3 = \frac{4}{3} \times \frac{1}{8} \pi a^3 = \frac{1}{6} \pi a^3 \)

 

Short Answer Questions-I

 

Question. The volume of a right circular cylinder with its height equal to the radius is \( 25 \frac{1}{7} \text{ cm}^3 \). Find the height of the cylinder. \( \left[ \text{Use } \pi = \frac{22}{7} \right] \) 
Answer: We have, in a right circular cylinder
Height = Radius \( \implies h = r \)
and volume of cylinder \( = 25 \frac{1}{7} \text{ cm}^3 \)

\( \implies \pi r^2 h = \frac{176}{7} \)

\( \implies \pi \times h^2 \times h = \frac{176}{7} \quad (\because h = r) \)

\( \implies \frac{22}{7} \times h^3 = \frac{176}{7} \)

\( \implies h^3 = \frac{176}{22} = 8 \implies h^3 = (2)^3 \implies h = 2 \text{ cm} \)
\( \therefore \) Height of the cylinder \( = 2 \text{ cm} \)

 

Question. If the total surface area of a solid hemisphere is \( 462 \text{ cm}^2 \), find its volume. \( \left[ \text{Take } \pi = \frac{22}{7} \right] \) 
Answer: Given, total surface area of solid hemisphere \( = 462 \text{ cm}^2 \)

\( \implies 3\pi r^2 = 462 \text{ cm}^2 \)
\( 3 \times \frac{22}{7} \times r^2 = 462 \)
\( r^2 = 49 \implies r = 7 \text{ cm} \)
Volume of solid hemisphere \( = \frac{2}{3} \pi r^3 \)
\( = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = 718.67 \text{ cm}^3 \)

 

Question. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer. 
Answer: Radius of hemispherical bowl, \( R = \frac{36}{2} = 18 \text{ cm} \)
Radius of cylindrical bottle, \( r = 3 \text{ cm} \)
Let height of cylindrical bottle \( = h \)
Since, 10% liquid is wasted, therefore only 90% liquid is filled into 72 cylindrical bottles.
\( \therefore \) Volume of 72 cylindrical bottles = 90% of the volume in bowl

\( \implies 72 \times \pi r^2 h = 90\% \text{ of } \frac{2}{3} \pi R^3 \)
\( 72 \times \pi \times 3 \times 3 \times h = \frac{90}{100} \times \frac{2}{3} \times \pi \times 18 \times 18 \times 18 \)

\( \implies h = \frac{90 \times 2 \times \pi \times 18 \times 18 \times 18}{100 \times 3 \times \pi \times 72 \times 3 \times 3} \)

\( \implies h = \frac{27}{5} = 5.4 \text{ cm} \)

 

Short Answer Questions-II

Question. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?
Answer: speed of water in canal \( = 25 \text{ km/hr.} \)
in 40 min \( = \frac{40}{60} = \frac{2}{3} \text{ hr.} \)
length of water \( = 25 \times \frac{2}{3} = \frac{50}{3} \text{ km} = \frac{50000}{3} \text{ m} \)
volume of water in canal in 40 minutes = volume of water for irrigation.
\( \frac{54}{10} \times \frac{18}{10} \times \frac{50000}{3} \text{ m}^3 = \frac{10}{100} \times l \times b \text{ m}^3 \)
\( 324 \times 5000 = l \times b \)
\( 1620000 = l \times b \)
Area irrigated in 40 minutes is \( 1620000 \text{ m}^2 \)
\( = \frac{1620000}{10000} = 162 \text{ hectares} \)
\( = 1.62 \text{ km}^2 \) or 162 hectares. [Topper's Answer 2017]

 

Question. 500 persons are taking dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is \( 0.04 \text{ m}^3 \)? 
Answer: Let \( h \) m be the rise of water level in the pond.
\( \therefore \) Displacement of water by 500 persons \( = l \times b \times h \)
\( \implies \) \( 500 \times 0.04 = 80 \times 50 \times h \)
\( \implies \) \( 20 = 4000 h \)
\( \therefore h = \frac{20}{4000} = \frac{1}{200} \text{ m} = \frac{100}{200} \text{ cm} = \frac{1}{2} \text{ cm} = 0.5 \text{ cm} \)

 

Question. A sphere of diameter 6 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 4 cm. Find the diameter of the cylindrical vessel. 
Answer: Volume of sphere \( = \frac{4}{3} \pi(3)^3 \text{ cm}^3 \)
Volume of water rise in cylinder \( = \pi r^2 h = \pi r^2 4 \quad [\because h = 4 \text{ cm}] \)
\( \therefore \pi r^2 4 = \frac{4}{3} \pi(3)^3 \)
\( \implies \) \( r^2 = \frac{27}{3} = 9 \)
\( \implies \) \( r = 3 \text{ cm} \)

 

Question. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?
Answer: Conical heap of rice:
Dimensions: diameter = 24m, height 3.5m \( \implies \) radius = 12m.
Volume of cone \( = \frac{1}{3} \pi r^2 h \) cu. units.
\( = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 \text{ cu.m.} = 132 \times 4 = 528 \text{ cu.m.} \)
The volume of the rice heap is \( 528 \text{ cu. m.} \)
Area of cloth required = Curved surface area.
CSA of cone \( = \pi r l \text{ sq. units} \) where \( l = \sqrt{h^2 + r^2} \text{ units.} \)
Finding \( l \): \( l = \sqrt{h^2 + r^2} \text{ units} = \sqrt{3.5^2 + 12^2} \text{ m} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ m.} \)
The area of canvas cloth required is \( 471.428571 \text{ m}^2 \).
\( \implies \) CSA \( = \pi r l \text{ sq. units} = \frac{22}{7} \times 12.5 \times 12 = \frac{22 \times 150}{7} = \frac{3300}{7} = 471.428571 \text{ m}^2 \). 

 

Question. The \( \frac{3}{4} \text{th} \) part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer: Let the height of cylindrical vessel be \( h \) cm.
According to question,
\( \frac{3}{4} \times \text{volume of conical vessel} = \text{volume of cylindrical vessel} \)
\( \implies \) \( \frac{3}{4} \times \frac{1}{3} \times \pi \times 5 \times 5 \times 24 = \pi \times 10 \times 10 \times h \)
\( \implies \) \( h = \frac{3}{2} \text{ cm} \) or 1.5 cm

 

Question. A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. \( \left[ \text{Use } \pi = \frac{22}{7} \right] \) 
Answer: Radius & height of conical vessel \( = 5 \text{ cm} \) & \( 24 \text{ cm} \) resp.
Volume of cone \( = \frac{1}{3} \pi r^2 h \)
Volume of cone \( = \frac{1}{3} \pi \times 25 \times 24 \text{ cm}^3 \)
Water is emptied of cylindrical vessel of \( r = 10 \text{ cm} \) & height \( h \)
Volume of cone = Volume of cylinder
\( \implies \) \( \frac{1}{3} \pi \times 25 \times 24 = \pi \times 10 \times 10 \times h \)
\( = \frac{200}{100} \text{ cm} = h \)
\( \implies \) \( 2 \text{ cm} = h \) [Topper's Answer 2016]

 

Question. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly, filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by \( 3 \frac{5}{9} \text{ cm} \). Find the diameter of the cylindrical vessel. 
Answer: Diameter of sphere \( = 12 \text{ cm} \)
Its radius \( = 6 \text{ cm} \)
Volume \( = \frac{4 \pi \times 6^3}{3} \text{ cm}^3 \)
It is submerged into water in cylindrical vessel, then water level rise by \( 3 \frac{5}{9} \text{ cm} = \frac{32}{9} \text{ cm} \)
Volume submerged = Volume rise
Let radius of cylinder be \( r \) cm
\( \implies \) \( \frac{4}{3} \pi \times 6^3 = \pi \times r^2 \times \frac{32}{9} \)
\( \implies \) \( \frac{216 \times 3 \times 4 \times 9}{3 \times 32} = r^2 \)
\( \implies \) \( 81 = r^2 \)
\( \implies \) \( r = 9 \text{ cm} \)
Diameter \( = 2r = 9 \text{ cm} \times 2 = 18 \text{ cm} \). [Topper's Answer 2016]

 

Question. The radius and height of a solid right circular cone are in the ratio \( 5 : 12 \). If its volume is \( 314 \text{ cm}^3 \), find its total surface area. [ Take \( \pi = 3.14 \) ] 
Answer: Given \( r : h = 5 : 12 \)
Let \( r = 5x \) and \( h = 12x \)
Volume of cone \( = \frac{1}{3} \pi r^2 h \)
\( 314 = \frac{1}{3} \times 3.14 (5x)^2 \times 12x \)
\( \implies \) \( x^3 = \frac{314 \times 3}{3.14 \times 25 \times 12} \)
\( \implies \) \( x^3 = 1 \)
\( \implies \) \( x = 1 \)
So, the value of \( r = 5 \text{ cm} \) and \( h = 12 \text{ cm} \)
Now, \( l = \sqrt{(12)^2 + (5)^2} = 13 \text{ cm} \)
TSA of cone \( = \pi r (l + r) \)
\( = 3.14 \times 5 (13 + 5) = 3.14 \times 90 = 282.6 \text{ cm}^2 \)

 

Long Answer Questions

Each of the following questions are of 5 marks.

 

Question. A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be distributed to 10 children in equal cones having hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone. 
Answer: Volume of ice-cream in the cylinder \( = \pi (6)^2 \cdot 15 \text{ cm}^3 \)
Volume of ice-cream in one cone \( = \frac{1}{3} \pi r^2 \cdot 4r + \frac{2}{3} \pi r^3 \text{ cm}^3 \quad (\text{Given } h = 4r) \)
\( = 2 \pi r^3 \text{ cm}^3 \)
\( \implies \) \( 10(2 \pi r^3) = \pi (6)^2 \times 15 \)
\( \implies \) \( r^3 = (3)^3 \)
\( \implies \) \( r = 3 \text{ cm} \).

 

Question. A right triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of double cone thus generated. (Use \( \pi = 3.14 \)).
Answer: In the given Fig. 12.25, \( \Delta PQR \) is a right triangle, where \( PQ = 3 \text{ cm}, PR = 4 \text{ cm} \) and \( QR = 5 \text{ cm} \).
Let \( OQ = x \implies OR = 5 - x \) and \( OP = y \)
Now in right angled-triangle \( POQ \), we have
\( PQ^2 = OQ^2 + OP^2 \) (By Pythagoras Theorem)
\( \implies \) \( (3)^2 = x^2 + y^2 \)
\( \implies \) \( y^2 = 9 - x^2 \) ...(i)
Also from right angled triangle \( POR \), we have
\( OP^2 + OR^2 = PR^2 \)
\( \implies \) \( y^2 + (5 - x)^2 = (4)^2 \)
\( \implies \) \( y^2 = 16 - (5 - x)^2 \) ...(ii)
From (i) and (ii), we get
\( 9 - x^2 = 16 - (5 - x)^2 \)
\( \implies \) \( 9 - x^2 = 16 - (25 + x^2 - 10x) \)
\( \implies \) \( 9 - x^2 = -9 - x^2 + 10x \)
\( \implies \) \( 10x = 18 \)
\( \implies \) \( x = \frac{9}{5} \)
\( \therefore OR = 5 - x = 5 - \frac{9}{5} = \frac{16}{5} \)
Now putting \( x = \frac{9}{5} \) in (i), we get
\( y^2 = 9 - \left( \frac{9}{5} \right)^2 = 9 - \frac{81}{25} = \frac{144}{25} \)
\( \implies \) \( y = \frac{12}{5} \)
\( \therefore OP = y = \frac{12}{5} \)
Now for the cone \( PQM \), radius \( OP = \frac{12}{5} \text{ cm} \), height \( OQ = \frac{9}{5} \text{ cm} \)
\( \therefore \text{Volume} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{12}{5} \right)^2 \times \frac{9}{5} = \frac{432\pi}{125} \text{ cm}^3 \)
Also for the cone \( PRM \), radius \( OP = \frac{12}{5} \text{ cm} \), height \( OR = \frac{16}{5} \text{ cm} \)
\( \therefore \text{Volume} = \frac{1}{3} \pi \left( \frac{12}{5} \right)^2 \times \frac{16}{5} = \frac{768\pi}{125} \text{ cm}^3 \)
Hence total volume, i.e., volume of the double cone
\( = \left( \frac{432\pi}{125} + \frac{768\pi}{125} \right) = \frac{1200\pi}{125} = 9.6 \times 3.14 \text{ cm}^3 = 30.144 \text{ cm}^3 \)

 

Question. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour? 
Answer: For pipe, \( r = 1 \text{ cm} \)
Length of water flowing in 1 sec, \( h = 0.7 \text{ m} = 70 \text{ cm} \)
Cylindrical Tank, \( R = 40 \text{ cm} \), rise in water level \( = H \)
Volume of water flowing in 1 sec \( = \pi r^2 h = \pi \times 1 \times 1 \times 70 = 70 \pi \)
Volume of water flowing in 60 sec \( = 70 \pi \times 60 \)
Volume of water flowing in 30 minutes \( = 70\pi \times 60 \times 30 \)
Volume of water in Tank \( = \pi R^2 H = \pi \times 40 \times 40 \times H \)
Volume of water in Tank = Volume of water flowing in 30 minutes
\( \pi \times 40 \times 40 \times H = 70\pi \times 60 \times 30 \)
\( H = 78.75 \text{ cm} \) 

 

Key Notes

Surface Areas and Volumes

• Surface Area of a Cuboid and a Cube
• Surface Area of a Right Circular Cylinder
• Surface Area of a Right Circular Cone
• Surface Area of a Sphere
• Volume of a Cuboid
• Volume of a Cylinder
• Volume of a Right Circular Cone
• Volume of a Sphere

 

Polyhedrons Shapes:

• Cube:
  • Cube whose edge = \( a \)
  • Diagonal of Cube = \( \sqrt{3}a \)
  • Lateral Surface Area of Cube = \( 4a^2 \)
  • Total Surface Area of Cube = \( 6a^2 \)
  • Volume of Cube = \( a^3 \)
• Cuboid:
  • Cuboid whose length = \( l \), breadth = \( b \) and height = \( h \)
  • Diagonal of Cuboid = \( \sqrt{l^2 + b^2 + h^2} \)
  • Lateral Surface Area of Cuboid = \( 2(l + b) h \)
  • Total Surface Area of Cuboid = \( 2 (lb + bh + hl) \)
  • Volume of Cuboid = \( lbh \)

 

Non-polyhedrons:

• Cylinder:
  • Cylinder whose radius = \( r \), height = \( h \)
  • Curved Surface Area of Cylinder = \( 2\pi rh \)
  • Total Surface Area of Cylinder = \( 2\pi r(r + h) \)
  • Volume of Cylinder = \( \pi r^2 h \)
• Cone:
  • Cone having height = \( h \), radius = \( r \) and slant height = \( l \)
  • Slant height of Cone \( (l) = \sqrt{r^2 + h^2} \)
  • Curved Surface Area of Cone = \( \pi rl \)
  • Total Surface Area of Cone = \( \pi r(r + l) \)
  • Volume of Cone = \( \frac{1}{3} \pi r^2 h \)
• Sphere:
  • Sphere whose radius = \( r \)
  • Surface Area of a Sphere = \( 4\pi r^2 \)
  • Volume of Sphere = \( \frac{4}{3} \pi r^3 \)
• Hemisphere:
  • Hemisphere whose radius = \( r \)
  • Curved Surface Area of Hemisphere = \( 2\pi r^2 \)
  • Total Surface Area of Hemisphere = \( 3\pi r^2 \)
  • Volume of Hemisphere = \( \frac{2}{3} \pi r^3 \)

 

 

Please click on below link to download CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set A