SOLUTION: A heavy cube of side 8cm is placed vertically in a cylindrical tank of radius 7cm which contains water. Calculate the rise in the water level if the original depth of water was:

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Question 1205029: A heavy cube of side 8cm is placed vertically in a cylindrical tank of radius 7cm which contains water.
Calculate the rise in the water level if the original depth of water was:
a) 10cm
b) 2cm
(leave your answers in fractional form)
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
heavy cube of side 8cm is placed vertically in a cylindrical tank of radius 7cm which contains water.
Calculate the rise in the water level if the original depth of water was:
a) 10cm
b) 2cm
If the height of water is above length of cube then increase in height will be same for any height of water.
Volume of cube = l^3 = 8^3= 512 cm^3
Volume of cylinder per cm height = pi*7^2 = 49 *pi
Increase in height = 512/(49*pi) = 3.326 cm
When the height of water is 2 cm
The cube will sink only 2 cm in height
Volume of sunk portion = 8*8*2=128 cm^3
Increase in height = 128/949*pi) =0.424 cm
you round it off as desired or convert to fraction









Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
A heavy cube of side 8cm is placed vertically in a cylindrical tank of radius 7cm which contains water.
Calculate the rise in the water level if the original depth of water was:
a) 10cm
b) 2cm
~~~~~~~~~~~~~~~~~


        Regarding this problem and its solution in the post by @mananth,  I'd like to make two notices.

        First notice to  (1)  is that the solution by  @mananth is correct  ONLY  IF  the cylindrical tank
        has enough height in order for the displaced water does not flow out of the cylinder.
        Otherwise, the rise in the water level will be limited by the height of the cylinder.
        The problem says nothing about it,  and it is the problem's  FAULT.

        Second notice is that the solution by  @mananth for part  2)  is  FATALLY  INCORRECT.
        See my correct solution below.


                                Solution to part 2 

The original volume of the water in the tank is pi%2Ar%5E2%2Ah = 3.14159%2A7%5E2%2A2 = 307.87582 cm^3.


The area of the horizontal section of the tank, occupied by the water after placing the solid cube
is  pi%2Ar%5E2-8%5E2 = 3.14159%2A7%5E2-8%5E2 = 89.93791 cm^2.


The final level of the water is then  307.87582%2F89.93791 = 3.423204075 cm,  or 3.423 cm after rounding.


Thus the rise of the water level is the difference  3.423 - 2 = 1.423 cm.

Solved correctly.

----------------------

Again,  the solution by  @mananth for part  2)  is  CONCEPTUALLY  INCORRECT,
since he/she incorrectly determines the volume of the displaced water,
which is of fundamental importance in this problem.