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

Options

A. `FEP^(-2)`
B. `FEP^(2)`
C. `F^(2)EP^(-1)`
D. `F^(-1)EP^(-2)`

Answer

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

### Step 1: Write the dimension of acceleration
Acceleration \(a\) is defined as the change in velocity per unit time. The dimension of acceleration is given by:
\[ [a] = L^1 T^{-2} \]
where \(L\) represents length and \(T\) represents time.

### Step 2: Express acceleration in terms of fundamental units
We express acceleration in terms of the fundamental units \(E\), \(P\), and \(F\):
\[ a = E^a P^b F^c \]
where \(a\), \(b\), and \(c\) are the powers to which \(E\), \(P\), and \(F\) are raised.

### Step 3: Write the dimensions of \(E\), \(P\), and \(F\)
Next, we need to express the dimensions of energy, momentum, and force:
- Energy \(E\) has the dimension:
\[ [E] = M^1 L^2 T^{-2} \]
- Momentum \(P\) has the dimension:
\[ [P] = M^1 L^1 T^{-1} \]
- Force \(F\) has the dimension:
\[ [F] = M^1 L^1 T^{-2} \]

### Step 4: Substitute dimensions into the equation
Now we substitute these dimensions into our equation:
\[ L^1 T^{-2} = (M^1 L^2 T^{-2})^a (M^1 L^1 T^{-1})^b (M^1 L^1 T^{-2})^c \]

### Step 5: Expand the right side
Expanding the right side gives:
\[ L^1 T^{-2} = M^{a+b+c} L^{2a+b+c} T^{-2a-b-2c} \]

### Step 6: Compare the dimensions
Now we compare the dimensions on both sides:
1. For mass \(M\):
\[ 0 = a + b + c \quad \text{(1)} \]
2. For length \(L\):
\[ 1 = 2a + b + c \quad \text{(2)} \]
3. For time \(T\):
\[ -2 = -2a - b - 2c \quad \text{(3)} \]

### Step 7: Solve the equations
From equation (1):
\[ b + c = -a \quad \text{(4)} \]
Substituting equation (4) into equation (2):
\[ 1 = 2a + (-a) \quad \Rightarrow \quad 1 = a \quad \Rightarrow \quad a = 1 \]
Substituting \(a = 1\) into equation (4):
\[ b + c = -1 \quad \Rightarrow \quad b = -1 - c \quad \text{(5)} \]
Now substituting \(a = 1\) into equation (3):
\[ -2 = -2(1) - b - 2c \quad \Rightarrow \quad -2 = -2 - b - 2c \quad \Rightarrow \quad 0 = -b - 2c \quad \Rightarrow \quad b = -2c \quad \text{(6)} \]

### Step 8: Solve for \(b\) and \(c\)
From equations (5) and (6):
Substituting \(b = -2c\) into (5):
\[ -2c + c = -1 \quad \Rightarrow \quad -c = -1 \quad \Rightarrow \quad c = 1 \]
Then substituting \(c = 1\) back into (6):
\[ b = -2(1) = -2 \]

### Final Result
Thus, we have:
\[ a = 1, \quad b = -2, \quad c = 1 \]
Therefore, the dimensions of acceleration in the new system are:
\[ E^1 P^{-2} F^1 \]