If energy 'E' momentum 'P' and force 'F' are choosen as fundamental units.
Dimension of mass in new system is

Options

A. `E^(-1)PF^(0)`
B. `EP^(2)F^(-1)`
C. `E^(-1)P^(2)F^(0)`
D. `E^(-1)P^(0)F^(2)`

Answer

To find the dimension of mass in a new system where energy \(E\), momentum \(P\), and force \(F\) are chosen as fundamental units, we can follow these steps:

### Step 1: Express mass in terms of \(E\), \(P\), and \(F\)
Assume that mass \(M\) can be expressed as:
\[ M = E^A \cdot P^B \cdot F^C \]
where \(A\), \(B\), and \(C\) are the powers to which \(E\), \(P\), and \(F\) are raised.

### Step 2: Write the dimensions of \(E\), \(P\), and \(F\)
1. **Energy \(E\)**: The dimension of energy is the same as work, which is given by:
\[ [E] = [F \cdot d] = [M^1 L^1 T^{-2}] = M^1 L^2 T^{-2} \]

2. **Momentum \(P\)**: The dimension of momentum is given by:
\[ [P] = [M \cdot v] = [M^1 L^1 T^{-1}] = M^1 L^1 T^{-1} \]

3. **Force \(F\)**: The dimension of force is given by:
\[ [F] = [M \cdot a] = [M^1 L^1 T^{-2}] = M^1 L^1 T^{-2} \]

### Step 3: Substitute the dimensions into the equation
Substituting the dimensions into our expression for mass:
\[ M = (M^1 L^2 T^{-2})^A \cdot (M^1 L^1 T^{-1})^B \cdot (M^1 L^1 T^{-2})^C \]

### Step 4: Combine the dimensions
This gives:
\[ M = M^{A+B+C} \cdot L^{2A + B + C} \cdot T^{-2A - B - 2C} \]

### Step 5: Set up equations for the dimensions
Now we equate the dimensions on both sides:
1. For mass \(M\):
\[ A + B + C = 1 \quad \text{(1)} \]

2. For length \(L\):
\[ 2A + B + C = 0 \quad \text{(2)} \]

3. For time \(T\):
\[ -2A - B - 2C = 0 \quad \text{(3)} \]

### Step 6: Solve the equations
From equation (1):
\[ C = 1 - A - B \]

Substituting \(C\) into equation (2):
\[ 2A + B + (1 - A - B) = 0 \implies A + 1 = 0 \implies A = -1 \]

Now substitute \(A = -1\) into equation (1):
\[ -1 + B + C = 1 \implies B + C = 2 \quad \text{(4)} \]

Substituting \(A = -1\) into equation (3):
\[ -2(-1) - B - 2C = 0 \implies 2 - B - 2C = 0 \implies B + 2C = 2 \quad \text{(5)} \]

### Step 7: Solve equations (4) and (5)
From equation (4):
\[ B + C = 2 \implies C = 2 - B \]

Substituting into equation (5):
\[ B + 2(2 - B) = 2 \implies B + 4 - 2B = 2 \implies -B + 4 = 2 \implies B = 2 \]

Then substituting \(B = 2\) back into equation (4):
\[ 2 + C = 2 \implies C = 0 \]

### Step 8: Conclusion
Now we have:
\[ A = -1, \quad B = 2, \quad C = 0 \]
Thus, the dimension of mass in the new system is:
\[ M = E^{-1} \cdot P^{2} \cdot F^{0} \]

### Final Answer
The dimension of mass in the new system is given by:
\[ M = \frac{P^2}{E} \]