CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set B

Points to Remember

• 1. Volume of frustum of a cone = \(\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)\)

• 2. CSA of a frustum of a cone = \(\pi l (r_1 + r_2)\), where, \(l = \sqrt{h^2 + (r_1 - r_2)^2}\)
• 3. Total surface area of frustum of a cone = \(\pi l(r_1 + r_2) + \pi(r_1^2 + r_2^2)\)

Multiple Choice Questions

Question. The surface area of two spheres are in the ratio 16 : 9. The ratio of their volumes is :
(a) 64 : 27
(b) 16 : 9
(c) 4 : 3
(d) \(16^3 : 9^3\)
Answer: a

Question. If the radius of the base of a right-circular cylinder is halved keeping the height same then the ratio of the volume of the cylinder thus obtained to the volume of the original cylinder is :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer: c

Question. The radius of the largest circular cone that can be cut out from a cube of edge 4.2 cm is :
(a) 4.2
(b) 2.1
(c) 8.4
(d) 1.05
Answer: b

Question. The volume of frustum of a cone is given by :
(a) \(\frac{4}{3} \pi (R^3 - r^3)\)
(b) \(\pi (R + r)l\)
(c) \(\frac{1}{3} \pi h (R^2 + r^2 + Rr)\)
(d) \(\frac{1}{3} \pi (R^2 - r^2)h\)
Answer: c

Fill in the Blanks

Question. If the height of a cone is doubled and its radius is tripled, then its volume will become ........... times.
Answer: 18

Question. When a sphere is cut into two equal parts, then each part is called a ............. .
Answer: Hemisphere

Question. The total surface area of hemisphere of radius a is ...........
Answer: \(3\pi a^2\)

Question. The volume of a spherical shell is given by .......... .
Answer: \(\frac{4}{3} \pi (R^3 - r^3)\)

True/False

Question. During conversion of a solid from one shape to another, the volume of new shape will increase.
Answer: False. During conversion of a solid from one shape to another, volume remains same.

Question. Total surface area of a frustum is \(\pi (R + r)l\).
Answer: False. Total surface of a frustum is \(\pi(R + r)l + \pi R^2 + \pi r^2\)

Question. If the radii and heights of a cylinder and a cone is same, then the volume of cone is \(\frac{1}{3}\) of volume of cylinder.
Answer: True

Very Short Answer Type Questions

Question. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas. [2018 C]
Answer: Let \(a_1\) and \(a_2\) be the sides of two cubes respectively. Then,
\(\frac{a_1^3}{a_2^3} = \frac{1}{27}\) [Given]
\(\Rightarrow \left(\frac{a_1}{a_2}\right)^3 = \left(\frac{1}{3}\right)^3\)
\(\Rightarrow \frac{a_1}{a_2} = \frac{1}{3}\)
Now, ratio of their surface areas is given as,
\(\frac{6a_1^2}{6a_2^2} = \left(\frac{a_1}{a_2}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\).

Question. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? [2017 Delhi]
Answer: Let radius of hemisphere be r units.
Volume of hemisphere = S.A. of hemisphere (Given)
\(\frac{2}{3} \pi r^3 = 3 \pi r^2\)
\(\Rightarrow r = \frac{9}{2}\) or diameter = 9 units.

Question. A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two are equal then find the ratio of the radius and the slant height of the conical part.
Answer: Given, Radius of hemisphere = Radius of cone = \(r\)
Also, curved surface area of hemisphere = Curved surface area of cone
\(\Rightarrow 2\pi r^2 = \pi rl\) [\(l\) = slant height of the cone]
\(\Rightarrow 2r = l\)
\(\Rightarrow r : l = 1 : 2\)
So, the required ratio is 1 : 2.

Question. Two cubes, each of volume 27 cm³ are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Answer: Given, volume of the each cube = 27 cm³
\(\Rightarrow a^3 = 27 \Rightarrow a^3 = (3)^3 \Rightarrow a = 3\) cm
Thus, length of the larger solid = (3 + 3) cm = 6 cm
Width of the larger solid = 3 cm
Height of the larger solid = 3 cm
Thus, surface area of the resultant cuboid
= \(2(6 \times 3 + 3 \times 3 + 6 \times 3)\) cm²
= \(2(18 + 9 + 18)\) cm²
= \(2(45)\) cm²
= 90 cm².

Question. A cylinder and a cone are of the same base radius and height. Calculate the ratio of the volume of the cylinder and the cone.
Answer: Given, Radius of cylinder = Radius of cone = \(r\) cm
and Height of cylinder = Height of cone = \(h\) cm
Thus, \(\frac{\text{Volume of cylinder}}{\text{Volume of cone}} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h} = \frac{3}{1}\)
Hence, Volume of cylinder : Volume of cone = 3 : 1.

Question. The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, find the height of the frustum.
Answer: Given, slant height of frustum, \( l = 5 \) cm
Let radii of two circular ends of frustum be \( R, r \) cm.
Then, \( R - r = 4 \) cm (Given)
Let the height of the cone be \( h \) cm.
We know, \( l = \sqrt{h^2 + (R-r)^2} \)
\( \Rightarrow 5 = \sqrt{h^2 + (4)^2} \) cm
\( \Rightarrow 25 = h^2 + (4)^2 \) cm
\( \Rightarrow 25 = h^2 + 16 \) cm
\( \Rightarrow h^2 = 25 - 16 \)
\( \Rightarrow h^2 = 9 \)
\( \Rightarrow h = 3 \) cm.

Short Answer Type Questions-I

Question. The volume of a right circular cylinder with its height equal to the radius is \( 25 \frac{1}{7} \) cm\(^3\). Find the height of the cylinder. Use \( \pi = \frac{22}{7} \)
Answer: Given :
Height of cylinder = Radius of cylinder
\( \Rightarrow h = r \)
And, Volume of cylinder = \( 25 \frac{1}{7} \text{ cm}^3 = \frac{176}{7} \text{ cm}^3 \)
We know, Volume of cylinder = \( \pi r^2 h \)
\( = \pi h^3 \) [ \(\because h = r\) ]
\( \Rightarrow \frac{176}{7} = \frac{22}{7} h^3 \)
\( \Rightarrow h^3 = 8 \)
\( \Rightarrow h = 2 \).
Hence, the height of the cylinder is 2 cm.

Question. Find the ratio of volumes of a cube to that of a sphere which will exactly fit inside the cube.
Answer: Let the side of cube be \( a \) units and radius of sphere be \( r \) units.
Then, Diameter of sphere = Side of cube
\( \Rightarrow 2r = a \)
Now, \( \frac{\text{Volume of cube}}{\text{Volume of sphere}} = \frac{a^3}{\frac{4}{3} \pi r^3} \)
\( = \frac{3 a^3}{4 \pi r^3} = \frac{3(2r)^3}{4 \pi r^3} \)
\( = \frac{3 \times 8r^3}{4 \pi r^3} = \frac{6}{\pi} \)

Question. The volume of a hemisphere is 2425.5 cm\(^3\). Find its curved surface area. Use \( \pi = \frac{22}{7} \)
Answer: Given, volume = 2425.5 cm\(^3\)
Let the radius of hemisphere be \( r \).
Thus, \( \frac{2}{3} \pi r^3 = 2425.5 \)
\( \frac{2}{3} \times \frac{22}{7} \times r^3 = \frac{24255}{100} \)
\( \Rightarrow \frac{44}{21} r^3 = \frac{4851}{2} \)
\( \Rightarrow \frac{4}{21} r^3 = \frac{441}{2} \)
\( \Rightarrow 8r^3 = 441 \times 21 \)
\( \Rightarrow 8r^3 = 21^3 \)
\( \Rightarrow (2r)^3 = (21)^3 \)
\( \Rightarrow r = \frac{21}{2} \) cm
Thus, the curved surface area
\( = 2\pi \left( \frac{21}{2} \right)^2 \) cm\(^2\)
\( = 2 \left( \frac{22}{7} \right) \left( \frac{21}{2} \right) \left( \frac{21}{2} \right) \) cm\(^2\)
\( = (33)(21) \) cm\(^2\)
\( = 693 \) cm\(^2\).

Question. If the total surface area of a solid hemisphere is 462 cm\(^2\), find its volume. Use \( \pi = \frac{22}{7} \)
Answer: Given, total surface area of a solid hemisphere = 462 cm\(^2\)
Let its radius be \( r \).
Thus \( 3 \pi r^2 = 462 \)
\( \Rightarrow \left( \frac{22}{7} \right) r^2 = 154 \)
\( \Rightarrow \left( \frac{1}{7} \right) r^2 = 7 \)
\( \Rightarrow r^2 = (7)^2 \)
\( \Rightarrow r = 7 \) cm
So, volume of the hemisphere = \( \frac{2}{3} \pi r^3 \)
\( = \frac{2}{3} \left( \frac{22}{7} \right) (7)^3 \) cm\(^3\)
\( = \frac{2}{3} \left( \frac{22}{1} \right) (7)^2 \) cm\(^3\)
\( = \frac{44}{3} (49) \) cm\(^3\)
\( = 718.67 \) cm\(^3\).

Question. A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.
Answer: Given, height of cone = 20 cm
Radius of base of cone = 5 cm
Thus, Volume of cone = \( \frac{1}{3} \pi (5)^2 \cdot 20 \) cm\(^3\)
Let the radius of the sphere be \( r \).
Now, the volume of the cone = Volume of the sphere
Thus \( \frac{1}{3} \pi (5)^2 \cdot 20 = \frac{4}{3} \pi (r)^3 \)
\( \Rightarrow (5)^2 \cdot 20 = 4(r)^3 \)
\( \Rightarrow (r)^3 = (5)^2 \cdot 5 \)
\( \Rightarrow r = 5 \) cm
Thus, diameter of the sphere = \( 2 \times 5 \) cm = 10 cm.

Question. 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.
Answer: Given : \( l = 4 \) cm
Let \( R \) and \( r \) be the radii of two circular ends of frustum, respectively.
Then, \( 2\pi R = 18 \) cm and \( 2\pi r = 6 \) cm
\( \Rightarrow \pi R = 9 \) cm and \( \pi r = 3 \) cm
Now, Curved surface area of frustum = \( (\pi R + \pi r)l \)
\( = (9 + 3) \times 4 \)
\( = 12 \times 4 \)
\( = 48 \) cm\(^2\).

Question. The dimensions of a metallic cuboid are 100 cm \(\times\) 80 cm \(\times\) 64 cm. It is melted and recast into a cube. Find the surface area of the cube.
Answer: Given, the dimensions of a cuboid = 100 cm \(\times\) 80 cm \(\times\) 64 cm
Let the side of a cube be \( a \) cm.
Thus, \( a^3 = 100 \times 80 \times 64 \) cm\(^3\)
\( \Rightarrow a^3 = 1000 \times 8 \times 64 \) cm\(^3\)
\( \Rightarrow a^3 = 10^3 \times 8^3 \) cm\(^3\)
\( \Rightarrow a = 10 \times 8 \) cm = 80 cm
\(\therefore\) Total surface area of cube = \( 6a^2 \)
\( = 6 \times (80)^2 \) cm\(^2\)
\( = 38400 \) cm\(^2\).

Question. Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm respectively are melted to form a single solid sphere. Find the diameter of the resulting sphere.
Answer: Given, radii of the smaller spheres are 3 cm, 4 cm and 5 cm.
Let the radius of the larger sphere be \( r \) cm.
Thus, volume of the larger sphere = Volume of the three smaller spheres taken together
\( \Rightarrow \frac{4}{3} \pi(r)^3 = \frac{4}{3} \pi(3)^3 + \frac{4}{3} \pi(4)^3 + \frac{4}{3} \pi(5)^3 \)
\( \Rightarrow \frac{4}{3} \pi(r)^3 = \frac{4}{3} \pi[(3)^3 + (4)^3 + (5)^3] \)
\( \Rightarrow (r)^3 = [27 + 64 + 125] \)
\( \Rightarrow (r)^3 = (216) \)
\( \Rightarrow (r)^3 = (6)^3 \)
\( \Rightarrow r = 6 \) cm
Thus, diameter of the largest sphere = 12 cm.

Short Answer Type Questions-II

Question. The area of a circular play ground is 22176 cm\(^2\). Find the cost of fencing this ground at the rate of ₹ 50 per metre.
Answer: Given : Area of play ground = 22176 cm\(^2\)
Let the radius of play ground be \( r \) cm.
\(\therefore\) Area of play ground = \( \pi r^2 \)
\( \Rightarrow 22176 = \frac{22}{7} \times r^2 \)
\( \Rightarrow r^2 = \frac{22176 \times 7}{22} \)
\( \Rightarrow r^2 = 7056 \)
\( \Rightarrow r = \sqrt{7056} = 84 \) cm = 0.84 m
Now, circumference of play ground = \( 2\pi r \)
\( = 2 \times \frac{22}{7} \times 0.84 = 5.28 \) m
Rate of fencing = ₹ 50 per metre
\(\therefore\) Total of fencing the circular ground = \( 50 \times 5.28 \)
= ₹ 264

Question. A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
Answer: Given, Radius of sphere = 10.5 cm
Radius of each cone = 3.5 cm
Height of each cone = 3 cm
Thus, Number of cones = \( \frac{\text{Volume of sphere}}{\text{Volume of cone}} \)
\( = \frac{\frac{4}{3}\pi(10.5)^3}{\frac{1}{3}\pi(3.5)^2(3)} \)
\( = \frac{4(10.5)^3}{(3.5)^2(3)} \)
\( = \frac{36(10.5)}{(3)} \)
\( = 12(10.5) \)
= 126.

Question. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Answer: Given, radius of small sphere \( r = 3 \) cm
Then,
Volume of small sphere = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3}\pi(3)^3 \)
\( = 36\pi \text{ cm}^3 \)
Density of small sphere = \( \frac{\text{Mass of sphere}}{\text{Volume of sphere}} \)
\( = \frac{1}{36\pi} \text{ kg/cm}^3 \)
\(\because\) Both spheres are made by same metal, then their densities will be same.
Let radius of bigger sphere = \( r' \) then,
Density of bigger sphere = \( \frac{\text{Mass of bigger sphere}}{\text{Volume of bigger sphere}} \)
\( \frac{1}{36\pi} = \frac{7}{4/3\pi (r')^3} \)
\( (r')^3 = 189 \)
Then according to question, we have,
Volume of bigger sphere + Volume of smaller sphere = Volume of new sphere.
\( \frac{4}{3}\pi(r')^3 + \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \)
\( (r')^3 + r^3 = R^3 \)
\( 189 + 27 = R^3 \)
\( 216 = R^3 \)
\( R = 6 \)
Radius of new sphere is 6 cm.
So, diameter is 12 cm.

Question. The radii of two circular ends of a bucket of height 15 cm are 14 cm and \( r \) cm (\( r < 14 \) cm). If the volume of the bucket is 5390 cm\(^3\), find the value of \( r \). Use \( \pi = \frac{22}{7} \)
Answer: Given, height of bucket = 15 cm
Larger radius (\( R \)) = 14 cm
Smaller radius = \( r \) cm
and Volume of bucket = 5390 cm\(^3\)
Since, \( r < 14 \) cm, the bucket is in the shape of frustum.
\(\therefore \frac{\pi}{3}[R^2 + r^2 + R \times r]h = 5390 \)
\( \Rightarrow \frac{\pi}{3}[(14)^2 + (r)^2 + (14)(r)]15 = 5390 \)
\( \Rightarrow \frac{22}{21}[196 + r^2 + 14r]15 = 5390 \)
\( \Rightarrow \frac{2}{21}[196 + r^2 + 14r]15 = 490 \)
\( \Rightarrow \frac{1}{7}[196 + r^2 + 14r]5 = 245 \)
\( \Rightarrow \frac{1}{7}[196 + r^2 + 14r] = 49 \)
\( \Rightarrow [196 + r^2 + 14r] = 343 \)
\( \Rightarrow r^2 + 14r - 147 = 0 \)
\( \Rightarrow r^2 + 21r - 7r - 147 = 0 \)
\( \Rightarrow r(r + 21) - 7(r + 21) = 0 \)
\( \Rightarrow (r + 21)(r - 7) = 0 \)
\( \Rightarrow r = 7, -21 \)
Since, radius cannot be negative, so \( r = 7 \) cm.

Question. An open metal bucket is in the shape of a frustum of height 21 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the cost of milk that can completely fill the bucket at ₹ 30 per litre. Use \( \pi = \frac{22}{7} \)
Answer: Given, height of the frustum = 21 cm
Radius of lower end = 10 cm
Radius of upper end = 20 cm
Cost of milk per litre = ₹ 30
Now, volume of the frustum = \( \frac{\pi}{3} [R_1^2 + R_2^2 + R_1 R_2]H \)
\( = \frac{1}{3} \times \frac{22}{7} \times [(10)^2 + (20)^2 + (10)(20)]21 \text{ cm}^3 \)
\( = \frac{22}{21} [100 + 400 + 200]21 \text{ cm}^3 \)
\( = \frac{22}{21} [700]21 \text{ cm}^3 \)
\( = 22(700) \text{ cm}^3 \)
\( = 15400 \text{ cm}^3 \)
\( = 15.4 \text{ litre} \) [\(\because 1 \text{ litre} = 1000 \text{ cm}^3\)]
\(\therefore\) The cost of the milk in completely filling up the frustum = ₹ \( 30 \times 15.4 = \text{₹ } 462 \).

Question. The dimensions of a solid iron cuboid are 4.4 m \(\times\) 2.6 m \(\times\) 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.
Answer: Volume of cuboid = \( 4.4 \times 2.6 \times 1 \text{ m}^3 \)
Inner radius of cylindrical pipe '\( r \)' = 30 cm
Thickness of pipe = 5 cm
\(\therefore\) Outer radius of cylinder '\( R \)' = \( 30 + 5 = 35 \text{ cm} = \frac{35}{100} \text{ m} \)
Let the length of pipe be \( h \).
\(\therefore\) Volume of material used = \( \pi(R^2 - r^2)h \)
\( = \frac{22}{7} \left[ \left( \frac{35}{100} \right)^2 - \left( \frac{30}{100} \right)^2 \right] h \)
\( = \frac{22}{7} \left( \frac{65}{100} \right) \left( \frac{5}{100} \right) h \)
Now, \( \frac{22}{7} \left( \frac{65}{100} \right) \left( \frac{5}{100} \right) h = 4.4 \times 2.6 \)
\( \Rightarrow h = \frac{7 \times 4.4 \times 2.6 \times 100 \times 100}{22 \times 65 \times 5} \)
\( \Rightarrow h = 112 \) m

Question. A hemispherical tank full of water is emptied by a pipe at the rate of \( \frac{25}{7} \) l/s. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m?
Answer: Given, diameter of the tank = 3 m
\(\therefore\) Radius = 1.5 m
Rate of flow of water = \( \frac{25}{7} \text{ litre/s} \)
Thus, Volume = \( \frac{2}{3}\pi(1.5)^3 \text{ m}^3 \)
\( = 2.25 \pi \text{ m}^3 \)
\( = 2250 \pi \text{ litre} \) [\(\because 1 \text{ m}^3 = 1000 \text{ litre}\)]
Thus, half of the volume of the tank = \( 1125\pi \text{ litre} \)
Thus, the time taken to empty half of the tank
\( = \frac{1125\pi}{\frac{25}{7}} \text{ sec.} \)
\( = \frac{1125 \times \frac{22}{7}}{\frac{25}{7}} \text{ sec.} \)
\( = \frac{1125 \times 22}{25} \text{ sec.} \)
= 990 sec
= 16 minutes 30 seconds.

Question. A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys each in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.
Answer: Given : Diameter of cylinder = 12 cm
\( \Rightarrow \) Radius of cylinder, \( r = \frac{12}{2} \text{ cm} = 6 \text{ cm} \)
Height of cylinder, \( h = 15 \) cm
Radius of cone, \( R = 3 \) cm
Height of cone, \( H = 9 \) cm

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Question. A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys each in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.
Answer: Given : Diameter of cylinder = 12 cm
\( \Rightarrow \) Radius of cylinder, \( r = \frac{12}{2} \) cm = 6 cm
Height of cylinder, \( h = 15 \) cm
Radius of cone, \( R = 3 \) cm
Height of cone, \( H = 9 \) cm
Now, Number of cones formed
\( = \frac{\text{Volume of metallic cylinder}}{\text{Volume of a cone}} \)
\( = \frac{\pi r^2 h}{\frac{1}{3}\pi R^2 H} = \frac{3r^2 h}{R^2 H} \)
\( = \frac{3 \times 6 \times 6 \times 15}{3 \times 3 \times 9} \)
= 20

Question. The largest possible sphere is carved out of a solid wooden cube of side 7 cm. Find the volume of the wood left. Use \( \pi = \frac{22}{7} \)
Answer: Length of each side of the cube = 7 cm
\(\therefore\) Largest possible diameter of the sphere = 7 cm
\(\therefore\) Radius of the sphere = 3.5 cm
Thus, Volume of cube = \( (7)^3 \text{ cm}^3 \)
= 343 cm\(^3\)
and Volume of sphere = \( \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 \text{ cm}^3 \)
\( = \frac{539}{3} \text{ cm}^3 \)
\(\therefore\) Remaining volume of wood
\( = \left( 343 - \frac{539}{3} \right) \text{ cm}^3 \)
\( = \frac{1029 - 539}{3} \text{ cm}^3 \)
= 163.33 cm\(^3\).

Passage Based Questions

Question. Read the following passage and answer the questions that follows :
Mr. Sharma told 4 students to write the formula of spherical shell on the black board. Four students wrote respectively given below.
(a) \( 4\pi (R^2 - r^2) \)
(b) \( \pi(R^3 - r^2) \)
(c) \( 4\pi (R^3 - r^3) \)
(d) \( \frac{4}{3}\pi (R^3 - r^3) \)

Question. How many students wrote wrong Formula?
Answer: Three students write wrong formula

Question. What is the correct formula and which student write the correct formula?
Answer: Correct formula is \( \frac{4}{3}\pi (R^3 - r^3) \)

Question. If Mr. Sharma says write the formula of total surface area of cylinder then write this formula.
Answer: Total surface area of cylinder = \( 2\pi r (h + r) \)

Long Answer Type Questions

Question. Three cubes of a metal whose edges are in the ratio of 3 : 4 : 5 are melted and converted into a single cube whose diagonal is \( 12\sqrt{3} \) cm. Find the edges of the three cubes.
Answer: Given, diagonal of the larger cube = \( 12\sqrt{3} \) cm
Ratio of the sides of the three smaller cube = 3 : 4 : 5
Let the common factor between the edges of the smaller cubes be \( x \) cm.
Sum of ratios of the edges of the smaller cubes = 3 + 4 + 5 = 12
Thus, Edge of 1st cube = \( \frac{3x}{12} \)
Edge of 2nd cube = \( \frac{4x}{12} \)
and Edge of 3rd cube = \( \frac{5x}{12} \)
Thus, volume of the 1st cube = \( \left( \frac{3x}{12} \right)^3 = \frac{27x^3}{1728} \text{ cm}^3 \)
Volume of the 2nd cube = \( \left( \frac{4x}{12} \right)^3 = \frac{64x^3}{1728} \text{ cm}^3 \)
and Volume of the 3rd cube = \( \left( \frac{5x}{12} \right)^3 = \frac{125x^3}{1728} \text{ cm}^3 \)
Now, diagonal of the larger cube = \( 12\sqrt{3} = a\sqrt{3} \)
Thus, side of the larger cube (\( a \)) = 12
Volume of the larger cube = \( (12)^3 = 1728 \text{ cm}^3 \)
Thus, \( \frac{27x^3}{1728} + \frac{64x^3}{1728} + \frac{125x^3}{1728} = 1728 \)
\( \Rightarrow \frac{27x^3 + 64x^3 + 125x^3}{1728} = 1728 \)
\( \Rightarrow \frac{216x^3}{1728} = 1728 \)
\( \Rightarrow \frac{x^3}{8} = 1728 \)
\( \Rightarrow \left( \frac{x}{2} \right)^3 = (12)^3 \)
\( \Rightarrow \frac{x}{2} = 12 \)
\( \Rightarrow x = 24 \) cm
Thus, Edge of 1st cube = \( \frac{3 \times 24}{12} = 6 \) cm
Edge of 2nd cube = \( \frac{4 \times 24}{12} = 8 \) cm
and Edge of 3rd cube = \( \frac{5 \times 24}{12} = 10 \) cm.

Question. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area and the capacity of the vessel. Use \( \pi = \frac{22}{7} \)
Answer: Given, diameter of hemispherical bowl = diameter of hollow cylinder = 14 cm
Thus, radius of hemispherical bowl = radius of hollow cylinder = 7 cm
Thus, height of hemispherical bowl = 7 cm
and height of the hollow cylinder = (13 – 7) cm = 6 cm
Thus, total surface area of the vessel = Curved surface area of hemispherical bowl + Curved surface area of cylindrical vessel
\( = 2\pi r^2 + 2\pi rh \)
\( = 2\pi(7)^2 + 2\pi(7) (6) \)
\( = 2\pi(7) [7 + 6] \)
\( = 2 \left( \frac{22}{7} \right) (7) [7 + 6] \)
= 2(22) (13)
= 572 cm\(^2\)
and capacity (volume) = \( \frac{2}{3}\pi r^3 + \pi r^2h \)
\( = \frac{2}{3}\pi(7)(7)(7) + \pi(7)(7)(6) \)
\( = \frac{49}{3} \left( \frac{22}{7} \right) (14 + 18) \text{ cm}^3 \)
\( = \frac{7}{3} \left( \frac{22}{1} \right) (32) \text{ cm}^3 \)
= 1642.67 cm\(^3\).

Question. Water in a canal, 6 m wide and 1.5 m deep is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes if 8 cm of standing water is needed for irrigation ?
Answer: Given, Width of canal = 6 m
Depth of canal = 1.5 m
Speed of water flow = 4 km/hr = \( 4 \times \frac{5}{18} \) m/s
\( = \frac{10}{9} \) m/s
Time taken to irrigate = 10 minutes = 600 seconds
Height of standing water = 8 cm = 0.08 m
Thus, volume of water passing through the canal in 1 second
\( = 6 \times 1.5 \times \frac{10}{9} \text{ m}^3 \)
\( = 2 \times 1.5 \times \frac{10}{3} \text{ m}^3 \)
\( = 2 \times 0.5 \times 10 \text{ m}^3 \)
\( = 10 \text{ m}^3 \)
Thus, volume of water that passes through in 600 seconds
\( = 600 \times 10 \text{ m}^3 \)
\( = 6000 \text{ m}^3 \)
\(\therefore\) Area irrigated in 10 minutes
\( = \frac{6000}{0.08} \text{ m}^2 \)
\( = 75000 \text{ m}^2 \).

Question. A solid sphere of diameter 28 cm is melted and recast into smaller solid cones each of diameter \( 4 \frac{2}{3} \) cm and height 3 cm. Find the number of cones so formed. Use \( \pi = \frac{22}{7} \)
Answer: Given, Diameter of sphere = 28 cm
\(\therefore\) Radius of sphere = 14 cm
Diameter of small 1 cone \( = 4 \frac{2}{3} \text{ cm} = \frac{14}{3} \text{ cm} \)
\(\therefore\) Radius of small 1 cone \( = 2 \frac{1}{3} \text{ cm} = \frac{7}{3} \text{ cm} \)
Height of 1 small cone = 3 cm
Volume of the sphere \( = \frac{4}{3} \pi(14)^3 \text{ cm}^3 \)
\( = \frac{88}{21} (2744) \text{ cm}^3 \)
Volume of 1 cone \( = \frac{1}{3} \pi \left(\frac{7}{3}\right)^2 \times 3 \text{ cm}^3 \)
\( = \frac{22}{7} \left(\frac{7}{3}\right)^2 \text{ cm}^3 \)
\( = \frac{154}{9} \text{ cm}^3 \)
Thus, number of cones \( = \frac{\frac{88}{21} \times 2744}{\frac{154}{9}} \)
= 672.

Question. A hemispherical depression is cut out from one face of a cubical block of side 21 cm such that the diameter of the hemisphere is equal to the edge of the cube. Find the volume and the total surface area of the remaining solid. Use \( \pi = \frac{22}{7} \)
Answer: Given, Side of the cube = 21 cm
Diameter of the hemisphere = 21 cm
Thus, radius of the hemisphere = 10.5 cm
Volume of the cube \( = (21)^3 \text{ cm}^3 \)
\( = 9261 \text{ cm}^3 \)
Volume of the hemisphere \( = \frac{2}{3} \pi(10.5)^3 \text{ cm}^3 \)
\( = 2425.5 \text{ cm}^3 \)
Thus, volume of the remaining block
\( = (9261 - 2425.5) \text{ cm}^3 \)
\( = 6835.5 \text{ cm}^3 \)
Total surface area of the cube \( = 6(21)^2 \text{ cm}^2 \)
Curved surface area of the hemisphere \( = 2\pi(10.5)^2 \text{ cm}^2 \)
Thus, total surface area of the remaining solid
= Total surface area of cube + Curved surface area of the hemisphere – Area of the top face of the hemisphere
\( = [6(21)^2 + 2\pi(10.5)^2 - \pi(10.5)^2] \text{ cm}^2 \)
\( = \left[ 6(21)^2 + \left(\frac{22}{7}\right)(10.5)^2 \right] \text{ cm}^2 \)
\( = [2646 + 346.5] \text{ cm}^2 \)
\( = 2992.5 \text{ cm}^2 \).

Question. A container shaped like a cylinder having diameter 12 cm and height 15 cm is full of ice-cream. This 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 that can be filled with ice-cream. Use \( \pi = \frac{22}{7} \)
Answer: Given, Diameter of cylinder = 12 cm
\(\therefore\) Radius of cylinder = 6 cm
Height of cylinder = 15 cm
Height of cone = 12 cm
Diameter of cone = 6 cm
\(\therefore\) Radius of cone = 3 cm
Diameter of hemisphere = 6 cm
\(\therefore\) Radius of hemisphere = 3 cm
Hence, volume of cylinder \( = \pi (6)^2 \times 15 \text{ cm}^3 \)
\( = \pi (36)15 \text{ cm}^3 \)
\( = 540 \pi \text{ cm}^3 \)
Volume of cone \( = \frac{1}{3} \pi(3)^2 \times 12 \text{ cm}^3 \)
\( = 36 \pi \text{ cm}^3 \)
Volume of hemisphere \( = \frac{2}{3} \pi(3)^3 \text{ cm}^3 \)
\( = 18 \pi \text{ cm}^3 \)
\(\therefore\) Volume of ice-cream in each cone \( = (36\pi + 18\pi) \text{ cm}^3 \)
\( = 54\pi \text{ cm}^3 \)
Thus, number of ice-cream cones
\( = \frac{\text{Total volume of ice-cream in the cylinder}}{\text{Total volume of ice-cream in each cone}} \)
\( = \frac{540\pi}{54\pi} = 10 \)
Hence, the number of ice-cream cones are 10.

Question. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, in how much time will the tank be completely filled ?
Answer: Internal diameter of the pipe = 20 cm
\(\therefore\) Internal radius of pipe = 10 cm
Diameter of the tank = 10 m = 1000 cm
\(\therefore\) Radius of tank = 500 cm
Height = 2 m = 200 cm
Rate of flow of water = 4 km/hr \( = 4 \times \frac{5}{18} \) m/s
\( = \frac{10}{9} \text{ m/s} = \frac{1000}{9} \text{ cm/s} \)
Now, volume of water flowing through the pipe in 1 second
\( = \pi(10)^2 \times \frac{1000}{9} \text{ cm}^3 \)
\( = \pi \left(\frac{100000}{9}\right) \text{ cm}^3 \)
Volume of the tank \( = \pi(500)^2 (200) \text{ cm}^3 \)
\( = \pi 50000000 \text{ cm}^3 \)
Thus, time taken to fill the tank
\( = \frac{\pi 50000000}{\pi \left(\frac{100000}{9}\right)} \text{ sec} \)
\( = \frac{500}{\frac{1}{9}} \text{ sec} \)
\( = 4500 \text{ sec} = 75 \text{ min.} \)
= 1 hour 15 minutes.

Question. A tent is in the form of a right circular cylinder of base diameter 4.2 m and height 4 m surmounted by a right circular cone of the same base radius and height 2.8 m. Find the capacity of the tent and the cost of canvas used in making the tent at ₹ 100 per square metre.
Answer: Given, height of the cylinder = 4 m
Height of the cone = 2.8 m
Base diameter of cylinder = Base diameter of cone = 4.2 m.
\(\therefore\) Radius of cylindrical structure = Radius of conical structure = 2.1 m.
Rate of canvas = ₹ 100/per m\(^2\)
\(\therefore\) Capacity of the tent
\( = \left[ \pi(2.1)^2 \times 4 + \frac{1}{3}\pi(2.1)^2 \times 2.8 \right] \text{ m}^3 \)
\( = \pi(2.1)^2 \left[ 4 + \frac{1}{3}(2.8) \right] \text{ m}^3 \)
\( = \frac{22}{21}(2.1)^2 [12 + 2.8] \text{ m}^3 \)
\( = \frac{22}{100}(21) [14.8] \text{ m}^3 \)
\( = 68.38 \text{ m}^3 \)
Thus, total surface area of the tent = Curved surface area of the (cylindrical structure + conical structure)
\( = \pi[2(2.1)(4) + 2.1(\sqrt{(2.1)^2 + (2.8)^2})] \text{ m}^2 \)
\( = \pi(2.1)(8 + \sqrt{4.41 + 7.84}) \text{ m}^2 \)
\( = (2.1)\pi(8 + \sqrt{12.25}) \text{ m}^2 \)
\( = (2.1)\left(\frac{22}{7}\right)(8 + 3.5) \text{ m}^2 \)
\( = (0.3)(22)(11.5) \text{ m}^2 \)
\( = \frac{66}{10} (11.5) \text{ m}^2 \)
\( = 75.9 \text{ m}^2 \)
Thus, the cost of canvas = ₹ \( 100 \times 75.9 \)
= ₹ 7590.

Question. A tent consists of a frustum of a cone surmounted by a cone. If the diameters of the upper and lower circular ends of the frustum are 14 m and 26 m respectively, the height of the frustum is 8 m and the slant height of the surmounted conical portion is 12 m, find the area of the canvas required to make the tent assuming that the radii of the upper circular end and the base of the surmounted conical portion being equal. [Use \( \pi = 3.14 \)]
Answer: Given, diameter of the upper circular end = 14 m.
∴ Radius of the upper end = 7 m
Diameter of the lower circular end = 26 m
∴ Radius of the lower end = 13 m
Height of frustum = 8 m
Slant height of the cone = 12 m
Also Radius of the cone = 7 m
Hence, slant height of the frustum
\( = \sqrt{(8)^2 + (13 - 7)^2} \)
\( = \sqrt{(8)^2 + (6)^2} \)
\( = \sqrt{64 + 36} \)
\( = \sqrt{100} = 10 \text{ m} \)
∴ Total area of the canvas required
= Lateral surface area of the frustum + Curved surface area of the cone
\( = [\pi(13 + 7)10 + \pi(7)12] \text{ cm}^2 \)
\( = \pi[(20)10 + (7)12] \text{ cm}^2 \)
\( = \pi[200 + 84] \text{ cm}^2 \)
\( = 284 \pi \text{ cm}^2 \)
\( = 284(3.14) \text{ cm}^2 \)
\( = 891.76 \text{ cm}^2 \).

Question. A milk container made of metal and in the shape of a frustum has a volume of \( 10459 \frac{3}{7} \text{ cm}^3 \). The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal used in making the container at the rate of ₹ 1.40 per cm\(^2\). Use \( \pi = \frac{22}{7} \)
Answer: Given,
Volume of the frustum \( = 10459 \frac{3}{7} \text{ cm}^3 = \frac{73216}{7} \text{ cm}^3 \)
Radius of the lower circular end = 8 cm
Radius of the upper circular end = 20 cm
Rate of metal sheet = ₹ 1.40 per cm\(^2\)
Let the height of the frustum be \( h \).
Thus, \( \frac{\pi}{3} [(8)^2 + (20)^2 + 8(20)]h = \frac{73216}{7} \)
\( \Rightarrow \frac{22}{21} [64 + 400 + 160]h = \frac{73216}{7} \)
\( \Rightarrow \frac{22}{3} [624]h = 73216 \)
\( \Rightarrow 22[208]h = 73216 \)
\( \Rightarrow h = 16 \text{ cm} \)
Now, let \( l \) be the slant height.
Thus, \( l = \sqrt{(16)^2 + (20 - 8)^2} \)
\( = \sqrt{(16)^2 + (12)^2} \)
\( = \sqrt{256 + 144} \)
\( = \sqrt{400} = 20 \text{ cm} \)
Thus, the surface area that needs to be covered with metal
= Lateral surface area + Area of the lower end
\( = [\pi(20 + 8)20 + \pi(8)^2] \text{ cm}^2 \)
\( = \pi[(28)(20) + 64] \text{ cm}^2 \)
\( = \pi(624) \text{ cm}^2 \)
Thus, the total cost of the metal sheet
\( = Rs (1.40 \times 624\pi) \)
\( = Rs \left[ 1.40 \times 624 \times \frac{22}{7} \right] \)
= Rs 2745.60.

Case Study

Snehal throws a party in her house. For the party she has to serve snacks and juice to her guests. She has a big jar of diameter 14 cm and height 42 cm filled with juice. She has two type of glasses for serving juice of cylindrical shape. Glass A with diameter 7 cm and height 10 cm and glass B with diameter 7 cm and height 12 cm. Based on the above information answer the questions that follows :

Question. What is the capacity of juice jar?
(a) 7,468 cm\(^3\)
(b) 6,468 cm\(^3\)
(c) 4,785 cm\(^3\)
(d) 5,276 cm\(^3\)
Answer: (b) Since, jar is of cylindrical shape. So the capacity of jar is given by volume of cylinder.
Volume of cylinder = \( \pi r^2h \)
Radius of jar \( (r) = \frac{14}{2} = 7 \text{ cm} \).
Height of jar \( (h) = 42 \text{ cm} \)
Volume \( = \frac{22}{7} \times 7 \times 7 \times 42 = 6,468 \text{ cm}^3 \).
Hence correct option is (b).

Question. If the juice is to served in glass B. Then how many guest will be served?
(a) 14
(b) 10
(c) 15
(d) 12
Answer: (a) Dimensions of glass B are
Radius of glass \( = \frac{7}{2} \text{ cm.} \)
Height of glass = 12 cm
Capacity of each glass \( = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12 = 462 \text{ cm}^3 \)
Number of guests served \( = \frac{\text{Volume of juice jar}}{\text{Volume of each glass}} = \frac{6468}{462} = 14 \).
Correct option is (a).

Question. Which glass has more capacity glass A or glass B?
(a) Glass A
(b) Glass B
(c) Both have same capacity
(d) Cannot be determined.
Answer: (b) Capacity of glass B is 462 cm\(^3\)
Capacity of glass A \( = \pi r^2h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 10 = 385 \text{ cm}^3 \)
Correct option is (b) glass B has more capacity.

Question. Approximately how many guests can be served using glass A?
(a) 17 guests
(b) 14 guests
(c) 19 guests
(d) 18 guests
Answer: (a) Volume of juice jar = 6468 cm\(^3\)
Volume of glass A = 385 cm\(^3\)
Number of guests served \( = \frac{6468}{385} = 16.8 \approx 17 \)
Correct option is (a) 17 guests.

Question. If there are 40 guests in the party and Snehal has 2 jars full of juice. Then upto what height she has to fill the glass? So that juice can be served to all the guests in the party.
(a) 8.4 cm
(b) 8 cm
(c) 9 cm
(d) 7.4 cm
Answer: (a) Number of guests = 40
Capacity of each jar = 6468 cm\(^3\)
Capacity of two jars \( = 2 \times 6468 \text{ cm}^3 = 12,936 \text{ cm}^3 \)
Since, radius of both the glass is same, then she can serve in any of the glass.
Let, height upto which she can serve = \( h \text{ cm} \)
diameter of each glass = 7 cm
Radius of each glass \( (r) = \frac{7}{2} \text{ cm.} \)
\( 12936 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h \times 40 \)
\( h = \frac{12936}{11 \times 7 \times 20} = 8.4 \text{ cm} \)

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true, but statement 2 is not the correct explanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false, but statement 2 is true.

Question. Statement 1 : Total surface area of the cylinder having radius of the base 21 cm and height 30 cm is 6732 cm\(^2\).
Statement 2 : If r be the radius and h be the height of the cylinder, then total surface is \( 2\pi rh + 2\pi r^2 \).
Answer: (A) In statement 1, the cylinder having radius \( (r) = 21 \) cm and height \( (h) = 30 \) cm
Area of the cylinder = \( 6732 \text{ cm}^2 \).
Total surface area of the cylinder \( = (2\pi rh + 2\pi r^2) \)
\( = 2 \times \frac{22}{7} \times 21 \times 30 + 2 \times \frac{22}{7} \times 21^2 \)
\( = 3960 + 2772 = 6732 \text{ cm}^2 \)
∴ Statement 1 is true.
Total surface area of the cylinder \( = 2\pi rh + 2\pi r^2 \).
Statement 2 is used in statement 1.
So, statement 2 is also true.

Question. Statement 1 : If r be the radius and h be the height of the cone, then slant height \( = \sqrt{h^2 + r^2} \).
Statement 2 : If the height of a cone is 24 cm and diameter of the base is 14 cm, then the slant height of the cone is 15 cm.
Answer: (C) The slant height of the cone is \( \sqrt{h^2 + r^2} \)
So, the statement 1 is true.
In statement 2, the cone whose diameter \( (d) \) is 14 cm and height \( (h) \) is 24 cm.
Then, \( r = \frac{d}{2} = \frac{14}{2} = 7 \text{ cm} \)
Slant height of the cone is \( = \sqrt{h^2 + r^2} \)
\( = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm} \)
In statement 2, the given slant height of the cone is 15 cm.
∴ Statement 2 is false.