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A cylindrical tank with diameter 20in. Is half filled with water. How much will the water level in the tank rise if you place a metallic ball with radius 4in. In the tank? What causes the water level in the tank to rise? Which volume formula should be used?

Guest Apr 8, 2015

Best Answer 

 #1
avatar+26406 
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The ball will displace its own volume of water, namely (4/3)*pi*4^3 cubic inches.

This will appear as an increase in height, h, of the water in the tank such that pi*(20^2)/4*h =  (4/3)*pi*4^3

or h = (4/3)*4^4/20^2 in.

 

$${\mathtt{h}} = {\frac{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{4}}}}{{{\mathtt{20}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{h}} = {\mathtt{0.853\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

 

so h ≈ 0.853 in.

 

 

(Note: The cross-sectional area of the tank is given by pi*diameter^2/4).

Alan  Apr 8, 2015
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1+0 Answers

 #1
avatar+26406 
+5
Best Answer

The ball will displace its own volume of water, namely (4/3)*pi*4^3 cubic inches.

This will appear as an increase in height, h, of the water in the tank such that pi*(20^2)/4*h =  (4/3)*pi*4^3

or h = (4/3)*4^4/20^2 in.

 

$${\mathtt{h}} = {\frac{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{4}}}}{{{\mathtt{20}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{h}} = {\mathtt{0.853\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

 

so h ≈ 0.853 in.

 

 

(Note: The cross-sectional area of the tank is given by pi*diameter^2/4).

Alan  Apr 8, 2015

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