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`

Explanation

To solve the problem, we need to analyze the given equation for the position of a body with acceleration \( a \):

\[ x = K a^m t^n \]

where \( x \) is the position, \( a \) is the acceleration, \( t \) is time, and \( K \) is a constant.

### Step 1: Identify the dimensions of each quantity

1. **Position \( x \)** has the dimension of length:
\[ [x] = L^1 \]

2. **Acceleration \( a \)** is defined as the rate of change of velocity, which is the change in displacement over time squared:
\[ [a] = L^1 T^{-2} \]

3. **Time \( t \)** has the dimension:
\[ [t] = T^1 \]

### Step 2: Write the dimensions of the right-hand side

Now, we can express the dimensions of the right-hand side of the equation:

\[ [K a^m t^n] = K [a]^m [t]^n = K (L^1 T^{-2})^m (T^1)^n \]

This simplifies to:

\[ = K L^m T^{-2m} T^n = K L^m T^{n - 2m} \]

### Step 3: Equate the dimensions

Since \( K \) is a constant, we can assume it has no dimensions (or its dimensions will cancel out). Therefore, we can set the dimensions of both sides equal to each other:

\[ L^1 = K L^m T^{n - 2m} \]

This gives us two equations by comparing the powers of \( L \) and \( T \):

1. From the length dimension:
\[ m = 1 \]

2. From the time dimension:
\[ n - 2m = 0 \]

### Step 4: Solve for \( n \)

Substituting \( m = 1 \) into the second equation:

\[ n - 2(1) = 0 \implies n - 2 = 0 \implies n = 2 \]

### Conclusion

Thus, the dimensions of \( m \) and \( n \) are:

\[ m = 1, \quad n = 2 \]

### Final Answer

- \( m = 1 \)
- \( n = 2 \)