Measurement of volume - class-VIII

$1000l=1m^3$. $l$ is used to measure the volume of a fluid and $m^3$ is generally used to express volume of a solid.

$1m=100cm=10^2{cm}$

SI unit of volume is $m^3$ which can also be written as $cubicmetre$ and $metre^3$. It is a big unit, the smaller units are also in use.

The space occupied by a body is called its volume. SI unit of volume is $m^3$.

The volume of liquids is expressed in $litre(l)$.

It is true that the volume of a stone (or any irregular shaped body) can be determined by the displacement method using a measuring cylinder. Volume of the stone is equal to the difference of final level of water in the mesuring jar and the initial level.

Volume is generally measured in liter, $1\ liter=10^{-3}m^3$

If the length, breadth and height are equal, a cube is formed. Let $a$ be the length of each side of the cube, then its volume $V$ is equal to $a \times a \times a$, or ${a}^{3}$.

Volume of a cube of side a each, $V  = {a}^{3}$

The volume of a cuboid of length $\left(l\right)$, breadth $\left(b\right)$ and height $\left(h\right)$, is the product of length, breadth and height.

Volume of a cuboid $=$ Length $\times$ breadth $\times$ Height

The space occupied by a substance is its volume. The $SI$ unit for volume is the cubic metre $\left({m}^{3}\right)$.

Sub-multiples of the cubic metre, the cubic centimetre and the cubic millimetre are used for smaller measurements.

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.

Water displacement is the standard method for measuring the volume of an irregularly shaped solid. The object is submerged in a graduated cylinder, and the volume of water displaced equals the volume of the object.

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  $V = \dfrac{m}{d}$
$\implies \ V = \dfrac{5}{7850} = 6.36\times 10^{-4} \ m^3$

Side of cube  $a = 7.203 \ m$
Volume of cube  $V = a^3$
$\implies \ V = (7.203)^3 = 373.714 \ m^3$
Since the value of side of cube has 4 significant figures, thus the value of volume of cube must also have 4 significant figure.
Thus volume of cube  $V = 373.7 \ m^3$

Mass of solid block  $m = 2 \ kg$
Density of solid block   $d = 5.2 \ kg/m^3$
Volume of solid block  $V = \dfrac{m}{d}$
$\implies \ V = \dfrac{2}{5.2} = 0.3846 \ m^3$

Measuring cylinder, pipette and measuring flask are used to measure the volume of the given liquid whereas vernier calliper is used measure the length of the object.

Let n number of grains are in the vessel.

So, volume of cuboid vessel is equal to n times volume of spheres. i.e.

$ 1\times 0.6\times 0.5=n\left( \dfrac{4}{3}\times 3.14\times {{\left( \dfrac{3}{2000} \right)}^{3}} \right) $

$ 0.3=n\dfrac{4}{3}\times 3.14\times 3.375\times {{10}^{-9}} $

$ 0.3=n\times 14.13\times {{10}^{-9}} $

$ n=\dfrac{0.3}{14.13\times {{10}^{-9}}} $

$ n=0.021\times {{10}^{9}}\, $

$ n=2\times {{10}^{7}}$

The zero error (−0.20) can be removed but its uncertainty (± 0.02) must be added to the measurement uncertainty.

S.I. unit of volume is $m^3$. 

A Burette  is a device used in analytical chemistry for the dispensing of variable, measured amounts of a chemical solution. A volumetric burette delivers measured volumes of liquid.

We know that $1$ litre $=10^3 cm^3$

The given statement is true that the measuring jar is graduated in ml from bottom to top.

SI Unit of volume is $m^3$ 

When water freezes into ice, its density decreases, meaning its volume increases (it expands). This expansion is a unique property of water. Statement 1 is true. Statement 2 is false because ice floats with about 90% of its volume submerged, not 50%. Statement 3 is false as a general rule.

Let M_w be the mass of wood and M_c be the mass of concrete. For the combination to float completely submerged, total buoyant force equals total weight: (M_w + M_c)g = (V_w + V_c)rho_w g. Using volumes in terms of mass and specific gravities (0.5 and 2.5), we solve for the ratio of masses to get 3/5.

$PV=mRT$

Volume of the object is defined as the space occupied by the object.

$1ml=10^{-3}L=10^{-3}\times10^{-3}m^3=10^{-6}m^3$

We know that: $1 m^3=1000\ liter$

We know that $1 m^3=1000\ liter=1000\times 1000ml$

Volume$=$Length$\times$ breadth$\times$ depth