A rod of mass `m` and length `l` is lying on a horizontal table. Work done in making it stand on one end will be

Options

A. mgl
B. `(mgl)/(2)`
C. `(mgl)/(4)`
D. 2mgl

Explanation

To find the work done in making a rod of mass `m` and length `l` stand on one end, we can follow these steps:

### Step 1: Understand the Problem
We have a rod lying horizontally on a table, and we want to calculate the work done to lift it to a vertical position.

### Step 2: Identify Initial Potential Energy
When the rod is lying horizontally, the center of mass is at its midpoint. Since the height (h) of the center of mass from the ground is zero in this position, the initial potential energy (PE_initial) can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
Here, \( h = 0 \):
\[ PE_{\text{initial}} = mg \cdot 0 = 0 \]

### Step 3: Identify Final Potential Energy
When the rod is standing vertically on one end, the center of mass is at a height of \( \frac{l}{2} \) (half the length of the rod). The final potential energy (PE_final) can be calculated as:
\[ PE_{\text{final}} = mgh \]
Here, \( h = \frac{l}{2} \):
\[ PE_{\text{final}} = mg \cdot \frac{l}{2} = \frac{mgl}{2} \]

### Step 4: Calculate Work Done
The work done (W) in lifting the rod can be found by calculating the change in potential energy:
\[ W = PE_{\text{final}} - PE_{\text{initial}} \]
Substituting the values we found:
\[ W = \frac{mgl}{2} - 0 = \frac{mgl}{2} \]

### Conclusion
The work done in making the rod stand on one end is:
\[ W = \frac{mgl}{2} \]