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(D) The volume of the submerged cuboid is $V = 22\, cm 	imes 16\, cm 	imes 7\, cm = 2464\, cm^3$.When the cuboid is submerged,the volume of water di

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$4$

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(D) The volume of the submerged cuboid is $V = 22\, cm 	imes 16\, cm 	imes 7\, cm = 2464\, cm^3$.When the cuboid is submerged,the volume of water displaced is equal to the volume of the cuboid.Let the rise in the water level be $h\, cm$.The volume of the displaced water in the cylindrical container is given by $\pi r^2 h$,where $r = 14\, cm$.So,$\pi r^2 h = 2464$.Using $\pi = \frac{22}{7}$,we have $\frac{22}{7} 	imes 14 	imes 14 	imes h = 2464$.$22 	imes 2 	imes 14 	imes h = 2464$.$616 	imes h = 2464$.$h = \frac{2464}{616} = 4\, cm$.Therefore,the rise in the water level is $4\, cm$.

$A$ cylindrical container of radius $14\, cm$ is partially filled with water. $A$ solid cuboid of iron with dimensions $22\, cm \times 16\, cm \times 7\, cm$ is submerged in it. Find the rise in the level of the water when the solid is completely submerged. (in $cm$)

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