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# maths

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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

#2
+98061
+10

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

[8000pi + 1500]  = pi * (20)^2 * h

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

The water rises  ≈  1.2 cm

Jan 19, 2017

#1
+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
+98061
+10

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

[8000pi + 1500]  = pi * (20)^2 * h

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

The water rises  ≈  1.2 cm

CPhill Jan 19, 2017