
The relative density of oak wood is $0.64$ . What is its density in?
$a.$ the S I system?
$b.$ the C G S system?
Answer
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Hint : The relative density of a substance is defined as the ratio of density of substance to the density of water at \[{4^0}C\] . In this problem first we will find the density of substance in S I unit and then we need to obtain the density of substance in C G S unit. We should know the S I unit of density of substance is $kg{m^{ - 3}}$ and the C G S unit is $gc{m^{ - 3}}$ .
Formula used:
$ \Rightarrow $ Relative density = $\dfrac{{Density{\text{ }}of{\text{ }}substance}}{{Density{\text{ }}of{\text{ }}water{\text{ }}at\;{4^0}C}}$ …………$\left( 1 \right)$
Complete step-by-step solution:
We know that, Density of Water at ${4^0}C$ = $1000kg{m^{ - 3}}$ .
Given, Relative density = 0.64
$a.$ In S I system:-
The density of the substance can be calculated by using the formula,
Relative density = $\dfrac{{Density{\text{ }}of{\text{ }}substance}}{{Density{\text{ }}of{\text{ }}water{\text{ }}at\;{4^0}C}}$
On substituting the given data in above equation, we get
$0.64 = \dfrac{{Density{\text{ }}of{\text{ }}substance}}{{1000}}$
On simplifying the above equation, we get
Density of substance = $0.64 \times 1000$
Density of substance in S I unit = $640kg{m^{ - 3}}$ ………. $\left( 2 \right)$ .
$b.$ In C G S system:-
On Converting S I unit to C G S unit, by using
$1kg = 1000g$ and $1m = 100cm$ ………. $\left( 3 \right)$
Substituting equation $\left( 3 \right)$ in equation $\left( 2 \right)$ , we get
Density of substance = $\dfrac{{640 \times 1000}}{{{{\left( {{{10}^2}} \right)}^3}}}$
Density of substance = $\dfrac{{640 \times 1000}}{{{{10}^6}}}$
On simplifying the above equation, we have
Density of substance = $640 \times {10^3} \times {10^{ - 6}}$
Density of substance = $640 \times {10^{ - 3}}$
Density of substance C G S = $0.640gc{m^{ - 3}}$.
Note:Relative density is a physical quantity which is dimensionless. That is it doesn’t not have any S.I unit. It should be noted that the density of water is $1000kg{m^{ - 3}}$ at ${4^0}C$ . Relative density or specific density is nearly measured with respect to water at its densest because water will be having anomalous expansion from ${0^0}C$ to ${4^0}C$ .
Formula used:
$ \Rightarrow $ Relative density = $\dfrac{{Density{\text{ }}of{\text{ }}substance}}{{Density{\text{ }}of{\text{ }}water{\text{ }}at\;{4^0}C}}$ …………$\left( 1 \right)$
Complete step-by-step solution:
We know that, Density of Water at ${4^0}C$ = $1000kg{m^{ - 3}}$ .
Given, Relative density = 0.64
$a.$ In S I system:-
The density of the substance can be calculated by using the formula,
Relative density = $\dfrac{{Density{\text{ }}of{\text{ }}substance}}{{Density{\text{ }}of{\text{ }}water{\text{ }}at\;{4^0}C}}$
On substituting the given data in above equation, we get
$0.64 = \dfrac{{Density{\text{ }}of{\text{ }}substance}}{{1000}}$
On simplifying the above equation, we get
Density of substance = $0.64 \times 1000$
Density of substance in S I unit = $640kg{m^{ - 3}}$ ………. $\left( 2 \right)$ .
$b.$ In C G S system:-
On Converting S I unit to C G S unit, by using
$1kg = 1000g$ and $1m = 100cm$ ………. $\left( 3 \right)$
Substituting equation $\left( 3 \right)$ in equation $\left( 2 \right)$ , we get
Density of substance = $\dfrac{{640 \times 1000}}{{{{\left( {{{10}^2}} \right)}^3}}}$
Density of substance = $\dfrac{{640 \times 1000}}{{{{10}^6}}}$
On simplifying the above equation, we have
Density of substance = $640 \times {10^3} \times {10^{ - 6}}$
Density of substance = $640 \times {10^{ - 3}}$
Density of substance C G S = $0.640gc{m^{ - 3}}$.
Note:Relative density is a physical quantity which is dimensionless. That is it doesn’t not have any S.I unit. It should be noted that the density of water is $1000kg{m^{ - 3}}$ at ${4^0}C$ . Relative density or specific density is nearly measured with respect to water at its densest because water will be having anomalous expansion from ${0^0}C$ to ${4^0}C$ .
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