Force is given by the expression F = A cos (Bx) + C cos (Dt), where x is displacement and t is time. The dimension of `(D)/(B)` is same as that of

Options

A. velocity
B. velocity gradient
C. Angular velocity
D. Angular momentum

Answer

To solve the problem, we need to analyze the given expression for force and determine the dimensions of \( D \) and \( B \) to find the ratio \( \frac{D}{B} \).

### Step-by-Step Solution:

1. **Understand the Expression for Force**:
The force is given by:
\[ F = A \cos(Bx) + C \cos(Dt) \]
Here, \( x \) is displacement and \( t \) is time.

2. **Identify the Dimensions of Force**:
The dimension of force \( F \) is given by:
\[ [F] = [M][L][T^{-2}] \]
where \( [M] \) is mass, \( [L] \) is length, and \( [T] \) is time.

3. **Analyze the Cosine Terms**:
The cosine function is dimensionless. Therefore, the arguments of the cosine functions \( Bx \) and \( Dt \) must also be dimensionless:
\[ [Bx] = [D] = 1 \]
This means:
\[ [B][x] = 1 \quad \text{and} \quad [D][t] = 1 \]

4. **Determine the Dimensions of \( B \) and \( D \)**:
- Since \( x \) has the dimension of length \( [L] \):
\[ [B] = [L^{-1}] \]
- Since \( t \) has the dimension of time \( [T] \):
\[ [D] = [T^{-1}] \]

5. **Calculate the Ratio \( \frac{D}{B} \)**:
Now, we can find the dimensions of the ratio \( \frac{D}{B} \):
\[ \frac{D}{B} = \frac{[T^{-1}]}{[L^{-1}]} = [T^{-1}][L^{1}] = \frac{L}{T} \]

6. **Identify the Dimension**:
The dimension \( \frac{L}{T} \) corresponds to velocity.

### Conclusion:
Thus, the dimension of \( \frac{D}{B} \) is the same as that of velocity.

### Final Answer:
The dimension of \( \frac{D}{B} \) is the same as that of **velocity**.

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