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Found in: Page 170

### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

# Here are two examples of floating objects: (a) A block of wood 20 cm long by 10 cm wide by 6 cm high has a density of 0.7 g/cm3 and floats in water (whose density is 1.0 g/cm3). How far below the surface of water is the bottom of the block. Explain your reasoning?

The bottom of the block is 4.2 cm below the surface of water.

See the step by step solution

## Step 1: Identification of the given data

The given data is listed below as-

• The dimensions of the wooden block are, $20 \mathrm{cm}×10 \mathrm{cm}×6 \mathrm{cm}$
• The density of the wood is, ${\rho }_{wood}=0.7 \frac{\mathrm{g}}{{\mathrm{cm}}^{3}}$
• The density of water is, ${\rho }_{water}=1 \frac{\mathrm{g}}{{\mathrm{cm}}^{3}}$

## Step 2: Significance of the displacement of water

The displacement of water occurs when a wooden block floats on water. The mass of the water is equal to the displacement of water.

The concept of the displacement of water gives the depth of block.

## Step 3: Determination of the depth of block

The equation of mass of the block is given by displacement of the water

The density is given by-

$\rho =\frac{mass}{Volume}\phantom{\rule{0ex}{0ex}}mass=\rho ×Volume$

Now, mass of water= mass of wood

${m}_{water}={m}_{wood}\phantom{\rule{0ex}{0ex}}={\rho }_{wood}.{V}_{wood}$

For ${\rho }_{wood}=0.7\frac{g}{c{m}^{3}}$and

${V}_{wood}=0.20×0.10×0.06\phantom{\rule{0ex}{0ex}}=0.0012{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}{m}_{wood}=0.7×{10}^{3}\left(\frac{\mathrm{kg}}{{\mathrm{m}}^{3}}\right)×0.0012\left({\mathrm{m}}^{3}\right)\phantom{\rule{0ex}{0ex}}=0.84\mathrm{kg}$

Now, find the volume of water by density of water

role="math" localid="1658901709608"

The displaced water takes the shape of the cube same as the shape of a block with an area,

$A=0.20×0.10×h$

The height is the depth of the block which is given by:

$h=\frac{{V}_{water}}{A}\phantom{\rule{0ex}{0ex}}=\frac{0.84×{10}^{-3}}{0.20×0.10m}\phantom{\rule{0ex}{0ex}}=0.042\mathrm{m}\phantom{\rule{0ex}{0ex}}=4.2\mathrm{cm}$

Thus, the block is 4.2 cm below the surface of water .