Answer
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Hint: Relative density is a dimensionless quantity and it denotes the ratio of density of a material with the density of water in any system of units. The density of water in the CGS system of units is 1 grams per cubic centimeters.
Complete step by step solution:
Since the relative density is the ratio of density of a material and density of water. The relative density of mercury can be written as
\[R{D_{mercury}} = \dfrac{{{d_{mercury}}}}{{{d_{water}}}}\] ,
Where dmercury denotes density of mercury and water denotes density of water,
Putting the value of \[{d_{water}} = 1g/c{m^3}\], and given value of \[R{D_{mercury}} = 13.6\], we get
\[13.6 = \dfrac{{{d_{mercury}}}}{{1g/c{m^3}}}\],
Cross multiplication gives
\[{d_{mercury}} = \dfrac{{13.6g}}{{c{m^3}}}\],
Converting grams to Kilograms by using $1g = {10^{ - 3}}Kg$ ,
And centimeters into meters using $1cm = {10^{ - 2}}m$ , we get
\[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{ - 3}}Kg}}{{{{({{10}^{ - 2}}m)}^3}}}\],
This gives \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{ - 3}}Kg}}{{{{10}^{ - 6}}{m^3}}}\],
Using the exponents of 10, \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{6 - 3}}Kg}}{{{m^3}}}\],
Therefore, \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^3}Kg}}{{{m^3}}}\],
\[{d_{mercury}} = 13.6 \times {10^3}Kg{m^{ - 3}}\].
The density in SI units as mentioned in the question was $X \times {10^3}kg{m^{ - 3}}$ , comparing it to the calculated result gives us
$X = 13.6$ .
Therefore, the correct answer to the question is option : C
Note: Alternately, if one knew the density of water in SI units to be ${10^3}kg{m^{ - 3}}$, it could have been used in the formula of relative density to calculate the density of mercury in a more easy manner.
Similarly another term that is used in place of relative density is specific gravity, Specific gravity is also a dimensionless quantity and it also denotes the ratio of density of a material with the density of water in any system of units.
Complete step by step solution:
Since the relative density is the ratio of density of a material and density of water. The relative density of mercury can be written as
\[R{D_{mercury}} = \dfrac{{{d_{mercury}}}}{{{d_{water}}}}\] ,
Where dmercury denotes density of mercury and water denotes density of water,
Putting the value of \[{d_{water}} = 1g/c{m^3}\], and given value of \[R{D_{mercury}} = 13.6\], we get
\[13.6 = \dfrac{{{d_{mercury}}}}{{1g/c{m^3}}}\],
Cross multiplication gives
\[{d_{mercury}} = \dfrac{{13.6g}}{{c{m^3}}}\],
Converting grams to Kilograms by using $1g = {10^{ - 3}}Kg$ ,
And centimeters into meters using $1cm = {10^{ - 2}}m$ , we get
\[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{ - 3}}Kg}}{{{{({{10}^{ - 2}}m)}^3}}}\],
This gives \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{ - 3}}Kg}}{{{{10}^{ - 6}}{m^3}}}\],
Using the exponents of 10, \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^{6 - 3}}Kg}}{{{m^3}}}\],
Therefore, \[{d_{mercury}} = \dfrac{{13.6 \times {{10}^3}Kg}}{{{m^3}}}\],
\[{d_{mercury}} = 13.6 \times {10^3}Kg{m^{ - 3}}\].
The density in SI units as mentioned in the question was $X \times {10^3}kg{m^{ - 3}}$ , comparing it to the calculated result gives us
$X = 13.6$ .
Therefore, the correct answer to the question is option : C
Note: Alternately, if one knew the density of water in SI units to be ${10^3}kg{m^{ - 3}}$, it could have been used in the formula of relative density to calculate the density of mercury in a more easy manner.
Similarly another term that is used in place of relative density is specific gravity, Specific gravity is also a dimensionless quantity and it also denotes the ratio of density of a material with the density of water in any system of units.
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