A tank measuring $22 * 9 * 10$ . Its half full of water. A solid metal sphere with radius 3cm is placed into the tank. Assuming that the sphere sinks to the bottom of the tank calculate the rise in height of water in cms

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Solution

Volume of tank = volume of sphere
$/ 22 \times 9 \times h = \frac{4}{/ 3} \times \frac{/ 22}{7} \times (3)^{/ 3}$
$h = \frac{4 \times / 3^{2}}{7 \times 9}$
$h = \frac{4}{7} c m$
$h = 0.571 c m$

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A tank measuring 22∗9∗10. Its half full of water. A solid metal sphere with radius 3cm is placed into the tank. Assuming that the sphere sinks to the bottom of the tank calculate the rise in height of water in cms

Volume of tank = volume of sphere /22×9×h=4/3×/227×(3)/3 h=4×/327×9 h=47cm h=0.571cm

A tank measuring $22 * 9 * 10$ . Its half full of water. A solid metal sphere with radius 3cm is placed into the tank. Assuming that the sphere sinks to the bottom of the tank calculate the rise in height of water in cms

Volume of tank = volume of sphere
$/ 22 \times 9 \times h = \frac{4}{/ 3} \times \frac{/ 22}{7} \times (3)^{/ 3}$
$h = \frac{4 \times / 3^{2}}{7 \times 9}$
$h = \frac{4}{7} c m$
$h = 0.571 c m$