Question

The density of iron is $7800\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$. Its relative density will be

• A. $7.8$
• B. $3.9$
• C. $15.6$
• D. $0.78$

Answer 1

The correct option is A

$7.8$

Step 1. Given data:

Relative density of iron = $7800\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$

We know that, the density of water = $1000\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$

Step 2. Calculations:

Relative density, or specific gravity, may be defined as the ratio of the density (mass of a unit volume) of a substance to the density of given reference material.

According to the definition of relative density,

We know the relative density of the given material = $\frac{Densityofmaterial}{Densityofwater}$

The relative density of iron is equal to the density of iron divided by the density of water

Relative density of iron = $\raisebox{1ex}{$7800$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.=7.8$

Therefore, the relative density of iron is $7.8$

So, the correct option is option A.