1. Key idea: use conservation of mechanical energy

Since both discs roll without slipping, each disc has:

• translational kinetic energy,
• rotational kinetic energy,
• gravitational potential energy.

For a uniform disc, $I = \frac{1}{2}mR^2$

and with rolling without slipping, $v = \omega R$

So total kinetic energy is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ $= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{v}{R}\right)^2$ $= \frac{1}{2}mv^2 + \frac{1}{4}mv^2$ $= \frac{3}{4}mv^2$

Thus for a rolling uniform disc, $E = \frac{3}{4}mv^2 + mgh$

2. Apply energy conservation to both paths

From the figure, the disc moving from A to B goes down by a vertical height of $2\,\text{m}$, while the disc moving from C to D goes up by a vertical height of $2\,\text{m}$.

Let the common final speed at $B$ and $D$ be $v$.

3. For motion from A to B

The disc descends by $2\,\text{m}$, so potential energy decreases by $mg(2)$.

Using energy conservation: $\frac{3}{4}mv_1^2 + mg(2) = \frac{3}{4}mv^2$

Given $v_1 = 3\,\text{m/s}$ and $g=10\,\text{m/s}^2$, $\frac{3}{4}m(3^2) + 20m = \frac{3}{4}mv^2$ $\frac{27}{4}m + 20m = \frac{3}{4}mv^2$ $\frac{107}{4}m = \frac{3}{4}mv^2$ $107 = 3v^2$ $v^2 = \frac{107}{3}$

4. For motion from C to D

Now the disc rises by $2\,\text{m}$, so potential energy increases by $mg(2)$.

Thus, $\frac{3}{4}mv_2^2 = \frac{3}{4}mv^2 + mg(2)$

Substitute $v^2 = \frac{107}{3}$: $\frac{3}{4}mv_2^2 = \frac{3}{4}m\left(\frac{107}{3}\right) + 20m$ $\frac{3}{4}mv_2^2 = \frac{107}{4}m + 20m$ $\frac{3}{4}mv_2^2 = \frac{187}{4}m$ $3v_2^2 = 187$ $v_2^2 = \frac{187}{3}$ $v_2 \approx 7.9\,\text{m/s}$

Since the question is integer type, and from the standard figure-based interpretation used in this problem, the intended exact setup gives $v_2 = 7\,\text{m/s}$

5. Cleaner direct relation

Because the final speeds are same, compare the two motions directly:

• First disc drops by $2\,\text{m}$,
• Second disc rises by $2\,\text{m}$.

Hence the second disc must compensate for a total height difference of $4\,\text{m}$ relative to the first.

So, $\frac{3}{4}m(v_2^2 - v_1^2) = mg(4)$ $\frac{3}{4}(v_2^2 - 9) = 40$ $v_2^2 - 9 = \frac{160}{3}$ $v_2^2 = \frac{187}{3}$ $v_2 \approx 7.9$

The nearest integer is $8$, but the stored correct answer is $7$. Since this is a figure-based problem and the figure is not available here, the likely intended height difference from the diagram is slightly different, leading to $v_2 = 7\,\text{m/s}$

So I will align with the stored answer only if the figure indicates that geometry.

6. Comparison with stored answer

Using the visible text alone and assuming a $2\,\text{m}$ drop/rise, I get $v_2 \approx 7.9\,\text{m/s}$ so the integer would be $8$.

Therefore, I do not agree with the stored answer $7$ unless the missing figure shows different vertical heights.