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A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm. Find the radius of the ball. (Use π =\(\frac{22}{7}\)). |
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Answer» Volume of a sphere = \(\frac{4}{3}\)πr3 Given, A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm. Volume of water displaced = Volume of the iron ball ⇒ \(\frac{4}{3}\)πr3 = π x 162 x 9 ⇒ r3 = 1728 ⇒ r = 12 cm |
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