Find the total surface area of a hemispherical bowl whose volume is 7.065 m3. (Take \(\\pi = 3.14\).)
21.195 m2
Volume of a hemisphere is \(V = \\dfrac{2}{3}\\pi r^{3}\).
Substitute \(V = 7.065\) and \(\\pi = 3.14\): \(7.065 = \\dfrac{2}{3}(3.14) r^{3}\), giving \(r^{3} = \\dfrac{7.065 \\times 3}{2 \\times 3.14} = 3.375\), so \(r = 1.5\\ \\text{m}\).
Total surface area of a hemispherical bowl is \(3\\pi r^{2}\).
\(3 \\times 3.14 \\times (1.5)^{2} = 3 \\times 3.14 \\times 2.25 = 21.195\\ \\text{m}^{2}\).
Hence, the total surface area is 21.195 m2.
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Important Questions from 3-D Mensuration
Find the total surface area of a closed cylinder having a base radius of 70 m and a height of 110 m. [Use π = \(22\over7\)]
- The curved surface area of a cylinder is half of its total surface area. If its height is 195 cm, then find its diameter (in cm).
The diameter of the base and slant height of a right circular cone are 30 cm and 113 cm, respectively. Find the volume (in cm³) of the given cone.
(Use $\pi = \frac{22}{7}$)- A cylindrical rod has an curved surface area of $4,900 \text{ cm}^2$. If the length of the rod is 97 cm, then the radius (in cm) of the rod, correct to two places of decimal, is:
Take $\pi = \frac{22}{7}$ - The volume (in $m^3$) of a cube, each of whose edges is 42 m, is:
