Question

The density of mercury is $13.6gm/cc$ . Convert it into S.I. units:
\[
  A.{\text{ }}13.6 \times {10^3}kg/{m^3} \\
  B.{\text{ }}13.6 \times {10^2}kg/{m^3} \\
  C.{\text{ }}13.6 \times {10^4}kg/{m^3} \\
  D.{\text{ }}13.6 \times {10^5}kg/{m^3} \\
 \]

Answer 1

Hint: In order to convert the given units in S.I. unit, we have to convert it in $kg/{m^3}$ , so we will use some basic conversion as 1 kg=1000 g and 1 m=100 cm, by using it we will get the required answer.

Formula Used- $1kg = 1000g,1m = 100cm$

Complete step-by-step solution -

Dimensions of Density are $M{L^{ - 3}}$
We know that $1kg = 1000g$ and $1m = 100cm$
So simultaneously we have:
$
  \because 1kg = 1000g \\
   \Rightarrow 1g = {10^{ - 3}}kg \\
 $
And also we have:
$
  \because 1m = 100cm \\
   \Rightarrow 1cm = {10^{ - 2}}m \\
 $
Given that the density of mercury is $13.6gm/cc$ so, let us convert it to standard unit $kg/{m^3}$ by the use of conversion terms shown above.
So, density of mercury:
\[
   = 13.6gm/cc \\
   = 13.6 \times \dfrac{{1g}}{{1c{m^3}}} \\
   = 13.6 \times \dfrac{{1g}}{{{{\left( {1cm} \right)}^3}}} \\
   = 13.6 \times \dfrac{{{{10}^{ - 3}}kg}}{{{{\left( {{{10}^{ - 2}}m} \right)}^3}}} \\
   = 13.6 \times \dfrac{{{{10}^{ - 3}}kg}}{{{{10}^{ - 6}}{m^3}}} \\
   = 13.6 \times \dfrac{{{{10}^3}kg}}{{{m^3}}} \\
   = 13.6 \times {10^3}\dfrac{{kg}}{{{m^3}}} \\
   = 13.6 \times {10^3}kg/{m^3} \\
 \]
Hence, the density of mercury in S.I. unit is \[13.6 \times {10^3}kg/{m^3}\] .
So, the correct answer is option A.

Note- Density is a measurement comparing the volume of matter an object has. An object with a certain volume with more matter has high density. An object of the same volume has a low density, with little matter. Density is determined by dividing an object's mass by its volume. In order to solve such problems students must follow the basic step still there are formulas for direct conversion from one unit to another which can be used as well.