The term `(1//2)rhov^(2)` occurs in Bernoulli's equation, with `rho` being the density of a fluid and v its speed. The dimensions of this term are

Options

A. `[M^(-1)L^(5)T^(2)]`
B. `[MLT^(2)]`
C. `[ML^(-1)T^(-2)]`
D. `[M^(-1)L^(9)T^(-2)]`

Answer

To find the dimensions of the term \((\frac{1}{2})\rho v^2\) in Bernoulli's equation, we will break it down step by step.

### Step 1: Identify the components of the term
The term we are analyzing is \((\frac{1}{2})\rho v^2\), where:
- \(\rho\) is the density of the fluid.
- \(v\) is the speed of the fluid.

### Step 2: Write the dimensions of density (\(\rho\))
Density (\(\rho\)) is defined as mass per unit volume. The dimensions of mass are \([M]\) and the dimensions of volume are \([L^3]\). Therefore, the dimensions of density are:
\[ [\rho] = \frac{[M]}{[L^3]} = [M L^{-3}] \]

### Step 3: Write the dimensions of speed (\(v\))
Speed (\(v\)) is defined as distance traveled per unit time. The dimensions of distance are \([L]\) and the dimensions of time are \([T]\). Therefore, the dimensions of speed are:
\[ [v] = \frac{[L]}{[T]} = [L T^{-1}] \]

### Step 4: Calculate the dimensions of \(v^2\)
To find the dimensions of \(v^2\), we square the dimensions of speed:
\[ [v^2] = [L T^{-1}]^2 = [L^2 T^{-2}] \]

### Step 5: Combine the dimensions of \(\rho\) and \(v^2\)
Now, we can combine the dimensions of \(\rho\) and \(v^2\) to find the dimensions of the term \((\frac{1}{2})\rho v^2\):
\[ [\rho v^2] = [M L^{-3}] \cdot [L^2 T^{-2}] = [M L^{-3} \cdot L^2 T^{-2}] = [M L^{-1} T^{-2}] \]

### Step 6: Conclusion
Thus, the dimensions of the term \((\frac{1}{2})\rho v^2\) are:
\[ [M L^{-1} T^{-2}] \]