Suppose y represents the work done and x the power, then dimensions of `(d^(2)y)/(dx^(2))` will be

Options

A. `[M^(-1)L^(-2)T^(4)]`
B. `[M^(2)L^(-3)T^(-2)]`
C. `[M^(-2)L^(-4)T^(4)]`
D. `[ML^(3)T^(-6)]`

Answer

To find the dimensions of \(\frac{d^2y}{dx^2}\) where \(y\) represents work done and \(x\) represents power, we can follow these steps:

### Step 1: Identify the dimensions of work and power
- Work done (\(y\)) has the dimension of:
\[ [y] = [\text{Work}] = M L^2 T^{-2} \]
- Power (\(x\)) has the dimension of:
\[ [x] = [\text{Power}] = M L^2 T^{-3} \]

### Step 2: Write the expression for \(\frac{d^2y}{dx^2}\)
The second derivative \(\frac{d^2y}{dx^2}\) can be expressed as:
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} \cdot \frac{dy}{dx} \]
This means we need to find the dimensions of \(\frac{dy}{dx}\) first.

### Step 3: Find the dimension of \(\frac{dy}{dx}\)
Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{[y]}{[x]} \]
Substituting the dimensions we found:
\[ \frac{dy}{dx} = \frac{M L^2 T^{-2}}{M L^2 T^{-3}} = \frac{M L^2 T^{-2}}{M L^2 T^{-3}} = T \]
Thus, the dimension of \(\frac{dy}{dx}\) is:
\[ \left[\frac{dy}{dx}\right] = T \]

### Step 4: Find the dimension of \(\frac{d^2y}{dx^2}\)
Now, we can find \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} \cdot \frac{dy}{dx} = T \cdot T = T^2 \]

### Step 5: Final expression for \(\frac{d^2y}{dx^2}\)
Since we have \( \frac{d^2y}{dx^2} = \frac{M L^2 T^{-2}}{(M L^2 T^{-3})^2} \), we can express it as:
\[ \frac{d^2y}{dx^2} = \frac{M L^2 T^{-2}}{M^2 L^4 T^{-6}} = \frac{M^1 L^2 T^{-2}}{M^2 L^4 T^{-6}} = M^{-1} L^{-2} T^{4} \]

### Conclusion
Thus, the dimensions of \(\frac{d^2y}{dx^2}\) are:
\[ [M^{-1} L^{-2} T^{4}] \]