Tag: learning how to measure

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.