+0  
 
0
398
2
avatar+502 

A cylindrical can of internal radius 20 cm stands upright on a flat surface. It contains water to a depth of 20 cm. Calculate the rise in the level of the water when a brick of brick of volume 1500cm is immersed in the water.

 Jan 19, 2017

Best Answer 

 #2
avatar+98061 
+10

Original volume....  =   pi * r^2 * h  =  pi * (20)^2 * (20)   =  8000pi cm^3

 

Adding  1500xm^3...we have that

 

[8000pi + 1500]  = pi * (20)^2 * h   

 

[8000pi + 1500] / [ pi *(20)^2  ]  = h  ≈  21.19 cm

 

The water rises  ≈  1.2 cm

 

 

 

cool cool cool

 Jan 19, 2017
 #1
avatar+99246 
+5

 

Hi Rauhan

A cylindrical can of internal radius 20 cm stands upright on a flat surface. It contains water to a depth of 20 cm. Calculate the rise in the level of the water when a brick of brick of volume 1500cm3  is immersed in the water.

 

A really important thing to remember is that  1cm^3 holds 1mL of water.

so

a 1500cm^3 brick will displace 1500ml of water.

The original height of the water is irrelavant.  Unless the water overflows.

 

How much height is this?

\(\pi r^2 h=Volume\\ \pi *20^2*h=1500\\ h=\frac{1500}{400\pi}\\ h=\frac{15}{4\pi}\;cm\)

 

h=1.193662073189215 cm

 

The water will rise approx 1.2cm

 Jan 19, 2017
 #2
avatar+98061 
+10
Best Answer

Original volume....  =   pi * r^2 * h  =  pi * (20)^2 * (20)   =  8000pi cm^3

 

Adding  1500xm^3...we have that

 

[8000pi + 1500]  = pi * (20)^2 * h   

 

[8000pi + 1500] / [ pi *(20)^2  ]  = h  ≈  21.19 cm

 

The water rises  ≈  1.2 cm

 

 

 

cool cool cool

CPhill Jan 19, 2017

29 Online Users

avatar
avatar
avatar
avatar
avatar
avatar