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**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**

Young`s modulus is defined as the ratio of the stress and the strain

Mathematically, $young`s{\text{ modulus = }}\dfrac{{Stress}}{{strain}}$

Stress is defined as the force per unit area

Mathematically,$Stress = \dfrac{{force}}{{area}}$

Strain is defined as the ratio of the change in the length and original legth

Mathematically,$Strain = \dfrac{{\Delta l}}{l}$

Now $young`s{\text{ modulus = }}\dfrac{{Stress}}{{strain}}$

Putting the value of stress and strain in above given formula

$young`s{\text{ modulus = }}\dfrac{{\dfrac{{force}}{{area}}}}{{\dfrac{{\Delta l}}{l}}}$

So

$young`s{\text{ modulus = }}\dfrac{{force \times l}}{{area \times \Delta l}}$

$force = mass \times acceleration$

$

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 young`s modulus will be-

$\dfrac{{{\text{dimension of force }}}}{{\dim ension{\text{ of area}}}} = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}$

So dimension for young`s modulus will be:$M{L^{ - 1}}{T^{ - 2}}$

**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.

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