Courses
Courses for Kids
Free study material
Offline Centres
More

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

seo-qna
Last updated date: 03rd Mar 2024
Total views: 21k
Views today: 0.21k
IVSAT 2024
Answer
VerifiedVerified
21k+ views
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.