{"id":8866,"date":"2021-12-05T21:16:26","date_gmt":"2021-12-05T15:46:26","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8866"},"modified":"2022-01-16T17:11:35","modified_gmt":"2022-01-16T11:41:35","slug":"volume-of-sphere-and-hemisphere","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/volume-of-sphere-and-hemisphere\/","title":{"rendered":"Volume of Sphere and Hemisphere – Formula and Examples"},"content":{"rendered":"
Here you will learn what is the formula for volume of sphere and hemisphere with examples based on it.<\/p>\n
Let’s begin – <\/p>\n
\nVolume of sphere = \\({4\\over 3}\\pi r^3\\)<\/p>\n
where r is the radius of sphere.<\/p>\n<\/blockquote>\n
(ii) Volume of Hemisphere Formula<\/strong><\/h4>\n
\nVolume of Hemisphere = \\({2\\over 3}\\pi r^3\\)<\/p>\n
where r is the radius of hemisphere.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Find the volume of a sphere of radius 11.2 cm.<\/span> <\/p>\n
Solution<\/span><\/strong> : Here, radius = 11.2 cm<\/p>\n
Volume of Sphere = \\({4\\over 3}\\pi r^3\\)<\/p>\n
= \\({4\\over 3} \\times {22\\over 7} \\times (11.3)^3\\) = 5888.72 \\(cm^3\\)<\/p>\n
Example<\/span><\/strong> : A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?<\/span> <\/p>\n
Solution<\/span><\/strong> : Here, radius = 3.5 cm<\/p>\n
The volume of water the bowl can contain <\/span>= \\({2\\over 3}\\pi r^3\\)<\/p>\n
= \\({2\\over 3} \\times {22\\over 7} \\times (3.5)^3\\) = 89.8 \\(cm^3\\)<\/p>\n
Example<\/span><\/strong> : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per \\(cm^3\\)<\/span>, find the mass of the shot-putt.<\/span> <\/p>\n
Solution<\/span><\/strong> : Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.<\/span> <\/p>\n
Now, Volume of the Sphere <\/span>= \\({4\\over 3}\\pi r^3\\)<\/p>\n
= \\({4\\over 3} \\times {22\\over 7} \\times (4.9)^3\\) = 493 \\(cm^3\\)<\/p>\n
Further mass of 1 \\(cm^3\\) of metal is 7.8 g.<\/p>\n
Therefore, mass of the shot-putt = 7.8 \\(\\times\\) 493 = 3845.44 gm = 3.85 kg<\/p>\n\n\n