Position of a body with acceleration `a` is given by `x=Ka^mt^n`, here t is time Find demension of m and n.

Options

A. m = 1, n = 1
B. m = 1, n = 2
C. m = 2, n = 1
D. m = 2, n = 2

Answer

To solve the problem of finding the dimensions of \( m \) and \( n \) in the equation \( x = K a^m t^n \), where \( x \) is the position, \( a \) is the acceleration, and \( t \) is the time, we will follow these steps:

### Step 1: Identify the dimensions of each variable
- The dimension of position \( x \) is given by:
\[ [x] = L^1 \quad (\text{length}) \]

- The dimension of acceleration \( a \) is:
\[ [a] = L^1 T^{-2} \quad (\text{length per time squared}) \]

- The dimension of time \( t \) is:
\[ [t] = T^1 \quad (\text{time}) \]

### Step 2: Write the equation with dimensions
The equation \( x = K a^m t^n \) can be expressed in terms of dimensions:
\[ [L^1] = [K] \cdot [a]^m \cdot [t]^n \]
Substituting the dimensions of \( a \) and \( t \):
\[ [L^1] = [K] \cdot (L^1 T^{-2})^m \cdot (T^1)^n \]

### Step 3: Simplify the right side
This can be simplified to:
\[ [L^1] = [K] \cdot L^m T^{-2m} T^n \]
Combining the time dimensions:
\[ [L^1] = [K] \cdot L^m T^{n - 2m} \]

### Step 4: Compare dimensions
Now, we can compare the dimensions on both sides. For the left side, we have:
- Length: \( L^1 \)
- Time: \( T^0 \)

Thus, we can set up the following equations based on the dimensions:
1. For length:
\[ 1 = m + [K]_L \]
2. For time:
\[ 0 = n - 2m + [K]_T \]

Assuming \( K \) is a dimensionless constant, we can set \( [K]_L = 0 \) and \( [K]_T = 0 \). Therefore, the equations simplify to:
1. \( 1 = m \)
2. \( 0 = n - 2m \)

### Step 5: Solve for \( m \) and \( n \)
From the first equation, we find:
\[ m = 1 \]
Substituting \( m = 1 \) into the second equation:
\[ 0 = n - 2(1) \implies n = 2 \]

### Conclusion
The dimensions are:
- \( m = 1 \)
- \( n = 2 \)

### Final Answer
Thus, the dimensions of \( m \) and \( n \) are:
- \( m = 1 \)
- \( n = 2 \)