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(C) The volume of the sphere with radius $r_s = 3 \, cm$ is given by $V_s = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \pi (3)^3 = 36 \pi \, cm^3$.When the s

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$\frac{9}{4}$

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(C) The volume of the sphere with radius $r_s = 3 \, cm$ is given by $V_s = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \pi (3)^3 = 36 \pi \, cm^3$.When the sphere is fully immersed,the volume of water displaced is equal to the volume of the sphere.Let the rise in the water level in the cylindrical vessel be $h$. The volume of the displaced water in the cylinder is $V_w = A_c 	imes h$,where $A_c$ is the cross-sectional area of the cylinder.The radius of the cylinder is $r_c = 4 \, cm$,so $A_c = \pi r_c^2 = \pi (4)^2 = 16 \pi \, cm^2$.Equating the volumes: $16 \pi 	imes h = 36 \pi$.Solving for $h$: $h = \frac{36 \pi}{16 \pi} = \frac{9}{4} \, cm$.

$A$ cylindrical vessel of radius $4 \, cm$ contains water. $A$ solid sphere of radius $3 \, cm$ is lowered into the water until it is completely immersed. The water level in the vessel will rise by (in $cm$)

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