Sunita had a hemispherical bowl of radius 𝑟.She made a conical vessel of radius 𝑟 with a tinsheet.(i) find the height of the conical vessel so that it canhold the water same as that of the hemisphericalbowl.(ii) If the radius of the cone formed in the above part is 14 cm, then find how much sheet isused?(iii) If the height of the conical vessel is doubled, how much more water can it hold than thehemispherical bowl?
Question
Sunita had a hemispherical bowl of radius 𝑟.She made a conical vessel of radius 𝑟 with a tinsheet.(i) find the height of the conical vessel so that it canhold the water same as that of the hemisphericalbowl.(ii) If the radius of the cone formed in the above part is 14 cm, then find how much sheet isused?(iii) If the height of the conical vessel is doubled, how much more water can it hold than thehemispherical bowl?
Solution
(i) The volume of a hemisphere is given by the formula (2/3)πr³ and the volume of a cone is given by the formula (1/3)πr²h. If the conical vessel can hold the same amount of water as the hemispherical bowl, then their volumes are equal. Therefore, we can set the two formulas equal to each other and solve for h (the height of the cone):
(2/3)πr³ = (1/3)πr²h
Solving for h, we get h = 2r.
(ii) The surface area of a cone (excluding the base) is given by the formula πrl, where l is the slant height of the cone. Since the cone is formed from a hemisphere, the slant height of the cone is equal to the radius of the hemisphere. Therefore, the amount of sheet used is πr².
If the radius of the cone is 14 cm, then the amount of sheet used is π(14 cm)² = 616 cm².
(iii) If the height of the conical vessel is doubled, its new volume is (1/3)πr²(2h) = (2/3)πr²h. Since h = 2r, the new volume is (2/3)πr²(2*2r) = (8/3)πr³.
The difference in volume between the new conical vessel and the hemispherical bowl is (8/3)πr³ - (2/3)πr³ = (6/3)πr³ = 2πr³. Therefore, the conical vessel can hold 2πr³ more water than the hemispherical bowl when its height is doubled.
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