Two solid hemispheres of radii 6 cm and 8 cm respectively are melted and recast into a single hemisphere. What is the approximate radius of the new hemisphere formed?
9 cm
Volume is conserved on melting and recasting. Each hemisphere has volume \(\dfrac{2}{3}\pi r^{3}\), so the total \(\dfrac{2}{3}\pi\) factor cancels, and we just add cubes of radii.
Step 1 — Sum of cubes:
\(R^{3} = 6^{3} + 8^{3} = 216 + 512 = 728\)
Step 2 — Take the cube root:
\(R = \sqrt[3]{728}\)
Since \(9^{3} = 729\), we have \(R \approx 8.999 \approx 9 \text{ cm}\).
Hence the new hemisphere's radius is approximately 9 cm. The answer is option A.
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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:
