If the velocity `(V)` , acceleration `(A)` , and force `(F)` are taken as fundamental quantities instead of mass `(M)` , length `(L) , and time (T)` , the dimensions of young's modulus `(Y)` would be

Options

A. `FA^(2)V^(-4)`
B. `FA^(2)V^(-5)`
C. `FA^(2)V^(-3)`
D. `FA^(2)V^(-2)`

Explanation

To find the dimensions of Young's modulus (Y) when velocity (V), acceleration (A), and force (F) are taken as fundamental quantities, we will follow these steps:

### Step 1: Understand the relationship
Young's modulus (Y) is defined as the ratio of stress to strain. The dimensions of stress are force per unit area, and strain is a dimensionless quantity. Therefore, we need to express Young's modulus in terms of the fundamental quantities V, A, and F.

### Step 2: Write the proportionality
Assume that Young's modulus (Y) can be expressed as:
\[ Y \propto V^A A^B F^C \]
where A, B, and C are the powers to which the fundamental quantities are raised.

### Step 3: Write the dimensions of the quantities
We need to express the dimensions of Y, V, A, and F in terms of M (mass), L (length), and T (time):
- The dimension of force (F) is given by \( [F] = M L T^{-2} \).
- The dimension of velocity (V) is given by \( [V] = L T^{-1} \).
- The dimension of acceleration (A) is given by \( [A] = L T^{-2} \).

### Step 4: Substitute the dimensions
Now, substituting the dimensions into the proportionality:
\[ [Y] = K [V]^A [A]^B [F]^C \]
Substituting the dimensions:
\[ [Y] = K (L T^{-1})^A (L T^{-2})^B (M L T^{-2})^C \]

### Step 5: Expand the dimensions
Expanding the right-hand side:
\[ [Y] = K L^{A + B + C} M^C T^{-A - 2B - 2C} \]

### Step 6: Set the dimensions of Young's modulus
The dimension of Young's modulus is given by:
\[ [Y] = M L^{-1} T^{-2} \]

### Step 7: Compare dimensions
Now we can set the dimensions equal to each other:
1. For mass (M): \( C = 1 \)
2. For length (L): \( A + B + C = -1 \)
3. For time (T): \( -A - 2B - 2C = -2 \)

### Step 8: Solve the equations
From the first equation, we have \( C = 1 \).
Substituting \( C = 1 \) into the second equation:
\[ A + B + 1 = -1 \Rightarrow A + B = -2 \] (Equation 1)

Substituting \( C = 1 \) into the third equation:
\[ -A - 2B - 2 = -2 \Rightarrow -A - 2B = 0 \Rightarrow A + 2B = 0 \] (Equation 2)

Now, we can solve Equations 1 and 2:
From Equation 2, we have \( A = -2B \).
Substituting into Equation 1:
\[ -2B + B = -2 \Rightarrow -B = -2 \Rightarrow B = 2 \]
Now substituting \( B = 2 \) back into \( A = -2B \):
\[ A = -2(2) = -4 \]

### Step 9: Final expression for Young's modulus
Now we have:
- \( A = -4 \)
- \( B = 2 \)
- \( C = 1 \)

Thus, we can express Young's modulus as:
\[ Y = K V^{-4} A^{2} F^{1} \]

### Conclusion
The dimensions of Young's modulus (Y) when velocity (V), acceleration (A), and force (F) are taken as fundamental quantities are:
\[ Y \propto V^{-4} A^{2} F^{1} \]