Answer
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Hint: This question is based on the concept of specific gravity’s dimension For these types of questions you should know the basic rules to find out the dimensional formula of physical quantities of specific gravity.
Complete Step-by-Step solution:
According to the question we are asked about the dimensions of specific gravity. So, we should know what specific gravity is; Specific gravity can be defined as the ratio between the density of a substance and the density of a given reference substance. It is also known as relative density. The formula of specific density is = density of a substance/density of a given reference substance.
Therefore specific gravity = density of substance/density of water at ${4^ \circ }C$ .
So, we know that the formula of density which is = $\dfrac{{Mass}}{{Volume}}$
The dimensional formula of the volume is [${L^3}$] and that of mass is [M].
Therefore, the dimensional formula of density can be written as:$\dfrac{M}{{{L^3}}} = M{L^{ - 3}}$
Similarly the dimensional formula of the density of water is also $M{L^{ - 3}}$ .
Thus, the formula of specific gravity is = density of substance/ density of water ${4^ \circ }C$ .
So, the dimensional formula of specific gravity will be= $\dfrac{{M{L^{ - 3}}}}{{M{L^{ - 3}}}} = M^0L^0$ .
Therefore the units of density cancel out each other and the specific gravity becomes dimensionless.
Hence, M=0, L=0, T=0.
So, option A is the correct option.
Note: A dimensional formula can be explained as physical quantities in terms of fundamental quantities. Fundamental quantities are Mass, Length and Time knowing the Dimensional formula of specific gravity we can answer the given question.
Complete Step-by-Step solution:
According to the question we are asked about the dimensions of specific gravity. So, we should know what specific gravity is; Specific gravity can be defined as the ratio between the density of a substance and the density of a given reference substance. It is also known as relative density. The formula of specific density is = density of a substance/density of a given reference substance.
Therefore specific gravity = density of substance/density of water at ${4^ \circ }C$ .
So, we know that the formula of density which is = $\dfrac{{Mass}}{{Volume}}$
The dimensional formula of the volume is [${L^3}$] and that of mass is [M].
Therefore, the dimensional formula of density can be written as:$\dfrac{M}{{{L^3}}} = M{L^{ - 3}}$
Similarly the dimensional formula of the density of water is also $M{L^{ - 3}}$ .
Thus, the formula of specific gravity is = density of substance/ density of water ${4^ \circ }C$ .
So, the dimensional formula of specific gravity will be= $\dfrac{{M{L^{ - 3}}}}{{M{L^{ - 3}}}} = M^0L^0$ .
Therefore the units of density cancel out each other and the specific gravity becomes dimensionless.
Hence, M=0, L=0, T=0.
So, option A is the correct option.
Note: A dimensional formula can be explained as physical quantities in terms of fundamental quantities. Fundamental quantities are Mass, Length and Time knowing the Dimensional formula of specific gravity we can answer the given question.
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