Question

A glass ball of density 2600 kg\ m^ 3 is coated with a thick layer of was of density 800 kg\m^3. If the combination floats in water completely submersed find the ratio of the volume of wax with that of the glass ball

Answer 1

Answer:8:1

Explanation:

Answer 2

The ratio of the volume of wax with that of the glass ball is 8.

Explanation:

Given that,

Density of glass ball = 2600 kg/m³

Density of wax =800 kg/m³

Let $\rho_{g}$ and $V_{g}$ be the density and volume of the glass.

Let $\rho_{w}$ and $V_{w}$ be the density and volume of the water.

Let $\rho_{x}$ and $V_{x}$ be the density and volume of the wax.

We need to calculate the ratio of the volume of wax with that of the glass ball

Using buoyant force

$F_{g}+F_{x}=F_{w}$

$(m_{g}+m_{x})g=m_{w}g$

$(\rho_{g}V_{g}+\rho_{x}V_{x})g=\rho_{w}V_{w}g$

$\rho_{g}V_{g}+\rho_{x}V_{x}=\rho_{w}V_{w}$

We know ,

$V_{w}=V_{g}+V_{x}$

$\rho_{g}V_{g}+\rho_{x}V_{x}=\rho_{w}(V_{g}+V_{x})$

$(\rho_{g}-\rho_{w})V_{g}=(\rho_{w}-\rho_{x})V_{x}$

$\dfrac{V_{g}}{V_{x}}=\dfrac{\rho_{w}-\rho_{x}}{\rho_{g}-\rho_{w}}$

Put the value into the formula

$\dfrac{V_{g}}{V_{x}}=\dfrac{1000-800}{2600-1000}$

$\dfrac{V_{g}}{V_{x}}=\dfrac{1}{8}$

$\dfrac{V_{x}}{V_{g}}=8$

Hence, The ratio of the volume of wax with that of the glass ball is 8.

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Topic : Density and volume

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