Tag: measurement of area and volume

To find the volume of a regular shaped object ,first its dimensions are measured using either a metre scale, a vernier callipers or a screw gauge and then volume is calculated using corresponding relations.

Measuring flask consists of a long narrow neck provided with a stopper. Different capacities (50 ml, 100 ml, 250 ml etc) of standard measuring flasks are available.

Option D is the definition of Volume. It is only a physical property of any object and has nothing to do with the chemical properties. Since density is mass/ volume, it is necessary to find density.

Volume of a liquid or an irregular solid object is measured with the help of burettes, measuring flasks and measuring cylinders.

A liquid or an irregular shaped object does not have a geometric shape. So, by measuring its certain dimensions, volume may not be calculated. Hence a meter scale can not be used to measure the volume of a liquid or an irregular shaped object.

Initial volume level   $V _i = 55.2 \ ml$
Final volume level   $V _f = 75.3 \ ml$
Thus, volume of ring    $V = V _f-V _i$
$\implies \ V = 75.3-55.2 = 20.1 \ ml$

Diameter of cylindrical vessel  $d = 1.06 \ m$
Height of cylindrical vessel  $h = 7.2 \ m$
Volume   $V = \dfrac{\pi d^2 h}{4}$
$\implies \ V = \dfrac{\pi\times (1.06)^2\times 7.2}{4} = 6.354 \ m^3$
Since the number $7.2$ has two significant figures and $1.06$ has three significant figures.
So the final answer must have 2 significant figures.
Thus volume of cylinder   $V = 6.3 \ m^3$

The significant figures are the number of digits in a value, that are significant. In $1.06m$, the number of significant figures is 3. In the answer also we must maintain 3 significant figures only. Hence option B is correct.

Mass of steel rod $m = 5 \ kg$
Density of steel rod   $d = 7850 \ kg/m^3$
Volume of steel rod  $V = \dfrac{m}{d}$
$\implies \ V = \dfrac{5}{7850} = 6.36\times 10^{-4} \ m^3$