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A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball. |
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Answer» Given, the radius of the cylindrical tube, r = 12 cm Level of water raised in tube, h = 6.75 cm Volume of cylinder = πr2 h = π (12)2 × 6.75 cm3 = π x 122 x 6.75 cm3----------(i) let r be the radius of a spherical shell balls volume of the sphere = \(\frac{4}{3}πr^3\)-------(ii) ∵ volume of cylinder = volume of spherical ball ∴ π x 122 x 6.75 = \(\frac{4}{3}πr^3\) ⇒ r3 = \(\frac{{12}^2\times{6.75}}{\frac{4}{3}\times{π}}\) ⇒ r = 9 cm ∴ radius of spherical ball, r = 9 cm |
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