Force on a particle in one-dimensional motion is given by `F=Av+Bt+(Cx)/(At+D)`, where F= force, v= speed, t= time, x= position and A, B, C and D are constants. Dimension of C will be-

Options

A. `M^(2)L^(-2)T^(0)`
B. `ML^(-1)T^(0)`
C. `M^(2)L^(0)T^(-2)`
D. None of these

Explanation

To find the dimension of the constant \( C \) in the given force equation:

\[ F = Av + Bt + \frac{Cx}{At + D} \]

we will analyze the dimensions of each term in the equation.

### Step 1: Determine the dimension of force \( F \)

The dimension of force \( F \) is given by:

\[ [F] = M L T^{-2} \]

### Step 2: Determine the dimension of speed \( v \)

Speed \( v \) is defined as distance over time, so its dimension is:

\[ [v] = L T^{-1} \]

### Step 3: Determine the dimension of constant \( A \)

From the term \( Av \), we have:

\[ [A] \cdot [v] = [F] \]

Substituting the dimensions we have:

\[ [A] \cdot (L T^{-1}) = M L T^{-2} \]

Solving for \( [A] \):

\[ [A] = \frac{M L T^{-2}}{L T^{-1}} = M T^{-1} \]

### Step 4: Determine the dimension of constant \( B \)

From the term \( Bt \), we have:

\[ [B] \cdot [t] = [F] \]

Substituting the dimensions we have:

\[ [B] \cdot T = M L T^{-2} \]

Solving for \( [B] \):

\[ [B] = \frac{M L T^{-2}}{T} = M L T^{-3} \]

### Step 5: Determine the dimension of position \( x \)

The dimension of position \( x \) is simply:

\[ [x] = L \]

### Step 6: Analyze the term \( \frac{Cx}{At + D} \)

Since \( At + D \) is in the denominator, we need to ensure that the dimensions of \( At + D \) match the dimensions of \( F \) when added.

The dimension of \( At \) is:

\[ [A] \cdot [t] = (M T^{-1}) \cdot T = M \]

Thus, the dimension of \( D \) must also be \( M \) for the addition to be valid.

### Step 7: Set up the equation for \( C \)

Now we can analyze the term \( \frac{Cx}{At + D} \):

The dimension of \( \frac{Cx}{At + D} \) must equal the dimension of \( F \):

\[ \frac{[C] \cdot [x]}{[At + D]} = [F] \]

Substituting the dimensions we have:

\[ \frac{[C] \cdot L}{M} = M L T^{-2} \]

### Step 8: Solve for the dimension of \( C \)

Rearranging gives:

\[ [C] \cdot L = M^2 L T^{-2} \]

Thus:

\[ [C] = \frac{M^2 L T^{-2}}{L} = M^2 T^{-2} \]

### Conclusion

The dimension of \( C \) is:

\[ [C] = M^2 L^0 T^{-2} \]

### Final Answer

The dimension of \( C \) is \( M^2 L^0 T^{-2} \).

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