1. Motion in air

The ball is dropped from rest from a height of $4.9\,\text{m}$ above the water.

Using $s = \frac{1}{2}gt^2$ with $s=4.9\,\text{m}$ and $g=9.8\,\text{m/s}^2$:

$4.9 = \frac{1}{2}(9.8)t^2 = 4.9t^2$

So, $t^2=1 \Rightarrow t=1\,\text{s}$

Hence, time taken to reach water surface = $1\,\text{s}$.

2. Velocity when it hits water

Using $v = gt$ we get $v = 9.8 \times 1 = 9.8\,\text{m/s}$

So the ball hits the water with speed $v = 9.8\,\text{m/s}$

3. Motion inside water

It is given that after entering water, the ball sinks with constant velocity $v$, i.e. with speed $9.8\,\text{m/s}$.

Total time from release to reaching the bottom = $4.0\,\text{s}$.

Therefore, time spent inside water: $t_{\text{water}} = 4.0 - 1.0 = 3.0\,\text{s}$

4. Depth of lake

Since speed in water is constant, $\text{depth} = v\,t_{\text{water}} = 9.8 \times 3 = 29.4\,\text{m}$

5. Option check

The depth of the lake is $29.4\,\text{m}$ which corresponds to Option B.