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A right circular cylinder is partially filled with water. Two iron spherical balls are completely immersed in the water so that the height of the water in the cylinder rises by 4 cm. If the radius of one ball is half of the other and the diameter of the cylinder is 18 cm, then the radii of the spherical balls are
6 cm and 12 cm
4 cm and 8 cm
3 cm and 6 cm
2 cm and 4 cm
total volume of 2 balls = volume of cylinder of height = 4 cm 4/3πr^31+43πr^32=π9^2×4 r1=2r2 given 4/3π(8r^32+r^32)=π×9×9×4 9r^32=9×9×3 r2=3 r1=2r2=2×3=6
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error
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