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For a National Board Exam Review:

A floating cylinder 8cm in diameter and weighing 950 grams is placed in a cylindrical container 20cm in diameter and partially full of water. The increase in the depth of water in the container due to placing the float in it is:

Answer is 3cm

So I try:

$${ P_1 = \frac{0.950kg * 9.81 \frac{m}{s^2}}{ \frac{\pi}{4} * { (20cm * \frac{1m}{100cm}})^2 } = 296.649 \frac{N}{m^2}}$$

$${ \gamma = 9.81 \frac{kN}{m^3} }$$

$${ P_2 = P_{atm} = 101325 \frac{N}{m^2}}$$

$${ P_1 + \gamma h = P_2 }$$

$${h = 10.29m }$$

What is my error? Am I using wrong approach? Wrong formula?

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  • $\begingroup$ @ james: Are the units in your question correct? Placing a float that is 8m in diameter into a 20 cm cylinder can't be done. Also, in your maths, what does 9081 m/s2 refer to? Also, a diagram would help. Are the float & cylinder siting on the circular base or on the elongated side? $\endgroup$
    – Fred
    Aug 23, 2015 at 3:13
  • $\begingroup$ sorry typo errors. $\endgroup$
    – james
    Aug 23, 2015 at 3:23

2 Answers 2

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The first part of your maths is OK.

$P_1$ = $296.649$ $Pa$

Now, pressure at the base of a fluid is:

$P$ $=$ $\rho gh$

Rearrange the equation & $h$ $=$ $P/\rho g$

Using your numbers & the fact that the density of water is $1$ $g/cm^3$ or $1000$ $kg/m^3$,

$h$ $=$ $296.649/(1000)(9.81)$ = $0.0302$ $m$ = $3$ $cm$

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  • $\begingroup$ why use the diameter of the larger cylinder instead of the smaller one? $\endgroup$
    – james
    Aug 23, 2015 at 3:49
  • $\begingroup$ The weight of the float acts as a surcharge to the weight of water in the large cylinder, this will increase the pressure on the base of the large cylinder (the base of the cylinder must bear more weight). Because water is incompressible the affect of the weight of the float is to force the water to move up the sides of the large cylinder (the only direction the water can move), increasing the depth/height of the water. $\endgroup$
    – Fred
    Aug 23, 2015 at 4:16
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20cm diameter gives a cross sectional area of $\pi r^2$ or 314.15 cm2.

8cm diameter uses the same formula for a cross sectional area of 50.26 cm2.

The free space around the float in which the water may rise is the difference in cross sectional areas, 314.15-50.26 = 263.886 cm2.

The float, uh, floats, so the water that is displaced must weigh the same as the float, or 950g.

950g / 1 g/cm3 = 950 cm3.

950cm3 / 263.886cm2 = 3.6cm.

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