d
(d)

after reaching ground aal the potential energy will bw converted to kinrtic enetgy,

for rolling object,\(K \cdot E=\frac{m v^2}{2}+\frac{I w^2}{2}\)

where \(I=\)Moment of inertia \(=\frac{m r^2}{2}\)

\(\omega =\) angular velocity \(= \frac{v}{r}\) 

let the two disc have mass \(m\) and \(2m\) and radius \(r\) and \(8r\) for disc \(1.\)

\(PE=KE\)

\(m g h=\frac{m v^2}{2}+\left(\frac{1}{2}\right) \times \frac{m v^2}{2} \times\left(\frac{v}{r}\right)^2\)

\(m g h=\frac{m v^2}{2}+\frac{m v^2}{4}\)

\(m g h=\frac{3 m v^2}{4}\)

\(\therefore v=\sqrt{\frac{4 g h}{3}}\)

similarly for disc \(2\),let the speed be \(v^{\prime}\)

\(2 m g h=\frac{2 m v^{.2}}{2}+\frac{1}{2}\left(2 m \times \frac{64 r^2}{2}\right) \times\left(\frac{v^{\prime}}{64 r^2}\right)\)

\(2 m g h=\frac{3 m v^{\prime 2}}{2}\)

\(v^{\prime}=\sqrt{\frac{4 g h}{3}}\)

Hence the ratio of theair speed will be \(v : v^{\prime}\) \(=1:1\)