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 will follow these steps:

### Step 1: Understand Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain.

### Step 2: Define Stress and Strain
- **Stress** is defined as force per unit area.
- **Strain** is defined as the change in length divided by the original length, which is a dimensionless quantity (unitless).

### Step 3: Write the Formula for Young's Modulus
Thus, we can express Young's modulus as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\text{Force}/\text{Area}}{\text{Strain}} \]
Since strain is dimensionless, we can simplify this to:
\[ Y = \frac{\text{Stress}}{1} = \text{Stress} \]

### Step 4: Determine the Dimensions of Stress
Stress can be expressed as:
\[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \]
Now, we need to find the dimensions of force and area.

### Step 5: Dimensions of Force and Area
- The dimension of force (F) is given by:
\[ [F] = MLT^{-2} \]
- The dimension of area (A) is given by:
\[ [A] = L^2 \]

### Step 6: Substitute the Dimensions into the Stress Formula
Now we can substitute the dimensions of force and area into the stress formula:
\[ \text{Stress} = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} \]

### Step 7: Simplify the Expression
When we simplify this, we get:
\[ \text{Stress} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

### Step 8: Conclusion
Since Young's modulus has the same dimensions as stress, we conclude:
\[ [Y] = ML^{-1}T^{-2} \]

### Final Answer
Thus, the dimension of Young's modulus is:
\[ [Y] = ML^{-1}T^{-2} \]

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