A cylindrical can with a horizontal base and | vedclass

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(C) Let the radius of the base of the cylinder be $r = 3.5 \, cm$.Since the sphere fits exactly into the can,the radius of the sphere is also $r = 3.5

Vedclass · Accepted Answer

$\frac{7}{3} \, cm$

Vedclass · Answer

(C) Let the radius of the base of the cylinder be $r = 3.5 \, cm$.Since the sphere fits exactly into the can,the radius of the sphere is also $r = 3.5 \, cm$,and the height of the water level after placing the sphere is $H = 2r = 7 \, cm$.The volume of the water in the can is equal to the total volume of the cylinder up to height $H$ minus the volume of the sphere.Volume of water $= \pi r^2 (2r) - \frac{4}{3} \pi r^3 = 2 \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$.Let the initial depth of the water be $h$. The volume of this water is $\pi r^2 h$.Equating the volumes: $\pi r^2 h = \frac{2}{3} \pi r^3$.Solving for $h$: $h = \frac{2}{3} r = \frac{2}{3} 	imes 3.5 = \frac{7}{3} \, cm$.

$A$ cylindrical can with a horizontal base and an internal radius of $3.5 \, cm$ contains sufficient water so that when a solid sphere is placed inside,the water just covers the sphere. The sphere fits exactly into the can. The depth of the water in the can before the sphere was added is:

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