Question

A sample of an unknown material weighs 290 N in air and 190N when immersed in alcohol of specific gravity 0.700.

(a) What is the volume of the material?

Answer 1

The volume of the unknown material is approximately
`\( 0.0144 \, \text{m}^3 \).`

To find the volume of the unknown material, we can use Archimedes' principle, which states that the buoyant force acting on a submerged object is equal to the weight of the fluid it displaces.

The weight loss in the fluid (in this case, alcohol) is equal to the buoyant force acting on the material when it is immersed. The weight loss is the difference between the weight in air and the weight in the fluid.

Let's denote the weight in air as
`\( W_{\text{air}} = 290 \, \text{N} \)` and the weight in the fluid as
`\( W_{\text{fluid}} = 190 \, \text{N} \).`

The weight loss
`(\( \Delta W \))` can be calculated as follows:

`\[ \Delta W = W_{\text{air}} - W_{\text{fluid}} \]`

`\[ \Delta W = 290 \, \text{N} - 190 \, \text{N} \]`

`\[ \Delta W = 100 \, \text{N} \]`

Now, the buoyant force acting on the material is equal to the weight loss, and it can be calculated using the formula:

`\[ \text{Buoyant force} = \text{Weight loss} = \Delta W \]`

Now, the buoyant force can also be expressed in terms of the volume of the material
`(\( V \)),` the density of the fluid
`(\( \rho_{\text{fluid}} \)),`and the acceleration due to gravity
`(\( g \)):`

`\[ \text{Buoyant force} = \rho_{\text{fluid}} \cdot V \cdot g \]`

Since we're dealing with the weight loss, we can rewrite this as:

`\[ \Delta W = \rho_{\text{fluid}} \cdot V \cdot g \]`

Now, we need to find the density of the fluid
`(\( \rho_{\text{fluid}} \))`. The specific gravity
`(\( SG \))` is the ratio of the density of the fluid to the density of water. Since the density of water is approximately
`\( 1000 \, \text{kg/m}^3 \),` we can find
`\( \rho_{\text{fluid}} \)`as follows:

`\[ \rho_{\text{fluid}} = SG \cdot \rho_{\text{water}} \]`

`\[ \rho_{\text{fluid}} = 0.700 \cdot 1000 \, \text{kg/m}^3 \]`

`\[ \rho_{\text{fluid}} = 700 \, \text{kg/m}^3 \]`

Now, we can rearrange the buoyant force equation to solve for the volume (\( V \)):

`\[ V = \frac{\Delta W}{\rho_{\text{fluid}} \cdot g} \]`

Substitute the values:

`\[ V = \frac{100 \, \text{N}}{700 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2} \]`

`\[ V \approx 0.0144 \, \text{m}^3 \]`

So, the volume of the unknown material is approximately
`\( 0.0144 \, \text{m}^3 \).`