Which of the following does not have dimensions of force?

Options

A. Weight
B. Rate of change of momentum
C. Work per unit length
D. Work done per unit charge

Answer

To solve the question "Which of the following does not have dimensions of force?", we will analyze each option provided and determine their dimensions. The dimensions of force are given by the formula:

\[ \text{Force} = \text{mass} \times \text{acceleration} \]
\[ \text{Dimensions of Force} = [M^1 L^1 T^{-2}] \]

Now, let's evaluate each option:

### Step 1: Analyze Weight
- **Weight** is defined as the force exerted by gravity on an object.
- It can be calculated using the formula:
\[ \text{Weight} = \text{mass} \times \text{gravity} \]
- The dimensions of mass are \([M^1]\) and the dimensions of gravity (acceleration) are \([L^1 T^{-2}]\).
- Therefore, the dimensions of weight are:
\[ [M^1][L^1 T^{-2}] = [M^1 L^1 T^{-2}] \]
- **Conclusion**: Weight has dimensions of force.

### Step 2: Analyze Rate of Change of Momentum
- **Rate of change of momentum** is defined as:
\[ \text{Force} = \frac{dp}{dt} \]
- Momentum (\(p\)) is given by:
\[ p = \text{mass} \times \text{velocity} \]
- The dimensions of momentum are:
\[ [M^1][L^1 T^{-1}] = [M^1 L^1 T^{-1}] \]
- Therefore, the dimensions of rate of change of momentum are:
\[ \frac{[M^1 L^1 T^{-1}]}{[T^1]} = [M^1 L^1 T^{-2}] \]
- **Conclusion**: Rate of change of momentum has dimensions of force.

### Step 3: Analyze Work per Unit Length
- **Work** is defined as:
\[ \text{Work} = \text{Force} \times \text{displacement} \]
- The dimensions of work are:
\[ [M^1 L^1 T^{-2}][L^1] = [M^1 L^2 T^{-2}] \]
- Work per unit length is:
\[ \frac{[M^1 L^2 T^{-2}]}{[L^1]} = [M^1 L^1 T^{-2}] \]
- **Conclusion**: Work per unit length has dimensions of force.

### Step 4: Analyze Work Done per Unit Charge
- **Work done per unit charge** can be expressed as:
\[ \frac{\text{Work}}{\text{Charge}} \]
- We know:
\[ \text{Work} = [M^1 L^2 T^{-2}] \]
- The dimensions of charge (\(Q\)) can be expressed as:
\[ Q = \text{Current} \times \text{Time} = [I^1 T^1] \]
- Therefore, the dimensions of work done per unit charge are:
\[ \frac{[M^1 L^2 T^{-2}]}{[I^1 T^1]} = [M^1 L^2 T^{-3} I^{-1}] \]
- **Conclusion**: Work done per unit charge does not have dimensions of force.

### Final Answer
The correct answer is **option 4: Work done per unit charge**, as it does not have the dimensions of force.

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