Question

A 10cm×10cm×10cm wood block with a density of 730 kg/m3 floats in water.

What is the distance from the top of the block to the water if the water is fresh?

If it's seawater? Suppose that *ρ*=1030kg/m3.

Answer #1

Density of fresh water, = 1000 kg/m^3

Density of sea water, = 1030 kg/m^3 (Given)

Volume of the wooden block, V = 10 cm x 10 cm x 10 cm = 1000 cm^3 = 1000 x 10^-6 m^3 = 10^-3 m^3

So, mass of wooden block, m = V*730 kg = 10^-3 * 730 kg = 0.730 kg

(a) When the wooden block floats in fresh water -

Suppose h meter is the distance of the top from water.

So, volume of displaced water = 0.10 x 0.10 x h m^3

Apply Arhimedes principle -

0.10 x 0.10 x (0.10 - h) x 1000 = 0.730

=> 0.10 - h = 0.730 / (0.10 x 0.10 x 1000) = 0.073

=> h = 0.10 - 0.073 = 0.027 m = 2.7 cm

(b) When the block floats in sea water -

We have -

0.10 x 0.10 x (0.10 - h') x 1030 = 0.730

=> 0.10 - h' = 0.730 / (0.10 x 0.10 x 1030) = 0.0709

=> h' = 0.10 - 0.0709 = 0.0291 m = 2.91 cm (Answer)

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