A spherical steel ball released at the top of along column of glycerin of length `l` falls through a distance `l//2` with accelerated motion and the remaining distance `l//2` with uniform velocity let `t_(1)` and `t_(2)` denote the times taken to cover the first and second half and `w_(1)` and `w_(2)` are the work done against gravity in the two halves, then compare times and work done.

Options

A. `t_(1) lt t_(2), W_(1) gt W_(2)`
B. `t_(1) gt t_(2), W_(1) lt W_(2)`
C. `t_(1) = t_(2), W_(1) = W_(2)`
D. `t_(1) gt t_(2), W_(1) = W_(2)`

Answer

To solve the problem, we need to analyze the motion of the spherical steel ball as it falls through a column of glycerin. The ball falls through a distance of \( \frac{l}{2} \) with accelerated motion and then through the remaining \( \frac{l}{2} \) with uniform velocity. We will denote the time taken to cover the first half as \( t_1 \) and the time taken to cover the second half as \( t_2 \). The work done against gravity in the two halves will be denoted as \( w_1 \) and \( w_2 \).

### Step 1: Analyze the first half of the fall
- The ball falls through the first half of the distance \( \frac{l}{2} \) with accelerated motion.
- Since the ball is released from rest, we can use the equations of motion to determine the time taken \( t_1 \).
- For an object falling under gravity \( g \) with an initial velocity of 0, the distance covered in time \( t_1 \) is given by:
\[ \frac{l}{2} = \frac{1}{2} g t_1^2 \]
- Rearranging this gives:
\[ t_1^2 = \frac{l}{g} \quad \Rightarrow \quad t_1 = \sqrt{\frac{l}{g}} \]

### Step 2: Analyze the second half of the fall
- In the second half, the ball falls through the distance \( \frac{l}{2} \) with uniform velocity \( v \).
- The average velocity during this phase is \( v \), and the time taken \( t_2 \) is given by:
\[ \frac{l}{2} = v t_2 \quad \Rightarrow \quad t_2 = \frac{l}{2v} \]

### Step 3: Compare the times \( t_1 \) and \( t_2 \)
- We know that the average velocity in the first half is less than \( v \) (since the ball is accelerating), which implies:
\[ t_1 > t_2 \]
- Therefore, we conclude:
\[ t_1 > t_2 \]

### Step 4: Calculate the work done against gravity
- The work done against gravity in both halves can be calculated as follows:
- The work done \( w_1 \) in the first half is:
\[ w_1 = mgh_1 = mg \left(\frac{l}{2}\right) = \frac{mgl}{2} \]
- The work done \( w_2 \) in the second half is:
\[ w_2 = mgh_2 = mg \left(\frac{l}{2}\right) = \frac{mgl}{2} \]
- Thus, we find that:
\[ w_1 = w_2 \]

### Final Comparison
- From our analysis, we have:
\[ t_1 > t_2 \quad \text{and} \quad w_1 = w_2 \]

### Conclusion
- The final results are:
- \( t_1 > t_2 \)
- \( w_1 = w_2 \)