Question 1205029
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A heavy cube of side 8 cm is placed vertically in a cylindrical tank of radius 7 cm which contains water.
Calculate the rise in the water level if the original depth of water was:
    (a)   10 cm
    (b)   2 cm
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(a)  In this case, the entire cube is wholly submerged into the water in the tank.

     The water level rises over the entire base area of the cylindrical tank.

     The raised water represents the volume of the displaced water in the tank by the solid cube.


             Use the law of the volume of water conservation.


     To find the rise for question (a), we should divide the volume of the cube,  {{{8^3}}} cm^3

     by the area of the base of the cylinder

         the rise = {{{8^3/(pi*r^2)}}} = {{{8^3/(3.14159*7^2)}}} = 3.326 cm.


     <U>ANSWER to question (a)</U>.  The rise of the water level is 3.326 cm.




(b)  In this case, the cube is only partly submerged into the water in the tank.

     The water level rises over the part of the base area of the cylindrical tank.

     This part of the area where the water rises is the entire area of the base of the tank 
     minus the area of the base of the cube, which is only partially submerged.


             Use the law of the volume of water conservation.


     To find the new level of water for question (b), we should divide the volume of the water in the tank,  
 
     which is  {{{pi*r^2*2}}} cm^3, by the (area of the tank base MINUS area of the cube base)


         the new level = {{{(3.14159*7^2*2)/(3.14159*7^2-8^2)}}} = 3.4232 cm  (rounded).


     Thus the raise of the water level is  3.4232 - 2 = 1.4232 centimeters.

     The new level is still lower than the height of the cube, so our calculations make sense.


     <U>ANSWER to question (b)</U>.  The rise of the water level is 3.4232 cm.
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Solved.