A uniform rod of mass 2kg and length `l` is lying on a horizontal surface. If the work done in raising one end of the rod through an angle `45^(@)` is `'W'`, then the work done in raising it further `45^(@)` is

Options

A. `W`
B. `sqrt(2)W`
C. `(W)/(sqrt(2))`
D. `(sqrt(2)-1)W`

Explanation

To solve the problem, we need to calculate the work done in raising one end of a uniform rod further after it has already been raised through an angle of \(45^\circ\).

### Step-by-Step Solution:

1. **Understanding the Initial Setup:**
- We have a uniform rod of mass \(m = 2 \, \text{kg}\) and length \(l\) lying horizontally.
- The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from either end.

2. **Work Done in Raising the Rod to \(45^\circ\):**
- When one end of the rod is raised to an angle of \(45^\circ\), the height of the center of mass (h1) can be calculated using trigonometry.
- The height \(h_1\) when the rod is at \(45^\circ\) is given by:
\[ h_1 = \frac{l}{2} \sin(45^\circ) = \frac{l}{2} \cdot \frac{1}{\sqrt{2}} = \frac{l}{2\sqrt{2}} \]
- The work done \(W\) in raising the rod to this position against gravity is:
\[ W = mgh_1 = mg \cdot \frac{l}{2\sqrt{2}} = 2g \cdot \frac{l}{2\sqrt{2}} = \frac{gl}{\sqrt{2}} \]

3. **Raising the Rod from \(45^\circ\) to \(90^\circ\):**
- Now we need to calculate the work done \(W'\) in raising the rod from \(45^\circ\) to \(90^\circ\).
- The height \(h_2\) when the rod is vertical (at \(90^\circ\)) is:
\[ h_2 = \frac{l}{2} \]
- The work done \(W'\) in raising the rod from \(45^\circ\) to \(90^\circ\) is:
\[ W' = mg h_2 = mg \cdot \frac{l}{2} = 2g \cdot \frac{l}{2} = gl \]

4. **Finding the Relationship Between \(W'\) and \(W\):**
- Now we can express \(W'\) in terms of \(W\):
\[ W' = mg h_2 = mg \cdot \frac{l}{2} = 2g \cdot \frac{l}{2} = gl \]
- From the previous calculation, we have:
\[ W = \frac{gl}{\sqrt{2}} \]
- Therefore, the ratio of \(W'\) to \(W\) is:
\[ \frac{W'}{W} = \frac{gl}{\frac{gl}{\sqrt{2}}} = \sqrt{2} \]
- Thus, we can express \(W'\) as:
\[ W' = W \cdot \sqrt{2} \]

### Final Answer:
The work done in raising the rod further from \(45^\circ\) to \(90^\circ\) is:
\[ W' = W \cdot \sqrt{2} \]