If the relative density of gold is 19.3 and the density of water is \(10^6\) g/m3, what will be the density of gold in SI units?
19300 kg/m3
Relative density is defined as the ratio of a substance's density to the density of water: \(\text{Relative density} = \dfrac{\rho_{gold}}{\rho_{water}}\).
Given \(\rho_{water} = 10^6\ \text{g/m}^3\), the density of gold is: \(\rho_{gold} = 19.3 \times 10^6\ \text{g/m}^3 = 1.93 \times 10^7\ \text{g/m}^3\).
To convert to SI units (kg/m3), divide by 1000 since 1 kg = 1000 g: \(\rho_{gold} = \dfrac{1.93 \times 10^7}{10^3}\ \text{kg/m}^3 = 1.93 \times 10^4\ \text{kg/m}^3 = 19300\ \text{kg/m}^3\).
Hence, the density of gold in SI units is 19300 kg/m3.
Important Questions from Units and Dimensional Formula
Parsec is a unit of ______.
'Torr' is a unit of _______.
Electron-volt is a unit of ______.
Which of the following is the SI unit for measuring the amount of a substance?
Weber per second is equivalent to ______.
