The dimensions of `K` in the equation `W=1/2Kx^(2)` is

Options

A. `[M^1L^0T^(-2)]`
B. `[M^0L^1 T^(-1)]`
C. `[M^1L^1T^(-2)]`
D. `[M^1 L^0 T^(-1)]`

Answer

To find the dimensions of \( K \) in the equation \( W = \frac{1}{2} K x^2 \), we will follow these steps:

### Step 1: Identify the quantities involved
In the equation, \( W \) represents work done, and \( x \) represents distance.

### Step 2: Write down the dimensions of work and distance
- Work \( W \) is measured in joules (J). The dimensional formula for work is:
\[ [W] = [\text{Force}] \times [\text{Distance}] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] \]
Thus, the dimensions of work are \( [W] = [M L^2 T^{-2}] \).

- Distance \( x \) is measured in meters (m), so its dimensions are:
\[ [x] = [L] \]

### Step 3: Substitute the dimensions into the equation
The equation can be rearranged to solve for \( K \):
\[ K = \frac{W}{x^2} \]

### Step 4: Substitute the dimensions of \( W \) and \( x \) into the equation for \( K \)
Substituting the dimensions we found:
\[ [K] = \frac{[W]}{[x]^2} = \frac{[M L^2 T^{-2}]}{[L]^2} \]

### Step 5: Simplify the dimensions
Now, simplify the expression:
\[ [K] = \frac{[M L^2 T^{-2}]}{[L^2]} = [M L^2 T^{-2}] \times [L^{-2}] = [M L^{2-2} T^{-2}] = [M L^0 T^{-2}] \]
This simplifies to:
\[ [K] = [M T^{-2}] \]

### Final Answer
Thus, the dimensions of \( K \) are:
\[ [K] = [M^1 L^0 T^{-2}] \]