Volume is a physical quantity that characterizes the property of an object to occupy part of the space.

The unit of volume in the International System of Units (\(SI\)) is the cubic meter (\(м^3\)). There are multiple and divided units of volume: $$1dm^3 = 0.1m х 0.1m х 0.1m = 0.001m^3$$ $$1cm^3 = 0.001dm^3 = 0.000 001m^3$$ Litre is a non-systemic unit of volume (\(L\)): $$1L = 1dm^3$$ $$1mL = 1cm^3$$

Volumes of solids, liquids and bulk materials can be determined by direct measurements using a measuring cylinder, a measuring vessel.

To measure the volume of liquid or bulk material with a measuring cylinder, we need:

To measure the volume of a solid with a measuring vessel, we need:

The volume of water \(V\) displaced by the object is equal to the volume of the object. If the object has the correct geometric shape, its volume can be determined by indirect measurements: measure the linear dimensions of the object using a ruler and calculate the volume of the object according to the corresponding mathematical formulas. For example, the volume \(V\) of an object having the shape of a rectangular parallelepiped is calculated by the formula: $$V = ldh$$ where \(l\) – is the length; \(d\) – is the width; \(h\) – is the height of the object.

The absolute error, during one direct measurement, is equal to the value of the scale division of the measuring device.

Subject. Measuring of volumes of solid objects, liquids and bulk materials.

Target: to measure the volumes of solids (regular and irregular forms), liquids and loose materials.

Equipment: measuring cylinder, graded glass, ruler (treadmill), threads, an object of irregular geometric form (object 1), object of parallelepiped form (object 2), water, millet, sand.

The results of the measurements I’m writing down into the table:

Table 1

Table 2

1. I’m determining the value of the scale division of the measuring cylinder and the value of the scale division of the ruler: $$ C_{m.c.} = \frac{\FormInput{name}}{\FormInput{name}} = \ \FormInput{name}. \qquad C_{rul} = \frac{\FormInput{name}}{\FormInput{name}} = \ \FormInput{name}. $$

2. Using a measuring vessel I’m measuring the volume of bulk materials:

3. I’m measuring the volume of a solid object of irregular geometric shape (object 1) by direct measurements: \( V_{Ob1} = V_2 - V_1; V_{Ob1} = \FormInput{name} - \FormInput{name} = \FormInput{name} \).

4. I’m measuring the volume of a solid object of irregular geometric shape (object 2) by direct measurements: \( V_{Ob2} = V_2 - V_1; V_{Ob2} = \FormInput{name} - \FormInput{name} = \FormInput{name} \).

5. I’m measuring the volume of a solid object of the correct geometric shape (object 2) by indirect measurements: \( V_{Ob2} = ldh; l = \FormInput{name}; d =\FormInput{name} ; h =\FormInput{name} ; \) \( V_{Ob2} = \FormInput{name} \FormInput{name} \FormInput{name} = \FormInput{name} \).

6. I’m estimating the absolute and relative errors of measurement results: \(V = V_{meas} ± \Delta V; where \Delta V \) - an absolute error; Sand: \( V_{sand} = \FormInput{name} ± \FormInput{name}\); Millet: \(V_{millet} = \FormInput{name} ± \FormInput{name}\); Water: \(V_{water} = \FormInput{name} ± \FormInput{name} \).

7. Analyzing the results of the experiment:

Notes

Do not worry and wait a bit. The results of the work are sent to the specified address.