The formula `W=(F+2Ma)v^(n)`, where W is the work done, F is the force, M is the mass, a is the acceleration and v is the velocity can be made dimensionally correct for

Options

A. `n=0`
B. `n=1`
C. `n=-1`
D. no value of n

Answer

To determine the value of \( n \) for which the formula \( W = (F + 2Ma)v^n \) is dimensionally correct, we will analyze the dimensions of each term in the equation.

### Step 1: Identify the dimensions of each variable
1. **Work (W)**: The dimension of work is given by:
\[ [W] = M^1 L^2 T^{-2} \]
2. **Force (F)**: The dimension of force is:
\[ [F] = M^1 L^1 T^{-2} \]
3. **Mass (M)**: The dimension of mass is:
\[ [M] = M^1 \]
4. **Acceleration (a)**: The dimension of acceleration is:
\[ [a] = L^1 T^{-2} \]
5. **Velocity (v)**: The dimension of velocity is:
\[ [v] = L^1 T^{-1} \]

### Step 2: Combine the dimensions of the terms in the formula
The term \( 2Ma \) can be analyzed as follows:
\[ [2Ma] = [M][a] = (M^1)(L^1 T^{-2}) = M^1 L^1 T^{-2} \]

Now, we can combine the dimensions of \( F \) and \( 2Ma \):
\[ [F + 2Ma] = [F] = M^1 L^1 T^{-2} \]

### Step 3: Analyze the entire expression
Now, substituting back into the equation:
\[ W = (F + 2Ma)v^n \]
The dimensions of the right-hand side become:
\[ [F + 2Ma] \cdot [v^n] = (M^1 L^1 T^{-2})(L^1 T^{-1})^n = M^1 L^{1+n} T^{-2-n} \]

### Step 4: Set the dimensions equal for dimensional consistency
For the equation to be dimensionally correct, the dimensions of both sides must be equal:
\[ M^1 L^2 T^{-2} = M^1 L^{1+n} T^{-2-n} \]

### Step 5: Equate the dimensions
1. For mass \( M \):
\[ 1 = 1 \quad \text{(This is satisfied)} \]
2. For length \( L \):
\[ 2 = 1 + n \implies n = 1 \]
3. For time \( T \):
\[ -2 = -2 - n \implies n = 0 \]

### Step 6: Conclusion
From the equations derived, we see that we have two conflicting results for \( n \):
- From the length dimension, \( n = 1 \)
- From the time dimension, \( n = 0 \)

Since there is no single value of \( n \) that satisfies both conditions simultaneously, we conclude that there is **no value of \( n \)** for which the equation is dimensionally correct.

### Final Answer
The formula \( W = (F + 2Ma)v^n \) can be made dimensionally correct for **no value of \( n \)**.

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