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A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the balls ?

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Correct Answer - 9 cm
Volume of water displaced `= {pi xx (12)^(2) xx 6.75} cm^(3)`
Volume of the ball `= (4)/(3) pi r^(3)`
`:. (4)/(3) pi r^(3) = pi xx 144 xx 6.75 rArr r^(3) = (144 xx 6.75 xx 3)/(4) = 729 = (9)^(3) rArr r= 9 cm`
`:.` radius of the ball is 9 cm

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