\(v_{0}=r \omega ; \quad \therefore v_{0} \alpha r\)
For a wheel (uniform ring) the distance of a point on the rim in level with the center 
\(\text { [i.e., radius] is } \sqrt{2} r\)
\(\therefore \text { The speed of the center is } \sqrt{2} v_{0}\)

Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

The moments of inertia of two discs are I_a and I_b respectively The angular velocity of the disc A is \(\omega\).
The sum of kinetic energies of two discs before coming in contact is \(k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}+\frac{1}{2} I_{b} \omega_{2}^{2}\)
\(\text { But angular velocity of the disc be is } \omega_{2}=0 \ \text {(rest)}\)
\(\therefore k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}\)
The final kinetic energy of the two discs system \(k_{2}=\frac{1}{2} \frac{I_{a}^{2} \omega_{1}^{2}}{I_{a}+I_{b}}\)
The loss of kinetic energy is
\(k_{1}-k_{2} =\frac{1}{2} I_{a} \omega^{2}-\frac{1}{2}\left[\frac{I a^{2} \omega_{1}^{2}}{I_{a}+I_{2 b}}\right] \)
\(=\frac{1}{2} \frac{\left[I_{1}\left(I_{a}+I_{b}\right) \omega^{2}-I_{a}^{2} \omega^{2}\right]}{I_{a}+I_{b}} \)
\(k_{1}-k_{2} =\frac{1}{2} \frac{I_{a} b}{\left(I_{a}+I_{b}\right)} \omega^{2} \)

\(K \cdot E=\frac{1}{2} I \omega^{2} ; \quad L=I \omega ; \quad K \cdot E=\frac{2^{2}}{2^{2}} \)
\(\therefore K \cdot E \alpha L^{2} \quad E_{1} \alpha L_{1}^{2} \quad E_{2} \alpha L_{2}^{2}\)
\(\frac{E_{1}}{E_{2}}=\left(\frac{L_{1}}{L_{2}}\right)^{2} \)
\(\text { Here } E_{1}=E \quad E_{2}=\frac{E}{2} \)
\(L_{1}=L \quad \quad L_{2}=? \)
\(\frac{E}{\frac{E}{2}}=\left(\frac{L}{L_{2}}\right)^{2} \)
\(\frac{2 E}{E}=\left(\frac{L}{L_{2}}\right)^{2}\left(\frac{L_{1}}{L_{2}}\right)^{2}=2 \)
\(\therefore \frac{L}{L_{2}}=\sqrt{2} \)
\(L_{2}=\frac{L}{\sqrt{2}} \)