Which of the following equations is dimensionally incorrect ?
(E= energy, U= potential energy, P=momentum, m= mass, v= speed)

Options

A. `E=U+(P^(2))/(2M)`
B. `E=mv^(2)+(P^(2))/(m)`
C. `2E=(U)/(2)-(1)/(2)mv^(2)`
D. `E=(P^(2)U)/(2mv^(2))`

Answer

To determine which of the given equations is dimensionally incorrect, we will analyze the dimensions of each equation step by step. The quantities involved are:

- \( E \) = energy
- \( U \) = potential energy
- \( P \) = momentum
- \( m \) = mass
- \( v \) = speed

### Step 1: Analyze the dimensions of each term

1. **Energy (E) and Potential Energy (U)**:
- The dimension of energy (E) is given by:
\[ [E] = [U] = [M L^2 T^{-2}] \]

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

### Step 2: Evaluate each equation

#### Equation A: \( E = E + \frac{P^2}{2m} \)

- The term \( \frac{P^2}{2m} \):
\[ P^2 = (M L T^{-1})^2 = M^2 L^2 T^{-2} \]
\[ \frac{P^2}{m} = \frac{M^2 L^2 T^{-2}}{M} = M L^2 T^{-2} \]
- Thus, the dimensions of \( E \) and \( \frac{P^2}{2m} \) are the same:
\[ [E] = [E] + [\frac{P^2}{2m}] \implies [M L^2 T^{-2}] = [M L^2 T^{-2}] + [M L^2 T^{-2}] \]
- This equation is dimensionally correct.

#### Equation B: \( E = mv^2 + \frac{P^2}{m} \)

- The term \( mv^2 \):
\[ mv^2 = M (L T^{-1})^2 = M L^2 T^{-2} \]
- The term \( \frac{P^2}{m} \) was calculated previously as:
\[ \frac{P^2}{m} = M L^2 T^{-2} \]
- Thus, both terms have the same dimension:
\[ [E] = [mv^2] + [\frac{P^2}{m}] \implies [M L^2 T^{-2}] = [M L^2 T^{-2}] + [M L^2 T^{-2}] \]
- This equation is dimensionally correct.

#### Equation C: \( E = E - \frac{1}{2} mv^2 \)

- The term \( \frac{1}{2} mv^2 \) has the same dimension as kinetic energy:
\[ \frac{1}{2} mv^2 = M L^2 T^{-2} \]
- Thus, the dimensions are:
\[ [E] = [E] - [\frac{1}{2} mv^2] \implies [M L^2 T^{-2}] = [M L^2 T^{-2}] - [M L^2 T^{-2}] \]
- This equation is dimensionally correct.

#### Equation D: \( E = \frac{P^2 U}{2m v^2} \)

- The term \( \frac{P^2 U}{2m v^2} \):
- We already know \( P^2 \) has dimensions \( M^2 L^2 T^{-2} \).
- The dimension of \( U \) (potential energy) is \( M L^2 T^{-2} \).
- Therefore,
\[ P^2 U = (M^2 L^2 T^{-2})(M L^2 T^{-2}) = M^3 L^4 T^{-4} \]
- The dimension of \( v^2 \) is:
\[ v^2 = (L T^{-1})^2 = L^2 T^{-2} \]
- Thus, the dimension of the entire term is:
\[ \frac{P^2 U}{2m v^2} = \frac{M^3 L^4 T^{-4}}{M L^2 T^{-2}} = M^2 L^2 T^{-2} \]
- The dimensions of \( E \) are \( M L^2 T^{-2} \), but the dimensions of the right side are \( M^2 L^2 T^{-2} \), which do not match.

### Conclusion
The equation that is dimensionally incorrect is:
\[ \text{Option D: } E = \frac{P^2 U}{2m v^2} \]