Answer- The above question is from the chapter 'Surface Areas and Volumes'.
Concepts used: 1) Melting a solid and then recasting it into another solid form equals both of the volumes of the solids.
Volume of cylinder = Volume of sphere
2) Volume of cylinder = πr²h
3) Volume of sphere =
4) Diameter (d) = 2r
Given question:A metallic solid cylinder of height 1 m and base radius 0.75 m is melted and cast into a solid sphere. The diameter of the sphere (in m) so formed is _____ .
Solution:Volume of cylinder = Volume of sphere
πr²h =
0.75 × 0.75 × 1 =
= R³
R³ =
R³ = 0.421875
R = ∛0.421875
R = 0.75 m
d = 2R
d = 2 × 0.75
d = 1.5 m
∴ The diameter of the sphere (in m) so formed is 1.5 m .
Answers & Comments
Answer- The above question is from the chapter 'Surface Areas and Volumes'.
Concepts used: 1) Melting a solid and then recasting it into another solid form equals both of the volumes of the solids.
Volume of cylinder = Volume of sphere
2) Volume of cylinder = πr²h
3) Volume of sphere =![\frac{4}{3} \pi r^{3} \frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20)
4) Diameter (d) = 2r
Given question: A metallic solid cylinder of height 1 m and base radius 0.75 m is melted and cast into a solid sphere. The diameter of the sphere (in m) so formed is _____ .
Solution: Volume of cylinder = Volume of sphere
πr²h =![\frac{4}{3} \pi R^{3} \frac{4}{3} \pi R^{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R%5E%7B3%7D%20)
0.75 × 0.75 × 1 =![\frac{4}{3} \times R^{3} \frac{4}{3} \times R^{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%20R%5E%7B3%7D%20)
R³ =![\frac{16875}{40000} \frac{16875}{40000}](https://tex.z-dn.net/?f=%20%5Cfrac%7B16875%7D%7B40000%7D%20)
R³ = 0.421875
R = ∛0.421875
R = 0.75 m
d = 2R
d = 2 × 0.75
d = 1.5 m
∴ The diameter of the sphere (in m) so formed is 1.5 m .