Exercise 10.4 Solutions
Exercise 10.4 Solutions
Problem 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.
Diagram Description: A sphere with radius 4.2 cm is transformed into a cylinder with radius 6 cm. The volume remains the same during this transformation.
Solution:
Volume of sphere = Volume of cylinder
\(\frac{4}{3}\pi r^3 = \pi R^2 h\)
\(\frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h\)
\(\frac{4}{3} \times 74.088 = 36 h\)
\(98.784 = 36 h\)
\(h = \frac{98.784}{36} = 2.744 \text{ cm}\)
Answer: The height of the cylinder is 2.744 cm.
Problem 2: Three metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Diagram Description: Three separate spheres with radii 6 cm, 8 cm, and 10 cm are combined to form one larger sphere. The total volume of the three spheres equals the volume of the new sphere.
Solution:
Total volume = Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3
\(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6)^3 + \frac{4}{3}\pi (8)^3 + \frac{4}{3}\pi (10)^3\)
\(r^3 = 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728\)
\(r = \sqrt[3]{1728} = 12 \text{ cm}\)
Answer: The radius of the resulting sphere is 12 cm.
Problem 3: A 20 m deep well of diameter 7 m is dug and the earth got by digging is evenly spread out to form a rectangular platform of base 22 m × 14 m. Find the height of the platform.
Diagram Description: A cylindrical well with diameter 7 m and depth 20 m is dug. The excavated earth forms a rectangular platform with dimensions 22 m × 14 m × height h m.
Solution:
Volume of earth dug = Volume of well = \(\pi r^2 h = \pi (3.5)^2 \times 20\)
Volume of platform = \(22 \times 14 \times h\)
\(\pi (3.5)^2 \times 20 = 22 \times 14 \times h\)
\(\frac{22}{7} \times 12.25 \times 20 = 308 h\)
\(770 = 308 h\)
\(h = \frac{770}{308} = 2.5 \text{ m}\)
Answer: The height of the platform is 2.5 m.
Problem 4: A well of diameter 14 m is dug 15 m deep. The earth taken out of it has been spread evenly to form circular embankment all around the wall of width 7 m. Find the height of the embankment.
Diagram Description: A cylindrical well with diameter 14 m and depth 15 m is dug. The excavated earth forms a circular embankment (a ring-shaped structure) around the well with width 7 m and height h m.
Solution:
Volume of earth dug = \(\pi (7)^2 \times 15\)
Outer radius of embankment = 7 m (well radius) + 7 m (width) = 14 m
Volume of embankment = \(\pi (14^2 – 7^2) h = \pi (196 – 49) h = 147\pi h\)
\(\pi \times 49 \times 15 = 147\pi h\)
\(h = \frac{49 \times 15}{147} = 5 \text{ m}\)
Answer: The height of the embankment is 5 m.
Problem 5: A container shaped 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, making a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Diagram Description: A cylindrical container with diameter 12 cm and height 15 cm contains ice cream. The ice cream is to be distributed into cones with height 12 cm and diameter 6 cm, each topped with a hemispherical scoop (radius 3 cm).
Solution:
Volume of ice cream in cylinder = \(\pi (6)^2 \times 15 = 540\pi \text{ cm}^3\)
Volume of one cone = \(\frac{1}{3}\pi (3)^2 \times 12 = 36\pi \text{ cm}^3\)
Volume of hemisphere = \(\frac{2}{3}\pi (3)^3 = 18\pi \text{ cm}^3\)
Total volume per cone = \(36\pi + 18\pi = 54\pi \text{ cm}^3\)
Number of cones = \(\frac{540\pi}{54\pi} = 10\)
Answer: 10 cones can be filled with the ice cream.
Problem 6: How many silver coins, 1.75 cm in diameter and thickness 2 mm, need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Diagram Description: Multiple cylindrical coins (each with diameter 1.75 cm and thickness 2 mm) are melted to form a rectangular cuboid with dimensions 5.5 cm × 10 cm × 3.5 cm.
Solution:
Volume of one coin = \(\pi (0.875)^2 \times 0.2 \approx 0.481 \text{ cm}^3\)
Volume of cuboid = \(5.5 \times 10 \times 3.5 = 192.5 \text{ cm}^3\)
Number of coins = \(\frac{192.5}{0.481} \approx 400\)
Answer: Approximately 400 coins are needed.
Problem 7: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, \(\frac{1}{4}\) of the water flows out. Find the number of lead shots dropped into the vessel.
Diagram Description: An inverted cone-shaped vessel with height 8 cm and top radius 5 cm is completely filled with water. Small spherical lead shots (each with radius 0.5 cm) are dropped into the vessel, displacing 1/4 of the water volume.
Solution:
Volume of cone = \(\frac{1}{3}\pi (5)^2 \times 8 = \frac{200}{3}\pi \text{ cm}^3\)
Volume of water displaced = \(\frac{1}{4} \times \frac{200}{3}\pi = \frac{50}{3}\pi \text{ cm}^3\)
Volume of one lead shot = \(\frac{4}{3}\pi (0.5)^3 = \frac{\pi}{6} \text{ cm}^3\)
Number of lead shots = \(\frac{\frac{50}{3}\pi}{\frac{\pi}{6}} = 100\)
Answer: 100 lead shots were dropped into the vessel.
Problem 8: A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 \(\frac{2}{3}\) cm and height 3 cm. Find the number of cones so formed.
Diagram Description: A large metallic sphere with diameter 28 cm is melted and reshaped into multiple smaller cones. Each cone has diameter 14/3 cm (4 2/3 cm) and height 3 cm.
Solution:
Radius of sphere = 14 cm
Volume of sphere = \(\frac{4}{3}\pi (14)^3 = \frac{10976}{3}\pi \text{ cm}^3\)
Radius of each cone = \(\frac{14}{6} = \frac{7}{3} \text{ cm}\)
Volume of one cone = \(\frac{1}{3}\pi \left(\frac{7}{3}\right)^2 \times 3 = \frac{49}{9}\pi \text{ cm}^3\)
Number of cones = \(\frac{\frac{10976}{3}\pi}{\frac{49}{9}\pi} = \frac{10976}{3} \times \frac{9}{49} = 672\)
Answer: 672 cones can be formed.
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M.Sc,B.Ed:
SDVR ZPHS B.Gangaram
Sathupally(md)
Khammam(dt)
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