Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds $v_{1}$ and $v_{2},$ respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and $v_{1} = 3 m/s$ then $v_{2}$ in $m/s$ is $(g = 10 m/s^{2})$

Step 1: Express the total kinetic energy of a rolling disc.

For a uniform disc of mass $m$ and radius $r$ rolling without slipping, the relationship between linear velocity $v$ and angular velocity $\omega$ is $v = \omega r$.

$$I = \frac{1}{2}mr^2$$

The total kinetic energy is the sum of translational and rotational kinetic energies.

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ $$KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$ $$KE = \frac{3}{4}mv^2$$

Step 2: Apply conservation of mechanical energy for both surfaces.

Since static friction does no work during rolling without slipping, mechanical energy is conserved.

For surface AB with a height drop of $h_1 = 30\text{ m}$ and initial speed $v_1 = 3\text{ m/s}$:

$$\frac{3}{4}mv_1^2 + mgh_1 = \frac{3}{4}mV^2$$ $$\frac{3}{4}v_1^2 + gh_1 = \frac{3}{4}V^2$$

For surface CD with a height drop of $h_2 = 27\text{ m}$ and initial speed $v_2$:

$$\frac{3}{4}v_2^2 + gh_2 = \frac{3}{4}V^2$$

Step 3: Solve for the initial speed $v_2$.

Equate the two expressions since the final speed $V$ is the same for both discs.

$$\frac{3}{4}v_2^2 + gh_2 = \frac{3}{4}v_1^2 + gh_1$$ $$\frac{3}{4}(v_2^2 - v_1^2) = g(h_1 - h_2)$$

Substitute the given values $v_1 = 3\text{ m/s}$, $h_1 = 30\text{ m}$, $h_2 = 27\text{ m}$, and $g = 10\text{ m/s}^2$.

$$\frac{3}{4}(v_2^2 - 3^2) = 10(30 - 27)$$ $$\frac{3}{4}(v_2^2 - 9) = 30$$ $$v_2^2 - 9 = 40$$ $$v_2^2 = 49$$ $$v_2 = 7\text{ m/s}$$

Answer: $7$