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Question: The physical quantity that does not have the dimensional formula \([ML^{-1}T^{-2}]\) is (A). Forc...

The physical quantity that does not have the dimensional formula [ML−1T−2][ML^{-1}T^{-2}] is
(A). Force
(B). Pressure
(C). Stress
(D). Modulus of elasticity
(E). Energy density

Explanation

Solution

Hint: The dimensional formula is given as MaLbTcM^{a}L^{b}T^{c} where a,b,c are the dimensions. Dimensions of a physical quantity are the powers to which the fundamental units, such as mass, length, and time.

Complete step-by-step solution -
Dimensions of a physical quantity are the powers to which the fundamental units, such as mass, length, and time, arise to obtain one unit of that quantity. Dimensional analysis is the practice of identifying the dimensions of the physical quantities. The dimensional formula is given as MaLbTcM^{a}L^{b}T^{c} where a,b,c are the dimensions.
The dimensional formulas are given below:
a.Force
We know thatF=maF=ma where mm is mass and aa is the acceleration,
then we can writeF=[ML0T0]×[M0LT−2]=[MLT−2]F=[ML^{0}T^{0}]\times [M^{0}LT^{-2}]=[MLT^{-2}]
Hence, dimensions of force is given as F=[M1L1T−2]F=[M^{1}L^{1}T^{-2}]
b.Pressure:
We know that P=FAP=\dfrac{F}{A} where FF is force, and AA is area
Then P=maA=[MLT−2][M0L2T0]=[ML−1T−2]P=\dfrac{ma}{A}=\dfrac{[MLT^{-2}]}{[M^{0}L^{2}T^{0}]}=[ML^{-1}T^{-2}]
Hence, dimensions of pressure is given as P=[M1L−1T−2]P=[M^{1}L^{-1}T^{-2}]
c.Stress:
We know that S=FAS=\dfrac{F}{A} where FF is force, and AA is area
Then S=maA=[MLT−2][M0L2T0]=[ML−1T−2]S=\dfrac{ma}{A}=\dfrac{[MLT^{-2}]}{[M^{0}L^{2}T^{0}]}=[ML^{-1}T^{-2}]
Hence, dimensions of stress is given as S=[M1L−1T−2]S=[M^{1}L^{-1}T^{-2}]
d.Modulus of elasticity:
We know that modulus  of  elasticity=linear  stresslinear  strainmodulus\; of\; elasticity=\dfrac{linear\; stress}{linear\; strain}
Then M.E=linear  stresslinear  strain=[ML−1T−2][M0L0T0]=[ML−1T−2]M.E=\dfrac{linear \;stress}{linear\; strain}=\dfrac{[ML^{-1}T^{-2}]}{[M^{0}L^{0}T^{0}]}=[ML^{-1}T^{-2}]
Hence, dimensions of modulus of elasticity is given as M.E=[M1L−1T−2]M.E=[M^{1}L^{-1}T^{-2}]
e.Energy density:
We know that energy  density=energyvolumeenergy\; density=\dfrac{energy}{volume}
Then E.D=[ML2T−2][M0L3T0]=[M1L−1T−2]E.D=\dfrac{[ML^{2}T{-2}]}{[M^{0}L^{3}T^{0}]}=[M^{1}L^{-1}T^{-2}]
Hence, dimensions of energy density is given as E.D=[M1L−1T−2]E.D=[M^{1}L^{-1}T^{-2}]
Clearly, expect force, every other quantity is of the form [M1L−1T−2][M^{1}L^{-1}T^{-2}]

Additional Information:
There are two types of dimensions: dimensional constants and dimensional variables. The physical quantities which have dimensions and have a fixed value are called dimensional constants. Whereas, Dimensional variables are those physical quantities which have dimensions and do not have a fixed value. Only similar physical quantities can be added or subtracted, thus two quantities having different dimensions cannot be added together. The application of dimensional analysis is given by Fourier as follows:
Verify the correctness of a physical equation,
Derive a relationship between physical quantities
Converting the units of a physical quantity from one system to another system.
Dimensional analysis is also used to deduce the relation between two or more physical quantities.
However, this has a few drawbacks also:
Dimensionless quantities cannot be determined by this method.
In the case of physical quantities which are dependent upon more than three physical quantities. This method cannot be used.

Note: Only similar physical quantities can be added or subtracted, thus two quantities having different dimensions cannot be added together. It is an expression that relates derived quantity to fundamental quantities. But it is not related to the magnitude of the derived quantity.