Chapter 12: Surface Areas and Volumes
Exercise 12.1
Q1
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Step 1: Find the side of the cube.
Given volume of one cube = \( 64 \text{ cm}^3 \).
Let the side of the cube be \( a \).
\[ a^3 = 64 \Rightarrow a = \sqrt[3]{64} = 4 \text{ cm} \]
Step 2: Find dimensions of the resulting cuboid.
When two cubes are joined end to end:
- Length (\(l\)) = \( 4 + 4 = 8 \text{ cm} \)
- Breadth (\(b\)) = \( 4 \text{ cm} \)
- Height (\(h\)) = \( 4 \text{ cm} \)
Step 3: Calculate Total Surface Area.
\[ \text{TSA} = 2(lb + bh + hl) \]
\[ = 2(8 \times 4 + 4 \times 4 + 4 \times 8) \]
\[ = 2(32 + 16 + 32) = 2(80) = 160 \text{ cm}^2 \]
Answer: \( 160 \text{ cm}^2 \)
Q2
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.
Solution:
Given:
- Diameter of hemisphere = 14 cm \(\Rightarrow\) Radius (\(r\)) = 7 cm.
- Total height of vessel = 13 cm.
- Height of cylindrical part (\(h\)) = Total height - Radius of hemisphere = \( 13 - 7 = 6 \text{ cm} \).
Calculation:
Inner Surface Area = CSA of Cylinder + CSA of Hemisphere
\[ = 2\pi rh + 2\pi r^2 \]
\[ = 2\pi r(h + r) \]
\[ = 2 \times \frac{22}{7} \times 7 (6 + 7) \]
\[ = 44 \times 13 = 572 \text{ cm}^2 \]
Answer: \( 572 \text{ cm}^2 \)
Q3
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.
Solution:
Given:
- Radius (\(r\)) = 3.5 cm.
- Total height of toy = 15.5 cm.
- Height of cone (\(h\)) = Total height - Radius of hemisphere = \( 15.5 - 3.5 = 12 \text{ cm} \).
Step 1: Find Slant Height (\(l\)) of the cone.
\[ l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (12)^2} \]
\[ l = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm} \]
Step 2: Calculate Total Surface Area.
TSA = CSA of Cone + CSA of Hemisphere
\[ = \pi rl + 2\pi r^2 \]
\[ = \pi r(l + 2r) \]
\[ = \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5) \]
\[ = 11 (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2 \]
Answer: \( 214.5 \text{ cm}^2 \)
Q4
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.
Solution:
Part 1: Greatest Diameter
The hemisphere is mounted on the cube. To fit, the greatest diameter is equal to the side of the cube.
Greatest Diameter = 7 cm.
Radius (\(r\)) = 3.5 cm.
Part 2: Surface Area of Solid
The base of the hemisphere covers a circular part of the cube's top face.
SA = TSA of Cube + CSA of Hemisphere - Area of Base of Hemisphere
\[ = 6l^2 + 2\pi r^2 - \pi r^2 \]
\[ = 6l^2 + \pi r^2 \]
\[ = 6(7)^2 + \frac{22}{7}(3.5)^2 \]
\[ = 6(49) + \frac{22}{7}(12.25) \]
\[ = 294 + 22 \times 1.75 = 294 + 38.5 = 332.5 \text{ cm}^2 \]
Answer: Greatest diameter is 7 cm. Surface area is \( 332.5 \text{ cm}^2 \).
Q5
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.
Solution:
Given:
Edge of cube = \(l\).
Diameter of hemisphere = \(l\) \(\Rightarrow\) Radius \(r = \frac{l}{2}\).
Calculation:
Surface Area = TSA of Cube + CSA of Hemisphere - Area of circular opening
\[ = 6l^2 + 2\pi r^2 - \pi r^2 \]
\[ = 6l^2 + \pi r^2 \]
\[ = 6l^2 + \pi \left(\frac{l}{2}\right)^2 \]
\[ = 6l^2 + \frac{\pi l^2}{4} \]
\[ = \frac{l^2}{4} (24 + \pi) \]
Answer: \( \frac{l^2}{4} (24 + \pi) \text{ units}^2 \)
Q6
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.
Solution:
Given:
- Diameter of capsule = 5 mm \(\Rightarrow r = 2.5 \text{ mm}\).
- Total length = 14 mm.
- Length of cylindrical part (\(h\)) = Total length - 2(radius of hemisphere)
- \( h = 14 - (2.5 + 2.5) = 14 - 5 = 9 \text{ mm}\).
Calculation:
Total Surface Area = CSA of Cylinder + 2 \(\times\) CSA of Hemisphere
\[ = 2\pi rh + 2(2\pi r^2) \]
\[ = 2\pi r(h + 2r) \]
\[ = 2 \times \frac{22}{7} \times 2.5 (9 + 5) \]
\[ = \frac{44}{7} \times 2.5 \times 14 \]
\[ = 44 \times 2.5 \times 2 = 220 \text{ mm}^2 \]
Answer: \( 220 \text{ mm}^2 \)
Q7
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 ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
Given:
- Cylinder: Height (\(h\)) = 2.1 m, Diameter = 4 m \(\Rightarrow r = 2\) m.
- Cone: Slant height (\(l\)) = 2.8 m, Radius (\(r\)) = 2 m.
Step 1: Calculate Canvas Area.
Area of canvas = CSA of Cylinder + CSA of Cone
\[ = 2\pi rh + \pi rl = \pi r(2h + l) \]
\[ = \frac{22}{7} \times 2 (2 \times 2.1 + 2.8) \]
\[ = \frac{44}{7} (4.2 + 2.8) = \frac{44}{7} \times 7 = 44 \text{ m}^2 \]
Step 2: Calculate Cost.
Cost = Area \(\times\) Rate = \( 44 \times 500 = 22000 \)
Answer: Area is \( 44 \text{ m}^2 \). Cost is ₹ 22,000.
Q8
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.
Solution:
Given:
- Height (\(h\)) = 2.4 cm.
- Diameter = 1.4 cm \(\Rightarrow r = 0.7\) cm.
Step 1: Find Slant Height (\(l\)).
\[ l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} \]
\[ l = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5 \text{ cm} \]
Step 2: Calculate Total Surface Area.
The total surface area includes the CSA of the cylinder, the CSA of the conical cavity, and the area of the top circular base.
TSA = CSA of Cylinder + CSA of Cone + Area of Top Base
\[ = 2\pi rh + \pi rl + \pi r^2 \]
\[ = \pi r (2h + l + r) \]
\[ = \frac{22}{7} \times 0.7 (2 \times 2.4 + 2.5 + 0.7) \]
\[ = 2.2 (4.8 + 2.5 + 0.7) \]
\[ = 2.2 (8.0) = 17.6 \text{ cm}^2 \]
Rounding to the nearest cm²: 18 cm².
Answer: \( 18 \text{ cm}^2 \)
Q9
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. 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.
Solution:
Given:
- Height (\(h\)) = 10 cm.
- Radius (\(r\)) = 3.5 cm.
Calculation:
Total Surface Area = CSA of Cylinder + CSA of two Hemispheres
\[ = 2\pi rh + 2(2\pi r^2) \]
\[ = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r) \]
\[ = 2 \times \frac{22}{7} \times 3.5 (10 + 2 \times 3.5) \]
\[ = 22 (10 + 7) = 22 \times 17 \]
\[ = 374 \text{ cm}^2 \]
Answer: \( 374 \text{ cm}^2 \)
Exercise 12.2
Q1
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 \(\pi\).
Solution:
Given:
- Radius of hemisphere (\(r\)) = 1 cm.
- Radius of cone (\(r\)) = 1 cm.
- Height of cone (\(h\)) = \(r\) = 1 cm.
Calculation:
Total Volume = Volume of Cone + Volume of Hemisphere
\[ = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \]
\[ = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 \]
\[ = \frac{1}{3}\pi + \frac{2}{3}\pi = \frac{3}{3}\pi = \pi \text{ cm}^3 \]
Answer: \( \pi \text{ cm}^3 \)
Q2
Rachel made a model shaped like a cylinder with two cones attached at its two ends. 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.
Solution:
Given:
- Diameter = 3 cm \(\Rightarrow r = 1.5\) cm.
- Total length = 12 cm.
- Height of each cone (\(h_{cone}\)) = 2 cm.
- Height of cylinder (\(h_{cyl}\)) = \( 12 - 2 - 2 = 8 \) cm.
Calculation:
Total Volume = Volume of Cylinder + 2 \(\times\) Volume of Cone
\[ = \pi r^2 h_{cyl} + 2 \times \frac{1}{3} \pi r^2 h_{cone} \]
\[ = \pi r^2 (h_{cyl} + \frac{2}{3} h_{cone}) \]
\[ = \frac{22}{7} \times (1.5)^2 \left(8 + \frac{2}{3} \times 2\right) \]
\[ = \frac{22}{7} \times 2.25 \left(8 + \frac{4}{3}\right) = \frac{22}{7} \times 2.25 \times \frac{28}{3} \]
\[ = 22 \times 0.75 \times 4 = 66 \text{ cm}^3 \]
Answer: \( 66 \text{ cm}^3 \)
Q3
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.
Solution:
Given:
- Total length = 5 cm. Diameter = 2.8 cm \(\Rightarrow r = 1.4\) cm.
- Height of cylindrical part (\(h\)) = \( 5 - 1.4 - 1.4 = 2.2 \) cm.
Volume of 1 Gulab Jamun:
\[ V = \text{Vol Cylinder} + 2 \times \text{Vol Hemisphere} \]
\[ = \pi r^2 h + \frac{4}{3} \pi r^3 \]
\[ = \pi r^2 \left(h + \frac{4}{3} r\right) \]
\[ = \frac{22}{7} \times (1.4)^2 \left(2.2 + \frac{4}{3} \times 1.4\right) \]
\[ = 6.16 \times (2.2 + 1.867) = 6.16 \times 4.067 \approx 25.05 \text{ cm}^3 \]
Total Syrup Volume:
Volume of 45 gulab jamuns = \( 45 \times 25.05 = 1127.25 \text{ cm}^3 \).
Quantity of syrup = 30% of Total Volume = \( \frac{30}{100} \times 1127.25 \approx 338 \text{ cm}^3 \).
Answer: Approximately \( 338 \text{ cm}^3 \).
Q4
A pen stand made of wood is in the shape of a cuboid with four conical depressions. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Solution:
Step 1: Volume of Cuboid.
\[ V_{cuboid} = l \times b \times h = 15 \times 10 \times 3.5 = 525 \text{ cm}^3 \]
Step 2: Volume of 4 Conical Depressions.
For one cone: \( r = 0.5 \text{ cm}, h = 1.4 \text{ cm} \).
\[ V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4 \]
\[ = \frac{1}{3} \times 22 \times 0.25 \times 0.2 = \frac{1.1}{3} \approx 0.366 \text{ cm}^3 \]
Volume of 4 cones = \( 4 \times 0.366 = 1.464 \text{ cm}^3 \).
Step 3: Volume of Wood.
\[ V_{wood} = V_{cuboid} - V_{4 cones} = 525 - 1.46 = 523.54 \text{ cm}^3 \]
Answer: \( 523.53 \text{ cm}^3 \) (approx)
Q5
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.
Solution:
Step 1: Volume of Water (Cone).
\[ V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3} \]
Step 2: Water flowed out.
Water displaced = \( \frac{1}{4} \times \text{Volume of cone} = \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \text{ cm}^3 \).
Step 3: Volume of 1 Lead Shot (Sphere).
\[ V_{sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{0.5\pi}{3} \]
Step 4: Calculate Number of Shots (\(n\)).
Total volume displaced = \( n \times \) Volume of one shot
\[ \frac{50\pi}{3} = n \times \frac{0.5\pi}{3} \]
\[ n = \frac{50}{0.5} = 100 \]
Answer: 100 lead shots.
Q6
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 cm³ of iron has approximately 8g mass. (Use \(\pi = 3.14\))
Solution:
Step 1: Volume of Larger Cylinder.
Height (\(H\)) = 220 cm, Radius (\(R\)) = 12 cm.
\[ V_1 = \pi R^2 H = 3.14 \times (12)^2 \times 220 = 3.14 \times 144 \times 220 = 99475.2 \text{ cm}^3 \]
Step 2: Volume of Smaller Cylinder.
Height (\(h\)) = 60 cm, Radius (\(r\)) = 8 cm.
\[ V_2 = \pi r^2 h = 3.14 \times (8)^2 \times 60 = 3.14 \times 64 \times 60 = 12057.6 \text{ cm}^3 \]
Step 3: Total Volume and Mass.
Total Volume = \( 99475.2 + 12057.6 = 111532.8 \text{ cm}^3 \).
Mass = \( 111532.8 \times 8 \text{ g} = 892262.4 \text{ g} = 892.26 \text{ kg} \).
Answer: 892.26 kg
Q7
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.
Solution:
Step 1: Volume of Solid (Cone + Hemisphere).
Radius \(r = 60\), Cone height \(h_{cone} = 120\).
\[ V_{solid} = \text{Vol Cone} + \text{Vol Hemisphere} \]
\[ = \frac{1}{3}\pi r^2 h_{cone} + \frac{2}{3}\pi r^3 \]
\[ = \frac{1}{3}\pi (60)^2 (120) + \frac{2}{3}\pi (60)^3 \]
\[ = 144000\pi + 144000\pi = 288000\pi \text{ cm}^3 \]
Step 2: Volume of Cylinder.
Radius \(r = 60\), Height \(h_{cyl} = 180\).
\[ V_{cyl} = \pi r^2 h_{cyl} = \pi (60)^2 (180) = 648000\pi \text{ cm}^3 \]
Step 3: Volume of Water Left.
\[ V_{water} = V_{cyl} - V_{solid} = 648000\pi - 288000\pi = 360000\pi \text{ cm}^3 \]
\[ = 360000 \times \frac{22}{7} \approx 1131428.57 \text{ cm}^3 \approx 1.131 \text{ m}^3 \]
Answer: \( 1.131 \text{ m}^3 \)
Q8
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 \(\pi = 3.14\).
Solution:
Step 1: Volume of Spherical Part.
Diameter = 8.5 cm \(\Rightarrow\) Radius \(R = 4.25\) cm.
\[ V_{sphere} = \frac{4}{3}\pi R^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 \]
\[ \approx \frac{4}{3} \times 3.14 \times 76.7656 \approx 321.39 \text{ cm}^3 \]
Step 2: Volume of Cylindrical Neck.
Length \(h = 8\) cm, Diameter = 2 cm \(\Rightarrow\) Radius \(r = 1\) cm.
\[ V_{cyl} = \pi r^2 h = 3.14 \times (1)^2 \times 8 = 25.12 \text{ cm}^3 \]
Step 3: Total Volume.
Total Volume = \( 321.39 + 25.12 = 346.51 \text{ cm}^3 \).
The child found 345 cm³. Since \( 346.51 \neq 345 \), she is incorrect.
Answer: She is incorrect. The correct volume is \( 346.51 \text{ cm}^3 \).