d
If disk slips on inclined plane, then it's aceleration

\(a_{1}=g \sin \theta\)

\(L=\frac{1}{2} a_{1} t_{1}^{2}\)

\(\Rightarrow t_{1}=\sqrt{\frac{2 L}{a_{1}}}\ldots \ldots({i} )\)

If disk rols on inclined plane, its acceleration,

\(a_{2} =\frac{g \sin \theta}{1+\frac{I}{m R^{2}}}\)

\(a_{2} =\frac{g \sin \theta}{1+\frac{m R^{2}}{2 m R^{2}}}\)

\(a_{2} =\frac{2}{3} g \sin \theta\)

Now \({L}=\frac{1}{2} {a}_{2} \cdot {t}_{2}^{2}\)

\(\Rightarrow {t}_{2}=\sqrt{\frac{2 {L}}{{a}_{2}}} \quad \ldots \ldots({ii} )\)

Now \(\frac{t_{2}}{t_{1}}=\sqrt{\frac{a_{1}}{a_{2}}}=\sqrt{\frac{3}{2}}\)

\(\Rightarrow {x}=2\)