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Question: Density of mercury is \(13.6g.c{{m}^{-3}}\) , then the value in MKS system is _______...

Density of mercury is 13.6g.cm−313.6g.c{{m}^{-3}} , then the value in MKS system is _______

Explanation

Solution

Hint: This problem can be solved by converting each unit in the combination of units used to represent the value of density into the MKS system. That is, we will have to convert gg into KgKg and cm−3c{{m}^{-3}} into m−3{{m}^{-3}}. However, the physical value of the density must remain the same and hence proper conversion should be made for the units.
Formula used:
1g=10−3Kg1g={{10}^{-3}}Kg
1cm=10−2m1cm={{10}^{-2}}m
∴1cm3=(10−2)3m=10−6m\therefore 1c{{m}^{3}}={{\left( {{10}^{-2}} \right)}^{3}}m={{10}^{-6}}m

Complete step by step answer:
The density of a body refers to the mass contained in a body per unit volume. Hence, the unit of density is essentially that ofmassvolume\dfrac{\text{mass}}{\text{volume}}.
Now, in the question, we are given the unit of density as g.cm−3g.c{{m}^{-3}}which is the combination of the unit of mass (g)\left( g \right)and the unit of volume (cm−3)\left( c{{m}^{-3}} \right)in the CGS system. To convert the unit into the MKS system we will have to convert each unit making up the unit of density into its respective MKS unit. That is, we will have to change gg toKgKg and cm−3c{{m}^{-3}} tom−3{{m}^{-3}}.
However, we have to change in such a way so that the physical value of the density does not change.
Hence, we will have to apply the proper conversions, which are as follows.

1g=10−3Kg1g={{10}^{-3}}Kg --(1)
1cm=10−2m1cm={{10}^{-2}}m
∴1cm3=(10−2)3m3=10−6m3\therefore 1c{{m}^{3}}={{\left( {{10}^{-2}} \right)}^{3}}{{m}^{3}}={{10}^{-6}}{{m}^{3}}
∴1cm−3=(10−2)−3m−3=106m−3\therefore 1c{{m}^{-3}}={{\left( {{10}^{-2}} \right)}^{-3}}{{m}^{-3}}={{10}^{6}}{{m}^{-3}}--(2)

We are given that the density of mercury in the CGS system is 13.6g.cm−313.6g.c{{m}^{-3}}. we are required to convert it into the MKS system. Therefore, using (1) and (2) to do the required conversion, we get,

13.6gm.cm−3=13.6×(10−3Kg)×(106m−3)13.6gm.c{{m}^{-3}}=13.6\times \left( {{10}^{-3}}Kg \right)\times \left( {{10}^{6}}{{m}^{-3}} \right)
∴13.6g.cm−3=13.6×10−3×106Kg.m−3=13.6×103Kg.m−3\therefore 13.6g.c{{m}^{-3}}=13.6\times {{10}^{-3}}\times {{10}^{6}}Kg.{{m}^{-3}}=13.6\times {{10}^{3}}Kg.{{m}^{-3}}

Hence, the required value of the density is 13.6×103Kg.m−313.6\times {{10}^{3}}Kg.{{m}^{-3}}.

Note: Students can easily convert between cmcmand mmbut often face problems while converting the higher powers of these units especially if they are in the denominator (like cm−3c{{m}^{-3}} andm−3{{m}^{-3}}). They often make mistakes while doing this conversion. The best way to avoid this is to approach in a step by step manner as shown above.Often questions give much more complicated units to convert between systems. The correct approach to solving these problems will be to first break down the unit into the fundamental units and then apply the conversions between fundamental units of the two systems. Directly converting a complex unit from one system to another makes the question very hard. For example when converting between dyne and Newton, it is better to break down the unit into the fundamental units of mass, length and time using the definition of force.