If `F` is the force, `v` is the velocity and `A` is the area, considered as fundamental quantity . Find the dimension of youngs modulus.

Options

A. `F^1A^0V^(-1)`
B. `F^1A^1V^(-1)`
C. `F^1A^2V^(3)`
D. `F^1A^(-1)V^0`

Answer

To find the dimension of Young's modulus, we need to understand that Young's modulus (Y) is defined as the ratio of stress to strain.

1. **Understanding Stress**:
- Stress is defined as force (F) per unit area (A).
- Therefore, the formula for stress can be expressed as:
\[ \text{Stress} = \frac{F}{A} \]

2. **Finding the Dimensions of Force and Area**:
- The dimension of force (F) is given as:
\[ [F] = MLT^{-2} \]
where M is mass, L is length, and T is time.
- The dimension of area (A) is given as:
\[ [A] = L^2 \]

3. **Calculating the Dimension of Stress**:
- Now, substituting the dimensions of force and area into the stress formula:
\[ [\text{Stress}] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

4. **Understanding Strain**:
- Strain is a dimensionless quantity as it is the ratio of change in length to original length.

5. **Finding the Dimension of Young's Modulus**:
- Since Young's modulus is defined as stress divided by strain, and strain is dimensionless, the dimension of Young's modulus is the same as that of stress:
\[ [Y] = [\text{Stress}] = ML^{-1}T^{-2} \]

6. **Final Expression**:
- Therefore, the dimension of Young's modulus can be expressed in terms of the fundamental quantities given in the question:
\[ [Y] = F^1 A^{-1} V^0 \]
- Here, we have used F for force, A for area, and V for velocity, where the exponent of V is zero since it does not appear in the expression for Young's modulus.

Thus, the final result is:
\[ [Y] = F^1 A^{-1} V^0 \]