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Question: The dimensional formula for young's modulus is: A. \([M{L^{ - 1}}{T^{ - 2]}}\) B. \([{M^0}L{T^{ ...

The dimensional formula for young's modulus is:
A. [ML−1T−2][M{L^{ - 1}}{T^{ - 2]}}
B. [M0LT−2][{M^0}L{T^{ - 2}}]
C. [MLT−2][ML{T^{ - 2}}]
D. [ML2T−2][M{L^2}{T^{ - 2}}]

Explanation

Solution

Hint In this type of question we have to define the unit of any physical quantity .After defining the unit we have to arrange it in the form of a fundamental unit like in mass, length and time.
For this type of question we have appropriate knowledge of the formula of that definition:

Complete Step by step solution
Youngs modulus is defined as the ratio of the stress and the strain Mathematically, $youngs{\text{ modulus = }}\dfrac{{Stress}}{{strain}}StressisdefinedastheforceperunitareaMathematically, Stress is defined as the force per unit area Mathematically,Stress = \dfrac{{force}}{{area}}StrainisdefinedastheratioofthechangeinthelengthandoriginallegthMathematically, Strain is defined as the ratio of the change in the length and original legth Mathematically,Strain = \dfrac{{\Delta l}}{l}Now Nowyoungs{\text{ modulus = }}\dfrac{{Stress}}{{strain}}$ Putting the value of stress and strain in above given formula $youngs{\text{ modulus = }}\dfrac{{\dfrac{{force}}{{area}}}}{{\dfrac{{\Delta l}}{l}}}So So youngs{\text{ modulus = }}\dfrac{{force \times l}}{{area \times \Delta l}}$ $force = mass \times acceleration$ $ acceleration = \dfrac{{velocity}}{{time}} \\\ velocity = \dfrac{{dis\tan ce}}{{time}} \\\ $ $area = length \times length$ So after seeing above equation If we write – for mass=M, for length=L and for time =T So dimension for velocity will be$L{T^{ - 1}}$ Dimension for acceleration will be $L{T^{ - 2}}$ Similar dimension for force will be $ML{T^{ - 2}}$ Dimension for area will be ${L^2}$ So dimension for youngs modulus will be-
dimension of force dim⁡ension of area=MLT−2L2=ML−1T−2\dfrac{{{\text{dimension of force }}}}{{\dim ension{\text{ of area}}}} = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}
So dimension for young`s modulus will be:ML−1T−2M{L^{ - 1}}{T^{ - 2}}

Hence answer number A will be the correct option.

Note Dimension is used to check the unit of similar quantities . After knowing the dimension we can formulate that physical quantity and after which we can define that quantity .
Dimension is also used to change the physical quantity from one unit system to another.