The dimension of young's modulus,E is given by

To find the dimension of Young's Modulus ($E$), we need to look at its definition and the dimensions of the quantities involved.

1. Analyze the Concept

Young's Modulus ($E$) is defined as the ratio of Stress to Strain:

• Stress is defined as Force ($F$) per unit Area ($A$): $\text{Stress} = \frac{F}{A}$
• Strain is defined as the change in length ($\Delta L$) divided by the original length ($L$): $\text{Strain} = \frac{\Delta L}{L}$

2. Break Down the Dimensions

• Force ($F$): From Newton's Second Law ($F = ma$), the dimension is $[M][L][T^{-2}]$.
• Area ($A$): Area is length squared, so the dimension is $[L^2]$.
• Stress: $\frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}]$.
• Strain: Since it is a ratio of two lengths ($\frac{L}{L}$), it is dimensionless. It has no units and no dimensions ($[M^0L^0T^0]$).

3. Calculate the Dimension of Young's Modulus

Since $E = \frac{\text{Stress}}{\text{Strain}}$, and Strain is dimensionless:

4. Conclusion

Comparing our result $[ML^{-1}T^{-2}]$ to the given options:

• A: $MLT^{-1}$
• B: $MT^{-2}$
• C: $ML^{-1}T^{-2}$
• D: $M^{-1}T$

The calculated dimension matches option C.

Exam Tip

Always remember that Stress, Pressure, and Young's Modulus all share the same dimension: $[ML^{-1}T^{-2}]$. If you ever forget the formula for Young's Modulus, just remember that it has the same units as Pressure (Newtons per square meter, $N/m^2$), which makes it easy to derive the dimensions from there!